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</div><h2>HL Paper 1</h2><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the sum of the infinite geometric sequence 27, −9, 3, −1, ... .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Use mathematical induction to prove that for \(n \in {\mathbb{Z}^ + }\) ,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[a + ar + a{r^2} + ... + a{r^{n - 1}} = \frac{{a(1 - {r^n})}}{{1 - r}}.\]</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = - \frac{1}{3}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({S_\infty } = \frac{{27}}{{1 + \frac{1}{3}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({S_\infty } = \frac{{81}}{4}\,\,\,\,\,( = 20.25)\) <strong><em>A1 N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Attempting to show that the result is true for <em>n</em> = 1 <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">LHS = <em>a</em> <strong>and</strong> \({\text{RHS}} = \frac{{a(1 - r)}}{{1 - r}} = a\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence the result is true for <em>n</em> = 1</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Assume it is true for <em>n</em> = <em>k</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a + ar + a{r^2} + ... + a{r^{k - 1}} = \frac{{a(1 - rk)}}{{1 - r}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider <em>n</em> = <em>k</em> + 1:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a + ar + a{r^2} + ... + a{r^{k - 1}} + a{r^k} = \frac{{a(1 - {r^k})}}{{1 - r}} + a{r^k}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{a(1 - {r^k}) + a{r^k}(1 - r)}}{{1 - r}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{a - a{r^k} + a{r^k} - a{r^{k + 1}}}}{{1 - r}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for an equivalent correct intermediate step.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{a - a{r^{k + 1}}}}{{1 - r}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{a(1 - {r^{k + 1}})}}{{1 - r}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Illogical attempted proofs that use the result to be proved would gain </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1A0A0</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for the last three above marks.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The result is true for \(n = k \Rightarrow \) it is true for \(n = k + 1\) </span><strong style="font-family: 'times new roman', times; font-size: medium;">and</strong><span style="font-family: 'times new roman', times; font-size: medium;"> as it is true for \(n = 1\), the result is proved by mathematical induction. </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>R1 N0</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> To obtain the final </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>R1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> mark a reasonable attempt must have been made to prove the </span><em style="font-family: 'times new roman', times; font-size: medium;">k</em><span style="font-family: 'times new roman', times; font-size: medium;"> + 1 step.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[7 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) was correctly answered by the majority of candidates, although a few found <em>r</em> = –3.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (b) was often started off well, but a number of candidates failed to initiate the <em>n</em> = <em>k</em> + 1 step in a satisfactory way. A number of candidates omitted the ‘P(1) is true’ part of the concluding statement.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">Find integer values of \(m\) and \(n\) for which</p>
<p class="p1">\[m - n{\log _3}2 = 10{\log _9}6\]</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1"><strong>METHOD 1</strong></p>
<p class="p2">\(m - n{\log _3}2 = 10{\log _9}6\)</p>
<p class="p2"><span class="Apple-converted-space">\(m - n{\log _3}2 = 5{\log _3}6\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(m = {\log _3}\left( {{6^5}{2^n}} \right)\) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\({3^m}{2^{ - n}} = {6^5} = {3^5} \times {2^5}\) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(m = 5,{\text{ }}n = - 5\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>First <strong><em>M1 </em></strong>is for any correct change of base, second <strong><em>M1 </em></strong>for writing as a single logarithm, third <strong><em>M1 </em></strong><span class="s2">is for writing 6 </span>as \(2 \times 3\).</p>
<p class="p3"> </p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p2">\(m - n{\log _3}2 = 10{\log _9}6\)</p>
<p class="p2"><span class="Apple-converted-space">\(m - n{\log _3}2 = 5{\log _3}6\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(m - n{\log _3}2 = 5{\log _3}3 + 5{\log _3}2\) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(m - n{\log _3}2 = 5 + 5{\log _3}2\) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(m = 5,{\text{ }}n = - 5\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>First <strong><em>M1 </em></strong>is for any correct change of base, second <strong><em>M1 </em></strong><span class="s2">for writing 6 </span>as \(2 \times 3\) and third <strong><em>M1 </em></strong>is for forming an expression without \({\log _3}3\).</p>
<p class="p3"> </p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">The first stage on this question was to change base, so each logarithm was written in the same base. Some candidates chose to move to base 10 or base e, rather than the more obvious base 3, but a few still successfully reached the correct answer having done this. A large majority though did not seem to know how to change the base of a logarithm.</p>
<p class="p1">Simplifying the expression further was a struggle for many candidates.</p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Expand \({(x + h)^3}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence find the derivative of \(f(x) = {x^3}\) from first principles.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({(x + h)^3} = {x^3} + 3{x^2}h + 3x{h^2} + {h^3}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p1"><span class="s1"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{{(x + h)}^3} - {x^3}}}{h}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p1">\( = \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3x{h^2} + {h^3} - {x^3}}}{h}\)</p>
<p class="p1">\( = \mathop {\lim }\limits_{h \to 0} (3{x^2} + 3xh + {h^2})\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">\( = 3{x^2}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Do not award final A1 on FT if \( = 3{x^2}\) is not obtained</p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Final A1 can only be obtained if previous A1 is given</p>
<p class="p3"><em><strong>[3 marks]</strong></em></p>
<p class="p3"><em><strong>Total [5 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Well done although some did not use the binomial expansion.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Fine by those who knew what first principles meant, not by the others.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">The fifth term of an arithmetic sequence is equal to <span class="s1">6 </span>and the sum of the first <span class="s1">12 </span>terms is <span class="s1">45</span>.</p>
<p class="p1">Find the first term and the common difference.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1">use of either \({u_n} = {u_1} + (n - 1)d\) or \({S_n} = \frac{n}{2}\left( {2{u_1} + (n - 1)d} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\({u_1} + 4d = 6\) </span><strong><em>(A1)</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(\frac{{12}}{2}(2{u_1} + 11d) = 45\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2">\( \Rightarrow 4{u_1} + 22d = 15\)</p>
<p class="p1">attempt to solve simultaneous equations <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p3">\(4(6 - 4d) + 22d = 15\)</p>
<p class="p3"><span class="Apple-converted-space">\(6d = - 9 \Rightarrow d = - 1.5\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p3"><span class="Apple-converted-space">\({u_1} = 12\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">Most candidates tackled this question through the use of the standard formula for arithmetic series. Others attempted a variety of trial and improvement approaches with varying degrees of success.</p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The complex numbers \({z_1} = 2 - 2{\text{i}}\) and \({{\text{z}}_2} = 1 - \sqrt 3 {\text{i}}\) are represented by the points A and B respectively on an Argand diagram. Given that O is the origin,</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find AB, giving your answer in the form \(a\sqrt {b - \sqrt 3 } \) , where <em>a</em> , \(b \in {\mathbb{Z}^ + }\) .<br></span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Calculate \({\rm{A\hat OB}}\) in terms of \(\pi \).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{AB}} = \sqrt {{1^2} + {{(2 - \sqrt 3 )}^2}} \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt {8 - 4\sqrt 3 } \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2\sqrt {2 - \sqrt 3 } \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]<br></em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\arg {z_1} = - \frac{\pi }{4}{\text{ }}\arg {z_2} = - \frac{\pi }{3}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow \(\frac{\pi }{4}\) and \(\frac{\pi }{3}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow degrees at this stage.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\rm{A\hat OB}} = \frac{\pi }{3} - \frac{\pi }{4}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{\pi }{{12}}{\text{ (accept }} - \frac{\pi }{{12}})\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow <strong><em>FT</em></strong> for final <strong><em>A1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 2</strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to use scalar product or cosine rule <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos {\rm{A\hat OB}} = \frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\rm{A\hat OB}} = \frac{\pi }{{12}}\) <strong><em>A1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">It was disappointing to note the lack of diagram in many solutions. Most importantly the lack of understanding of the notation <em>AB</em> was apparent. Teachers need to make sure that students are aware of correct notation as given in the outline. A number used the cosine rule but then confused the required angle or sides.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">It was disappointing to note the lack of diagram in many solutions. Most importantly the lack of understanding of the notation <em>AB</em> was apparent. Teachers need to make sure that students are aware of correct notation as given in the outline. A number used the cosine rule but then confused the required angle or sides.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>Let \(z = 1 - \cos 2\theta - {\text{i}}\sin 2\theta ,{\text{ }}z \in \mathbb{C},{\text{ }}0 \leqslant \theta \leqslant \pi \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Solve \(2\sin (x + 60^\circ ) = \cos (x + 30^\circ ),{\text{ }}0^\circ \leqslant x \leqslant 180^\circ \).</p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the modulus and argument of \(z\) in terms of \(\theta \). Express each answer in its simplest form.</p>
<div class="marks">[9]</div>
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence find the cube roots of \(z\) in modulus-argument form.</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(2\sin (x + 60^\circ ) = \cos (x + 30^\circ )\)</p>
<p>\(2(\sin x\cos 60^\circ + \cos x\sin 60^\circ ) = \cos x\cos 30^\circ - \sin x\sin 30^\circ \) <strong><em>(M1)(A1)</em></strong></p>
<p>\(2\sin x \times \frac{1}{2} + 2\cos x \times \frac{{\sqrt 3 }}{2} = \cos x \times \frac{{\sqrt 3 }}{2} - \sin x \times \frac{1}{2}\) <strong><em>A1</em></strong></p>
<p>\( \Rightarrow \frac{3}{2}\sin x = - \frac{{\sqrt 3 }}{2}\cos x\)</p>
<p>\( \Rightarrow \tan x = - \frac{1}{{\sqrt 3 }}\) <strong><em>M1</em></strong></p>
<p>\( \Rightarrow x = 150^\circ \) <strong><em>A1</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>EITHER</strong></p>
<p>choosing two appropriate angles, for example 60° and 45° <strong><em>M1</em></strong></p>
<p>\(\sin 105^\circ = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \) and</p>
<p>\(\cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \) <strong><em>(A1)</em></strong></p>
<p>\(\sin 105^\circ + \cos 105^\circ = \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}\) <strong><em>A1</em></strong></p>
<p>\( = \frac{1}{{\sqrt 2 }}\) <strong><em>AG</em></strong></p>
<p><strong>OR</strong></p>
<p>attempt to square the expression <strong><em>M1</em></strong></p>
<p>\({(\sin 105^\circ + \cos 105^\circ )^2} = {\sin ^2}105^\circ + 2\sin 105^\circ \cos 105^\circ + {\cos ^2}105^\circ \)</p>
<p>\({(\sin 105^\circ + \cos 105^\circ )^2} = 1 + \sin 210^\circ \) <strong><em>A1</em></strong></p>
<p>\( = \frac{1}{2}\) <strong><em>A1</em></strong></p>
<p>\(\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}\) <strong><em>AG</em></strong></p>
<p> </p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>EITHER</strong></p>
<p>\(z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta \)</p>
<p>\(\left| z \right| = \sqrt {{{(1 - \cos 2\theta )}^2} + {{(\sin 2\theta )}^2}} \) <strong><em>M1</em></strong></p>
<p>\(\left| z \right| = \sqrt {1 - 2\cos 2\theta + {{\cos }^2}2\theta + {{\sin }^2}2\theta } \) <strong><em>A1</em></strong></p>
<p>\( = \sqrt 2 \sqrt {(1 - \cos 2\theta )} \) <strong><em>A1</em></strong></p>
<p>\( = \sqrt {2(2{{\sin }^2}\theta )} \)</p>
<p>\( = 2\sin \theta \) <strong><em>A1</em></strong></p>
<p>let \(\arg (z) = \alpha \)</p>
<p>\(\tan \alpha = - \frac{{\sin 2\theta }}{{1 - \cos 2\theta }}\) <strong><em>M1</em></strong></p>
<p>\( = \frac{{ - 2\sin \theta \cos \theta }}{{2{{\sin }^2}\theta }}\) <strong><em>(A1)</em></strong></p>
<p>\( = - \cot \theta \) <strong><em>A1</em></strong></p>
<p>\(\arg (z) = \alpha = - \arctan \left( {\tan \left( {\frac{\pi }{2} - \theta } \right)} \right)\) <strong><em>A1</em></strong></p>
<p>\( = \theta - \frac{\pi }{2}\) <strong><em>A1</em></strong></p>
<p><strong>OR</strong></p>
<p>\(z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta \)</p>
<p>\( = 2{\sin ^2}\theta - 2{\text{i}}\sin \theta \cos \theta \) <strong><em>M1A1</em></strong></p>
<p>\( = 2\sin \theta (\sin \theta - {\text{i}}\cos \theta )\) <strong><em>(A1)</em></strong></p>
<p>\( = - 2{\text{i}}\sin \theta (\cos \theta + {\text{i}}\sin \theta )\) <strong><em>M1A1</em></strong></p>
<p>\( = 2\sin \theta \left( {\cos \left( {\theta - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\theta - \frac{\pi }{2}} \right)} \right)\) <strong><em>M1A1</em></strong></p>
<p>\(\left| z \right| = 2\sin \theta \) <strong><em>A1</em></strong></p>
<p>\(\arg (z) = \theta - \frac{\pi }{2}\) <strong><em>A1</em></strong></p>
<p><strong><em>[9 marks]</em></strong></p>
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempt to apply De Moivre’s theorem <strong><em>M1</em></strong></p>
<p>\({(1 - \cos 2\theta - {\text{i}}\sin 2\theta )^{\frac{1}{3}}} = {2^{\frac{1}{3}}}{(\sin \theta )^{\frac{1}{3}}}\left[ {\cos \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right)} \right]\) <strong><em>A1A1A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> <strong><em>A1 </em></strong>for modulus, <strong><em>A1 </em></strong>for dividing argument of \(z\) by 3 and <strong><em>A1 </em></strong>for \(2n\pi \).</p>
<p> </p>
<p>Hence cube roots are the above expression when \(n = - 1,{\text{ }}0,{\text{ }}1\). Equivalent forms are acceptable. <strong><em>A1</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.ii.</div>
</div>
<br><hr><br><div class="specification">
<p>Consider the distinct complex numbers \(z = a + {\text{i}}b,\,\,w = c + {\text{i}}d\), where \(a,\,b,\,c,\,d \in \mathbb{R}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the real part of \(\frac{{z + w}}{{z - w}}\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the value of the real part of \(\frac{{z + w}}{{z - w}}\) when \(\left| z \right| = \left| w \right|\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{{z + w}}{{z - w}} = \frac{{\left( {a + c} \right) + {\text{i}}\left( {b + d} \right)}}{{\left( {a - c} \right) + {\text{i}}\left( {b - d} \right)}}\)</p>
<p>\( = \frac{{\left( {a + c} \right) + {\text{i}}\left( {b + d} \right)}}{{\left( {a - c} \right) + {\text{i}}\left( {b - d} \right)}} \times \frac{{\left( {a - c} \right) - {\text{i}}\left( {b - d} \right)}}{{\left( {a - c} \right) - {\text{i}}\left( {b - d} \right)}}\) <em><strong>M1A1</strong></em></p>
<p>real part \( = \frac{{\left( {a + c} \right)\left( {a - c} \right) + \left( {b + d} \right)\left( {b - d} \right)}}{{{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}} = \left( {\frac{{{a^2} - {c^2} + {b^2} - {d^2}}}{{{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}}} \right)\) <em><strong>A1A1</strong></em></p>
<p><strong>Note</strong>: Award <em><strong>A1</strong> </em>for numerator, <em><strong>A1</strong> </em>for denominator.</p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left| z \right| = \left| w \right| \Rightarrow {a^2} + {b^2} = {c^2} + {d^2}\) <em><strong>R1</strong></em></p>
<p>hence real part = 0 <em><strong>A1</strong></em></p>
<p><strong>Note:</strong> Do not award <em><strong>R0A1</strong></em>.</p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the complex number \(\omega = \frac{{z + {\text{i}}}}{{z + 2}}\), where \(z = x + {\text{i}}y\) and \({\text{i}} = \sqrt { - 1} \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) If \(\omega = {\text{i}}\), determine <em>z</em> in the form \(z = r\,{\text{cis}}\,\theta \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Prove that \(\omega = \frac{{({x^2} + 2x + {y^2} + y) + {\text{i}}(x + 2y + 2)}}{{{{(x + 2)}^2} + {y^2}}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) <strong>Hence</strong> show that when \(\operatorname{Re} (\omega) = 1\) the points \((x,{\text{ }}y)\) lie on a straight line, \({l_1}\), and write down its gradient.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) Given \(\arg (z) = \arg (\omega) = \frac{\pi }{4}\), find \(\left| z \right|\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{z + {\text{i}}}}{{z + 2}} = {\text{i}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z + {\text{i}} = {\text{i}}z + 2{\text{i}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((1 - {\text{i}})z = {\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{\text{i}}}{{1 - {\text{i}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{{\text{cis}}\left( {\frac{\pi }{2}} \right)}}{{\sqrt 2 \,{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{4\pi }}{4}} \right)} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{ - 1 + {\text{i}}}}{2}{\text{ }}\left( { = - \frac{1}{2} + \frac{1}{2}{\text{i}}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{i}} = \frac{{x + {\text{i}}(y + 1)}}{{x + 2 + {\text{i}}y}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x + {\text{i}}(y + 1) = - y + {\text{i}}(x + 2)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = - y;{\text{ }}x + 2 = y + 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solving, \(x = - \frac{1}{2};{\text{ }}y = \frac{1}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = - \frac{1}{2} + \frac{1}{2}{\text{i}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> fort the correct modulus and <strong><em>A1</em></strong> for the correct argument, but the final answer must be in the form \({\text{r}}\,{\text{cis}}\,\theta \). Accept 135° for the argument.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) substituting \(z = x + {\text{i}}y\) to obtain \(w = \frac{{x + (y + 1){\text{i}}}}{{(x + 2) + y{\text{i}}}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">use of \((x + 2) - y{\text{i}}\) to rationalize the denominator <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\omega = \frac{{x(x + 2) + y(y + 1) + {\text{i}}\left( { - xy + (y + 1)(x + 2)} \right)}}{{{{(x + 2)}^2} + {y^2}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{({x^2} + 2x + {y^2} + y) + {\text{i}}(x + 2y + 2)}}{{{{(x + 2)}^2} + {y^2}}}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \(\operatorname{Re} \omega = \frac{{{x^2} + 2x + {y^2} + y}}{{{{(x + 2)}^2} + {y^2}}} = 1\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {x^2} + 2x + {y^2} + y = {x^2} + 4x + 4 + {y^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow y = 2x + 4\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">which has gradient <em>m</em> = 2 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) <strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\arg (z) = \frac{\pi }{4} \Rightarrow x = y{\text{ (and }}x,{\text{ }}y > 0)\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\omega = \frac{{2{x^2} + 3x}}{{{{(x + 2)}^2} + {x^2}}} + \frac{{{\text{i}}(3x + 2)}}{{{{(x + 2)}^2} + {x^2}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">if \(\arg (\omega) = \theta \Rightarrow \tan \theta = \frac{{3x + 2}}{{2{x^2} + 3x}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{3x + 2}}{{2{x^2} + 3x}} = 1\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\arg (z) = \frac{\pi }{4} \Rightarrow x = y{\text{ (and }}x,{\text{ }}y > 0)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\arg (w) = \frac{\pi }{4} \Rightarrow {x^2} + 2x + {y^2} + y = x + 2y + 2\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solve simultaneously <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({x^2} + 2x + {x^2} + x = x + 2x + 2\) (or equivalent) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>THEN</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({x^2} = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = 1{\text{ (as }}x > 0)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A0</em></strong> for <em>x</em> = ±1.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| z \right| = \sqrt 2 \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow <strong><em>FT</em></strong> from incorrect values of <em>x</em>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [19 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates knew what had to be done in (a) but algebraic errors were fairly common. Parts (b) and (c) were well answered in general. Part (d), however, proved beyond many candidates who had no idea how to convert the given information into mathematical equations.</span></p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find three distinct roots of the equation \(8{z^3} + 27 = 0,{\text{ }}z \in \mathbb{C}\) <span class="s1">giving your answers in modulus-argument form.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">The roots are represented by the vertices of a triangle in an Argand diagram.</p>
<p class="p1">Show that the area of the triangle is \(\frac{{27\sqrt 3 }}{{16}}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\({z^3} = - \frac{{27}}{8} = \frac{{27}}{8}(\cos \pi + {\text{i}}\sin \pi )\) <strong><em>M1(A1)</em></strong></p>
<p>\( = \frac{{27}}{8}\left( {\cos (\pi + 2n\pi ) + {\text{i}}\sin (\pi + 2n\pi )} \right)\) (<strong><em>A1)</em></strong></p>
<p>\(z = \frac{3}{2}\left( {\cos \left( {\frac{{\pi + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\pi + 2n\pi }}{3}} \right)} \right)\) <strong><em>M1</em></strong></p>
<p>\({z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)\),</p>
<p>\({z_2} = \frac{3}{2}(\cos \pi + {\text{i}}\sin \pi )\),</p>
<p>\({z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)\). <strong><em>A2</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept \( - \frac{\pi }{3}\) as the argument for \({z_3}\).</p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for \(2\) correct roots<em>.</em></p>
<p> </p>
<p><strong>Note: </strong>Allow solutions expressed in Eulerian \((r{e^{{\text{i}}\theta }})\) form.</p>
<p> </p>
<p><strong>Note: </strong>Allow use of degrees in mod-arg (r-cis) form only.</p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>\(8{z^3} + 27 = 0\)</p>
<p>\( \Rightarrow z = - \frac{3}{2}\) so \((2z + 3)\) is a factor</p>
<p>Attempt to use long division or factor theorem: <strong><em>M1</em></strong></p>
<p>\( \Rightarrow 8{z^3} + 27 = (2z + 3)(4{z^2} - 6z + 9)\)</p>
<p>\( \Rightarrow 4{z^2} - 6z + 9 = 0\) <strong><em>A1</em></strong></p>
<p>Attempt to solve quadratic: <strong><em>M1</em></strong></p>
<p>\(z = \frac{{3 \pm 3\sqrt 3 {\text{i}}}}{4}\) <strong><em>A1</em></strong></p>
<p>\({z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)\),</p>
<p>\({z_2} = \frac{3}{2}(\cos \pi + {\text{i}}\sin \pi )\),</p>
<p>\({z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)\). <strong><em>A2</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept \( - \frac{\pi }{3}\) as the argument for \({z_3}\).</p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for \(2\) correct roots<em>.</em></p>
<p> </p>
<p><strong>Note: </strong>Allow solutions expressed in Eulerian \((r{e^{{\text{i}}\theta }})\) form.</p>
<p> </p>
<p><strong>Note: </strong>Allow use of degrees in mod-arg (r-cis) form only.</p>
<p> </p>
<p><strong>METHOD 3</strong></p>
<p>\(8{z^3} + 27 = 0\)</p>
<p>Substitute \(z = x + {\text{i}}y\) <strong><em>M1</em></strong></p>
<p>\(8({x^3} + 3{\text{i}}{x^2}y - 3x{y^2} - {\text{i}}{y^3}) + 27 = 0\)</p>
<p>\( \Rightarrow 8{x^3} - 24x{y^2} + 27 = 0\) and \(24{x^2}y - 8{y^3} = 0\) <strong><em>A1</em></strong></p>
<p>Attempt to solve simultaneously: <strong><em>M1</em></strong></p>
<p>\(8y(3{x^2} - {y^2}) = 0\)</p>
<p>\(y = 0,{\text{ }}y = x\sqrt 3 ,{\text{ }}y = - x\sqrt 3 \)</p>
<p>\( \Rightarrow \left( {x = - \frac{3}{2},{\text{ }}y = 0} \right),{\text{ }}x = \frac{3}{4},{\text{ }}y = \pm \frac{{3\sqrt 3 }}{4}\) <strong><em>A1</em></strong></p>
<p>\({z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)\),</p>
<p>\({z_2} = \frac{3}{2}(\cos \pi + {\text{i}}\sin \pi )\),</p>
<p>\({z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)\). <strong><em>A2</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept \( - \frac{\pi }{3}\) as the argument for \({z_3}\)<em>.</em></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for \(2\) correct roots<em>.</em></p>
<p> </p>
<p><strong>Note: </strong>Allow solutions expressed in Eulerian \((r{e^{{\text{i}}\theta }})\) form.</p>
<p> </p>
<p><strong>Note: </strong>Allow use of degrees in mod-arg (r-cis) form only.</p>
<p><em>[6 marks]</em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>EITHER</strong></p>
<p class="p2"><span class="s1">Valid attempt to use </span>\({\text{area}} = 3\left( {\frac{1}{2}ab\sin C} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2">\( = 3 \times \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times \frac{{\sqrt 3 }}{2}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>A1 </em></strong>for correct sides, <strong><em>A1 </em></strong><span class="s2">for correct sin \(C\)</span>.</p>
<p class="p3"> </p>
<p class="p1"><strong>OR</strong></p>
<p class="p2"><span class="s1">Valid attempt to use </span>\({\text{area}} = \frac{1}{2}{\text{base}} \times {\text{height}}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2">\({\text{area}} = \frac{1}{2} \times \left( {\frac{3}{4} + \frac{3}{2}} \right) \times \frac{{6\sqrt 3 }}{4}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span><em>A1 </em></strong>for correct height, <strong><em>A1 </em></strong>for correct base.</p>
<p class="p3"> </p>
<p class="p1"><strong>THEN</strong></p>
<p class="p2">\( = \frac{{27\sqrt 3 }}{{16}}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p2"><span class="s1"><strong><em>[3 marks]</em></strong></span></p>
<p class="p2"><span class="s1"><strong><em>Total [9 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the complex numbers \(z = 1 + 2{\text{i}}\) and \(w = 2 + a{\text{i}}\) , where \(a \in \mathbb{R}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find <em>a</em> when</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\left| w \right| = 2\left| z \right|;\) ;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({\text{Re}}(zw) = 2\operatorname{Im} (zw)\) .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\left| z \right| = \sqrt 5 \) and \(\left| w \right| = \sqrt {4 + {a^2}} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| w \right| = 2\left| z \right|\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sqrt {4 + {a^2}} = 2\sqrt 5 \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to solve equation <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M0</em></strong> if modulus is not used.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a = \pm 4\) <strong><em>A1A1</em></strong> <strong><em>N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(zw = (2 - 2a) + (4 + a){\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">forming equation \(2 - 2a = 2(4 + a)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a = - \frac{3}{2}\) <strong><em>A1 N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates made good attempts to answer this question. Weaker candidates did not get full marks due to difficulties recognizing the notation and working with modulus of a complex number.</span></p>
</div>
<br><hr><br><div class="question">
<p>In the following Argand diagram the point A represents the complex number \( - 1 + 4{\text{i}}\) and the point B represents the complex number \( - 3 + 0{\text{i}}\). The shape of ABCD is a square. Determine the complex numbers represented by the points C and D.</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2017-08-09_om_06.11.20.png" alt="M17/5/MATHL/HP1/ENG/TZ2/05"></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>C represents the complex number \(1 - 2{\text{i}}\) <strong><em>A2</em></strong></p>
<p>D represents the complex number \(3 + 2{\text{i}}\) <strong><em>A2</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p class="p1">Use mathematical induction to prove that \(n({n^2} + 5)\) <span class="s1">is divisible by 6 </span>for \(n \in {\mathbb{Z}^ + }\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1"><span class="s1">let \({\text{P}}(n)\) </span>be the proposition that \(n({n^2} + 5)\) <span class="s1">is divisible by 6 </span>for \(n \in {\mathbb{Z}^ + }\)</p>
<p class="p2">consider P(1)<span class="s2">:</span></p>
<p class="p1">when \(n = 1,{\text{ }}n({n^2} + 5) = 1 \times ({1^2} + 5) = 6\) <span class="s1">and so P(1) </span>is true <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="s1">assume \({\text{P}}(k)\) </span>is true <em>ie</em>, \(k({k^2} + 5) = 6m\) where \(k,{\text{ }}m \in {\mathbb{Z}^ + }\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Do not award <strong><em>M1 </em></strong>for statements such as “let \(n = k\)<span class="s1">”.</span></p>
<p class="p2">consider \({\text{P}}(k + 1)\):</p>
<p class="p2"><span class="Apple-converted-space">\((k + 1)\left( {{{(k + 1)}^2} + 5} \right)\) </span><span class="s2"><strong><em>M1</em></strong></span></p>
<p class="p5">\( = (k + 1)({k^2} + 2k + 6)\)</p>
<p class="p5"><span class="Apple-converted-space">\( = {k^3} + 3{k^2} + 8k + 6\) </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p5"><span class="Apple-converted-space">\( = ({k^3} + 5k) + (3{k^2} + 3k + 6)\) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p5"><span class="Apple-converted-space">\( = k({k^2} + 5) + 3k(k + 1) + 6\) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2">\(k(k + 1)\) is even hence all three terms are divisible by 6 <span class="Apple-converted-space"> </span><span class="s2"><strong><em>R1</em></strong></span></p>
<p class="p2">\({\text{P}}(k + 1)\) is true whenever \({\text{P}}(k)\) is true and P(1) is true, so \({\text{P}}(n)\) <span class="s2">is true for \(n \in {\mathbb{Z}^ + }\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></span></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>To obtain the final <strong><em>R1</em></strong>, four of the previous marks must have been awarded.</p>
<p class="p1"><strong><em>[8 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">This proved to be a good discriminator. The average candidate seemed able to work towards \({\text{P}}(k + 1) = {k^3} + 3{k^2} + 8k + 6\), and a number made some further progress.</p>
<p class="p1">Unfortunately, even otherwise good candidates are still writing down incorrect or incomplete induction statements, such as ‘Let \(n = k\)’ rather than ‘Suppose true for \(n = k\)’ (or equivalent).</p>
<p class="p1">It was also noted than an increasing number of candidates this session assumed ‘\({\text{P}}(n)\) to be true’ before going to consider \({\text{P}}(n + 1)\). Showing a lack of understanding of the induction argument, these approaches scored very few marks.</p>
</div>
<br><hr><br><div class="specification">
<p>It is given that \({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x\) where \(r,\,x \in {\mathbb{R}^ + }\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Express \(y\) in terms of \(x\). Give your answer in the form \(y = p{x^q}\), where <em>p</em> , <em>q</em> are constants.</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>The region <em>R</em>, is bounded by the graph of the function found in part (b), the <em>x</em>-axis, and the lines \(x = 1\) and \(x = \alpha \) where \(\alpha > 1\). The area of <em>R</em> is \(\sqrt 2 \).</p>
<p>Find the value of \(\alpha \).</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)\) <em><strong> M1A1</strong></em></p>
<p>\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}\) <em><strong>M1</strong></em></p>
<p>\( = \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}\) <em><strong>A1</strong></em></p>
<p>\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<p> </p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)</p>
<p>\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0\) <em><strong>M1</strong></em></p>
<p>\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0\) <em><strong>M1</strong></em></p>
<p>\({\text{lo}}{{\text{g}}_2}\,y = - \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}\)</p>
<p>\({\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)\) <em><strong>M1A1</strong></em></p>
<p>\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\) <em><strong>A1</strong></em></p>
<p><strong>Note</strong>: For the final <em><strong>A</strong></em> mark, \(y\) must be expressed in the form \(p{x^q}\).</p>
<p><em><strong>[5 marks]</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)</p>
<p>\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0\) <strong><em>M1</em></strong></p>
<p>\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0\) <em><strong>M1</strong></em></p>
<p>\({\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0\) <em><strong>M1</strong></em></p>
<p>\(\sqrt 2 xy = 1\) <strong><em>A1</em></strong></p>
<p>\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\) <em><strong>A1</strong></em></p>
<p><strong>Note</strong>: For the final <em><strong>A</strong></em> mark, \(y\) must be expressed in the form \(p{x^q}\).</p>
<p><em><strong>[5 marks]</strong></em></p>
<p> </p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the area of <em>R</em> is \(\int\limits_1^\alpha {\frac{1}{{\sqrt 2 }}} {x^{ - 1}}{\text{d}}x\) <em><strong>M1</strong></em></p>
<p>\( = \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha \) <em><strong>A1</strong></em></p>
<p>\( = \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha \) <em><strong>A1</strong></em></p>
<p>\(\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha = \sqrt 2 \) <em><strong>M1</strong></em></p>
<p>\(\alpha = {{\text{e}}^2}\) <em><strong>A1</strong></em></p>
<p><strong>Note:</strong> Only follow through from part (b) if \(y\) is in the form \(y = p{x^q}\)</p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">An 81 metre rope is cut into <em>n</em> pieces of increasing lengths that form an arithmetic sequence with a common difference of <em>d</em> metres. Given that the lengths of the shortest and longest pieces are 1.5 metres and 7.5 metres respectively, find the values of <em>n</em> and <em>d</em> .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(81 = \frac{n}{2}(1.5 + 7.5)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow n = 18\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(1.5 + 17d = 7.5\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow d = \frac{6}{{17}}\) <strong><em>A1</em></strong> <strong><em>N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 18.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">There were many totally correct solutions to this question, but a number of candidates found two simultaneous equations and then spent a lot of time and working trying, often unsuccessfully, to solve these equations.</span></p>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Given that \(\frac{z}{{z + 2}} = 2 - {\text{i}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> , \(z \in \mathbb{C}\) , find z in the form \(a + {\text{i}}b\) .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong><br><span style="font-family: times new roman,times; font-size: medium;">\(z = \left( {2 - {\text{i}}} \right)\left( {z + 2} \right)\) <em><strong>M1</strong></em></span><br><span style="font-family: times new roman,times; font-size: medium;">\( = 2z + 4 - {\text{i}}z - 2{\text{i}}\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\(z\left( {1 - {\text{i}}} \right) = - 4 + 2{\text{i}}\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\(z = \frac{{ - 4 + 2{\text{i}}}}{{1 - {\text{i}}}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em><br><span style="font-family: times new roman,times; font-size: medium;">\(z = \frac{{ - 4 + 2{\text{i}}}}{{1 - {\text{i}}}} \times \frac{{1 + {\text{i}}}}{{1 + {\text{i}}}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em><br><span style="font-family: times new roman,times; font-size: medium;">\( = - 3 - {\text{i}}\) <em><strong>A1</strong></em></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 2</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">let \(z = a + {\text{i}}b\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{a + {\text{i}}b}}{{a + {\text{i}}b + 2}} = 2 - {\text{i}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em><br><span style="font-family: times new roman,times; font-size: medium;">\(a + {\text{i}}b = \left( {2 - i} \right)\left( {\left( {a + 2} \right) + {\text{i}}b} \right)\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\(a + {\text{i}}b = 2\left( {a + 2} \right) + 2b{\text{i}} - {\text{i}}\left( {a + 2} \right) + b\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\(a + {\text{i}}b = 2a + b + 4 + \left( {2b - a - 2} \right){\text{i}}\)</span><br><span style="font-family: times new roman,times; font-size: medium;">attempt to equate real and imaginary parts <em><strong>M1</strong></em></span><br><span style="font-family: times new roman,times; font-size: medium;">\(a = 2a + b + 4\left( { \Rightarrow a + b + 4 = 0} \right)\)</span><br><span style="font-family: times new roman,times; font-size: medium;">and \(b = 2b - a - 2\left( { \Rightarrow - a + b - 2 = 0} \right)\) <em><strong>A1</strong></em></span></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><em style="font-family: 'times new roman', times; font-size: medium;"><strong>Al</strong></em><span style="font-family: 'times new roman', times; font-size: medium;"> for two correct equations.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = - 1\); \(a = - 3\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(z = - 3 - {\text{i}}\)</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">A number of different methods were adopted in this question with some candidates working through their method to a correct answer. However many other candidates either stopped with \(z\) still expressed as a quotient of two complex numbers or made algebraic mistakes.</span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \((4 - 5{\text{i}})m + 4n = 16 + 15{\text{i}}\) , where \({{\text{i}}^2} = - 1\), find <em>m </em>and <em>n </em>if</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>m </em>and <em>n </em>are real numbers;</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>m </em>and <em>n </em>are conjugate complex numbers.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to equate real and imaginary parts <strong><em>M1<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">equate real parts: \(4m + 4n = 16\); equate imaginary parts: \( -5m = 15\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow m = -3,{\text{ }}n = 7\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[3 marks]</span><br></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(m = x + {\text{i}}y,{\text{ }}n = x - {\text{i}}y\) <strong><em>M1</em></strong></span></p>
<p style="font: normal normal normal 11px/normal Times; text-align: left; margin: 0px;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow (4 - 5{\text{i}})(x + {\text{i}}y) + 4(x - {\text{i}}y) = 16 + 15{\text{i}}\)</span></p>
<p style="font: normal normal normal 11px/normal Times; text-align: left; margin: 0px;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 4x - 5{\text{i}}x + 4{\text{i}}y + 5y + 4x - 4{\text{i}}y = 16 + 15{\text{i}}\)</span></p>
<p style="font: normal normal normal 11px/normal Times; text-align: left; margin: 0px;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to equate real and imaginary parts <strong><em>M1<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(8x + 5y = 16,{\text{ }} -5x = 15\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow x = -3,{\text{ }}y = 8\) <strong> <em>A1</em></strong></span></p>
<p style="font: normal normal normal 11px/normal Times; text-align: left; margin: 0px;"><span style="font-family: 'times new roman', times; font-size: medium;">\(( \Rightarrow m = -3 + 8{\text{i}},{\text{ }}n = -3 - 8{\text{i}})\)</span></p>
<p style="font: normal normal normal 11px/normal Times; text-align: left; margin: 0px;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[4 marks]</span><br></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) was generally well answered. In (b), however, some candidates put \(m = a + {\text{i}}b\) and \(n = c + {\text{i}}d\) which gave four equations for two unknowns so that no further progress could be made.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) was generally well answered. In (b), however, some candidates put \(m = a + {\text{i}}b\) and \(n = c + {\text{i}}d\) which gave four equations for two unknowns so that no further progress could be made.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the complex number \(z = \cos \theta + {\text{i}}\sin \theta \).</span></p>
</div>
<div class="specification">
