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</div><h2>SL Paper 1</h2><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^3} - {x^3}\) for which \(y = 1\) when \(x = 0\). Use Euler’s method with a step length of \(0.1\) to find an approximation for the value of \(y\) when \(x = 0.4\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">use of \(y \to y + h\frac{{{\text{d}}y}}{{{\text{d}}x}}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<table style="height: 137px; width: 417px;" border="0">
<tbody>
<tr>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">\(x\)</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">\(y\)</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">\({\text{d}}y{\text{/d}}x\)</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">\(h{\text{d}}y{\text{/d}}x\)</span></td>
<td> </td>
</tr>
<tr>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;"> 0.1</span></td>
<td><strong style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;"><em>(A1)</em></strong></td>
</tr>
<tr>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0.1</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1.1</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1.33</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0.133</span></td>
<td><strong style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;"><em>A1</em></strong></td>
</tr>
<tr>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0.2</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1.233</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1.866516337</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0.1866516337</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;"> </span><strong style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;"><em>A1</em></strong></td>
</tr>
<tr>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0.3</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1.419651634</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">2.834181181</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0.283418118</span></td>
<td><strong style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;"><em>A1</em></strong></td>
</tr>
<tr>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">0.4</span></td>
<td><span style="color: #3f3f3f; font-family: 'times new roman', times; font-size: medium; line-height: normal;">1.703069752</span></td>
<td> </td>
<td> </td>
<td><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>(A1)</em></strong></span></td>
</tr>
</tbody>
</table>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>After the first line, award <strong><em>A1 </em></strong>for each subsequent \(y\) value, provided it is correct to 3sf.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">approximate value of \(y(0.4) = 1.70\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept \(1.7\) or any answers that round to \(1.70\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p>Use the integral test to determine whether or not \(\sum\limits_{n = 2}^\infty {\frac{1}{{n{{\left( {{\text{ln}}\,n} \right)}^2}}}} \) converges.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>let \(u = {\text{ln}}\,x\) <em><strong>(M1)</strong></em></p>
<p>\( \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{1}{x}\)</p>
<p>\(\int {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \int {\frac{1}{{{u^2}}}} \,{\text{d}}u\) <em><strong>(A1)</strong></em></p>
<p>\( = - \frac{1}{u} = - \frac{1}{{{\text{ln}}\,x}}\) <em><strong>(A1)</strong></em></p>
<p>\(\int\limits_2^m {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \left[ { - \frac{1}{{{\text{ln}}\,x}}} \right]_2^m\) <em><strong>M1</strong></em></p>
<p>\( = \left[ { - \frac{1}{{{\text{ln}}\,m}} + - \frac{1}{{{\text{ln}}\,2}}} \right]\) <em><strong>A1</strong></em></p>
<p>as \(m \to \infty \), \( - \frac{1}{{{\text{ln}}\,m}} \to 0\) <em><strong>(A1)</strong></em></p>
<p>\(\int\limits_2^\infty {\frac{1}{{x{{\left( {{\text{ln}}\,x} \right)}^2}}}} \,{\text{d}}x = \frac{1}{{{\text{ln}}\,2}}\) and hence the series converges <em><strong>R1</strong></em></p>
<p><em><strong>[7 marks</strong><strong>]</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>Let</p>
<p>\({I_n} = \int_1^\infty {{x^n}{{\text{e}}^{ - x}}{\text{d}}x} \) where \(n \in \mathbb{N}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Using l’Hôpital’s rule, show that</p>
<p>\(\mathop {\lim }\limits_{x \to \infty } {x^n}{{\text{e}}^{ - x}} = 0\) where \(n \in \mathbb{N}\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that, for \(n \in {\mathbb{Z}^ + }\),</p>
<p>\[{I_n} = \alpha {{\text{e}}^{ - 1}} + \beta n{I_{n - 1}}\]</p>
<p>where \(\alpha \), \(\beta \) are constants to be determined.</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the value of \({I_3}\), giving your answer as a multiple of \({{\text{e}}^{ - 1}}\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>consider \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}}\) <strong><em>M1</em></strong></p>
<p>its value is \(\frac{\infty }{\infty }\) so we use l’Hôpital’s rule</p>
<p>\(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n{x^{n - 1}}}}{{{{\text{e}}^x}}}\) <strong><em>(A1)</em></strong></p>
<p>its value is still \(\frac{\infty }{\infty }\) so we need to differentiate numerator and denominator a further \(n - 1\) times <strong><em>(R1)</em></strong></p>
<p>this gives \(\mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{{\text{e}}^x}}}\) <strong><em>A1</em></strong></p>
<p>since the numerator is finite and the denominator \( \to \infty \), the limit is zero <strong><em>AG</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>attempt at integration by parts \(\left( {{I_n} = - \int_1^\infty {{x^n}{\text{d}}({{\text{e}}^{ - x}})} } \right)\) <strong><em>M1</em></strong></p>
<p>\({I_n} = - [{x^n}{{\text{e}}^{ - x}}]_1^\infty + n\int_1^\infty {{x^{n - 1}}{{\text{e}}^{ - x}}{\text{d}}x} \) <strong><em>A1A1</em></strong></p>
<p>\( = {{\text{e}}^{ - 1}} + n{I_{n - 1}}\) <strong><em>A1</em></strong></p>
<p>\(\alpha = \beta = 1\)</p>
<p><strong><em>[9 marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({I_3} = {{\text{e}}^{ - 1}} + 3{I_2}\) <strong><em>M1</em></strong></p>
<p>\( = {{\text{e}}^{ - 1}} + 3({{\text{e}}^{ - 1}} + 2{I_1})\) <strong><em>A1</em></strong></p>
<p>\( = 4{{\text{e}}^{ - 1}} + 6({{\text{e}}^{ - 1}} + {I_0})\) <strong><em>A1</em></strong></p>
<p>\( = 4{{\text{e}}^{ - 1}} + 6{{\text{e}}^{ - 1}} + 6\int_1^\infty {{{\text{e}}^{ - x}}{\text{d}}x} \)</p>
<p>\( = 10{{\text{e}}^{ - 1}} - 6[{{\text{e}}^{ - x}}]_1^\infty \) <strong><em>A1</em></strong></p>
<p>\( = 16{{\text{e}}^{ - 1}}\) <strong><em>A1</em></strong></p>
<p><strong><em>[9 marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<br><hr><br><div class="specification">
<p>Given that \(y\) is a function of \(x\), the function \(z\) is given by \(z = \frac{{y - x}}{{y + x}}\), where \(x \in \mathbb{R},\,\,x \ne 3,\,\,y + x \ne 0\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{{{\left( {y + x} \right)}^2}}}\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right)\).</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>Show that the differential equation \(f\left( x \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right) = {y^2} - {x^2}\) can be written as \(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z\).</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>Hence show that the solution to the differential equation \(\left( {x - 3} \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right) = {y^2} - {x^2}\) given that \(x = 4\) when \(y = 5\) is \(\frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}\).</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>\(z = \frac{{y - x}}{{y + x}}\)</p>
