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</div><h2>SL Paper 2</h2><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 \(\frac{{\rm{d}}}{{{\rm{d}}\theta }}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta )) = 2{\sec ^3}\theta \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> (ii) Hence write down \(\int {{{\sec }^3}\theta {\rm{d}}\theta } \) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Consider the differential equation \((1 + {x^2})\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + xy = 1 + {x^2}\) </span><span style="font-family: times new roman,times; font-size: medium;">given that \(y = 1\) </span><span style="font-family: times new roman,times; font-size: medium;">when \(x = 0\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Use Euler’s method with a step length of \(0.1\) to find an approximate value </span><span style="font-family: times new roman,times; font-size: medium;">for <strong><em>y</em></strong> when \(x = 0.3\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Find an integrating factor for determining the exact solution of the </span><span style="font-family: times new roman,times; font-size: medium;">differential equation.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iii) Find the solution of the equation in the form \(y = f(x)\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iv) To how many significant figures does the approximation found in part (i) </span><span style="font-family: times new roman,times; font-size: medium;">agree with the exact value of \(y\) when \(x = 0.3\) ?</span></p>
<div class="marks">[24]</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) \(\frac{{\rm{d}}}{{{\rm{d}}\theta }}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta ))\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = {\sec ^3}\theta + \sec \theta {\tan ^2}\theta + \frac{{\sec \theta \tan \theta + {{\sec }^2}\theta }}{{\sec \theta + \tan \theta }}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>M1</strong></em> for a valid attempt to differentiate either term.</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;">\( = {\sec ^3}\theta + \sec \theta ({\sec ^2}\theta - 1) + \sec \theta \) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 2{\sec ^3}\theta \) <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) \(\int {{{\sec }^3}\theta {\rm{d}}\theta } = \frac{1}{2}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta ))( + C)\) </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;"> [5 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) \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 1 - \frac{{xy}}{{1 + {x^2}}}\) <strong><em> A1</em></strong></span></p>
<p><img src="images/mack.png" alt></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept tabular values correct to 3 significant figures.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(y \approx 1.27\) when \(x = 0.3\) <strong><em>A1</em></strong></span></p>
<p> </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) consider the equation in the form</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{xy}}{{1 + {x^2}}} = 1\) <strong><em>(M1)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the integrating factor <strong><em>I</em></strong> is given by</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(I = \exp \int {\left( {\frac{x}{{1 + {x^2}}}} \right)} {\rm{d}}x\) </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; font-size: medium;">\( = \exp \left( {\frac{1}{2}\ln (1 + {x^2})} \right)\) </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;">\( = \sqrt {1 + {x^2}} \) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept also the fact that the integrating factor for the original </span><span style="font-family: times new roman,times; font-size: medium;">equation is \(\frac{1}{{\sqrt {1 + {x^2}} }}\) .</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;">(iii) consider the equation in the form</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\sqrt {1 + {x^2}} \frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{xy}}{{\sqrt {1 + {x^2}} }} = \sqrt {1 + {x^2}} \) </span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>(M1)</strong></em></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;">\(y\sqrt {1 + {x^2}} = \int {\sqrt {1 + {x^2}} {\rm{d}}x} \) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">to integrate the right hand side, put \(x = \tan \theta \) , \({\rm{d}}x = {\sec ^2}\theta {\rm{d}}\theta \) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\int {\sqrt {1 + {x^2}} } {\rm{d}}x = \int {\sqrt {1 + {{\tan }^2}\theta } } .{\sec ^2}\theta {\rm{d}}\theta \) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \int {{{\sec }^3}\theta {\rm{d}}\theta } \) <em><strong> A1</strong></em></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{2}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta ))\) </span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{2}\left( {x\sqrt {1 + {x^2}} + \ln (x + \sqrt {1 + {x^2}} )} \right)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the solution to the differential equation is therefore</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(y\sqrt {1 + {x^2}} = \frac{1}{2}\left( {x\sqrt {1 + {x^2}} + \ln \left( {x + \sqrt {1 + {x^2}} } \right)} \right) + C\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Do not penalize the omission of <strong><em>C</em></strong> at this stage.</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;">\(y = 1\) when \(x = 0\) gives \(C = 1\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the solution is \(y = \frac{1}{{2\sqrt {1 + {x^2}} }}\left( {x\sqrt {1 + {x^2}} + \ln \left( {x + \sqrt {1 + {x^2}} } \right)} \right) + \frac{1}{{\sqrt {1 + {x^2}} }}\) <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;">(iv) when \(x = 0.3\) , \(y = 1.249 \ldots \) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the approximation is only correct to 1 significant figure <strong><em>A1</em></strong></span></p>