<p><span style="font-family: 'times new roman', times; font-size: medium;">The region <em>S</em> is bounded by the curve \(y = \sin x{\cos ^2}x\) and the <em>x</em>-axis between \(x = 0\) and \(x = \frac{\pi }{2}\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Use De Moivre’s theorem to show that \({z^n} + {z^{ - n}} = 2\cos n\theta ,{\text{ }}n \in {\mathbb{Z}^ + }\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Expand \({\left( {z + {z^{ - 1}}} \right)^4}\).</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence show that \({\cos ^4}\theta = p\cos 4\theta + q\cos 2\theta + r\), where \(p,{\text{ }}q\) and \(r\) are constants to </span><span style="font-family: 'times new roman', times; font-size: medium;">be determined.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Show that \({\cos ^6}\theta = \frac{1}{{32}}\cos 6\theta + \frac{3}{{16}}\cos 4\theta + \frac{{15}}{{32}}\cos 2\theta + \frac{5}{{16}}\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Hence find the value of \(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta } \).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>S</em> is rotated through \(2\pi \) radians about the <em>x</em>-axis. Find the value of the volume generated.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">f.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">(i) Write down an expression for the constant term in the expansion of \({\left( {z + {z^{ - 1}}} \right)^{2k}}\), \(k \in {\mathbb{Z}^ + }\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Hence determine an expression for \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta } \) in terms of <em>k</em>.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">g.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^n} + {z^{ - n}} = \cos n\theta + i\sin n\theta + \cos ( - n\theta ) + i\sin ( - n\theta )\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \cos n\theta + \cos n\theta + i\sin n\theta - i\sin n\theta \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2\cos n\theta \) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({\left( {z + {z^{ - 1}}} \right)^4} = {z^4} + 4{z^3}\left( {\frac{1}{z}} \right) + 6{z^2}\left( {\frac{1}{{{z^2}}}} \right) + 4z\left( {\frac{1}{{{z^3}}}} \right) + \frac{1}{{{z^4}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept \({\left( {z + {z^{ - 1}}} \right)^4} = 16{\cos ^4}\theta \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {z + {z^{ - 1}}} \right)^4} = \left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 4\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 6\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(2\cos \theta )^4} = 2\cos 4\theta + 8\cos 2\theta + 6\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1 </em></strong>for RHS, <strong><em>A1 </em></strong>for LHS, independent of the <strong><em>M1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\cos ^4}\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\cos ^4}\theta = {\left( {\frac{{\cos 2\theta + 1}}{2}} \right)^2}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{4}({\cos ^2}2\theta + 2\cos 2\theta + 1)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{4}\left( {\frac{{\cos 4\theta + 1}}{2} + 2\cos 2\theta + 1} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\cos ^4}\theta = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {z + {z^{ - 1}}} \right)^6} = {z^6} + 6{z^5}\left( {\frac{1}{z}} \right) + 15{z^4}\left( {\frac{1}{{{z^2}}}} \right) + 20{z^3}\left( {\frac{1}{{{z^3}}}} \right) + 15{z^2}\left( {\frac{1}{{{z^4}}}} \right) + 6z\left( {\frac{1}{{{z^5}}}} \right) + \frac{1}{{{z^6}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {z + {z^{ - 1}}} \right)^6} = \left( {{z^6} + \frac{1}{{{z^6}}}} \right) + 6\left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 15\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 20\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(2\cos \theta )^6} = 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1 </em></strong>for RHS, <strong><em>A1 </em></strong>for LHS, independent of the <strong><em>M1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({\cos ^6}\theta = \frac{1}{{32}}\cos 6\theta + \frac{3}{{16}}\cos 4\theta + \frac{{15}}{{32}}\cos 2\theta + \frac{5}{{16}}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept a purely trigonometric solution as for (c).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta = \int_0^{\frac{\pi }{2}} {\left( {\frac{1}{{32}}\cos 6\theta + \frac{3}{{16}}\cos 4\theta + \frac{{15}}{{32}}\cos 2\theta + \frac{5}{{16}}} \right){\text{d}}\theta } } \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left[ {\frac{1}{{192}}\sin 6\theta + \frac{3}{{64}}\sin 4\theta + \frac{{15}}{{64}}\sin 2\theta + \frac{5}{{16}}\theta } \right]_0^{\frac{\pi }{2}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{5\pi }}{{32}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{V}} = \pi \int_0^{\frac{\pi }{2}} {{{\sin }^2}x{{\cos }^4}x{\text{d}}x} \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \pi \int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x - \pi \int_0^{\frac{\pi }{2}} {{{\cos }^6}x{\text{d}}x} } \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x} = \frac{{3\pi }}{{16}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{V}} = \frac{{3{\pi ^2}}}{{16}} - \frac{{5{\pi ^2}}}{{32}} = \frac{{{\pi ^2}}}{{32}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Follow through from an incorrect <em>r</em> in (c) provided the final answer is positive.</span></p>
<div class="question_part_label">f.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">(i) constant term = \(\left( \begin{array}{c}2k\\k\end{array} \right)\) \( = \frac{{(2k)!}}{{k!k!}} = \frac{{(2k)!}}{{{{(k!)}^2}}}{\text{ (accept }}C_k^{2k})\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({2^{2k}}\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta = \frac{{(2k)!\pi }}{{{{(k!)}^2}}}\frac{\pi }{2}} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta = \frac{{(2k)!\pi }}{{{2^{2k + 1}}{{(k!)}^2}}}} \) \(\left( {{\rm{or}}\frac{{\left( \begin{array}{c}2k\\k\end{array} \right)\pi }}{{{2^{2k + 1}}}}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">g.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part a) has appeared several times before, though with it again being a ‘show that’ question, some candidates still need to be more aware of the need to show every step in their working, including the result that \(\sin ( - n\theta ) = - \sin (n\theta )\).</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part b) was usually answered correctly.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part c) was again often answered correctly, though some candidates often less successfully utilised a trig-only approach rather than taking note of part b).</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part d) was a good source of marks for those who kept with the spirit of using complex numbers for this type of question. Some limited attempts at trig-only solutions were seen, and correct solutions using this approach were extremely rare.</span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part e) was well answered, though numerical slips were often common. A small number integrated \(\sin n\theta \) as \(n\cos n\theta \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">A large number of candidates did not realise the help that part e) inevitably provided for part f). Some correctly expressed the volume as \(\pi \int {{{\cos }^4}x{\text{d}}x - \pi \int {{{\cos }^6}x{\text{d}}x} } \) and thus gained the first 2 marks but were able to progress no further. Only a small number of able candidates were able to obtain the correct answer of \(\frac{{{\pi ^2}}}{{32}}\).</span></p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">f.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part g) proved to be a challenge for the vast majority, though it was pleasing to see some of the highest scoring candidates gain all 3 marks.</span></p>
<div class="question_part_label">g.</div>
</div>
<br><hr><br><div class="specification">
<p>The geometric sequence <em>u</em><sub>1</sub>, <em>u</em><sub>2</sub>, <em>u</em><sub>3</sub>, … has common ratio <em>r.</em></p>
<p>Consider the sequence \(A = \left\{ {{a_n} = {\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right|{\text{:}}\,n \in {\mathbb{Z}^ + }} \right\}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that <em>A</em> is an arithmetic sequence, stating its common difference<em> d</em> in terms of <em>r</em>.</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>A particular geometric sequence has <em>u</em><sub>1</sub> = 3 and a sum to infinity of 4.</p>
<p>Find the value of <em>d</em>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>state that \({u_n} = {u_1}{r^{n - 1}}\) (or equivalent) <em><strong>A1</strong></em></p>
<p>attempt to consider \({{a_n}}\) and use of at least one log rule <em><strong>M1</strong></em></p>
<p>\({\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right| = {\text{lo}}{{\text{g}}_2}\left| {{u_1}} \right| + \left( {n - 1} \right){\text{lo}}{{\text{g}}_2}\left| r \right|\) <em><strong>A1</strong></em></p>
<p>(which is an AP) with \(d = {\text{lo}}{{\text{g}}_2}\left| r \right|\) (and 1<sup>st</sup> term \({\text{lo}}{{\text{g}}_2}\left| {{u_1}} \right|\)) <em><strong> A1</strong></em></p>
<p>so A is an arithmetic sequence <em><strong>AG</strong></em></p>
<p><strong>Note:</strong> Condone absence of modulus signs.</p>
<p><strong>Note:</strong> The final <em><strong>A</strong></em> mark may be awarded independently.</p>
<p><strong>Note:</strong> Consideration of the first two or three terms only will score <em><strong>M0</strong></em>.</p>
<p><em><strong>[4 marks]</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>consideration of \(\left( {d = } \right){a_{n + 1}} - {a_n}\) <em><strong>M1</strong></em></p>
<p>\(\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| {{u_{n + 1}}} \right| - {\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right|\)</p>
<p>\(\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right|\) <em><strong>M1</strong></em></p>
<p>\(\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| r \right|\) <em><strong>A1</strong></em></p>
<p>which is constant <em><strong>R1</strong></em></p>
<p><strong>Note:</strong> Condone absence of modulus signs.</p>
<p><strong>Note:</strong> The final <em><strong>A</strong></em> mark may be awarded independently.</p>
<p><strong>Note:</strong> Consideration of the first two or three terms only will score <em><strong>M0</strong></em>.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempting to solve \(\frac{3}{{1 - r}} = 4\) <em><strong>M1</strong></em></p>
<p>\(r = \frac{1}{4}\) <em><strong>A1</strong></em></p>
<p>\(d = - \,2\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the following equations, where <em>a </em>, \(b \in \mathbb{R}:\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x + 3y + (a - 1)z = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(2x + 2y + (a - 2)z = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(3x + y + (a - 3)z = b.\)</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">If each of these equations defines a plane, show that, for any value of <em>a </em>, the planes do not intersect at a unique point.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the value of <em>b </em>for which the intersection of the planes is a straight line.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\det \left( {\begin{array}{*{20}{c}}<br> 1&3&{a - 1} \\ <br> 2&2&{a - 2} \\ <br> 3&1&{a - 3} <br>\end{array}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1\left( {2(a - 3) - (a - 2)} \right) - 3\left( {2(a - 3) - 3(a - 2)} \right) + (a - 1)(2 - 6)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(or equivalent) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= 0 (therefore there is no unique solution) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span><span style="font-family: 'Helvetica Neue', Arial, 'Lucida Grande', 'Lucida Sans Unicode', sans-serif; font-size: 11px;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><strong>METHOD 2</strong></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>1&3&{a - 1}\\<br>2&2&{a - 2}\\<br>3&1&{a - 3}<br>\end{array}} \right|\left. {\begin{array}{*{20}{c}}<br>1\\<br>1\\<br>b<br>\end{array}} \right):\left( {\begin{array}{*{20}{c}}<br>1&3&{a - 1}\\<br>0&{ - 4}&{ - a}\\<br>0&{ - 8}&{ - 2a}<br>\end{array}} \right|\left. {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>{b - 3}<br>\end{array}} \right)\) <strong><em><strong>M1A1</strong></em></strong></span></p>
<p style="font: normal normal normal 11px/normal Helvetica; display: inline !important; margin: 0px;"><span style="font-family: 'times new roman', times; font-size: medium;">\(:\left( {\begin{array}{*{20}{c}}<br>1&3&{a - 1}\\<br>0&{ - 4}&{ - a}\\<br>0&0&0<br>\end{array}} \right|\left. {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>{b - 1}<br>\end{array}} \right)\) (and 3 zeros imply no unique solution) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: times new roman,times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p> </p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>1&3&{a - 1}\\<br>2&2&{a - 2}\\<br>3&1&{a - 3}<br>\end{array}} \right|\left. {\begin{array}{*{20}{c}}<br>1\\<br>1\\<br>b<br>\end{array}} \right):\left( {\begin{array}{*{20}{c}}<br>1&3&{a - 1}\\<br>0&{ - 4}&{ - a}\\<br>0&{ - 8}&{ - 2a}<br>\end{array}} \right|\left. {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>{b - 3}<br>\end{array}} \right)\) <em><strong>M1A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(:\left( {\begin{array}{*{20}{c}}<br>1&3&{a - 1}\\<br>0&{ - 4}&{ - a}\\<br>0&0&0<br>\end{array}} \right|\left. {\begin{array}{*{20}{c}}<br>1\\<br>{ - 1}\\<br>{b - 1}<br>\end{array}} \right)\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><em>b</em> = 1 <em><strong>A1 N2</strong></em></span></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">for an attempt to use row operations.</span></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[4 marks]</em></strong></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 2</strong></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><em>b</em> = 1 <em><strong>A4</strong></em></span></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A4 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">only if “ </span><em style="font-family: 'times new roman', times; font-size: medium;">b </em><span style="font-family: 'times new roman', times; font-size: medium;">−1 ” seen in (a).</span></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">The best candidates used row reduction correctly in part a) and were hence able to deduce <em>b</em> = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">The best candidates used row reduction correctly in part a) and were hence able to deduce <em>b</em> = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the complex numbers \({z_1} = 1 + \sqrt 3 {\text{i, }}{z_2} = 1 + {\text{i}}\) and \(w = \frac{{{z_1}}}{{{z_2}}}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By expressing \({z_1}\) and \({z_2}\) in modulus-argument form write down the modulus of \(w\);</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By expressing \({z_1}\) and \({z_2}\) in modulus-argument form write down the argument of \(w\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the smallest positive integer value of \(n\), such that \({w^n}\) is a real number.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\({z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)\) and \({z_2} = \sqrt 2 {\text{cis}}\left( {\frac{\pi }{4}} \right)\) <em><strong>A1A1</strong></em></p>
<p> </p>
<p><strong>Note:</strong> Award <em><strong>A1A0 </strong></em>for correct moduli and arguments found, but not written in mod-arg form.</p>
<p> </p>
<p>\(\left| w \right| = \sqrt 2 \) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)\) and \({z_2} = \sqrt 2 {\text{cis}}\left( {\frac{\pi }{4}} \right)\) <em><strong>A1A1</strong></em></p>
<p> </p>
<p><strong>Note:</strong> Award <em><strong>A1A0 </strong></em>for correct moduli and arguments found, but not written in mod-arg form.</p>
<p> </p>
<p>\(\arg w = \frac{\pi }{{12}}\) <em><strong>A1</strong></em></p>
<p> </p>
<p><strong>Notes:</strong> Allow <em><strong>FT </strong></em>from incorrect answers for \({z_1}\) and \({z_2}\) in modulus-argument form.</p>
<p> </p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>EITHER</strong></p>
<p>\(\sin \left( {\frac{{\pi n}}{{12}}} \right) = 0\) <em><strong>(M1)</strong></em></p>
<p><strong>OR</strong></p>
<p>\(\arg ({w^n}) = \pi \) <em><strong>(M1)</strong></em></p>
<p>\(\frac{{n\pi }}{{12}} = \pi \)</p>
<p><strong>THEN</strong></p>
<p>\(\therefore n = 12\) <em><strong>A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the complex numbers</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2}\) and \({z_2} = - 1 + \sqrt 3 {\text{i }}\)</span><span style="font-family: 'times new roman', times; font-size: medium;"> .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-size: medium; font-family: 'times new roman', times;">(i) Write down \({z_1}\) in Cartesian form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-size: medium; font-family: 'times new roman', times;">(ii) Hence determine \({({z_1} + {z_2})^ * }\) in Cartesian form.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Write \({z_2}\) in modulus-argument form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Hence solve the equation \({z^3} = {z_2}\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Let \(z = r\,{\text{cis}}\theta \) , where \(r \in {\mathbb{R}^ + }\) <span style="font: 7.0px Times;"> </span>and \(0 \leqslant \theta < 2\pi \) . Find all possible values of <em>r </em>and \(\theta \)<span style="font: 12.5px Times;"> ,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) if \({z^2} = {(1 + {z_2})^2}\);</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) if \(z = - \frac{1}{{{z_2}}}\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the smallest positive value of <em>n </em>for which \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} \in {\mathbb{R}^ + }\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">(i) \({z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2} \Rightarrow {z_1} = - 2\sqrt 3 {\text{i}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong> </strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({z_1} + {z_2} = - 2\sqrt 3 {\text{i}} - 1 + \sqrt 3 {\text{i}} = - 1 - \sqrt 3 {\text{i}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({({z_1} + {z_2})^ * } = - 1 + \sqrt 3 {\text{i}}\) <em><strong>A1<br></strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong>[3 marks]</strong></em></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<div><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(\left| {{z_2}} \right| = 2\)</span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;">\(\tan \theta = - \sqrt 3 \) <em><strong>(M1)</strong></em></span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;">\({z_2}\) lies on the second quadrant</span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;">\(\theta = \arg {z_2} = \frac{{2\pi }}{3}\)</span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;">\({z_2} = 2{\text{cis}}\frac{{2\pi }}{3}\) <em><strong>A1A1</strong></em></span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong> </strong></em></span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;">(ii) attempt to use De Moivre’s theorem <em><strong>M1</strong></em></span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \sqrt[3]{2}\,{\text{cis}}\frac{{\frac{{2\pi }}{3} + 2k\pi }}{3},{\text{ }}k = 0{\text{, 1 and 2}}\)</span></div>
<div><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \sqrt[3]{2}\,{\text{cis}}\frac{{2\pi }}{9},\,\sqrt[3]{2}\,{\text{cis}}\frac{{8\pi }}{9},\,\sqrt[3]{2}\,{\text{cis}}\frac{{14\pi }}{9}\left( { = \sqrt[3]{2}\,{\text{cis}}\left( {\frac{{ - 4\pi }}{9}} \right)} \right)\) <em><strong>A1A1</strong></em></span></div>
<div><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><em style="font-family: 'times new roman', times; font-size: medium;"><strong>A1</strong></em><span style="font-family: 'times new roman', times; font-size: medium;"> for modulus, </span><em style="font-family: 'times new roman', times; font-size: medium;"><strong>A1</strong></em><span style="font-family: 'times new roman', times; font-size: medium;"> for arguments.</span></div>
<div> </div>
<div><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow equivalent forms for <em>z</em> .</span></div>
<div> </div>
<div><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong>[6 marks]</strong></em></span></div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">(i)</span> <span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^2} = {\left( {1 - 1 + \sqrt 3 {\text{i}}} \right)^2} = - 3\left( { \Rightarrow z = \pm \sqrt 3 {\text{i}}} \right)\) <em><strong>M1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \sqrt 3 \,{\text{cis}}\frac{\pi }{2}{\text{ or }}{z_1} = \sqrt 3 \,{\text{cis}}\frac{{3\pi }}{2}\left( { = \sqrt 3 \,{\text{cis}}\left( {\frac{{ - \pi }}{2}} \right)} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so \(r = \sqrt 3 {\text{ and }}\theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}\left( { = \frac{{ - \pi }}{2}} \right)\)</span><span style="font-family: 'Helvetica Neue', Arial, 'Lucida Grande', 'Lucida Sans Unicode', sans-serif; font-size: 11px;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept \(r\,{\text{cis}}(\theta )\) form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"> <strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 2</strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^2} = {\left( {1 - 1 + \sqrt 3 {\text{i}}} \right)^2} = - 3 \Rightarrow {z^2} = 3{\text{cis}}\left( {(2n + 1)\pi } \right)\) <em style="font-style: italic;"><strong style="font-weight: bold;">M1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"> <span style="font-family: 'times new roman', times; font-size: medium;">\({r^2} = 3 \Rightarrow r = \sqrt 3 \) </span><strong style="font-weight: bold;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(2\theta = (2n + 1)\pi \Rightarrow \theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}{\text{ (as }}0 \leqslant \theta < 2\pi )\) <em><strong>A1</strong></em></span><em style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></em></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept \(r\,{\text{cis}}(\theta )\) form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = - \frac{1}{{2{\text{cis}}\frac{{2\pi }}{3}}} \Rightarrow z = \frac{{{\text{cis}}\pi }}{{2{\text{cis}}\frac{{2\pi }}{3}}}\) <em style="font-style: italic;"><strong style="font-weight: bold;">M1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so \(r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}\) <em><strong>A1A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z_1} = - \frac{1}{{ - 1 + \sqrt 3 {\text{i}}}} \Rightarrow {z_1} = - \frac{{ - 1 - \sqrt 3 {\text{i}}}}{{\left( { - 1 + \sqrt 3 {\text{i}}} \right)\left( { - 1 - \sqrt 3 {\text{i}}} \right)}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{1 + \sqrt 3 {\text{i}}}}{4} \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so \(r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{z_1}}}{{{z_2}}} = \sqrt 3 \,{\text{cis}}\frac{{5\pi }}{6}\) (<strong><em>A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} = {\sqrt 3 ^n}{\text{cis}}\frac{{5n\pi }}{6}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">equating imaginary part to zero and attempting to solve <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">obtain <em>n </em>= 12 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Working which only includes the argument is valid.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding <em>n</em> = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding <em>n</em> = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding <em>n</em> = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding <em>n</em> = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.</span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) The sum of the first six terms of an arithmetic series is 81. The sum of its first eleven terms is 231. Find the first term and the common difference.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) The sum of the first two terms of a geometric series is 1 and the sum of its first four terms is 5. If all of its terms are positive, find the first term and the common ratio.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) The \({r^{{\text{th}}}}\) term of a new series is defined as the product of the \({r^{{\text{th}}}}\) term of the arithmetic series and the \({r^{{\text{th}}}}\) term of the geometric series above. Show that the \({r^{{\text{th}}}}\) term of this new series is \((r + 1){2^{r - 1}}\) .</span></p>
<div class="marks">[14]</div>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Using mathematical induction, prove that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[\sum\limits_{r = 1}^n {(r + 1){2^{r - 1}} = n{2^n},{\text{ }}n \in {\mathbb{Z}^ + }.} \]</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \({S_6} = 81 \Rightarrow 81 = \frac{6}{2}(2a + 5d)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 27 = 2a + 5d\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({S_{11}} = 231 \Rightarrow 231 = \frac{{11}}{2}(2a + 10d)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 21 = a + 5d\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solving simultaneously, <em>a</em> = 6 , <em>d</em> = 3 <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(a + ar = 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a + ar + a{r^2} + a{r^3} = 5\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow (a + ar) + a{r^2}(1 + r) = 5\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 1 + a{r^2} \times \frac{1}{a} = 5\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">obtaining \({r^2} - 4 = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow r = \pm 2\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = 2\,\,\,\,\,\)(since all terms are positive) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a = \frac{1}{3}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \({\text{AP }}{r^{{\text{th}}}}{\text{ term is }}3r + 3\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{GP }}{r^{{\text{th}}}}{\text{ term is }}\frac{1}{3}{2^{r - 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(3(r + 1) \times \frac{1}{3}{2^{r - 1}} = (r + 1){2^{r - 1}}\) <strong><em>M1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [14 marks]</em></strong></span></p>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">prove: \({P_n}:\sum\limits_{r = 1}^n {(r + 1){2^{r - 1}} = n{2^n},{\text{ }}n \in {\mathbb{Z}^ + }.} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">show true for <em>n</em> = 1 , <em>i.e.</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{LHS}} = 2 \times {2^0} = 2 = {\text{RHS}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">assume true for <em>n</em> = <em>k</em> , <em>i.e.</em> <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sum\limits_{r = 1}^k {(r + 1){2^{r - 1}} = k{2^k},{\text{ }}k \in {\mathbb{Z}^ + }} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">consider <em>n</em> = <em>k</em> +1</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sum\limits_{r = 1}^{k + 1} {(r + 1){2^{r - 1}} = k{2^k} + (k + 2){2^k}} \) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^k}(k + k + 2)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2(k + 1){2^k}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = (k + 1){2^{k + 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence true for <em>n</em> = <em>k</em> + 1</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({P_{k + 1}}\) is true whenever \({P_k}\) is true, and \({P_1}\)is true, therefore \({P_n}\) is true <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for \(n \in {\mathbb{Z}^ + }\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Parts (a), (b) and (c) were answered successfully by a large number of candidates. Some, however, had difficulty with the arithmetic.</span></p>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In part (d) many candidates showed little understanding of sigma notation and proof by induction. There were cases of circular reasoning and using <em>n</em>, <em>k</em> and <em>r</em> randomly. A concluding sentence almost always appeared, even if the proof was done incorrectly, or not done at all.</span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p>The 1st, 4th and 8th terms of an arithmetic sequence, with common difference \(d\), \(d \ne 0\), are the first three terms of a geometric sequence, with common ratio \(r\). Given that the 1st term of both sequences is 9 find</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>the value of \(d\);</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>the value of \(r\);</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong>EITHER</strong></p>
<p>the first three terms of the geometric sequence are \(9\), \(9r\) and \(9{r^2}\) <strong><em>(M1)</em></strong></p>
<p>\(9 + 3d = 9r( \Rightarrow 3 + d = 3r)\) and \(9 + 7d = 9{r^2}\) <strong><em>(A1)</em></strong></p>
<p>attempt to solve simultaneously <strong><em>(M1)</em></strong></p>
<p>\(9 + 7d = 9{\left( {\frac{{3 + d}}{3}} \right)^2}\)</p>
<p><strong>OR</strong></p>
<p>the \({{\text{1}}^{{\text{st}}}}\), \({{\text{4}}^{{\text{th}}}}\) and \({{\text{8}}^{{\text{th}}}}\) terms of the arithmetic sequence are</p>
<p>\(9,{\text{ }}9 + 3d,{\text{ }}9 + 7d\) <strong><em>(M1)</em></strong></p>
<p>\(\frac{{9 + 7d}}{{9 + 3d}} = \frac{{9 + 3d}}{9}\) <strong><em>(A1)</em></strong></p>
<p>attempt to solve <strong><em>(M1)</em></strong></p>
<p><strong>THEN</strong></p>
<p>\(d = 1\) <strong><em>A1</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(r = \frac{4}{3}\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Accept answers where a candidate obtains \(d\) by finding \(r\) first. The first two marks in either method for part (a) are awarded for the same ideas and the third mark is awarded for attempting to solve an equation in \(r\).</p>
<p> </p>
<p><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\sin \left( {\theta + \frac{\pi }{2}} \right) = \cos \theta \).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Consider \(f(x) = \sin (ax)\) where \(a\) is a constant. Prove by mathematical induction that \({f^{(n)}}(x) = {a^n}\sin \left( {ax + \frac{{n\pi }}{2}} \right)\) where \(n \in {\mathbb{Z}^ + }\) and \({f^{(n)}}(x)\) represents the \({{\text{n}}^{{\text{th}}}}\) derivative of \(f(x)\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\sin \left( {\theta + \frac{\pi }{2}} \right) = \sin \theta \cos \frac{\pi }{2} + \cos \theta \sin \frac{\pi }{2}\) <strong><em>M1</em></strong></p>
<p>\( = \cos \theta \) <strong><em>AG</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept a transformation/graphical based approach.</p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">consider \(n = 1,{\text{ }}f'(x) = a\cos (ax)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="s1">since </span>\(\sin \left( {ax + \frac{\pi }{2}} \right) = \cos ax\) then the proposition is true for \(n = 1\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">assume that the proposition is true for \(n = k\) so \({f^{(k)}}(x) = {a^k}\sin \left( {ax + \frac{{k\pi }}{2}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({f^{(k + 1)}}(x) = \frac{{{\text{d}}\left( {{f^{(k)}}(x)} \right)}}{{{\text{d}}x}}\;\;\;\left( { = a\left( {{a^k}\cos \left( {ax + \frac{{k\pi }}{2}} \right)} \right)} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( = {a^{k + 1}}\sin \left( {ax + \frac{{k\pi }}{2} + \frac{\pi }{2}} \right)\) (using part (a)) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( = {a^{k + 1}}\sin \left( {ax + \frac{{(k + 1)\pi }}{2}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">given that the proposition is true for \(n = k\) then we have shown that the proposition is true for \(n = k + 1\). Since we have shown that the proposition is true for \(n = 1\) then the proposition is true for all \(n \in {\mathbb{Z}^ + }\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award final <strong><em>R1 </em></strong>only if all prior <em><strong>M</strong></em> and <em><strong>R</strong></em> marks have been awarded.</p>
<p class="p3"><em><strong>[7 marks]</strong></em></p>
<p class="p3"><em><strong>Total [8 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Solve the equation \(2 - {\log _3}(x + 7) = {\log _{\tfrac{1}{3}}}2x\) .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({\log _3}\left( {\frac{9}{{x + 7}}} \right) = {\log _3}\frac{1}{{2x}}\) <strong><em>M1M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>M1 </em></strong>for changing to single base, <strong><em>M1 </em></strong>for incorporating the 2 into a log and <strong><em>A1 </em></strong>for a correct equation with maximum one log expression each side.</span></p>
<p> </p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(x + 7 = 18x\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(x = \frac{7}{{17}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks] </em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Some good solutions to this question and few candidates failed to earn marks on the question. Many were able to change the base of the logs, and many were able to deal with the 2, but of those who managed both, poor algebraic skills were often evident. Many students attempted to change the base into base 10, resulting in some complicated algebra, few of which managed to complete successfully. </span></p>
</div>
<br><hr><br><div class="question">
<p class="p1">Solve the equation \({4^x} + {2^{x + 2}} = 3\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1">attempt to form a quadratic in \({2^x}\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\({({2^x})^2} + 4 \bullet {2^x} - 3 = 0\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\({2^x} = \frac{{ - 4 \pm \sqrt {16 + 12} }}{2}{\text{ }}\left( { = - 2 \pm \sqrt 7 } \right)\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\({2^x} = - 2 + \sqrt 7 {\text{ }}\left( {{\text{as }} - 2 - \sqrt 7 < 0} \right)\) </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(x = {\log _2}\left( { - 2 + \sqrt 7 } \right){\text{ }}\left( {x = \frac{{\ln \left( { - 2 + \sqrt 7 } \right)}}{{\ln 2}}} \right)\) </span><strong><em>A1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: </strong>Award <strong><em>R0 A1 </em></strong>if final answer is \(x = {\log _2}\left( { - 2 + \sqrt 7 } \right)\).</p>
<p class="p2"> </p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down and simplify the expansion of \({(2 + x)^4}\) <span class="s1">in ascending powers of \(x\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence find the exact value of \({(2.1)^4}\)<span class="s1">.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({(2 + x)^4} = {2^4} + 4 \bullet {2^3}x + 6 \bullet {2^2}{x^2} + 4 \bullet 2{x^3} + {x^4}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1(A1)</em></strong></span></p>
<p class="p2"> </p>
<p class="p1"><span class="s1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>M1 </em></strong></span>for an expansion, by whatever method, giving five terms in any order.</p>
<p class="p3"> </p>
<p class="p1">\( = 16 + 32x + 24{x^2} + 8{x^3} + {x^4}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"> </p>
<p class="p4"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>M1A1A0 </em></strong>for correct expansion not given in ascending powers of \(x\).</p>
<p class="p4"><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">let \(x = 0.1\) (in the binomial expansion) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">\({2.1^4} = 16 + 3.2 + 0.24 + 0.008 + 0.0001\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\( = 19.4481\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>At most one of the marks can be implied.</p>
<p class="p3"><em><strong>[3 marks]</strong></em></p>
<p class="p3"><em><strong>Total [6 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">A geometric sequence has first term <em>a</em>, common ratio <em>r</em> and sum to infinity 76. A second geometric sequence has first term <em>a</em>, common ratio \({r^3}\) and sum to infinity 36.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find <em>r</em>.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for the first series \(\frac{a}{{1 - r}} = 76\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for the second series \(\frac{a}{{1 - {r^3}}} = 36\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to eliminate <em>a</em> <em>e.g.</em> \(\frac{{76(1 - r)}}{{1 - {r^3}}} = 36\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">simplify and obtain \(9{r^2} + 9r - 10 = 0\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Only award the </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> if a quadratic is seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">obtain \(r = \frac{{12}}{{18}}\) and \(- \frac{{30}}{{18}}\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>(A1)</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = \frac{{12}}{{18}}\left( { = \frac{2}{3} = 0.666 \ldots } \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A0</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> if the extra value of </span><em style="font-family: 'times new roman', times; font-size: medium;">r</em><span style="font-family: 'times new roman', times; font-size: medium;"> is given in the final answer.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>Total [7 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Almost all candidates obtained the cubic equation satisfied by the common ratio of the first sequence, but few were able to find its roots. One of the roots was \(r = 1\).</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The sum of the first two terms of a geometric series is 10 and the sum of the first four terms is 30.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that the common ratio \(r\) satisfies \({r^2} = 2\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Given \(r = \sqrt 2 \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> (i) find the first term;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) find the sum of the first ten terms.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a + ar = 10\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a + ar + a{r^2} + a{r^3} = 30\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a + ar = 10 \Rightarrow a{r^2} + a{r^3} = 10{r^2}\) <strong>or</strong> \(a{r^2} + a{r^3} = 20\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(10 + 10{r^2} = 30\) <strong>or</strong> \({r^2}(a + ar) = 20\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {r^2} = 2\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{a(1 - {r^2})}}{{1 - r}} = 10\) and \(\frac{{a(1 - {r^4})}}{{1 - r}} = 30\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{1 - {r^4}}}{{1 - {r^2}}} = 3\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">leading to either \(1 + {r^2} = 3{\text{ (or }}{r^4} - 3{r^2} + 2 = 0)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {r^2} = 2\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) \(a + a\sqrt 2 = 10\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow a = \frac{{10}}{{1 + \sqrt 2 }}\) <strong>or</strong> \(a = 10\left( {\sqrt 2 - 1} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({S_{10}} = \frac{{10}}{{1 + \sqrt 2 }}\left( {\frac{{{{\sqrt 2 }^{10}} - 1}}{{\sqrt 2 - 1}}} \right){\text{ }}( = 10 \times 31)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 310\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">This question was invariably answered very well. Candidates showed some skill in algebraic manipulation to derive the given answer in part a). Poor attempts at part b) were a rarity, though the final mark was sometimes lost after a correctly substituted equation was seen but with little follow-up work.</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(z = \cos \theta + i\sin \theta \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use de Moivre’s theorem to find the value of \({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use mathematical induction to prove that</p>
<p class="p1">\[{(\cos \theta - {\text{i}}\sin \theta )^n} = \cos n\theta - {\text{i}}\sin n\theta {\text{ for }}n \in {\mathbb{Z}^ + }.\]</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find an expression in terms of \(\theta \)<span class="s1"> for \({(z)^n} + {(z{\text{*}})^n},{\text{ }}n \in {\mathbb{Z}^ + }\) </span>where \(z{\text{*}}\) is the complex conjugate of \(z\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Show that \(zz{\text{*}} = 1\)<span class="s1">.</span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>Write down the binomial expansion of \({(z + z{\text{*}})^3}\) <span class="s2">in terms of \(z\) and \(z{\text{*}}\).</span></p>
<p class="p2">(iii) <span class="Apple-converted-space"> </span>Hence show that \(\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \).</p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence solve \(4{\cos ^3}\theta - 2{\cos ^2}\theta - 3\cos \theta + 1 = 0\) for \(0 \leqslant \theta < \pi \).</p>
<div class="marks">[6]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3} = \cos \pi + {\text{i}}\sin \pi \) </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = - 1\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">show the expression is true for \(n = 1\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">assume true for \(n = k,{\text{ }}{(\cos \theta - {\text{i}}\sin \theta )^k} = \cos k\theta - {\text{i}}\sin k\theta \) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Do not accept “let \(n = k\)” or “assume \(n = k\)”, assumption of truth must be present.</p>
<p class="p1">\({(\cos \theta - {\text{i}}\sin \theta )^{k + 1}} = {(\cos \theta - {\text{i}}\sin \theta )^k}(\cos \theta - {\text{i}}\sin \theta )\)</p>
<p class="p1"><span class="Apple-converted-space">\( = (\cos k\theta - {\text{i}}\sin k\theta )(\cos \theta - {\text{i}}\sin \theta )\) </span><strong><em>M1</em></strong></p>
<p class="p3"><span class="Apple-converted-space">\( = \cos k\theta \cos \theta - \sin k\theta \sin \theta - {\text{i}}(\cos k\theta \sin \theta + \sin k\theta \cos \theta )\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>A1 </em></strong>for any correct expansion.</p>
<p class="p1"><span class="Apple-converted-space">\( = \cos \left( {(k + 1)\theta } \right) - {\text{i}}\sin \left( {(k + 1)\theta } \right)\) </span><strong><em>A1</em></strong></p>
<p class="p1">therefore if true for \(n = k\) true for \(n = k + 1\), true for \(n = 1\), so true for all \(n( \in {\mathbb{Z}^ + })\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>To award the final <strong><em>R </em></strong>mark the first 4 marks must be awarded.</p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({(z)^n} + {(z{\text{*}})^n} = {(\cos \theta + {\text{i}}\sin \theta )^n} + {(\cos \theta - {\text{i}}\sin \theta )^n}\)</p>
<p class="p1"><span class="Apple-converted-space">\( = \cos n\theta + {\text{i}}\sin n\theta + \cos n\theta - {\text{i}}\sin n\theta = 2\cos (n\theta )\) </span><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p2"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \(zz* = (\cos \theta + {\text{i}}\sin \theta )(\cos \theta - {\text{i}}\sin \theta )\)</span></p>
<p class="p1"><span class="Apple-converted-space">\( = {\cos ^2}\theta + {\sin ^2}\theta \) </span><strong><em>A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = 1\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p4"><span class="s1"><strong>Note: <span class="Apple-converted-space"> </span></strong></span>Allow justification starting with \(|z| = 1\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> \({(z + z{\text{*}})^3} = {z^3} + 3{z^2}z{\text{*}} + 3z{({z^*})^2} + (z{\text{*}})3\left( { = {z^3} + 3z + 3z{\text{*}} + {{(z{\text{*}})}^3}} \right)\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(iii) <span class="Apple-converted-space"> \({(z + z{\text{*}})^3} = {(2\cos \theta )^3}\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\({z^3} + 3z + 3z{\text{*}} + {(z{\text{*}})^3} = 2\cos 3\theta + 6\cos \theta \) </span><strong><em>M1A1</em></strong></p>
<p class="p6"><span class="Apple-converted-space">\(\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span><em>M1 </em></strong>is for using \(zz{\text{*}} = 1\), this might be seen in d(ii).</p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(4{\cos ^3}\theta - 2{\cos ^2}\theta - 3\cos \theta + 1 = 0\)</p>
<p class="p1">\(4{\cos ^3}\theta - 3\cos \theta = 2{\cos ^2}\theta - 1\)</p>
<p class="p1"><span class="Apple-converted-space">\(\cos (3\theta ) = \cos (2\theta )\) </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span><em>A1 </em></strong>for \(\cos (3\theta )\) and <strong><em>A1 </em></strong>for \(\cos (2\theta )\).</p>
<p class="p3"><span class="Apple-converted-space">\(\theta = 0\) </span><strong><em>A1</em></strong></p>
