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</div><h2>HL Paper 3</h2><div class="specification">
<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 integral \({I_n}\) is defined by \({I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x,{\text{ for }}n \in \mathbb{N}} \) .</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: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \({I_0} = \frac{1}{2}(1 + {{\text{e}}^{ - \pi }})\) .</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 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;">By letting \(y = x - n\pi \) , show that \({I_n} = {{\text{e}}^{ - n\pi }}{I_0}\) .</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: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence determine the exact value of \(\int_0^\infty {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} \) .</span></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 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;">\({I_0} = \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x} \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><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 \({I_0} = \int_0^\pi {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} \)</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;">Attempt at integration by parts, even if inappropriate modulus signs are present. </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;">\( = - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\cos x{\text{d}}x} \) <strong>or</strong> \( = - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\cos x{\text{d}}x} \) <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;">\( = - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x} \) <strong>or</strong> \( = - \left[ {{{\text{e}}^{ - x}}\sin x + {{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x} \) <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;">\( = - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - {I_0}\) <strong>or</strong> \( - \left[ {{{\text{e}}^{ - x}}\sin x + {{\text{e}}^{ - x}}\cos x} \right]_0^\pi - {I_0}\) <strong><em>M1</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;"> Do not penalise absence of limits at this stage</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;">\({I_0} = {{\text{e}}^{ - \pi }} + 1 - {I_0}\) </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;">\({I_0} = \frac{1}{2}(1 + {{\text{e}}^{ - \pi }})\) <strong><em>AG</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;"> If modulus signs are used around cos </span><em style="font-family: 'times new roman', times; font-size: medium;">x</em><span style="font-family: 'times new roman', times; font-size: medium;"> , award no accuracy marks but do not penalise modulus signs around sin </span><em style="font-family: 'times new roman', times; font-size: medium;">x</em><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;"><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>[6 marks]</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: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} \)</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 use the substitution \(y = x - n\pi \) <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;">(putting \(y = x - n\pi \) , \({\text{d}}y = {\text{d}}x\) and \(\left[ {n\pi ,{\text{ }}(n + 1)\pi } \right] \to [0,{\text{ }}\pi ]\))</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;">so \({I_n} = \int_0^\pi {{{\text{e}}^{ - (y + n\pi )}}|\sin (y + n\pi )|{\text{d}}y} \) <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;">\( = {{\text{e}}^{ - n\pi }}\int_0^\pi {{{\text{e}}^{ - y}}|\sin (y + n\pi )|{\text{d}}y} \) <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;">\( = {{\text{e}}^{ - n\pi }}\int_0^\pi {{{\text{e}}^{ - y}}\sin y{\text{d}}y} \) <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;">\( = {{\text{e}}^{ - n\pi }}{I_0}\) <strong><em>AG</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>[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;">\(\int_0^\infty {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} = \sum\limits_{n = 0}^\infty {{I_n}} \) <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;">\( = \sum\limits_{n = 0}^\infty {{{\text{e}}^{ - n\pi }}{I_0}} \) <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 \(\sum \) term is an infinite geometric series with common ratio \({{\text{e}}^{ - \pi }}\) <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;">therefore</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;">\(\int_0^\infty {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} = \frac{{{I_0}}}{{1 - {{\text{e}}^{ - \pi }}}}\) <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;">\( = \frac{{1 + {{\text{e}}^{ - \pi }}}}{{2(1 - {{\text{e}}^{ - \pi }})}}{\text{ }}\left( { = \frac{{{{\text{e}}^\pi } + 1}}{{2({{\text{e}}^\pi } - 1)}}} \right)\) <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;"><strong><em>[5 marks]</em></strong></span></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 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;">Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.