<p>\( \Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{{\left( {y + x} \right)\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} - 1} \right) - \left( {y - x} \right)\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} + 1} \right)}}{{{{\left( {y + x} \right)}^2}}}\) <em><strong>M1A1</strong></em></p>
<p>\( \Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{{y\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y - x - y\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y + x}}{{{{\left( {y + x} \right)}^2}}}\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{{{\left( {y + x} \right)}^2}}}\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right)\) <em><strong> AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(f\left( x \right)\left( {\frac{{{{\left( {y + x} \right)}^2}}}{2}} \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = {y^2} - {x^2}\) <em><strong>(M1)</strong></em></p>
<p>\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2\frac{{\left( {y - x} \right)\left( {y + x} \right)}}{{{{\left( {y + x} \right)}^2}}}\) <em><strong>A1</strong></em></p>
<p>\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2\frac{{\left( {y - x} \right)}}{{\left( {y + x} \right)}} = 2z\) <em><strong>AG</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><strong>METHOD 1</strong></p>
<p>\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z\)</p>
<p>\(\frac{1}{z}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{f\left( x \right)}}\)</p>
<p>\(\frac{1}{z}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{x - 3}}\) <em><strong>M1A1</strong></em></p>
<p><strong>EITHER</strong></p>
<p>\( \Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + c\) <em><strong>A1</strong></em></p>
<p>when \(y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}\) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow c = {\text{ln}}\frac{1}{9}\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + {\text{ln}}\frac{1}{9}\)</p>
<p>\( \Rightarrow {\text{ln}}\,z = {\text{ln}}{\left( {x - 3} \right)^2} - {\text{ln}}\,9\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow {\text{ln}}\,z = {\text{ln}}{\left( {\frac{{x - 3}}{3}} \right)^2}\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow z = {\left( {\frac{{x - 3}}{3}} \right)^2}\)</p>
<p><strong>OR</strong></p>
<p>\( \Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + {\text{ln}}\,c\) <em><strong>A1</strong></em></p>
<p>\(z = c{\left( {x - 3} \right)^2}\) <em><strong>M1A1</strong></em></p>
<p>when \(y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}\) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow c = \frac{1}{9}\) <em><strong>A1</strong></em></p>
<p><strong>THEN</strong></p>
<p>\( \Rightarrow \frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}\) <em><strong>AG</strong></em></p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z\)</p>
<p>\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} - 2z = 0\)</p>
<p>\(\frac{{{\text{d}}z}}{{{\text{d}}x}} - \frac{{2z}}{{x - 3}} = 0\) <em><strong>M1</strong></em></p>
<p>integrating factor is \({e^{\int {\frac{{ - 2}}{{x - 3}}} {\text{d}}x}}\) <em><strong>A1</strong></em></p>
<p>\({e^{\int {\frac{{ - 2}}{{x - 3}}} {\text{d}}x}} = {e^{ - 2\,{\text{ln}}\left( {x - 3} \right)}}\)</p>
<p>\( = \frac{1}{{{{\left( {x - 3} \right)}^2}}}\) <em><strong>A1</strong></em></p>
<p>hence \(\frac{{\text{d}}}{{{\text{d}}x}}\left[ {\frac{z}{{{{\left( {x - 3} \right)}^2}}}} \right] = 0\) <em><strong>M1</strong></em></p>
<p>\(z = A{\left( {x - 3} \right)^2}\) <em><strong>A1</strong></em></p>
<p>when when \(y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}\) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow A = \frac{1}{9}\) <em><strong>A1</strong></em></p>
<p>\( \Rightarrow \frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}\) <em><strong>AG</strong></em></p>
<p><em><strong>[7 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p>Find the interval of convergence of the series \(\sum\limits_{k = 1}^\infty {\frac{{{{(x - 3)}^k}}}{{{k^2}}}} \).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>ratio test \(\mathop {\lim }\limits_{k \to \infty } \left| {\frac{{{u_{k + 1}}}}{{{u_k}}}} \right| = \mathop {\lim }\limits_{k \to \infty } \left| {\frac{{{{(x - 3)}^{k + 1}}{k^2}}}{{{{(k + 1)}^2}{{(x - 3)}^k}}}} \right|\) <strong><em>M1A1</em></strong></p>
<p>\(\mathop {\lim }\limits_{k \to \infty } \left| {(x - 3)\frac{{{k^2}}}{{{{(k + 1)}^2}}}} \right|\)</p>
<p><strong>Note: </strong>Condone absence of limits and modulus signs in above.</p>
<p> </p>
<p>\(\left| {x - 3} \right|\mathop {\lim }\limits_{k \to \infty } \left| {{{\left( {\frac{1}{{1 + \frac{1}{k}}}} \right)}^2}} \right| - \left| {x - 3} \right|\) <strong><em>A1</em></strong></p>
<p>for convergence \(\left| {x - 3} \right| < 1\) <em>(</em><strong><em>M1)</em></strong></p>
<p>\( \Rightarrow - 1 < x - 3 < 1\)</p>
<p>\( \Rightarrow 2 < x < 4\) <strong><em>(A1)</em></strong></p>
<p>now we need to test end points. <strong><em>(M1)</em></strong></p>
<p>when \(x = 4\) we have \(\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}} \) which is a convergent series <strong><em>R1</em></strong></p>
<p>when \(x = 2\) we have \(\sum\limits_{k = 1}^\infty {\frac{{{{( - 1)}^k}}}{{{k^2}}} = - 1 + \frac{1}{{{2^2}}} - \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \ldots } \) which is convergent <strong><em>R1</em></strong></p>
<p>(alternating series/absolutely converging series)</p>
<p>hence the interval of convergence is \([2,{\text{ }}4]\) <strong><em>A1</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">This question was well answered in general although the presentation was sometimes poor.</p>
</div>
<br><hr><br><div class="specification">
<p class="p1">The function \(f:\mathbb{R} \to \mathbb{R}\) is defined by \(f:x \to \left\{ {\begin{array}{*{20}{c}} { - 3x + 1}&{{\text{for }}x < 0} \\ 1&{{\text{for }}x = 0} \\ {2{x^2} - 3x + 1}&{{\text{for }}x > 0} \end{array}} \right.\).</p>
<p class="p1">By considering limits prove that \(f\) is</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">continuous at \(x = 0\);</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">differentiable at \(x = 0\).</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 class="p1">consider \(\mathop {\lim }\limits_{h \to 0 + } f(0 + h) = \mathop {\lim }\limits_{h \to 0 + } (2{h^2} - 3h + 1)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = 1 = f(0)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(\mathop {\lim }\limits_{h \to 0 - } f(0 + h) = \mathop {\lim }\limits_{h \to 0 - } ( - 3h + 1)\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = 1 = f(0)\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">hence \(f\) is continuous at \(x = 0\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"> </p>
<p class="p1"><strong>Note: </strong>The \( = f(0)\) needs only to be seen once.</p>