<p class="p3">or \(3\theta = 2\pi - 2\theta {\text{ }}({\text{or }}3\theta = 4\pi - 2\theta )\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p4"><span class="Apple-converted-space">\(\theta = \frac{{2\pi }}{5},{\text{ }}\frac{{4\pi }}{5}\) </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Do not accept solutions via factor theorem or other methods that do not follow “hence”.</p>
<p class="p3"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was well done by most candidates who correctly applied de Moivre’s theorem.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was poorly done, which was surprising as it is very similar to the proof of de Moivre’s theorem which is stated as being required in the course guide. Many candidates spotted that they needed to use trigonometric identities but fell down through not being able to set out the proof in a logical form.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was well done by the majority of candidates.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(d) parts (i) and (ii) were well done by the candidates, who were able to successfully use trigonometrical identities and the binomial theorem.</p>
<p class="p1">(d)(iii) This is a familiar technique that has appeared in several recent past papers and was successfully completed by many of the better candidates. Some candidates though neglected the instruction ‘hence’ and tried to derive the expression using trigonometric identities.</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Again some candidates ignored ‘hence’ and tried to form a polynomial equation. Many candidates obtained the solution \(\cos (2\theta ) = \cos (3\theta )\) and hence the solution \(\theta = 0\). Few were able to find the other solutions which can be obtained from consideration of the unit circle or similar methods.</p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Express each of the complex numbers \({z_1} = \sqrt 3 + {\text{i, }}{z_2} = - \sqrt 3 + {\text{i}}\) and \({z_3} = - 2{\text{i}}\) in modulus-argument form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Hence show that the points in the complex plane representing \({z_1}\), \({z_2}\) and \({z_3}\) form the vertices of an equilateral triangle.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Show that \({\text{z}}_1^{3n} + z_2^{3n} = 2z_3^{3n}\) where \(n \in \mathbb{N}\).</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) State the solutions of the equation \({z^7} = 1\) for \(z \in \mathbb{C}\), giving them in modulus-argument form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) If <em>w</em> is the solution to \({z^7} = 1\) with least positive argument, determine the argument of 1 + <em>w</em>. Express your answer in terms of \(\pi \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Show that \({z^2} - 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\) is a factor of the polynomial \({z^7} - 1\). State the two other quadratic factors with real coefficients.</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \({z_1} = 2{\text{cis}}\left( {\frac{\pi }{6}} \right),{\text{ }}{z_2} = 2{\text{cis}}\left( {\frac{{5\pi }}{6}} \right),{\text{ }}{z_3} = 2{\text{cis}}\left( { - \frac{\pi }{2}} \right){\text{ or }}2{\text{cis}}\left( {\frac{{3\pi }}{2}} \right)\) <strong><em>A1A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Accept modulus and argument given separately, or the use of exponential (Euler) form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Accept arguments given in rational degrees, except where exponential form is used.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the points lie on a circle of radius 2 centre the origin </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">differences are all \(\frac{{2\pi }}{3}(\bmod 2\pi )\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \) points equally spaced \( \Rightarrow \) triangle is equilateral <strong><em>R1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Accept an approach based on a clearly marked diagram.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) \({\text{z}}_1^{3n} + z_2^{3n} = {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right) + {2^{3n}}{\text{cis}}\left( {\frac{{5n\pi }}{2}} \right)\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(2z_3^{3n} = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{9n\pi }}{2}} \right) = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\) <strong><em>A1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[9 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) attempt to obtain <strong>seven</strong> solutions in modulus argument form <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = {\text{cis}}\left( {\frac{{2k\pi }}{7}} \right),{\text{ }}k = 0,{\text{ }}1 \ldots 6\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) <em>w</em> has argument \(\frac{{2\pi }}{7}\) and 1 + <em>w</em> has argument \(\phi \),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">then \(\tan (\phi ) = \frac{{\sin \left( {\frac{{2\pi }}{7}} \right)}}{{1 + \cos \left( {\frac{{2\pi }}{7}} \right)}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{2\sin \left( {\frac{\pi }{7}} \right)\cos \left( {\frac{\pi }{7}} \right)}}{{2{{\cos }^2}\left( {\frac{\pi }{7}} \right)}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \tan \left( {\frac{\pi }{7}} \right) \Rightarrow \phi = \frac{\pi }{7}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Accept alternative approaches.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) since roots occur in conjugate pairs, </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>(R1)</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^7} - 1\) has a quadratic factor \(\left( {z - {\text{cis}}\left( {\frac{{2\pi }}{7}} \right)} \right) \times \left( {z - {\text{cis}}\left( { - \frac{{2\pi }}{7}} \right)} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {z^2} - 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">other quadratic factors are \({z^2} - 2z\cos \left( {\frac{{4\pi }}{7}} \right) + 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">and \({z^2} - 2z\cos \left( {\frac{{6\pi }}{7}} \right) + 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[9 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) A disappointingly large number of candidates were unable to give the correct arguments for the three complex numbers. Such errors undermined their efforts to tackle parts (ii) and (iii).</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates were successful in part (i), but failed to capitalise on that – in particular, few used the fact that roots of \({z^7} - 1 = 0\) come in complex conjugate pairs.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the complex numbers \({z_1} = 2{\text{cis}}150^\circ \) and \({z_2} = - 1 + {\text{i}}\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Calculate \(\frac{{{z_1}}}{{{z_2}}}\) giving your answer both in modulus-argument form and Cartesian form.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Using your results, find the exact value of tan 75° , giving your answer in the form \(a + \sqrt b \) , a , \(b \in {\mathbb{Z}^ + }\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">in Cartesian form</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z_1} = 2 \times - \frac{{\sqrt 3 }}{2} + 2 \times \frac{1}{2}{\text{i}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \sqrt 3 + {\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{z_1}}}{{{z_2}}} = \frac{{ - \sqrt 3 + {\text{i}}}}{{ - 1 + {\text{i}}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\left( { - \sqrt 3 + {\text{i}}} \right)}}{{( - 1 + {\text{i}})}} \times \frac{{( - 1 - {\text{i}})}}{{( - 1 - {\text{i}})}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{1 + \sqrt 3 }}{2} + \frac{{\left( {\sqrt 3 - 1} \right)}}{2}{\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">in modulus-argument form</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({{\text{z}}_2} = \sqrt 2 {\text{cis}}135^\circ \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{z_1}}}{{{z_2}}} = \frac{{2{\text{cis}}150^\circ }}{{\sqrt 2 {\text{cis}}135^\circ }}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt 2 {\text{cis}}15^\circ \) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">equating the two expressions for \(\frac{{{z_1}}}{{{z_2}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos 15^\circ = \frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sin 15^\circ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\tan 75^\circ = \frac{{\cos 15^\circ }}{{\sin 15^\circ }} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2 + \sqrt 3 \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\frac{1}{{\sqrt n + \sqrt {n + 1} }} = \sqrt {n + 1} - \sqrt n \) where \(n \ge 0,{\text{ }}n \in \mathbb{Z}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence show that \(\sqrt 2 - 1 < \frac{1}{{\sqrt 2 }}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Prove, by mathematical induction, that \(\sum\limits_{r = 1}^{r = n} {\frac{1}{{\sqrt r }} > \sqrt n } \) for \(n \ge 2,{\text{ }}n \in \mathbb{Z}\).</p>
<div class="marks">[9]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{1}{{\sqrt n + \sqrt {n + 1} }} = \frac{1}{{\sqrt n + \sqrt {n + 1} }} \times \frac{{\sqrt {n + 1} - \sqrt n }}{{\sqrt {n + 1} - \sqrt n }}\) <strong><em>M1</em></strong></p>
<p>\( = \frac{{\sqrt {n + 1} - \sqrt n }}{{(n + 1) - n}}\) <strong><em>A1</em></strong></p>
<p>\( = \sqrt {n + 1} - \sqrt n \) <strong><em>AG</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\sqrt 2 - 1 = \frac{1}{{\sqrt 2 + \sqrt 1 }}\) <strong><em>A2</em></strong></p>
<p>\( < \frac{1}{{\sqrt 2 }}\) <strong><em>AG</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider the case \(n = 2\): required to prove that \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \) <strong><em>M1</em></strong></p>
<p>from part (b) \(\frac{1}{{\sqrt 2 }} > \sqrt 2 - 1\)</p>
<p>hence \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \) is true for \(n = 2\) <strong><em>A1</em></strong></p>
<p>now assume true for \(n = k:\sum\limits_{r = 1}^{r = k} {\frac{1}{{\sqrt r }} > \sqrt k } \) <strong><em>M1</em></strong></p>
<p>\(\frac{1}{{\sqrt 1 }} + \ldots + \frac{{\sqrt 1 }}{{\sqrt k }} > \sqrt k \)</p>
<p>attempt to prove true for \(n = k + 1:\frac{1}{{\sqrt 1 }} + \ldots + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \) (<strong><em>M1)</em></strong></p>
<p>from assumption, we have that \(\frac{1}{{\sqrt 1 }} + \ldots + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt k + \frac{1}{{\sqrt {k + 1} }}\) <strong><em>M1</em></strong></p>
<p>so attempt to show that \(\sqrt k + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \) <strong>(<em>M1)</em></strong></p>
<p><strong>EITHER</strong></p>
<p>\(\frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} - \sqrt k \) <strong><em>A1</em></strong></p>
<p>\(\frac{1}{{\sqrt {k + 1} }} > \frac{1}{{\sqrt k + \sqrt {k + 1} }}\), (from part a), which is true <strong><em>A1</em></strong></p>
<p><strong>OR</strong></p>
<p>\(\sqrt k + \frac{1}{{\sqrt {k + 1} }} = \frac{{\sqrt {k + 1} \sqrt k + 1}}{{\sqrt {k + 1} }}\) <strong><em>A1</em></strong></p>
<p>\( > \frac{{\sqrt k \sqrt k + 1}}{{\sqrt {k + 1} }} = \sqrt {k + 1} \) <strong><em>A1</em></strong></p>
<p><strong>THEN</strong></p>
<p>so true for \(n = 2\) and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \ge 2\) <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>R1 </em></strong>only if all previous <strong><em>M </em></strong>marks have been awarded.</p>
<p><em><strong>[9 marks]</strong></em></p>
<p><em><strong>Total [13 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Prove by mathematical induction \(\sum\limits_{r = 1}^n {r(r!) = (n + 1)! - 1} \), \(n \in {\mathbb{Z}^ + }\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(n = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">LHS \( = 1 \times 1! = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">RHS \( = (1 + 1)! - 1 = 2 - 1 = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence true for \(n = 1\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">assume true for \(n = k\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sum\limits_{r = 1}^k {r(r!) = (k + 1)! - 1} \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sum\limits_{r = 1}^{k + 1} {r(r!) = (k + 1)! - 1} + (k + 1) \times (k + 1)!\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = (k + 1)!(1 + k + 1) - 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = (k + 1)!(k + 2) - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = (k + 2)! - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence if true for \(n = k\), true for \(n = k + 1\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">since the result is true for \(n = 1\) and \({\text{P}}(k) \Rightarrow {\text{P}}(k + 1)\) the result is proved by mathematical induction \(\forall n \in {\mathbb{Z}^ + }\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This question was done poorly on a number of levels. Many students knew the structure of induction but did not show that they understood what they were doing. The general notation was poor for both the induction itself and the sigma notation.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In noting the case for \(n = 1\) too many stated the equation rather than using the LHS and RHS separately and concluding with a statement. There were also too many who did not state the conclusion for this case.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many did not state the assumption for \(n = k\) as an assumption.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most stated the equation for \(n = k + 1\) and worked with the equation. Also common was the lack of sigma and inappropriate use of <em>n</em> and <em>k</em> in the statement. There were some very nice solutions however.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The final conclusion was often not complete or not considered which would lead to the conclusion that the student did not really understand what induction is about.</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1">The cubic equation \({x^3} + p{x^2} + qx + c = 0\)<span class="s1">, has roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \)</span>. By expanding \((x - \alpha )(x - \beta )(x - \gamma )\) show that</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) \(p = - (\alpha + \beta + \gamma )\);</p>
<p>(ii) \(q = \alpha \beta + \beta \gamma + \gamma \alpha \);</p>
<p>(iii) \(c = - \alpha \beta \gamma \).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>It is now given that \(p = - 6\) and \(q = 18\) for parts (b) and (c) below.</p>
<p>(i) In the case that the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) form an arithmetic sequence, show that one of the roots is \(2\).</p>
<p>(ii) Hence determine the value of \(c\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">In another case the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) <span class="s1">form a geometric sequence. Determine the value of \(c\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>(i)-(iii) given the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \), we have</p>
<p>\({x^3} + p{x^2} + qx + c = (x - \alpha )(x - \beta )(x - \gamma )\) <strong><em>M1</em></strong></p>
<p>\( = \left( {{x^2} - (\alpha + \beta )x + \alpha \beta } \right)(x - \gamma )\) <strong><em>A1</em></strong></p>
<p>\( = {x^3} - (\alpha + \beta + \gamma ){x^2} + (\alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma \) <strong><em>A1</em></strong></p>
<p>comparing coefficients:</p>
<p>\(p = - (\alpha + \beta + \gamma )\) <strong><em>AG</em></strong></p>
<p>\(q = (\alpha \beta + \beta \gamma + \gamma \alpha )\) <strong><em>AG</em></strong></p>
<p>\(c = - \alpha \beta \gamma \) <strong><em>AG</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>(i) Given \( - \alpha - \beta - \gamma = - 6\)</p>
<p>And \(\alpha \beta + \beta \gamma + \gamma \alpha = 18\)</p>
<p>Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \)</p>
<p>So \(\beta - \alpha = \gamma - \beta \) <strong><em>M1</em></strong></p>
<p>or \(2\beta = \alpha + \gamma \)</p>
<p>Attempt to solve simultaneous equations: <strong><em>M1</em></strong></p>
<p>\(\beta + 2\beta = 6\) <strong><em>A1</em></strong></p>
<p>\(\beta = 2\) <strong><em>AG</em></strong></p>
<p>(ii) \(\alpha + \gamma = 4\)</p>
<p>\(2\alpha + 2\gamma + \alpha \gamma = 18\)</p>
<p>\( \Rightarrow {\gamma ^2} - 4\gamma + 10 = 0\)</p>
<p>\( \Rightarrow \gamma = \frac{{4 \pm {\text{i}}\sqrt {24} }}{2}\) <strong><em>(A1)</em></strong></p>
<p>Therefore \(c = - \alpha \beta \gamma = - \left( {\frac{{4 + {\text{i}}\sqrt {24} }}{2}} \right)\left( {\frac{{4 - {\text{i}}\sqrt {24} }}{2}} \right)2 = - 20\) <strong><em>A1</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>(i) let the three roots be \(\alpha ,{\text{ }}\alpha - d,{\text{ }}\alpha + d\) <strong><em>M1</em></strong></p>
<p>adding roots <strong><em>M1</em></strong></p>
<p>to give \(3\alpha = 6\) <strong><em>A1</em></strong></p>
<p>\(\alpha = 2\) <strong><em>AG</em></strong></p>
<p>(ii) \(\alpha \) is a root, so \({2^3} - 6 \times {2^2} + 18 \times 2 + c = 0\) <strong><em>M1</em></strong></p>
<p>\(8 - 24 + 36 + c = 0\)</p>
<p>\(c = - 20\) <strong><em>A1</em></strong></p>
<p><strong>METHOD 3</strong></p>
<p>(i) let the three roots be \(\alpha ,{\text{ }}\alpha - d,{\text{ }}\alpha + d\) <strong><em>M1</em></strong></p>
<p>adding roots <strong><em>M1</em></strong></p>
<p>to give \(3\alpha = 6\) <strong><em>A1</em></strong></p>
<p>\(\alpha = 2\) <strong><em>AG</em></strong></p>
<p>(ii) \(q = 18 = 2(2 - d) + (2 - d)(2 + d) + 2(2 + d)\) <strong><em>M1</em></strong></p>
<p>\({d^2} = - 6 \Rightarrow d = \sqrt 6 {\text{i}}\)</p>
<p>\( \Rightarrow c = - 20\) <strong><em>A1</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>Given \( - \alpha - \beta - \gamma = - 6\)</p>
<p>And \(\alpha \beta + \beta \gamma + \gamma \alpha = 18\)</p>
<p>Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \).</p>
<p>So \(\frac{\beta }{\alpha } = \frac{\gamma }{\beta }\) <strong><em>M1</em></strong></p>
<p>or \({\beta ^2} = \alpha \gamma \)</p>
<p>Attempt to solve simultaneous equations: <strong><em>M1</em></strong></p>
<p>\(\alpha \beta + \gamma \beta + {\beta ^2} = 18\)</p>
<p>\(\beta (\alpha + \beta + \gamma ) = 18\)</p>
<p>\(6\beta = 18\)</p>
<p>\(\beta = 3\) <strong><em>A1</em></strong></p>
<p>\(\alpha + \gamma = 3,{\text{ }}\alpha = \frac{9}{\gamma }\)</p>
<p>\( \Rightarrow {\gamma ^2} - 3\gamma + 9 = 0\)</p>
<p>\( \Rightarrow \gamma = \frac{{3 \pm {\text{i}}\sqrt {27} }}{2}\) (<strong><em>A1)(A1)</em></strong></p>
<p>Therefore \(c = - \alpha \beta \gamma = - \left( {\frac{{3 + {\text{i}}\sqrt {27} }}{2}} \right)\left( {\frac{{3 - {\text{i}}\sqrt {27} }}{2}} \right)3 = - 27\) <strong><em>A1</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>let the three roots be \(a,{\text{ }}ar,{\text{ }}a{r^2}\) <strong><em>M1</em></strong></p>
<p>attempt at substitution of \(a,{\text{ }}ar,{\text{ }}a{r^2}\) and \(p\) and \(q\) into equations from (a) <strong><em>M1</em></strong></p>
<p>\(6 = a + ar + a{r^2}\left( { = a(1 + r + {r^2})} \right)\) <strong><em>A1</em></strong></p>
<p>\(18 = {a^2}r + {a^2}{r^3} + {a^2}{r^2}\left( { = {a^2}r(1 + r + {r^2})} \right)\) <strong><em>A1</em></strong></p>
<p>therefore \(3 = ar\) <strong><em>A1</em></strong></p>
<p>therefore \(c = - {a^3}{r^3} = - {3^3} = - 27\) <strong><em>A1</em></strong></p>
<p><strong><em>[6 marks]</em></strong></p>
<p><strong><em>Total [14 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(\{ {u_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be an arithmetic sequence with first term equal to \(a\) and common difference of \(d\), where \(d \ne 0\). Let another sequence \(\{ {v_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be defined by \({v_n} = {2^{{u_n}}}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Show that \(\frac{{{v_{n + 1}}}}{{{v_n}}}\) <span class="s1">is a constant.</span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>Write down the first term of the sequence \(\{ {v_n}\} \).</p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>Write down a formula for \({v_n}\) in terms of \(a\), \(d\) and \(n\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Let \({S_n}\) be the sum of the first \(n\) terms of the sequence \(\{ {v_n}\} \).</p>
<p>(i) Find \({S_n}\), in terms of \(a\), \(d\) and \(n\).</p>
<p>(ii) Find the values of \(d\) for which \(\sum\limits_{i = 1}^\infty {{v_i}} \) exists.</p>
<p>You are now told that \(\sum\limits_{i = 1}^\infty {{v_i}} \) does exist and is denoted by \({S_\infty }\).</p>
<p>(iii) Write down \({S_\infty }\) in terms of \(a\) and \(d\) .</p>
<p>(iv) Given that \({S_\infty } = {2^{a + 1}}\) find the value of \(d\) .</p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Let \(\{ {w_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), <span class="s1">be a geometric sequence with first term equal to \(p\) and common ratio \(q\), where \(p\) and \(q\) </span>are both greater than zero. Let another sequence \(\{ {z_n}\} \) be defined by \({z_n} = \ln {w_n}\).</p>
<p class="p1">Find \(\sum\limits_{i = 1}^n {{z_i}} \) giving your answer in the form \(\ln k\) <span class="s1">with \(k\) in terms of \(n\), \(p\) and \(q\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span></span><strong>METHOD 1</strong></p>
<p class="p2">\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{{u_{n + 1}}}}}}{{{2^{{u_n}}}}}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>M1</em></strong></span></p>
<p class="p2">\( = {2^{{u_{n + 1}} - {u_n}}} = {2^d}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p2">\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{a + nd}}}}{{{2^{a + (n - 1)d}}}}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>M1</em></strong></span></p>
<p class="p2">\( = {2^d}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>\( = {2^a}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Accept \( = {2^{{u_1}}}\).</p>
<p class="p3"> </p>
<p class="p2">(iii) <span class="Apple-converted-space"> </span><span class="s2"><strong>EITHER</strong></span></p>
<p class="p2">\({v_n}\) is a GP with first term \({2^a}\) and common ratio \({2^d}\)</p>
<p class="p2">\({v_n} = {2^a}{({2^d})^{(n - 1)}}\)</p>
<p class="p1"><strong>OR</strong></p>
<p class="p2">\({u_n} = a + (n - 1)d\) as it is an AP</p>
<p class="p1"><strong>THEN</strong></p>
<p class="p2">\({v_n} = {2^a}^{ + (n - 1)d}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="s2"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>(i) \({S_n} = \frac{{{2^a}\left( {{{({2^d})}^n} - 1} \right)}}{{{2^d} - 1}} = \frac{{{2^a}({2^{dn}} - 1)}}{{{2^d} - 1}}\) <strong><em>M1A1</em></strong></p>
<p><strong>Note: </strong>Accept either expression.</p>
<p> </p>
<p>(ii) for sum to infinity to exist need \( - 1 < {2^d} < 1\) <strong><em>R1</em></strong></p>
<p>\( \Rightarrow \log {2^d} < 0 \Rightarrow d\log 2 < 0 \Rightarrow d < 0\) <strong><em>(M1)A1</em></strong></p>
<p><strong>Note: </strong>Also allow graph of \({2^d}\).</p>
<p> </p>
<p>(iii) \({S_\infty } = \frac{{{2^a}}}{{1 - {2^d}}}\) <strong><em>A1</em></strong></p>
<p>(iv) \(\frac{{{2^a}}}{{1 - {2^d}}} = {2^{a + 1}} \Rightarrow \frac{1}{{1 - {2^d}}} = 2\) <strong><em>M1</em></strong></p>
<p>\( \Rightarrow 1 = 2 - {2^{d + 1}} \Rightarrow {2^{d + 1}} = 1\)</p>
<p>\( \Rightarrow d = - 1\) <strong><em>A1</em></strong></p>
<p><strong><em>[8 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\({w_n} = p{q^{n - 1}},{\text{ }}{z_n} = \ln p{q^{n - 1}}\) <strong><em>(A1)</em></strong></p>
<p>\({z_n} = \ln p + (n - 1)\ln q\) <strong><em>M1A1</em></strong></p>
<p>\({z_{n + 1}} - {z_n} = (\ln p + n\ln q) - (\ln p + (n - 1)\ln q) = \ln q\)</p>
<p>which is a constant so this is an AP</p>
<p>(with first term \(\ln p\) and common difference \(\ln q\))</p>
<p>\(\sum\limits_{i = 1}^n {{z_i} = \frac{n}{2}\left( {2\ln p + (n - 1)\ln q} \right)} \) <strong><em>M1</em></strong></p>
<p>\( = n\left( {\ln p + \ln {q^{\left( {\frac{{n - 1}}{2}} \right)}}} \right) = n\ln \left( {p{q^{\left( {\frac{{n - 1}}{2}} \right)}}} \right)\) <strong><em>(M1)</em></strong></p>
<p>\( = \ln \left( {{p^n}{q^{\frac{{n(n - 1)}}{2}}}} \right)\) <strong><em>A1</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>\(\sum\limits_{i = 1}^n {{z_i} = \ln p + \ln pq + \ln p{q^2} + \ldots + \ln p{q^{n - 1}}} \) <strong><em>(M1)A1</em></strong></p>
<p>\( = \ln \left( {{p^n}{q^{\left( {1 + 2 + 3 + \ldots + (n - 1)} \right)}}} \right)\) <strong><em>(M1)A1</em></strong></p>
<p>\( = \ln \left( {{p^n}{q^{\frac{{n(n - 1)}}{2}}}} \right)\) <strong><em>(M1)A1</em></strong></p>
<p><strong><em>[6 marks]</em></strong></p>
<p><strong><em>Total [18 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Method of first part was fine but then some algebra mistakes often happened. The next two parts were generally good.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Given that (a) indicated that there was a common ratio a disappointing number thought it was an AP. Although some good answers in the next parts, there was also some poor notational misunderstanding with the sum to infinity still involving \(n\).</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Not enough candidates realised that this was an AP.</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Express \(\frac{1}{{{{(1 - {\text{i}}\sqrt 3 )}^3}}}{\text{ in the form }}\frac{a}{b}{\text{ where }}a,{\text{ }}b \in \mathbb{Z}\) .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = 2,{\text{ }}\theta = - \frac{\pi }{3}\) <strong><em>(A1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore {(1 - {\text{i}}\sqrt 3 )^{ - 3}} = {2^{ - 3}}{\left( {\cos \left( { - \frac{\pi }{3}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{3}} \right)} \right)^{ - 3}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{8}(\cos \pi + {\text{i}}\sin \pi )\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \frac{1}{8}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((1 - \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 1 - 2\sqrt 3 {\text{i}} - 3\,\,\,\,\,( = - 2 - 2\sqrt 3 {\text{i}})\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(( - 2 - 2\sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = - 8\) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 3</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Attempt at Binomial expansion <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(1 - \sqrt 3 {\text{i}})^3} = 1 + 3( - \sqrt 3 {\text{i}}) + 3{( - \sqrt 3 {\text{i}})^2} + {( - \sqrt 3 {\text{i}})^3}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1 - 3\sqrt 3 {\text{i}} - 9 + 3\sqrt 3 {\text{i}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - 8\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates made a meaningful attempt at this question using a variety of different, but correct methods. Weaker candidates sometimes made errors with the manipulation of the square roots, but there were many fully correct solutions.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the value of <em>k </em>if \({\sum\limits_{r = 1}^\infty{k\left( {\frac{1}{3}} \right)}^r} = 7\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({u_1} = \frac{1}{3}k{\text{ , }}r = \frac{1}{3}\) <strong><em>(A1) (A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>\(7 = \frac{{\frac{1}{3}k}}{{1 - \frac{1}{3}}}\) M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>\(k = 14\) A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">The question was well done generally. Those that did make mistakes on the question usually had the first term wrong, but did understand to use the formula for an infinite geometric series.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \({z_1} = 2\) and \({z_2} = 1 + \sqrt 3 {\text{i}}\) are roots of the cubic equation \({z^3} + b{z^2} + cz + d = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">where <em>b</em>, <em>c</em>, \(d \in \mathbb{R}\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) write down the third root, \({z_3}\), of the equation;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) find the values of <em>b</em>, <em>c</em> and <em>d</em> ;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) write \({z_2}\) and \({z_3}\) in the form \(r{{\text{e}}^{{\text{i}}\theta }}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(1 - \sqrt 3 {\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {z - (1 + \sqrt 3 {\text{i)}}} \right)\left( {z - (1 - \sqrt 3 {\text{i)}}} \right) = {z^2} - 2z + 4\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(p(z) = (z - 2)({z^2} - 2z + 4)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {z^3} - 4{z^2} + 8z - 8\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore \(b = - 4,{\text{ }}c = 8,{\text{ }}d = - 8\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">relating coefficients of cubic equations to roots</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - b = 2 + 1 + \sqrt 3 {\text{i}} + 1 - \sqrt 3 {\text{i}} = 4\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(c = 2(1 + \sqrt 3 {\text{i}}) + 2(1 - \sqrt 3 {\text{i}}) + (1 + \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 8\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - d = 2(1 + \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 8\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(b = - 4,{\text{ }}c = 8,{\text{ }}d = - 8\) <strong><em>A1A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \({z_2} = 2{{\text{e}}^{\frac{{{\text{i}}\pi }}{3}}},{\text{ }}{z_3} = 2{{\text{e}}^{ - \frac{{{\text{i}}\pi }}{3}}}\) <strong><em>A1A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> for modulus,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A1</em></strong> for each argument.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Parts a) and c) were done quite well by many but the method used in b) often lead to tedious and long algebraic manipulations in which students got lost and so did not get to the correct solution. Many did not give the principal argument in c).</span></p>
</div>
<br><hr><br><div class="specification">
<p>Consider the function \({f_n}(x) = (\cos 2x)(\cos 4x) \ldots (\cos {2^n}x),{\text{ }}n \in {\mathbb{Z}^ + }\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine whether \({f_n}\) is an odd or even function, justifying your answer.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By using mathematical induction, prove that</p>
<p style="text-align: center;">\({f_n}(x) = \frac{{\sin {2^{n + 1}}x}}{{{2^n}\sin 2x}},{\text{ }}x \ne \frac{{m\pi }}{2}\) where \(m \in \mathbb{Z}\).</p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence or otherwise, find an expression for the derivative of \({f_n}(x)\) with respect to \(x\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that, for \(n > 1\), the equation of the tangent to the curve \(y = {f_n}(x)\) at \(x = \frac{\pi }{4}\) is \(4x - 2y - \pi = 0\).</p>
<div class="marks">[8]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>even function <strong><em>A1</em></strong></p>
<p>since \(\cos kx = \cos ( - kx)\) <strong>and</strong> \({f_n}(x)\) is a product of even functions <strong><em>R1</em></strong></p>
<p><strong>OR</strong></p>
<p>even function <strong><em>A1</em></strong></p>
<p>since \((\cos 2x)(\cos 4x) \ldots = \left( {\cos ( - 2x)} \right)\left( {\cos ( - 4x)} \right) \ldots \) <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Do not award <strong><em>A0R1</em></strong>.</p>
<p> </p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider the case \(n = 1\)</p>
<p>\(\frac{{\sin 4x}}{{2\sin 2x}} = \frac{{2\sin 2x\cos 2x}}{{2\sin 2x}} = \cos 2x\) <strong><em>M1</em></strong></p>
<p>hence true for \(n = 1\) <strong><em>R1</em></strong></p>
<p>assume true for \(n = k\), <em>ie</em>, \((\cos 2x)(\cos 4x) \ldots (\cos {2^k}x) = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\) <strong><em>M1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Do not award <strong><em>M1 </em></strong>for “let \(n = k\)” or “assume \(n = k\)” or equivalent.</p>
<p> </p>
<p>consider \(n = k + 1\):</p>
<p>\({f_{k + 1}}(x) = {f_k}(x)(\cos {2^{k + 1}}x)\) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\cos {2^{k + 1}}x\) <strong><em>A1</em></strong></p>
<p>\( = \frac{{2\sin {2^{k + 1}}x\cos {2^{k + 1}}x}}{{{2^{k + 1}}\sin 2x}}\) <strong><em>A1</em></strong></p>
<p>\( = \frac{{\sin {2^{k + 2}}x}}{{{2^{k + 1}}\sin 2x}}\) <strong><em>A1</em></strong></p>
<p>so \(n = 1\) true and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \in {\mathbb{Z}^ + }\) <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> To obtain the final <strong><em>R1</em></strong>, all the previous <strong><em>M </em></strong>marks must have been awarded.</p>
<p> </p>
<p><strong><em>[8 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>attempt to use \(f’ = \frac{{vu' - uv'}}{{{v^2}}}\) (or correct product rule) <strong><em>M1</em></strong></p>
<p>\({f’_n}(x) = \frac{{({2^n}\sin 2x)({2^{n + 1}}\cos {2^{n + 1}}x) - (\sin {2^{n + 1}}x)({2^{n + 1}}\cos 2x)}}{{{{({2^n}\sin 2x)}^2}}}\) <strong><em>A1A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>A1 </em></strong>for correct numerator and <strong><em>A1 </em></strong>for correct denominator.</p>
<p> </p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{\left( {{2^n}\sin \frac{\pi }{2}} \right)\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right) - \left( {\sin {2^{n + 1}}\frac{\pi }{4}} \right)\left( {{2^{n + 1}}\cos \frac{\pi }{2}} \right)}}{{{{\left( {{2^n}\sin \frac{\pi }{2}} \right)}^2}}}\) <strong><em>(M1)(A1)</em></strong></p>
<p>\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{({2^n})\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right)}}{{{{({2^n})}^2}}}\) <strong><em>(A1)</em></strong></p>
<p>\( = 2\cos {2^{n + 1}}\frac{\pi }{4}{\text{ }}( = 2\cos {2^{n - 1}}\pi )\) <strong><em>A1</em></strong></p>
<p>\({f’_n}\left( {\frac{\pi }{4}} \right) = 2\) <strong><em>A1</em></strong></p>
<p>\({f_n}\left( {\frac{\pi }{4}} \right) = 0\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> This <strong><em>A </em></strong>mark is independent from the previous marks.</p>
<p> </p>
<p>\(y = 2\left( {x - \frac{\pi }{4}} \right)\) <strong><em>M1A1</em></strong></p>
<p>\(4x - 2y - \pi = 0\) <strong><em>AG</em></strong></p>
<p><strong><em>[8 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<p>Use the method of mathematical induction to prove that \({4^n} + 15n - 1\) is divisible by 9 for \(n \in {\mathbb{Z}^ + }\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>let \(P(n)\) be the proposition that \({4^n} + 15n - 1\) is divisible by 9</p>
<p>showing true for \(n = 1\) <strong><em>A1</em></strong></p>
<p><em>ie</em>\(\,\,\,\,\,\)for \(n = 1,{\text{ }}{4^1} + 15 \times 1 - 1 = 18\)</p>
<p>which is divisible by 9, therefore \(P(1)\) is true</p>
<p>assume \(P(k)\) is true so \({4^k} + 15k - 1 = 9A,{\text{ }}(A \in {\mathbb{Z}^ + })\) <strong><em>M1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Only award <strong><em>M1 </em></strong>if “truth assumed” or equivalent.</p>
<p> </p>
<p>consider \({4^{k + 1}} + 15(k + 1) - 1\)</p>
<p>\( = 4 \times {4^k} + 15k + 14\)</p>
<p>\( = 4(9A - 15k + 1) + 15k + 14\) <strong><em>M1</em></strong></p>
<p>\( = 4 \times 9A - 45k + 18\) <strong><em>A1</em></strong></p>
<p>\( = 9(4A - 5k + 2)\) which is divisible by 9 <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>R1 </em></strong>for either the expression or the statement above.</p>
<p> </p>
<p>since \(P(1)\) is true and \(P(k)\) true implies \(P(k + 1)\) is true, therefore (by the principle of mathematical induction) \(P(n)\) is true for \(n \in {\mathbb{Z}^ + }\) <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Only award the final <strong><em>R1 </em></strong>if the 2 <strong><em>M1</em></strong>s have been awarded.</p>
<p> </p>
<p><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Expand and simplify \({\left( {{x^2} - \frac{2}{x}} \right)^4}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {{x^2} - \frac{2}{x}} \right)^4} = {({x^2})^4} + 4{({x^2})^3}\left( { - \frac{2}{x}} \right) + 6{({x^2})^2}{\left( { - \frac{2}{x}} \right)^2} + 4({x^2}){\left( { - \frac{2}{x}} \right)^3} + {\left( { - \frac{2}{x}} \right)^4}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {x^8} - 8{x^5} + 24{x^2} - \frac{{32}}{x} + \frac{{16}}{{{x^4}}}\) <strong><em>A3</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Deduct one <strong><em>A</em></strong> mark for each incorrect or omitted term.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates solved this question correctly with most candidates who explained how they obtained their coefficients using Pascal’s triangle rather than the combination formula.</span></p>
</div>
<br><hr><br><div class="question">
<p class="p1"><span class="s1">Expand \({(3 - x)^4}\) </span>in ascending powers of \(x\) and simplify your answer.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1">\({(3 - x)^4} = {1.3^4} + {4.3^3}( - x) + {6.3^2}{( - x)^2} + 4.3{( - x)^3} + 1{( - x)^4}\) or equivalent <span class="s1">(<strong><em>M1)(A1)</em></strong></span></p>
<p class="p1">\( = 81 - 108x + 54{x^2} - 12{x^3} + {x^4}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span><em>A1 </em></strong>for ascending powers, <strong><em>A1 </em></strong>for correct coefficients including signs.</p>
<p class="p3"> </p>
<p class="p3"><em><strong>[4 marks]</strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Consider the following sequence of equations.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \(1 \times 2 = \frac{1}{3}(1 \times 2 \times 3),\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \(1 \times 2 + 2 \times 3 = \frac{1}{3}(2 \times 3 \times 4),\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \(1 \times 2 + 2 \times 3 + 3 \times 4 = \frac{1}{3}(3 \times 4 \times 5),\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> \( \ldots {\text{ .}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Formulate a conjecture for the \({n^{{\text{th}}}}\) equation in the sequence.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Verify your conjecture for <em>n</em> = 4 .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) A sequence of numbers has the \({n^{{\text{th}}}}\) term given by \({u_n} = {2^n} + 3,{\text{ }}n \in {\mathbb{Z}^ + }\). Bill conjectures that all members of the sequence are prime numbers. Show that Bill’s conjecture is false.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Use mathematical induction to prove that \(5 \times {7^n} + 1\) is divisible by 6 for all \(n \in {\mathbb{Z}^ + }\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) (i) \(1 \times 2 + 2 \times 3 + \ldots + n(n + 1) = \frac{1}{3}n(n + 1)(n + 2)\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) LHS = 40; RHS = 40 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) the sequence of values are:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">5, 7, 11, 19, 35 … or an example <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">35 is not prime, so Bill’s conjecture is false <strong><em>R1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \({\text{P}}(n):5 \times {7^n} + 1\) is divisible by 6</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{P}}(1):36\) is divisible by \(6 \Rightarrow {\text{P}}(1)\) true <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">assume \({\text{P}}(k)\) is true \((5 \times {7^k} + 1 = 6r)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do <strong>not</strong> award <strong><em>M1</em></strong> for statement starting ‘let <em>n</em> = <em>k</em>’.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Subsequent marks are independent of this <strong><em>M1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">consider \(5 \times {7^{k + 1}} + 1\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 7(6r - 1) + 1\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 6(7r - 1) \Rightarrow {\text{P}}(k + 1)\) is true <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">P(1) true and \({\text{P}}(k)\) true \( \Rightarrow {\text{P}}(k + 1)\) true, so by MI \({\text{P}}(n)\) is true for all \(n \in {\mathbb{Z}^ + }\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Only award <strong><em>R1</em></strong> if there is consideration of P(1), \({\text{P}}(k)\) and \({\text{P}}(k + 1)\) in the final statement.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Only award <strong><em>R1</em></strong> if at least one of the two preceding <strong><em>A</em></strong> marks has been awarded.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [10 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Although there were a good number of wholly correct solutions to this question, it was clear that a number of students had not been prepared for questions on conjectures. The proof by induction was relatively well done, but candidates often showed a lack of rigour in the proof. It was fairly common to see students who did not appreciate the idea that \({\text{P}}(k)\) is assumed not given and this was penalised. Also it appeared that a number of students had been taught to write down the final reasoning for a proof by induction, even if no attempt of a proof had taken place. In these cases, the final reasoning mark was not awarded.</span></p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Show that \({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = \frac{{2\cos n\theta }}{{{{\cos }^n}\theta }},\;\;\;\cos \theta \ne 0\).</p>
<p>(ii) Hence verify that \({\text{i}}\tan \frac{{3\pi }}{8}\) is a root of the equation \({(1 + z)^4} + {(1 - z)^4} = 0,\;\;\;z \in \mathbb{C}\).</p>
<p>(iii) State another root of the equation \({(1 + z)^4} + {(1 - z)^4} = 0,\;\;\;z \in \mathbb{C}\).</p>
<div class="marks">[10]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Use the double angle identity \(\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\) to show that \(\tan \frac{\pi }{8} = \sqrt 2 - 1\).</p>
<p>(ii) Show that \(\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1\).</p>
<p>(iii) Hence find the value of \(\int_0^{\frac{\pi }{8}} {\frac{{2\cos 4x}}{{{{\cos }^2}x}}{\text{d}}x} \).</p>
<div class="marks">[13]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>(i) <strong> METHOD 1</strong></p>
<p>\({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = {\left( {1 + {\text{i}}\frac{{\sin \theta }}{{\cos \theta }}} \right)^n} + {\left( {1 - {\text{i}}\frac{{\sin \theta }}{{\cos \theta }}} \right)^n}\) <strong><em>M1</em></strong></p>
<p>\( = {\left( {\frac{{\cos \theta + i\sin \theta }}{{\cos \theta }}} \right)^n} + {\left( {\frac{{\cos \theta - i\sin \theta }}{{\cos \theta }}} \right)^n}\) <strong><em>A1</em></strong></p>
<p>by de Moivre’s theorem (<strong><em>M1)</em></strong></p>
<p>\({\left( {\frac{{\cos \theta + i\sin \theta }}{{\cos \theta }}} \right)^n} = \frac{{\cos n\theta + i\sin n\theta }}{{{{\cos }^n}\theta }}\) <strong><em>A1</em></strong></p>
<p>recognition that \(\cos \theta - i\sin \theta \) is the complex conjugate of \(\cos \theta + i\sin \theta \) (<strong><em>R1)</em></strong></p>
<p>use of the fact that the operation of complex conjugation commutes with the operation of raising to an integer power:</p>
<p>\({\left( {\frac{{\cos \theta - i\sin \theta }}{{\cos \theta }}} \right)^n} = \frac{{\cos n\theta - i\sin n\theta }}{{{{\cos }^n}\theta }}\) <strong><em>A1</em></strong></p>
<p>\({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = \frac{{2\cos n\theta }}{{{{\cos }^n}\theta }}\) <strong><em>AG</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>\({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = {(1 + {\text{i}}\tan \theta )^n} + {\left( {1 + {\text{i}}\tan ( - \theta )} \right)^n}\) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{{{{(\cos \theta + i\sin \theta )}^n}}}{{{{\cos }^n}\theta }} + \frac{{{{\left( {\cos ( - \theta ) + i\sin ( - \theta )} \right)}^n}}}{{{{\cos }^n}\theta }}\) <strong><em>M1A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>M1 </em></strong>for converting to cosine and sine terms.</p>