<p> </p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[24 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;">Most candidates managed to solve (a) successfully although some solutions required a page or more to complete with candidates rewriting \(\sec \theta \) and \(\tan \theta \) in terms of \(\sin \theta \) and \(\cos \theta \) which increased the complexity of the problem and sometimes led to algebraic errors. </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;">Most candidates made a good attempt at (b) (i), those candidates who gave their solution in tabular form being most successful. In (b)(ii), most candidates found the correct integrating factor but many were unable to solve the differential equation in (b)(iii) with some failing to see that the result in (a) was intended as a hint for an appropriate substitution. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that the improper integral \(\int_0^\infty {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\)</span><span style="font-family: times new roman,times; font-size: medium;"> is convergent.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Use the integral test to deduce that the series \(\sum\limits_{n = 0}^\infty {\frac{1}{{{n^2} + 1}}} \)</span><span style="font-family: times new roman,times; font-size: medium;"> is convergent, </span><span style="font-family: times new roman,times; font-size: medium;">giving reasons why this test can be applied.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) Show that the series \(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{{n^2} + 1}}} \) is convergent.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> (ii) If the sum of the above series is \(S\), show that \(\frac{3}{5} < S < \frac{2}{3}\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">For the series \(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{n^2} + 1}}} \)</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) determine the radius of convergence;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) determine the interval of convergence using your answers to (b) and (c).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) consider \(\int_0^R {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\) <strong><em>M1 </em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \left[ {\arctan (x)} \right]_0^R = \arctan (R)\) </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; font-size: medium;">\(\mathop {\lim }\limits_{R \to \infty } \arctan (R) = \frac{\pi }{2}\) </span><span style="font-family: times new roman,times; font-size: medium;">(a finite number) <strong><em>R1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence the improper integral is convergent <em><strong>AG</strong> </em></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) the terms of the series are positive <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the terms are decreasing <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the terms tend to zero <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">by the integral test, the series converges <em><strong>AG </strong></em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong> </strong></span></em></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[6 marks]</strong> </span></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) the absolute values of the terms are monotonically decreasing <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">to zero <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the series converges by the alternating series test <em><strong>R1AG </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept absolute convergence, with reference to part (b)(ii) \( \Rightarrow \) </span><span style="font-family: times new roman,times; font-size: medium;">convergence.</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) statement that successive partial sums bound the total sum <strong><em>R1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S > \frac{1}{1} - \frac{1}{2} + \frac{1}{5} - \frac{1}{{10}} = \frac{3}{5}\) <strong><em>A1</em></strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S < \frac{1}{1} - \frac{1}{2} + \frac{1}{5} - \frac{1}{{10}} + \frac{1}{{17}} = 0.6588\) <strong><em>A1</em></strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S < 0.6588 < \frac{2}{3}\) <strong><em>AG </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </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">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) consider \(\left| {\frac{{\frac{{{x^{n + 1}}}}{{{{(n + 1)}^2} + 1}}}}{{\frac{{{x^n}}}{{{n^2} + 1}}}}} \right|\) <strong><em>M1</em></strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \left| {\frac{{x({n^2} + 1)}}{{{{(n + 1)}^2} + 1}}} \right|\) <strong><em>A1</em> </strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( \to \left| x \right|\) as \(n \to \infty \) <strong><em>A1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore radius of convergence \( = 1\) <em><strong>A1</strong> </em></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) interval of convergence \( = \left[ { - 1,1} \right]\) <em><strong>A1A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: <em><strong>A1</strong></em> for [\( - 1\), and <em><strong>A1</strong></em> for \(1\)].