<p class="p1"> </p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">consider</p>
<p class="p1"><span class="Apple-converted-space">\(\mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{f(0 + h) - f(0)}}{h}} \right) = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{2{h^2} - 3h + 1 - 1}}{h}} \right)\) </span><strong><em>M1A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = \mathop {\lim }\limits_{h \to 0 + } \left( {\frac{{2{h^2} - 3h}}{h}} \right) = - 3\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\(\mathop {\lim }\limits_{h \to 0 - } \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0 - } \frac{{ - 3h + 1 - 1}}{h} = - 3\) </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p1">hence \(f\) is differentiable at \(x = 0\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[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">Again this was a reasonably successful question for many candidates with full marks often being awarded. However a significant minority were let down by giving very informal and descriptive answers which only gained partial marks. As the command term in the question is “prove” there is a need for a degree of formality and an explicit use of limits was expected.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Again this was a reasonably successful question for many candidates with full marks often being awarded. However a significant minority were let down by giving very informal and descriptive answers which only gained partial marks. As the command term in the question is “prove” there is a need for a degree of formality and an explicit use of limits was expected.</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;">Consider the infinite series \(S = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n + 1}}\sin } \left( {\frac{1}{n}} \right)\) .</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 the series is conditionally convergent but not absolutely convergent.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(S > 0.4\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sin \left( {\frac{1}{n}} \right)\) decreases as <strong><em>n</em></strong> increases <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(\sin \left( {\frac{1}{n}} \right) \to 0\) as \(n \to \infty \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so using the alternating series test, the series is conditionally convergent <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">comparing (the absolute series) with the harmonic series:</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\sum\limits_{n = 1}^\infty {\frac{1}{n}} \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {\frac{1}{n}} \right)}}{{\frac{1}{n}}} = 1\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since the harmonic series is divergent, it follows by the limit comparison </span><span style="font-family: times new roman,times; font-size: medium;">theorem that the given series is not absolutely convergent <strong><em>R1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence the series is conditionally convergent <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [6 marks]</span></em></strong></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;">successive partial sums are</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(0.841…\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(0.362…\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(0.689…\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(0.441…\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since \(S\) lies between any pair of successive partial sums, it follows that \(S\) lies </span><span style="font-family: times new roman,times; font-size: medium;">between \(0.441…\) and \(0.689…\) <strong><em>R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">and is therefore greater than \(0.4\) <strong><em>AG</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Use of the facts that the error is always less than the modulus of the next </span><span style="font-family: times new roman,times; font-size: medium;">term, or the sequence of even partial sums gives lower bounds are equally </span><span style="font-family: times new roman,times; font-size: medium;">acceptable.</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>
<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>The function \(f\) is defined by</p>
<p>\[f(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}} + 2\cos x}}{4},{\text{ }}x \in \mathbb{R}.\]</p>
</div>
<div class="specification">
<p>The random variable \(X\) has a Poisson distribution with mean \(\mu \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \({f^{(4)}}x = f(x)\);</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By considering derivatives of \(f\), determine the first three non-zero terms of the Maclaurin series for \(f(x)\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Write down a series in terms of \(\mu \) for the probability \(p = {\text{P}}[X \equiv 0(\bmod 4)]\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(p = {{\text{e}}^{ - \mu }}f(\mu )\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the numerical value of \(p\) when \(\mu = 3\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.iii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(f’(x) = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}} - 2\sin x}}{4}\) <strong><em>(A1)</em></strong></p>
<p>\(f’’(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}} - 2\cos x}}{4}\) <strong><em>(A1)</em></strong></p>
<p>\(f’’’(x) = \frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}} + 2\sin x}}{4}\) <strong><em>(A1)</em></strong></p>
<p>\({f^{(4)}}(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ - x}} + 2\cos x}}{4} = f(x)\) <strong><em>AG</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>therefore,</p>
<p>\(f(0) = 1\) and \({f^{(4)}}(0) = 1\) <strong><em>(A1)</em></strong></p>
<p>\(f’(0) = f''(0) = f'''(0) = 0\) <strong><em>(A1)</em></strong></p>
<p>the sequence of derivatives repeats itself so the next non-zero derivative is \({f^{(8)}}(0) = 1\) <strong><em>(A1)</em></strong></p>
<p>the MacLaurin series is \(1 + \frac{{{x^4}}}{{4!}} + \frac{{{x^8}}}{{8!}}( + \ldots )\) <strong><em>(M1)A1</em></strong></p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(p = {\text{P}}(X = 0) + {\text{P}}(X = 4) + {\text{P}}(X = 8) + \ldots \) <strong><em>(M1)</em></strong></p>
<p>\( = \frac{{{{\text{e}}^{ - \mu }}{\mu ^0}}}{{0!}} + \frac{{{{\text{e}}^{ - \mu }}{\mu ^4}}}{{4!}} + \frac{{{{\text{e}}^{ - \mu }}{\mu ^8}}}{{8!}} + \ldots \) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(p = {{\text{e}}^{ - \mu }}\left( {1 + \frac{{{\mu ^4}}}{{4!}} + \frac{{{\mu ^8}}}{{8!}} + \ldots } \right)\) <strong><em>A1</em></strong></p>
<p>\( = {{\text{e}}^{ - \mu }}f(\mu )\) <strong><em>AG</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(p = {{\text{e}}^{ - 3}}\left( {\frac{{{{\text{e}}^3} + {{\text{e}}^{ - 3}} + 2\cos 3}}{4}} \right)\) <strong><em>(M1)</em></strong></p>
<p>\( = 0.226\) <strong><em>A1</em></strong></p>
<p><strong><em>[??? marks]</em></strong></p>