<p> </p>
<p>use of de Moivre’s theorem <strong><em>(M1)</em></strong></p>
<p>\( = \frac{1}{{{{\cos }^n}\theta }}\left( {\cos n\theta + {\text{i}}\sin n\theta + \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta )} \right)\) <strong><em>A1</em></strong></p>
<p>\( = \frac{{2\cos n\theta }}{{{{\cos }^2}\theta }}\;\;\;{\text{as}}\;\;\;\cos ( - n\theta ) = \cos n\theta \;\;\;{\text{and}}\;\;\;\sin ( - n\theta ) = - \sin n\theta \) <strong><em>R1AG</em></strong></p>
<p>(ii) \({\left( {1 + {\text{i}}\tan \frac{{3\pi }}{8}} \right)^4} + {\left( {1 - {\text{i}}\tan \frac{{3\pi }}{8}} \right)^4} = \frac{{2\cos \left( {4 \times \frac{{3\pi }}{8}} \right)}}{{{{\cos }^4}\frac{{3\pi }}{8}}}\) <strong><em>(A1)</em></strong></p>
<p>\( = \frac{{2\cos \frac{{3\pi }}{2}}}{{{{\cos }^4}\frac{{3\pi }}{8}}}\) <strong><em>A1</em></strong></p>
<p>\( = 0\;\;\;{\text{as}}\;\;\;\cos \frac{{3\pi }}{2} = 0\) <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>The above working could involve theta and the solution of \(\cos (4\theta ) = 0\).</p>
<p> </p>
<p>so \({\text{i}}\tan \frac{{3\pi }}{8}\) is a root of the equation <strong><em>AG</em></strong></p>
<p>(iii) either \( - {\text{i}}\tan \frac{{3\pi }}{8}\;\;\;{\text{or}}\;\;\; - {\text{i}}\tan \frac{\pi }{8}\;\;\;{\text{or}}\;\;\;{\text{i}}\tan \frac{\pi }{8}\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept \({\text{i}}\tan \frac{{5\pi }}{8}\;\;\;{\text{or}}\;\;\;{\text{i}}\tan \frac{{7\pi }}{8}\).</p>
<p>Accept \( - \left( {1 + \sqrt 2 } \right){\text{i}}\;\;\;{\text{or}}\;\;\;\left( {1 - \sqrt 2 } \right){\text{i}}\;\;\;{\text{or}}\;\;\;\left( { - 1 + \sqrt 2 } \right){\text{i}}\).</p>
<p><em><strong>[10 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>(i) \(\tan \frac{\pi }{4} = \frac{{2\tan \frac{\pi }{8}}}{{1 - {{\tan }^2}\frac{\pi }{8}}}\) <strong><em>(M1)</em></strong></p>
<p>\({\tan ^2}\frac{\pi }{8} + 2\tan \frac{\pi }{8} - 1 = 0\) <strong><em>A1</em></strong></p>
<p>let \(t = \tan \frac{\pi }{8}\)</p>
<p>attempting to solve \({t^2} + 2t - 1 = 0\;\;\;{\text{for}}\;\;\;t\) <strong><em>M1</em></strong></p>
<p>\(t = - 1 \pm \sqrt 2 \) <strong><em>A1</em></strong></p>
<p>\(\frac{\pi }{8}\) is a first quadrant angle and tan is positive in this quadrant, so</p>
<p>\(\tan \frac{\pi }{8} > 0\) <strong><em>R1</em></strong></p>
<p>\(\tan \frac{\pi }{8} = \sqrt 2 - 1\) <strong><em>AG</em></strong></p>
<p>(ii) \(\cos 4x = 2{\cos ^2}2x - 1\) <strong><em>A1</em></strong></p>
<p>\( = 2{\left( {2{{\cos }^2}x - 1} \right)^2} - 1\) <strong><em>M1</em></strong></p>
<p>\( = 2\left( {4{{\cos }^4}x - 4{{\cos }^2}x + 1} \right) - 1\) <strong><em>A1</em></strong></p>
<p>\( = 8{\cos ^4}x - 8{\cos ^2}x + 1\) <strong><em>AG</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept equivalent complex number derivation.</p>
<p> </p>
<p>(iii) \(\int_0^{\frac{\pi }{8}} {\frac{{2\cos 4x}}{{{{\cos }^2}x}}{\text{d}}x = 2} \int_0^{\frac{\pi }{8}} {\frac{{8{{\cos }^4}x - 8{{\cos }^2}x + 1}}{{{{\cos }^2}x}}{\text{d}}x} \)</p>
<p>\( = 2\int_0^{\frac{\pi }{8}} {8{{\cos }^2}x - 8 + {{\sec }^2}x{\text{d}}x} \) <strong><em>M1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>The <strong><em>M1 </em></strong>is for an integrand involving no fractions.</p>
<p> </p>
<p>use of \({\cos ^2}x = \frac{1}{2}(\cos 2x + 1)\) <strong><em>M1</em></strong></p>
<p>\( = 2\int_0^{\frac{\pi }{8}} {4\cos 2x - 4 + {{\sec }^2}x{\text{d}}x} \) <strong><em>A1</em></strong></p>
<p>\( = [4\sin 2x - 8x + 2\tan x]_0^{\frac{\pi }{8}}\) <strong><em>A1</em></strong></p>
<p>\( = 4\sqrt 2 - \pi - 2\;\;\;\)(or equivalent) <strong><em>A1</em></strong></p>
<p><strong><em>[13 marks]</em></strong></p>
<p><strong><em>Total [23 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Fairly successful.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Most candidates attempted to use the hint. Those who doubled the angle were usually successful – but many lost the final mark by not giving a convincing reason to reject the negative solution to the intermediate quadratic equation. Those who halved the angle got nowhere.</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>The majority of candidates obtained full marks.</p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>This was poorly answered, few candidates realising that part of the integrand could be re-expressed using \(\frac{1}{{{{\cos }^2}x}} = {\sec ^2}x\), which can be immediately integrated.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the equation \(9{x^3} - 45{x^2} + 74x - 40 = 0\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Write down the numerical value of the sum and of the product of the roots of this equation.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The roots of this equation are three consecutive terms of an arithmetic sequence.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Taking the roots to be \(\alpha {\text{ , }}\alpha \pm \beta \) , solve the equation.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{sum}} = \frac{{45}}{9},{\text{ product}} = \frac{{40}}{9}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">it follows that \(3\alpha = \frac{{45}}{9}\) and \(\alpha ({\alpha ^2} - {\beta ^2}) = \frac{{40}}{9}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solving, \(\alpha = \frac{5}{3}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{5}{3}\left( {\frac{{25}}{9} - {\beta ^2}} \right) = \frac{{40}}{9}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\beta = ( \pm )\frac{1}{3}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the other two roots are 2, \(\frac{4}{3}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that the following system of equations has an infinite number of solutions.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> \(x + y + 2z = - 2\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> \(3x - y + 14z = 6\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> \(x + 2y = - 5\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The system of equations represents three planes in space.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find the parametric equations of the line of intersection of the three planes.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) <strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}\left| \begin{array}{c} - 2\\6\\ - 5\end{array} \right.} \right) \to \left( {\begin{array}{*{20}{c}}1&1&2\\0&1&{ - 2}\\0&0&0\end{array}\left| \begin{array}{c} - 2\\ - 3\\0\end{array} \right.} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">row of zeroes implies infinite solutions, (or equivalent). <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1 </em></strong>for any attempt at row reduction.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}} \right| = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><span style="background-color: #f7f7f7;">\(\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}} \right| = 0\)</span> with one valid point <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(x + y + 2z = - 2\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(3x - y + 14z = 6\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(x + 2y = - 5\) \( \Rightarrow x = - 5 - 2y\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">substitute \(x = - 5 - 2y\) into the first two equations:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 5 - 2y + y + 2z = - 2\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(3( - 5 - 2y) - y + 14z = 6\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( - y + 2z = 3\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 7y + 14z = 21\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">the latter two equations are equivalent (by multiplying by 7) therefore an infinite number of solutions. <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">for example, \(7 \times {{\text{R}}_1} - {{\text{R}}_2}\) gives \(4x + 8y = - 20\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">this equation is a multiple of the third equation, therefore an infinite</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">number of solutions. <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) let \(y = t\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">then \(x = - 5 - 2t\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{t + 3}}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(x = t\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">then \(y = \frac{{ - 5 - t}}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{1 - t}}{4}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(z = t\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">then \(x = 1 - 4t\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = - 3 + 2t\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to find cross product of two normal vectors:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em>: \(\left| {\begin{array}{*{20}{c}}i&j&k\\1&1&2\\1&2&0\end{array}} \right| = - 4i + 2j + k\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = 1 - 4t\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = - 3 + 2t\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = t\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(or equivalent)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that \(\sin 2nx = \sin \left( {(2n + 1)x} \right)\cos x - \cos \left( {(2n + 1)x} \right)\sin x\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <strong>Hence</strong> prove, by induction, that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[\cos x + \cos 3x + \cos 5x + \ldots + \cos \left( {(2n - 1)x} \right) = \frac{{\sin 2nx}}{{2\sin x}},\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for all \(n \in {\mathbb{Z}^ + }{\text{, }}\sin x \ne 0\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Solve the equation \(\cos x + \cos 3x = \frac{1}{2},{\text{ }}0 < x < \pi \).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\sin (2n + 1)x\cos x - \cos (2n + 1)x\sin x = \sin (2n + 1)x - x\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0px 0px 0px 30px; font: 26px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sin 2nx\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) if <em>n</em> = 1 <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{LHS}} = \cos x\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{RHS}} = \frac{{\sin 2x}}{{2\sin x}} = \frac{{2\sin x\cos x}}{{2\sin x}} = \cos x\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so LHS = RHS and the statement is true for <em>n</em> = 1 <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">assume true for <em>n</em> = <em>k</em> <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Only award <strong><em>M1</em></strong> if the word <strong>true</strong> appears.</span></p>
<p style="margin: 0px 0px 0px 30px; font: 26px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> Do <strong>not</strong> award <strong><em>M1</em></strong> for ‘let <em>n</em> = <em>k</em>’ only.</span></p>
<p style="margin: 0px 0px 0px 30px; font: 26px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> Subsequent marks are independent of this <strong><em>M1</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so \(\cos x + \cos 3x + \cos 5x + \ldots + \cos (2k - 1)x = \frac{{\sin 2kx}}{{2\sin x}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">if <em>n</em> = <em>k</em> + 1 then</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos x + \cos 3x + \cos 5x + \ldots + \cos (2k - 1)x + \cos (2k + 1)x\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sin 2kx}}{{2\sin x}} + \cos (2k + 1)x\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sin 2kx + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sin (2k + 1)x\cos x - \cos (2k + 1)x\sin x + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sin (2k + 1)x\cos x + \cos (2k + 1)x\sin x}}{{2\sin x}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sin (2k + 2)x}}{{2\sin x}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sin 2(k + 1)x}}{{2\sin x}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so if true for <em>n</em> = <em>k</em>, then also true for <em>n</em> = <em>k</em> + <em>1</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">as true for <em>n</em> = 1 then true for all \(n \in {\mathbb{Z}^ + }\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Final <strong><em>R1</em></strong> is independent of previous work.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[12 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \(\frac{{\sin 4x}}{{2\sin x}} = \frac{1}{2}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sin 4x = \sin x\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(4x = x \Rightarrow x = 0\) but this is impossible</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(4x = \pi - x \Rightarrow x = \frac{\pi }{5}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(4x = 2\pi + x \Rightarrow x = \frac{{2\pi }}{3}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(4x = 3\pi - x \Rightarrow x = \frac{{3\pi }}{5}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for not including any answers outside the domain <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award the first <strong><em>M1A1</em></strong> for correctly obtaining \(8{\cos ^3}x - 4\cos x - 1 = 0\) or equivalent and subsequent marks as appropriate including the answers \(\left( { - \frac{1}{2},\frac{{1 \pm \sqrt 5 }}{4}} \right)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [20 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This question showed the weaknesses of many candidates in dealing with formal proofs and showing their reasoning in a logical manner. In part (a) just a few candidates clearly showed the result and part (b) showed that most candidates struggle with the formality of a proof by induction. The logic of many solutions was poor, though sometimes contained correct trigonometric work. Very few candidates were successful in answering part (c) using the unit circle. Most candidates attempted to manipulate the equation to obtain a cubic equation but made little progress. A few candidates guessed \(\frac{{2\pi }}{3}\) as a solution but were not able to determine the other solutions.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the cube roots of i in the form \(a + b{\text{i}}\), where \(a,{\text{ }}b \in \mathbb{R}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{i}} = \cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z_1} = {{\text{i}}^{\frac{1}{3}}} = {\left( {\cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}} \right)^{\frac{1}{3}}} = \cos \frac{\pi }{6} + {\text{i}}\sin \frac{\pi }{6}\,\,\,\,\,\left( { = \frac{{\sqrt 3 }}{2} + \frac{1}{2}{\text{i}}} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z_2} = \cos \frac{{5\pi }}{6} + {\text{i}}\sin \frac{{5\pi }}{6}\,\,\,\,\,\left( { = - \frac{{\sqrt 3 }}{2} + \frac{1}{2}{\text{i}}} \right)\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z_3} = \cos \left( { - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{2}} \right) = - {\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Accept exponential and cis forms for intermediate results, but not the final roots.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Accept the method based on expanding \({({\text{a}} + {\text{b}})^3}\). </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for attempt, </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for equating real and imaginary parts, </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for finding a = 0 and \({\text{b}} = \frac{1}{2}\), then </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1A1A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for the roots.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">A varied response. Many knew how to solve this standard question in the most efficient way. A few candidates expanded \({(a + ib)^3}\) and solved the resulting fairly simple equations. A disappointing minority of candidates did not know how to start.</span></p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Factorize \({z^3} + 1\) into a linear and quadratic factor.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(\gamma = \frac{{1 + {\text{i}}\sqrt 3 }}{2}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Show that \(\gamma \) is one of the cube roots of −1.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Show that \({\gamma ^2} = \gamma - 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Hence find the value of \({(1 - \gamma )^6}\).</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">using the factor theorem <em>z</em> +1 is a factor <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^3} + 1 = (z + 1)({z^2} - z + 1)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^3} = - 1 \Rightarrow {z^3} + 1 = (z + 1)({z^2} - z + 1) = 0\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solving \({z^2} - z + 1 = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \frac{{1 \pm \sqrt {1 - 4} }}{2} = \frac{{1 \pm {\text{i}}\sqrt 3 }}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore one cube root of −1 is \(\gamma \) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\gamma ^2} = \left( {{{\frac{{1 + i\sqrt 3 }}{2}}^2}} \right) = \frac{{ - 1 + i\sqrt 3 }}{2}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2} \times \frac{{1 + i\sqrt 3 }}{2} = \frac{{ - 1 - 3}}{4}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= −1 AG</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 3</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\gamma = \frac{{1 + i\sqrt 3 }}{2} = {e^{i\frac{\pi }{3}}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\gamma ^3} = {e^{i\pi }} = - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">as \(\gamma \) is a root of \({z^2} - z + 1 = 0\) then \({\gamma ^2} - \gamma + 1 = 0\) <strong><em>M1R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore {\gamma ^2} = \gamma - 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for the use of \({z^2} - z + 1 = 0\) in any way.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Award <strong><em>R1</em></strong> for a correct reasoned approach.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\gamma - 1 = \frac{{1 + i\sqrt 3 }}{2} - 1 = \frac{{ - 1 + i\sqrt 3 }}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(1 - \gamma )^6} = {( - {\gamma ^2})^6}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {(\gamma )^{12}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {({\gamma ^3})^4}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {( - 1)^4}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(1 - \gamma )^6}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1 - 6\gamma + 15{\gamma ^2} - 20{\gamma ^3} + 15{\gamma ^4} - 6{\gamma ^5} + {\gamma ^6}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for attempt at binomial expansion.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">use of any previous result </span><em style="font-family: 'times new roman', times; font-size: medium;">e.g.</em><span style="font-family: 'times new roman', times; font-size: medium;"> \( = 1 - 6\gamma + 15{\gamma ^2} + 20 - 15\gamma + 6{\gamma ^2} + 1\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong></p>
<p style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; font: 19px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">= 1 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica; min-height: 23.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[9 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In part a) the factorisation was, on the whole, well done.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (b) was done well by most although using a substitution method rather than the result above. This used much m retime than was necessary but was successful. A number of candidates did not use the previous results in part (iii) and so seemed to not understand the use of the ‘hence’.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>An arithmetic sequence \({u_1}{\text{, }}{u_2}{\text{, }}{u_3} \ldots \) has \({u_1} = 1\) and common difference \(d \ne 0\). Given that \({u_2}{\text{, }}{u_3}\) and \({u_6}\) are the first three terms of a geometric sequence</p>
</div>
<div class="specification">
<p>Given that \({u_N} = - 15\)</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>find the value of \(d\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>determine the value of \(\sum\limits_{r = 1}^N {{u_r}} \).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>use of \({u_n} = {u_1} + (n - 1)d\) <strong><em>M1</em></strong></p>
<p>\({(1 + 2d)^2} = (1 + d)(1 + 5d)\) (or equivalent) <strong><em>M1A1</em></strong></p>
<p>\(d = - 2\) <strong><em>A1</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(1 + (N - 1) \times - 2 = - 15\)</p>
<p>\(N = 9\) <strong><em>(A1)</em></strong></p>
<p>\(\sum\limits_{r = 1}^9 {{u_r}} = \frac{9}{2}(2 + 8 \times - 2)\) <strong><em>(M1)</em></strong></p>
<p>\( = - 63\) <strong><em>A1</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p>If \({z_1} = a + a\sqrt 3 i\) and \({z_2} = 1 - i\), where a is a real constant, express \({z_1}\) and \({z_2}\) in the form \(r\,{\text{cis}}\,\theta \), and hence find an expression for \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6}\) in terms of a and i.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({z_1} = 2a{\text{cis}}\left( {\frac{\pi }{3}} \right){\text{, }}{z_2} = \sqrt 2 {\text{ cis}}\left( { - \frac{\pi }{4}} \right)\) <strong><em>M1 A1 A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = \frac{{{2^6}{a^6}{\text{cis(0)}}}}{{{{\sqrt 2 }^6}{\text{cis}}\left( {\frac{\pi }{2}} \right)}}\left( { = 8{a^6}{\text{cis}}\left( { - \frac{\pi }{2}} \right)} \right)\) <strong> <em>M1 A1 A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = {\left( {\frac{{2a}}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{7\pi }}{{12}}} \right)} \right)^6}\) <strong> <em>M1 A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 8{a^6}{\text{cis}}\left( { - \frac{\pi }{2}} \right)\) <strong style="font-weight: bold;"><em style="font-style: italic;">A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong style="font-weight: bold;">THEN</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - 8{a^6}{\text{i}}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept equivalent angles, in radians or degrees. </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Accept alternate answers without cis e.g. \({\text{ = }}\frac{{8{a^6}}}{{\text{i}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Most students had an idea of what to do but were frequently let down in their calculations of the modulus and argument. The most common error was to give the argument of \({z_2}\) as \(\frac{{3\pi }}{4}\), failing to recognise that it should be in the fourth quadrant. There were also errors seen in the algebraic manipulation, in particular forgetting to raise the modulus to the power 6.</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(w = \cos \frac{{2\pi }}{7} + {\text{i}}\sin \frac{{2\pi }}{7}\).</p>
</div>
<div class="specification">
<p class="p1">Consider the quadratic equation \({z^2} + bz + c = 0\) where \(b,{\text{ }}c \in \mathbb{R},{\text{ }}z \in \mathbb{C}\). The roots of this equation are \(\alpha \) and \(\alpha *\) where \(\alpha *\) is the complex conjugate of \(\alpha \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Verify that \(w\) is a root of the equation \({z^7} - 1 = 0,{\text{ }}z \in \mathbb{C}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Expand \((w - 1)(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6})\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Hence deduce that \(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6} = 0\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down the roots of the equation \({z^7} - 1 = 0,{\text{ }}z \in \mathbb{C}\) in terms of \(w\) and plot these roots on an Argand diagram.</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Given that \(\alpha = w + {w^2} + {w^4}\), show that \(\alpha * = {w^6} + {w^5} + {w^3}\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Find the value of \(b\) and the value of \(c\).</p>
<div class="marks">[10]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Using the values for \(b\) and \(c\) obtained in part (d)(ii), find the imaginary part of \(\alpha \), giving your answer in surd form.</p>
<div class="marks">[4]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>EITHER</strong></p>
<p class="p2"><span class="Apple-converted-space">\({w^7} = {\left( {\cos \frac{{2\pi }}{7} + {\text{i}}\sin \frac{{2\pi }}{7}} \right)^7}\) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = \cos 2\pi + {\text{i}}\sin 2\pi \) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = 1\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">so \(w\) is a root <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p2"><span class="Apple-converted-space">\({z^7} = 1 = \cos (2\pi k) + {\text{i}}\sin (2\pi k)\) </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(z = \cos \left( {\frac{{2\pi k}}{7}} \right) + {\text{i}}\sin \left( {\frac{{2\pi }}{7}} \right)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(k = 1 \Rightarrow z = \cos \left( {\frac{{2\pi }}{7}} \right) + {\text{i}}\sin \left( {\frac{{2\pi }}{7}} \right)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">so \(w\) is a root <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \((w - 1)(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6})\)</span></p>
<p class="p1"><span class="Apple-converted-space">\( = w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6} + {w^7} - 1 - w - {w^2} - {w^3} - {w^4} - {w^5} - {w^6}\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p1"><span class="Apple-converted-space">\( = {w^7} - 1{\text{ }}( = 0)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> \({w^7} - 1 = 0\)</span> and \(w - 1 \ne 0\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">so \(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6} = 0\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p2"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">the roots are \(1,{\text{ }}w,{\text{ }}{w^2},{\text{ }}{w^3},{\text{ }}{w^4},{\text{ }}{w^5}\) <span class="s1">and \({w^6}\)</span></p>
<p class="p1"><span class="s1"><img src="images/Schermafbeelding_2017-01-27_om_17.06.34.png" alt="M16/5/MATHL/HP1/ENG/TZ2/12.c/M"></span></p>
<p class="p1"> </p>
<p class="p1"><span class="s1">7 </span>points equidistant from the origin <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2">approximately correct angular positions for \(1,{\text{ }}w,{\text{ }}{w^2},{\text{ }}{w^3},{\text{ }}{w^4},{\text{ }}{w^5}\) <span class="s2">and \({w^6}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></span></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Condone use of <span class="s1"><em>cis </em></span>notation for the final two <strong><em>A </em></strong>marks.</p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>For the final <strong><em>A </em></strong>mark there should be one root in the first quadrant, two in the second, two in the third, one in the fourth, and one on the real axis.</p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \(\alpha * = (w + {w^2} + {w^4})*\)</span></p>
<p class="p1"><span class="Apple-converted-space">\( = w* + ({w^2})* + ({w^4})*\) </span><strong><em>A1</em></strong></p>
<p class="p1">since \( = w* = {w^6},{\text{ }}({w^2})* = {w^5}\) and \(({w^4})* = {w^3}\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow \alpha * = {w^6} + {w^5} + {w^3}\) </span><strong><em>AG</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> \(b = - (\alpha + \alpha *)\)</span> (using sum of roots (or otherwise)) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(b = - (w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6})\) </span><strong><em>(A1)</em></strong></p>
<p class="p2">\( = - ( - 1)\)</p>
<p class="p2"><span class="Apple-converted-space">\( = 1\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">\(c = \alpha \alpha *\) (using product of roots (or otherwise)) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2">\(c = (w + {w^2} + {w^4})({w^6} + {w^5} + {w^3})\)</p>
<p class="p1"><strong>EITHER</strong></p>
<p class="p2"><span class="Apple-converted-space">\( = {w^{10}} + {w^9} + {w^8} + 3{w^7} + {w^6} + {w^5} + {w^4}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = ({w^6} + {w^5} + {w^4} + {w^3} + {w^2} + w) + 3\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = 3 - 1\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p1"><strong>OR</strong></p>
<p class="p2"><span class="Apple-converted-space">\( = {w^{10}} + {w^9} + {w^8} + 3{w^7} + {w^6} + {w^5} + {w^4}\left( { = {w^4}(1 + w + {w^3})({w^3} + {w^2} + 1)} \right)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = {w^4}({w^6} + {w^5} + {w^4} + {w^2} + w + 1 + 3{w^3})\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2">\( = {w^4}({w^6} + {w^5} + {w^4} + {w^3} + {w^2} + w + 1 + 2{w^3})\)</p>
<p class="p2"><span class="Apple-converted-space">\( = {w^4}(2{w^3})\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p1"><strong>THEN</strong></p>
<p class="p2"><span class="Apple-converted-space">\( = 2\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[10 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\({z^2} + z + 2 = 0 \Rightarrow z = \frac{{ - 1 \pm {\text{i}}\sqrt 7 }}{2}\) </span><strong><em>M1A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(\operatorname{Im} (w + {w^2} + {w^4}) > 0\) </span><span class="s1"><strong><em>R1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(\operatorname{Im} \alpha = \frac{{\sqrt 7 }}{2}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Final <strong><em>A </em></strong>mark is independent of previous <strong><em>R </em></strong>mark.</p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">The majority of candidates scored full marks in part (a).</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">The majority of candidates scored full marks in part (b)(i). It was expected to see \(w - 1 \ne 0\) stated for the (b)(ii) mark, though some did appreciate this.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (c), the roots were required to be stated in terms of \(w\). This was sometimes ignored, thankfully not too frequently. Clear Argand diagrams were not often seen, and candidates’ general presentation in this area could be improved. Having said this, most scripts were awarded at least 2 of 3 marks available.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part (d) proved to be a good discriminator for the better candidates. The product and sum of roots formulae now seem to be better appreciated, and while only the best scored full marks, a good number were able to demonstrate the result \(b = 1\).</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (e), of those candidates who reached this far in the paper, most were able to pick up two or three marks, albeit from sometimes following through incorrect work. A correct reason for choosing \({\text{i}}\sqrt 7 \) over \( - {\text{i}}\sqrt 7 \) was necessary, but rarely, if ever seen.</p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(\omega \) be one of the non-real solutions of the equation \({z^3} = 1\).</p>
</div>
<div class="specification">
<p class="p1">Consider the complex numbers \(p = 1 - 3{\text{i}}\) and \(q = x + (2x + 1){\text{i}}\), where \(x \in \mathbb{R}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Determine the value of</p>
<p class="p1">(i) <span class="Apple-converted-space"> \(1 + \omega + {\omega ^2}\)</span>;</p>
<p class="p1">(ii) <span class="Apple-converted-space"> \(1 + \omega {\text{*}} + {(\omega {\text{*}})^2}\)</span>.</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \((\omega - 3{\omega ^2})({\omega ^2} - 3\omega ) = 13\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Find the values of \(x\) </span>that satisfy the equation \(\left| p \right| = \left| q \right|\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Solve the inequality \(\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2}\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span><strong>METHOD 1</strong></p>
<p class="p1"><span class="Apple-converted-space">\(1 + \omega + {\omega ^2} = \frac{{1 - {\omega ^3}}}{{1 - \omega }} = 0\) </span><strong><em>A1</em></strong></p>
<p class="p1">as \(\omega \ne 1\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p1"><span class="s1">solutions of \(1 - {\omega ^3} = 0\)</span> are \(\omega = 1,{\text{ }}\omega {\text{ = }}\frac{{ - 1 \pm \sqrt 3 {\text{i}}}}{2}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2">verification that the sum of these roots is 0 <span class="Apple-converted-space"> </span><span class="s2"><strong><em>R1</em></strong></span></p>
<p class="p1">(ii) <span class="Apple-converted-space"> \(1 + \omega {\text{*}} + {(\omega {\text{*}})^2} = 0\)</span> <span class="Apple-converted-space"> </span><strong><em>A2</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\((\omega - 3{\omega ^2})({\omega ^2} - 3\omega ) = - 3{\omega ^4} + 10{\omega ^3} - 3{\omega ^2}\) </span><strong><em>M1A1</em></strong></p>
<p class="p1"><strong>EITHER</strong></p>
<p class="p1"><span class="Apple-converted-space">\( = - 3{\omega ^2}({\omega ^2} + \omega + 1) + 13{\omega ^3}\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( = - 3{\omega ^2} \times 0 + 13 \times 1\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1"><span class="Apple-converted-space">\( = - 3\omega + 10 - 3{\omega ^2} = - 3({\omega ^2} + \omega + 1) + 13\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( = - 3 \times 0 + 13\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1"><span class="s1">substitution by \(\omega = \frac{{ - 1 \pm \sqrt 3 {\text{i}}}}{2}\)</span> in any form <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">numerical values of each term seen <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong>THEN</strong></p>
<p class="p1"><span class="Apple-converted-space">\( = 13\) </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(\left| p \right| = \left| q \right| \Rightarrow \sqrt {{1^2} + {3^2}} = \sqrt {{x^2} + {{(2x + 1)}^2}} \) </span><strong><em>(M1)(A1)</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(5{x^2} + 4x - 9 = 0\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\((5x + 9)(x - 1) = 0\) </span><strong><em>(M1)</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(x = 1,{\text{ }}x = - \frac{9}{5}\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(pq = (1 - 3{\text{i}})\left( {x + (2x + 1){\text{i}}} \right) = (7x + 3) + (1 - x){\text{i}}\) </span><strong><em>M1A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2} \Rightarrow (7x + 3) + 8 < {(1 - x)^2}\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow {x^2} - 9x - 10 > 0\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow (x + 1)(x - 10) > 0\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(x < - 1,{\text{ }}x > 10\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Expand and simplify \({\left( {x - \frac{2}{x}} \right)^4}\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence determine the constant term in the expansion \((2{x^2} + 1){\left( {x - \frac{2}{x}} \right)^4}\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {x - \frac{2}{x}} \right)^4} = {x^4} + 4{x^3}\left( { - \frac{2}{x}} \right) + 6{x^2}{\left( { - \frac{2}{x}} \right)^2} + 4x{\left( { - \frac{2}{x}} \right)^3} + {\left( { - \frac{2}{x}} \right)^4}\) <strong><em>(A2)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></span><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>(A1) </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">for 3 or 4 correct terms. </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept combinatorial expressions, <em>e.g.</em> \(\left( {\begin{array}{*{20}{c}}<br> 4 \\ <br> 2 <br>\end{array}} \right)\) for 6.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {x^4} - 8{x^2} + 24 - \frac{{32}}{{{x^2}}} + \frac{{16}}{{{x^4}}}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[3 marks]</span><br></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">constant term from expansion of \((2{x^2} + 1){\left( {x - \frac{2}{x}} \right)^4} = -64 + 24 = -40\) <strong><em>A2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>A1 </em></strong>for –64 or 24 seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[2 marks]</span><br></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">It was disappointing to see many candidates expanding \({\left( {x - \frac{2}{x}} \right)^4}\) by first expanding \({\left( {x - \frac{2}{x}} \right)^2}\) and then either squaring the result or multiplying twice by \(\left( {x - \frac{2}{x}} \right)\), processes which often resulted in arithmetic errors being made. Candidates at this level are expected to be sufficiently familiar with Pascal’s Triangle to use it in this kind of problem. In (b), some candidates appeared not to understand the phrase ‘constant term’.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">It was disappointing to see many candidates expanding \({\left( {x - \frac{2}{x}} \right)^4}\) by first expanding \({\left( {x - \frac{2}{x}} \right)^2}\) and then either squaring the result or multiplying twice by \(\left( {x - \frac{2}{x}} \right)\), processes which often resulted in arithmetic errors being made. Candidates at this level are expected to be sufficiently familiar with Pascal's Triangle to use it in this kind of problem. In (b), some candidates appeared not to understand the phrase "constant term".</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the value of \(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \(\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi \) <span class="s1">where \(k \in \mathbb{Z}\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use the principle of mathematical induction to prove that</p>
<p class="p1">\(\sin x + \sin 3x + \ldots + \sin (2n - 1)x = \frac{{1 - \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi \) where \(k \in \mathbb{Z}\).</p>
<div class="marks">[9]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence or otherwise solve the equation \(\sin x + \sin 3x = \cos x\) <span class="s1">in the interval \(0 < x < \pi \).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}\) </span><strong><em>(M1)A1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: </strong>Award <strong><em>M1 </em></strong>for 5 equal terms with \) + \) or \( - \) signs.</p>
<p class="p2"> </p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \frac{{1 - (1 - 2{{\sin }^2}x)}}{{2\sin x}}\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p1"><span class="Apple-converted-space">\( \equiv \frac{{2{{\sin }^2}x}}{{2\sin x}}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="Apple-converted-space">\( \equiv \sin x\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p2"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">let \({\text{P}}(n):\sin x + \sin 3x + \ldots + \sin (2n - 1)x \equiv \frac{{1 - \cos 2nx}}{{2\sin x}}\)</p>
<p class="p1">if \(n = 1\)</p>
<p class="p1">\({\text{P}}(1):\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x\) which is true (as proved in part (b)) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="s1">assume \({\text{P}}(k)\)</span> true, \(\sin x + \sin 3x + \ldots + \sin (2k - 1)x \equiv \frac{{1 - \cos 2kx}}{{2\sin x}}\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Notes: </strong>Only award <strong><em>M1 </em></strong>if the words “assume” and “true” appear. Do not award <strong><em>M1 </em></strong>for “let \(n = k\)<em>” </em>only. Subsequent marks are independent of this <strong><em>M1</em></strong><em>.</em></p>
<p class="p3"> </p>
<p class="p4">consider \({\text{P}}(k + 1)\)<span class="s2">:</span></p>
<p class="p1">\({\text{P}}(k + 1):\sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x \equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}\)</p>
<p class="p1"><span class="Apple-converted-space">\(LHS = \sin x + \sin 3x + \ldots + \sin (2k - 1)x + \sin (2k + 1)x\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \equiv \frac{{1 - \cos 2kx}}{{2\sin x}} + \sin (2k + 1)x\) </span><strong><em>A1</em></strong></p>
<p class="p1">\( \equiv \frac{{1 - \cos 2kx + 2\sin x\sin (2k + 1)x}}{{2\sin x}}\)</p>
<p class="p1"><span class="Apple-converted-space">\( \equiv \frac{{1 - \cos 2kx + 2\sin x\cos x\sin 2kx + 2{{\sin }^2}x\cos 2kx}}{{2\sin x}}\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \equiv \frac{{1 - \left( {(1 - 2{{\sin }^2}x)\cos 2kx - \sin 2x\sin 2kx} \right)}}{{2\sin x}}\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \equiv \frac{{1 - (\cos 2x\cos 2kx - \sin 2x\sin 2kx)}}{{2\sin x}}\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \equiv \frac{{1 - \cos (2kx + 2x)}}{{2\sin x}}\) </span><strong><em>A1</em></strong></p>
<p class="p1">\( \equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}\)</p>
<p class="p1">so if true for \(n = k\) , then also true for \(n = k + 1\)</p>
<p class="p1">as true for \(n = 1\) then true for all \(n \in {\mathbb{Z}^ + }\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: </strong>Accept answers using transformation formula for product of sines if steps are shown clearly.</p>
<p class="p2"> </p>
<p class="p1"><strong>Note: </strong>Award <strong><em>R1 </em></strong>only if candidate is awarded at least 5 marks in the previous steps.</p>
<p class="p2"> </p>
<p class="p1"><strong><em>[9 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>EITHER</strong></p>
<p class="p1"><span class="Apple-converted-space">\(\sin x + \sin 3x = \cos x \Rightarrow \frac{{1 - \cos 4x}}{{2\sin x}} = \cos x\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow 1 - \cos 4x = 2\sin x\cos x,{\text{ }}(\sin x \ne 0)\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow 1 - (1 - 2{\sin ^2}2x) = \sin 2x\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow \sin 2x(2\sin 2x - 1) = 0\) </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="s1">\( \Rightarrow \sin 2x = 0\) or \(\sin 2x = \frac{1}{2}\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(2x = \pi ,{\text{ }}2x = \frac{\pi }{6}\) and \(2x = \frac{{5\pi }}{6}\)</p>
<p class="p1"><strong>OR</strong></p>
<p class="p1"><span class="Apple-converted-space">\(\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x\) </span><strong><em>M1A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow (2\sin 2x - 1)\cos x = 0,{\text{ }}(\sin x \ne 0)\) </span><strong><em>M1A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( \Rightarrow \sin 2x = \frac{1}{2}\) of \(\cos x = 0\) </span><strong><em>A1</em></strong></p>
<p class="p1">\(2x = \frac{\pi }{6},{\text{ }}2x = \frac{{5\pi }}{6}\) and \(x = \frac{\pi }{2}\)</p>
<p class="p1"><strong>THEN</strong></p>
<p class="p1"><span class="s1">\(\therefore x = \frac{\pi }{2},{\text{ }}x = \frac{\pi }{{12}}\) and \(x = \frac{{5\pi }}{{12}}\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: </strong>Do not award the final <strong><em>A1 </em></strong>if extra solutions are seen.</p>
<p class="p2"> </p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">A box contains four red balls and two white balls. Darren and Marty play a game by each taking it in turn to take a ball from the box, without replacement. The first player to take a white ball is the winner.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Darren plays first, find the probability that he wins.</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">The game is now changed so that the ball chosen is replaced after each turn.</p>
<p class="p1">Darren still plays first.</p>
<p class="p1">Show that the probability of Darren winning has not changed.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">probability that Darren wins \({\text{P}}(W) + {\text{P}}(RRW) + {\text{P}}(RRRRW)\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Only award <strong><em>M1 </em></strong>if three terms are seen or are implied by the following numerical equivalent.</p>
