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</div><h2>SL Paper 1</h2><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(\sin \theta = \frac{2}{{\sqrt {13} }}\) , where \(\frac{\pi }{2} < \theta < \pi \) .</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 \(\cos \theta \) .</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><span style="font-family: times new roman,times; font-size: medium;">Find \(\tan 2\theta \) .</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><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of choosing \({\sin ^2}\theta + {\cos ^2}\theta = 1\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\cos ^2}\theta = \frac{9}{{13}}\) , \(\cos \theta = \pm \frac{3}{{\sqrt {13} }}\) , \(\cos \theta = \sqrt {\frac{9}{{13}}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\cos \theta = - \frac{3}{{\sqrt {13} }}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: If no working shown, award <em><strong>N1</strong></em> for \(\frac{3}{{\sqrt {13} }}\) .</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;">approach involving Pythagoras’ theorem <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({2^2} + {x^2} = 13\) , </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/jam.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding third side equals 3 <em><strong> (A1)</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\cos \theta = - \frac{3}{{\sqrt {13} }}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: If no working shown, award <em><strong>N1</strong></em> for \(\frac{3}{{\sqrt {13} }}\) .</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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution into \(\sin 2\theta \) (seen anywhere) <em><strong> (A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution into \(\cos 2\theta \) (seen anywhere) <em><strong> (A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - {\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\) , \(2{\left( { - \frac{3}{{\sqrt {13} }}} \right)^2} - 1\) , \(1 - 2{\left( {\frac{2}{{\sqrt {13} }}} \right)^2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">valid attempt to find \(\tan 2\theta \) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{2\left( {\frac{2}{{\sqrt {13} }}} \right)\left( { - \frac{3}{{\sqrt {13} }}} \right)}}{{{{\left( { - \frac{3}{{\sqrt {13} }}} \right)}^2} - {{\left( {\frac{2}{{\sqrt {13} }}} \right)}^2}}}\) , \(\frac{{2\left( { - \frac{2}{3}} \right)}}{{1 - {{\left( { - \frac{2}{3}} \right)}^2}}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{\frac{{(2)(2)( - 3)}}{{13}}}}{{\frac{9}{{13}} - \frac{4}{{13}}}}\) , \(\frac{{ - \frac{{12}}{{{{\left( {\sqrt {13} } \right)}^2}}}}}{{\frac{{18}}{{13}} - 1}}\) , \(\frac{{ - \frac{{12}}{{13}}}}{{\frac{5}{{13}}}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\tan 2\theta = - \frac{{12}}{5}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N4</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: If students find answers for \(\cos \theta \) which are not in the range \([ - 1{\text{, }}1]\), </span><span style="font-family: times new roman,times; font-size: medium;">award full <em><strong>FT</strong></em> in (b) for correct <em><strong>FT</strong></em> working shown.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [5 marks]</span></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><span style="font-family: times new roman,times; font-size: medium;">While the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle in the second quadrant will have a negative value. </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;">In part (b), many candidates incorrectly tried to calculate \(\tan 2\theta \) as \(2 \times \tan \theta \) , rather than using the double-angle identities. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = \frac{{\cos x}}{{\sin x}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> , for \(\sin x \ne 0\) .</span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">In the following table, \(f'\left( {\frac{\pi }{2}} \right) = p\) and \(f''\left( {\frac{\pi }{2}} \right) = q\) . The table also gives approximate values of \(f'(x)\) and \(f''(x)\) near \(x = \frac{\pi }{2}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/batman.png" alt></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;">Use the quotient rule to show that \(f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}\) .</span></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><span style="font-family: times new roman,times; font-size: medium;">Find \(f''(x)\) .</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;">Find the value of <em>p</em> and of <em>q</em>.</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Use information from the table to explain why there is a point of inflexion on the </span><span style="font-family: times new roman,times; font-size: medium;">graph of <em>f</em> where \(x = \frac{\pi }{2}\) .</span></p>
<div class="marks">[2]</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="color: #000000; font-family: Times New Roman; font-size: medium;"> <span style="color: #3f3f3f;">\(\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x\) </span></span><span style="font-family: times new roman,times; font-size: medium;">, \(\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = - \sin x\) (seen anywhere) <em><strong>(A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using the quotient rule <em><strong>M1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{\sin x( - \sin x) - \cos x(\cos x)}}{{{{\sin }^2}x}}\) , \(\frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(f'(x) = \frac{{ - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1</strong> </span></em></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>AG N0</strong> </span></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [5 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;"><strong>METHOD 1</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">appropriate approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(f'(x) = - {(\sin x)^{ - 2}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(f''(x) = 2({\sin ^{ - 3}}x)(\cos x)\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1A1 N3</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for \(2{\sin ^{ - 3}}x\) , <em><strong>A1</strong></em> for \(\cos x\) . </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">derivative of \({\sin ^2}x = 2\sin x\cos x\) (seen anywhere) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of choosing quotient rule <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(u = - 1\) , \(v = {\sin ^2}x\) , \(f'' = \frac{{{{\sin }^2}x \times 0 - ( - 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(f''(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N3</span></strong></em></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;">evidence of substituting \(\frac{\pi }{2}\) <em><strong>M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{ - 1}}{{{{\sin }^2}\frac{\pi }{2}}}\) , \(\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(p = - 1\) </span><span style="font-family: times new roman,times; font-size: medium;">, \(q = 0\) <em><strong>A1A1 N1N1</strong> </em></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;">second derivative is zero, second derivative changes sign <em><strong>R1R1 N2</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">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many candidates comfortably applied the quotient rule, although some did not completely </span><span style="font-family: times new roman,times; font-size: medium;">show that the Pythagorean identity achieves the numerator of the answer given. Whether </span><span style="font-family: times new roman,times; font-size: medium;">changing to \( - {(\sin x)^{ - 2}}\) , or applying the quotient rule a second time, most candidates </span><span style="font-family: times new roman,times; font-size: medium;">neglected the chain rule in finding the second derivative.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Whether </span><span style="font-family: times new roman,times; font-size: medium;">changing to \( - {(\sin x)^{ - 2}}\) , or applying the quotient rule a second time, most candidates </span><span style="font-family: times new roman,times; font-size: medium;">neglected the chain rule in finding the second derivative.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Those who knew the trigonometric </span><span style="font-family: times new roman,times; font-size: medium;">ratios at </span><span style="font-family: times new roman,times; font-size: medium;">\(\frac{\pi }{2}\)</span><span style="font-family: times new roman,times; font-size: medium;">typically found the values of <em>p</em> and of <em>q</em>, sometimes in follow-through from an </span><span style="font-family: times new roman,times; font-size: medium;">incorrect \(f''(x)\) .</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Few candidates gave two reasons from the table that supported the existence </span><span style="font-family: times new roman,times; font-size: medium;">of a point of inflexion. Most stated that the second derivative is zero and neglected to consider </span><span style="font-family: times new roman,times; font-size: medium;">the sign change to the left and right of <em>q</em>. Some discussed a change of concavity, but without </span><span style="font-family: times new roman,times; font-size: medium;">supporting this statement by referencing the change of sign in \(f''(x)\) , so no marks were </span><span style="font-family: times new roman,times; font-size: medium;">earned.</span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(f(t) = a\cos b(t - c) + d\) , \(t \ge 0\) . Part of the graph of \(y = f(t)\) is given below.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/evening.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">When \(t = 3\) , there is a maximum value of 29, at M.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> When \(t = 9\) , there is a minimum value of 15.</span></p>
<p> </p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the value of <em>a</em>.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Show that \(b = \frac{\pi }{6}\) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) Find the value of <em>d</em>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iv) Write down a value for <em>c</em>.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">a(i), (ii), (iii) and (iv).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The transformation <em>P</em> is given by a horizontal stretch of a scale factor of \(\frac{1}{2}\) </span><span style="font-family: times new roman,times; font-size: medium;">, followed </span><span style="font-family: times new roman,times; font-size: medium;">by a translation of \(\left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 10}<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Let \({M'}\) be the image of M under <em>P</em>. Find the coordinates of \({M'}\) .</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><span style="font-family: times new roman,times; font-size: medium;">The graph of <em>g</em> is the image of the graph of <em>f</em> under <em>P</em>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Find \(g(t)\) in the form \(g(t) = 7\cos B(t - c) + D\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">The graph of <em>g</em> is the image of the graph of <em>f</em> under <em>P</em>.</span></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Give a full geometric description of the transformation that maps the graph of <em>g </em></span><span style="font-family: times new roman,times; font-size: medium;">to the graph of <em>f</em> .</span></p>
<div class="marks">[3]</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) attempt to substitute <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(a = \frac{{29 - 15}}{2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(a = 7\) (accept \(a = - 7\) ) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\text{period}} = 12\) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = \frac{{2\pi }}{{12}}\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = \frac{\pi }{6}\) <em><strong>AG N0</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) attempt to substitute <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(d = \frac{{29 + 15}}{2}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(d = 22\) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iv) \(c = 3\) (accept \(c = 9\) from \(a = - 7\) ) <em><strong>A1 N1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Other correct values for <em>c</em> can be found, \(c = 3 \pm 12k\) , \(k \in \mathbb{Z}\) . </span></p>
<p><strong><em> <span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></em></strong></p>
<div class="question_part_label">a(i), (ii), (iii) and (iv).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">stretch takes 3 to 1.5 <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">translation maps \((1.5{\text{, }}29)\) to \((4.5{\text{, }}19)\) (so \({M'}\) is \((4.5{\text{, }}19)\)) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N2</strong> </span></em></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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(g(t) = 7\cos \frac{\pi }{3}\left( {t - 4.5} \right) + 12\) <strong><em>A1A2A1 N4</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for \(\frac{\pi }{3}\) , <em><strong>A2</strong></em> for 4.5, <em><strong>A1</strong></em> for 12.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Other correct values for <em>c</em> can be found, \(c = 4.5 \pm 6k\) , \(k \in \mathbb{Z}\) . </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 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;">translation \(\left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>{10}<br>\end{array}} \right)\) <em><strong> (A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">horizontal stretch of a scale factor of 2 <strong><em>(A1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">completely correct description, in correct order <strong><em>A1 N3</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. translation \(\left( {\begin{array}{*{20}{c}}<br>{ - 3}\\<br>{10}<br>\end{array}} \right)\) then horizontal stretch of a scale factor of 2</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">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 was the most difficult on the paper. Where candidates attempted this question, part (a) was answered satisfactorily. </span></p>
<div class="question_part_label">a(i), (ii), (iii) and (iv).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Few answered part (b) correctly as most could not interpret the horizontal stretch. </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;">Few answered part (b) correctly as most could not interpret the horizontal stretch. As a result, there were many who were unable to answer part (c) although follow through marks were often obtained from incorrect answers in both parts (a) and (b). The link between the answer in (b) and the value of <em>C</em> in part (c) was lost on all but the most attentive. </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;">In part (d), some candidates could name the transformations required, although only a handful provided the correct order of the transformations to return the graph to its original state. </span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Consider \(g(x) = 3\sin 2x\) . </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;">Write down the period of <em>g</em>.</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><span style="font-family: times new roman,times; font-size: medium;">On the diagram below, sketch the curve of <em>g</em>, for \(0 \le x \le 2\pi \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/cake.png" alt></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;">Write down the number of solutions to the equation \(g(x) = 2\) , for \(0 \le x \le 2\pi \) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</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;">\({\text{period}} = \pi \) <em><strong> A1 N1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p> <img src="images/cake2.png" alt><span style="font-family: times new roman,times; font-size: medium;"><em><strong> A1A1A1 N3</strong></em></span></p>
<p> </p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for amplitude of 3, <em><strong>A1</strong></em> for <strong>their</strong> period, </span><span style="font-family: times new roman,times; font-size: medium;"><em><strong>A1</strong></em> for a sine curve passing through \((0{\text{, }}0)\) and \((0{\text{, }}2\pi )\) .</span></p>
<p align="LEFT"><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of appropriate approach <em><strong> (M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. line \(y = 2\) on graph, discussion of number of solutions in the domain</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">4 (solutions) <em><strong>A1 N2</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">c.</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;">Many candidates were unable to write down the period of the function.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many candidates were unable to write down the period of the function. However, they were </span><span style="font-family: times new roman,times; font-size: medium;">often then able to go and correctly sketch the graph with the correct period.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The final part was </span><span style="font-family: times new roman,times; font-size: medium;">poorly done with many candidates finding the number of zeros instead of the intersection with </span><span style="font-family: times new roman,times; font-size: medium;">the line \(y = 2\) .</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = {(\sin x + \cos x)^2}\) .</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;">Show that \(f(x)\) can be expressed as \(1 + \sin 2x\) .</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;">The graph of <em>f</em> is shown below for \(0 \le x \le 2\pi \) .</span></p>
<p><span style="font-family: TimesNewRomanPSMT;"><br><img src="images/pirate.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(g(x) = 1 + \cos x\) . On the same set of axes, sketch the graph of <em>g</em> for </span><span style="font-family: times new roman,times; font-size: medium;">\(0 \le x \le 2\pi \) .</span></p>
<p> </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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The graph of <em>g</em> can be obtained from the graph of <em>f</em> under a horizontal stretch of </span><span style="font-family: times new roman,times; font-size: medium;">scale factor <em>p</em> followed by a translation by the vector \(\left( {\begin{array}{*{20}{c}}<br>k\\<br>0<br>\end{array}} \right)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Write down the value of <em>p</em> and a possible value of <em>k</em> .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</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;">attempt to expand <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \((\sin x + \cos x)(\sin x + \cos x)\) ; at least 3 terms </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct expansion <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\sin ^2}x + 2\sin x\cos x + {\cos ^2}x\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(f(x) = 1 + \sin 2x\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>AG N0</strong> </span></em></p>
<p><em> <span style="font-family: times new roman,times; font-size: medium;"><strong>[2 marks]</strong> </span></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;"><br><img src="images/wittering.png" alt></span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A1A1 N2</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for correct sinusoidal shape with period \(2\pi \) and range \([0{\text{, }}2]\), <em><strong>A1</strong></em> for minimum in circle. </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;">\(p = 2\) , \(k = - \frac{\pi }{2}\) <em><strong>A1A1 N2</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">c.</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;">Simplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidates were certainly helped by the fact that it was a "show that" 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;">More candidates had difficulty with part (b) with many assuming the first graph was \(1 + \sin (x)\) and hence sketching a horizontal translation of \(\pi /2\) for the graph of <em>g</em>; some attempts were not even sinusoidal. While some candidates found the stretch factor <em>p</em> correctly or from follow-through on their own graph, very few successfully found the value and direction for the translation. </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) certainly served as a discriminator between the grade 6 and 7 candidates. </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = \sqrt 3 {{\rm{e}}^{2x}}\sin x + {{\rm{e}}^{2x}}\cos x\) , for \(0 \le x \le \pi \) . Solve </span><span style="font-family: times new roman,times; font-size: medium;">the </span><span style="font-family: times new roman,times; font-size: medium;">equation \(f(x) = 0\) .</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;">\({{\rm{e}}^{2x}}\left( {\sqrt 3 \sin x + \cos x} \right) = 0\) <em><strong>(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{e}}^{2x}} = 0\) not possible (seen anywhere) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">simplifying </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sqrt 3 \sin x + \cos x = 0\) , \(\sqrt 3 \sin x = - \cos x\) , \(\frac{{\sin x}}{{ - \cos x}} = \frac{1}{{\sqrt 3 }}\) <em><strong>A1</strong></em> </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;">\(\tan x = - \frac{1}{{\sqrt 3 }}\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{{5\pi }}{6}\) <strong><em>A2 N4</em></strong> </span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;">OR</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">sketch of \(30^\circ \) , \(60^\circ \) , \(90^\circ \) triangle with sides \(1\), \(2\), \(\sqrt 3 \) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">work leading to \(x = \frac{{5\pi }}{6}\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">verifying \(\frac{{5\pi }}{6}\) satisfies equation <em><strong>A1 N4</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 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;">Those who realized \({{\rm{e}}^{2x}}\) was a common factor usually earned the first four marks. Few could reason with the given information to solve the equation from there. There were many candidates who attempted some fruitless algebra that did not include factorization. </span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(f(x) = 3\sin \left( {\frac{\pi }{2}x} \right)\), for \(0 \leqslant x \leqslant 4\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Write down the amplitude of \(f\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Find the period of \(f\).</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">On the following grid sketch the graph of \(f\).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2017-02-01_om_05.59.27.png" alt="M16/5/MATME/SP1/ENG/TZ1/03.b"></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">(i) <span class="Apple-converted-space"> </span>3 <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N1</em></strong></span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>valid attempt to find the period <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\frac{{2\pi }}{b},{\text{ }}\frac{{2\pi }}{{\frac{\pi }{2}}}\)</p>