</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;">[6 marks] </span></strong></em></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Although the various parts of this question were algebraically uncomplicated, many candidates revealed their lack of understanding of the necessary rigour required in the analysis of limits, improper integrals and the testing of series for convergence. In (b)(i), the upper limit in the integral was often taken as infinity, without any mention of an underlying limiting process.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Although the various parts of this question were algebraically uncomplicated, many candidates revealed their lack of understanding of the necessary rigour required in the analysis of limits, improper integrals and the testing of series for convergence.</span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Although the various parts of this question were algebraically uncomplicated, many candidates revealed their lack of understanding of the necessary rigour required in the analysis of limits, improper integrals and the testing of series for convergence. Many candidates were more confident with part (d) than with the other parts of the question.<br></span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p>Consider the differential equation</p>
<p style="text-align: center;">\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\sec ^2}x,{\text{ }}0 \leqslant x < \frac{\pi }{2}\), given that \(y = 1\) when \(x = 0\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>By considering integration as the reverse of differentiation, show that for</p>
<p>\(0 \leqslant x < \frac{\pi }{2}\)</p>
<p>\[\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C.} \]</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>Hence, using integration by parts, show that</p>
<p>\[\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C.} \]</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>Find an integrating factor and hence solve the differential equation, giving your answer in the form \(y = f(x)\).</p>
<div class="marks">[9]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Starting with the differential equation, show that</p>
<p>\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x.\]</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence, by using your calculator to draw two appropriate graphs or otherwise, find the \(x\)-coordinate of the point of inflection on the graph of \(y = f(x)\).</p>
<div class="marks">[4]</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>\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\ln (\sec x + \tan x)} \right) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}}\) <strong><em>M1</em></strong></p>
<p>\( = \sec x\) <strong><em>A1</em></strong></p>
<p>therefore \(\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C} \) <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>\(\int {{{\sec }^3}x{\text{d}}x = \int {\sec x \times {{\sec }^2}x{\text{d}}x} } \) <strong><em>M1</em></strong></p>
<p>\( = \sec x\tan x - \int {\sec x{{\tan }^2}x{\text{d}}x} \) <strong><em>A1A1</em></strong></p>
<p>\( = \sec x\tan x - \int {\sec x({{\sec }^2}x - 1){\text{d}}x} \) <strong><em>A1</em></strong></p>
<p>\( = \sec x\tan x - \int {{{\sec }^3}x{\text{d}}x + \int {\sec x{\text{d}}x} } \)</p>
<p>\( = \sec x\tan x - \int {{{\sec }^3}x{\text{d}}x + \ln (\sec x + \tan x)} \) <strong><em>A1</em></strong></p>
<p>\(2\int {{{\sec }^3}x{\text{d}}x = \left( {\sec x\tan x + \ln (\sec x + \tan x)} \right)} \) <strong><em>A1</em></strong></p>
<p>therefore</p>
<p>\(\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C} \) <strong><em>AG</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>\({\text{int factor}} = {{\text{e}}^{\int {\tan x{\text{d}}x} }}\) <strong><em>(M1)</em></strong></p>
<p>\( = {{\text{e}}^{\ln \sec x}}\) <strong><em>(A1)</em></strong></p>
<p>\( = \sec x\) <strong><em>A1</em></strong></p>
<p>the differential equation can be written as</p>
<p>\(\frac{{\text{d}}}{{{\text{d}}x}}(y\sec x) = 2{\sec ^3}x\) <strong><em>M1A1</em></strong></p>
<p>integrating,</p>
<p>\(y\sec x = \sec x\tan x + \ln (\sec x + \tan x) + C\) <strong><em>A1</em></strong></p>
<p>putting \(x = 0,{\text{ }}y = 1,\) <strong><em>M1</em></strong></p>
<p>\(C = 1\) <strong><em>A1</em></strong></p>
<p>the solution is \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\) <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>differentiating the differential equation,</p>
<p>\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) <strong><em>A1A1</em></strong></p>
<p>\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + (2{\sec ^2}x - y\tan x)\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) <strong><em>A1</em></strong></p>
<p>\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x\) <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>at a point of inflection, \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0\) so \(y = 2{\sec ^2}x\tan x\) <strong><em>(M1)</em></strong></p>
<p>therefore the point of inflection can be found as the point of intersection of the graphs of \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\)</p>
<p>and \(y = 2{\sec ^2}x\tan x\) <strong><em>(M1)</em></strong></p>
<p>drawing these graphs on the calculator, \(x = 0.605\) <strong><em>A2</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="specification">
<p class="p1">Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x + y - 1\) <span class="s1">with boundary condition \(y = 1\) when \(x = 0\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Using Euler’s method with increments of \(0.2\)</span>, find an approximate value for \(y\) when \(x = 1\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Explain how Euler’s method could be improved to provide a better approximation.</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Solve the differential equation to find an exact value for \(y\) when \(x = 1\).</p>