</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;">Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.</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;">Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">In this question you may assume that \(\arctan x\) is continuous and differentiable for \(x \in \mathbb{R}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Consider the infinite geometric series</p>
<p>\[1 - {x^2} + {x^4} - {x^6} + \ldots \;\;\;\left| x \right| < 1.\]</p>
<p>Show that the sum of the series is \(\frac{1}{{1 + {x^2}}}\).</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>Hence show that an expansion of \(\arctan x\) is \(\arctan x = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + \ldots \)</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">\(f\) is a continuous function defined on \([a,{\text{ }}b]\) and differentiable on \(]a,{\text{ }}b[\) with \(f'(x) > 0\) on \(]a,{\text{ }}b[\).</p>
<p class="p1">Use the mean value theorem to prove that for any \(x,{\text{ }}y \in [a,{\text{ }}b]\), if \(y > x\) then \(f(y) > f(x)\).</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 class="p1">(i) <span class="Apple-converted-space"> </span>Given \(g(x) = x - \arctan x\), prove that \(g'(x) > 0\), for \(x > 0\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Use the result from part (c) to prove that \(\arctan x < x\), for \(x > 0\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use the result from part (c) to prove that \(\arctan x > x - \frac{{{x^3}}}{3}\), for \(x > 0\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence show that \(\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(r = - {x^2},\;\;\;S = \frac{1}{{1 + {x^2}}}\) <strong><em>A1AG</em></strong></p>
<p><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{1}{{1 + {x^2}}} = 1 - {x^2} + {x^4} - {x^6} + \ldots \)</p>
<p><strong>EITHER</strong></p>
<p>\(\int {\frac{1}{{1 + {x^2}}}{\text{d}}x} = \int {1 - {x^2} + {x^4} - {x^6} + \ldots } {\text{d}}x\) <strong><em>M1</em></strong></p>
<p>\(\arctan x = c + x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + \ldots \) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Do not penalize the absence of <em>\(c\) </em>at this stage.</p>
<p> </p>
<p>when \(x = 0\) we have \(\arctan 0 = c\) hence \(c = 0\) <strong><em>M1A1</em></strong></p>
<p><strong>OR</strong></p>
<p>\(\int_0^x {\frac{1}{{1 + {t^2}}}{\text{d}}t = } \int_0^x {1 - {t^2} + {t^4}} - {t^6} + \ldots {\text{d}}t\) <strong><em>M1A1A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Allow <em>\(x\) </em>as the variable as well as the limit.</p>
<p><strong><em>M1 </em></strong>for knowing to integrate, <strong><em>A1 </em></strong>for each of the limits.</p>
<p> </p>
<p>\([\arctan t]_0^x = \left[ {t - \frac{{{t^3}}}{3} + \frac{{{t^5}}}{5} - \frac{{{t^7}}}{7} + \ldots } \right]_0^x\) <strong><em>A1</em></strong></p>
<p>hence \(\arctan x = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + \ldots \) <strong><em>AG</em></strong></p>
<p><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">applying the \(MVT\) to the function \(f\) on the interval \([x,{\text{ }}y]\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\(\frac{{f(y) - f(x)}}{{y - x}} = f'(c)\;\;\;({\text{for some }}c \in ]x,{\text{ }}y[)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\(\frac{{f(y) - f(x)}}{{y - x}} > 0\;\;\;({\text{as }}f'(c) > 0)\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">\(f(y) - f(x) > 0{\text{ as }}y > x\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">\( \Rightarrow f(y) > f(x)\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>If they use <em>\(x\) </em>rather than \(c\) they should be awarded <strong><em>M1A0R0</em></strong>, but could get the next <strong><em>R1</em></strong>.</p>