<p class="p1">period \( = 4\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></span></p>
<p class="p2"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><img src="images/Schermafbeelding_2017-02-01_om_06.02.18.png" alt="M16/5/MATME/SP1/ENG/TZ1/03.b/M"> <strong><em>A1A1A1A1 N4</em></strong></p>
<p class="p1"><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;">
<p class="p1">Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>Almost all candidates correctly stated the amplitude but then had difficulty finding the correct period. Few students faced problems in sketching the graph of the given function, even if they had found the wrong period, thus indicating a lack of understanding of the term ‘period’ in part a(ii). Most sketches were good although care should be taken to observe the given domain and to draw a neat curve.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">Given that \(\sin x = \frac{3}{4}\), </span>where \(x\) is an obtuse angle,</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">find the value of \(\cos x;\)</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 \(\cos 2x.\)</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">valid approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;\)<img src="images/Schermafbeelding_2016-01-13_om_06.28.54.png" alt>, \({\sin ^2}x + {\cos ^2}x = 1\)</p>
<p>correct working <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;{4^2} - {3^2},{\text{ }}{\cos ^2}x = 1 - {\left( {\frac{3}{4}} \right)^2}\)</p>
<p>correct calculation <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{{\sqrt 7 }}{4},{\text{ }}{\cos ^2}x = \frac{7}{{16}}\)</p>
<p>\(\cos x = - \frac{{\sqrt 7 }}{4}\) <strong><em>A1 N3</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>correct substitution (accept missing minus with cos) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;1 - 2{\left( {\frac{3}{4}} \right)^2},{\text{ }}2{\left( { - \frac{{\sqrt 7 }}{4}} \right)^2} - 1,{\text{ }}{\left( {\frac{{\sqrt 7 }}{4}} \right)^2} - {\left( {\frac{3}{4}} \right)^2}\)</p>
<p>correct working <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\;\;\;2\left( {\frac{7}{{16}}} \right) - 1,{\text{ }}1 - \frac{{18}}{{16}},{\text{ }}\frac{7}{{16}} - \frac{9}{{16}}\)</p>
<p>\(\cos 2x = - \frac{2}{{16}}\;\;\;\left( { = - \frac{1}{8}} \right)\) <strong><em>A1 N2</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;">
<p class="p1">Many candidates were able to find the cosine ratio of \(\frac{{\sqrt 7 }}{4}\) but did not take into account the information about the obtuse angle and seldom selected the negative answer. Finding \(\cos 2x\) proved easier; the most common error seen was \(\cos 2x = 2\cos x\).</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Many candidates were able to find the cosine ratio of \(\frac{{\sqrt 7 }}{4}\) but did not take into account the information about the obtuse angle and seldom selected the negative answer. Finding \(\cos 2x\) proved easier; the most common error seen was \(\cos 2x = 2\cos x\).</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><strong>Note: In this question, distance is in metres and time is in seconds.</strong></p>
<p>Two particles \({P_1}\) and \({P_2}\) start moving from a point A at the same time, along different straight lines.</p>
<p>After \(t\) seconds, the position of \({P_1}\) is given by <strong><em>r</em></strong> = \(\left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right)\).</p>
</div>
<div class="specification">
<p>Two seconds after leaving A, \({P_1}\) is at point B.</p>
</div>
<div class="specification">
<p>Two seconds after leaving A, \({P_2}\) is at point C, where \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 4 \end{array}} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the coordinates of A.</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>Find \(\overrightarrow {{\text{AB}}} \);</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>Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\cos {\rm{B\hat AC}}\).</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>Hence or otherwise, find the distance between \({P_1}\) and \({P_2}\) two seconds after they leave A.</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>recognizing \(t = 0\) at A <strong><em>(M1)</em></strong></p>
<p>A is \((4,{\text{ }} - 1,{\text{ }}3)\) <strong><em>A1</em></strong> <strong><em>N2</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><strong>METHOD 1</strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right),{\text{ }}(6,{\text{ }}3,{\text{ }} - 1)\)</p>
<p>correct approach to find \(\overrightarrow {{\text{AB}}} \) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} - {\text{A, }}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4 \\ { - 1} \\ 3 \end{array}} \right)\)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong>METHOD 2</strong></p>
<p>recognizing \(\overrightarrow {{\text{AB}}} \) is two times the direction vector <strong><em>(M1)</em></strong></p>
<p>correct working <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}} = 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 2} \end{array}} \right)\)</p>
<p>\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { - 4} \end{array}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</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>correct substitution <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{2^2} + {4^2} + {4^2}} ,{\text{ }}\sqrt {4 + 16 + 16} ,{\text{ }}\sqrt {36} \)</p>
<p>\(\left| {\overrightarrow {{\text{AB}}} } \right| = 6\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1 (vector approach)</strong></p>
<p>valid approach involving \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} ,{\text{ }}\frac{{\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{AC}}} }}{{{\text{AB}} \times {\text{AC}}}}\)</p>
<p>finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\) <strong><em>(A1)(A1)</em></strong></p>
<p>scalar product \(2(3) + 4(0) - 4(4){\text{ }}( = - 10)\)</p>
<p>\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + {0^2} + {4^2}} {\text{ }}( = 5)\)</p>
<p>substitution of <strong>their </strong>scalar product and magnitudes into cosine formula <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{6 + 0 - 16}}{{6\sqrt {{3^2} + {4^2}} }}\)</p>
<p>\({\text{cos}}\,B\hat AC = - \frac{{10}}{{30}}\left( { = - \frac{1}{3}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p> </p>
<p><strong>METHOD 2 (triangle approach)</strong></p>
<p>valid approach involving cosine rule <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{{\text{A}}{{\text{B}}^2} + {\text{A}}{{\text{C}}^2} - {\text{B}}{{\text{C}}^2}}}{{2 \times {\text{AB}} \times {\text{AC}}}}\)</p>
<p>finding lengths AC and BC <strong><em>(A1)(A1)</em></strong></p>
<p>\({\text{AC}} = 5,{\text{ BC}} = 9\)</p>
<p>substitution of <strong>their </strong>lengths into cosine formula <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC}} = \frac{{{5^2} + {6^2} - {9^2}}}{{2 \times 5 \times 6}}\)</p>
<p>\(\cos {\rm{B\hat AC}} = - \frac{{20}}{{60}}{\text{ }}\left( { = - \frac{1}{3}} \right)\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>Note:</strong> Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).</p>
<p> </p>
<p><strong>METHOD 1 (using cosine rule)</strong></p>
<p>recognizing need to find BC <strong><em>(M1)</em></strong></p>
<p>choosing cosine rule <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} - 2ab\cos {\text{C}}\)</p>
<p>correct substitution into RHS <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{B}}{{\text{C}}^2} = {(6)^2} + {(5)^2} - 2(6)(5)\left( { - \frac{1}{3}} \right),{\text{ }}36 + 25 + 20\)</p>
<p>distance is 9 <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p> </p>
<p><strong>METHOD 2 (finding magnitude of </strong>\(\overrightarrow {BC} \)<strong>) </strong></p>
<p>recognizing need to find BC <strong><em>(M1)</em></strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)attempt to find \(\overrightarrow {{\rm{OB}}} \) or \(\overrightarrow {{\rm{OC}}} \), \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { - 1} \end{array}} \right)\) or \(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}} 7 \\ { - 1} \\ 7 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \)</p>
<p>correct working <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { - 4} \\ 8 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{CB}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \\ { - 8} \end{array}} \right),{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}} = \sqrt {81} \)</p>
<p>distance is 9 <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p> </p>
<p><strong>METHOD 3 (finding coordinates and using distance formula)</strong></p>
<p>recognizing need to find BC <strong><em>(M1)</em></strong></p>
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)attempt to find coordinates of B or C, \({\text{B}}(6,{\text{ }}3,{\text{ }} - 1)\) or \({\text{C}}(7,{\text{ }} - 1,{\text{ }}7)\)</p>
<p>correct substitution into distance formula <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{BC}} = \sqrt {{{(6 - 7)}^2} + {{\left( {3 - ( - 1)} \right)}^2} + {{( - 1 - 7)}^2}} ,{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}} = \sqrt {81} \)</p>
<p>distance is 9 <strong><em>A1</em></strong> <strong><em>N2</em></strong></p>
<p><strong><em>[4 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.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</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><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = 6 + 6\sin x\) . Part of the graph of <em>f</em> is shown below.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/abba.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The shaded region is enclosed by the curve of <em>f</em> , the <em>x</em>-axis, and the <em>y</em>-axis.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Solve for \(0 \le x < 2\pi \)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(6 + 6\sin x = 6\) ;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(6 + 6\sin x = 0\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<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 exact value of the <em>x</em>-intercept of <em>f</em> , for \(0 \le x < 2\pi \) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The area of the shaded region is <em>k</em> . Find the value of <em>k</em> , giving your answer in </span><span style="font-family: times new roman,times; font-size: medium;">terms of \(\pi \) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . </span><span style="font-family: times new roman,times; font-size: medium;">The graph of <em>f</em> is transformed to the graph of <em>g</em>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Give a full geometric description of this transformation.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . </span><span style="font-family: times new roman,times; font-size: medium;">The graph of <em>f</em> is transformed to the graph of <em>g</em>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Given that \(\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x} = k\) and \(0 \le p < 2\pi \) , write down the two values of <em>p</em>.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\sin x = 0\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = 0\) , \(x = \pi \) <em><strong>A1A1 N2</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\sin x = - 1\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{{3\pi }}{2}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N1</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></strong></em></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{3\pi }}{2}\) <em><strong>A1 N1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</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;">evidence of using anti-differentiation <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct integral \(6x - 6\cos x\) (seen anywhere) <em><strong>A1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(6\left( {\frac{{3\pi }}{2}} \right) - 6\cos \left( {\frac{{3\pi }}{2}} \right) - ( - 6\cos 0)\) , \(9\pi - 0 + 6\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(k = 9\pi + 6\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1A1 N3</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">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">translation of \(\left( {\begin{array}{*{20}{c}}<br>{\frac{\pi }{2}}\\<br>0<br>\end{array}} \right)\) <em><strong>A1A1 N2</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">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">recognizing that the area under <em>g</em> is the same as the shaded region in <em>f</em> <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(p = \frac{\pi }{2}\) , \(p = 0\) <em><strong>A1A1 N3</strong></em></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">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many candidates again had difficulty finding the common angles in the trigonometric </span><span style="font-family: times new roman,times; font-size: medium;">equations. In part (a), some did not show sufficient working in solving the equations. Others </span><span style="font-family: times new roman,times; font-size: medium;">obtained a single solution in (a)(i) and did not find another. Some candidates worked in </span><span style="font-family: times new roman,times; font-size: medium;">degrees; the majority worked in radians.</span></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">While some candidates appeared to use their understanding of the graph of the original </span><span style="font-family: times new roman,times; font-size: medium;">function to find the <em>x</em>-intercept in part (b), most used their working from part (a)(ii) sometimes </span><span style="font-family: times new roman,times; font-size: medium;">with follow-through on an incorrect answer.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Most candidates recognized the need for integration in part (c) but far fewer were able to see </span><span style="font-family: times new roman,times; font-size: medium;">the solution through correctly to the end. Some did not show the full substitution of the limits, </span><span style="font-family: times new roman,times; font-size: medium;">having incorrectly assumed that evaluating the integral at 0 would be 0; without this working, </span><span style="font-family: times new roman,times; font-size: medium;">the mark for evaluating at the limits could not be earned. Again, many candidates had trouble </span><span style="font-family: times new roman,times; font-size: medium;">working with the common trigonometric values.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">While there was an issue in the wording of the question with the given domains, this did not </span><span style="font-family: times new roman,times; font-size: medium;">appear to bother candidates in part (d). This part was often well completed with candidates </span><span style="font-family: times new roman,times; font-size: medium;">using a variety of language to describe the horizontal translation to the right by \(\frac{\pi }{2}\) .</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;">Most candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majority tried to integrate the function <em>g</em> without success. Some candidates used sketches to find one or both values for <em>p</em>. The problem in the wording of the question did not appear to have been noticed by candidates in this part either.</span></p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">The following diagram shows triangle \(ABC\).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2015-12-14_om_04.52.14.png" alt></p>
<p>Let \(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} = - 5\sqrt 3 \) and \(\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right| = 10\). Find the area of triangle \(ABC\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>attempt to find \(\cos {\rm{C\hat AB}}\) (seen anywhere) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\cos \theta = \frac{{\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} }}{{\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|}}\)</p>
<p>\(\cos {\rm{C\hat AB}} = \frac{{ - 5\sqrt 3 }}{{10}}\;\;\;\left( { = - \frac{{\sqrt 3 }}{2}} \right)\) <strong><em>A1</em></strong></p>
<p>valid attempt to find \(\sin {\rm{C\hat AB}}\) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\)triangle, Pythagorean identity, \({\rm{C\hat AB}} = \frac{{5\pi }}{6},{\text{ }}150^\circ \)</p>
<p>\(\sin {\rm{C\hat AB}} = \frac{1}{2}\) <strong><em>(A1)</em></strong></p>
<p>correct substitution into formula for area <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\frac{1}{2} \times 10 \times \frac{1}{2},{\text{ }}\frac{1}{2} \times 10 \times \sin \frac{\pi }{6}\)</p>
<p>\({\text{area}} = \frac{{10}}{4}\;\;\;\left( { = \frac{5}{2}} \right)\) <strong><em>A1 N3</em></strong></p>
<p><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">The large majority of candidates were able to find the correct expression for \(\cos {\rm{C\hat AB}}\), but few recognized that an angle with a negative cosine will be obtuse, rather than acute, and many stated that \({\rm{C\hat AB}} = 30^\circ \). When substituting into the triangle area formula, a common error was to substitute \(5\sqrt 3 \) rather than 10, as many did not understand the relationship between the magnitude of a vector and the length of a line segment in the triangle formula.</p>
<p class="p1">Some of the G2 comments from schools suggested that it might have been easier for their students if this question were split into two parts. While we do tend to provide more support on the earlier questions in the paper, questions 6 and 7 are usually presented with little or no scaffolding. On these later questions, the candidates are often required to use knowledge from different areas of the syllabus within a single question.</p>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Solve \(\cos 2x - 3\cos x - 3 - {\cos ^2}x = {\sin ^2}x\) , for \(0 \le x \le 2\pi \) .</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;">evidence of substituting for \(\cos 2x\) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of substituting into \({\sin ^2}x + {\cos ^2}x = 1\) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct equation in terms of \(\cos x\) (seen anywhere) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2{\cos ^2}x - 1 - 3\cos x - 3 = 1\) , \(2{\cos ^2}x - 3\cos x - 5 = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of appropriate approach to solve <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. factorizing, quadratic formula </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">appropriate working <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \((2\cos x - 5)(\cos x + 1) = 0\) , \((2x - 5)(x + 1)\) , \(\cos x = \frac{{3 \pm \sqrt {49}}}{4}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct solutions to the equation </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos x = \frac{5}{2}\) , \(\cos x = - 1\) , \(x = \frac{5}{2}\) , \(x = - 1\) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = \pi \) <em><strong>A1 N4</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 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;">This question was quite difficult for most candidates. A number of students earned some credit for manipulating the equation with identities, but many earned no further marks due to algebraic errors. Many did not substitute for \(\cos 2x\) ; others did this substitution but then did nothing further. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Few candidates were able to get a correct equation in terms of \(\cos x\) and many who did get the equation didn't know what to do with it. Candidates who correctly solved the resulting quadratic usually found the one correct value of \(x\), earning full marks. </span></p>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(f:x \mapsto {\sin ^3}x\) .</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;">(i) Write down the range of the function <em>f</em> .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Consider \(f(x) = 1\) , \(0 \le x \le 2\pi \) . Write down the number of solutions to </span><span style="font-family: times new roman,times; font-size: medium;">this equation. Justify your answer.</span></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><span style="font-family: times new roman,times; font-size: medium;">Find \(f'(x)\) , giving your answer in the form \(a{\sin ^p}x{\cos ^q}x\) where \(a{\text{, }}p{\text{, }}q \in \mathbb{Z}\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}\) for \(0 \le x \le \frac{\pi }{2}\) </span><span style="font-family: times new roman,times; font-size: medium;">. Find the volume generated when the </span><span style="font-family: times new roman,times; font-size: medium;">curve of <em>g</em> is revolved through \(2\pi \) about the <em>x</em>-axis.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</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) range of <em>f</em> is \([ - 1{\text{, }}1]\) , \(( - 1 \le f(x) \le 1)\) <em><strong>A2 N2</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">justification for one solution on \([0{\text{, }}2\pi ]\) <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(x = \frac{\pi }{2}\) </span><span style="font-family: times new roman,times; font-size: medium;">, unit circle, sketch of \(\sin x\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">1 solution (seen anywhere) <em><strong>A1 N1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [5 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f'(x) = 3{\sin ^2}x\cos x\) <em><strong>A2 N2</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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">using \(V = \int_a^b {\pi {y^2}{\rm{d}}x} \) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x\) </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;">\( = \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x} \) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}\) \(\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) - {{\sin }^3}0} \right)} \right)\) <em><strong>A2</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using \(\sin \frac{\pi }{2} = 1\) </span><span style="font-family: times new roman,times; font-size: medium;">and \(\sin 0 = 0\) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\pi \left( {1 - 0} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(V = \pi \) <em><strong>A1 N1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [7 marks]</span></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;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was not done well by most candidates. No more than one-third of them could correctly give the range of \(f(x) = {\sin ^3}x\) and few could provide adequate justification for there being exactly one solution to \(f(x) = 1\) in the interval \([0{\text{, }}2\pi ]\) .</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 was not done well by most candidates.</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 was not done well by most candidates. No more than one-third of them could correctly give the range of \(f(x) = {\sin ^3}x\) and few could provide adequate justification for there being exactly one solution to \(f(x) = 1\) in the interval \([0{\text{, }}2\pi ]\) . Finding the derivative of this function also presented major problems, thus making part (c) of the question much more difficult. In spite of the formula for volume of revolution being given in the Information Booklet, fewer than half of the candidates could correctly put the necessary function and limits into \(\pi \int_a^b {{y^2}{\rm{d}}x} \) and fewer still could square \(\sqrt 3 \sin x{\cos ^{\frac{1}{2}}}x\) correctly. From those who did square correctly, the correct antiderivative was not often recognized. All manner of antiderivatives were suggested instead.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The diagram below shows part of the graph of a function \(f\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img style="display: block; margin-left: auto; margin-right: auto;" src="images/lap.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The graph has a maximum at A(\(1\), \(5\)) and a minimum at B(\(3\), \( -1\)) .</span></p>