<div class="marks">[9]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find the first three non-zero terms of the Maclaurin series for \(y\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Hence find an approximate value for \(y\) when \(x = 1\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space"><img src="images/Schermafbeelding_2015-12-17_om_13.59.09.png" alt> </span><strong><em>(M1)(A1)(A1)(A1)</em></strong></p>
<p class="p1"><strong>Note: </strong>Award <strong><em>M1 </em></strong>for equivalent of setting up first row of table, <strong><em>A1 </em></strong>for each of row <span class="s1">2</span>, <span class="s1">3 </span>and <span class="s1">5</span>.</p>
<p class="p2"> </p>
<p class="p1">approximate solution \(y = 1.98\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">make the increments smaller or any specific correct instruction – for example change increment from \(0.2\) to \(0.1\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y = 2x - 1\)</p>
<p>integrating factor is \({{\text{e}}^{\int { - 1{\text{d}}x} }} = {{\text{e}}^{ - x}}\) (<strong><em>M1)(A1)</em></strong></p>
<p>\(\frac{{\text{d}}}{{{\text{d}}x}}(y{{\text{e}}^{ - x}}) = {{\text{e}}^{ - x}}(2x - 1)\) <strong><em>M1</em></strong></p>
<p>attempt at integration by parts of \(\int {{{\text{e}}^{ - x}}(2x - 1){\text{d}}x} \) <strong><em>(M1)</em></strong></p>
<p>\( = - (2x - 1){{\text{e}}^{ - x}} + \int {2{{\text{e}}^{ - x}}{\text{d}}x} \) <strong><em>A1</em></strong></p>
<p>\( = - (2x - 1){{\text{e}}^{ - x}} - 2{{\text{e}}^{ - x}}( + c)\) <strong><em>A1</em></strong></p>
<p>\(y{{\text{e}}^{ - x}} = - (1 + 2x){{\text{e}}^{ - x}} + c\)</p>
<p>\(y = - (1 + 2x) + c{{\text{e}}^x}\)</p>
<p>when \(x = 0,{\text{ }}y = 1 \Rightarrow c = 2\) <strong><em>M1</em></strong></p>
<p>\(y = - (1 + 2x) + 2{{\text{e}}^{ - x}}\) <strong><em>A1</em></strong></p>
<p>when \(x = 1,{\text{ }}y = - 3 + 2{\text{e}}\) <strong><em>A1</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>(i) <strong>METHOD 1</strong></p>
<p>\(f(0) = 1,{\text{ }}f'(0) = 0\) <strong><em>A1</em></strong></p>
<p>\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 + \frac{{{\text{d}}y}}{{{\text{d}}x}} \Rightarrow {f^2}(0) = 2\) <strong><em>A1</em></strong></p>
<p>\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} \Rightarrow {f^3}(0) = 2\) <strong><em>A1</em></strong></p>
<p>hence \(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) <strong><em>A1</em></strong></p>
<p><strong>Note: </strong>Accuracy marks are independent of each other.</p>
<p> </p>
<p><strong>METHOD 2</strong></p>
<p>using Maclaurin series for \({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \) <strong><em>M1</em></strong></p>
<p>\(y = - 1 - 2x + 2\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots } \right)\) <strong><em>M1A1</em></strong></p>
<p>\(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) <strong><em>A1</em></strong></p>
<p> </p>
<p>(ii) when \(x = 1,{\text{ }}y = 1 + 1 + \frac{1}{3} = \frac{7}{3} = 2.33\) <strong><em>A1</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Most candidates were successful in applying Euler’s method and in explaining how it could be improved to provide a better approximation.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Most candidates were successful in applying Euler’s method and in explaining how it could be improved to provide a better approximation.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part c) many candidates successfully used an integrating factor to solve the differential equation but a significant minority were unable to make a meaningful start.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part d) produced many fully correct answers, but candidates sometimes used their own answers to part c) to derive the Maclaurin series rather than the given equation. In most cases this did not cause a problem but a small number of candidates produced an expression of such complexity that they were unable to differentiate to the required number of terms.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The function \(f(x)\) is defined by the series \(f(x) = 1 + \frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Write down the general term.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Find the interval of convergence.</span></p>
<div class="marks">[13]</div>
<div class="question_part_label">A.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;">Solve the differential equation \((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\) </span><span style="font-family: times new roman,times; font-size: medium;">, giving your answer in the form \(u = f(v)\) .</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 align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">the general term is \(\frac{{{{(x + 2)}^n}}}{{{3^n}n}}\) <strong><em>A1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[1 mark] </span></em></strong></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right) = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}}(n + 1)}} \times \frac{{{3^n}n}}{{{{(x + 2)}^n}}}} \right]\) <strong><em>M1A1A1 </em></strong></span></p>
<p style="margin-left: 30px;" align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{(x + 2){n^{}}}}{{3(n + 1)}}} \right]\) <strong><em>A1</em> </strong></span></p>