<div class="question_part_label">b.iii.</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.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">b.iii.</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;">Find the general solution of the differential equation \((1 - {x^2})\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 1 + xy\) , for \(\left| x \right| < 1\) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) Show that the solution \(y = f(x)\) that satisfies the condition \(f(0) = \frac{\pi }{2}\) is \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 - {x^2}} }}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Find \(\mathop {\lim }\limits_{x \to - 1} f(x)\) .</span></p>
<p align="LEFT"> </p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">rewrite in linear form <em><strong>M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \left( {\frac{{ - x}}{{1 - {x^2}}}} \right)y = \frac{1}{{1 - {x^2}}}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">attempt to find integrating factor <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(I = {e^{\int {\frac{{ - x}}{{1 - {x^2}}}} {\rm{d}}x}} = {e^{\frac{1}{2}\ln (1 - {x^2})}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \sqrt {1 - {x^2}} \) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">multiply by \(I\) and attempt to integrate <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(y{(1 - {x^2})^{\frac{1}{2}}} = \int {\frac{1}{{\sqrt {1 - {x^2}} }}} {\rm{d}}x\) <strong><em>(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(y{(1 - {x^2})^{\frac{1}{2}}} = \arcsin x + c\) <strong><em>A1 </em></strong></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.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) attempt to find <strong><em>c</em></strong> <strong><em>M1 </em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\frac{\pi }{2} = 0 + c\) </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,times; font-size: medium;">so \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 - {x^2}} }}\) <strong><em>AG</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\mathop {\lim }\limits_{x \to - 1} f(x) = \frac{0}{0}\) , so attempt l’Hôpital’s rule <em><strong>(M1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">consider \(\mathop {\lim }\limits_{x \to - 1} \frac{{\frac{1}{{{{(1 - {x^2})}^{\frac{1}{2}}}}}}}{{\frac{{ - x}}{{{{(1 - {x^2})}^{\frac{1}{2}}}}}}}\) <strong><em> A1A1</em> </strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = 1\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1 </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </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">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 wholly correct answers were seen to this question and for stronger candidates it proved to be a successful question. A small number of candidates failed to recognise part (a) as needing an integrating factor. More commonly, students left out the negative sign in the integrating factor or were unable to simplify the integrating factor.</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 good number of wholly correct answers were seen to this question and for stronger candidates it proved to be a successful question. In part (b) many students recognised the use of L’Hopital’s rule, but in a number of cases made errors in the differentiation.</span></p>
<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;">Calculate the following limit</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Calculate the following limit</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} - 1}}{{\ln (1 + x) - x}}\) .</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;">\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\ln 2}}{1}\) <em><strong>M1A1</strong></em></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = \ln 2\) <em><strong>A1</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"><strong><span style="font-family: times new roman,times; font-size: medium;">EITHER</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} - 1}}{{\ln (1 + x) - x}} = \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}}}{{\frac{1}{{1 + x}} - 1}}\) <em><strong>M1A1</strong></em></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}(1 + x)}}{{ - x}}\) <strong><em>A1A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = - 3\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">OR</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({(1 + {x^2})^{\frac{3}{2}}} - 1 = 1 + \frac{3}{2}{x^2} + \ldots - 1 = \frac{3}{2}{x^2} + \ldots \) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 + x) - x = x - \frac{1}{2}{x^2} + \ldots - x = - \frac{1}{2}{x^2} + \ldots \) <em><strong>M1A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Limit \( = - 3\) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">By evaluating successive derivatives at \(x = 0\) , find the Maclaurin series for </span><span style="font-family: times new roman,times; font-size: medium;">\(\ln \cos x\) up to and including the term in \({x^4}\) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Consider \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \cos x}}{{{x^n}}}\) </span><span style="font-family: times new roman,times; font-size: medium;">, where \(n \in \mathbb{R}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Using your result from (a), determine the set of values of \(n\) for which</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) the limit does not exist;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) the limit is zero;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iii) the limit is finite and non-zero, giving its value in this case.</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 at repeated differentiation (at least 2) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">let \(f(x) = \ln \cos x\) , \(f(0) = 0\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f'(x) = - \tan x\) , \(f'(0) = 0\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f''(x) = - {\sec ^2}x\) , \(f''(0) = - 1\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f'''(x) = - 2{\sec ^2}x\tan x\) , \(f'''(0) = 0\) <strong><em> A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({f^{iv}}(x) = - 2se{c^4}x - 4se{c^2}xta{n^2}x\) , \({f^{iv}}(0) = - 2\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the Maclaurin series is</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln \cos x = - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}} + \ldots \) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Allow follow-through on final <strong><em>A1</em></strong>.