<p><span style="font-size: medium;"><span style="font-family: times new roman,times;">The function \(f\)</span> <span style="font-family: times new roman,times;">can be written in the form \(f(x) = p\sin (qx) + r\) . Find the value of<br></span></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;">(a) \(p\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(b) \(q\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(c) \(r\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(p\)</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;">\(q\)</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><span style="font-family: times new roman,times; font-size: medium;">\(r\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</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;">(a) valid approach to find \(p\) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> amplitude \( = \frac{{{\rm{max}} - {\rm{min}}}}{2}\) , \(p = 6\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = 3\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></em></strong></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) valid approach to find \(q\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> period = 4 , \(q = \frac{{2\pi }}{{{\rm{period}}}}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(q = \frac{\pi }{2}\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"><br></span></em></strong><span style="font-family: times new roman,times; font-size: medium;">(c) valid approach to find \(r\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> axis = \(\frac{{{\rm{max}} + {\rm{min}}}}{2}\) , sketch of horizontal axis, \(f(0)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(r = 2\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">Total [6 marks]<br></span></em></strong></p>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">valid approach to find \(p\) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> amplitude \( = \frac{{{\rm{max}} - {\rm{min}}}}{2}\) , \(p = 6\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(p = 3\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </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;">valid approach to find \(q\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> period = 4 , \(q = \frac{{2\pi }}{{{\rm{period}}}}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(q = \frac{\pi }{2}\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></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;">valid approach to find \(r\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>eg</em> axis = \(\frac{{{\rm{max}} + {\rm{min}}}}{2}\) , sketch of horizontal axis, \(f(0)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(r = 2\) <strong><em>A1 N2 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">Total [6 marks]<br></span></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;">
<div dir="ltr" style="font-size: 13.28px; font-family: sans-serif; left: 96.032px; top: 442.147px; transform: scale(0.978235, 1); transform-origin: 0% 0% 0px;" data-font-name="Helvetica" data-canvas-width="593.7886576962807"><span style="font-family: times new roman,times; font-size: medium;">Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters \(p\),\(q\) and \(r\). The successful candidates were able to find the answers using the given points and their understanding of the different transformations.</span></div>
<div dir="ltr" style="font-size: 13.28px; font-family: sans-serif; left: 325.027px; top: 484.227px; transform: scale(1.03898, 1); transform-origin: 0% 0% 0px;" data-font-name="Helvetica" data-canvas-width="74.80624222939969"><span style="font-family: times new roman,times; font-size: medium;">Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between \(q\) and the period of the function. There were also some candidates who showed working such as\(\frac{{2\pi }}{b}\) without explaining what \(b\) represented.</span></div>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
<div style="font-size: 13.28px; font-family: sans-serif; left: 96.032px; top: 442.147px; transform: scale(0.978235, 1); transform-origin: 0% 0% 0px;" dir="ltr" data-font-name="Helvetica" data-canvas-width="593.7886576962807"><span style="font-family: times new roman,times; font-size: medium;">Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters \(p\),\(q\) and \(r\). The successful candidates were able to find the answers using the given points and their understanding of the different transformations.</span></div>
<div style="font-size: 13.28px; font-family: sans-serif; left: 325.027px; top: 484.227px; transform: scale(1.03898, 1); transform-origin: 0% 0% 0px;" dir="ltr" data-font-name="Helvetica" data-canvas-width="74.80624222939969"><span style="font-family: times new roman,times; font-size: medium;">Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between \(q\) and the period of the function. There were also some candidates who showed working such as\(\frac{{2\pi }}{b}\) without explaining what \(b\) represented.</span></div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<div style="font-size: 13.28px; font-family: sans-serif; left: 96.032px; top: 442.147px; transform: scale(0.978235, 1); transform-origin: 0% 0% 0px;" dir="ltr" data-font-name="Helvetica" data-canvas-width="593.7886576962807"><span style="font-family: times new roman,times; font-size: medium;">Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters \(p\),\(q\) and \(r\). The successful candidates were able to find the answers using the given points and their understanding of the different transformations.</span></div>
<div style="font-size: 13.28px; font-family: sans-serif; left: 325.027px; top: 484.227px; transform: scale(1.03898, 1); transform-origin: 0% 0% 0px;" dir="ltr" data-font-name="Helvetica" data-canvas-width="74.80624222939969"><span style="font-family: times new roman,times; font-size: medium;">Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between \(q\) and the period of the function. There were also some candidates who showed working such as\(\frac{{2\pi }}{b}\) without explaining what \(b\) represented.</span></div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<div style="font-size: 13.28px; font-family: sans-serif; left: 96.032px; top: 442.147px; transform: scale(0.978235, 1); transform-origin: 0% 0% 0px;" dir="ltr" data-font-name="Helvetica" data-canvas-width="593.7886576962807"><span style="font-family: times new roman,times; font-size: medium;">Many candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when they attempted to substitute into the equation of the function with the parameters \(p\),\(q\) and \(r\). The successful candidates were able to find the answers using the given points and their understanding of the different transformations.</span></div>
<div style="font-size: 13.28px; font-family: sans-serif; left: 325.027px; top: 484.227px; transform: scale(1.03898, 1); transform-origin: 0% 0% 0px;" dir="ltr" data-font-name="Helvetica" data-canvas-width="74.80624222939969"><span style="font-family: times new roman,times; font-size: medium;">Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between \(q\) and the period of the function. There were also some candidates who showed working such as\(\frac{{2\pi }}{b}\) without explaining what \(b\) represented.</span></div>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The following diagram shows a circle with centre \(O\) and a radius of \(10\) cm. Points \(A\), \(B\) and \(C\) lie on the circle.</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2016-01-13_om_05.32.02.png" alt></p>
<p class="p1">Angle \(AOB\) is \(1.2\) radians.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the length of \({\text{arc ACB}}\).</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">Find the perimeter of the shaded region.</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">correct substitution <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;10(1.2)\)</p>
<p class="p1">\(ACB\) is \(12{\text{ (cm)}}\) <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></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>valid approach to find major arc <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\;\;\;\)circumference \( - {\text{AB}}\), major angle \({\text{AOB}} \times {\text{radius}}\)</p>
<p>correct working for arc length <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\;\;\;2\pi (10) - 12,{\text{ }}10(2 \times 3.142 - 1.2),{\text{ }}2\pi (10) - 12 + 20\)</p>
<p>perimeter is \(20\pi + 8\;\;\;( = 70.8){\text{ (cm)}}\) <strong><em>A1 N2</em></strong></p>
<p><strong><em>[3 marks]</em></strong></p>
<p><strong><em>Total [5 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">Most candidates were able to find the minor arc length. Similarly most candidates successfully found the major arc length in part b) but did not go on to add the two radii. Quite a few candidates worked with decimal approximations, rather than in terms of \(\pi \).</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Most candidates were able to find the minor arc length. Similarly most candidates successfully found the major arc length in part b) but did not go on to add the two radii. Quite a few candidates worked with decimal approximations, rather than in terms of \(\pi \).</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>The following table shows the probability distribution of a discrete random variable \(A\), in terms of an angle \(\theta \).</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2017-08-11_om_09.10.36.png" alt="M17/5/MATME/SP1/ENG/TZ1/10"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\cos \theta = \frac{3}{4}\).</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>Given that \(\tan \theta > 0\), find \(\tan \theta \).</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 \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.</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>evidence of summing to 1 <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\sum {p = 1} \)</p>
<p>correct equation <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos \theta + 2\cos 2\theta = 1\)</p>
<p>correct equation in \(\cos \theta \) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos \theta + 2(2{\cos ^2}\theta - 1) = 1,{\text{ }}4{\cos ^2}\theta + \cos \theta - 3 = 0\)</p>
<p>evidence of valid approach to solve quadratic <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ - 1 \pm \sqrt {1 - 4 \times 4 \times ( - 3)} }}{8}\)</p>
<p>correct working, clearly leading to required answer <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\((4\cos \theta - 3)(\cos \theta + 1),{\text{ }}\frac{{ - 1 \pm 7}}{8}\)</p>
<p>correct reason for rejecting \(\cos \theta \ne - 1\) <strong><em>R1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta > 0\)</p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>R0 </em></strong>for \(\cos \theta \ne - 1\) without a reason.</p>
<p> </p>
<p>\(\cos \theta = \frac{3}{4}\) <em><strong>AG N0</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>valid approach <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)</p>
<p>correct working </p>
<p><strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)</p>
<p>\(\tan \theta = \frac{{\sqrt 7 }}{3}\) <strong><em>A1</em></strong> <strong><em>N2</em></strong></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>attempt to substitute either limits or the function into formula involving \({f^2}\) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)</p>
<p>correct substitution of both limits and function <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)</p>
<p>correct integration <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\tan x\)</p>
<p>substituting <strong>their </strong>limits into <strong>their </strong>integrated function and subtracting <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} - \tan \theta \)</p>
<p> </p>
<p><strong>Note:</strong> Award <strong><em>M0 </em></strong>if they substitute into original or differentiated function.</p>
<p> </p>
<p>\(\tan \frac{\pi }{4} = 1\) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(1 - \tan \theta \)</p>
<p>\(V = \pi - \frac{{\pi \sqrt 7 }}{3}\) <strong><em>A1</em></strong> <strong><em>N3</em></strong></p>
<p> </p>
<p><strong><em>[6 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" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(4 - \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin \theta + 3\) .</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;"><strong>Hence</strong>, solve the equation \(4 - \cos 2\theta + 5\sin \theta = 0\) for \(0 \le \theta \le 2\pi \) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">attempt to substitute \(1 - 2{\sin ^2}\theta \) for \(\cos 2\theta \) <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(4 - (1 - 2{\sin ^2}\theta ) + 5\sin\theta \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4 - \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin\theta + 3\) <em><strong>AG N0</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;">evidence of appropriate approach to solve <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. factorizing, quadratic formula</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \((2\sin \theta + 3)(\sin \theta + 1)\) , \((2x + 3)(x + 1) = 0\) , \(\sin x = \frac{{ - 5 \pm \sqrt 1 }}{4}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct solution \(\sin \theta = - 1\) (do not penalise for including \(\sin \theta = - \frac{3}{2}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\theta = \frac{{3\pi }}{2}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N3</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [5 marks]</span></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><span style="font-family: times new roman,times; font-size: medium;">In part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidates who worked backward, starting with the required answer and manipulating the equation in various ways. As this was a "show that" question, working backward from the given answer is not a valid method. </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;">In part (b), many candidates seemed to realize what was required by the word “hence”, though some had trouble factoring the quadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solution to the equation \(\sin \theta = - 1\) , and there were a few who did not realize that the equation \(\sin \theta = - \frac{3}{2}\) has no solution. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p>Solve \({\log _2}(2\sin x) + {\log _2}(\cos x) = - 1\), for \(2\pi < x < \frac{{5\pi }}{2}\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>correct application of \(\log a + \log b = \log ab\) <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\log _2}(2\sin x\cos x),{\text{ }}\log 2 + \log (\sin x) + \log (\cos x)\)</p>
<p>correct equation without logs <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(2\sin x\cos x = {2^{ - 1}},{\text{ }}\sin x\cos x = \frac{1}{4},{\text{ }}\sin 2x = \frac{1}{2}\)</p>
<p>recognizing double-angle identity (seen anywhere) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\log (\sin 2x),{\text{ }}2\sin x\cos x = \sin 2x,{\text{ }}\sin 2x = \frac{1}{2}\)</p>
<p>evaluating \({\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}{\text{ }}(30^\circ )\) <strong><em>(A1)</em></strong></p>
<p>correct working <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(x = \frac{\pi }{{12}} + 2\pi ,{\text{ }}2x = \frac{{25\pi }}{6},{\text{ }}\frac{{29\pi }}{6},{\text{ }}750^\circ ,{\text{ }}870^\circ ,{\text{ }}x = \frac{\pi }{{12}}\)<strong>and</strong> \(x = \frac{{5\pi }}{{12}}\), one correct final answer</p>
<p>\(x = \frac{{25\pi }}{{12}},{\text{ }}\frac{{29\pi }}{{12}}\) (do not accept additional values) <strong><em>A2</em></strong> <strong><em>N0</em></strong></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="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The straight line with equation \(y = \frac{3}{4}x\) </span><span style="font-family: times new roman,times; font-size: medium;">makes an acute angle \(\theta \) with the <em>x</em>-axis.</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;">Write down the value of \(\tan \theta \) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find the value of</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\sin 2\theta \) ;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\cos 2\theta \) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">b(i) and (ii).</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;">\(\tan \theta = \frac{3}{4}\) (do not accept \(\frac{3}{4}x\) ) <em><strong>A1 N1</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</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;">(i) \(\sin \theta = \frac{3}{5}\) , \(\cos \theta = \frac{4}{5}\) <em><strong>(A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sin 2\theta = 2\left( {\frac{3}{5}} \right)\left( {\frac{4}{5}} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\sin 2\theta = \frac{{24}}{{25}}\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N3</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) correct substitution <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos 2\theta = 1 - 2{\left( {\frac{3}{5}} \right)^2}\) , \({\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\cos 2\theta = \frac{7}{{25}}\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N1</strong> </span></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">b(i) and (ii).</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many candidates drew a diagram to correctly find \(\tan \theta \) , although few recognized that a line </span><span style="font-family: times new roman,times; font-size: medium;">through the origin can be expressed as \(y = x\tan \theta \) , with gradient \(\tan \theta \) , which is explicit in the </span><span style="font-family: times new roman,times; font-size: medium;">syllabus.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A surprising number </span><span style="font-family: times new roman,times; font-size: medium;">were unable to find the ratios for \(\sin \theta \) and \(\cos \theta \) from \(\tan \theta \) . It was not uncommon for </span><span style="font-family: times new roman,times; font-size: medium;">candidates to use unreasonable values, such as \(\sin \theta = 3\) and \(\cos \theta = 4\) , or to write nonsense </span><span style="font-family: times new roman,times; font-size: medium;">such as \(2\sin \frac{3}{5}\cos \frac{4}{5}\) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(\sin \theta = \frac{{\sqrt 5 }}{3}\), where \(\theta \) is acute.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find \(\cos \theta \).</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>Find \(\cos 2\theta \).</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">evidence of valid approach <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><span class="s1"><em>eg</em>\(\,\,\,\,\,\)</span>right triangle, \({\cos ^2}\theta = 1 - {\sin ^2}\theta \)</p>