<p class="p1"><em><strong>[4 marks]</strong></em></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"> </span>\(g(x) = x - \arctan x \Rightarrow g'(x) = 1 - \frac{1}{{1 + {x^2}}}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">this is greater than zero because \(\frac{1}{{1 + {x^2}}} < 1\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">so \(g'(x) > 0\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>(\(g\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on \(]0,{\text{ }}b[\) with \(g'(x) > 0\) on \(]0,{\text{ }}b[\) for all \(b \in \mathbb{R}\))</p>
<p class="p1"><span class="s1">(If </span>\(x \in [0,{\text{ }}b]\) then) from part (c) \(g(x) > g(0)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\(x - \arctan x > 0 \Rightarrow \arctan x < x\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">(as <em>\(b\) </em>can take any positive value it is true for all <span class="s2">\(x > 0\)) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></span></p>
<p class="p2"><span class="s2"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">let \(h(x) = \arctan x - \left( {x - \frac{{{x^3}}}{3}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">(\(h\) is a continuous function defined on \([0,{\text{ }}b]\) and differentiable on <span class="s1">\(]0,{\text{ }}b[\) with \(h'(x) > 0\) on \(]0,{\text{ }}b[\))</span></p>
<p class="p1">\(h'(x) = \frac{1}{{1 + {x^2}}} - (1 - {x^2})\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( = \frac{{1 - (1 - {x^2})(1 + {x^2})}}{{1 + {x^2}}} = \frac{{{x^4}}}{{1 + {x^2}}}\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\(h'(x) > 0\) hence \(({\text{for }}x \in [0,{\text{ }}b]){\text{ }}h(x) > h(0)( = 0)\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">\( \Rightarrow \arctan x > x - \frac{{{x^3}}}{3}\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: </strong>Allow correct working with \(h(x) = x - \frac{{{x^3}}}{3} - \arctan x\).</p>
<p class="p1"><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>use of \(x - \frac{{{x^3}}}{3} < \arctan x < x\) <strong><em>M1</em></strong></p>
<p>choice of \(x = \frac{1}{{\sqrt 3 }}\) <strong><em>A1</em></strong></p>
<p>\(\frac{1}{{\sqrt 3 }} - \frac{1}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\) <strong><em>M1</em></strong></p>
<p>\(\frac{8}{{9\sqrt 3 }} < \frac{\pi }{6} < \frac{1}{{\sqrt 3 }}\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award final <strong><em>A1 </em></strong>for a correct inequality with a single fraction on each side that leads to the final answer.</p>
<p> </p>
<p>\(\frac{{16}}{{3\sqrt 3 }} < \pi < \frac{6}{{\sqrt 3 }}\) <strong><em>AG</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<p><strong><em>Total [22 marks]</em></strong></p>
<div class="question_part_label">f.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Most candidates picked up this mark for realizing the common ratio was \( - {x^2}\).</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Quite a few candidates did not recognize the importance of ‘hence’ in this question, losing a lot of time by trying to work out the terms from first principles.</p>
<p class="p1">Of those who integrated the formula from part (a) only a handful remembered to include the ‘\( + c\)’ term, and to verify that this must be equal to zero.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Most candidates were able to achieve some marks on this question. The most commonly lost mark was through not stating that the inequality was unchanged when multiplying by \(y - x{\text{ as }}y > x\).</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">The first part of this question proved to be very straightforward for the majority of candidates.</p>
<p class="p1">In (ii) very few realized that they had to replace the lower variable in the formula from part (c) by zero.</p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Candidates found this part difficult, failing to spot which function was required.</p>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Many candidates, even those who did not successfully complete (d) (ii) or (e), realized that these parts gave them the necessary inequality.</p>
<div class="question_part_label">f.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Prove by induction that \(n! > {3^n}\), for \(n \ge 7,{\text{ }}n \in \mathbb{Z}\).</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>Hence use the comparison test to prove that the series \(\sum\limits_{r = 1}^\infty {\frac{{{2^r}}}{{r!}}} \) converges.</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">if \(n = 7\) <span class="s1">then </span>\(7! > {3^7}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">so true for \(n = 7\)</p>
<p class="p1">assume true for \(n = k\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="s1">so </span>\(k! > {3^k}\)</p>