</span></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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\ln \cos x}}{{{x^n}}} = - \frac{{{x^{2 - n}}}}{2} - \frac{{{x^{4 - n}}}}{{12}} + \ldots \) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) the limit does not exist if \(n > 2\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) the limit is zero if \(n < 2\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) if \(n = 2\) , the limit is \( - \frac{1}{2}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1A1</span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[5 marks]</span></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><span style="font-family: times new roman,times; font-size: medium;">A reasonable number of candidates achieved full marks on this question. However, in part (a) a number of candidates struggled to find the Maclaurin series accurately. It was not uncommon to see errors in finding the higher derivatives, which was often caused by not simplifying the answer for earlier derivatives. At this level, it is expected that candidates understand the importance of using the most efficient methods. </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 made significant progress or achieved full marks in part (b), provided that they realised the importance of recognizing that<span style="font-family: times new roman,times; font-size: medium;">\[\frac{{\ln \cos x}}{{{x^n}}} = - \frac{{{x^{2 - n}}}}{2} - \frac{{{x^{4 - n}}}}{{12}} + \ldots \]<br></span></span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Given that the series \(\sum\limits_{n = 1}^\infty {{u_n}} \) is convergent, where \({u_n} > 0\), show that the series \(\sum\limits_{n = 1}^\infty {u_n^2} \) is also convergent.</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>State the converse proposition.</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By giving a suitable example, show that it is false.</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>since \(\sum {{u_n}} \) is convergent, it follows that \(\mathop {\lim }\limits_{n \to \infty } {u_n} = 0\) <strong><em>R1</em></strong></p>
<p>therefore, there exists \(N\) such that for \(n \geqslant N,{\text{ }}{u_n} < 1\) <strong><em>R1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Accept as \(n\) gets larger, eventually \({u_n} < 1\).</p>
<p> </p>
<p>therefore (for \(n \geqslant N\)), \(u_n^2 < {u_n}\) <strong><em>R1</em></strong></p>
<p>by the comparison test, \(\sum {u_n^2} \) is convergent <strong><em>R1</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>the converse proposition is that if \(\sum {u_n^2} \) is convergent, then \(\sum {{u_n}} \) is also convergent <strong><em>A1</em></strong></p>
<p><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>a suitable counter-example is</p>
<p>\({u_n} = \frac{1}{n}{\text{ }}\left( {{\text{for which }}\sum {u_n^2} {\text{ is convergent but }}\sum {{u_n}} {\text{ is not convergent}}} \right)\) <strong><em>A1</em></strong></p>
<p><strong><em>[1 mark]</em></strong></p>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Sum the series \(\sum\limits_{r = 0}^\infty {{x^r}} \) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>Hence</strong>, using sigma notation, deduce a series for</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (a) \(\frac{1}{{1 + {x^2}}}\) ;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (b) \(\arctan x\) ;</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (c) \(\frac{\pi }{6}\) .</span></p>
<div class="marks">[11]</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 \(\sum\limits_{n = 1}^{100} {n! \equiv 3(\bmod 15)} \) .</span></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><span style="font-family: times new roman,times; font-size: medium;">(i) \(\sum\limits_{r = 0}^\infty {{x^r}} = 1 + x + {x^2} + {x^3} + {x^4} + \ldots = \frac{1}{{1 - x}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) (a) replacing <strong><em>x</em></strong> by \( - {x^2}\) gives <strong><em>(M1) </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{1}{{1 - ( - {x^2})}} = 1 + ( - {x^2}) + {( - {x^2})^2} + {( - {x^2})^3} + {( - {x^2})^4} + \ldots \) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{1}{{1 + {x^2}}} = 1 - {x^2} + {x^4} - {x^6} + {x^8} - \ldots \) <strong><em>(A1)</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \sum\limits_{r = 0}^\infty {{{( - 1)}^r}{x^{2r}}} \) <strong><em>A1 N2</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(b) \(\arctan x = \int {\frac{{{\rm{d}}x}}{{1 + {x^2}}}} = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} + \ldots + c\) <em><strong>M1A1</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(x = 0 \Rightarrow c = 0\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\arctan x = \sum\limits_{r = 0}^\infty {{{( - 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} \) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(c) by taking \(x = \frac{1}{{\sqrt 3 }}\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\arctan \frac{1}{{\sqrt 3 }} = \frac{\pi }{6} = \sum\limits_{r = 0}^\infty {\frac{{{{( - 1)}^r}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^{2r + 1}}}}{{2r + 1}}} \) <em><strong>A1 </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[11 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;">\(\sum\limits_{n = 1}^{100} {n! = 1! + 2! + 3! + 4! + 5! + \ldots } \) <strong><em>M1</em> </strong> </span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = 1 + 2 + 6 + 24 + 120 + \ldots \)</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( \equiv 1 + 2 + 6 + 24 + 0 + 0 + 0 + \ldots (\bmod 15)\) <strong><em>M1A1 </em></strong> </span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( \equiv 33(\bmod 15)\) <strong><em>A1 </em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( \equiv 3(\bmod 15)\) <strong><em>AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[4 marks] </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;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">In part (b) many did not recognize the sum of a simple geometric series to infinity and got involved in some heavy Maclaurin work thus wasting time. The clear instruction "Hence" was ignored by many candidates so that the question became more difficult and time consuming than it should have been. </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;">Part (c) proved not to be as difficult as expected. </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: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Assuming the Maclaurin series for \({{\text{e}}^x}\), determine the first three non-zero terms in the Maclaurin expansion of \(\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) The random variable \(X\) has a Poisson distribution with mean \(\mu \). Show that \({\text{P}}\left( {X \equiv 1(\bmod 2)} \right) = a + b{{\text{e}}^{c\mu }}\) where \(a\), \(b\) and \(c\) are constants whose values are to be found.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \frac{{{x^5}}}{{5!}} + \ldots \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({{\text{e}}^{ - x}} = 1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} - \frac{{{x^5}}}{{5!}} + \ldots \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{e}}^x} - {{\text{e}}^{ - x}}}}{2} = x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + \ldots \) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept any valid (otherwise) method.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) \({\text{P}}\left( {X \equiv 1(\bmod 2)} \right) = {\text{P}}(X = 1,{\text{ }}3,{\text{ }}5,{\text{ }} \ldots )\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{ - \mu }}\left( {\mu + \frac{{{\mu ^3}}}{{3!}} + \frac{{{\mu ^5}}}{{5!}} + \ldots } \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{{{\text{e}}^{ - \mu }}({{\text{e}}^\mu } - {{\text{e}}^{ - \mu }})}}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{1}{2} - \frac{1}{2}{{\text{e}}^{ - 2\mu }}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\left( {a = \frac{1}{2},{\text{ }}b = - \frac{1}{2},{\text{ }}c = - 2} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p class="p1">Use l’Hôpital’s rule to find \(\mathop {\lim }\limits_{x \to 0} (\csc x - \cot x)\).</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p class="p1">\(\mathop {\lim }\limits_{x \to 0} (\csc x - \cot x) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 - \cos x}}{{\sin x}}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{{\cos x}}} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\( = 0\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">This question was well answered in general although some of the weaker candidates differentiated the whole expression rather than the numerator and denominator separately. Some candidates wrote \(\cos {\text{ec}}x - \cot x = \frac{1}{{\sin x}} - \frac{1}{{\tan x}} = \frac{{\tan x - \sin x}}{{\sin x\tan x}}\) which is correct but it introduced an extra round of differentiation with opportunity for error.</p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>QUESTION 1</p>