<p class="p1">correct working <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><span class="s1"><em>eg</em>\(\,\,\,\,\,\)</span>missing side is 2<span class="s2">, \(\sqrt {1 - {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}} \)</span></p>
<p class="p3"><span class="s2">\(\cos \theta = \frac{2}{3}\) <span class="Apple-converted-space"> </span></span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p3"><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">correct substitution into formula for \(\cos 2\theta \) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(2 \times {\left( {\frac{2}{3}} \right)^2} - 1,{\text{ }}1 - 2{\left( {\frac{{\sqrt 5 }}{3}} \right)^2},{\text{ }}{\left( {\frac{2}{3}} \right)^2} - {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}\)</p>
<p class="p2"><span class="Apple-converted-space">\(\cos 2\theta = - \frac{1}{9}\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p2"><strong><em>[2 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><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = {\sin ^3}x + {\cos ^3}x\tan x,\frac{\pi }{2} < x < \pi \) .</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;">Show that \(f(x) = \sin x\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(\sin x = \frac{2}{3}\) </span><span style="font-family: times new roman,times; font-size: medium;">. Show that \(f(2x) = - \frac{{4\sqrt 5 }}{9}\)</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;">changing \(\tan x\) into \(\frac{{\sin x}}{{\cos x}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\sin ^3}x + {\cos ^3}x\frac{{\sin x}}{{\cos x}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">simplifying <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g \(\sin x({\sin ^2}x + {\cos ^2}x)\) , \({\sin ^3}x + \sin x - {\sin ^3}x\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f(x) = \sin x\) <em><strong>AG N0</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;">recognizing \(f(2x) = \sin 2x\) , seen anywhere <em><strong>(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using double angle identity \(\sin (2x) = 2\sin x\cos x\) , seen anywhere <strong><em>(M1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using Pythagoras with \(\sin x = \frac{2}{3}\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. sketch of right triangle, \({\sin ^2}x + {\cos ^2}x = 1\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos x = - \frac{{\sqrt 5 }}{3}\) (accept \(\frac{{\sqrt 5 }}{3}\) ) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f(2x) = 2\left( {\frac{2}{3}} \right)\left( { - \frac{{\sqrt 5 }}{3}} \right)\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f(2x) = - \frac{{4\sqrt 5 }}{9}\) <em><strong>AG N0</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></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><span style="font-family: times new roman,times; font-size: medium;">Not surprisingly, this question provided the greatest challenge in section A. In part (a), candidates were able to use the identity \(\tan x = \frac{{\sin x}}{{\cos x}}\) , but many could not proceed any further. </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 (b) was generally well done by those candidates who attempted it, the major error arising when the negative sign "magically" appeared in the answer. Many candidates could find the value of cos<em>x</em> but failed to observe that cosine is negative in the given domain. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The following diagram shows the graph of \(f(x) = a\sin (b(x - c)) + d\) , for \(2 \le x \le 10\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/deanna.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Use the graph to write down the value of</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) <em>a</em> ;</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) <em>c</em> ;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) <em>d</em> .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a(i), (ii) and (iii).</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 \(b = \frac{\pi }{4}\) .</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><span style="font-family: times new roman,times; font-size: medium;">Find \(f'(x)\) .</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;">At a point R, the gradient is \( - 2\pi \) . Find the <em>x</em>-coordinate of R.</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(a = 8\) <strong><em>A1 N1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(c = 2\) <em><strong>A1 N1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) \(d = 4\) <em><strong> A1 N1</strong></em></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">a(i), (ii) and (iii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">recognizing that period \( = 8\) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(8 = \frac{{2\pi }}{b}\) , \(b = \frac{{2\pi }}{8}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = \frac{\pi }{4}\) <em><strong>AG N0</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;">attempt to substitute <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(12 = 8\sin (b(4 - 2)) + 4\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sin 2b = 1\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = \frac{\pi }{4}\) <em><strong>AG N0</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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of attempt to differentiate or choosing chain rule <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos \frac{\pi }{4}(x - 2)\) , \(\frac{\pi }{4} \times 8\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(f'(x) = 2\pi \cos \left( {\frac{\pi }{4}(x - 2)} \right)\) (</span><span style="font-family: times new roman,times; font-size: medium;">accept \(2\pi \cos \frac{\pi }{4}(x - 2)\) ) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A2 N3</span></strong></em></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;">recognizing that gradient is \(f'(x)\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(f'(x) = m\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct equation <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 2\pi = 2\pi \cos \left( {\frac{\pi }{4}(x - 2)} \right)\) , \( - 1 = \cos \left( {\frac{\pi }{4}(x - 2)} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\cos ^{ - 1}}( - 1) = \frac{\pi }{4}(x - 2)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">using \({\cos ^{ - 1}}( - 1) = \pi \) (seen anywhere) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\pi = \frac{\pi }{4}(x - 2)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">simplifying <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(4 = (x - 2)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = 6\) <em><strong>A1 N4</strong></em></span></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>
<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) of this question proved challenging for most candidates. </span></p>
<div class="question_part_label">a(i), (ii) and (iii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Although a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize that this period could be found using the given coordinates of the maximum and minimum points. </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;">In part (c), not many candidates found the correct derivative using the chain rule. </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;">For part (d), a good number of candidates correctly set their expression equal to \( - 2\pi \) , but errors in their previous values kept most from correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here. </span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The expression \(6\sin x\cos x\) can be expressed in the form \(a\sin bx\) .</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 value of <em>a</em> and of <em>b</em> .</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><span style="font-family: times new roman,times; font-size: medium;">Hence or otherwise, solve the equation \(6\sin x\cos x = \frac{3}{2}\) , for \(\frac{\pi }{4} \le x \le \frac{\pi }{2}\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">recognizing double angle <em><strong> M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3 \times 2\sin x\cos x\) , \(3\sin 2x\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(a = 3\) , \(b = 2\) <em><strong>A1A1 N3</strong></em></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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">substitution \(3\sin 2x = \frac{3}{2}\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\sin 2x = \frac{1}{2}\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">finding the angle <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{\pi }{6}\) , \(2x = \frac{{5\pi }}{6}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{{5\pi }}{{12}}\) <em><strong>A1 N2</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A0</strong></em> if other values are included.</span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></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><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: TimesNewRomanPSMT;">Let \(\sin {100^ \circ } = m\)</span><span style="font-family: TimesNewRomanPSMT;">. Find an expression for </span></span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: TimesNewRomanPSMT;"><span style="font-family: times new roman,times; font-size: medium;">\(\cos {100^ \circ }\)</span> in terms of </span><em><span style="font-family: TimesNewRomanPS-ItalicMT;">m</span></em><span style="font-family: TimesNewRomanPS-ItalicMT;">.</span></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><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: TimesNewRomanPSMT;">Let \(\sin {100^ \circ } = m\) </span><span style="font-family: TimesNewRomanPSMT;">. Find an expression for </span></span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: TimesNewRomanPSMT;"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: TimesNewRomanPS-ItalicMT;">\(\tan {100^ \circ }\)</span></span> in terms of </span><em><span style="font-family: TimesNewRomanPS-ItalicMT;">m</span></em><span style="font-family: TimesNewRomanPS-ItalicMT;">.</span></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><span style="font-family: times new roman,times; font-size: medium;">Let \(\sin {100^ \circ } = m\). Find an expression for </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\sin {200^ \circ }\)</span> in terms of <em>m</em>.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</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;"><strong>Note: </strong>All answers must be given in terms of <em>m</em>. If a candidate makes an error that means there is no <em>m </em>in their answer, do not award the final <strong><em>A1FT </em></strong>mark.</span></p>
<p><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;">valid approach involving Pythagoras <strong><em>(M1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g. </em>\({\sin ^2}x + {\cos ^2}x = 1\) , labelled diagram <img src="data:image/png;base64,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" alt> <br></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working (may be on diagram) <strong><em>(A1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g. </em>\({m^2} + {(\cos 100)^2} = 1\) , \(\sqrt {1 - {m^2}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 100 = - \sqrt {1 - {m^2}} \) <em><strong>A1 N2</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong>[3 marks]<br></strong></em></span></p>
<p><br><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">valid approach involving tan identity <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g. </em>\(\tan = \frac{{\sin }}{{\cos }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em>e.g.</em> \(\cos 100 = \frac{{\sin 100}}{{\tan 100}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 100 = \frac{m}{{\tan 100}}\) <em><strong>A1 N2</strong></em></span></p>
<p><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><span style="font-family: times new roman,times;"><strong><span style="font-size: medium;">METHOD 1</span></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\tan 100 = - \frac{m}{{\sqrt {1 - {m^2}} }}\) (accept \(\frac{m}{{ - \sqrt {1 - {m^2}} }}\)) <em><strong>A1 N1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong>[1 mark]<br></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;"><strong>METHOD 2</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\tan 100 = \frac{m}{{\cos 100}}\) <em><strong>A1 N1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong>[1 mark]</strong></em></span></p>
<p><span style="font-family: times new roman,times;"> </span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;">METHOD 1</span></span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;">valid approach involving double angle formula </span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">(M1) </span></span></em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><em>e.g. </em></span></span><span style="font-size: medium;">\(\sin 2\theta = 2\sin \theta cos\theta \) </span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">\(\sin 200 = - 2m\sqrt {1 - {m^2}} \) (accept \(2m\left( { - \sqrt {1 - {m^2}} } \right)\)) <em><strong>A1 N2 </strong></em></span></span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><strong>Note: </strong></span></span><span style="font-size: medium;">If candidates find \(\cos 100 = \sqrt {1 - {m^2}} \) , award full </span><strong><em><span style="font-family: Times New Roman,Times New Roman; font-size: medium;"><span style="font-family: Times New Roman,Times New Roman; font-size: medium;">FT </span></span></em></strong><span style="font-size: medium;">in parts (b) and (c), even though the values may not have appropriate signs for the angles.</span></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-size: medium;"><em><strong>[2 marks]</strong></em><br></span></span></p>
<p> </p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">valid approach involving double angle formula <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sin 2\theta = 2\sin \theta \cos \theta \) , \(2m \times \frac{m}{{\tan 100}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sin 200 = \frac{{2{m^2}}}{{\tan 100}}( = 2m\cos 100)\) <em><strong>A1 N2</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">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-size: medium;"><span style="font-family: times new roman,times;">While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write \(\sqrt {1 - {m^2}} = 1 - m\) . </span><span style="font-family: times new roman,times;">In part (c), many candidates failed to use the double-angle </span><span style="font-family: times new roman,times;">identity. Many incorrectly assumed that because \(\sin {100^ \circ } = m\) , then \(\sin {200^ \circ } = 2m\) . In addition, some candidates did not seem to understand what writing an expression "in terms of <em>m</em></span><span style="font-family: times new roman,times;">" meant.<br></span></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-size: small;"><span style="font-family: times new roman,times; font-size: medium;">While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write \(\sqrt {1 - {m^2}} = 1 - m\) . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because \(\sin {100^ \circ } = m\) , then \(\sin {200^ \circ } = 2m\) . In addition, some candidates did not seem to understand what writing an expression "in terms of <em>m</em>" meant.</span></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;">While many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of an angle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. A common algebraic error in part (a) was for candidates to write \(\sqrt {1 - {m^2}} = 1 - m\) . In part (c), many candidates failed to use the double-angle identity. Many incorrectly assumed that because \(\sin {100^ \circ } = m\) , then \(\sin {200^ \circ } = 2m\) . In addition, some candidates did not seem to understand what writing an expression "in terms of <em>m</em>" meant.</span></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: 12.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(\int_\pi ^a {\cos 2x{\text{d}}x} = \frac{1}{2}{\text{, where }}\pi < a < 2\pi \). Find the value of \(a\).</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'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">correct integration (ignore absence of limits and “\(+C\)”) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">substituting limits into <strong>their </strong>integrated function and subtracting (in any order) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\frac{1}{2}\sin (2a) - \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) - \sin (2a)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sin (2\pi ) = 0\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">setting <strong>their </strong>result from an integrated function equal to \(\frac{1}{2}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">recognizing \({\sin ^{ - 1}}1 = \frac{\pi }{2}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">correct value <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> \(\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">\(a = \frac{{5\pi }}{4}\) <strong><em>A1 N3</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; color: #3f3f3f;"><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">
[N/A]
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The following diagram represents a large Ferris wheel, with a diameter of 100 metres.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/ferris.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let P be a point on the wheel. The wheel starts with P at the lowest point, </span><span style="font-family: times new roman,times; font-size: medium;">at ground level. The wheel rotates at a constant rate, in an anticlockwise </span><span style="font-family: times new roman,times; font-size: medium;">(counter-clockwise) direction. One revolution takes 20 minutes.</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let \(h(t)\) metres be the height of P above ground level after <em>t</em> minutes. Some values </span><span style="font-family: times new roman,times; font-size: medium;">of \(h(t)\) are given in the table below.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/garage.png" alt></span></p>
<p> </p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Write down the height of P above ground level after</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) 10 minutes;</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) 15 minutes.</span></p>
<p align="LEFT"> </p>
<div class="marks">[2]</div>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that \(h(8) = 90.5\).</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find \(h(21)\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b(i) and (ii).</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>Sketch</strong> the graph of <em>h</em> , for \(0 \le t \le 40\) .</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;">Given that <em>h</em> can be expressed in the form \(h(t) = a\cos bt + c\) , find <em>a</em> , <em>b</em> and <em>c</em> .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) 100 (metres) <em><strong>A1 N1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) 50 (metres) <strong><em>A1 N1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></em></strong></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) identifying symmetry with \(h(2) = 9.5\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">subtraction <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(100 - h(2)\) , \(100 - 9.5\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(h(8) = 90.5\) <strong><em>AG N0</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) recognizing period <strong><em> (M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(h(21) = h(1)\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(h(21) = 2.4\) <em><strong>A1 N2</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></strong></em></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/rush.png" alt></span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A1A1A1 N3</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for end points (0, 0) and (40, 0) , <strong><em>A1</em></strong> for range \(0 \le h \le 100\) , </span><span style="font-family: times new roman,times; font-size: medium;"><strong><em>A1</em></strong> for approximately correct sinusoidal shape, with two cycles.</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;">evidence of a quotient involving 20, \(2\pi \) or \({360^ \circ }\) to find <em>b</em> <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{2\pi }}{b} = 20\) , \(b = \frac{{360}}{{20}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(b = \frac{{2\pi }}{{20}}\) \(\left( { = \frac{\pi }{{10}}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">(accept \(b = 18\) if working in degrees) <strong><em>A1 N2</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(a = - 50\) , \(c = 50\) <em><strong>A2A1 N3</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [5 marks]</span></strong></em></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;">Nearly all candidates answered part (a) correctly, finding the height of the wheel at \(\frac{1}{2}\) and \(\frac{3}{4}\) of a revolution. </span></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">While many candidates were successful in part (b), there were many who tried to use right-angled triangles or find a function for height, rather than recognizing the symmetry of the wheel in its different positions and using the values given in the table. </span></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In part (c), most candidates were able to sketch a somewhat accurate representation of the height of the wheel over two full cycles. However, it seems that many candidates are not familiar with the shape of a sinusoidal wave, as many of the candidates' graphs were constructed of line segments, rather than a curve. </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;">For part (d), candidates were less successful in finding the parameters of the cosine function. Even candidates who drew accurate sketches were not always able to relate their sketch to the function. These candidates understood the context of the problem, that the position on the wheel goes up and down, but they did not relate this to a trigonometric function. Only a small number of candidates recognized that the value of <strong><em>a</em></strong> would be negative. Candidates should be aware that while working in degrees may be acceptable, the expectation is that radians will be used in these types of questions. </span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<p>Let \(f(x) = 15 - {x^2}\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\) and the rectangle OABC, where A is on the negative \(x\)-axis, B is on the graph of \(f\), and C is on the \(y\)-axis.</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2018-02-11_om_13.13.04.png" alt="N17/5/MATME/SP1/ENG/TZ0/06"></p>