<p style="margin-left: 30px;" align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{(x + 2)}}{3}\) since \( = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{{n^{}}}}{{n + 1}}} \right] = 1\) <em><strong>A1R1</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">the series is convergent if \(\left| {\frac{{(x + 2)}}{3}} \right| < 1\) <strong><em>R1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">then \( - 3 < x + 2 < 3 \Rightarrow - 5 < x < 1\) <strong><em>A1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">if \(x = - 5\) , series is \(1 - 1 + \frac{1}{2} - \frac{1}{3} + \ldots + \frac{{{{( - 1)}^n}}}{n} + \ldots \) which converges <em><strong>M1A1</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">if \(x = 1\) , series is \(1 + 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n} + \ldots \) which diverges <strong><em>M1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">the interval of convergence is \( - 5 \le x < 1\) <strong><em>A1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[13 marks] </span></em></strong></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\((u + 3{v^3})\frac{{{\rm{d}}v}}{{{\rm{d}}u}} = 2v\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} = \frac{{(u + 3{v^3})}}{{2v}} = \frac{u}{{2v}} + \frac{{3{v^2}}}{2}\) <strong><em>M1A1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{\rm{d}}u}}{{{\rm{d}}v}} - \frac{u}{{2v}} = \frac{{3{v^2}}}{2}\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">IF is \({{\rm{e}}^{\int {\frac{1}{{2v}}} {\rm{d}}v}} = {{\rm{e}}^{\frac{1}{2}\ln v}}\) <strong><em>M1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = {v^{\frac{1}{2}}}\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{u}{{\sqrt v }} = \int {\frac{{3{v^{\frac{3}{2}}}}}{2}} {\rm{d}}v\) <strong><em>M1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{3}{5}{v^{\frac{5}{2}}} + c\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(u = \frac{3}{5}{v^3} + c\sqrt v \) <strong><em>A1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[8 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;">In (a) the general term was usually found.</span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Part (b) was completed mostly except for testing the ends of the interval of convergence. </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">A surprising number of candidates started off their solution by saying "let \(x = u\) and \(y = v\) " as if the world suddenly changed when \(x\) and \(y\) were not being used in a differential equation. Some also after seeing \(u\) and \(v\) thought they had a homogeneous equation and got lost in a maze of algebra that lead nowhere. Find \(\frac{{{\rm{d}}u}}{{{\rm{d}}v}}\) by inverting the given expression was also something that only the best candidates were able to do. </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 \(f(x) = \ln (1 + \sin x)\) .</span></p>
</div>
<div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">When a scientist measures the concentration \(\mu \) of a solution, the measurement </span><span style="font-family: times new roman,times; font-size: medium;">obtained may be assumed to be a normally distributed random variable with mean </span><span style="font-family: times new roman,times; font-size: medium;">\(\mu \) and standard deviation \(1.6\).</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 \(f''(x) = \frac{{ - 1}}{{1 + \sin x}}\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">A.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 \(f(x)\) as far as the term in \({x^4}\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">A.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;">Deduce the Maclaurin series for \(\ln (1 - \sin x)\) as far as the term in \({x^4}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">By combining your two series, show that \(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">A.d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Hence, or otherwise, find \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \sec x}}{{x\sqrt x }}\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">A.e.</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;">He makes 5 independent measurements of the concentration of a particular </span><span style="font-family: times new roman,times; font-size: medium;">solution and correctly calculates the following confidence interval for \(\mu \) .</span></p>
<p style="text-align: center;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">[\(22.7\) , \(26.1\)]</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Determine the confidence level of this interval.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">B.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;">He is now given a different solution and is asked to determine a \(95\%\) confidence </span><span style="font-family: times new roman,times; font-size: medium;">interval for its concentration. The confidence interval is required to have a width </span><span style="font-family: times new roman,times; font-size: medium;">less than \(2\). Find the minimum number of independent measurements required.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">B.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;">\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f''(x) = \frac{{ - \sin x(1 + \sin x) - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{ - \sin x - 1}}{{{{(1 + \sin x)}^2}}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \frac{{ - 1}}{{1 + \sin x}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">AG</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">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({f^{iv}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\) </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;">\(f(0) = 0\) , \(f'(0) = 1\) , \(f''(0) = - 1\) , \(f'''(0) = 1\) , \({f^{iv}}(0) = - 2\) <strong><em>(A2)</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for 2 errors and <em><strong>A0</strong></em> for more than 2 errors.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 + \sin x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} + \ldots \) <strong><em>M1A1</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.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 - \sin x) = \ln (1 + \sin ( - x)) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} + \ldots \) <em><strong>M1A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></strong></em></p>