<div class="marks">[[N/A]]</div>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>QUESTION 1</p>
<div class="marks">[[N/A]]</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;">Using l’Hôpital’s Rule, determine the value of\[\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - x}}{{1 - \cos x}} .\]<br></span></p>
<div class="marks">[6]</div>
<div class="question_part_label">.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>MARKSCHEME 1</p>
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>MARKSCHEME 1</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;">\(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - x}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x - 1}}{{\sin x}}\) <strong><em>M1A1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">this still gives \(\frac{0}{0}\)</span></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;">repeat the process <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sec }^2}x\tan x}}{{\cos x}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 0\) <strong><em>A1</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;">\( = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}x}}{{\sin x}}\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{{{\cos }^2}x}}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 0\) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [6 marks]</span></em></strong></p>
<div class="question_part_label">.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">.</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;">(i) Find the range of values of \(n\) for which \(\int_1^\infty {{x^n}{\rm{d}}x} \)</span><span style="font-family: times new roman,times; font-size: medium;"> exists.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Write down the value of \(\int_1^\infty {{x^n}{\rm{d}}x} \)</span><span style="font-family: times new roman,times; font-size: medium;"> in terms of \(n\) , when it does exist.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Find the solution to the differential equation</span></p>
<p style="text-align: center;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\((\cos x - \sin x)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + (\cos x + \sin x)y = \cos x + \sin x\) ,<br></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">given that \(y = - 1\) when \(x = \frac{\pi }{2}\) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<div class="marks">[8]</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) \(\int_1^b {{x^n}{\rm{d}}x} = \left[ {\frac{{{x^{n + 1}}}}{{n + 1}}} \right]_1^b\) , \(n \ne - 1\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{{b^{n + 1}}}}{{n + 1}} - \frac{1}{{n + 1}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\int_1^b {{x^n}{\rm{d}}x} = \left[ {\ln x} \right]_1^b = \ln b\) when \(n = - 1\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">if \(n + 1 > 0,\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}} - \frac{1}{{n + 1}}} \right]\) </span><span style="font-family: times new roman,times; font-size: medium;">does not exist since \({b^{n + 1}}\) increases </span><span style="font-family: times new roman,times; font-size: medium;">without limit <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">if \(n + 1 < 0,\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}} - \frac{1}{{n + 1}}} \right]\) </span><span style="font-family: times new roman,times; font-size: medium;">exists since \({b^{n + 1}} \to 0\) as \(b \to \infty \) <strong><em>R1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">if \(n = - 1\) , \(\mathop {\lim }\limits_{b \to \infty } \left[ {\ln b} \right]\) </span><span style="font-family: times new roman,times; font-size: medium;">does not exist since \({\ln b}\) increases without limit <em><strong>R1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(so integral exists when \(n < - 1\) )</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\int_1^b {{x^n}{\rm{d}}x} = \frac{1}{{n + 1}}\) , \((n < - 1)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</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;"> [7 marks]</span></em></strong></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;">\((\cos x - \sin x)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + (\cos x + \sin x)y = \cos x + \sin x\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{\cos x + \sin x}}{{\cos x - \sin x}}y = \frac{{\cos x + \sin x}}{{\cos x - \sin x}}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">IF \({{\rm{e}}^{\int {\frac{{\cos x + \sin x}}{{\cos x - \sin x}}} {\rm{d}}x}} = {{\rm{e}}^{ - \ln (\cos x - \sin x)}} = \frac{1}{{\cos x - \sin x}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1A1</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{y}{{\cos x - \sin x}} = \int {\frac{{\cos x + \sin x}}{{{{(\cos x - \sin x)}^2}}}} {\rm{d}}x\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{1}{{\cos x - \sin x}} + k\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award the above <em><strong>A1</strong></em> even if \(k\) is missing.</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;">\(y = 1 + k(\cos x - \sin x)\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{\pi }{2}\) , \(y = - 1\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( - 1 = 1 + k( - 1)\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(k = 2\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(y = 1 + 2(\cos x - \sin x)\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: It is acceptable to solve the equation </span><span style="font-family: times new roman,times; font-size: medium;">using separation of variables.</span></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">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 was found to be the most difficult question on the paper. Whilst the question looked straightforward, indiscriminate use of the infinity symbol showed a lack of appreciation of the subtleties involved in the question. Many of the refinements required in each section were not considered. Knowledge of improper integrals was very poor. Perhaps integration is seen too easily as the application of a number of rules without much thought. </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;">Generally this was quite well done. However, many candidates did not realize that it could be solved using an integrating factor and used substitution or variables separable. While these last two approaches could work, the algebra involved soon became unmanageable. This led to many mistakes. </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 function \(f\) is defined by \(f(x) = {{\rm{e}}^x}\cos x\) .