<p>Find the \(x\)-coordinate of A that gives the maximum area of OABC.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>attempt to find the area of OABC <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{OA}} \times {\text{OC, }}x \times f(x),{\text{ }}f(x) \times ( - x)\)</p>
<p>correct expression for area in one variable <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{area}} = x(15 - {x^2}),{\text{ }}15x - {x^3},{\text{ }}{x^3} - 15x\)</p>
<p>valid approach to find maximum <strong>area</strong> (seen anywhere) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(A’(x) = 0\)</p>
<p>correct derivative <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(15 - 3{x^2},{\text{ }}(15 - {x^2}) + x( - 2x) = 0,{\text{ }} - 15 + 3{x^2}\)</p>
<p>correct working <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(15 = 3{x^2},{\text{ }}{x^2} = 5,{\text{ }}x = \sqrt 5 \)</p>
<p>\(x = - \sqrt 5 {\text{ }}\left( {{\text{accept A}}\left( { - \sqrt 5 ,{\text{ }}0} \right)} \right)\) <strong><em>A2 N3</em></strong></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="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;">Let \(f(x) = \sin \left( {x + \frac{\pi }{4}} \right) + k\). The graph of <em>f </em>passes through the point \(\left( {\frac{\pi }{4},{\text{ }}6} \right)\).</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;">Find the value of \(k\).</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 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find the minimum value of \(f(x)\).</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 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;">Let \(g(x) = \sin x\). The graph of <em>g </em>is translated to the graph of \(f\) by the vector </span><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}} p \\ q \end{array}} \right)\)</span><span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="line-height: normal;">.</span></span></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;">Write down the value of \(p\) and of \(q\).</span></p>
<div class="marks">[2]</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: 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 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to substitute both coordinates (in any order) into \(f\) <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;"><em>eg</em> \(f\left( {\frac{\pi }{4}} \right) = 6,{\text{ }}\frac{\pi }{4} = \sin \left( {6 + \frac{\pi }{4}} \right) + k\)</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;">correct working <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;"><em>eg</em> \(\sin \frac{\pi }{2} = 1,{\text{ }}1 + k = 6\)</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;">\(k = 5\) <strong><em>A1 N2</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;"><strong><em>[3 marks]</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>METHOD 2</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;">recognizing shift of \(\frac{\pi }{4}\) left means maximum at \(6\) <strong><em>R1)</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;">recognizing \(k\) is difference of maximum and amplitude <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;"><em>eg</em> \(6 - 1\)</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;">\(k = 5\) <strong><em>A1 N2</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;"><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: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">evidence of appropriate approach <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><em>eg</em> minimum value of \(\sin x\) is \( - 1,{\text{ }} - 1 + k,{\text{ }}f'(x) = 0,{\text{ }}\left( {\frac{{5\pi }}{4},{\text{ }}4} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">minimum value is \(4\) <strong><em>A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><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.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(p = - \frac{\pi }{4},{\text{ }}q = 5{\text{ }}\left( {{\text{accept \(\left( \begin{array}{c} - {\textstyle{\pi \over 4}}\\5\end{array} \right)\)}}} \right)\) <strong><em>A1A1 N2</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.5px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 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;">
[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 style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-size: medium; font-family: 'times new roman', times;">The following diagram shows a right-angled triangle, \(\rm{ABC}\), where \(\sin \rm{A} = \frac{5}{{13}}\).</span></p>
<p style="font: normal normal normal 21px/normal 'Times New Roman'; text-align: center; margin: 0px;"><br><img src="images/maths_1_2.png" 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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \(\cos A = \frac{{12}}{{13}}\).</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 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find \(\cos 2A\).</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: 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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">approach involving Pythagoras’ theorem <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;"><em>eg</em> \({5^2} + {x^2} = {13^2}\), labelling correct sides on triangle</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;">finding third side is 12 (may be seen on diagram) <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;">\(\cos A = \frac{{12}}{{13}}\) <strong><em>AG N0</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>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;">approach involving \({\sin ^2}\theta + {\cos ^2}\theta = 1\) <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;"><em>eg</em> \({\left( {\frac{5}{{13}}} \right)^2} + {\cos ^2}\theta = 1,{\text{ }}{x^2} + \frac{{25}}{{169}} = 1\)</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;">correct working <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;"><em>eg</em> \({\cos ^2}\theta = \frac{{144}}{{169}}\)</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;">\(\cos A = \frac{{12}}{{13}}\) <strong><em>AG N0</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>
<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 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">correct substitution into \(\cos 2\theta \) <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;"><em>eg</em> \(1 - 2{\left( {\frac{5}{{13}}} \right)^2},{\text{ }}2{\left( {\frac{{12}}{{13}}} \right)^2} - 1,{\text{ }}{\left( {\frac{{12}}{{13}}} \right)^2} - {\left( {\frac{5}{{13}}} \right)^2}\)</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;">correct working <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;"><em>eg</em> \(1 - \frac{{50}}{{169}},{\text{ }}\frac{{288}}{{169}} - 1,{\text{ }}\frac{{144}}{{169}} - \frac{{25}}{{169}}\)</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;">\(\cos 2A = \frac{{119}}{{169}}\) <strong><em>A1 N2</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>[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;">
[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><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = \cos 2x\) and \(g(x) = 2{x^2} - 1\) .</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 \(f\left( {\frac{\pi }{2}} \right)\) .</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;">Find \((g \circ f)\left( {\frac{\pi }{2}} \right)\) .</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><span style="font-family: times new roman,times; font-size: medium;">Given that \((g \circ f)(x)\) can be written as \(\cos (kx)\) , find the value of <em>k</em>, \(k \in \mathbb{Z}\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f\left( {\frac{\pi }{2}} \right) = \cos \pi \) </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;">\( = - 1\) <em><strong>A1 N2</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((g \circ f)\left( {\frac{\pi }{2}} \right) = g( - 1)\) \(( = 2{( - 1)^2} - 1)\) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(= 1\) <em><strong>A1 N2</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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((g \circ f)(x) = 2{(\cos (2x))^2} - 1\) \(( = 2{\cos ^2}(2x) - 1)\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of \(2{\cos ^2}\theta - 1 = \cos 2\theta \) (seen anywhere) <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((g \circ f)(x) = \cos 4x\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(k = 4\) <em><strong>A1 N2</strong></em></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>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">In part (a), a number of candidates were not able to evaluate \(\cos \pi \) , either leaving it or </span><span style="font-family: times new roman,times; font-size: medium;">evaluating it incorrectly.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Almost all candidates evaluated the composite function in part (b) in the given order, many </span><span style="font-family: times new roman,times; font-size: medium;">earning follow-through marks for incorrect answers from part (a). On both parts (a) and (b), </span><span style="font-family: times new roman,times; font-size: medium;">there were candidates who correctly used double-angle formulas to come up with correct </span><span style="font-family: times new roman,times; font-size: medium;">answers; while this is a valid method, it required unnecessary additional work.</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;">Candidates were not as successful in part (c). Many tried to use double-angle formulas, but either used the formula incorrectly or used it to write the expression in terms of \(\cos x\) and went no further. There were a number of cases in which the candidates "accidentally" came up with the correct answer based on errors or lucky guesses and did not earn credit for their final answer. Only a few candidates recognized the correct method of solution.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The following diagram shows a triangle <span class="s1">ABC </span>and a sector <span class="s1">BDC </span>of a circle with centre <span class="s1">B </span>and radius <span class="s1">6 cm</span>. The points <span class="s1">A </span>, <span class="s1">B </span>and <span class="s1">D </span>are on the same line.</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2017-02-02_om_09.25.23.png" alt="M16/5/MATME/SP1/ENG/TZ2/05"></p>
<p class="p1" style="text-align: center;"><span class="s1">\({\text{AB}} = 2\sqrt 3 {\text{ cm, BC}} = 6{\text{ cm, area of triangle ABC}} = 3\sqrt 3{\text{ c}}{{\text{m}}^{\text{2}}}{\rm{, A\hat BC}}\) </span>is obtuse.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \({\rm{A\hat BC}}\).</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 class="p1">Find the exact area of the sector <span class="s1">BDC</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"><strong>METHOD 1</strong></p>
<p class="p1">correct substitution into formula for area of triangle <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B,{\text{ }}6\sqrt 3 \sin B,{\text{ }}\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B = 3\sqrt 3 \)</p>
<p class="p1">correct working <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(6\sqrt 3 \sin B = 3\sqrt 3 ,{\text{ }}\sin B = \frac{{3\sqrt 3 }}{{\frac{1}{2}(6)2\sqrt 3 }}\)</p>
<p class="p1"><span class="Apple-converted-space">\(\sin B = \frac{1}{2}\) </span><strong><em>(A1)</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(\frac{\pi }{6}(30^\circ )\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p1"><span class="s2">\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\) <span class="Apple-converted-space"> </span></span><strong><em>A1 <span class="Apple-converted-space"> </span>N3</em></strong></p>
<p class="p1"><strong>METHOD 2</strong></p>
<p class="p2">(using height of triangle ABC by drawing perpendicular segment from C to AD<span class="s1">)</span></p>
<p class="p1">correct substitution into formula for area of triangle <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\frac{1}{2}\left( {2\sqrt 3 } \right)(h) = 3\sqrt 3 ,{\text{ }}h\sqrt 3 \)</p>
<p class="p1">correct working <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(h\sqrt 3 = 3\sqrt 3 \)</p>
<p class="p2">height of triangle is 3 <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\({\rm{C\hat BD}} = \frac{\pi }{6}(30^\circ )\) </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p1"><span class="s2">\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\) <span class="Apple-converted-space"> </span></span><strong><em>A1 <span class="Apple-converted-space"> </span>N3</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">recognizing supplementary angle <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\({\rm{C\hat BD}} = \frac{\pi }{6},{\text{ sector}} = \frac{1}{2}(180 - {\rm{A\hat BC)(}}{{\text{6}}^2})\)</p>
<p class="p1">correct substitution into formula for area of sector <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{\pi }{6} \times {6^2},{\text{ }}\pi ({6^2})\left( {\frac{{30}}{{360}}} \right)\)</p>
<p class="p1">\({\text{area}} = 3\pi {\text{ }}({\text{c}}{{\text{m}}^2})\) <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1"><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;">
<p class="p1">In part (a) of this question, the large majority of candidates substituted correctly into the area formula for the triangle, though algebraic errors kept some of them from simplifying the equation to \(\sin {\rm{A\hat BC = }}\frac{1}{2}\). Unfortunately, a number of candidates who got to this point often did not know the correct angles that correspond with this sine value.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (b), many candidates realized that \({\rm{C\hat BD}}\) was the supplement of \({\rm{A\hat BC}}\). However, at this point many candidates substituted 30°, or their follow-through angle in degrees, into the formula for the area of a sector found in the formula booklet, not understanding that this formula only works for angles in radians.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(\overrightarrow {{\text{OA}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 4 \end{array}} \right)\) and \(\overrightarrow {{\text{OB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 3 \end{array}} \right)\).</p>
</div>
<div class="specification">
<p class="p1">The point <span class="s1">C </span>is such that \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ { - 1} \end{array}} \right)\).</p>
</div>
<div class="specification">
<p class="p1">The following diagram shows triangle <span class="s1">ABC</span>. Let <span class="s1">D </span>be a point on <span class="s1">[BC]</span>, with acute angle \({\text{ADC}} = \theta \).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2017-03-03_om_05.46.25.png" alt="N16/5/MATME/SP1/ENG/TZ0/08.c.d.e"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Find \(\overrightarrow {{\text{AB}}} \).</p>
<p>(ii) Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).</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 the coordinates of <span class="s1">C </span><span class="s2">are \(( - 2,{\text{ }}1,{\text{ }}3)\).</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 class="p1">Write down an expression in terms of \(\theta \) <span class="s1">for</span></p>
<p class="p2">(i) <span class="Apple-converted-space"> </span>angle <span class="s2">ADB</span><span class="s1">;</span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>area of triangle <span class="s2">ABD</span>.</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"><span class="s1">Given that \(\frac{{{\text{area }}\Delta {\text{ABD}}}}{{{\text{area }}\Delta {\text{ACD}}}} = 3\)</span>, show that \(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\).</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 or otherwise, find the coordinates of point <span class="s1">D</span>.</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">(i) <span class="Apple-converted-space"> </span>valid approach to find \(\overrightarrow {{\text{AB}}} \)</p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OB}}} - \overrightarrow {{\text{OA}}} ,{\text{ }}\left( {\begin{array}{*{20}{c}} {4 - ( - 1)} \\ {1 - 0} \\ {3 - 4} \end{array}} \right)\)</p>
<p class="p2"><span class="Apple-converted-space">\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 5 \\ 1 \\ { - 1} \end{array}} \right)\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>valid approach to find \(\left| {\overrightarrow {{\text{AB}}} } \right|\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\sqrt {{{(5)}^2} + {{(1)}^2} + {{( - 1)}^2}} \)</p>
<p class="p2"><span class="s2">\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {27} \) <span class="Apple-converted-space"> </span></span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p2"><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">correct approach <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 1 \\ { - 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 4 \end{array}} \right)\)</p>
<p class="p3"><span class="s1">\(C\) </span>has coordinates \(( - 2,{\text{ }}1,{\text{ }}3)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></span></p>
<p class="p2"><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \({\rm{A\hat DB}} = \pi - \theta ,{\rm{ \hat D}} = 180 - \theta \)</span> <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N1</em></strong></span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>any correct expression for the area involving \(\theta \) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N1</em></strong></span></p>
<p class="p3"><em>eg</em>\(\,\,\,\,\,\)\({\text{area}} = \frac{1}{2} \times {\text{AD}} \times {\text{BD}} \times \sin (180 - \theta ),{\text{ }}\frac{1}{2}ab\sin \theta ,{\text{ }}\frac{1}{2}\left| {\overrightarrow {{\text{DA}}} } \right|\left| {\overrightarrow {{\text{DB}}} } \right|\sin (\pi - \theta )\)</p>
<p class="p3"><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"><span class="s1"><strong>METHOD 1 </strong></span>(using sine formula for area)</p>
<p class="p1">correct expression for the area of triangle ACD <span class="s1">(seen anywhere) <span class="Apple-converted-space"> </span></span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta \)</p>
<p class="p3">correct equation involving areas <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\frac{{\frac{1}{2}{\text{AD}} \times {\text{BD}} \times \sin (\pi - \theta )}}{{\frac{1}{2}{\text{AD}} \times {\text{DC}} \times \sin \theta }} = 3\)</p>
<p class="p1">recognizing that \(\sin (\pi - \theta ) = \sin \theta \) <span class="s1">(seen anywhere) <span class="Apple-converted-space"> </span></span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p3"><span class="s3">\(\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3\) </span>(seen anywhere) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p3">correct approach using ratio <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(3\overrightarrow {{\text{DC}}} + \overrightarrow {{\text{DC}}} = \overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{BC}}} = 4\overrightarrow {{\text{DC}}} \)</p>
<p class="p1">correct ratio \(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></span></p>
<p class="p3"><strong>METHOD 2 </strong>(Geometric approach)</p>
<p class="p3">recognising \(\Delta {\text{ABD}}\) and \(\Delta {\text{ACD}}\) have same height <span class="Apple-converted-space"> </span><span class="s2"><strong><em>(A1)</em></strong></span></p>
<p class="p1"><span class="s1"><em>eg</em>\(\,\,\,\,\,\)use of \(h\) </span>for both triangles, \(\frac{{\frac{1}{2}{\text{BD}} \times h}}{{\frac{1}{2}{\text{CD}} \times h}} = 3\)</p>
<p class="p3">correct approach <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A2</em></strong></span></p>
<p class="p1"><span class="s2"><em>eg</em>\(\,\,\,\,\,\)\({\text{BD}} = 3x\) </span>and \({\text{DC}} = x,{\text{ }}\frac{{{\text{BD}}}}{{{\text{DC}}}} = 3\)</p>
<p class="p3">correct working <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A2</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\({\text{BC}} = 4x,{\text{ BD}} + {\text{DC}} = 4{\text{DC, }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3x}}{{4x}},{\text{ }}\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{{3{\text{DC}}}}{{4{\text{DC}}}}\)</p>
<p class="p2"><span class="Apple-converted-space">\(\frac{{{\text{BD}}}}{{{\text{BC}}}} = \frac{3}{4}\) </span><strong><em>AG <span class="Apple-converted-space"> </span>N0</em></strong></p>
<p class="p2"><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">correct working (seen anywhere) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{BD}}} = \frac{3}{4}\overrightarrow {{\text{BC}}} ,{\text{ }}\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}\overrightarrow {{\text{CD}}} = \frac{1}{4}\overrightarrow {{\text{CB}}} \)</p>
<p class="p1">valid approach (seen anywhere) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(M1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \overrightarrow {{\text{BD}}} ,{\text{ }}\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right)\)</p>
<p class="p1">correct working to find \(x\)-coordinate <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(A1)</em></strong></span></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 3 \end{array}} \right) + \frac{3}{4}\left( {\begin{array}{*{20}{c}} { - 6} \\ 0 \\ 0 \end{array}} \right),{\text{ }}x = 4 + \frac{3}{4}( - 6),{\text{ }} - 2 + \frac{1}{4}(6)\)</p>