<p class="p4"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Accept equivalent tree diagram for method mark.</p>
<p class="p4"> </p>
<p class="p1">\( = \frac{2}{6} + \frac{4}{6} \bullet \frac{3}{5} \bullet \frac{2}{4} + \frac{4}{6} \bullet \frac{3}{5} \bullet \frac{2}{4} \bullet \frac{1}{3} \bullet \frac{2}{2}\;\;\;\left( { = \frac{1}{3} + \frac{1}{5} + \frac{1}{{15}}} \right)\) <strong><em>A2</em></strong></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span><em>A1 </em></strong>for two correct.</p>
<p class="p4"> </p>
<p class="p1">\( = \frac{3}{5}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>the probability that Darren wins is given by</p>
<p>\({\text{P}}(W) + {\text{P}}(RRW) + {\text{P}}(RRRRW) + \ldots \) <strong><em>(M1)</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept equivalent tree diagram with correctly indicated path for method mark.</p>
<p> </p>
<p>\({\text{P (Darren Win)}} = \frac{1}{3} + \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{1}{3} + \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{1}{3} + \ldots \)</p>
<p>or \( = \frac{1}{3}\left( {1 + \frac{4}{9} + {{\left( {\frac{4}{9}} \right)}^2} + \ldots } \right)\) <strong><em>A1</em></strong></p>
<p>\( = \frac{1}{3}\left( {\frac{1}{{1 - \frac{4}{9}}}} \right)\) <strong><em>A1</em></strong></p>
<p>\( = \frac{3}{5}\) <strong><em>AG</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>\({\text{P (Darren wins)}} = {\text{P}}\)</p>
<p>\({\text{P}} = \frac{1}{3} + \frac{4}{9}{\text{P}}\) <strong><em>M1A2</em></strong></p>
<p>\(\frac{5}{9}{\text{P}} = \frac{1}{3}\)</p>
<p>\({\text{P}} = \frac{3}{5}\) <strong><em>AG</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<p><strong><em>Total [7 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The sum of the first \(n\) terms of a sequence \(\{ {u_n}\} \) is given by \({S_n} = 3{n^2} - 2n\), where \(n \in {\mathbb{Z}^ + }\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down the value of \({u_1}\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the value of \({u_6}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Prove that \(\{ {u_n}\} \) </span>is an arithmetic sequence, stating clearly its common difference.</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\({u_1} = 1\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\({u_6} = {S_6} - {S_5} = 31\) </span><strong><em>M1A1</em></strong></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\({u_n} = {S_n} - {S_{n - 1}}\) </span><strong><em>M1</em></strong></p>
<p class="p1">\( = (3{n^2} - 2n) - \left( {3{{(n - 1)}^2} - 2(n - 1)} \right)\)</p>
<p class="p1">\( = (3{n^2} - 2n) - (3{n^2} - 6n + 3 - 2n + 2)\)</p>
<p class="p1"><span class="Apple-converted-space">\( = 6n - 5\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(d = {u_{n + 1}} - {u_n}\) </span><strong><em>R1</em></strong></p>
<p class="p1">\( = 6n + 6 - 5 - 6n + 5\)</p>
<p class="p1">\( = \left( {6(n + 1) - 5} \right) - (6n - 5)\)</p>
<p class="p1">\( = 6\) (constant) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Notes: </strong>Award <strong><em>R1 </em></strong>only if candidate provides a clear argument that proves that the difference between <strong>ANY </strong>two consecutive terms of the sequence is constant. Do not accept examples involving particular terms of the sequence nor circular reasoning arguments (<em>eg </em>use of formulas of APs to prove that it is an AP). Last <strong><em>A1 </em></strong>is independent of <strong><em>R1</em></strong>.</p>
<p class="p2"> </p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>A given polynomial function is defined as \(f(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_n}{x^n}\). The roots of the polynomial equation \(f(x) = 0\) are consecutive terms of a geometric sequence with a common ratio of \(\frac{1}{2}\) and first term 2.</p>
<p>Given that \({a_{n - 1}} = - 63\) and \({a_n} = 16\) find</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">the degree of the polynomial;</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">the value of \({a_0}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">the sum of the roots of the polynomial \( = \frac{{63}}{{16}}\) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1">\(2\left( {\frac{{1 - {{\left( {\frac{1}{2}} \right)}^n}}}{{1 - \frac{1}{2}}}} \right) = \frac{{63}}{{16}}\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>The formula for the sum of a geometric sequence must be equated to a value for the <strong><em>M1 </em></strong>to be awarded.</p>
<p class="p4"> </p>
<p class="p1">\(1 - {\left( {\frac{1}{2}} \right)^n} = \frac{{63}}{{64}} \Rightarrow {\left( {\frac{1}{2}} \right)^n} = \frac{1}{{64}}\)</p>
<p class="p1">\(n = 6\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="s1"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(\frac{{{a_0}}}{{{a_n}}} = 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}},{\text{ (}}{{\text{a}}_n} = 16)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\({a_0} = 16 \times 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}}\)</p>
<p class="p1">\({a_0} = {2^{ - 5}}\;\;\;\left( { = \frac{1}{{32}}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<p class="p1"><strong><em>Total [6 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p>Solve \({\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>\({\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) - 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)\)</p>
<p><strong>EITHER</strong></p>
<p>\({\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}\) <em><strong>M1</strong></em></p>
<p>\( = \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}\) <em><strong>A1</strong></em></p>
<p><strong>OR</strong></p>
<p>\(\left( {{\text{ln}}\,x - 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)\) <em><strong> M1A1</strong></em></p>
<p><strong>THEN</strong></p>
<p>\({\text{ln}}\,x = 2\,{\text{ln}}\,2\) or \( - {\text{ln}}\,2\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow x = 4\) or \(x = \frac{1}{2}\) <em><strong> (M1)A1</strong></em> </p>
<p><strong>Note:</strong> <em><strong>(M1)</strong></em> is for an appropriate use of a log law in either case, dependent on the previous <em><strong>M1</strong></em> being awarded, <strong>A1</strong> for both correct answers.</p>
<p>solution is \(\frac{1}{2} < x < 4\) <em><strong>A1</strong></em></p>
<p><em><strong>[6 marks]</strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p class="p1">A set of positive integers {\(1,2,3,4,5,6,7,8,9\)} is used to form a pack of nine cards.</p>
<p class="p1">Each card displays one positive integer without repetition from this set. Grace wishes to select four cards at random from this pack of nine cards.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the number of selections Grace could make if the largest integer drawn among the four cards is either a \(5\), a \(6\) or a \(7\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the number of selections Grace could make if at least two of the four integers drawn are even.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">use of the addition principle with \(3\) terms <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">to obtain \(^4{C_3}{ + ^5}{C_3}{ + ^6}{C_3}{\text{ }}( = 4 + 10 + 20)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">number of possible selections is \(34\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong>EITHER</strong></p>
<p class="p1">recognition of three cases: (\(2\) odd and \(2\) even or \(1\) odd and \(3\) even or \(0\) odd and \(4\) even) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p1">\(\left( {^5{C_2}{ \times ^4}{C_2}} \right) + \left( {^5{C_1}{ \times ^4}{C_3}} \right) + \left( {^5{C_0}{ \times ^4}{C_4}} \right)\;\;\;( = 60 + 20 + 1)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1">recognition to subtract the sum of \(4\) odd and \(3\) odd and \(1\) even from the total <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p1">\(^9{C_4}{ - ^5}{C_4} - \left( {^5{C_3}{ \times ^4}{C_1}} \right)\;\;\;( = 126 - 5 - 40)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)A1</em></strong></span></p>
<p class="p1"><strong>THEN</strong></p>
<p class="p1">number of possible selections is \(81\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<p class="p1"><strong><em>Total [7 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p>Solve the equation \({\log _2}(x + 3) + {\log _2}(x - 3) = 4\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>\({\log _2}(x + 3) + {\log _2}(x - 3) = 4\)</p>
<p>\({\log _2}({x^2} - 9) = 4\) <strong><em>(M1)</em></strong></p>
<p>\({x^2} - 9 = {2^4}{\text{ }}( = 16)\) <strong><em>M1A1</em></strong></p>
<p>\({x^2} = 25\)</p>
<p>\(x = \pm 5\) <strong><em>(A1)</em></strong></p>
<p>\(x = 5\) <strong><em>A1</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p>Find the solution of \({\log _2}x - {\log _2}5 = 2 + {\log _2}3\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>\({\log _2}x - {\log _2}5 = 2 + {\log _2}3\)</p>
<p>collecting at least two log terms <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\log _2}\frac{x}{5} = 2 + {\log _2}3{\text{ or }}{\log _2}\frac{x}{{15}} = 2\)</p>
<p>obtaining a correct equation without logs <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\frac{x}{5} = 12\)\(\,\,\,\)<strong>OR</strong>\(\,\,\,\)\(\frac{x}{{15}} = {2^2}\) <strong><em>(A1)</em></strong></p>
<p>\(x = 60\) <strong><em>A1</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the following system of equations:</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[x + y + z = 1\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[2x + 3y + z = 3\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[x + 3y - z = \lambda \]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">where \(\lambda \in \mathbb{R}\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that this system does not have a unique solution for any value of \(\lambda \) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Determine the value of \(\lambda \) for which the system is consistent.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) For this value of \(\lambda \) , find the general solution of the system.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">using row operations, <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">to obtain 2 equations in the same 2 variables <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for example \(y - z = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(2y - 2z = \lambda - 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the fact that one of the left hand sides is a multiple of the other left hand side indicates that the equations do not have a unique solution, or equivalent <strong><em>R1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(\lambda = 3\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) put \(z = \mu \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">then \(y = 1 + \mu \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">and \(x = - 2\mu \) or equivalent <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Solve the equation \({z^3} = 8{\text{i}},{\text{ }}z \in \mathbb{C}\) giving your answers in the form \(z = r(\cos \theta + {\text{i}}\sin \theta )\) <strong>and </strong>in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Consider the complex numbers \({z_1} = 1 + {\text{i}}\) and \({z_2} = 2\left( {\cos \left( {\frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right)\).</p>
<p>(i) Write \({z_1}\) in the form \(r(\cos \theta + {\text{i}}\sin \theta )\).</p>
<p>(ii) Calculate \({z_1}{z_2}\) and write in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).</p>
<p>(iii) Hence find the value of \(\tan \frac{{5\pi }}{{12}}\) in the form \(c + d\sqrt 3 \), where \(c,{\text{ }}d \in \mathbb{Z}\).</p>
<p>(iv) Find the smallest value \(p > 0\) such that \({({z_2})^p}\) is a positive real number.</p>
<div class="marks">[11]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong>Note: </strong>Accept answers and working in degrees, throughout.</p>
<p> </p>
<p>\({z^3} = 8\left( {\cos \left( {\frac{\pi }{2} + 2\pi k} \right) + {\text{i}}\sin \left( {\frac{\pi }{2} + 2\pi k} \right)} \right)\) <strong><em>(A1)</em></strong></p>
<p>attempt the use of De Moivre’s Theorem in reverse <strong><em>M1</em></strong></p>
<p>\(z = 2\left( {\cos \left( {\frac{\pi }{6}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right);{\text{ }}2\left( {\cos \left( {\frac{{5\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{5\pi }}{6}} \right)} \right);\)</p>
<p>\(2\left( {\cos \left( {\frac{{9\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{9\pi }}{6}} \right)} \right)\) <strong><em>A2</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept cis form.</p>
<p> </p>
<p>\(z = \pm \sqrt 3 + {\text{i}},{\text{ }} - 2{\text{i}}\) <strong><em>A2</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for two correct solutions in each of the two lines above.</p>
<p><em><strong>[6 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>Note: </strong>Accept answers and working in degrees, throughout.</p>
<p> </p>
<p>(i) \({z_1} = \sqrt 2 \left( {\cos \left( {\frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{4}} \right)} \right)\) <strong><em>A1A1</em></strong></p>
<p>(ii) \(\left( {{z_2} = \left( {\sqrt 3 + {\text{i}}} \right)} \right)\)</p>
<p>\({z_1}{z_2} = (1 + {\text{i}})\left( {\sqrt 3 + {\text{i}}} \right)\) <strong><em>M1</em></strong></p>
<p>\( = \left( {\sqrt 3 - 1} \right) + {\text{i}}\left( {1 + \sqrt 3 } \right)\) <strong><em>A1</em></strong></p>
<p>(iii) \({z_1}{z_2} = 2\sqrt 2 \left( {\cos \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right)} \right)\) <strong><em>M1A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Interpret “hence” as “hence or otherwise”.</p>
<p> </p>
<p>\(\tan \frac{{5\pi }}{{12}} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\) <strong><em>A1</em></strong></p>
<p>\( = 2 + \sqrt 3 \) <strong><em>M1A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award final <strong><em>M1 </em></strong>for an attempt to rationalise the fraction.</p>
<p> </p>
<p>(iv) \({z_2}^p = {2^p}\left( {{\text{cis}}\left( {\frac{{p\pi }}{6}} \right)} \right)\) <strong><em>(M1)</em></strong></p>
<p>\({z_2}^p\) is a positive real number when \(p = 12\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Accept a solution based on part (a).</p>
<p><em><strong>[11 marks]</strong></em></p>
<p><em><strong>Total [17 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(z = x + {\text{i}}y\) be any non-zero complex number.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Express \(\frac{1}{z}\) in the form \(u + {\text{i}}v\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) If \(z + \frac{1}{z} = k\) , \(k \in \mathbb{R}\) , show that either <em>y</em> = 0 or \({x^2} + {y^2} = 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Show that if \({x^2} + {y^2} = 1\) then \(\left| k \right| \leqslant 2\) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(w = \cos \theta + {\text{i}}\sin \theta \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Show that \({w^n} + {w^{ - n}} = 2\cos n\theta \) , \(n \in \mathbb{Z}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Solve the equation \(3{w^2} - w + 2 - {w^{ - 1}} + 3{w^{ - 2}} = 0\), giving the roots in the form \(x + {\text{i}}y\) .</span></p>
<div class="marks">[14]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(\frac{1}{z} = \frac{1}{{x + {\text{i}}y}} \times \frac{{x - {\text{i}}y}}{{x - {\text{i}}y}} = \frac{x}{{{x^2} + {y^2}}} - {\text{i}}\frac{y}{{{x^2} + {y^2}}}\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(z + \frac{1}{z} = x + \frac{x}{{{x^2} + {y^2}}} + {\text{i}}\left( {y - \frac{y}{{{x^2} + {y^2}}}} \right) = k\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for <em>k</em> to be real, \(y - \frac{y}{{{x^2} + {y^2}}} = 0 \Rightarrow y({x^2} + {y^2} - 1) = 0\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence, \(y = 0{\text{ or }}{x^2} + {y^2} - 1 = 0 \Rightarrow {x^2} + {y^2} = 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) when \({x^2} + {y^2} = 1,{\text{ }}z + \frac{1}{z} = 2x\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| x \right| \leqslant 1\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \left| k \right| \leqslant 2\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \({w^{ - n}} = \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta ) = \cos n\theta - {\text{i}}\sin n\theta \) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {w^n} + {w^{ - n}} = (\cos n\theta + {\text{i}}\sin n\theta ) + (\cos n\theta - {\text{i}}\sin n\theta ) = 2\cos n\theta \) <strong><em>M1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) (rearranging)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(3({w^2} + {w^{ - 2}}) - (w + {w^{ - 1}}) + 2 = 0\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 3(2\cos 2\theta ) - 2\cos \theta + 2 = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 2(3\cos 2\theta - \cos \theta + 1) = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 3(2{\cos ^2}\theta - 1) - \cos \theta + 1 = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 6{\cos ^2}\theta - \cos \theta - 2 = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow (3\cos \theta - 2)(2\cos \theta + 1) = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore \cos \theta = \frac{2}{3},{\text{ }}\cos \theta = - \frac{1}{2}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos \theta = \frac{2}{3} \Rightarrow \sin \theta = \pm \frac{{\sqrt 5 }}{3}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos \theta = - \frac{1}{2} \Rightarrow \sin \theta = \pm \frac{{\sqrt 3 }}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore w = \frac{2}{3} \pm \frac{{{\text{i}}\sqrt 5 }}{3}, - \frac{1}{2} \pm \frac{{{\text{i}}\sqrt 3 }}{2}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow <strong><em>FT</em></strong> from incorrect \(\cos \theta \) and/or \(\sin \theta \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[14 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">A large number of candidates did not attempt part (a), or did so unsuccessfully.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">It was obvious that many candidates had been trained to answer questions of the type in part (b), and hence of those who attempted it, many did so successfully. Quite a few however failed to find all solutions.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p>Determine the roots of the equation \({(z + 2{\text{i}})^3} = 216{\text{i}}\), \(z \in \mathbb{C}\), giving the answers in the form \(z = a\sqrt 3 + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{Z}\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><strong>METHOD 1</strong></p>
<p>\(216{\text{i}} = 216\left( {\cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}} \right)\) <strong><em>A1</em></strong></p>
<p>\(z + 2{\text{i}} = \sqrt[3]{{216}}{\left( {\cos \left( {\frac{\pi }{2} + 2\pi k} \right) = {\text{i}}\sin \left( {\frac{\pi }{2} + 2\pi k} \right)} \right)^{\frac{1}{3}}}\) <strong><em>(M1)</em></strong></p>
<p>\(z + 2{\text{i}} = 6\left( {\cos \left( {\frac{\pi }{6} + \frac{{2\pi k}}{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6} + \frac{{2\pi k}}{3}} \right)} \right)\) <strong><em>A1</em></strong></p>
<p>\({z_1} + 2{\text{i}} = 6\left( {\cos \frac{\pi }{6} + {\text{i}}\sin \frac{\pi }{6}} \right) = 6\left( {\frac{{\sqrt 3 }}{2} + \frac{{\text{i}}}{2}} \right) = 3\sqrt 3 + 3{\text{i}}\)</p>
<p>\({z_2} + 2{\text{i}} = 6\left( {\cos \frac{{5\pi }}{6} + {\text{i}}\sin \frac{{5\pi }}{6}} \right) = 6\left( {\frac{{ - \sqrt 3 }}{2} + \frac{{\text{i}}}{2}} \right) = - 3\sqrt 3 + 3{\text{i}}\)</p>
<p>\({z_3} + 2{\text{i}} = 6\left( {\cos \frac{{3\pi }}{2} + {\text{i}}\sin \frac{{3\pi }}{2}} \right) = - 6{\text{i}}\) <strong><em>A2</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1A0 </em></strong>for one correct root.</p>
<p> </p>
<p>so roots are \({z_1} = 3\sqrt 3 + {\text{i, }}{z_2} = - 3\sqrt 3 + {\text{i}}\) and \({z_3} = - 8{\text{i}}\) <strong><em>M1A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>M1 </em></strong>for subtracting 2i from their three roots.</p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>\({\left( {a\sqrt 3 + (b + 2){\text{i}}} \right)^3} = 216{\text{i}}\)</p>
<p>\({\left( {a\sqrt 3 } \right)^3} + 3{\left( {a\sqrt 3 } \right)^2}(b + 2){\text{i}} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} - {\text{i}}{(b + 2)^3} = 216{\text{i}}\) <strong><em>M1A1</em></strong></p>
<p>\({\left( {a\sqrt 3 } \right)^3} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} + {\text{i}}\left( {3{{\left( {a\sqrt 3 } \right)}^2}(b + 2) - {{(b + 2)}^3}} \right) = 216{\text{i}}\)</p>
<p>\({\left( {a\sqrt 3 } \right)^3} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} = 0\) and \(3{\left( {a\sqrt 3 } \right)^2}(b + 2) - {(b + 2)^3} = 216\) <strong><em>M1A1</em></strong></p>
<p>\(a\left( {{a^2} - {{(b + 2)}^2}} \right) = 0\) and \(9{a^2}(b + 2) - {(b + 2)^3} = 216\)</p>
<p>\(a = 0\) or \({a^2} = {(b + 2)^2}\)</p>
<p>if \(a = 0,{\text{ }} - {(b + 2)^3} = 216 \Rightarrow b + 2 = - 6\)</p>
<p>\(\therefore b = - 8\) <strong><em>A1</em></strong></p>
<p>\((a,{\text{ }}b) = (0,{\text{ }} - 8)\)</p>
<p>if \({a^2} = {(b + 2)^2},{\text{ }}9{(b + 2)^2}(b + 2) - {(b + 2)^3} = 216\)</p>
<p>\(8{(b + 2)^3} = 216\)</p>
<p>\({(b + 2)^3} = 27\)</p>
<p>\(b + 2 = 3\)</p>
<p>\(b = 1\)</p>
<p>\(\therefore {a^2} = 9 \Rightarrow a = \pm 3\)</p>
<p>\(\therefore (a,{\text{ }}b) = ( \pm 3,{\text{ }}1)\) <strong><em>A1A1</em></strong></p>
<p>so roots are \({z_1} = 3\sqrt 3 + {\text{i, }}{z_2} = - 3\sqrt 3 + {\text{i}}\) and \({z_3} = - 8{\text{i}}\)</p>
<p> </p>
<p><strong>METHOD 3</strong></p>
<p>\({(z + 2{\text{i}})^3} - {( - 6{\text{i}})^3} = 0\)</p>
<p>attempt to factorise: <strong><em>M1</em></strong></p>
<p>\(\left( {(z + 2{\text{i}}) - ( - 6{\text{i}})} \right)\left( {{{(z + 2{\text{i}})}^2} + (z + 2{\text{i}})( - 6{\text{i}}) + {{( - 6{\text{i}})}^2}} \right) = 0\) <strong><em>A1</em></strong></p>
<p>\((z + 8{\text{i}})({z^2} - 2{\text{i}}z - 28) = 0\) <strong><em>A1</em></strong></p>
<p>\(z + 8{\text{i}} = 0 \Rightarrow z = - 8{\text{i}}\) <strong><em>A1</em></strong></p>
<p>\({z^2} - 2{\text{i}}z - 28 = 0 \Rightarrow z = \frac{{2{\text{i}} \pm \sqrt { - 4 - (4 \times 1 \times - 28)} }}{2}\) <strong><em>M1</em></strong></p>
<p>\(z = \frac{{2{\text{i}} \pm \sqrt {108} }}{2}\)</p>
<p>\(z = \frac{{2{\text{i}} \pm 6\sqrt 3 }}{2}\)</p>
<p>\(z = {\text{i}} \pm 3\sqrt 3 \) <strong><em>A1A1</em></strong></p>
<p> </p>
<p>Special Case:</p>
<p><strong>Note: </strong>If a candidate recognises that \(\sqrt[3]{{216{\text{i}}}} = - 6{\text{i}}\) (anywhere seen), and makes no valid progress in finding three roots, award <strong><em>A1 </em></strong>only.</p>
<p> </p>
<p><strong><em>[7 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The complex number <em>z</em> is defined as \(z = \cos \theta + {\text{i}}\sin \theta \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) State de Moivre’s theorem.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Show that \({z^n} - \frac{1}{{{z^n}}} = 2{\text{i}}\sin (n\theta )\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Use the binomial theorem to expand \({\left( {z - \frac{1}{z}} \right)^5}\) giving your answer in simplified form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) Hence show that \(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(e) Check that your result in part (d) is true for \(\theta = \frac{\pi }{4}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(f) Find \(\int_0^{\frac{\pi }{2}} {{{\sin }^5}\theta {\text{d}}\theta } \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(g) Hence, with reference to graphs of circular functions, find \(\int_0^{\frac{\pi }{2}} {{{\cos }^5}\theta {\text{d}}\theta } \) , explaining your reasoning.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) any appropriate form, <em>e.g.</em> \({(\cos \theta + {\text{i}}\sin \theta )^n} = \cos (n\theta) + {\text{i}}\sin (n\theta)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({z^n} = \cos n\theta + {\text{i}}\sin n\theta \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{{{z^n}}} = \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta )\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \cos n\theta - {\text{i}}\sin (n\theta )\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore \({z^n} - \frac{1}{{{z^n}}} = 2{\text{i}}\sin (n\theta )\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \({\left( {z - \frac{1}{z}} \right)^5} = {z^5} + \left( {\begin{array}{*{20}{c}}<br> 5 \\ <br> 1 <br>\end{array}} \right){z^4}\left( { - \frac{1}{z}} \right) + \left( {\begin{array}{*{20}{c}}<br> 5 \\ <br> 2 <br>\end{array}} \right){z^3}{\left( { - \frac{1}{z}} \right)^2} + \left( {\begin{array}{*{20}{c}}<br> 5 \\ <br> 3 <br>\end{array}} \right){z^2}{\left( { - \frac{1}{z}} \right)^3} + \left( {\begin{array}{*{20}{c}}<br> 5 \\ <br> 4 <br>\end{array}} \right)z{\left( { - \frac{1}{z}} \right)^4} + {\left( { - \frac{1}{z}} \right)^5}\) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {z^5} - 5{z^3} + 10z - \frac{{10}}{z} + \frac{5}{{{z^3}}} - \frac{1}{{{z^5}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) \({\left( {z - \frac{1}{z}} \right)^5} = {z^5} - \frac{1}{{{z^5}}} - 5\left( {{z^3} - \frac{1}{{{z^3}}}} \right) + 10\left( {z - \frac{1}{z}} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(2{\text{i}}\sin \theta )^5} = 2{\text{i}}\sin 5\theta - 10{\text{i}}\sin 3\theta + 20{\text{i}}\sin \theta \) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(e) \(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{LHS}} = 16{\left( {\sin \frac{\pi }{4}} \right)^5}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 16{\left( {\frac{{\sqrt 2 }}{2}} \right)^5}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2\sqrt 2 \,\,\,\,\,\left( { = \frac{4}{{\sqrt 2 }}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{RHS}} = \sin \left( {\frac{{5\pi }}{4}} \right) - 5\sin \left( {\frac{{3\pi }}{4}} \right) + 10\sin \left( {\frac{\pi }{4}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \frac{{\sqrt 2 }}{2} - 5\left( {\frac{{\sqrt 2 }}{2}} \right) + 10\left( {\frac{{\sqrt 2 }}{2}} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1</em></strong> for attempted substitution.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2\sqrt 2 \,\,\,\,\,\left( { = \frac{4}{{\sqrt 2 }}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence this is true for \(\theta = \frac{\pi }{4}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(f) \(\int_0^{\frac{\pi }{2}} {{{\sin }^5}\theta {\text{d}}\theta = \frac{1}{{16}}\int_0^{\frac{\pi }{2}} {(\sin 5\theta - 5\sin 3\theta + 10\sin \theta ){\text{d}}\theta } } \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{{16}}\left[ { - \frac{{\cos 5\theta }}{5} + \frac{{5\cos 3\theta }}{3} - 10\cos \theta } \right]_0^{\frac{\pi }{2}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{{16}}\left[ {0 - \left( { - \frac{1}{5} + \frac{5}{3} - 10} \right)} \right]\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{8}{{15}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(g) \(\int_0^{\frac{\pi }{2}} {{{\cos }^5}\theta {\text{d}}\theta = \frac{8}{{15}}} \) , with appropriate reference to symmetry and graphs. <strong><em>A1R1R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award first <strong><em>R1</em></strong> for partially correct reasoning <em>e.g.</em> sketches of graphs of sin and cos.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Award second <strong><em>R1</em></strong> for fully correct reasoning involving \({\sin ^5}\) and \({\cos ^5}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[3 marks]</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [22 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many students in b) substituted for the second term (again not making the connection to part a)) on the LHS and multiplied by the conjugate, which some managed well but it is inefficient. The binomial expansion was done well even if students did not do the earlier part. The connection between d) and f) was missed by many which lead to some creative attempts at the integral. Very few attempted the last part and of those many attempted another integral, ignoring the hence, while others related to the graph of sin and cos but not to the particular graphs here.</span></p>
</div>
<br><hr><br><div class="question">
<p>Three girls and four boys are seated randomly on a straight bench. Find the probability that the girls sit together and the boys sit together.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><strong>METHOD 1</strong></p>
<p>total number of arrangements 7! <strong><em>(A1)</em></strong></p>
<p>number of ways for girls and boys to sit together \( = 3! \times 4! \times 2\) <strong><em>(M1)(A1)</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>M1A0 </em></strong>if the 2 is missing.</p>
<p> </p>
<p>probability \(\frac{{3! \times 4! \times 2}}{{7!}}\) <strong><em>M1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>M1 </em></strong>for attempting to write as a probability.</p>
<p> </p>
<p>\(\frac{{2 \times 3 \times 4! \times 2}}{{7 \times 6 \times 5 \times 4!}}\)</p>
<p>\( = \frac{2}{{35}}\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>A0 </em></strong>if not fully simplified.</p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>\(\frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} + \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4}\) <strong><em>(M1)A1A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Accept \(\frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} \times 2\) or \(\frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times 2\).</p>
<p> </p>
<p>\( = \frac{2}{{35}}\) <strong><em>(M1)A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>A0 </em></strong>if not fully simplified.</p>
<p> </p>
<p><strong><em>[5 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p class="p1">The following system of equations represents three planes in space.</p>
<p class="p1">\[x + 3y + z = - 1\]</p>
<p class="p1">\[x + 2y - 2z = 15\]</p>
<p class="p1">\[2x + y - z = 6\]</p>
<p class="p1">Find the coordinates of the point of intersection of the three planes.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1"><strong>EITHER</strong></p>
<p class="p1">eliminating a variable, \(x\), for example to obtain \(y + 3z = - 16\)<span class="s1"> and \( - 5y - 3z = 8\) <span class="Apple-converted-space"> </span></span><strong><em>M1A1</em></strong></p>
<p class="p1">attempting to find the value of one variable <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">point of intersection is \(( - 1,{\text{ }}2,{\text{ }} - 6)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1A1A1</em></strong></span></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1">attempting row reduction of relevant matrix, <em>eg</em><span class="s1">. <img src="images/Schermafbeelding_2017-01-27_om_09.53.16.png" alt="M16/5/MATHL/HP1/ENG/TZ2/01_01"> <span class="Apple-converted-space"> </span></span><strong><em>M1</em></strong></p>
<p class="p1">correct matrix with two zeroes in a column, <em>eg</em><span class="s1">. <img src="images/Schermafbeelding_2017-01-27_om_09.54.14.png" alt="M16/5/MATHL/HP1/ENG/TZ2/01_02"> <span class="Apple-converted-space"> </span></span><strong><em>A1</em></strong></p>
<p class="p1">further attempt at reduction <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">point of intersection is \(( - 1,{\text{ }}2,{\text{ }} - 6)\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1A1A1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Allow solution expressed as \(x = - 1,{\text{ }}y = 2,{\text{ }}z = - 6\) for final <strong><em>A </em></strong>marks.</p>
<p class="p3"> </p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">This provided a generally easy start for many candidates. Most successful candidates obtained their answer through row reduction of a suitable matrix. Those choosing an alternative method often made slips in their algebra.</p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Write down the expansion of \({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3}\) in the form \(a + {\text{i}}b\) , where \(a\) and \(b\) </span><span style="font-family: times new roman,times; font-size: medium;">are in terms of \({\sin \theta }\) and \({\cos \theta }\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Hence show that \(\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Similarly show that \(\cos 5\theta = 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta \) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Hence</strong> solve the equation \(\cos 5\theta + \cos 3\theta + \cos \theta = 0\) , where \(\theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">By considering the solutions of the equation \(\cos 5\theta = 0\) , show that </span><span style="font-family: times new roman,times; font-size: medium;">\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \)</span><span style="font-family: times new roman,times; font-size: medium;"> and state the value of \(\cos \frac{{7\pi }}{{10}}\)</span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = {\cos ^3}\theta + 3{\cos ^2}\theta \left( {{\text{i}}\sin \theta } \right) + 3\cos \theta {\left( {{\text{i}}\sin \theta } \right)^2} + {\left( {{\text{i}}\sin \theta } \right)^3}\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {\cos ^3}\theta - 3\cos \theta {\sin ^2}\theta + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta - {{\sin }^3}\theta } \right)\) <em><strong>A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">from De Moivre’s theorem</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = \cos 3\theta + {\text{i}}\sin 3\theta \) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 3\theta + {\text{i}}\sin 3\theta = \left( {{{\cos }^3}\theta - 3\cos \theta {{\sin }^2}\theta } \right) + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta - {{\sin }^3}\theta } \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">equating real parts <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 3\theta = {\cos ^3}\theta - 3\cos \theta {\sin ^2}\theta \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {\cos ^3}\theta - 3\cos \theta \left( {1 - {{\cos }^2}\theta } \right)\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {\cos ^3}\theta - 3\cos \theta + 3{\cos ^3}\theta \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 4{\cos ^3}\theta - 3\cos \theta \) <em><strong>AG</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Do not award marks if part (a) is not used.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^5} = \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\cos ^5}\theta + 5{\cos ^4}\theta \left( {{\text{i}}\sin \theta } \right) + 10{\cos ^3}\theta {\left( {{\text{i}}\sin \theta } \right)^2} + 10{\cos ^2}\theta {\left( {{\text{i}}\sin \theta } \right)^3} + 5\cos \theta {\left( {{\text{i}}\sin \theta } \right)^4} + {\left( {{\text{i}}\sin \theta } \right)^5}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(A1)</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">from De Moivre’s theorem</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 5\theta = {\cos ^5}\theta - 10{\cos ^3}\theta {\sin ^2}\theta + 5\cos \theta {\sin ^4}\theta \) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {\cos ^5}\theta - 10{\cos ^3}\theta \left( {1 - {{\cos }^2}\theta } \right) + 5\cos \theta {\left( {1 - {{\cos }^2}\theta } \right)^2}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {\cos ^5}\theta - 10{\cos ^3}\theta + 10{\cos ^5}\theta + 5\cos \theta - 10{\cos ^3}\theta + 5{\cos ^5}\theta \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\therefore \cos 5\theta = 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta \) <em><strong>AG</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> If compound angles used in (b) and (c), then marks can be allocated in (c) only.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 5\theta + \cos 3\theta + \cos \theta \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \left( {16{{\cos }^5}\theta - 20{{\cos }^3}\theta + 5\cos \theta } \right) + \left( {4{{\cos }^3}\theta - 3\cos \theta } \right) + \cos \theta = 0\) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(16{\cos ^5}\theta - 16{\cos ^3}\theta + 3\cos \theta = 0\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \theta \left( {16{{\cos }^4}\theta - 16{{\cos }^2}\theta + 3} \right) = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \theta \left( {4{{\cos }^2}\theta - 3} \right)\left( {4{{\cos }^2}\theta - 1} \right) = 0\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\therefore \cos \theta = 0\)</span><span style="font-family: times new roman,times; font-size: medium;">; \( \pm \frac{{\sqrt 3 }}{2}\); \( \pm \frac{1}{2}\) <em><strong> A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\therefore \theta = \pm \frac{\pi }{6}\); \(\pm \frac{\pi }{3}\); \( \pm \frac{\pi }{2}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></strong></em></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 5\theta = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(5\theta = ...\frac{\pi }{2}\); \(\left( {\frac{{3\pi }}{2};\frac{{5\pi }}{2}} \right)\); \(\frac{{7\pi }}{2}\); \(...\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">(M1)</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\theta = ...\frac{\pi }{{10}}\); \(\left( {\frac{{3\pi }}{{10}};\frac{{5\pi }}{{10}}} \right)\); </span><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{7\pi }}{10}\); \(...\)</span> <em><strong><span style="font-family: times new roman,times; font-size: medium;">(M1)</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> These marks can be awarded for verifications later in the question.</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">now consider \(16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta = 0\) </span><em style="font-family: 'times new roman', times; font-size: medium;"><strong>M1</strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \theta \left( {16{{\cos }^4}\theta - 20{{\cos }^2}\theta + 5} \right) = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\cos ^2}\theta = \frac{{20 \pm \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}\)</span><span style="font-family: times new roman,times; font-size: medium;">; \(\cos \theta = 0\)</span><span style="font-family: times new roman,times; font-size: medium;"> </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \theta = \pm \sqrt {\frac{{20 \pm \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{20 + \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}} \) since max value of cosine \( \Rightarrow \) angle closest to zero</span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong><strong><span style="font-family: times new roman,times; font-size: medium;">R1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{4.5 + 4\sqrt {25 - 4\left( 5 \right)} }}{{4.8}}} = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos \frac{{7\pi }}{{10}} = - \sqrt {\frac{{5 - \sqrt 5 }}{8}} \) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A1A1</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></strong></em></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).</span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).</span></p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \(y = \frac{1}{{1 - x}}\), use mathematical induction to prove that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = \frac{{n!}}{{{{(1 - x)}^{n + 1}}}},{\text{ }}n \in {\mathbb{Z}^ + }\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">proposition is true for <em>n</em> = 1 since \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{{{(1 - x)}^2}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{1!}}{{{{(1 - x)}^2}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Must see the 1! for the </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;">.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">assume true for </span><em style="font-family: 'times new roman', times; font-size: medium;">n</em><span style="font-family: 'times new roman', times; font-size: medium;"> = </span><em style="font-family: 'times new roman', times; font-size: medium;">k</em><span style="font-family: 'times new roman', times; font-size: medium;">, \(k \in {\mathbb{Z}^ + }\), </span><em style="font-family: 'times new roman', times; font-size: medium;">i.e.</em><span style="font-family: 'times new roman', times; font-size: medium;"> \(\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = \frac{{k!}}{{{{(1 - x)}^{k + 1}}}}\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">consider \(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{{\text{d}}\left( {\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}}} \right)}}{{{\text{d}}x}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = (k + 1)k!{(1 - x)^{ - (k + 1) - 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{(k + 1)!}}{{{{(1 - x)}^{k + 2}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence, \({{\text{P}}_{k + 1}}\) is true whenever \({{\text{P}}_{k}}\) is true, and \({{\text{P}}_1}\) is true, and therefore the proposition is true for all positive integers <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> The final </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>R1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> is only available if at least 4 of the previous marks have been awarded.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[7 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates were awarded good marks for this question. A disappointing minority thought that the \((k + 1)\)th derivative was the \((k)\)th derivative multiplied by the first derivative. Providing an acceptable final statement remains a perennial issue. </span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">Consider the expansion of \({(1 + x)^n}\) </span>in ascending powers of \(x\), where \(n \geqslant 3\).</p>
</div>
<div class="specification">
<p class="p1">The coefficients of the second, third and fourth terms of the expansion are consecutive terms of an arithmetic sequence.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down the first four terms of the expansion.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Show that \({n^3} - 9{n^2} + 14n = 0\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Hence find the value of \(n\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(1,{\text{ }}nx,{\text{ }}\frac{{n(n - 1)}}{2}{x^2},{\text{ }}\frac{{n(n - 1)(n - 2)}}{6}{x^3}\) </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>A1 </em></strong>for the first two terms and <strong><em>A1 </em></strong>for the next two terms.</p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Accept \(\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right)\) notation.</p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Allow the terms seen in the context of an arithmetic sum.</p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Allow unsimplified terms, <em>eg</em><span class="s2">, those including powers of 1 </span>if seen.</p>
<p class="p3"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span><strong>EITHER</strong></p>
<p class="p1">using \({u_3} - {u_2} = {u_4} - {u_3}\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(\frac{{n(n - 1)}}{2} - n = \frac{{n(n - 1)(n - 2)}}{6} - \frac{{n(n - 1)}}{2}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">attempting to remove denominators and expanding (or vice versa) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="s2">\(3{n^2} - 9n = {n^3} - 6{n^2} + 5n\) </span>(or equivalent, <em>eg</em><span class="s2">, \(6{n^2} - 12n = {n^3} - 3{n^2} + 2n\)</span>) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong>OR</strong></p>
<p class="p1">using \({u_2} + {u_4} = 2{u_3}\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(n + \frac{{n(n - 1)(n - 2)}}{6} = n(n - 1)\) </span><strong><em>(A1)</em></strong></p>