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Adding, <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 - {\sin ^2}x) = \ln {\cos ^2}x\) <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;">\( = - {x^2} - \frac{{{x^4}}}{6} + \ldots \) <strong><em>A1</em></strong></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>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) <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">A.d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\ln \sec x}}{{x\sqrt x }} = \frac{{\sqrt x }}{2} + \frac{{{x^2}\sqrt x }}{{12}} + \ldots \) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Limit \( = 0\) <em><strong>A1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks]</span></strong></em></p>
<div class="question_part_label">A.e.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Interval width \( = 26.1 - 22.7 = 3.4\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">So \(3.4 = 2z \times \frac{{1.6}}{{\sqrt 5 }}\) </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;">\(z = 2.375 \ldots \) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Probability \( = 0.9912\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Confidence level \( = 2 \times 0.4912 = 98.2\% \) <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.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(z\)-value \( = 1.96\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">We require</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(2 \times \frac{{1.96 \times 1.6}}{{\sqrt n }} < 2\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Whence \(n > 9.83\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">So we need \(n = 10\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Accept \( = \) signs throughout.</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.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.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">A.e.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">B.b.</div>
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \({S_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} \) .</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, for \(n \ge 2\) , \({S_{2n}} > {S_n} + \frac{1}{2}\) .</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;">Deduce that \({S_{2m + 1}} > {S_2} + \frac{m}{2}\) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Hence show that the sequence \(\left\{ {{S_n}} \right\}\) is divergent.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({S_{2n}} = {S_n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \ldots + \frac{1}{{2n}}\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">M1</span></em></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( > {S_n} + \frac{1}{{2n}} + \frac{1}{{2n}} + \ldots + \frac{1}{{2n}}\) <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {S_n} + \frac{1}{2}\) <em><strong>AG</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Replacing \(n\) by \(2n\),</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({S_{4n}} > {S_{2n}} + \frac{1}{2}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( > {S_n} + 1\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Continuing this process,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({S_{8n}} > {S_n} + \frac{3}{2}\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">In general,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({S_{{2^m}n}} > {S_n} + \frac{m}{2}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>M1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Putting \(n = 2\) <strong><em> M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({S_{{2^{m + 1}}}} > {S_2} + \frac{m}{2}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>AG</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">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Consider the (large) number \(N\). <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Then, \({S_{2m + 1}} > N\) if \({S_2} + \frac{m}{2} > N\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">i.e. if \(m > 2(N - {S_2})\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">This establishes the divergence. <strong><em>AG</em></strong></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y}\), where \(y \ne 0\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the general solution of the differential equation, expressing your answer in the <span class="s1">form \(f(x,{\text{ }}y) = c\), where \(c\) is a constant.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span></span>Hence find the particular solution passing through the points \((1,{\rm{ \pm }}\sqrt 2 )\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Sketch the graph of your solution and name the type of curve represented.</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Write down the particular solution passing through the points \((1,{\text{ }} \pm 1)\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Give a geometrical interpretation of this solution in relation to part (b).</p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find the general solution of the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{x}{y} + \frac{y}{x}\), where \(xy \ne 0\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Find the particular solution passing through the point \((1,{\text{ }}\sqrt 2 )\).</p>