<p class="p3">D is \(\left( { - \frac{1}{2},{\text{ }}1,{\text{ }}3} \right)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1 <span class="Apple-converted-space"> </span>N3</em></strong></span></p>
<p class="p2"><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;">
[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>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The diagram below shows part of the graph of \(f(x) = a\cos (b(x - c)) - 1\) , where \(a > 0\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/huw.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The point \({\rm{P}}\left( {\frac{\pi }{4},2} \right)\) is a maximum point and the point \({\rm{Q}}\left( {\frac{{3\pi }}{4}, - 4} \right)\) is a minimum point.</span></p>
<p> </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 value of <em>a</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that the period of <em>f</em> is \(\pi \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence, find the value of <em>b</em> .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Given that \(0 < c < \pi \) , write down the value of <em>c</em> .</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">c.</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;">evidence of valid approach <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{{\text{max }}y{\text{ value}} - {\text{min }}y{\text{ value}}}}{2}\) , distance from \(y = - 1\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(a = 3\) <em><strong>A1 N2</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;">(i) evidence of valid approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. finding difference in <em>x</em>-coordinates, \(\frac{\pi }{2}\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of doubling <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(2 \times \left( {\frac{\pi }{2}} \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({\text{period}} = \pi \) <strong><em>AG N0</em></strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) evidence of valid approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(b = \frac{{2\pi }}{\pi }\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = 2\) <em><strong>A1 N2</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 marks] </span></strong></em></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(c = \frac{\pi }{4}\) <em><strong> A1 N1</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark] </span></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;">
<p><span style="font-family: times new roman,times; font-size: medium;">A pleasing number of candidates correctly found the values of <em>a</em>,<em> b</em>, and <em>c</em> for this sinusoidal graph. </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;">A pleasing number of candidates correctly found the values of <em>a</em>, <em>b</em>, and <em>c</em> for this sinusoidal graph. Some candidates had trouble showing that the period was \(\pi \) , either incorrectly adding the given \(\pi /4\) and \(3\pi /4\) or using the value of <em>b</em> that they found first for part (b)(ii). </span></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">A pleasing number of candidates correctly found the values of <em>a</em>, <em>b</em>, and <em>c</em> for this sinusoidal graph. Some candidates had trouble showing that the period was \(\pi \) , either incorrectly adding the given \(\pi /4\) and \(\pi /3\) or using the value of <em>b</em> that they found first for part (b)(ii). </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;"> Let \(p = \sin 40^\circ \) and \(q = \cos 110^\circ \) . Give your answers to the following in terms of <em>p</em> and/or <em>q .</em></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;">Write down an expression for </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \(\sin 140^\circ \) ; </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> (ii) \(\cos 70^\circ \) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Find an expression for \(\cos 140^\circ \) .</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;">Find an expression for \(\tan 140^\circ \) .</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\sin 140^\circ = p\) <em><strong>A1 N1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\cos 70^\circ = - q\) <em><strong>A1 N1</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(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of using \({\sin ^2}\theta + {\cos ^2}\theta = 1\) <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. diagram, \(\sqrt {1 - {p^2}} \) (seen anywhere)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\cos 140^\circ = \pm \sqrt {1 - {p^2}} \) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\cos 140^\circ = - \sqrt {1 - {p^2}} \) <em><strong>A1 N2</strong></em></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 2</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of using \(\cos 2\theta = 2{\cos ^2}\theta - 1\) <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\cos 140^\circ = 2{\cos ^2}70 - 1\) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos 140^\circ = 2{( - q)^2} - 1\) \(( = 2{q^2} - 1)\) <em><strong>A1 N2</strong></em></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 align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\tan 140^\circ = \frac{{\sin 140^\circ }}{{\cos 140^\circ }} = - \frac{p}{{\sqrt {1 - {p^2}} }}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N1</span></strong></em></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 2</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\tan 140^\circ = \frac{p}{{2{q^2} - 1}}\) <em><strong>A1 N1</strong></em></span></p>
<p align="LEFT"><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</span></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;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">This was one of the most difficult problems for the candidates. Even the strongest candidates </span><span style="font-family: times new roman,times; font-size: medium;">had a hard time with this one and only a few received any marks at all.</span></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many did not appear </span><span style="font-family: times new roman,times; font-size: medium;">to know the relationships between trigonometric functions of supplementary angles and that </span><span style="font-family: times new roman,times; font-size: medium;">the use of \({\sin ^2}x + {\cos ^2}x = 1\) results in a \( \pm \) value. The application of a double angle formula </span><span style="font-family: times new roman,times; font-size: medium;">also seemed weak.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">This was one of the most difficult problems for the candidates. Even the strongest candidates </span><span style="font-family: times new roman,times; font-size: medium;">had a hard time with this one and only a few received any marks at all. Many did not appear </span><span style="font-family: times new roman,times; font-size: medium;">to know the relationships between trigonometric functions of supplementary angles and that </span><span style="font-family: times new roman,times; font-size: medium;">the use of \({\sin ^2}x + {\cos ^2}x = 1\) results in a \( \pm \) value. The application of a double angle formula </span><span style="font-family: times new roman,times; font-size: medium;">also seemed weak.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p>The following diagram shows triangle ABC, with \({\text{AB}} = 3{\text{ cm}}\), \({\text{BC}} = 8{\text{ cm}}\), and \({\rm{A\hat BC = }}\frac{\pi }{3}\).</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2018-02-11_om_09.17.57.png" alt="N17/5/MATME/SP1/ENG/TZ0/04"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \({\text{AC}} = 7{\text{ cm}}\).</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 shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2018-02-11_om_10.50.00.png" alt="N17/5/MATME/SP1/ENG/TZ0/04.b"></p>
<p>Find the exact perimeter of this shape.</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>evidence of choosing the cosine rule <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} - ab\cos C\)</p>
<p>correct substitution into RHS of cosine rule <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({3^2} + {8^2} - 2 \times 3 \times 8 \times \cos \frac{\pi }{3}\)</p>
<p>evidence of correct value for \(\cos \frac{\pi }{3}\) (may be seen anywhere, including in cosine rule) <strong><em>A1</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\cos \frac{\pi }{3} = \frac{1}{2},{\text{ A}}{{\text{C}}^2} = 9 + 64 - \left( {48 \times \frac{1}{2}} \right),{\text{ }}9 + 64 - 24\)</p>
<p>correct working clearly leading to answer <strong><em>A1</em></strong></p>
<p>e<em>g</em>\(\,\,\,\,\,\)\({\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49} \)</p>
<p>\({\text{AC}} = 7{\text{ (cm)}}\) <strong><em>AG N0</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award no marks if the only working seen is \({\text{A}}{{\text{C}}^2} = 49\) or \({\text{AC}} = \sqrt {49} \) (or similar).</p>
<p> </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>correct substitution for semicircle <strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{semicircle}} = \frac{1}{2}(2\pi \times 3.5),{\text{ }}\frac{1}{2} \times \pi \times 7,{\text{ }}3.5\pi \)</p>
<p>valid approach (seen anywhere) <strong><em>(M1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{perimeter}} = {\text{AB}} + {\text{BC}} + {\text{semicircle, }}3 + 8 + \left( {\frac{1}{2} \times 2 \times \pi \times \frac{7}{2}} \right),{\text{ }}8 + 3 + 3.5\pi \)</p>
<p>\(11 + \frac{7}{2}\pi {\text{ }}( = 3.5\pi + 11){\text{ (cm)}}\) <strong><em>A1 N2</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>The following diagram shows triangle PQR.</p>
<p style="text-align: center;"><img src="images/Schermafbeelding_2017-08-11_om_09.36.55.png" alt="M17/5/MATME/SP1/ENG/TZ1/03"></p>
<p>Find PR.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><strong>METHOD 1 </strong></p>
<p>evidence of choosing the sine rule <em><strong>(M1)</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}}\)</p>
<p>correct substitution <em><strong>A1</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\frac{x}{{\sin 30}} = \frac{{13}}{{\sin 45}},{\text{ }}\frac{{13\sin 30}}{{\sin 45}}\)</p>
<p>\(\sin 30 = \frac{1}{2},{\text{ }}\sin 45 = \frac{1}{{\sqrt 2 }}\) <em><strong>(A1)(A1)</strong></em></p>
<p>correct working <em><strong>A1</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{{13}}{{\frac{1}{{\sqrt 2 }}}},{\text{ }}\frac{1}{2} \times 13 \times \frac{2}{{\sqrt 2 }},{\text{ }}13 \times \frac{1}{2} \times \sqrt 2 \)</p>
<p>correct answer <em><strong>A1</strong></em> <em><strong>N3</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)</p>
<p><strong>METHOD 2 (using height of Δ</strong><strong>PQR</strong><strong>)</strong></p>
<p>valid approach to find height of ΔPQR <em><strong>(M1)</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\sin 30 = \frac{x}{{13}},{\text{ }}\cos 60 = \frac{x}{{13}}\)</p>
<p>\(\sin 30 = \frac{1}{2}{\text{ or }}\cos 60 = \frac{1}{2}\) <em><strong>(A1)</strong></em></p>
<p>\({\text{height}} = 6.5\) <em><strong>A1</strong></em></p>
<p>correct working <em><strong>A1</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\sin 45 = \frac{{6.5}}{{{\text{PR}}}},{\text{ }}\sqrt {{{6.5}^2} + {{6.5}^2}} \)</p>
<p>correct working <em><strong>(A1)</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\(\sin 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\cos 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\sqrt {\frac{{169 \times 2}}{4}} \)</p>
<p>correct answer <em><strong>A1</strong></em> <em><strong>N3</strong></em></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)</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><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = {{\rm{e}}^{ - 3x}}\) and \(g(x) = \sin \left( {x - \frac{\pi }{3}} \right)\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Write down</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(f'(x)\) ;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(g'(x)\) .</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;">Let \(h(x) = {{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right)\) . Find the exact value of \(h'\left( {\frac{\pi }{3}} \right)\) .</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><span style="font-family: times new roman,times; font-size: medium;">(i) \( - 3{{\rm{e}}^{ - 3x}}\) <em><strong> A1 N1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\cos \left( {x - \frac{\pi }{3}} \right)\) <em><strong>A1 N1</strong></em> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 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;">evidence of choosing product rule <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(uv' + vu'\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct expression <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 3{{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3x}}\cos \left( {x - \frac{\pi }{3}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">complete correct substitution of \(x = \frac{\pi }{3}\) <em><strong>(A1)</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 3{{\rm{e}}^{ - 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(h'\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ - \pi }}\) <em><strong>A1 N3</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[4 marks]</span></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><span style="font-family: times new roman,times; font-size: medium;">A good number of candidates found the correct derivative expressions in (a). Many applied the product rule, although with mixed success. </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;">Often the substitution of \({\frac{\pi }{3}}\) was incomplete or not done at all. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Given that \(\cos A = \frac{1}{3}\) </span><span style="font-family: times new roman,times; font-size: medium;">and \(0 \le A \le \frac{\pi }{2}\) </span><span style="font-family: times new roman,times; font-size: medium;">, find \(\cos 2A\) .</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><span style="font-family: times new roman,times; font-size: medium;">Given that \(\sin B = \frac{2}{3}\) and \(\frac{\pi }{2} \le B \le \pi \) , find \(\cos B\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of choosing the formula \(\cos 2A = 2{\cos ^2}A - 1\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: If they choose another correct formula, do not award the <em><strong>M1</strong></em> unless there is </span><span style="font-family: times new roman,times; font-size: medium;">evidence of finding \({\sin ^2}A = 1 - \frac{1}{9}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong></em></span></p>
<p> <span style="font-family: times new roman,times; font-size: medium;">e.g.\(\cos 2A = {\left( {\frac{1}{3}} \right)^2} - \frac{8}{9}\) , \(\cos 2A = 2 \times {\left( {\frac{1}{3}} \right)^2} - 1\)</span></p>
<p> <span style="font-family: times new roman,times; font-size: medium;">\(\cos 2A = - \frac{7}{9}\) <em><strong>A1 N2</strong></em></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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using \({\sin ^2}B + {\cos ^2}B = 1\) <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\left( {\frac{2}{3}} \right)^2} + {\cos ^2}B = 1\) , \(\sqrt {\frac{5}{9}} \) </span><span style="font-family: times new roman,times; font-size: medium;">(seen anywhere),</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\cos B = \pm \sqrt {\frac{5}{9}} \) \(\left( { = \pm \frac{{\sqrt 5 }}{3}} \right)\) </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;"> \(\cos B = - \sqrt {\frac{5}{9}} \) \(\left( { = - \frac{{\sqrt 5 }}{3}} \right)\) <em><strong>A1 N2</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;">diagram <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong><br><img src="images/bryan.png" alt></strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">for finding third side equals \(\sqrt 5 \) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos B = - \frac{{\sqrt 5 }}{3}\) <em><strong>A1 N2</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">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">This question was very poorly done, and knowledge of basic trigonometric identities and </span><span style="font-family: times new roman,times; font-size: medium;">values of trigonometric functions of obtuse angles seemed distinctly lacking. Candidates who </span><span style="font-family: times new roman,times; font-size: medium;">recognized the need of an identity for finding \(\cos 2A\) given \(\cos A\) seldom chose the most </span><span style="font-family: times new roman,times; font-size: medium;">appropriate of the three and even when they did often used it incorrectly with expressions </span><span style="font-family: times new roman,times; font-size: medium;">such as \(2{\cos ^2}\frac{1}{9} - 1\) .</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">This question was very poorly done, and knowledge of basic trigonometric identities and </span><span style="font-family: times new roman,times; font-size: medium;">values of trigonometric functions of obtuse angles seemed distinctly lacking. Candidates who </span><span style="font-family: times new roman,times; font-size: medium;">recognized the need of an identity for finding \(\cos 2A\) given \(\cos A\) seldom chose the most </span><span style="font-family: times new roman,times; font-size: medium;">appropriate of the three and even when they did often used it incorrectly with expressions </span><span style="font-family: times new roman,times; font-size: medium;">such as \(2{\cos ^2}\frac{1}{9} - 1\) .</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(f(x) = 3\sin (\pi x)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down the amplitude of \(f\).</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 period of \(f\).</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">On the following grid, sketch the graph of \(y = f(x)\), for \(0 \le x \le 3\).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2016-01-22_om_07.47.02.png" alt></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">amplitude is 3 <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N1</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">valid approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\;\;\;{\text{period}} = \frac{{2\pi }}{\pi },{\text{ }}\frac{{360}}{\pi }\)</p>
<p class="p1">period is 2 <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><strong><em><img src="images/Schermafbeelding_2016-01-22_om_07.52.36.png" alt> A1</em></strong></p>
<p class="p1"><strong><em>A1A1A1 <span class="Apple-converted-space"> </span>N4</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>A1 </em></strong>for sine curve starting at (<span class="s1">0</span>, <span class="s1">0</span>) and correct period.</p>
<p class="p1">Only if this <strong><em>A1 </em></strong>is awarded, award the following for points in circles:</p>
<p class="p1"><strong><em>A1 </em></strong>for correct <em>x</em>-intercepts;</p>
<p class="p1"><strong><em>A1 </em></strong>for correct max and min points;</p>
<p class="p1"><strong><em>A1</em></strong> for correct domain.</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>The first two terms of an infinite geometric sequence are <em>u</em><sub>1</sub> = 18 and <em>u</em><sub>2</sub> = 12sin<sup>2</sup> <em>θ</em> , where 0 < <em>θ</em> < 2\(\pi \) , and <em>θ</em> ≠ \(\pi \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find an expression for <em>r</em> in terms of <em>θ</em>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the possible values of <em>r</em>.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).</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>Find the values of <em>θ</em> which give the greatest value of the sum.</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>valid approach <em><strong>(M1)</strong></em></p>
<p><em>eg</em> \(\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}\)</p>
<p>\(r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)\) <em><strong>A1 N2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>recognizing that sin<em>θ</em> is bounded <strong><em> (M1)</em></strong></p>
<p><em>eg</em> 0 <em>≤</em> sin<sup>2</sup> <em>θ </em>≤ 1, −1 ≤<em> sinθ </em>≤ 1, −1 <<em> sinθ </em>< 1</p>
<p>0 < r ≤ \(\frac{2}{3}\) <em><strong> A2 N3</strong></em></p>
<p><strong>Note:</strong> If working shown, award <em><strong>M1A1</strong></em> for correct values with incorrect inequality sign(s).<br>If no working shown, award <em><strong>N1</strong></em> for correct values with incorrect inequality sign(s).</p>
<p><em><strong>[</strong></em><em><strong>3 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>correct substitution into formula for infinite sum <em><strong>A1</strong></em></p>
<p><em>eg </em>\(\frac{{18}}{{1 - \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)</p>
<p>evidence of choosing an appropriate rule for cos 2<em>θ</em> (seen anywhere) <em><strong>(M1)</strong></em></p>
<p><em>eg </em>cos 2<em>θ </em>= 1<em> </em>− 2 sin<sup>2</sup> <em>θ</em></p>
<p>correct substitution of identity/working (seen anywhere) <em><strong>(A1)</strong></em></p>
<p><em>eg</em> \(\frac{{18}}{{1 - \frac{2}{3}\left( {\frac{{1 - {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 - 2\left( {\frac{{1 - {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 - 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)</p>
<p>correct working that clearly leads to the given answer <em><strong>A1</strong></em></p>
<p><em>eg</em> \(\frac{{18 \times 3}}{{2 + \left( {1 - 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 - \left( {1 - {\text{cos}}\,2\theta } \right)}}\)</p>
<p>\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\) <em><strong>AG N0</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p> </p>
<p><strong>METHOD 1 </strong>(using differentiation)</p>
<p>recognizing \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0\) (seen anywhere) <em><strong>(M1)</strong></em></p>
<p>finding any correct expression for \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}\) <em><strong>(A1)</strong></em></p>
<p><em>eg </em>\(\frac{{0 - 54 \times \left( { - 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, - 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ - 2}}\,\left( { - 2\,{\text{sin}}\,2\,\theta } \right)\)</p>