<p class="p1">attempting to remove denominators and expanding (or vice versa) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="s3">\(6n + {n^3} - 3{n^2} + 2n = 6{n^2} - 6n\) </span>(or equivalent) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><strong>THEN</strong></p>
<p class="p2"><span class="Apple-converted-space">\({n^3} - 9{n^2} + 14n = 0\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1">(ii) <span class="Apple-converted-space"> \(n(n - 2)(n - 7) = 0\)</span> <span class="s2">or \((n - 2)(n - 7) = 0\) <span class="Apple-converted-space"> </span></span><strong><em>(A1)</em></strong></p>
<p class="p1">\(n = 7\) only (as \(n \geqslant 3\)) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was another question that was very well answered by most candidates.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was another question that was very well answered by most candidates. Some struggled in part (b) by attempting to find an expression for \(d\), a common difference, then substituting this in to further equations, where algebra tended to falter. The most fruitful technique was to apply \({u_3} - {u_2} = {u_4} - {u_3}\). Good presentation often helped candidates reach the final result. Correct factorisation was more often seen than not in the final section, though a small number thought it judicious to guess the correct answer(s) here.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>Chloe and Selena play a game where each have four cards showing capital letters A, B, C and D.<br>Chloe lays her cards face up on the table in order A, B, C, D as shown in the following diagram.</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2018-02-07_om_14.39.35.png" alt="N17/5/MATHL/HP1/ENG/TZ0/10"></p>
<p>Selena shuffles her cards and lays them face down on the table. She then turns them over one by one to see if her card matches with Chloe’s card directly above.<br>Chloe wins if <strong>no</strong> matches occur; otherwise Selena wins.</p>
</div>
<div class="specification">
<p>Chloe and Selena repeat their game so that they play a total of 50 times.<br>Suppose the discrete random variable <em>X </em>represents the number of times Chloe wins.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the probability that Chloe wins the game is \(\frac{3}{8}\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the mean of <em>X</em>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the variance of <em>X</em>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>number of possible “deals” \( = 4! = 24\) <strong><em>A1</em></strong></p>
<p>consider ways of achieving “no matches” (Chloe winning):</p>
<p>Selena could deal B, C, D (<em>ie</em>, 3 possibilities)</p>
<p>as her first card <strong><em>R1</em></strong></p>
<p>for each of these matches, there are only 3 possible combinations for the remaining 3 cards <strong><em>R1</em></strong></p>
<p>so no. ways achieving no matches \( = 3 \times 3 = 9\) <strong><em>M1A1</em></strong></p>
<p>so probability Chloe wins \( = \frac{9}{{23}} = \frac{3}{8}\) <strong><em>A1AG</em></strong></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>number of possible “deals” \( = 4! = 24\) <strong><em>A1</em></strong></p>
<p>consider ways of achieving a match (Selena winning)</p>
<p>Selena card A can match with Chloe card A<em>, </em>giving 6 possibilities for this happening <strong><em>R1</em></strong></p>
<p>if Selena deals B as her first card, there are only 3 possible combinations for the remaining 3 cards. Similarly for dealing C and dealing D <strong><em>R1</em></strong></p>
<p>so no. ways achieving one match is \( = 6 + 3 + 3 + 3 = 15\) <strong><em>M1A1</em></strong></p>
<p>so probability Chloe wins \( = 1 - \frac{{15}}{{24}} = \frac{3}{8}\) <strong><em>A1AG</em></strong></p>
<p> </p>
<p><strong>METHOD 3</strong></p>
<p>systematic attempt to find number of outcomes where Chloe wins (no matches)</p>
<p>(using tree diag. or otherwise) <strong><em>M1</em></strong></p>
<p>9 found <strong><em>A1</em></strong></p>
<p>each has probability \(\frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times 1\) <strong><em>M1</em></strong></p>
<p>\( = \frac{1}{{24}}\) <strong><em>A1</em></strong></p>
<p>their 9 multiplied by their \(\frac{1}{{24}}\) <strong><em>M1A1</em></strong></p>
<p>\( = \frac{3}{8}\) <strong><em>AG</em></strong></p>
<p> </p>
<p><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(X \sim {\text{B}}\left( {50,{\text{ }}\frac{3}{8}} \right)\) <strong><em>(M1)</em></strong></p>
<p>\(\mu = np = 50 \times \frac{3}{8} = \frac{{150}}{8}{\text{ }}\left( { = \frac{{75}}{4}} \right){\text{ }}( = 18.75)\) <strong><em>(M1)A1</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({\sigma ^2} = np(1 - p) = 50 \times \frac{3}{8} \times \frac{5}{8} = \frac{{750}}{{64}}{\text{ }}\left( { = \frac{{375}}{{32}}} \right){\text{ }}( = 11.7)\) <strong><em>(M1)A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the complex numbers \(u = 2 + 3{\text{i}}\) and \(v = 3 + 2{\text{i}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Given that \(\frac{1}{u} + \frac{{1}}{v} = \frac{{10}}{w}\), express <em>w </em>in the form \(a + b{\text{i, }}a,{\text{ }}b \in \mathbb{R}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find \(w\)* and express it in the form \(r{e^{{\text{i}}\theta }}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{{2 + 3{\text{i}}}} + \frac{1}{{3 + 2{\text{i}}}} = \frac{{2 - 3{\text{i}}}}{{4 + 9}} + \frac{{3 - 2{\text{i}}}}{{9 + 4}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{10}}{w} = \frac{{5 - 5{\text{i}}}}{{13}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(w = \frac{{130}}{{5 - 5{\text{i}}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{130 \times 5 \times (1 + {\text{i}})}}{{50}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(w = 13 + 13{\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{{2 + 3{\text{i}}}} + \frac{1}{{3 + 2{\text{i}}}} = \frac{{3 + 2{\text{i}} + 2 + 3{\text{i}}}}{{(2 + 3{\text{i}})(3 + 2{\text{i}})}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{10}}{w} = \frac{{5 + 5{\text{i}}}}{{13{\text{i}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{w}{{10}} = \frac{{13{\text{i}}}}{{5 + 5{\text{i}}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(w = \frac{{130{\text{i}}}}{{(5 + 5{\text{i}})}} \times \frac{{(5 - 5{\text{i}})}}{{(5 - 5{\text{i}})}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{650 + 650{\text{i}}}}{{50}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 13 + 13{\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <em>w</em>* \( = 13 - 13{\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \sqrt {338} {e^{ - \frac{\pi }{4}{\text{i}}}}{\text{ }}\left( { = 13\sqrt 2 {e^{ - \frac{\pi }{4}{\text{i}}}}} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept \(\theta = \frac{{7\pi }}{4}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> Do not accept answers for \(\theta \) given in degrees.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">An arithmetic sequence has first term <em>a</em> and common difference <em>d</em>, \(d \ne 0\) . The \({{\text{3}}^{{\text{rd}}}}\), \({{\text{4}}^{{\text{th}}}}\) and \({{\text{7}}^{{\text{th}}}}\) terms of the arithmetic sequence are the first three terms of a </span><span style="font-family: 'times new roman', times; font-size: medium;">geometric sequence.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \(a = - \frac{3}{2}d\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that the \({{\text{4}}^{{\text{th}}}}\) term of the geometric sequence is the \({\text{1}}{{\text{6}}^{{\text{th}}}}\) term of the arithmetic sequence.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let the first three terms of the geometric sequence be given by \({u_1}\) , \({u_1}r\) , \({u_1}{r^2}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore {u_1} = a + 2d\) , \({u_1}r = a + 3d\) and \({u_1}{r^2} = a + 6d\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{a + 6d}}{{a + 3d}} = \frac{{a + 3d}}{{a + 2d}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({a^2} + 8ad + 12{d^2} = {a^2} + 6ad + 9{d^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">2<em>a</em> + 3<em>d</em> = 0</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a = - \frac{3}{2}d\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]<br></em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({u_1} = \frac{d}{2}\) , \({u_1}r = \frac{{3d}}{2}\) , \(\left( {{u_1}{r^2} = \frac{{9d}}{2}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">r = 3 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">geometric \({{\text{4}}^{{\text{th}}}}\) term \({u_1}{r^3} = \frac{{27d}}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">arithmetic \({\text{1}}{{\text{6}}^{{\text{th}}}}\) term \(a + 15d = - \frac{3}{2}d + 15d\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{27d}}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept alternative methods.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p>Use the principle of mathematical induction to prove that</p>
<p>\(1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + n{\left( {\frac{1}{2}} \right)^{n - 1}} = 4 - \frac{{n + 2}}{{{2^{n - 1}}}}\), where \(n \in {\mathbb{Z}^ + }\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>if \(n = 1\)</p>
<p>\({\text{LHS}} = 1\,{\text{;}}\,\,{\text{RHS}} = 4 - \frac{3}{{{2^0}}} = 4 - 3 = 1\) <em> <strong>M1</strong></em></p>
<p>hence true for \(n = 1\)</p>
<p>assume true for \(n = k\) <em><strong>M1</strong></em></p>
<p><strong>Note:</strong> Assumption of truth must be present. Following marks are not dependent on the first two <em><strong>M1</strong> </em>marks.</p>
<p>so \(1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + k{\left( {\frac{1}{2}} \right)^{k - 1}} = 4 - \frac{{k + 2}}{{{2^{k - 1}}}}\)</p>
<p>if \(n = k + 1\)</p>
<p>\(1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + k{\left( {\frac{1}{2}} \right)^{k - 1}} + \left( {k + 1} \right){\left( {\frac{1}{2}} \right)^k}\)</p>
<p>\( = 4 - \frac{{k + 2}}{{{2^{k - 1}}}} + \left( {k + 1} \right){\left( {\frac{1}{2}} \right)^k}\) <em><strong>M1A1</strong></em></p>
<p>finding a common denominator for the two fractions <em><strong>M1</strong></em></p>
<p>\( = 4 - \frac{{2\left( {k + 2} \right)}}{{{2^k}}} + \frac{{k + 1}}{{{2^k}}}\)</p>
<p>\( = 4 - \frac{{2\left( {k + 2} \right) - \left( {k + 1} \right)}}{{{2^k}}} = 4 - \frac{{k + 3}}{{{2^k}}}\left( { = 4 - \frac{{\left( {k + 1} \right) + 2}}{{{2^{\left( {k + 1} \right) - 1}}}}} \right)\) <em><strong>A1</strong></em></p>
<p>hence if true for \(n = k\) then also true for \(n = k + 1\), as true for \(n = 1\), so true (for all \(n \in {\mathbb{Z}^ + }\)) <em><strong>R1</strong></em></p>
<p><strong>Note:</strong> Award the final <em><strong>R1</strong> </em>only if the first four marks have been awarded.</p>
<p><em><strong>[7 marks]</strong></em></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Expand \({(2 - 3x)^5}\) in ascending powers of <em>x</em>, simplifying coefficients.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">clear attempt at binomial expansion for exponent 5 <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^5} + 5 \times {2^4} \times ( - 3x) + \frac{{5 \times 4}}{2} \times {2^3} \times {( - 3x)^2} + \frac{{5 \times 4 \times 3}}{6} \times {2^2} \times {( - 3x)^3}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( + \frac{{5 \times 4 \times 3 \times 2}}{{24}} \times 2 \times {( - 3x)^4} + {( - 3x)^5}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Only award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> if binomial coefficients are seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 32 - 240x + 720{x^2} - 1080{x^3} + 810{x^4} - 243{x^5}\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A2</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for correct moduli of coefficients and powers. </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for correct signs.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>Total [4 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Generally well done. The majority of candidates obtained a quintic with correct alternating signs. A few candidates made arithmetic errors. A small number of candidates multiplied out the linear expression, often correctly.</span></p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>The random variable \(X\) has the Poisson distribution \({\text{Po}}(m)\). Given that \({\text{P}}(X > 0) = \frac{3}{4}\), find the value of \(m\) in the form \(\ln a\) where \(a\) is an integer.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>The random variable \(Y\) has the Poisson distribution \({\text{Po}}(2m)\). Find \({\text{P}}(Y > 1)\) in the form \(\frac{{b - \ln c}}{c}\) where \(b\) and \(c\) are integers.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\({\text{P(}}X > 0) = 1 - {\text{P(}}X = 0)\) <strong><em>(M1)</em></strong></p>
<p>\( \Rightarrow 1 - {{\text{e}}^{ - m}} = \frac{3}{4}\) or equivalent <strong><em>A1</em></strong></p>
<p>\( \Rightarrow m = \ln 4\) <strong><em>A1</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({\text{P}}(Y > 1) = 1 - {\text{P}}(Y = 0) - {\text{P}}(Y = 1)\) <strong><em>(M1)</em></strong></p>
<p>\( = 1 - {{\text{e}}^{ - 2\ln 4}} - {{\text{e}}^{ - 2\ln 4}} \times 2\ln 4\) <strong><em>A1</em></strong></p>
<p>recognition that \(2\ln 4 = \ln 16\) <strong><em>(A1)</em></strong></p>
<p>\({\text{P}}(Y > 1) = \frac{{15 - \ln 16}}{{16}}\) <strong><em>A1</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">Use mathematical induction to prove that \((2n)! \ge {2^n}{(n!)^2},{\text{ }}n \in {\mathbb{Z}^ + }\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1">let \({\text{P}}(n)\) be the proposition that \((2n)! \ge {2^n}{(n!)^2},{\text{ }}n \in {\mathbb{Z}^ + }\)</p>
<p class="p1">consider \({\text{P}}(1)\):</p>
<p class="p1">\(2! = 2\) and \({2^1}{(1!)^2} = 2\) so \({\text{P}}(1)\) is true <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">assume \({\text{P}}(k)\) is true <em>ie</em> \((2k)! \ge {2^k}{(k!)^2},{\text{ }}n \in {\mathbb{Z}^ + }\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Do not award <strong><em>M1 </em></strong>for statements such as “let \(n = k\)”.</p>
<p class="p2"> </p>
<p class="p1">consider \({\text{P}}(k + 1)\):</p>
<p class="p1">\(\left( {2(k + 1)} \right)! = (2k + 2)(2k + 1)(2k)!\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\(\left( {2(k + 1)} \right)! \ge (2k + 2)(2k + 1){(k!)^2}{2^k}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( = 2(k + 1)(2k + 1){(k!)^2}{2^k}\)</p>
<p class="p1">\( > {2^{k + 1}}(k + 1)(k + 1){(k!)^2}\;\;\;{\text{since}}\;\;\;2k + 1 > k + 1\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">\( = {2^{k + 1}}{\left( {(k + 1)!} \right)^2}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\({\text{P}}(k + 1)\) is true whenever \({\text{P}}(k)\) is true and \({\text{P}}(1)\) is true, so \({\text{P}}(n)\) is true for \(n \in {\mathbb{Z}^ + }\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>To obtain the final <strong><em>R1</em></strong>, four of the previous marks must have been awarded.</p>
<p class="p2"> </p>
<p class="p3"><strong><em>[7 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">An easy question, but many candidates exhibited discomfort and poor reasoning abilities. The difficulty for most was that the proposition was expressed in terms of an inequality. Hopefully, as most publishers of IB textbooks have realised, inequalities in such questions are within the syllabus.</p>
</div>
<br><hr><br><div class="question">
<p>Prove by mathematical induction that \(\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right)\), where \(n \in \mathbb{Z},n \geqslant 3\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>\(\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right)\)</p>
<p>show true for \(n = 3\) <strong><em>(M1)</em></strong></p>
<p>\({\text{LHS}} = \left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) = 1\) \(\,\,\,\) \({\text{RHS}} = \left( {\begin{array}{*{20}{c}} 3 \\ 3 \end{array}} \right) = 1\) <strong><em>A1</em></strong></p>
<p>hence true for \(n = 3\)</p>
<p>assume true for \(n = k:\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {k - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} k \\ 3 \end{array}} \right)\) <strong><em>M1</em></strong></p>
<p>consider for \(n = k + 1:\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {k - 1} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right)\) <strong><em>(M1)</em></strong></p>
<p>\( = \left( {\begin{array}{*{20}{c}} k \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right)\) <strong><em>A1</em></strong></p>
<p>\( = \frac{{k!}}{{(k - 3)!3!}} + \frac{{k!}}{{(k - 2)!2!}}\,\,\,\left( { = \frac{{k!}}{{3!}}\left[ {\frac{1}{{(k - 3)!}} + \frac{3}{{(k - 2)!}}} \right]} \right)\) or any correct expression with a visible common factor <strong><em>(A1)</em></strong></p>
<p>\( = \frac{{k!}}{{3!}}\left[ {\frac{{k - 2 + 3}}{{(k - 2)!}}} \right]\) or any correct expression with a common denominator <strong><em>(A1)</em></strong></p>
<p>\( = \frac{{k!}}{{3!}}\left[ {\frac{{k + 1}}{{(k - 2)!}}} \right]\)</p>
<p> </p>
<p><strong>Note:</strong> At least one of the above three lines or equivalent must be seen.</p>
<p> </p>
<p>\( = \frac{{(k + 1)!}}{{3!(k - 2)!}}\) or equivalent <strong><em>A1</em></strong></p>
<p>\( = \left( {\begin{array}{*{20}{c}} {k + 1} \\ 3 \end{array}} \right)\)</p>
<p>Result is true for \(k = 3\). If result is true for \(k\) it is true for \(k + 1\). Hence result is true for all \(k \geqslant 3\). Hence proved by induction. <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> In order to award the <strong><em>R1 </em></strong>at least <strong><em>[5 marks] </em></strong>must have been awarded.</p>
<p> </p>
<p><strong><em>[9 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p>Find the coefficient of \({x^8}\) in the expansion of \({\left( {{x^2} - \frac{2}{x}} \right)^7}\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>each term is of the form \(\left( {\begin{array}{*{20}{c}} 7 \\ r \end{array}} \right){({x^2})^{7 - r}}{\left( {\frac{{ - 2}}{x}} \right)^\prime }\) <strong><em>(M1)</em></strong></p>
<p>\( = \left( {\begin{array}{*{20}{c}} 7 \\ r \end{array}} \right){x^{14 - 2r}}{( - 2)^r}{x^{ - r}}\)</p>
<p>so \(14 - 3r = 8\) <strong><em>(A1)</em></strong></p>
<p>\(r = 2\)</p>
<p>so require \(\left( {\begin{array}{*{20}{c}} 7 \\ 2 \end{array}} \right){({x^2})^{}}^5{\left( {\frac{{ - 2}}{x}} \right)^2}\) (or simply \(\left( {\begin{array}{*{20}{c}} 7 \\ 2 \end{array}} \right){( - 2)^2}\)) <strong><em>A1</em></strong></p>
<p>\( = 21 \times 4\)</p>
<p>\( = 84\) <em><strong>A1</strong></em></p>
<p> </p>
<p><strong>Note:</strong> Candidates who attempt a full expansion, including the correct term, may only be awarded <strong><em>M1A0A0A0</em></strong>.</p>
<p> </p>
<p><strong><em>[4 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Solve the equation \({4^{x - 1}} = {2^x} + 8\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^{2x - 2}} = {2^x} + 8\) <strong><em>(M1)</em></strong> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{4}{2^{2x}} = {2^x} + 8\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^{2x}} - 4 \times {2^x} - 32 = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(({2^x} - 8)({2^x} + 4) = 0\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^x} = 8 \Rightarrow x = 3\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Notes:</strong> Do not award final <strong><em>A1</em></strong> if more than 1 solution is given.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Very few candidates knew how to solve this equation. A significant number guessed the answer using trial and error after failed attempts to solve it. A number of misconceptions were identified involving properties of logarithms and exponentials.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">A geometric sequence \(\left\{ {{u_n}} \right\}\), with complex terms, is defined by \({u_{n + 1}} = (1 + {\text{i}}){u_n}\) and \({u_1} = 3\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Find the fourth term of the sequence, giving your answer in the form \(x + y{\text{i, }}x,{\text{ }}y \in \mathbb{R}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find the sum of the first 20 terms of \(\left\{ {{u_n}} \right\}\), giving your answer in the form \(a \times (1 + {2^m})\) where \(a \in \mathbb{C}\) and \(m \in \mathbb{Z}\) are to be determined.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">A second sequence \(\left\{ {{v_n}} \right\}\) is defined by \({v_n} = {u_n}{u_{n + k}},{\text{ }}k \in \mathbb{N}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) Show that \(\left\{ {{v_n}} \right\}\) is a geometric sequence.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) State the first term.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> (iii) Show that the common ratio is independent of <em>k</em>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">A third sequence \(\left\{ {{w_n}} \right\}\) is defined by \({w_n} = \left| {{u_n} - {u_{n + 1}}} \right|\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(d) (i) Show that \(\left\{ {{w_n}} \right\}\) is a geometric sequence.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) State the geometrical significance of this result with reference to points on the complex plane.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(r = 1 + {\text{i}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({u_4} = 3{(1 + {\text{i}})^3}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - 6 + 6{\text{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({S_{20}} = \frac{{\left( {{{(1 + {\text{i}})}^{20}} - 1} \right)}}{{\text{i}}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{3\left( {{{(2i{\text{)}}}^{10}} - 1} \right)}}{{\text{i}}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Only one of the two <strong><em>M1</em></strong>s can be implied. Other algebraic methods may be seen.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{3\left( { - {2^{10}} - 1} \right)}}{{\text{i}}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 3{\text{i}}\left( {{2^{10}} + 1} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({v_n} = \left( {3{{(1 + {\text{i}})}^{n - 1}}} \right)\left( {3{{(1 + {\text{i}})}^{n - 1 + k}}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(9{(1 + {\text{i}})^k}{(1 + i)^{2n - 2}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 9{(1 + {\text{i}})^k}{\left( {{{(1 + i)}^2}} \right)^{n - 1}}\left( { = 9{{(1 + {\text{i}})}^k}{{(2{\text{i}})}^{n - 1}}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">this is the general term of a geometrical sequence <strong><em>R1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Notes:</strong> Do not accept the statement that the product of terms in a geometric sequence is also geometric unless justified further.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> If the final expression for \({v_n}\) is \(9{(1 + {\text{i}})^k}{(1 + i)^{2n - 2}}\) award <strong><em>M1A1R0</em></strong>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{u_{n + 1}}{u_{n + k + 1}}}}{{{u_n}{u_{n + k}}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = (1 + {\text{i}})(1 + {\text{i}})\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">this is a constant, hence sequence is geometric <strong><em>R1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not allow methods that do not consider the general term.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(9{(1 + {\text{i}})^k}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) common ratio is \({(1 + i)^2}{\text{ }}( = 2i)\) (which is independent of <em>k</em>) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) (i) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({w_n}\left| {3{{(1 + i)}^{n - 1}} - 3{{(1 + {\text{i}})}^n}} \right|\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 3{\left| {1 + i} \right|^{n - 1}}\left| {1 - (1 + {\text{i)}}} \right|\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 3{\left| {1 + i} \right|^{n - 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( { = 3{{\left( {\sqrt 2 } \right)}^{n - 1}}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">this is the general term for a geometric sequence <strong><em>R1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({w_n} = \left| {{u_n} - (1 + {\text{i}}){u_n}} \right|\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left| {{u_n}} \right|\left| { - {\text{i}}} \right|\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left| {{u_n}} \right|\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left| {3{{(1 + {\text{i}})}^{n - 1}}} \right|\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 3{\left| {(1 + {\text{i}})} \right|^{n - 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( { = 3{{\left( {\sqrt 2 } \right)}^{n - 1}}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">this is the general term for a geometric sequence <strong><em>R1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not allow methods that do not consider the general term.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) distance between successive points representing \({u_n}\) in the complex plane forms a geometric sequence <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Various possibilities but must mention distance between successive points.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [17 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">On Saturday, Alfred and Beatrice play 6 different games against each other. In each game, one of the two wins. The probability that Alfred wins any one of these games is \(\frac{2}{3}\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that the probability that Alfred wins exactly 4 of the games is \(\frac{{80}}{{243}}\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Explain why the total number of possible outcomes for the results of the 6 games is 64.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) By expanding \({(1 + x)^6}\) and choosing a suitable value for <em>x</em>, prove</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[64 = \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 3 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 5 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right)\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) State the meaning of this equality in the context of the 6 games played.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The following day Alfred and Beatrice play the 6 games again. Assume that the probability that Alfred wins any one of these games is still \(\frac{2}{3}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find an expression for the probability Alfred wins 4 games on the first day and 2 on the second day. Give your answer in the form \({\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> r <br>\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^s}{\left( {\frac{1}{3}} \right)^t}\) where the values of <em>r</em>, <em>s</em> and <em>t</em> are to be found.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Using your answer to (c) (i) and 6 similar expressions write down the probability that Alfred wins a total of 6 games over the two days as the sum of 7 probabilities.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Hence prove that \(\left( {\begin{array}{*{20}{c}}<br> {12} \\ <br> 6 <br>\end{array}} \right) = {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 3 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 5 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right)^2}\).</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Alfred and Beatrice play <em>n</em> games. Let <em>A</em> denote the number of games Alfred wins. The expected value of <em>A</em> can be written as \({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}<br> n \\ <br> r <br>\end{array}} \right)} \frac{{{a^r}}}{{{b^n}}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find the values of <em>a</em> and <em>b</em>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) By differentiating the expansion of \({(1 + x)^n}\), prove that the expected number of games Alfred wins is \(\frac{{2n}}{3}\).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(B\left( {6,\frac{2}{3}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(p(4) = \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right){\left( {\frac{2}{3}} \right)^4}{\left( {\frac{1}{3}} \right)^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right) = 15\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 15 \times \frac{{{2^4}}}{{{3^6}}} = \frac{{80}}{{243}}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) 2 outcomes for each of the 6 games or \({2^6} = 64\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({(1 + x)^6} = \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right)x + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right){x^2} + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 3 <br>\end{array}} \right){x^3} + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right){x^4} + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 5 <br>\end{array}} \right){x^5} + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right){x^6}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept \(^n{C_r}\) notation or \(1 + 6x + 15{x^2} + 20{x^3} + 15{x^4} + 6{x^5} + {x^6}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica; min-height: 30.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">setting <em>x</em> = 1 in both sides of the expression <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not award <strong><em>R1</em></strong> if the right hand side is not in the correct form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\(64 = \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 3 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 5 <br>\end{array}} \right) + \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right)\)</span> <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) the total number of outcomes = number of ways Alfred can win no games, plus the number of ways he can win one game <em>etc.</em> <strong><em>R1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Let \({\text{P}}(x,{\text{ }}y)\) be the probability that Alfred wins <em>x</em> games on the first day and <em>y</em> on the second.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{P(4, 2)}} = \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^4} \times {\left( {\frac{1}{3}} \right)^2} \times \left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^2} \times {\left( {\frac{1}{3}} \right)^4}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) or </span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\)</span> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>r</em> = 2 or 4, <em>s</em> = <em>t</em> = 6</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) P(Total = 6) =</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">P(0, 6) + P(1, 5) + P(2, 4) + P(3, 3) + P(4, 2) + P(5, 1) + P(6, 0) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6} + ... + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right)^2}{\left( {\frac{2}{3}} \right)^6}{\left( {\frac{1}{3}} \right)^6}\) <strong><em>A2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{{2^6}}}{{{3^{12}}}}\left( {{{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 3 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 5 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right)}^2}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept any valid sum of 7 probabilities.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) use of \(\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> i <br>\end{array}} \right) = \left( {\begin{array}{*{20}{l}}<br> 6 \\ <br> {6 - i} <br>\end{array}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(can be used either here or in (c)(ii))</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">P(wins 6 out of 12) \( = \left( {\begin{array}{*{20}{c}}<br> {12} \\ <br> 6 <br>\end{array}} \right) \times {\left( {\frac{2}{3}} \right)^6} \times {\left( {\frac{1}{3}} \right)^6} = \frac{{{2^6}}}{{{3^{12}}}}\left( {\begin{array}{*{20}{c}}<br> {12} \\ <br> 6 <br>\end{array}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{{2^6}}}{{{3^{12}}}}\left( {{{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 3 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 5 <br>\end{array}} \right)}^2} + {{\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right)}^2}} \right) = \frac{{{2^6}}}{{{3^{12}}}}\left( {\begin{array}{*{20}{c}}<br> {12} \\ <br> 6 <br>\end{array}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore \({\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 0 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 1 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 2 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 3 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 4 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 5 <br>\end{array}} \right)^2} + {\left( {\begin{array}{*{20}{c}}<br> 6 \\ <br> 6 <br>\end{array}} \right)^2} = \left( {\begin{array}{*{20}{c}}<br> {12} \\ <br> 6 <br>\end{array}} \right)\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[9 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \({\text{E}}(A) = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}<br> n \\ <br> r <br>\end{array}} \right)} {\left( {\frac{2}{3}} \right)^r}{\left( {\frac{1}{3}} \right)^{n - r}} = \sum\limits_{r = 0}^n {r\left( {\begin{array}{*{20}{c}}<br> n \\ <br> r <br>\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(<em>a</em> = 2, <em>b</em> = 3) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> <strong><em>M0A0</em></strong> for <em>a</em> = 2, <em>b</em> = 3 without any method.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica; min-height: 29.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(n{(1 + x)^{n - 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}<br> n \\ <br> r <br>\end{array}} \right)} r{x^{r - 1}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(sigma notation not necessary)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(if sigma notation used also allow lower limit to be <em>r</em> = 0)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let <em>x</em> = 2 <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(n{3^{n - 1}} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}<br> n \\ <br> r <br>\end{array}} \right)} r{2^{r - 1}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">multiply by 2 and divide by \({3^n}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{2n}}{3} = \sum\limits_{r = 1}^n {\left( {\begin{array}{*{20}{c}}<br> n \\ <br> r <br>\end{array}} \right)} r\frac{{{2^r}}}{{{3^n}}}\left( { = \sum\limits_{r = 0}^n {\left( {\begin{array}{*{20}{c}}<br> n \\ <br> r <br>\end{array}} \right)} \frac{{{2^r}}}{{{3^n}}}} \right)\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[6 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(a) Candidates need to be aware how to work out binomial coefficients without a calculator</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(b) (ii) A surprising number of candidates chose to work out the values of all the binomial coefficients (or use Pascal’s triangle) to make a total of 64 rather than simply putting 1 into the left hand side of the expression.<br></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question linked the binomial distribution with binomial expansion and coefficients and was generally well done.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(d) This was poorly done. Candidates were not able to manipulate expressions given using sigma notation.<br></span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The function <em>f</em> is defined by \(f(x) = {{\text{e}}^x}\sin x\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \(f''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Obtain a similar expression for \({f^{(4)}}(x)\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Suggest an expression for \({f^{(2n)}}(x)\), \(n \in {\mathbb{Z}^ + }\), and prove your conjecture using mathematical induction.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f''(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2{{\text{e}}^x}\cos x\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f'''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({f^{(4)}}(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) + 2{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right) - 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 4{{\text{e}}^x}\cos \left( {x + \frac{\pi }{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 4{{\text{e}}^x}\sin (x + \pi )\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the conjecture is that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({f^{(2n)}}(x) = {2^n}{{\text{e}}^x}\sin \left( {x + \frac{{n\pi }}{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for <em>n</em> = 1, this formula gives</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f''(x) = 2{{\text{e}}^x}\sin \left( {x + \frac{\pi }{2}} \right)\) which is correct <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let the result be true for <em>n</em> = <em>k</em> , \(\left( {i.e.{\text{ }}{f^{(2k)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">consider \({f^{(2k + 1)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({f^{\left( {2(k + 1)} \right)}}(x) = {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) + {2^k}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right) - {2^k}{{\text{e}}^x}\sin \left( {x + \frac{{k\pi }}{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^{k + 1}}{{\text{e}}^x}\cos \left( {x + \frac{{k\pi }}{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^{k + 1}}{{\text{e}}^x}\sin \left( {x + \frac{{(k + 1)\pi }}{2}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the result is proved by induction. <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award the final </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>R1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> only if the two </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> marks have been awarded.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[8 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Expand and simplify \({\left( {\frac{x}{y} - \frac{y}{x}} \right)^4}\)<span style="font: 7.0px Helvetica;">.</span></span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\left( {\frac{x}{y} - \frac{y}{x}} \right)^4} = {\left( {\frac{x}{y}} \right)^4} + 4{\left( {\frac{x}{y}} \right)^3}\left( { - \frac{y}{x}} \right) + 6{\left( {\frac{x}{y}} \right)^2}{\left( { - \frac{y}{x}} \right)^2} + 4\left( {\frac{x}{y}} \right){\left( { - \frac{y}{x}} \right)^3} + {\left( { - \frac{y}{x}} \right)^4}\) <strong> <em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">for attempt to expand and </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">for correct unsimplified expansion.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{{x^4}}}{{{y^4}}} - 4\frac{{{x^2}}}{{{y^2}}} + 6 - 4\frac{{{y^2}}}{{{x^2}}} + \frac{{{y^4}}}{{{x^4}}}\,\,\,\,\,\left( { = \frac{{{x^8} - 4{x^6}{y^2} + 6{x^4}{y^4} - 4{x^2}{y^6} + {y^8}}}{{{x^4}{y^4}}}} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">for powers, </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">for coefficients and signs.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Final two <strong><em>A </em></strong>marks are independent of first <strong><em>A </em></strong>mark.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">This was generally very well answered. Those who failed to gain full marks often made minor sign slips. A surprising number obtained the correct simplified expression, but continued to rearrange their expressions, often doing so incorrectly. Fortunately, there were no penalties for doing so.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"> </p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">If <em>w</em> = 2 + 2i , find the modulus and argument of <em>w</em>.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given \(z = \cos \left( {\frac{{5\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{5\pi }}{6}} \right)\), find in its simplest form \({w^4}{z^6}\).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">modulus \( = \sqrt 8 \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">argument \( = \frac{\pi }{4}\) (accept 45°) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> <strong><em>A0</em></strong> if extra values given.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({w^4}{z^6} - 64{e^{\pi {\text{i}}}} \times {e^{5\pi {\text{i}}}}\) <strong><em>(A1)(A1)</em></strong></span><span style="font-family: 'Helvetica Neue', Arial, 'Lucida Grande', 'Lucida Sans Unicode', sans-serif; font-size: 27px;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Allow alternative notation.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 64{e^{6\pi {\text{i}}}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 64\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({w^4} = - 64\) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^6} = - 1\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({w^4}{z^6} = 64\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Those who tackled this question were generally very successful. A few, with varying success, tried to work out the powers of the complex numbers by multiplying the Cartesian form rather than using de Moivre’s Theorem.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Those who tackled this question were generally very successful. A few, with varying success, tried to work out the powers of the complex numbers by multiplying the Cartesian form rather than using de Moivre’s Theorem.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given the complex numbers \({z_1} = 1 + 3{\text{i}}\) and \({z_2} = - 1 - {\text{i}}\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Write down the exact values of \(\left| {{z_1}} \right|\) and \(\arg ({z_2})\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the minimum value of \(\left| {{z_1} + \alpha{z_2}} \right|\), where \(\alpha \in \mathbb{R}\).</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {{z_1}} \right| = \sqrt {10} ;{\text{ }}\arg ({z_2}) = - \frac{{3\pi }}{4}{\text{ }}\left( {{\text{accept }}\frac{{5\pi }}{4}} \right)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| {{z_1} + \alpha{z_2}} \right| = \sqrt {{{(1 - \alpha )}^2} + {{(3 - \alpha )}^2}} \) or the squared modulus <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to minimise \(2{\alpha ^2} - 8\alpha + 10\) or their quadratic or its half or its square root <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">obtain \(\alpha = 2\) at minimum <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">state \(\sqrt 2 \) as final answer <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Disappointingly, few candidates obtained the correct argument for the second complex number, mechanically using arctan(1) but not thinking about the position of the number in the complex plane.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates obtained the correct quadratic or its square root, but few knew how to set about minimising it.