</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;">Show that \(f''(x) = - 2{{\rm{e}}^x}\sin x\) .</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><span style="font-family: times new roman,times; font-size: medium;">Determine the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">By differentiating your series, determine the Maclaurin series for \({{\rm{e}}^x}\sin x\) up to </span><span style="font-family: times new roman,times; font-size: medium;">the term in \({x^3}\) .</span></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f'(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f''(x) = {{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\sin x - {{\rm{e}}^x}\cos x\) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = - 2{{\rm{e}}^x}\sin x\) <strong><em>AG</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.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f'''(x) = - 2{{\rm{e}}^x}\sin x - 2{{\rm{e}}^x}\cos x\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({f^{IV}}(x) = - 4{{\rm{e}}^x}\cos x\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f(0) = 1\), \(f'(0) = 1\) ,\(f''(0) = 0\), \(f'''(0) = - 2\), \({f^{IV}}(0) = - 4\) <strong><em>(A1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the Maclaurin series is</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{e}}^x}\cos x = 1 + x - \frac{{{x^3}}}{3} - \frac{{{x^4}}}{6} + \ldots \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept multiplication of series method.</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>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">differentiating,</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({{\rm{e}}^x}\cos x - {{\rm{e}}^x}\sin x = 1 - {x^2} - \frac{{2{x^3}}}{3} + \ldots \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({{\rm{e}}^x}\sin x = 1 + {x^{}} - \frac{{{x^3}}}{3} + \ldots - 1 + {x^2} + \frac{{2{x^3}}}{3} + \ldots \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = x + {x^2} + \frac{{{x^3}}}{3} + \ldots \) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [4 marks]</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;">
[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 class="p1">Differentiate the expression \({x^2}\tan y\) with respect to \(x\), where \(y\) is a function of \(x\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence solve the differential equation \({x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\sin 2y = {x^3}{\cos ^2}y\) given that \(y = 0\) when \(x = 1\). Give your answer in the form \(y = f(x)\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(\frac{{\text{d}}}{{{\text{d}}x}}({x^2}\tan y) = 2x\tan y + {x^2}{\sec ^2}y\frac{{{\text{d}}y}}{{{\text{d}}x}}\) <span class="Apple-converted-space"> </span><strong><em>M1A1A1</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2x\sin y\cos y = {x^3}{\cos ^2}y\) <strong><em>A1</em></strong></p>
<p>\( \Rightarrow {x^2}{\sec ^2}y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2x\tan y = {x^3}\) <strong><em>M1A1</em></strong></p>
<p>\( \Rightarrow \frac{{\text{d}}}{{{\text{d}}x}}({x^2}\tan y) = {x^3}\) <strong><em>A1</em></strong></p>
<p>\({x^2}\tan y = \frac{{{x^4}}}{4} + c\) <strong><em>A1</em></strong></p>
<p><strong>Note: </strong>Condone the omission of \(c\)<em> </em>in the line above</p>
<p> </p>
<p>when \(x = 1,{\text{ }}y = 0 \Rightarrow c = - \frac{1}{4}\) <strong><em>M1</em></strong></p>
<p>\(\tan y = \frac{{{x^2}}}{4} - \frac{1}{{4{x^2}}}\)</p>
<p>\(y = \arctan \left( {\frac{{{x^2}}}{4} - \frac{1}{{4{x^2}}}} \right)\) <strong><em>A1</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 made the connection between (a) and (b) and went on to solve the differential equation correctly. Candidates who failed to make the connection usually tried to write the equation in the form required for the use of an integrating factor but this led nowhere.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Many candidates made the connection between (a) and (b) and went on to solve the differential equation correctly. Candidates who failed to make the connection usually tried to write the equation in the form required for the use of an integrating factor but this led nowhere.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Solve the differential equation \(x\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + 2y = \sqrt {1 + {x^2}} \)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">given that \(y = 1\) when \(x = \sqrt 3 \) .</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;">Rewrite the equation in the form</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{2}{x}y = \frac{{\sqrt {1 + {x^2}} }}{x}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Integrating factor \( = {{\rm{e}}^{\int {\left( {\frac{2}{x}} \right)} {\rm{d}}x}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = {{\rm{e}}^{2\ln x}}\) <strong><em> (A1)</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = {x^2}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The equation becomes</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\rm{d}}}{{{\rm{d}}y}}(y{x^2}) = x\sqrt {1 + {x^2}} \) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\(y{x^2} = \frac{1}{3}{(1 + {x^2})^{\frac{3}{2}}} + C\) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\(3 = \frac{1}{3} \times 8 + C \to C = \frac{1}{3}\) <em><strong>M1A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left[ {{\text{giving }}y{x^2} = \frac{1}{3}\left( {{{\left( {1 + {x^2}} \right)}^{\frac{3}{2}}} + 1} \right)} \right]\)</span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [9 marks]</span></em></strong></p>
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<h2 style="margin-top: 1em">Examiners report</h2>
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[N/A]
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<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the infinite series \(S = \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{{2^{2n}}\left( {2{n^2} - 1} \right)}}} \).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Determine the radius of convergence.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Determine the interval of convergence.</span></p>
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<h2 style="margin-top: 1em">Markscheme</h2>