<p>correct working <em><strong> (A1)</strong></em></p>
<p><em>eg </em> sin 2<em>θ</em> = 0</p>
<p>any correct value for sin<sup>−1</sup>(0) (seen anywhere) <em><strong>(A1)</strong></em></p>
<p><em>eg </em> 0, \(\pi \), … , sketch of sine curve with <em>x</em>-intercept(s) marked both correct values for 2<em>θ</em> (ignore additional values) <em><strong>(A1)</strong></em></p>
<p>2<em>θ </em>= \(\pi \), 3\(\pi \) (accept values in degrees)</p>
<p>both correct answers \(\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}\) <em><strong>A1 N4</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>A0</strong></em> if either or both correct answers are given in degrees.<br>Award <em><strong>A0 </strong></em>if additional values are given.</p>
<p> </p>
<p><strong>METHOD 2 </strong>(using denominator)</p>
<p>recognizing when S<sub>∞</sub> is greatest <em><strong>(M1)</strong></em></p>
<p><em>eg</em> 2 + cos 2<em>θ </em>is a minimum, 1−<em>r</em> is smallest<br>correct working <em><strong>(A1)</strong></em></p>
<p><em>eg </em>minimum value of 2 + cos 2<em>θ </em>is 1, minimum <em>r</em> = \(\frac{2}{3}\)</p>
<p>correct working <em><strong>(A1)</strong></em></p>
<p><em>eg </em>\({\text{cos}}\,2\,\theta = - 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta = 1\)</p>
<p><strong>EITHER</strong> (using cos 2<em>θ</em>)</p>
<p>any correct value for cos<sup>−1</sup>(−1) (seen anywhere) <em><strong>(A1)</strong></em></p>
<p><em>eg </em>\(\pi \), 3\(\pi \), … (accept values in degrees), sketch of cosine curve with <em>x</em>-intercept(s) marked</p>
<p>both correct values for 2<em>θ </em>(ignore additional values) <em><strong>(A1)</strong></em></p>
<p>2<em>θ </em>= \(\pi \), 3\(\pi \) (accept values in degrees)</p>
<p><strong>OR</strong> (using sin<em>θ</em>)</p>
<p>sin<em>θ = </em>±1 (A1)</p>
<p>sin<sup>−1</sup>(1) = \(\frac{\pi }{2}\) (accept values in degrees) (seen anywhere) <em><strong>A1</strong></em></p>
<p><strong>THEN</strong></p>
<p>both correct answers \(\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}\) <em><strong>A1 N4</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>A0</strong></em> if either or both correct answers are given in degrees.<br>Award <em><strong>A0 </strong></em>if additional values are given.</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="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The following diagram shows the graph of \(f(x) = a\cos (bx)\) , for \(0 \le x \le 4\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/M12P1TZ2Q3.jpg" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">There is a minimum point at P(2, − 3) and a maximum point at Q(4, 3) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Write down the value of <em>a</em> .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find the value of <em>b</em> .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<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 gradient of the curve at P.</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><span style="font-family: times new roman,times; font-size: medium;">Write down the equation of the normal to the curve at P.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</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) \(a = 3\) <em><strong>A1 N1</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>METHOD 1</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to find period <strong><em>(M1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. 4 , \(b = 4\) , \(\frac{{2\pi }}{b}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = \frac{{2\pi }}{4}\left( { = \frac{\pi }{2}} \right)\) <strong><em>A1 N2</em></strong> </span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[3 marks]</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to substitute coordinates <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(3\cos (2b) = - 3\) , \(3\cos (4b) = 3\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(b = \frac{{2\pi }}{4}\left( { = \frac{\pi }{2}} \right)\) <em><strong>A1 N2</strong></em> </span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[3 marks]</strong> </span></em></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">0 <em><strong>A1 N1</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[1 mark] </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;">recognizing that normal is perpendicular to tangent <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({m_1} \times {m_2} = - 1\) , \(m = - \frac{1}{0}\) , sketch of vertical line on diagram </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = 2\) (do not accept 2 or \(y = 2\) ) <em><strong>A1 N2</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">c.</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;">In part (a), many candidates were able to successfully write down the value of <em>a</em> as instructed by inspecting the graph and seeing the amplitude of the function is 3. Many also used a formulaic approach to reach the correct answer. When finding the value of <em>b</em>, there were many candidates who thought <em>b</em> was the period of the function, rather than \(\frac{{2\pi }}{{{\rm{period}}}}\) . </span></p>
<div class="question_part_label">a(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In part (b), the directions asked candidates to write down the gradient of the curve at the local minimum point P. However, many candidates spent a good deal of time finding the derivative of the function and finding the value of the derivative for the given value of <em>x</em>, rather than simply stating that the gradient of a curve at a minimum point is zero. </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;">For part (c), finding the equation of the normal to the curve, many candidates tried to work with algebraic equations involving negative reciprocal gradients, rather than recognizing that the equation of the vertical line was \(x = 2\) . There were also candidates who had trouble expressing the correct equation of a line parallel to the <em>y</em>-axis.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The following diagram shows a semicircle centre O, diameter [AB], with radius 2.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Let P be a point on the circumference, with \({\rm{P}}\widehat {\rm{O}}{\rm{B}} = \theta \) radians.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/three_mins.png" alt></span></p>
</div>
<div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let <em>S</em> be the total area of the two segments shaded in the diagram below.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/flo.png" alt></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 area of the triangle OPB, in terms of \(\theta \) .</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;">Explain why the area of triangle OPA is the same as the area triangle OPB.</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;">Show that \(S = 2(\pi - 2\sin \theta )\) .</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;">Find the value of \(\theta \) when <em>S</em> is a local minimum, justifying that it is a minimum.</span></p>
<div class="marks">[8]</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;">Find a value of \(\theta \) for which <em>S</em> has its greatest value.</span></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of using area of a triangle <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. </span><span style="font-family: times new roman,times; font-size: medium;">\(A = \frac{1}{2} \times 2 \times 2 \times \sin \theta \)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(A = 2\sin \theta \) <em><strong>A1 N2</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 align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}\widehat {\rm{O}}{\rm{A = }}\pi - \theta \) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({\text{area }}\Delta {\rm{OPA}} = \frac{1}{2}2 \times 2 \times \sin (\pi - \theta )\) \(( = 2\sin (\pi - \theta ))\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since \(\sin (\pi - \theta ) = \sin \theta \) <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then both triangles have the same area <em><strong> AG N0</strong></em></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 2</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">triangle OPA has the same height and the same base as triangle OPB <em><strong>R3</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then both triangles have the same area <em><strong>AG N0</strong></em></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;">area semicircle \( = \frac{1}{2} \times \pi {(2)^2}\) \(( = 2\pi )\) </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;"> \({\text{area }}\Delta {\rm{APB}} = 2\sin \theta + 2\sin \theta \) \(( = 4\sin \theta )\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(S{\text{ = area of semicircle}} - {\text{area }}\Delta {\rm{APB}}\) \(( = 2\pi - 4\sin \theta )\) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(S = 2(\pi - 2\sin \theta )\) <em><strong>AG N0</strong></em></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><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to differentiate <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{{\rm{d}}S}}{{{\rm{d}}\theta }} = - 4\cos \theta \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">setting derivative equal to 0 <em><strong> (M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct equation <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 4\cos \theta = 0\) , \(\cos \theta = 0\) , \(4\cos \theta = 0\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\theta = \frac{\pi }{2}\) </span> <em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N3</span></strong></em></p>
<p><strong> <span style="font-family: times new roman,times; font-size: medium;">EITHER</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using second derivative <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(S''(\theta ) = 4\sin \theta \) </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;"> \(S''\left( {\frac{\pi }{2}} \right) = 4\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it is a minimum because \(S''\left( {\frac{\pi }{2}} \right) > 0\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">R1 N0</span></strong></em></p>
<p><strong> <span style="font-family: times new roman,times; font-size: medium;">OR</span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of using first derivative <em><strong>(M1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> for \(\theta < \frac{\pi }{2},S'(\theta ) < 0\) (may use diagram) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">for \(\theta > \frac{\pi }{2},S'(\theta ) > 0\) (may use diagram) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it is a minimum since the derivative goes from negative to positive <em><strong>R1 N0</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;">\(2\pi - 4\sin \theta \) is minimum when \(4\sin \theta \) is a maximum <em><strong>R3</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4\sin \theta \) is a maximum when \(\sin \theta = 1\) <em><strong>(A2)</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\theta = \frac{\pi }{2}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A3 N3</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">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><em>S</em> is greatest when \(4\sin \theta \) is smallest (or equivalent) <em><strong>(R1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\theta = 0\) (or \(\pi \) ) <em><strong>A1 N2</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">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Most candidates could obtain the area of triangle OPB as equal to \(2\sin \theta \) , though \(2\theta \) was </span><span style="font-family: times new roman,times; font-size: medium;">given quite often as the area.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A minority recognized the equality of the sines of supplementary angles and </span><span style="font-family: times new roman,times; font-size: medium;">the term complementary was frequently used instead of supplementary. Only a handful of </span><span style="font-family: times new roman,times; font-size: medium;">candidates used the simple equal base and altitude argument.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Many candidates seemed to </span><span style="font-family: times new roman,times; font-size: medium;">see why \(S = 2(\pi - 2\sin \theta )\) but the arguments presented for showing why this result was true </span><span style="font-family: times new roman,times; font-size: medium;">were not very convincing in many cases. Explicit evidence of why the area of the semicircle </span><span style="font-family: times new roman,times; font-size: medium;">was \(2\pi \) was often missing as was an explanation for \(2(2\sin \theta )\) </span><span style="font-family: times new roman,times; font-size: medium;">and for subtraction.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Only a small number of candidates recognized the fact <em>S</em> would be minimum when sin was </span><span style="font-family: times new roman,times; font-size: medium;">maximum, leading to a simple non-calculus solution. Those who chose the calculus route </span><span style="font-family: times new roman,times; font-size: medium;">often had difficulty finding the derivative of <em>S</em>, failing in a significant number of cases to </span><span style="font-family: times new roman,times; font-size: medium;">recognize that the derivative of a constant is 0, and also going through painstaking application </span><span style="font-family: times new roman,times; font-size: medium;">of the product rule to find the simple derivative. When it came to justify a minimum, there was </span><span style="font-family: times new roman,times; font-size: medium;">evidence in some cases of using some form of valid test, but explanation of the test being </span><span style="font-family: times new roman,times; font-size: medium;">used was generally poor.</span></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Candidates who answered part (d) correctly generally did well in </span><span style="font-family: times new roman,times; font-size: medium;">part (e) as well, though answers outside the domain of \(\theta \) were frequently seen.</span></p>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">Let \(f(x) = 6x\sqrt {1 - {x^2}} \)</span>, for \( - 1 \leqslant x \leqslant 1\)<span class="s1">, and \(g(x) = \cos (x)\)</span>, for \(0 \leqslant x \leqslant \pi \)<span class="s1">.</span></p>
<p class="p2">Let \(h(x) = (f \circ g)(x)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write \(h(x)\) in the form \(a\sin (bx)\), where \(a,{\text{ }}b \in \mathbb{Z}\).</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 class="p1">Hence find the range of \(h\).</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">attempt to form composite in any order <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p2"><em>eg</em>\(\,\,\,\,\,\)\(f\left( {g(x)} \right),{\text{ }}\cos \left( {6x\sqrt {1 - {x^2}} } \right)\)</p>
<p class="p1">correct working <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(6\cos x\sqrt {1 - {{\cos }^2}x} \)</p>
<p class="p1">correct application of Pythagorean identity (do not accept \({\sin ^2}x + {\cos ^2}x = 1\)) <span class="Apple-converted-space"> </span><strong><em>(A1)</em></strong></p>
<p><em>eg</em>\(\,\,\,\,\,\)\({\sin ^2}x = 1 - {\cos ^2}x,{\text{ }}6\cos x\sin x,{\text{ }}6\cos x \left| \sin x\right|\)</p>
<p class="p2">valid approach (do not accept \(2\sin x\cos x = \sin 2x\)<span class="s1">) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></span></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)\(3(2\cos x\sin x)\)</p>
<p class="p1"><span class="Apple-converted-space">\(h(x) = 3\sin 2x\) </span><strong><em>A1 <span class="Apple-converted-space"> </span>N3</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">valid approach <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1"><em>eg</em>\(\,\,\,\,\,\)amplitude \( = 3\), sketch with max and min \(y\)-values labelled, \( - 3 < y < 3\)</p>
<p class="p1">correct range <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N2</em></strong></p>
<p class="p2"><span class="s1"><em>eg</em>\(\,\,\,\,\,\)\( - 3 \leqslant y \leqslant 3\)</span>, \([ - 3,{\text{ }}3]\) from \( - 3\) to 3</p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Do not award <strong><em>A1 </em></strong>for \( - 3 < y < 3\) <span class="s2">or for “between \( - 3\)</span> and <span class="s2">3”.</span></p>
<p class="p1"><strong><em>[2 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">In part (a), nearly all candidates found the correct composite function in terms of \(\cos x\), though many did not get any further than this first step in their solution to the question. While some candidates seemed to recognize the need to use trigonometric identities, most were unsuccessful in finding the correct expression in the required form.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (b), very few candidates were able to provide the correct range of the function.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The diagram shows two concentric circles with centre O.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/222.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The radius of the smaller circle is 8 cm and the radius of the larger circle is 10 cm.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Points A, B and C are on the circumference of the larger circle such that \({\rm{A}}\widehat {\rm{O}}{\rm{B}}\) is \(\frac{\pi }{3}\) </span><span style="font-family: times new roman,times; font-size: medium;">radians.</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 length of the arc ACB .</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;">Find the area of the shaded region.</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><span style="font-family: times new roman,times; font-size: medium;">correct substitution in \(l = r\theta \) <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(10 \times \frac{\pi }{3}\) , \(\frac{1}{6} \times 2\pi \times 10\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> arc length</span><span style="font-family: Times New Roman; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;"> \( = \frac{{20\pi }}{6}\) \(\left( { = \frac{{10\pi }}{3}} \right)\)</span> </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2</span></strong></em></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;">area of large sector \( = \frac{1}{2} \times {10^2} \times \frac{\pi }{3}\) \(\left( { = \frac{{100\pi }}{6}} \right)\) </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;">area of small sector \( = \frac{1}{2} \times {8^2} \times \frac{\pi }{3}\) \(\left( { = \frac{{64\pi }}{6}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of valid approach (seen anywhere) <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. subtracting areas of two sectors, \(\frac{1}{2} \times \frac{\pi }{3}({10^2} - {8^2})\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area shaded <span lang="EN-US">\( = 6\pi \) (accept \(\frac{{36\pi }}{6}\) , etc.) </span></span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N3</span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [4 marks]</span></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><span style="font-family: times new roman,times; font-size: medium;">This question was very well done by the majority of candidates. Some candidates correctly substituted the values into the formulas, but failed to do the calculations and write their answers in finished form. </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;">Nearly all used the correct method of subtracting the sector areas in part (b), though multiplying with fractions proved challenging for some candidates. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The vertices of the triangle PQR are defined by the position vectors</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}}<br>4\\<br>{ - 3}\\<br>1<br>\end{array}} \right)\) , \(\overrightarrow {{\rm{OQ}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>{ - 1}\\<br>2<br>\end{array}} \right)\) and \(\overrightarrow {{\rm{OR}}} = \left( {\begin{array}{*{20}{c}}<br>6\\<br>{ - 1}\\<br>5<br>\end{array}} \right)\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\overrightarrow {{\rm{PQ}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">;</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\overrightarrow {{\rm{PR}}} \) </span><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><span style="font-family: times new roman,times; font-size: medium;"> Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .</span></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .</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><span style="font-family: times new roman,times; font-size: medium;">(i) evidence of approach <em><strong>(M1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} - {\rm{P}}\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}<br>{ - 1}\\<br>2\\<br>1<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 N2 </span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\overrightarrow {{\rm{PR}}} = \left( {\begin{array}{*{20}{c}}<br>2\\<br>2\\<br>4<br>\end{array}} \right)\) <em><strong>A1 N1</strong> </em></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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><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;">choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \) <em><strong>(A1)(A1)</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding \(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\) <em><strong>(A1) (A1)(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} = - 2 + 4 + 4( = 6)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( - 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting into formula for angle between two vectors <em><strong>M1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6 \times \sqrt {24} }}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">simplifying to expression clearly leading to \(\frac{1}{2}\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{6}{{\sqrt 6 \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\)</span> <em><strong> AG N0</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of choosing cosine rule (seen anywhere) <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\overrightarrow {{\rm{QR}}} = \left( {\begin{array}{*{20}{c}}<br>3\\<br>0\\<br>3<br>\end{array}} \right)\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \) <em><strong>(A1)(A1)(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} - {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6 \times \sqrt {24} }}\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 - 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) <em><strong> AG N0</strong> </em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 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;">(i) <strong>METHOD 1</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of appropriate approach <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting correctly <em><strong>(A1) </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 - {{\left( {\frac{1}{2}} \right)}^2}} \)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>A1 N3</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>METHOD 2</strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of approach </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. drawing a right triangle, finding the missing side <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\) <em><strong>A1 N3</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) evidence of appropriate approach <em><strong> (M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. attempt to substitute into \(\frac{1}{2}ab\sin C\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">correct substitution </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. area \( = \frac{1}{2}\sqrt 6 \times \sqrt {24} \times \frac{{\sqrt 3 }}{2}\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">area \( = 3\sqrt 3 \) <em><strong> A1 N2</strong> </em></span></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">c.</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;">Combining the vectors in (a) was generally well done, although some candidates reversed the subtraction, while others calculated the magnitudes. </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;">Many candidates successfully used scalar product and magnitude calculations to complete part (b). Alternatively, some used the cosine rule, and often achieved correct results. Some assumed the triangle was a right-angled triangle and thus did not earn full marks. Although PQR is indeed right-angled, in a “show that” question this attribute must be directly established. </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;">Many candidates attained the value for sine in (c) with little difficulty, some using the Pythagorean identity, while others knew the side relationships in a 30-60-90 triangle. Unfortunately, a good number of candidates then used the side values of \(1,2,\sqrt 3 \) to find the area of PQR , instead of the magnitudes of the vectors found in (a). Furthermore, the "hence" command was sometimes neglected as the value of sine was expected to be used in the approach. </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Solve the equation \(2\cos x = \sin 2x\) , for \(0 \le x \le 3\pi \) .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">using double-angle identity (seen anywhere) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sin 2x = 2\sin x\cos x\) , \(2\cos x = 2\sin x\cos x\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">evidence of valid attempt to solve equation <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(0 = 2\sin x\cos x - 2\cos x\) , \(2\cos x(1 - \sin x) = 0\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\cos x = 0\) , \(\sin x = 1\) <em><strong>A1A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{\pi }{2}\) , \(x = \frac{{3\pi }}{2}\) , \(x = \frac{{5\pi }}{2}\) <em><strong>A1A1A1 N4</strong></em></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 2</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/i20.png" alt></span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A1A1M1A1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Notes</strong>: Award <em><strong>A1</strong></em> for sketch of \(\sin 2x\) , <em><strong>A1</strong></em> for a sketch of \(2\cos x\) , <em><strong>M1</strong></em> for at least one </span><span style="font-family: times new roman,times; font-size: medium;">intersection point seen, and <em><strong>A1</strong></em> for 3 approximately correct intersection points. Accept sketches drawn outside \(\left[ {0,3\pi } \right]\) , even those with more than 3 intersections.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{\pi }{2}\) , \(x = \frac{{3\pi }}{2}\) , \(x = \frac{{5\pi }}{2}\) <em><strong>A1A1A1 N4</strong></em></span></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 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;">By far the most common error was to “cancel” the \(\cos x\) and find only two of the three solutions. It was disappointing how few candidates solved this by setting factors equal to zero. Some candidates wrote all three answers from \(\sin x = 1\) , which only earned two of the three final marks. On a brighter note, many candidates found the \(\frac{{5\pi }}{2}\) , which showed an appreciation for the period of the function as well as the domain restriction. A handful of candidates cleverly sketched both graphs and used the intersections to find the three solutions. </span></p>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . The following diagram shows the graph of \(f\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/movie.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The \(y\)-intercept is at (\(0\), \(1\)) , there is a minimum point at A (\(p\), \(q\)) and a maximum </span><span style="font-family: times new roman,times; font-size: medium;">point at B.</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 \(f'(x)\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Hence</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) show that \(q = - 2\) ;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) verify that A is a minimum point.</span></p>
<div class="marks">[10]</div>
<div class="question_part_label">b(i) and (ii).</div>
</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 maximum value of \(f(x)\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The function \(f(x)\) can be written in the form \(r\cos (x - a)\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Write down the value of <em>r</em> and of <em>a</em> .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f'(x) = - \sin x + \sqrt 3 \cos x\) <em><strong>A1A1 N2</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) at A, \(f'(x) = 0\) <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">correct working <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\sin x = \sqrt 3 \cos x\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\tan x = \sqrt 3 \) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">attempt to substitute <strong>their</strong> <em>x</em> into \(f(x)\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - \frac{1}{2} + \sqrt 3 \left( { - \frac{{\sqrt 3 }}{2}} \right)\)<br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"></span><span style="font-family: times new roman,times; font-size: medium;">correct working that clearly leads to \( - 2\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - \frac{1}{2} - \frac{3}{2}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span><span style="font-family: times new roman,times; font-size: medium;">\(q = - 2\) <em><strong>AG N0</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1A1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(f'(\pi ) = 0 - \sqrt 3 \) , \(f'(2\pi ) = 0 + \sqrt 3 \) </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f'(x)\) changes sign from negative to positive <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so A is a minimum <em><strong>AG N0</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[10 marks]</span></strong></em></p>
<div class="question_part_label">b(i) and (ii).</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">max when \(x = \frac{\pi }{3}\) </span><em><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;">correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\) </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;">e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">max value is 2 <em><strong>A1 N1</strong></em></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(r = 2\) , \(a = \frac{\pi }{3}\) <em><strong>A1A1 N2</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">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(i) and (ii).</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><span style="font-family: times new roman,times; font-size: medium;">A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/bike.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The point P(<em>x</em> , <em>y</em>) is a vertex of the rectangle and also lies on the circle. The angle </span><span style="font-family: times new roman,times; font-size: medium;">between (OP) and the <em>x</em>-axis is \(\theta \) radians, where \(0 \le \theta \le \frac{\pi }{2}\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Write down an expression in terms of \(\theta \) for</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(x\) ;</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(y\) .</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;">Let the area of the rectangle be <em>A</em>.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(A = 18\sin 2\theta \) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence, find the exact value of \(\theta \) which maximizes the area of </span><span style="font-family: times new roman,times; font-size: medium;">the rectangle.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) Use the second derivative to justify that this value of \(\theta \) does give </span><span style="font-family: times new roman,times; font-size: medium;">a maximum.</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><span style="font-family: times new roman,times; font-size: medium;">(i) \(x = 3\cos \theta \) <em><strong>A1 N1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(y = 3\sin \theta \) <em><strong> A1 N1</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;">finding area <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(A = 4 \times 3\sin \theta \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta \times 3\sin \theta \) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(A = 18(2\sin \theta \cos \theta )\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(A = 18\sin 2\theta \) <em><strong>AG N0</strong> </em></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;">(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \) <em><strong>A2 N2 </strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) for setting derivative equal to 0 <em><strong> (M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \(36\cos 2\theta = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(2\theta = \frac{\pi }{2}\) <em><strong> (A1)</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\theta = \frac{\pi }{4}\) <em><strong>A1 N2</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) valid reason (seen anywhere) <em><strong>R1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f''(x) < 0\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} = - 72\sin 2\theta \) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">evidence of substituting \(\frac{\pi }{4}\) <em><strong> M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">e.g. \( - 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( - 72\sin \left( {\frac{\pi }{2}} \right)\) , \( - 72\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\theta = \frac{\pi }{4}\) produces the maximum area <em><strong>AG N0</strong> </em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[8 marks]</strong> </span></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;">
<p><span style="font-family: times new roman,times; font-size: medium;">Candidates familiar with the circular nature of sine and cosine found part (a) accessible. However, a good number of candidates left this part blank, which suggests that there was difficulty interpreting the meaning of the<em> x</em> and <em>y</em> in the diagram. </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 with answers from (a) could begin part (b), but many worked backwards and thus earned no marks. In a "show that" question, a solution cannot begin with the answer given. The area of the rectangle could be found by using \(2x \times 2y\) , or by using the eight small triangles, but it was essential that the substitution of the double-angle formula was shown before writing the given answer. </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;">As the area function was given in part (b), many candidates correctly found the derivative in (c) and knew to set this derivative to zero for a maximum value. Many gave answers in degrees, however, despite the given domain in radians. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Although some candidates found the second derivative function correctly, few stated that the second derivative must be negative at a maximum value. Simply calculating a negative value is not sufficient for a justification. </span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(h(x) = \frac{{6x}}{{\cos x}}\) . Find \(h'(0)\) .</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 1 (quotient)</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">derivative of numerator is 6 <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">derivative of denominator is \( - \sin x\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">attempt to substitute into quotient rule <em><strong>(M1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">correct substitution <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{(\cos x)(6) - (6x)( - \sin x)}}{{{{(\cos x)}^2}}}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">substituting \(x = 0\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \(\frac{{(\cos 0)(6) - (6 \times 0)( - \sin 0)}}{{{{(\cos 0)}^2}}}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(h'(0) = 6\) <strong><em>A1 N2</em></strong></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">METHOD 2 (product)</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(h(x) = 6x \times {(\cos x)^{ - 1}}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">derivative of 6<em>x</em> is 6 <strong><em>(A1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">derivative of \({(\cos x)^{ - 1}}\) is \(( - {(\cos x)^{ - 2}}( - \sin x))\) <strong><em>(A1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">attempt to substitute into product rule <strong>(M1)</strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">correct substitution <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \((6x)( - {(\cos x)^{ - 2}}( - \sin x)) + (6){(\cos x)^{ - 1}}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">substituting \(x = 0\) <em><strong> (A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">e.g. \((6 \times 0)( - {(\cos 0)^{ - 2}}( - \sin 0)) + (6){(\cos 0)^{ - 1}}\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(h'(0) = 6\) <em><strong>A1 N2</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 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;">The majority of candidates were successful in using the quotient rule, and were able to earn most of the marks for this question. However, there were a large number of candidates who substituted correctly into the quotient rule, but then went on to make mistakes in simplifying this expression. These algebraic errors kept the candidates from earning the final mark for the correct answer. A few candidates tried to use the product rule to find the derivative, but they were generally not as successful as those who used the quotient rule. It was pleasing to note that most candidates did know the correct values for the sine and cosine of zero. </span></p>
</div>
<br><hr><br><div class="question">
<p>The following diagram shows a circle with centre O and radius<em> r</em> cm.</p>
<p><img src="data:image/png;base64,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"></p>
<p>The points A and B lie on the circumference of the circle, and \({\text{A}}\mathop {\text{O}}\limits^ \wedge {\text{B}}\) = <em>θ</em>. The area of the shaded sector AOB is 12 cm<sup>2</sup> and the length of arc AB is 6 cm.</p>
<p>Find the value of <em>r</em>.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>evidence of correctly substituting into circle formula (may be seen later) <em><strong>A1A1</strong></em><br><em>eg </em>\(\frac{1}{2}\theta {r^2} = 12,\,\,r\theta = 6\)</p>
<p>attempt to eliminate one variable <em><strong>(M1)</strong></em><br><em>eg</em> \(r = \frac{6}{\theta },\,\,\theta = \frac{1}{r},\,\,\frac{{\frac{1}{2}\theta {r^2}}}{{r\theta }} = \frac{{12}}{6}\)</p>
<p>correct elimination <em><strong>(A1)</strong></em><br><em>eg </em> \(\frac{1}{2} \times \frac{6}{r} \times {r^2} = 12,\,\,\frac{1}{2}\theta \times {\left( {\frac{6}{\theta }} \right)^2} = 12,\,\,A = \frac{1}{2} \times {r^2} \times \frac{l}{r},\,\,\frac{{{r^2}}}{{2r}} = 2\)</p>
<p>correct equation <em><strong>(A1)</strong></em><br><em>eg</em> \(\frac{1}{2} \times 6r = 12,\,\,\frac{1}{2} \times \frac{{36}}{\theta } = 12,\,\,12 = \frac{1}{2} \times {r^2} \times \frac{6}{r}\)</p>
<p>correct working <em><strong>(A1)</strong></em><br><em>eg</em> \(3r = 12,\,\,\frac{{18}}{\theta } = 12,\,\,\frac{r}{2} = 2,\,\,24 = 6r\)</p>
<p><em>r</em> = 4 (cm) <em><strong>A1 N2</strong></em></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>Six equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.<br>This is shown in the following diagram.</p>
<p><img src="data:image/png;base64,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"></p>
<p>The vectors <em><strong>p</strong></em> , <em><strong>q</strong></em> and <em><strong>r</strong></em> are shown on the diagram.</p>
<p>Find <em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p><strong>METHOD 1 </strong>(using |<em><strong>p</strong></em>| |2<em><strong>q</strong></em>| cos<em>θ</em>)</p>
<p>finding <em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r (A1)</strong></em></p>
<p><em>eg </em> 2<em><strong>q</strong></em>, <img src="data:image/png;base64,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"></p>
<p>| <em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r </strong></em>| = 2 × 3 (= 6) (seen anywhere) <em><strong>A1</strong></em></p>
<p>correct angle between <em><strong>p</strong></em> and <em><strong>q</strong></em> (seen anywhere) <em><strong>(A1)</strong></em></p>
<p>\(\frac{\pi }{3}\) (accept 60°)</p>
<p>substitution of <strong>their</strong> values <em><strong>(M1)</strong></em></p>
<p><em>eg</em> 3 × 6 × cos\(\left( {\frac{\pi }{3}} \right)\)</p>
<p>correct value for cos\(\left( {\frac{\pi }{3}} \right)\) (seen anywhere) <em><strong>(A1)</strong></em></p>
<p><em>eg</em> \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)</p>
<p><em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>) = 9 <em><strong> A1 N3</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong> (scalar product using distributive law)</p>
<p>correct expression for scalar distribution <em><strong>(A1)</strong></em></p>
<p>eg <em><strong>p</strong></em>• <em><strong>p</strong></em> + <em><strong>p</strong></em>•<em><strong>q</strong></em> + <em><strong>p</strong></em>•<em><strong>r</strong></em></p>
<p>three correct angles between the vector pairs (seen anywhere) <em><strong>(A2)</strong></em></p>
<p><em>eg </em> 0° between <em><strong>p</strong></em> and <em><strong>p</strong></em>, \(\frac{\pi }{3}\) between <em><strong>p</strong></em> and <em><strong>q</strong></em>, \(\frac{{2\pi }}{3}\) between <em><strong>p</strong></em> and <em><strong>r</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>A1</strong> </em>for only two correct angles.</p>
<p>substitution of <strong>their</strong> values <em><strong>(M1)</strong></em></p>
<p><em>eg </em> 3.3.cos0 +3.3.cos\(\frac{\pi }{3}\) + 3.3.cos120</p>
<p>one correct value for cos0, cos\(\left( {\frac{\pi }{3}} \right)\) or cos\(\left( {\frac{2\pi }{3}} \right)\) (seen anywhere) <em><strong>A1</strong></em></p>
<p><em>eg </em>\(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)</p>
<p><em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>) = 9 <em><strong> A1 N3</strong></em></p>
<p> </p>
<p><strong>METHOD 3</strong> (scalar product using relative position vectors)</p>
<p>valid attempt to find one component of <em><strong>p</strong></em> or <em><strong>r</strong></em> <em><strong>(M1)</strong></em></p>
<p><em>eg </em> sin 60 = \(\frac{x}{3}\), cos 60 = \(\frac{x}{3}\), one correct value \(\frac{3}{2},\,\,\frac{{3\sqrt 3 }}{2},\,\,\frac{{ - 3\sqrt 3 }}{2}\)</p>
<p>one correct vector (two or three dimensions) (seen anywhere) <em><strong>A1</strong></em></p>
<p><em>eg </em>\(p = \left( \begin{gathered}<br> \,\,\,\frac{3}{2} \hfill \\<br> \frac{{3\sqrt 3 }}{2} \hfill \\ <br>\end{gathered} \right),\,\,q = \left( \begin{gathered}<br> 3 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right),\,\,r = \left( \begin{gathered}<br> \,\,\,\,\frac{3}{2} \hfill \\<br> - \frac{{3\sqrt 3 }}{2} \hfill \\<br> \,\,\,\,0 \hfill \\ <br>\end{gathered} \right)\)</p>
<p>three correct vectors <em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r </strong></em>= 2<em><strong>q</strong></em> <em><strong>(A1)</strong></em></p>
<p><em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r </strong></em>= \(\left( \begin{gathered}<br> 6 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right)\) or \(\left( \begin{gathered}<br> 6 \hfill \\<br> 0 \hfill \\<br> 0 \hfill \\ <br>\end{gathered} \right)\) (seen anywhere, including scalar product) <em><strong>(A1)</strong></em></p>
<p>correct working <em><strong>(A1)</strong></em><br><em>eg </em>\(\left( {\frac{3}{2} \times 6} \right) + \left( {\frac{{3\sqrt 3 }}{2} \times 0} \right),\,\,9 + 0 + 0\)</p>
<p><em><strong>p</strong></em>•(<em><strong>p</strong></em> + <em><strong>q</strong></em> + <em><strong>r</strong></em>) = 9 <em><strong> A1 N3</strong></em></p>
<p><strong><em>[6 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
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