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(w = \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that <em>w</em> is a root of the equation \({z^5} - 1 = 0\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Show that \((w - 1)({w^4} + {w^3} + {w^2} + w + 1) = {w^5} - 1\) and deduce that \({w^4} + {w^3} + {w^2} + w + 1 = 0\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) <strong>Hence</strong> show that \(\cos \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} = - \frac{1}{2}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) <strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({w^5} = {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^5}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \cos 2\pi + {\text{i}}\sin 2\pi \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence <em>w</em> is a root of \({z^5} - 1 = 0\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Solving \({z^5} = 1\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = \cos \frac{{2\pi }}{5}n + {\text{i}}\sin \frac{{2\pi }}{5}n{\text{ ,}}\,\,\,\,\,n = {\text{0, 1, 2, 3, 4}}\). <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(n = 1{\text{ gives }}\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}\) which is <em>w</em> <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \((w - 1)(1 + w + {w^2} + {w^3} + {w^4}) = w + {w^2} + {w^3} + {w^4} + {w^5} - 1 - w - {w^2} - {w^3} - {w^4}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {w^5} - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Since \({w^5} - 1 = 0\) <strong>and</strong> \(w \ne 1{\text{ , }}{w^4} + {w^3} + {w^2} + w + 1 = 0\). <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \(1 + w + {w^2} + {w^3} + {w^4} = \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(1 + \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5} + {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^2} + {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^3} + {\left( {\cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5}} \right)^4}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1 + \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} + {\text{i}}\sin \frac{{4\pi }}{5} + \cos \frac{{6\pi }}{5} + {\text{i}}\sin \frac{{6\pi }}{5} + \cos \frac{{8\pi }}{5} + {\text{i}}\sin \frac{{8\pi }}{5}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1 + \cos \frac{{2\pi }}{5} + {\text{i}}\sin \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} + {\text{i}}\sin \frac{{4\pi }}{5} + \cos \frac{{4\pi }}{5} - {\text{i}}\sin \frac{{4\pi }}{5} + \cos \frac{{2\pi }}{5} - {\text{i}}\sin \frac{{2\pi }}{5}\) <strong><em>M1A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1</em></strong> for attempting to replace \({6\pi }\) and \({8\pi }\) by \({4\pi }\) and \({2\pi }\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Award <strong><em>A1</em></strong> for correct cosine terms and <strong><em>A1</em></strong> for correct sine terms.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1 + 2\cos \frac{{4\pi }}{5} + 2\cos \frac{{2\pi }}{5} = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Correct methods involving equating real parts, use of conjugates or reciprocals are also accepted.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\cos \frac{{2\pi }}{5} + \cos \frac{{4\pi }}{5} = - \frac{1}{2}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Use of cis notation is acceptable throughout this question.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [12 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Parts (a) and (b) were generally well done, although very few stated that \(w \ne 1\) in (b). Part (c), the last question on the paper was challenging. Those candidates who gained some credit correctly focussed on the real part of the identity and realise that different cosine were related.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The mean of the first ten terms of an arithmetic sequence is 6. The mean of the first twenty terms of the arithmetic sequence is 16. Find the value of the \({15^{{\text{th}}}}\) term of the sequence.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(5(2a + 9d) = 60{\text{ (or }}2a + 9d = 12)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(10(2a + 19d) = 320{\text{ (or }}2a + 19d = 32)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">solve simultaneously to obtain <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a = - 3,{\text{ }}d = 2\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the \({15^{{\text{th}}}}\) term is \( - 3 + 14 \times 2 = 25\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> <strong><em>FT</em></strong> the final <strong><em>A1</em></strong> on the values found in the penultimate line.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">with an AP the mean of an even number of consecutive terms equals the mean of the middle terms <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{a_{10}} + {a_{11}}}}{2} = 16\,\,\,\,\,{\text{(or }}{a_{10}} + {a_{11}} = 32)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{a_5} + {a_6}}}{2} = 6\,\,\,\,\,{\text{(or }}{a_5} + {a_6} = 12)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({a_{10}} - {a_5} + {a_{11}} - {a_6} = 20\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(5d + 5d = 20\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(d = 2{\text{ and }}a = - 3\,\,\,\,\,{\text{(or }}{a_5} = 5{\text{ or }}{a_{10}} = 15)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the \({15^{{\text{th}}}}\) term is \( - 3 + 14 \times 2 = 25\,\,\,\,\,{\text{(or }}5 + 10 \times 2 = 25{\text{ or }}15 + 5 \times 2 = 25)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> <strong><em>FT</em></strong> the final <strong><em>A1</em></strong> on the values found in the penultimate line.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates had difficulties with this question with the given information often translated into incorrect equations.</span></p>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">A geometric sequence \({u_1}\) , \({u_2}\) , \({u_3}\) , \(...\) has \({u_1} = 27\) and a sum to infinity of \(\frac{{81}}{2}\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Find the common ratio of the geometric sequence.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">An arithmetic sequence </span><span style="font-family: times new roman,times; font-size: medium;">\({v_1}\) , \({v_2}\) , \({v_3}\) , \(...\) is such that \({v_2} = {u_2}\) and \({v_4} = {u_4}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Find the greatest value of \(N\) such that \(\sum\limits_{n = 1}^N {{v_n}} > 0\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\({u_1} = 27\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{81}}{2} = \frac{{27}}{{1 - r}}\) <em><strong>M1</strong></em></span><br><span style="font-family: times new roman,times; font-size: medium;">\(r = \frac{1}{3}\) <em><strong> A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong>[2 marks]</strong></em></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\({v_2} = 9\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\({v_4} = 1\)</span><br><span style="font-family: times new roman,times; font-size: medium;">\(2d = - 8 \Rightarrow d = - 4\) <em><strong>(A1)</strong></em></span><br><span style="font-family: times new roman,times; font-size: medium;">\({v_1} = 13\) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{N}{2}\left( {2 \times 13 - 4\left( {N - 1} \right)} \right) > 0\) </span><span style="font-family: times new roman,times; font-size: medium;"> (accept equality) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{N}{2}\left( {30 - 4N} \right) > 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(N\left( {15 - 2N} \right) > 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(N < 7.5\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> (M1)</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(N = 7\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> \(13 + 9 + 5 + 1 - 3 - 7 - 11 > 0 \Rightarrow N = 7\) or equivalent receives full marks.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong>[5 marks]</strong></em><br></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Using the definition of a derivative as \(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\) , show that the derivative of \(\frac{1}{{2x + 1}}{\text{ is }}\frac{{ - 2}}{{{{(2x + 1)}^2}}}\).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Prove by induction that the \({n^{{\text{th}}}}\) derivative of \({(2x + 1)^{ - 1}}\) is \({( - 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}\).</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(f(x) = \frac{1}{{2x + 1}}\) and using the result \(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{\frac{1}{{2(x + h) + 1}} - \frac{1}{{2x + 1}}}}{h}} \right)\) <strong> <em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{[2x + 1] - [2(x + h) + 1]}}{{h[2(x + h) + 1][2x + 1]}}} \right)\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - 2}}{{[2(x + h) + 1][2x + 1]}}} \right)\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow f'(x) = \frac{{ - 2}}{{{{(2x + 1)}^2}}}\) <strong> <em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[4 marks]</span><br></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(y = \frac{1}{{2x + 1}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">we want to prove that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = {( - 1)^n}\frac{{{2^n}n!}}{{{{(2x + 1)}^{n + 1}}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(n = 1 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = {( - 1)^1}\frac{{{2^1}1!}}{{{{(2x + 1)}^{1 + 1}}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2}}{{{{(2x + 1)}^2}}}\) which is the same result as part (a)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">hence the result is true for \(n = 1\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">assume the result is true for \(n = k{\text{ : }}\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = {( - 1)^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( - 1)}^k}\frac{{{2^k}k!}}{{{{(2x + 1)}^{k + 1}}}}} \right]\) <strong> <em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left[ {{{( - 1)}^k}{2^k}k!{{(2x + 1)}^{ - k - 1}}} \right]\) <strong> <em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^k}{2^k}k!( - k - 1){(2x + 1)^{ - k - 2}} \times 2\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^{k + 1}}{2^{k + 1}}(k + 1)!{(2x + 1)^{ - k - 2}}\) <strong> <em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = {( - 1)^{k + 1}}\frac{{{2^{k + 1}}(k + 1)!}}{{{{(2x + 1)}^{k + 2}}}}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">hence if the result is true for \(n = k\) , it is true for \(n = k + 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">since the result is true for \(n = 1\) , the result is proved by mathematical induction <strong><em>R1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Only award final <strong><em>R1 </em></strong>if all the <strong><em>M </em></strong>marks have been gained.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[9 marks]</span><br></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for \(n = k\) and then show that this leads to it being true for \(n = k + 1\). Many candidates just write ‘Let \(n = k\)’ which is of course meaningless. The conclusion is often of the form ‘True for \(n = 1,{\text{ }}n = k{\text{ and }}n = k + 1\) therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for \(n = k \Rightarrow \) true for \(n = k + 1\)’.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Even though the definition of the derivative was given in the question, solutions to (a) were often disappointing with algebraic errors fairly common, usually due to brackets being omitted or manipulated incorrectly. Solutions to the proof by induction in (b) were often poor. Many candidates fail to understand that they have to assume that the result is true for \(n = k\) and then show that this leads to it being true for \(n = k + 1\). Many candidates just write ‘Let \(n = k\)’ which is of course meaningless. The conclusion is often of the form ‘True for \(n = 1,{\text{ }}n = k{\text{ and }}n = k + 1\) therefore true by induction’. Credit is only given for a conclusion which includes a statement such as ‘True for \(n = k \Rightarrow \) true for \(n = k + 1\)’.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Prove by mathematical induction that \({n^3} + 11n\) is divisible by 3 for all \(n \in {\mathbb{Z}^ + }\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(n = 1:{\text{ }}{1^3} + 11 = 12\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 3 \times 4\) <strong>or </strong>a multiple of 3 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">assume the proposition is true for \(n = k\) \((ie{\text{ }}{k^3} + 11k = 3{\text{ m)}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Do not award <strong><em>M1 </em></strong>for statements with “Let \(n = k\)<span style="font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">”</span>.</span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">consider \(n = k + 1:\) \({(k + 1)^3} + 11(k + 1)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {k^3} + 3{k^2} + 3k + 1 + 11k + 11\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {k^3} + 11k + (3{k^2} + 3k + 12)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 3(m + {k^2} + k + 4)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept \({k^3} + 11k + 3({k^2} + k + 4)\) or statement that \({k^3} + 11k + (3{k^2} + 3k + 12)\) is a multiple of 3.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">true for \(n = 1\), and \(n = k{\text{ true }} \Rightarrow n = k + 1{\text{ true}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence true for all \(n \in {\mathbb{Z}^ + }\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Only award the final <strong><em>R1 </em></strong>if at least 4 of the previous marks have been achieved.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">It was pleasing to see a great many clear and comprehensive answers for this relatively straightforward induction question. The inductive step only seemed to pose problems for the very weakest candidates. As in previous sessions, marks were mainly lost by candidates writing variations on ‘Let \(n = k\)’, rather than ‘Assume true for \(n = k\)’. The final reasoning step still needs attention, with variations on ‘\(n = k + 1{\text{ true }} \Rightarrow n = k{\text{ true}}\)’ evident, suggesting that mathematical induction as a technique is not clearly understood.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The sum, \({S_n}\), of the first <em>n</em> terms of a geometric sequence, whose \({n^{{\text{th}}}}\) term is \({u_n}\), is given by</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{S_n} = \frac{{{7^n} - {a^n}}}{{{7^n}}},{\text{ where }}a > 0.\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Find an expression for \({u_n}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find the first term and common ratio of the sequence.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Consider the sum to infinity of the sequence.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Determine the values of <em>a</em> such that the sum to infinity exists.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Find the sum to infinity when it exists.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \({u_n} = {S_n} - {S_{n - 1}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{{7^n} - {a^n}}}{{{7^n}}} - \frac{{{7^{n - 1}} - {a^{n - 1}}}}{{{7^{n - 1}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({u_1} = 1 - \frac{a}{7}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({u_2} = 1 - \frac{{{a^2}}}{{{7^2}}} - \left( {1 - \frac{a}{7}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{a}{7}\left( {1 - \frac{a}{7}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">common ratio \( = \frac{a}{7}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({u_n} = 1 - {\left( {\frac{a}{7}} \right)^n} - 1 + {\left( {\frac{a}{7}} \right)^{n - 1}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {\left( {\frac{a}{7}} \right)^{n - 1}}\left( {1 - \frac{a}{7}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{7 - a}}{7}{\left( {\frac{a}{7}} \right)^{n - 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({u_1} = \frac{{7 - a}}{7}\), common ratio \( = \frac{a}{7}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) \(0 < a < 7\,\,\,\,\,{\text{(accep }}a < 7)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) 1 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \({u_n} = b{r^{n - 1}} = \left( {\frac{{7 - a}}{7}} \right){\left( {\frac{a}{7}} \right)^{n - 1}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) for a GP with first term <em>b</em> and common ratio <em>r</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({S_n} = \frac{{b(1 - {r^n})}}{{1 - r}} = \left( {\frac{b}{{1 - r}}} \right) - \left( {\frac{b}{{1 - r}}} \right){r^n}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">as \({S_n} = \frac{{{7^n} - {a^n}}}{{{7^n}}} = 1 - {\left( {\frac{a}{7}} \right)^n}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">comparing both expressions <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{b}{{1 - r}} = 1\) and \(r = \frac{a}{7}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(b = 1 - \frac{a}{7} = \frac{{7 - a}}{7}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({u_1} = b = \frac{{7 - a}}{7}\), common ratio \( = r = \frac{a}{7}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award method marks if the expressions for <em>b</em> and <em>r</em> are deduced in part (a).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) \(0 < a < 7\,\,\,\,\,{\text{(accept }}a < 7)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) 1 <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates found this question difficult. In (a), few seemed to realise that \({u_n} = {S_n} - {S_{n - 1}}\). In (b), few candidates realised that \({u_1} = {S_1}\) and in (c) that \({S_n}\) could be written as \(1 - {\left( {\frac{a}{7}} \right)^n}\) from which it follows immediately that the sum to infinity exists when a < 7 and is equal to 1.</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(y(x) = x{e^{3x}},{\text{ }}x \in \mathbb{R}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Prove by induction that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = n{3^{n - 1}}{{\text{e}}^{3x}} + x{3^n}{{\text{e}}^{3x}}\) for \(n \in {\mathbb{Z}^ + }\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the coordinates of any local maximum and minimum points on the graph of \(y(x)\).</p>
<p class="p1">Justify whether any such point is a maximum or a minimum.</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the coordinates of any points of inflexion on the graph of \(y(x)\). Justify whether any such point is a point of inflexion.</p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence sketch the graph of \(y(x)\), indicating clearly the points found in parts (c) and (d) and any intercepts with the axes.</p>
<div class="marks">[2]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = ({{\text{e}}^{3x}} + 3x{{\text{e}}^{3x}})\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p1"><span class="s1"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">let \(P(n)\) be the statement \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = n{3^{n - 1}}{{\text{e}}^{3x}} + x{3^n}{{\text{e}}^{3x}}\)</p>
<p class="p1">prove for \(n = 1\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p1"><span class="s1">\(LHS\) of </span>\(P(1)\) is \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) which is \(1 \times {{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}}\) and <span class="s1">\(RHS\)</span> is \({3^0}{{\text{e}}^{3x}} + x{3^1}{{\text{e}}^{3x}}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>R1</em></strong></span></p>
<p class="p1"><span class="s1">as </span>\({\text{LHS}} = {\text{RHS, }}P(1)\) is true</p>
<p class="p1">assume \(P(k)\) is true and attempt to prove \(P(k + 1)\) is true <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">assuming \(\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = k{3^{k - 1}}{{\text{e}}^{3x}} + x{3^k}{{\text{e}}^{3x}}\)</p>
<p class="p1">\(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}}} \right)\) <span class="Apple-converted-space"> </span><span class="s1">(<strong><em>M1)</em></strong></span></p>
<p class="p1">\( = k{3^{k - 1}} \times 3{{\text{e}}^{3x}} + 1 \times {3^k}{{\text{e}}^{3x}} + x{3^k} \times 3{{\text{e}}^{3x}}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">\( = (k + 1){3^k}{{\text{e}}^{3x}} + x{3^{k + 1}}{{\text{e}}^{3x}}\;\;\;\)<span class="s1">(as required) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></span></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: <span class="Apple-converted-space"> </span></strong>Can award the <strong><em>A </em></strong>marks independent of the <strong><em>M </em></strong>marks</p>
<p class="p4"> </p>
<p class="p1">since \(P(1)\) is true and \(P(k)\) is true \( \Rightarrow P(k + 1)\) is true</p>
<p class="p1">then (by <span class="s1">\(PMI\)</span>), \(P(n)\) is true \((\forall n \in {\mathbb{Z}^ + })\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>R1</em></strong></span></p>
<p class="p2"> </p>
<p class="p3"><strong>Note: </strong>To gain last <strong><em>R1 </em></strong>at least four of the above marks must have been gained.</p>
<p class="p3"><em><strong>[7 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({{\text{e}}^{3x}} + x \times 3{{\text{e}}^{3x}} = 0 \Rightarrow 1 + 3x = 0 \Rightarrow x = - \frac{1}{3}\) <strong><em>M1A1</em></strong></p>
<p>point is \(\left( { - \frac{1}{3},{\text{ }} - \frac{1}{{3{\text{e}}}}} \right)\) <strong><em>A1</em></strong></p>
<p><strong>EITHER</strong></p>
<p>\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\)</p>
<p>when \(x = - \frac{1}{3},{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} > 0\) therefore the point is a minimum <strong><em>M1A1</em></strong></p>
<p><strong>OR</strong></p>
<p class="p1"><img src="images/Schermafbeelding_2015-12-22_om_17.44.19.png" alt></p>
<p class="p1">nature table shows point is a minimum <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}}\) <strong><em>A1</em></strong></p>
<p>\(2 \times 3{{\text{e}}^{3x}} + x \times {3^2}{{\text{e}}^{3x}} = 0 \Rightarrow 2 + 3x = 0 \Rightarrow x = - \frac{2}{3}\) <strong><em>M1A1</em></strong></p>
<p>point is \(\left( { - \frac{2}{3},{\text{ }} - \frac{2}{{3{{\text{e}}^2}}}} \right)\) <strong><em>A1</em></strong></p>
<p class="p1"><img src="images/Schermafbeelding_2015-12-22_om_17.52.15.png" alt></p>
<p>since the curvature does change (concave down to concave up) it is a point of inflection <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Allow \({3^{{\text{rd}}}}\) derivative is not zero at \( - \frac{2}{3}\)</p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><img src="image.html" alt></p>
<p class="p1">(general shape including asymptote and through origin) <strong><em>A1</em></strong></p>
<p class="p1">showing minimum and point of inflection <strong><em>A1</em></strong></p>
<p class="p1"> </p>
<p class="p1"><strong>Note: </strong>Only indication of position of answers to (c) and (d) required, not coordinates.</p>
<p class="p1"><em><strong>[2 marks]</strong></em></p>
<p class="p1"><em><strong>Total [21 marks]</strong></em></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Well done.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">The logic of an induction proof was not known well enough. Many candidates used what they had to prove rather than differentiating what they had assumed. They did not have enough experience in doing Induction proofs.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Good, some forgot to test for min/max, some forgot to give the \(y\) value.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Again quite good, some forgot to check for change in curvature and some forgot the \(y\) value.</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Some accurate sketches, some had all the information from earlier parts but could not apply it. The asymptote was often missed.</p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The first three terms of a geometric sequence are \(\sin x,{\text{ }}\sin 2x\) and \(4\sin x{\cos ^2}x,{\text{ }} - \frac{\pi }{2} < x < \frac{\pi }{2}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Find the common ratio <em>r</em>.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Find the set of values of <em>x </em>for which the geometric series \(\sin x + \sin 2x + 4\sin x{\cos ^2}x + \ldots \) converges.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider \(x = \arccos \left( {\frac{1}{4}} \right),{\text{ }}x > 0\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Show that the sum to infinity of this series is \(\frac{{\sqrt {15} }}{2}\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\sin x,{\text{ }}\sin 2x{\text{ and }}4\sin x{\cos ^2}x\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = \frac{{2\sin x\cos x}}{{\sin x}} = 2\cos x\)<em> </em><strong>A1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica; min-height: 26.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept \(\frac{{\sin 2x}}{{\sin x}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left| r \right| < 1 \Rightarrow \left| {2\cos x} \right| < 1\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 1 < r < 1 \Rightarrow - 1 < 2\cos x < 1\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>THEN</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(0 < \cos x < \frac{1}{2}{\text{ for }} - \frac{\pi }{2} < x < \frac{\pi }{2}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - \frac{\pi }{2} < x < - \frac{\pi }{3}{\text{ or }}\frac{\pi }{3} < x < \frac{\pi }{2}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \({S_\infty } = \frac{{\sin x}}{{1 - 2\cos x}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({S_\infty } = \frac{{\sin \left( {\arccos \left( {\frac{1}{4}} \right)} \right)}}{{1 - 2\cos \left( {\arccos \left( {\frac{1}{4}} \right)} \right)}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\frac{{\sqrt {15} }}{4}}}{{\frac{1}{2}}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1 </em></strong>for correct numerator and <strong><em>A1 </em></strong>for correct denominator.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\sqrt {15} }}{2}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider \(a = {\log _2}3 \times {\log _3}4 \times {\log _4}5 \times \ldots \times {\log _{31}}32\). Given that \(a \in \mathbb{Z}\), find the value of <em>a</em>.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{\log 3}}{{\log 2}} \times \frac{{\log 4}}{{\log 3}} \times \ldots \times \frac{{\log 32}}{{\log 31}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{\log 32}}{{\log 2}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{5\log 2}}{{\log 2}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 5\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">hence \(a = 5\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept the above if done in a specific base <em>eg </em>\({\log _2}x\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p>Find the term independent of \(x\) in the binomial expansion of \({\left( {2{x^2} + \frac{1}{{2{x^3}}}} \right)^{10}}\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>attempt at binomial expansion, relevant row of Pascal’s triangle or use of general term with binomial coefficient must be seen <strong><em>(M1)</em></strong></p>
<p>term independent of \(x\) is \(\left( {\begin{array}{*{20}{c}}{10} \\4\end{array}} \right){(2{x^2})^6}{\left( {\frac{1}{{2{x^3}}}} \right)^4}\) (or equivalent) <strong><em>(A1)(A1)(A1)</em></strong></p>
<p> </p>
<p><strong>Notes:</strong> \(x\)’s may be omitted. Also accept \(\left( {\begin{array}{*{20}{c}}{10} \\6\end{array}} \right)\) or 210.</p>
<p> </p>
<p>\( = 840\) <strong><em>A1</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that <em>z </em>is the complex number \(x + {\text{i}}y\) and that \(\left| {\,z\,} \right| + z = 6 - 2{\text{i}}\) , find the value of <em>x</em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">and the value of <em>y </em>.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\sqrt {{x^2} + {y^2}} + x + y{\text{i}} = 6 - 2{\text{i}}\) <strong><em>(A1)</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">equating real and imaginary parts <strong><em>M1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(y = - 2\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\sqrt {{x^2} + 4} + x = 6\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\({x^2} + 4 = {(6 - x)^2}\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\( - 32 = - 12x \Rightarrow x = \frac{8}{3}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p> </p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">There were some good solutions to this question, but those who failed to complete the question failed at a variety of different points. Many did not know the definition of the modulus of a complex number and so could not get started at all. Many then did not think to equate real and imaginary parts, and then many failed to solve the resulting irrational equation to be able to find <em>x</em>. </span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider a function <em>f </em>, defined by \(f(x) = \frac{x}{{2 - x}}{\text{ for }}0 \leqslant x \leqslant 1\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find an expression for \((f \circ f)(x)\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman',times; font-size: medium;">Let \({F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}}\), where \(0 \leqslant x \leqslant 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman',times; font-size: medium;">Use mathematical induction to show that for any \(n \in {\mathbb{Z}^ + }\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman',times; font-size: medium;">\[\underbrace {(f \circ f \circ ... \circ f)}_{n{\text{ times}}}(x) = {F_n}(x)\] .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \({F_{ - n}}(x)\) is an expression for the inverse of \({F_n}\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) State \({F_n}(0){\text{ and }}{F_n}(1)\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Show that \({F_n}(x) < x\) , given 0 < <em>x </em>< 1, \(n \in {\mathbb{Z}^ + }\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) For \(n \in {\mathbb{Z}^ + }\) , let \({A_n}\) be the area of the region enclosed by the graph of \(F_n^{ - 1}\) , the <em>x</em>-axis and the line <em>x </em>= 1. Find the area \({B_n}\) of the region enclosed by \({F_n}\) and \(F_n^{ - 1}\) in terms of \({A_n}\) .<span style="font: 7.0px Helvetica;"><br></span></span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((f \circ f)(x) = f\left( {\frac{x}{{2 - x}}} \right) = \frac{{\frac{x}{{2 - x}}}}{{2 - \frac{x}{{2 - x}}}}\) <em><strong>M1A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((f \circ f)(x) = \frac{x}{{4 - 3x}}\) <em><strong> A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><em><strong><span style="font-family: 'times new roman', times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(P(n):\underbrace {(f \circ f \circ ... \circ f)}_{n{\text{ times}}}(x) = {F_n}(x)\)</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(P(1):{\text{ }}f(x) = {F_1}(x)\)</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(LHS = f(x) = \frac{x}{{2 - x}}{\text{ and }}RHS = {F_1}(x) = \frac{x}{{{2^1} - ({2^1} - 1)x}} = \frac{x}{{2 - x}}\) <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\therefore P(1){\text{ true}}\)</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">assume that <em>P</em>(<em>k</em>) is true, <em>i.e.</em>, \(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k times}}}(x) = {F_k}(x)\) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">consider \(P(k + 1)\)</span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER</strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k}} + 1{\text{ times}}}(x) = \left( {f \circ \underbrace {f \circ f \circ ... \circ f}_{{\text{k times}}}} \right)(x) = f\left( {{F_k}(x)} \right)\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\( = f\left( {\frac{x}{{{2^k} - ({2^k} - 1)x}}} \right) = \frac{{\frac{x}{{{2^k} - ({2^k} - 1)x}}}}{{2 - \frac{x}{{{2^k} - ({2^k} - 1)x}}}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{{2\left( {{2^k} - ({2^k} - 1)x} \right) - x}} = \frac{x}{{{2^{k + 1}} - ({2^{k + 1}} - 2)x - x}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR </strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\(\underbrace {(f \circ f \circ ... \circ f)}_{{\text{k}} + 1{\text{ times}}}(x) = \left( {f \circ \underbrace {f \circ f \circ ... \circ f}_{{\text{k times}}}} \right)(x) = {F_k}(f(x))\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\( = {F_k}\left( {\frac{x}{{2 - x}}} \right) = \frac{{\frac{x}{{2 - x}}}}{{{2^k} - ({2^k} - 1)\frac{x}{{2 - x}}}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{{{2^{k + 1}} - {2^k}x - {2^k}x + x}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><strong>THEN</strong></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{{{2^{k + 1}} - ({2^{k + 1}} - 1)x}} = {F_{k + 1}}(x)\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><em>P</em>(<em>k</em>) true implies <em>P</em>(<em>k</em> + 1) true, <em>P</em>(1) true so <em>P</em>(<em>n</em>) true for all \(n \in {\mathbb{Z}^ + }\) <em><strong>R1</strong></em></span></p>
<p><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong>[8 marks]</strong></em></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = \frac{y}{{{2^n} - ({2^n} - 1)y}} \Rightarrow {2^n}x - ({2^n} - 1)xy = y\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {2^n}x = \left( {({2^n} - 1)x + 1} \right)y \Rightarrow y = \frac{{{2^n}x}}{{({2^n} - 1)x + 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(F_n^{ - 1}(x) = \frac{{{2^n}x}}{{({2^n} - 1)x + 1}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(F_n^{ - 1}(x) = \frac{x}{{\frac{{{2^n} - 1}}{{{2^n}}}x + \frac{1}{{{2^n}}}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(F_n^{ - 1}(x) = \frac{x}{{(1 - {2^{ - n}})x + {2^{ - n}}}}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(F_n^{ - 1}(x) = \frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}\) <strong> <em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt \({F_{ - n}}\left( {{F_n}(x)} \right)\) <em><strong> M1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {F_{ - n}}\left( {\frac{x}{{{2^n} - ({2^n} - 1)x}}} \right) = \frac{{\frac{x}{{{2^n} - ({2^n} - 1)x}}}}{{{2^{ - n}} - ({2^{ - n}} - 1)\frac{x}{{{2^n} - ({2^n} - 1)x}}}}\) <em><strong>A1A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{{{2^{ - n}}({2^n} - ({2^n} - 1)x) - ({2^{ - n}} - 1)x}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">marks for numerators and denominators</span><span style="font-family: 'times new roman', times; font-size: medium;">.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{1} = x\) <strong><em>A1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 3</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt \({F_n}\left( {{F_{ - n}}(x)} \right)\) <strong> M1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {F_n}\left( {\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}} \right) = \frac{{\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}}}{{{2^n} - ({2^n} - 1)\frac{x}{{{2^{ - n}} - ({2^{ - n}} - 1)x}}}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{{{2^n}({2^{ - n}} - ({2^{ - n}} - 1)x) - ({2^n} - 1)x}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note: </strong><span style="font-family: 'times new roman', times; font-size: medium;">Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1 </em></strong><span style="font-family: 'times new roman', times; font-size: medium;">marks for numerators and denominators.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{1} = x\) <strong><em>A1AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \({F_n}(0) = 0,{\text{ }}{F_n}(1) = 1\) <em><strong>A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong> </strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^n} - ({2^n} - 1)x - 1 = ({2^n} - 1)(1 - x)\) <em><strong>(M1)</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( > 0{\text{ if }}0 < x < 1{\text{ and }}n \in {\mathbb{Z}^ + }\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so \({2^n} - ({2^n} - 1)x > 1{\text{ and }}{F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}} < \frac{x}{1}( < x)\) <em><strong>R1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({F_n}(x) = \frac{x}{{{2^n} - ({2^n} - 1)x}} < x{\text{ for }}0 < x < 1{\text{ and }}n \in {\mathbb{Z}^ + }\) <em><strong>AG</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{x}{{{2^n} - ({2^n} - 1)x}} < x \Leftrightarrow {2^n} - ({2^n} - 1)x > 1\) <em><strong>(M1)</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Leftrightarrow ({2^n} - 1)x < {2^n} - 1\) <em><strong>A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Leftrightarrow x < \frac{{{2^n} - 1}}{{{2^n} - 1}} = 1\) true in the interval \(\left] {0,{\text{ }}1} \right[\) <em><strong>R1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong> </strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) \({B_n} = 2\left( {{A_n} - \frac{1}{2}} \right){\text{ }}( = 2{A_n} - 1)\) <em><strong>(M1)A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em><strong>[6 marks]</strong></em></span></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<div><span style="font-family: 'times new roman', times; font-size: medium;">Part a) proved to be an easy 3 marks for most candidates. </span></div>
<div> </div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Part b) was often answered well, and candidates were well prepared in this session for this type of question. Candidates still need to take care when showing explicitly that P(1) is true, and some are still writing ‘Let </span><em style="font-family: 'times new roman', times; font-size: medium;">n</em><span style="font-family: 'times new roman', times; font-size: medium;"> = </span><em style="font-family: 'times new roman', times; font-size: medium;">k</em><em style="font-family: 'times new roman', times; font-size: medium;">’ </em><span style="font-family: 'times new roman', times; font-size: medium;">which gains no marks. The inductive step was often well argued, and given in clear detail, though the final inductive reasoning step was incorrect, or appeared rushed, even from the better candidates. ‘True for </span><em style="font-family: 'times new roman', times; font-size: medium;">n</em><span style="font-family: 'times new roman', times; font-size: medium;"> =1, </span><em style="font-family: 'times new roman', times; font-size: medium;">n</em><span style="font-family: 'times new roman', times; font-size: medium;"> = </span><em style="font-family: 'times new roman', times; font-size: medium;">k</em><span style="font-family: 'times new roman', times; font-size: medium;"> and </span><em style="font-family: 'times new roman', times; font-size: medium;">n</em><span style="font-family: 'times new roman', times; font-size: medium;"> = </span><em style="font-family: 'times new roman', times; font-size: medium;">k</em><span style="font-family: 'times new roman', times; font-size: medium;"> + 1’ is still disappointingly seen, as were some even more unconvincing variations.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Part c) was again very well answered by the majority. A few weaker candidates attempted to find an inverse for the individual case n = 1 , but gained no credit for this.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">Part d) was not at all well understood, with virtually no candidates able to tie together the hints given by connecting the different parts of the question. Rash, and often thoughtless attempts were made at each part, though by this stage some seemed to be struggling through lack of time. The inequality part of the question tended to be ‘fudged’, with arguments seen by examiners being largely unconvincing and lacking clarity. A tiny number of candidates provided the correct answer to the final part, though a surprising number persisted with what should have been recognised as fruitless working – usually in the form of long-winded integration attempts.</span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The common ratio of the terms in a geometric series is \({2^x}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) State the set of values of <em>x</em> for which the sum to infinity of the series exists.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) If the first term of the series is 35, find the value of <em>x</em> for which the sum to infinity is 40.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(0 < {2^x} < 1\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x < 0\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \(\frac{{35}}{{1 - r}} = 40\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 40 - 40 \times r = 35\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow - 40 \times r = - 5\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow r = {2^x} = \frac{1}{8}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow x = {\log _2}\frac{1}{8}{\text{ }}( = - 3)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> The substitution \(r = {2^x}\) may be seen at any stage in the solution.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 32.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) was the first question that a significant majority of candidates struggled with. Only the best candidates were able to find the required set of values. However, it was pleasing to see that the majority of candidates made a meaningful start to part (b). Many candidates gained wholly correct answers to part (b).</span></p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">If <em>z</em> is a non-zero complex number, we define \(L(z)\) by the equation</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[L(z) = \ln \left| z \right| + {\text{i}}\arg (z),{\text{ }}0 \leqslant \arg (z) < 2\pi .\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that when <em>z</em> is a positive real number, \(L(z) = \ln z\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Use the equation to calculate</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(L( - 1)\) ;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(L(1 - {\text{i}})\) ;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) \(L( - 1 + {\text{i}})\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Hence show that the property \(L({z_1}{z_2}) = L({z_1}) + L({z_2})\) does not hold for all values of \({z_1}\) and \({z_2}\) .</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">Part A.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let <em>f</em> be a function with domain \(\mathbb{R}\) that satisfies the conditions,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(x + y) = f(x)f(y)\) , for all <em>x</em> and <em>y</em> and \(f(0) \ne 0\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that \(f(0) = 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Prove that \(f(x) \ne 0\) , for all \(x \in \mathbb{R}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Assuming that \(f'(x)\) exists for all \(x \in \mathbb{R}\) , use the definition of derivative to show that \(f(x)\) satisfies the differential equation \(f'(x) = k{\text{ }}f(x)\) , where \(k = f'(0)\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) Solve the differential equation to find an expression for \(f(x)\) .</span></p>
<div class="marks">[14]</div>
<div class="question_part_label">Part B.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\left| z \right| = z\) , \(\arg (z) = 0\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">so \(L(z) = \ln z\) <strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) \(L( - 1) = \ln 1 + {\text{i}}\pi = {\text{i}}\pi \) <strong><em>A1A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(L(1 - {\text{i}}) = \ln \sqrt 2 + {\text{i}}\frac{{7\pi }}{4}\) <strong><em>A1A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) \(L( - 1 + {\text{i}}) = \ln \sqrt 2 + {\text{i}}\frac{{3\pi }}{4}\) <strong><em>A1 N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) for comparing the product of two of the above results with the third <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">for stating the result \( - 1 + {\text{i}} = - 1 \times (1 - {\text{i}})\) and \(L( - 1 + {\text{i}}) \ne L( - 1) + L(1 - {\text{i}})\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">hence, the property \(L({z_1}{z_2}) = L({z_1}) + L({z_2})\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">does not hold for all values of \({z_1}\) and \({z_2}\) <strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Times; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [9 marks]</em></strong></span></p>