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<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) let \({T_n}\) denote the \(n{\text{th}}\) term</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">consider</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{{x^{(n + 1)}}}}{{{2^{2(n + 1)}}\left( {2{{[n + 1]}^2} - 1} \right)}} \times \frac{{{2^2}\left( {2{n^2} - 1} \right)}}{{{x^n}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{x}{{{2^2}}} \times \frac{{\left( {2{n^2} - 1} \right)}}{{\left( {2{{[n + 1]}^2} - 1} \right)}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \to \frac{x}{4}\) as \(n \to \infty \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so the radius of convergence is \(4\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) we need to consider \(x = \pm 4\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(S(4) = \sum\limits_{n = 1}^\infty {\frac{1}{{\left( {2{n^2} - 1} \right)}}} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(S(4) < \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} \) is convergent; therefore by the comparison test \(S(4)\) is convergent <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(S( - 4) = \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}}}{{\left( {2{n^2} - 1} \right)}}} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>EITHER</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">this series is convergent because it is absolutely convergent <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>OR</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">this series is alternating and is convergent <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>THEN</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the interval of convergence is therefore \(\left[ { - 4,{\text{ }}4} \right]\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>The final <strong><em>A1 </em></strong>is independent of any of the previous marks.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
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<h2 style="margin-top: 1em">Examiners report</h2>
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[N/A]
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<p><span style="font-family: times new roman,times; font-size: medium;">Solve the following differential equation\[(x + 1)(x + 2)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y = x + 1\]giving your answer in the form \(y = f(x)\) .</span></p>
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<h2 style="margin-top: 1em">Markscheme</h2>
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<p><span style="font-family: times new roman,times; font-size: medium;">Rewrite the equation in the form</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{y}{{(x + 1)(x + 2)}} = \frac{1}{{x + 2}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Integrating factor \( = \exp \left( {\int {\frac{{{\rm{d}}x}}{{(x + 1)(x + 2)}}} } \right)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \exp \left( {\int {\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right){\rm{d}}x} } \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \exp \ln \left( {\frac{{x + 1}}{{x + 2}}} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{x + 1}}{{x + 2}}\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Multiplying by the integrating factor,</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\frac{{x + 1}}{{x + 2}}} \right)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{y}{{{{(x + 2)}^2}}} = \frac{{x + 1}}{{{{(x + 2)}^2}}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{x + 2}}{{{{(x + 2)}^2}}} - \frac{1}{{{{(x + 2)}^2}}}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> Integrating,</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left( {\frac{{x + 1}}{{x + 2}}} \right)y = \ln (x + 2) + \frac{1}{{x + 2}} + C\) <strong><em>A1A1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(y = \left( {\frac{{x + 2}}{{x + 1}}} \right)\left\{ {\ln (x + 2) + \frac{1}{{x + 2}} + C} \right\}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [11 marks]</span></em></strong></p>
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<h2 style="margin-top: 1em">Examiners report</h2>
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[N/A]
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<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Given that \(\frac{{{\rm{d}}x}}{{{\rm{d}}y}} + 2y\tan x = \sin x\)</span><span style="font-family: times new roman,times; font-size: medium;"> , and \(y = 0\) when \(x = \frac{\pi }{3}\) </span><span style="font-family: times new roman,times; font-size: medium;">, find the maximum value of <strong><em>y</em></strong>.</span></p>
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<h2 style="margin-top: 1em">Markscheme</h2>
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<p><span style="font-family: times new roman,times; font-size: medium;">integrating factor \( = {{\rm{e}}^{\int {2\tan xdx} }}\) <strong><em>M1</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = {{\rm{e}}^{2\ln \sec x}}\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>A1 </em> </span></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = {\sec ^2}x\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>A1</em> </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it follows that </span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(y{\sec ^2}x = \int {\sin x{{\sec }^2}x{\rm{d}}x} \) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>M1</em> </span></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \int {\sec x\tan x{\rm{d}}x} \) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>(A1)</em> </span></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \sec x + C\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>A1</em> </span></strong></p>
<p><span style="font-family: times new roman,times; font-size: medium;">substituting, </span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(0 = 2 + C\) </span><span style="font-family: times new roman,times; font-size: medium;">so \(C = - 2\) <em><strong> M1A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the solution is </span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(y = \cos x - 2{\cos ^2}x\) </span><span style="font-family: times new roman,times; font-size: medium;"><strong><em>A1</em></strong> </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;">using a GDC </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">maximum value of \(y\) is \(0.125\) <strong><em>A2</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; font-size: medium;">\(y' = - \sin x + 4\sin x\cos x = 0\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>M1</em> </span></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( \Rightarrow \cos x = \frac{1}{4}\) </span><span style="font-family: times new roman,times; font-size: medium;">(or \(\sin x = 0\) which leads to a minimum) </span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( \Rightarrow y = \frac{1}{8}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1 </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [11 marks]</span></em></strong></p>
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<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p><span style="font-family: times new roman,times; font-size: medium;">Solutions were often disappointing with some candidates even unable to find the integrating factor because of an inability to integrate \(2\tan x\) . Some candidates who found the integrating factor correctly were then unable to integrate \(\sin x{\sec ^2}x\) and others omitted the constant of integration. Some of the candidates who obtained the correct expression for <strong><em>y</em></strong> failed to realise that the quickest way to find the maximum value was to plot the graph of \(y|) on their calculator. </span></p>
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