<p class="p1">consider \(n = k + 1\)</p>
<p class="p1">\((k + 1)! = (k + 1)k!\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( > (k + 1){3^k}\)</p>
<p class="p1">\( > 3.3k\;\;\;({\text{as }}k > 6)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( = {3^{k + 1}}\)</p>
<p class="p1">hence if true for \(n = k\) then also true for \(n = k + 1\). As true for \(n = 7\), so true for all \(n \ge 7\). <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>Do not award the <strong><em>R1 </em></strong>if the two <strong><em>M </em></strong>marks have not been awarded.</p>
<p class="p3"><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>consider the series \(\sum\limits_{r = 7}^\infty {{a_r}} \), where \({a_r} = \frac{{{2^r}}}{{r!}}\) <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award the <strong><em>R1 </em></strong>for starting at \(r = 7\)</p>
<p> </p>
<p>compare to the series \(\sum\limits_{r=7}^\infty {{b_r}} \) where \({b_r} = \frac{{{2^r}}}{{{3^r}}}\) <strong><em>M1</em></strong></p>
<p>\(\sum\limits_{r = 7}^\infty {{b_r}} \) is an infinite Geometric Series with \(r = \frac{2}{3}\) and hence converges <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award the <strong><em>A1 </em></strong>even if series starts at \(r = 1\).</p>
<p> </p>
<p>as \(r! > {3^r}\) so \((0 < ){a_r} < {b_r}\) for all \(r \ge 7\) <strong><em>M1R1</em></strong></p>
<p>as \(\sum\limits_{r = 7}^\infty {{b_r}} \) converges and \({a_r} < {b_r}\) so \(\sum\limits_{r = 7}^\infty {{a_r}} \) must converge</p>
<p> </p>
<p><strong>Note: </strong>Award the <strong><em>A1 </em></strong>even if series starts at \(r = 1\).</p>
<p> </p>
<p>as \(\sum\limits_{r = 1}^6 {{a_r}} \) is finite, so \(\sum\limits_{r = 1}^\infty {{a_r}} \) must converge <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>If the limit comparison test is used award marks to a maximum of <strong><em>R1M1A1M0A0R1</em></strong>.</p>
<p><em><strong>[6 marks]</strong></em></p>
<p><em><strong>Total [11 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: 34.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \(n! \geqslant {2^{n - 1}}\), for \(n \geqslant 1\).</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: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence use the comparison test to determine whether the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{n!}}} \) converges or diverges.</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: 34.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">for \(n \geqslant 1,{\text{ }}n! = n(n - 1)(n - 2) \ldots 3 \times 2 \times 1 \geqslant 2 \times 2 \times 2 \ldots 2 \times 2 \times 1 = {2^{n - 1}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 34.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow n! \geqslant {2^{n - 1}}{\text{ for }}n \geqslant 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 34.0px 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: 32.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(n! \geqslant {2^{n - 1}} \Rightarrow \frac{1}{{n!}} \leqslant \frac{1}{{{2^{n - 1}}}}{\text{ for }}n \geqslant 1\) <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;">\(\sum\limits_{n = 1}^\infty {\frac{1}{{{2^{n - 1}}}}} \) is a positive converging geometric series <strong><em>R1</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;">hence \(\sum\limits_{n = 1}^\infty {\frac{1}{{n!}}} \) converges by the comparison test <strong><em>R1</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>[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: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for <em>n</em> = 1, 2, 3 and therefore true for all <em>n</em>. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n - 1}}}}} \).</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: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) of this question was found challenging by the majority of candidates, a fairly common ‘solution’ being that the result is true for <em>n</em> = 1, 2, 3 and therefore true for all <em>n</em>. Some candidates attempted to use induction which is a valid method but no completely correct solution using this method was seen. Candidates found part (b) more accessible and many correct solutions were seen. The most common problem was candidates using an incorrect comparison test, failing to realise that what was required was a comparison between \(\sum {\frac{1}{{n!}}} \) and \(\sum {\frac{1}{{{2^{n - 1}}}}} \).</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br>