<div class="question_part_label">Part A.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) from \(f(x + y) = f(x)f(y)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for <em>x</em> = <em>y</em> = 0 <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">we have \(f(0 + 0) = f(0)f(0) \Leftrightarrow f(0) = {\left( {f(0)} \right)^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">as \(f(0) \ne 0\), this implies that \(f(0) = 1\) <strong><em>R1AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">from \(f(x + y) = f(x)f(y)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for <em>y</em> = –<em>x</em> , we have \(f(x - x) = f(x)f( - x) \Leftrightarrow f(0) = f(x)f( - x)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">as \(f(0) \ne 0\) this implies that \(f(x) \ne 0\) <strong><em>R1AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">suppose that, for a value of <em>x</em>, \(f(x) = 0\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">from \(f(x + y) = f(x)f(y)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for \(y = - x\), we have \(f(x - x) = f(x)f( - x) \Leftrightarrow f(0) = f(x)f( - x)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">substituting \(f(x)\) by 0 gives \(f(0) = 0\) which contradicts part (a) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore \(f(x) \ne 0\) for all <em>x. </em><strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) by the definition of derivative</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x)f(h) - f(x)f(0)}}{h}} \right)\) <strong><em>A1(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(h) - f(0)}}{h}} \right)f(x)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = f'(0)f(x)\,\,\,\,\,\left( { = k{\text{ }}f(x)} \right)\) <strong><em>AG N0</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) \(\int {\frac{{f'(x)}}{{f(x)}}{\text{d}}x = \int {k{\text{d}}x \Rightarrow \ln f(x) = kx + C} } \) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\ln f(0) = C \Rightarrow C = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(x) = {{\text{e}}^{kx}}\) <strong><em>A1 N1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1A0A0A0</em></strong> if no arbitrary constant <em>C</em> .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [14 marks]</em></strong></span></p>
<div class="question_part_label">Part B.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part A was answered well by a fair amount of candidates, with some making mistakes in calculating the arguments of complex numbers, as well as careless mistakes in finding the products of complex numbers.</span></p>
<div class="question_part_label">Part A.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part B proved demanding for most candidates, particularly parts (c) and (d). A surprising number of candidates did not seem to know what was meant by the ‘definition of derivative’ in part (c) as they attempted to use quotient rule rather than first principles.</span></p>
<div class="question_part_label">Part B.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The first terms of an arithmetic sequence are \(\frac{1}{{{{\log }_2}x}},{\text{ }}\frac{1}{{{{\log }_8}x}},{\text{ }}\frac{1}{{{{\log }_{32}}x}},{\text{ }}\frac{1}{{{{\log }_{128}}x}},{\text{ }} \ldots \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find <em>x</em> if the sum of the first 20 terms of the sequence is equal to 100.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(d = \frac{1}{{{{\log }_8}x}} - \frac{1}{{{{\log }_2}x}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{{{\log }_2}8}}{{{{\log }_2}x}} - \frac{1}{{{{\log }_2}x}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award this </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for a correct change of base anywhere in the question.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{2}{{{{\log }_2}x}}\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>(A1)</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{20}}{2}\left( {2 \times \frac{1}{{{{\log }_2}x}} + 19 \times \frac{2}{{{{\log }_2}x}}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{400}}{{{{\log }_2}x}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(100 = \frac{{400}}{{{{\log }_2}x}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({20^{{\text{th}}}}{\text{ term}} = \frac{1}{{{{\log }_{{2^{39}}}}x}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(100 = \frac{{20}}{2}\left( {\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_{{2^{39}}}}x}}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(100 = \frac{{20}}{2}\left( {\frac{1}{{{{\log }_2}x}} + \frac{{{{\log }_2}{2^{39}}}}{{{{\log }_2}x}}} \right)\) <strong><em>M1(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award this </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for a correct change of base anywhere in the question.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(100 = \frac{{400}}{{{{\log }_2}x}}\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>(A1)</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 3</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_8}x}} + \frac{1}{{{{\log }_{32}}x}} + \frac{1}{{{{\log }_{128}}x}} + \ldots \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{1}{{{{\log }_2}x}} + \frac{{{{\log }_2}8}}{{{{\log }_2}x}} + \frac{{{{\log }_2}32}}{{{{\log }_2}x}} + \frac{{{{\log }_2}128}}{{{{\log }_2}x}} + \ldots \) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award this </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for a correct change of base anywhere in the question.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{{{{\log }_2}x}}(1 + 3 + 5 + \ldots )\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{{{{\log }_2}x}}\left( {\frac{{20}}{2}(2 + 38)} \right)\) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(100 = \frac{{400}}{{{{\log }_2}x}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">There were plenty of good answers to this question. Those who realised they needed to make each log have the same base (and a great variety of bases were chosen) managed the question successfully.</span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Solve the equation \({8^{x - 1}} = {6^{3x}}\). Express your answer in terms of \(\ln 2\) and \(\ln 3\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^{3(x - 1)}} = {(2 \times 3)^{3x}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1 </em></strong>for writing in terms of 2 and 3.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^{3x}} \times {2^{ - 3}} = {2^{3x}} \times {3^{3x}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\({2^{ - 3}} = {3^{3x}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\ln \left( {{2^{ - 3}}} \right) = \ln \left( {{3^{3x}}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 3\ln 2 = 3x\ln 3\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = - \frac{{\ln 2}}{{\ln 3}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\ln {8^{x - 1}} = \ln {6^{3x}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((x - 1)\ln {2^3} = 3x\ln (2 \times 3)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(3x\ln 2 - 3\ln 2 = 3x\ln 2 + 3x\ln 3\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = - \frac{{\ln 2}}{{\ln 3}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 3</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\ln {8^{x - 1}} = \ln {6^{3x}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((x - 1)\ln 8 = 3x\ln 6\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = \frac{{\ln 8}}{{\ln 8 - 3\ln 6}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = \frac{{3\ln 2}}{{\ln \left( {\frac{{{2^3}}}{{{6^3}}}} \right)}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = - \frac{{\ln 2}}{{\ln 3}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p>Consider \(w = 2\left( {{\text{cos}}\frac{\pi }{3} + {\text{i}}\,{\text{sin}}\frac{\pi }{3}} \right)\)</p>
</div>
<div class="specification">
<p>These four points form the vertices of a quadrilateral, <em>Q</em>.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Express <em>w</em><sup>2</sup> and <em>w</em><sup>3</sup> in modulus-argument form.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Sketch on an Argand diagram the points represented by <em>w</em><sup>0</sup> , <em>w</em><sup>1</sup> , <em>w</em><sup>2</sup> and <em>w</em><sup>3</sup>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the area of the quadrilateral <em>Q</em> is \(\frac{{21\sqrt 3 }}{2}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Let \(z = 2\left( {{\text{cos}}\frac{\pi }{n} + {\text{i}}\,{\text{sin}}\frac{\pi }{n}} \right),\,\,n \in {\mathbb{Z}^ + }\). The points represented on an Argand diagram by \({z^0},\,\,{z^1},\,\,{z^2},\, \ldots \,,\,\,{z^n}\) form the vertices of a polygon \({P_n}\).</p>
<p>Show that the area of the polygon \({P_n}\) can be expressed in the form \(a\left( {{b^n} - 1} \right){\text{sin}}\frac{\pi }{n}\), where \(a,\,\,b\, \in \mathbb{R}\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\({w^2} = 4\text{cis}\left( {\frac{{2\pi }}{3}} \right){\text{;}}\,\,{w^3} = 8{\text{cis}}\left( \pi \right)\) <em><strong>(M1)A1A1</strong></em></p>
<p><strong>Note:</strong> Accept Euler form.</p>
<p><strong>Note:</strong> <em><strong>M1</strong></em> can be awarded for either both correct moduli or both correct arguments.</p>
<p><strong>Note:</strong> Allow multiplication of correct Cartesian form for <em><strong>M1</strong></em>, final answers must be in modulus-argument form.</p>
<p><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><img 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"> <em><strong>A1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>use of area = \(\frac{1}{2}ab\,\,{\text{sin}}\,C\) <em><strong>M1</strong></em></p>
<p>\(\frac{1}{2} \times 1 \times 2 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 2 \times 4 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 4 \times 8 \times {\text{sin}}\frac{\pi }{3}\) <em><strong>A1A1</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>A1</strong></em> for \(C = \frac{\pi }{3}\), <em><strong>A1</strong> </em>for correct moduli.</p>
<p>\( = \frac{{21\sqrt 3 }}{2}\) <em><strong> AG</strong></em></p>
<p><strong>Note:</strong> Other methods of splitting the area may receive full marks.</p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{1}{2} \times {2^0} \times {2^1} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^1} \times {2^2} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^2} \times {2^3} \times {\text{sin}}\frac{\pi }{n} + \, \ldots \, + \frac{1}{2} \times {2^{n - 1}} \times {2^n} \times {\text{sin}}\frac{\pi }{n}\) <em><strong>M1A1</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>M1</strong> </em>for powers of 2, <em><strong>A1</strong> </em>for any correct expression including both the first and last term.</p>
<p>\( = {\text{sin}}\frac{\pi }{n} \times \left( {{2^0} + {2^2} + {2^4} + \, \ldots \, + {2^{n - 2}}} \right)\)</p>
<p>identifying a geometric series with common ratio 2<sup>2</sup>(= 4) <em><strong>(</strong><strong>M1)A1</strong></em></p>
<p>\( = \frac{{1 - {2^{2n}}}}{{1 - 4}} \times {\text{sin}}\frac{\pi }{n}\) <em><strong>M1</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>M1</strong></em> for use of formula for sum of geometric series.</p>
<p>\( = \frac{1}{3}\left( {{4^n} - 1} \right){\text{sin}}\frac{\pi }{n}\) <em><strong>A1</strong></em></p>
<p><em><strong>[6 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the values of <em>n</em> such that \({\left( {1 + \sqrt 3 {\text{i}}} \right)^n}\) is a real number.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">changing to modulus-argument form</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>r</em> = 2</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\theta = \arctan \sqrt 3 = \frac{\pi }{3}\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 1 + {\sqrt 3 ^n} = {2^n}\left( {\cos \frac{{n\pi }}{3} + {\text{i}}\sin \frac{{n\pi }}{3}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">if \(\sin \frac{{n\pi }}{3} = 0 \Rightarrow n = \{ 0,{\text{ }} \pm 3,{\text{ }} \pm 6,{\text{ }} \ldots \} {\text{ }}\) <strong><em>(M1)A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\theta = \arctan \sqrt 3 = \frac{\pi }{3}\) <strong><em>(M1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><img 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" alt> <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(n \in \mathbb{R} \Rightarrow \frac{{n\pi }}{3} = k\pi ,{\text{ }}k \in \mathbb{Z}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow n = 3k,{\text{ }}k \in \mathbb{Z}\) <strong><em>A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Some candidates did not consider changing the number to modulus-argument form. Among those that did this successfully, many considered individual values of <em>n</em>, or only positive values. Very few candidates considered negative multiples of 3.</span></p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Use de Moivre’s theorem to find the roots of the equation \({z^4} = 1 - {\text{i}}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Draw these roots on an Argand diagram.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 19.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) If \({{\text{z}}_1}\) is the root in the first quadrant and \({{\text{z}}_2}\) is the root in the second quadrant, find \(\frac{{{{\text{z}}_2}}}{{{{\text{z}}_1}}}\) in the form <em>a</em> + i<em>b</em> .</span></p>
<div class="marks">[12]</div>
<div class="question_part_label">Part A.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Expand and simplify \((x - 1)({x^4} + {x^3} + {x^2} + x + 1)\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Given that <em>b</em> is a root of the equation \({z^5} - 1 = 0\) which does not lie on the real axis in the Argand diagram, show that \(1 + b + {b^2} + {b^3} + {b^4} = 0\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) If \(u = b + {b^4}\) and \(v = {b^2} + {b^3}\) show that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) <em>u</em> + <em>v</em> = <em>uv</em> = −1;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(u - v = \sqrt 5 \) , given that \(u - v > 0\) .</span></p>
<div class="marks">[13]</div>
<div class="question_part_label">Part B.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(z = {(1 - {\text{i}})^{\frac{1}{4}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(1 - {\text{i}} = r(\cos \theta + {\text{i}}\sin \theta )\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow r = \sqrt 2 \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\theta = - \frac{\pi }{4}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = {\left( {\sqrt 2 \left( {\cos \left( { - \frac{\pi }{4}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{4}} \right)} \right)} \right)^{\frac{1}{4}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {\left( {\sqrt 2 \left( {\cos \left( { - \frac{\pi }{4} + 2n\pi } \right) + {\text{i}}\sin \left( { - \frac{\pi }{4} + 2n\pi } \right)} \right)} \right)^{\frac{1}{4}}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^{\frac{1}{8}}}\left( {\cos \left( { - \frac{\pi }{{16}} + \frac{{n\pi }}{2}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{{16}} + \frac{{n\pi }}{2}} \right)} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^{\frac{1}{8}}}\left( {\cos \left( { - \frac{\pi }{{16}}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{{16}}} \right)} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>M1</em></strong> above for this line if the candidate has forgotten to add \(2\pi \) and no other solution given.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^{\frac{1}{8}}}\left( {\cos \left( {\frac{{7\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{7\pi }}{{16}}} \right)} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^{\frac{1}{8}}}\left( {\cos \left( {\frac{{15\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{15\pi }}{{16}}} \right)} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {2^{\frac{1}{8}}}\left( {\cos \left( { - \frac{{9\pi }}{{16}}} \right) + {\text{i}}\sin \left( { - \frac{{9\pi }}{{16}}} \right)} \right)\) <strong><em>A2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> for 2 correct answers. Accept any equivalent form.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><img 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" alt><span class="Apple-style-span" style="font-family: 'Helvetica Neue', Arial, 'Lucida Grande', 'Lucida Sans Unicode', sans-serif; font-size: 14px; line-height: 20px; display: inline; float: none;"><span class="Apple-style-span" style="font-family: 'times new roman', times; font-size: medium; display: inline; float: none; line-height: normal;"> <strong><em>A2</em></strong></span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica; min-height: 26.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> for roots being shown equidistant from the origin and one in each quadrant.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>A1</em></strong> for correct angular positions. It is not necessary to see written evidence of angle, but must agree with the diagram.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) \(\frac{{{z_2}}}{{{z_1}}} = \frac{{{2^{\frac{1}{8}}}\left( {\left( {\cos \frac{{15\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{15\pi }}{{16}}} \right)} \right)}}{{{2^{\frac{1}{8}}}\left( {\left( {\cos \frac{{7\pi }}{{16}}} \right) + {\text{i}}\sin \left( {\frac{{7\pi }}{{16}}} \right)} \right)}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {\text{i}}\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\text{(}} \Rightarrow a = 0,{\text{ }}b = 1)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong> </span></p>
<div class="question_part_label">Part A.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \((x - 1)({x^4} + {x^3} + {x^2} + x + 1)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {x^5} + {x^4} + {x^3} + {x^2} + x - {x^4} - {x^3} - {x^2} - x - 1\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {x^5} - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) <em>b</em> is a root</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(b) = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({b^5} = 1\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({b^5} - 1 = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((b - 1)({b^4} + {b^3} + {b^2} + b + 1) = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(b \ne 1\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(1 + b + {b^2} + {b^3} + {b^4} = 0\) as shown. <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) \(u + v = {b^4} + {b^3} + {b^2} + b = - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(uv = (b + {b^4})({b^2} + {b^3}) = {b^3} + {b^4} + {b^6} + {b^7}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Now \({b^5} = 1\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence \(uv = {b^3} + {b^4} + b + {b^2} = - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence \(u + v = uv = - 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \({(u - v)^2} = ({u^2} + {v^2}) - 2uv\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left( {{{(u + v)}^2} - 2uv} \right) - 2uv\,\,\,\,\,\left( { = {{(u + v)}^2} - 4uv} \right)\) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given \(u - v > 0\) </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(u - v = \sqrt {{{(u + v)}^2} - 4uv} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt {{{( - 1)}^2} - 4( - 1)} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt {1 + 4} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sqrt 5 \) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A0</em></strong> unless an indicator is given that \(u - v = - \sqrt 5 \) is invalid.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[8 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [13 marks]</em></strong></span></p>
<div class="question_part_label">Part B.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 18.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The response to Part A was disappointing. Many candidates did not know that they had to apply de Moivre’s theorem and did not appreciate that they needed to find four roots.</span></p>
<div class="question_part_label">Part A.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Helvetica;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part B started well for most candidates, but in part (b) many candidates did not appreciate the significance of b not lying on the real axis. A majority of candidates started (c) (i) and many fully correct answers were seen. Part (c) (ii) proved unsuccessful for all but the very best candidates.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px Helvetica;"> </p>
<div class="question_part_label">Part B.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider \(\omega= \cos \left( {\frac{{2\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{2\pi }}{3}} \right)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that</span></p>
<p style="margin: 0px 0px 0px 30px; font: 31px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> (i) \({\omega^3} = 1;\)</span></p>
<p style="margin: 0px 0px 0px 30px; font: 31px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) \(1 + \omega+ {\omega^2} = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) Deduce that \({{\text{e}}^{{\text{i}}\theta }} + {{\text{e}}^{{\text{i}}\left( {\theta + \frac{{2\pi }}{3}} \right)}} + {{\text{e}}^{{\text{i}}\left( {\theta + \frac{{4\pi }}{3}} \right)}} = 0\).</span></p>
<p style="margin: 0px 0px 0px 30px; font: 31px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) Illustrate this result for \(\theta = \frac{\pi }{2}\) on an Argand diagram.</span><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) Expand and simplify \(F(z) = (z - 1)(z - \omega)(z - {\omega^2})\) where <em>z</em> is a complex number.</span></p>
<p style="margin: 0px 0px 0px 30px; font: 31px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> (ii) Solve \(F(z) = 7\), giving your answers in terms of \(\omega\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) (i) \({\omega^3} = {\left( {\cos \left( {\frac{{2\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{2\pi }}{3}} \right)} \right)^3}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \cos \left( {x \times \frac{{2\pi }}{3}} \right) + {\text{i}}\sin \left( {3 \times \frac{{2\pi }}{3}} \right)\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \cos 2\pi + {\text{i}}\sin 2\pi \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(1 + \omega+ {\omega^2} = 1 + \cos \left( {\frac{{2\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{2\pi }}{3}} \right) + \cos \left( {\frac{{4\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{4\pi }}{3}} \right)\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 1 + - \frac{1}{2} + {\text{i}}\frac{{\sqrt 3 }}{2} - \frac{1}{2} - {\text{i}}\frac{{\sqrt 3 }}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 0\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) \({{\text{e}}^{{\text{i}}\theta }} + {{\text{e}}^{{\text{i}}\left( {\theta + \frac{{2\pi }}{3}} \right)}} + {{\text{e}}^{{\text{i}}\left( {\theta + \frac{{4\pi }}{3}} \right)}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{{\text{i}}\theta }} + {{\text{e}}^{{\text{i}}\theta }}{{\text{e}}^{{\text{i}}\left( {\frac{{2\pi }}{3}} \right)}} + {{\text{e}}^{{\text{i}}\theta }}{{\text{e}}^{{\text{i}}\left( {\frac{{4\pi }}{3}} \right)}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left( {{{\text{e}}^{{\text{i}}\theta }}\left( {1 + {{\text{e}}^{{\text{i}}\left( {\frac{{2\pi }}{3}} \right)}} + {{\text{e}}^{{\text{i}}\left( {\frac{{4\pi }}{3}} \right)}}} \right)} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{{\text{i}}\theta }}(1 + \omega+ {\omega^2})\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 0\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><img 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" alt><span class="Apple-style-span" style="font-family: 'Helvetica Neue', Arial, 'Lucida Grande', 'Lucida Sans Unicode', sans-serif; font-size: 14px; line-height: 20px; display: inline; float: none;"><span class="Apple-style-span" style="font-family: 'times new roman', times; font-size: medium; display: inline; float: none; line-height: normal;"> <strong><em>A1A1</em></strong></span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> for one point on the imaginary axis and another point marked with approximately correct modulus and argument. Award <strong><em>A1</em></strong> for third point marked to form an equilateral triangle centred on the origin.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) attempt at the expansion of at least two linear factors <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\((z - 1){z^2} - z(\omega+ {\omega^2}) + {\omega^3}\) or equivalent <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">use of earlier result <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(F(z) = (z - 1)({z^2} + z + 1) = {z^3} - 1\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) equation to solve is \({z^3} = 8\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z = 2,{\text{ }}2\omega,{\text{ }}2{\omega^2}\) <strong><em>A2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>A1</em></strong> for 2 correct solutions.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>Total [16 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Most candidates were able to make a meaningful start to part (a) with many fully correct answers seen. Part (b) was the exact opposite with the majority of candidates not knowing what was required and failing to spot the connection to part (a). Candidates made a reasonable start to part (c), but often did not recognise the need to use the result that \(1 + \omega+ {\omega ^2} = 0\). This meant that most candidates were unable to make any progress on part (c) (ii).</span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">The graph of a polynomial function <em>f </em>of degree 4 is shown below.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><img 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eYxPul88j/ffX3vxm+ZzwLv8OGD/YWXV2wy8fYr1XtPvfOFv3BNgySE6J50a5UqyzSBnRCkRfKMhhuzR/P3/acPHh+/mwkhxHx2+tv9thfM5teNuXv9Sxv8fHr8Ze3hi8HkUrat5b2/elGdvXu5+ENsMD97s/f5/s2bVH2DFARBjdM+gB0RpEXyNLKFZV3zi8Hze3eeDy5kMn4aHP/L6ew6H/OzN3t35WTGBvPg9ReVB4eDn4QQQvwSvP6q9vETLn1+rDc/e7N39157cLHwcwtByn5QdZ0PE+9v94//K5ht+TdrNhorBn8BrEeQFvU978bfxqoXwQU/9p98tv5XpV+C11/VKs/6k6tq3eyTEBeDwzt3v3h9RpE2ujz3/umw/RdL8b4K0s1B1e3mk++/eXj3/pZ3Elf/6a3BViFmZ/323r0nvbONTZLfTDxTC4iPIC1aXtEw8fYrn+2ve9t9MTi8s5CW1X4atB/cuIuaePuVxQL92H/yWe2JN0l77Rabz85+f/jdu9nF4PDOV2+Ci3PvVS/4RVwF6cagapzP9r7/9PNapRojSOsHW8XPp8dfLs1oLZGrBjk0HoiPIC2qO87iioYPp8f3K49enq58qbs8935zf+MrkhDiqjfLQVp8MbxVLCyYB6+/qNx9LOft5N/Vna9fXd+qfLL7oOflufeb+w8f/VmMIG0cbI31foTnmgM7IUgR+Y52cUXDhiDNz73W3qvTbfMIa4K0eIc0Oz1+RJCS+WTboOqS+eyst7/36jT49/3tQdo22CrzuPc6WP89wBOSgJ0QpMitFQ0ySKveBM9+eNM+HkwuY3zW5TfWtz4nQUruky2Dqktm717uPXtzdnHrLnWlrYOtHybes8VirdRsNHj2MBATQYosr2gQa+aQ5ueDf/z2zVm89+RbB36YQ0rhk42Dqkt+Pj3+zTeD87m4PWy60tbB1vnF4Pm9bV9dflfxLBMgDoIUWXFGw+0BoeUaXU4Gr15uXtcwf99/+vm9J72z2Qe5GfbGxiNW2aWwFKTZ6fGjtQe1PvxyP1r4kDxIN/6p4uSQpQ1AfAQpsurx0zenzOfng+d7yy908lfleWsPX6wex5ud9dt7V4c7eGezdZ8fO/pk7aDqsh/7Tz5bFaoNA25bB1tjBUmwtAGIjSBJ8o1s3/OWf2GHRcXzi8GL3bZmclJDOjvOIUVi3SFtHWyNNYckWNoAxEaQpNsrGj6a/fDmyaPt2y7n7/u/bvfP46x0EEKI+WT46skjzrJLY9dVdtfiBWnLYGuMVXYRTm0A4iBIkpx7XnsY5raDaeaTd73j3/aD+K+LnPadgQT7kIQQS0GSo3l7V+sdlqwdbI25L/oKpzYAcRAkqe26PFPNOIlOaljpYnD4dKeVjttPaljEAymAOAiS1Gw0bq1ogO6uD1eNOai61uW51/7L+FN588n/fPf1/W1n2S3hgRTAVgRJXL9/XbGiAXrL4rTv+eT0X1+9XBqO2yTuad9L5KNN+CYDNiBI4vrhNatXNEBj+j4PaSXGhYHNCJLYuqIBujIsSMPhkPXfwAYESQhx1OnwztVEhgVJsP4b2IggCQ7ANJZ5QZI344wOAysRJFY0mMu8IIVhuOqIKgBCEKTrFQ2j0Uj1hWBn5gVJXK//5vxv4DaCxPMBzGVkkOSxiWySBW4jSN2TLmcxG8rIIAk2yQJrEKSDVosVDYYyNUhymJh5S2AJQeKVwVymBkkI0XbduuNwkwQsKnmQWNFgNIODxE2SzoIgGI1Gfc9b+TEcDoMgYNo5DyUPku/7PBbAXAYHSVzfJKm+CgghxHg89n1f7pBf9WTh1R91x2m7bt/z2FiWlZIHSc4uq74KJGR2kDhJSLnxePy214siJLeI9T1vNBptaMx4PA6CoO953ZPuQau1+HuHwyHDsGmUPEisaDCa2UESQjQbDQ6tKl4YhsPhMOrQQavl+37icRL52eRqXVmm7kmXUZdkSh6kWqX6ttdTfRVIyPggcZNUsDAM+54ny9FsNHzfz3YqaDQaHXU68lW17br8y+6qzEGST6jhe8ZcxgdJcJNUlMUUtV0314VMS9njJSa+MgdJvj3l3tpcNgSJm6QCDIfDKEVFLkCIBgYPWi0WPsRR5iC97fVK+2e3gw1BEtwk5Wk6nbZdV96pqNreEd0tHXU6LBbfrMxBarsuKxqMZkmQ5E0Se5Iy5/u+LIHyv9swDOX737rj+L6v9mJ0VuYgccSl6SwJkhCi2WhwcEOGwjCUiwsOWi19BuXH47FcJt52XW6VViptkOSKBt6sGM2eIHFwQ4bG47GcudFzBa18vkDdcZg4vK20QRqNRjy603T2BElwul1G5PqFuuPofCBYdKvUPenyL76otEGSb1NUXwVSsSpI8iZJzzf1ppCzcVoN060ThqE8J+ag1WL4LlLaILVd96DVUn0VSMWqIAkh5HowXp6SiV7fDbrnkAWtOw5jNVJpgyQP+FB9FUjFtiDJh8kedTqqL8Q8skYmjoCNx2O5FJApJVHWIMn/8ZlCNp1tQRLXL6y8X95JVCPVF5JQGIZySokB23IGiRUNdrAwSGEY1h2H0eT45GywuTWSwjCUA7am/0FSKmeQ5PewcTf3WGJhkMT1dycDOHFEqxhUX0g2TL/VS6+cQZIP4lJ9FUjLziAJ9snGI/cSmrWKYSvZpKNOx6Y/VHzlDFKz0WDm2ALWBokl4FvJeZe649i3KNHE5YJZKWGQwjBkRYMdrA2SuF4Crv9+GlXk0XC2DmxGQ5Fla1IJgyTffeq8jxsx2Ryk6XRadxxO/11Jrkqye5SjnE0qYZDknLF9N/olZHOQxPV3KuctLpGpbjYa1r9Sl7BJJQxS96RbdxzVV4EMWB4kweqGVeSWnZIMZkZHOZTkz1vCIB20WgyE2MH+IMnxZbvHpnYip45Kddc4Go3kcbFlaFIJg8SKBmvYHyRx/RLMnKcox9TRSvJ4oTI0qWxBku84bV2bUzalCFIYhgzciTJNHa1UkiaVLUi+75dn/Nl6pQiSYOBu4bS3Mv+vGzXJ4jfUZQuSHP9QfRXIRlmCJGzfdrOV3Cta2j9+JHoYrq1/FWULUtt1WdFgjRIFKRq4K+F+BTmsUeYT3hZFN4tWNqlsQeJAFpuUKEhi4eg21RdSKLmQoWx/6s0sblKpgiT/jy7VklG7lStI4nqrbHneUslZk9IuZNjA1kcolSpIPAbJMqULkrg+464Mq8Dlo6GsX1eWmJWPUCpVkOT7S9VXgcyUMUjRy7Tdk0nRHQDvHzez7BFKpQpS23UZi7ZJGYMkLH0O0BJb50jyYFOTShWkuuOUeS+HfUoaJGH7wjP5Cst5KvFZ8wil8gRpOp3yTW6Z8gZJXL8G2bdEx6b3+0WSExKmN6k8QZK73RmRtkmpgyRsHNeiRmlY8LiK8gRJvoEw918Kt5U9SNFuWTvWoVGj9KImGbrmpTxBOup0eAySZcoeJGHRmZvUKCtGP0KpPEFqNhocGmQZgiTEQpMMfVMsqFHWgiAw9G1KSYIUhiErGuxDkK5E5+uYOCRNjfJg6K1zSYIkVzSUYXt7qRCkj0yc0LbyrAF9mPi4ipIESa5oMHdIAysRpBuiJhnxptjW09i0YtzjKkoSJFY0WIkgLTNlQlu+eTfohdJcZh0NXpIgNRsNzmiwD0FaIRqo0XbP3XA41PwKLRM1Sf9t1GUIEmc02Iograbz/Ud0yI3m93CWiZqk+XRdGYLEUydsRZDW0vAFaDweR5dk0MoLa4RhqP+CxjIEiTMabEWQNolegHTYt+/7vhym03/UyG6aN6kMQeKpE7YiSNspL8F4PJZruxmm04TOR4OXIUisLLUVQYolGitru26Rt0phGMrRibrjMIWrFW2PBrc+SHJLrIaTu0iPIO1AvgbJ5T0FvAzJO7PiK4iY9NxJbX2Q2BJrMYK0m+l0KkfP5C1LHq9EYRgOh0O5GbPZaLCUSGeySc1GQ5+hVOuDdNTpNBsN1VeBXBCkJIIgWMxSVm/WptPp215P3hU1Gw0GJYwQ7QnTpEnWB6nuONquKEFKBCm5KEu1SvWo0xkOh8lumKbTqe/7co5KDtBxZKRZtDqG1e4gjcdjJpAsRpDSkrc1coRNzii87fVGo9Hm16YgCHzf7550o9/YbDTe9nqMjBsqapLyNxN2B8n3fSaQLEaQMjMej/ueF90zRR9t140+otugxV/1fV+Hd9ZISZNjWO0OEhNIdiNIuQiCYDgc9j3vba+3GKS26/Y9r+95W2+hYCIdjmG1O0g6b0lGegQJyFLUJFX7xiwOEjuQrEeQgIypfWqixUF62+txhJ3dCBKQC1VH3lkcpINWiyPs7EaQgLzIJh11OkW+qbc1SPIZSBxhZzeCBORIjjIVebyQrUGSh2JwcIndCBKQr4KPvLM1SEedTt1xVF8F8kWQgNwVeeSdrUHixKAyIEhAEQo78s7KIMlnlrPg23oECShIMUfeWRkkuTyEBd/WI0hAcWSTcn2zb2WQ6o5z1OmovgrkjiABhYqePpxTk+wLEid8lwdBAooWHS/k+37mn9y+IHFAQ3kQJECBqEmZrxyzL0iM15UHQQKUyeN4IcuCxPq6UiFIgEqySW3XzWpIyrIgsb6uVAgSoFi2xwtZFiTG60qFIAHqZXi8kE1Bkn8typ8Kj8IQJEAL8sU3/bZZm4LUdl3OrysVggToIgiC9Ec5WBMknjdRQgQJ0Eh0vFDicSprgtT3vFqlWsBxtNAHQQL0Mh6Pm41G4rXO1gSp2WjwfNiyIUiAdqJtswmaZEeQ5Iwa24/KhiABOkp8lIMdQWI5QzkRJEBTYRgmOMrBgiDJ01T7nqf6QlA0ggRobdejHCwIkvwjT6dT1ReCohEkQHdyvVnMbbOmB0mu9uZp5eVEkAADREc5bF0GbXqQuD0qM4IEmGE4HMbZNmt0kOTtUdt1VV8I1CBIgDHibJs1Okjy9igIAtUXAjUIEmCSrdtmzQ0St0cgSIBhoi1KK895MzdI3B6BIAHmCcPwqNNZuRrN0CDJvUcsris5ggSYSt5SLC0HNzRIbddlcR0IEmAw3/eXloObGKTRaMTRDBAECTDd0nJw44IUhmGz0Wg2Gpk8wR1GI0iA8eRy8FqlKieWapWqQS/ub3s9nlMOiSABNphOp3/+xReyRgatDgiCgKXeiBAkYHez4A/Hz1rejyqvYeK1nrz6Q3AR/cTf/s3fREEyYtRODtbVHYe1DJAIErCb+eT7bx4+2P9uOJnLHw5fPfm8VqnWHrZ7p/LnkrmcvOvu36nWKtXana9ffh/MPn7F6y9Rufu4/fvTyaUQQoiLwHv++M7TN2dXTZJnsMqP//vggf4P/5aDdTyFDxGCBOxi/r7/9MHj43dXtZi9e/nws8fPv5/MLyeDF4/v/KZ/fpns885Of3coIzc/Hzzfq1W+ehP8cvUlnnzbDy4+Fuvhq9PZ/Pp3vYq+aBiG8iW+Vqk6f/qnm08YUk4eF3vU6ai+EGiEIAHxzS8Gz+/deT64kD34JXj9VS364fzszd7de+3BxcZPEcvE2688OBz8JMT8YtB9dfrz9S/8Erz+qlZ59PI0un36adB+sPhFZZCiE4b0XEstV2Gwsg5LCBIQ21Jy5mdv9u7WnniTq1++2acUX+Zi8Pz+U+98xaf5MPGe3fwS84vB83vR7dTCsu8wDOVu0/gP9yuGPPpo67HlKCGCBMQ28fYrn+1HaxkuBod3qvePTz9c/fjDxHtWW2hDIhfB96/2P11XteX7odtXtbSiQU4s1R1HkzPiooP4mDrCbQQJiOvD6fH9xeGyibdfuR2kxfG0nb/Ay0+vl8mtmo6an3utT2/9/M0u3l5iFwSB3KWkw/CdPO7I933VFwIdESQgrnhBWn+HtNib5Y9n/cnVp5lPTvvHX9+rVJfvhObv+0//4uXH+aQbn3ZDkMTC8N1Bq6VwjbWskSl7pFA8ggTE9eH0+P7VWgMhxO1VDLPT40dZzCHdnp0SQlycvf7228HKeaXtQZLkwTiuW0UAAAKVSURBVHd1xyn+BiUMQ2qErQgSENvSHFJ+q+zET4P2rz4uLheXk8Hx4esfVg8FbpxDWjIej+UUTtt1C7tViuaNqBE2I0hAbLeSMz/3Wnc+33/9w+xqm2rifUiLLieDF48fvhhcbYBdrtF88v23x/99fQ1rV9ltEG2h7Xte3gvwtj7iFogQJCC+pX1IQoj5LPAOH96tVaq1h8/70UE+E2+/Uq19jMpWP58ef3k9n7R3+Po/Tz/W6MXj5QmnhWHDNfuQtn696XQqZ5WajUZ+qYjW+Om8RRf6IEjALpZOatj4n14MXuR83t2NkxqknQ6yG41G8vYl8ywFQVD82CBMR5CAHc1+ePPkUXSW3Vrz9/1ft7MYwVvncvKuu//px7PspAQnqw6HwyhLvu+nHMSbTqdy/YKS1RMwGkECdrfttO/55F3v+Lf9IIv1DevcOu1bSnzUd5QlecTccDjctUyj0UgOA8r1C1odDwEjECTAKimfPREEQfekKzfSygG3vueNRqOVx/yEYRgEge/7R52O/C11x3nb6zFGh2QIEmCVrB6GNBqN3vZ6ch5o8aPZaLRdNypW9JPdky4rF5ASQQKskvnT+aLboL7n9T3vqNNpu+7bXk/+MAgC7oeQFYIEWMWUx8UCtxEkwCoECeYiSIBVCBLMRZAAqxAkmIsgAVYhSDAXQQKsQpBgLoIEWIUgwVwECbAKQYK5CBJgFYIEcxEkwCoECeYiSIBVCBLMRZAAqxAkmIsgAVYhSDAXQQKsQpBgLoIEWIUgwVwECbAKQYK5CBJgFYIEcxEkwCoECeYiSIBVCBLMRZAAqxAkmIsgAVYhSDAXQQKsQpBgLoIEWIUgwVwECbAKQYK5CBJgFYIEcxEkwCoECeb6XyjPnJTOQ7/xAAAAAElFTkSuQmCC" alt></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \({(x + {\text{i}}y)^2} = - 5 + 12{\text{i}},{\text{ }}x,{\text{ }}y \in \mathbb{R}\) . Show that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \({x^2} - {y^2} = - 5\) ;</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) \(xy = 6\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence find the two square roots of \( - 5 + 12{\text{i}}\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">For any complex number <em>z </em>, show that \({(z^*)^2} = ({z^2})^*\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence write down the two square roots of \( - 5 - 12{\text{i}}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">A.d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Explain why, of the four roots of the equation \(f(x) = 0\) , two are real and two are complex.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">The curve passes through the point \(( - 1,\, - 18)\) . Find \(f(x)\) in the form</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(x) = (x - a)(x - b)({x^2} + cx + d),{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}\)<em> </em>.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the two complex roots of the equation \(f(x) = 0\) in Cartesian form.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">B.c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Draw the four roots on the complex plane (the Argand diagram).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">B.d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Express each of the four roots of the equation in the form \(r{{\text{e}}^{{\text{i}}\theta }}\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">B.e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \({(x + {\text{i}}y)^2} = - 5 + 12{\text{i}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({x^2} + 2{\text{i}}xy + {{\text{i}}^2}{y^2} = - 5 + 12{\text{i}}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) equating real and imaginary parts <strong><em>M1<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({x^2} - {y^2} = - 5\) <strong> <em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(xy = 6\) <strong> <em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[2 marks]</span><br></em></strong></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">substituting <strong><em>M1<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>EITHER<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({x^2} - \frac{{36}}{{{x^2}}} = - 5\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({x^4} + 5{x^2} - 36 = 0\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({x^2} = 4,\, - 9\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = \pm 2\) and \(y = \pm 3\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>OR<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{36}}{{{y^2}}} - {y^2} = - 5\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({y^4} - 5{y^2} - 36 = 0\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({y^2} = 9,\, - 4\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({y^2} = \pm 3\)<strong> and</strong> \(x = \pm 2\) <strong> <em>(A1)</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept solution by inspection if completely correct.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>THEN<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">the square roots are \((2 + 3{\text{i}})\) and \(( - 2 - 3{\text{i}})\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[5 marks]</span><br></em></strong></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER<br></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">consider \(z = x + {\text{i}}y\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z^* = x - {\text{i}}y\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(z^*)^2} = {x^2} - {y^2} - 2{\text{i}}xy\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(({z^2}) = {x^2} - {y^2} + 2{\text{i}}xy\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(({z^2})^* = {x^2} - {y^2} - 2{\text{i}}xy\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(z^*)^2} = ({z^2})^*\) <strong> <em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR<br></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(z^* = r{{\text{e}}^{ - {\text{i}}\theta }}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(z^*)^2} = {r^2}{{\text{e}}^{ - 2{\text{i}}\theta }}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({z^2} = {r^2}{{\text{e}}^{2{\text{i}}\theta }}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(({z^2})^* = {r^2}{{\text{e}}^{ - 2{\text{i}}\theta }}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(z^*)^2} = ({z^2})^*\) <strong> <em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[3 marks]</span><br></em></strong></p>
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: 'times new roman', times; font-size: medium;">\((2 - 3{\text{i}})\) and \(( - 2 + 3{\text{i}})\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[2 marks]</span><br></em></strong></p>
<div class="question_part_label">A.d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">the graph crosses the <em>x</em>-axis twice, indicating two real roots <strong><em>R1<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">since the quartic equation has four roots and only two are real, the other two roots must be complex <em><strong>R1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(x) = (x + 4)(x - 2)({x^2} + cx + d)\) <strong> <em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(0) = - 32 \Rightarrow d = 4\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Since the curve passes through \(( - 1,\, - 18)\)<span style="font: 11.5px Times;">,<br></span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( - 18 = 3 \times ( - 3)(5 - c)\) <strong> <em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.5px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(c = 3\) <strong><em>A1<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence \(f(x) = (x + 4)(x - 2)({x^2} + 3x + 4)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[5 marks]</span><br></em></strong></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(x = \frac{{ - 3 \pm \sqrt {9 - 16} }}{2}\) <strong> <em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow x = - \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[2 marks]</span><br></em></strong></p>
<div class="question_part_label">B.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em><img 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" alt> A1A1<br></em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong> </strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept points or vectors on complex plane.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Award <strong><em>A1 </em></strong>for two real roots and <strong><em>A1 </em></strong>for two complex roots.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[2 marks]</span><br></em></strong></p>
<div class="question_part_label">B.d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">real roots are \(4{{\text{e}}^{{\text{i}}\pi }}\) and \(2{{\text{e}}^{{\text{i}}0}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">considering \( - \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(r = \sqrt {\frac{9}{4} + \frac{7}{4}} = 2\) <strong> <em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">finding \(\theta \) using \(\arctan \left( {\frac{{\sqrt 7 }}{3}} \right)\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\theta = \arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi {\text{ or }}\theta = \arctan \left( { - \frac{{\sqrt 7 }}{3}} \right) + \pi \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi } \right)}}{\text{ or}} \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{ - \sqrt 7 }}{3}} \right) + \pi } \right)}}\) <strong><em>A1</em></strong></span><strong style="font-family: 'times new roman', times; font-size: medium;"> </strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept arguments in the range \( - \pi {\text{ to }}\pi {\text{ or }}0{\text{ to }}2\pi \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Accept answers in degrees.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Times;"><strong><em><span style="font-family: 'times new roman', times; font-size: medium;">[6 marks]</span><br></em></strong></p>
<div class="question_part_label">B.e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.</span></p>
<div class="question_part_label">A.a.</div>
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<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.</span></p>
<div class="question_part_label">A.b.</div>
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<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.</span></p>
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.</span></p>
<div class="question_part_label">A.d.</div>
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<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial;"><span style="font-family: 'times new roman', times; font-size: medium;">In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the <em>x</em>-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of <em>a </em>and <em>b </em>correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.</span></p>
<div class="question_part_label">B.a.</div>
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<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the <em>x</em>-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of <em>a </em>and <em>b </em>correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.</span></p>
<div class="question_part_label">B.b.</div>
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<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the <em>x</em>-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of <em>a </em>and <em>b </em>correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.</span></p>
<div class="question_part_label">B.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the <em>x</em>-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of <em>a </em>and <em>b </em>correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.</span></p>
<div class="question_part_label">B.d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 10.0px Arial; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the <em>x</em>-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of <em>a </em>and <em>b </em>correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.</span></p>
<div class="question_part_label">B.e.</div>
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