<p class="p2">(iii) <span class="Apple-converted-space"> </span>Sketch the particular solution.</p>
<p class="p1">(iv) <span class="Apple-converted-space"> </span>The graph of the solution only contains points with \(\left| x \right| > a\).</p>
<p class="p1">Find the exact value of \(a,{\text{ }}a > 0\).</p>
<div class="marks">[12]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">attempt to separate the variables <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\int {y\frac{{{\text{d}}y}}{{{\text{d}}x}}{\text{d}}x = \int {x{\text{d}}x} } \) </span><strong><em>A1</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Accept \(\int {y{\text{d}}y = \int {x{\text{d}}x} } \).</p>
<p class="p2"> </p>
<p class="p3">obtain \(\frac{1}{2}{y^2} = \frac{1}{2}{x^2} + {\text{ constant }}( \Rightarrow {y^2} - {x^2} = c)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>substitute the coordinates for both points <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">\({( \pm \sqrt 2 )^2} - {1^2} = 1\)</p>
<p class="p1">obtain \({y^2} - {x^2} = 1\) or equivalent <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> <img src="images/Schermafbeelding_2017-02-09_om_08.36.33.png" alt="M16/5/FURMA/HP2/ENG/TZ0/04.b.ii/M"></span> <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span><em>A1 </em></strong>for general shape including two branches and symmetry;</p>
<p class="p1"><strong><em>A1 </em></strong>for values of the intercepts.</p>
<p class="p3"> </p>
<p class="p1">(rectangular) hyperbola <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \({y^2} - {x^2} = 0\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>the two straight lines \(y = \pm x\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">form the asymptotes to the hyperbola found above, or equivalent <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>the equation is homogeneous, so attempt to substitute \(y = vx\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">as a first step write \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x\frac{{{\text{d}}v}}{{{\text{d}}x}} + v\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>(</em></strong></span><span class="s2"><strong><em>A1</em></strong></span><span class="s1"><strong><em>)</em></strong></span></p>
<p class="p2">then \(x\frac{{{\text{d}}v}}{{{\text{d}}x}} + v = \frac{1}{v} + v\) <span class="Apple-converted-space"> </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1">attempt to solve the resulting separable equation <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(\int {v{\text{d}}v = \int {\frac{1}{x}{\text{d}}x} } \) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p1">obtain \(\frac{1}{2}{v^2} = \ln \left| x \right| + {\text{ constant}} \Rightarrow {y^2} = 2{x^2}\ln \left| x \right| + c{x^2}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>substituting the coordinates <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">obtain \(c = 2 \Rightarrow {y^2} = 2{x^2}\ln \left| x \right| + 2{x^2}\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(iii) <span class="Apple-converted-space"> <img src="images/Schermafbeelding_2017-02-09_om_09.05.46.png" alt="M16/5/FURMA/HP2/ENG/TZ0/04.d.iii/M"></span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">(iv) <span class="Apple-converted-space"> </span>since \({y^2} > 0\) and \({x^2} \ne 0\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p3"><span class="Apple-converted-space">\(\ln \left| x \right| > - 1 \Rightarrow \left| x \right| > {{\text{e}}^{ - 1}}\) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p3"><span class="Apple-converted-space">\(a = {{\text{e}}^{ - 1}}\) </span><span class="s2"><strong><em>A1</em></strong></span></p>
<p class="p4"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>The <strong><em>R1 </em></strong>may be awarded for a correct reason leading to subsequent correct work.</p>
<p class="p4"> </p>
<p class="p1"><strong><em>[12 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Although (a), (b) and (c) were generally well done, it was rare to see a completely satisfactory geometrical answer to part (c)(ii). A few candidates solved the differential equation as a homogeneous equation.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">For part (d) most candidates used the correct solution method for a homogeneous differential equation. A few found the algebra hard going in finding the particular solution. Most approaches to the final part were unsatisfactory, with a lack of proper consideration of the inequalities in the question.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Using a Taylor series, find a quadratic approximation for \(f(x) = \sin x\) centred about \(x = \frac{{3\pi }}{4}\).</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">When using this approximation to find angles between \(130^\circ\) and \(140^\circ\)<span class="s1">, find the maximum value of the Lagrange form of the error term.</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence find the largest number of decimal places to which \(\sin x\) can be estimated for angles between \(130^\circ\) and \(140^\circ\)<span class="s1">.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Explain briefly why the same maximum value of error term occurs for \(g(x) = \cos x\) </span>centred around \(\frac{\pi }{4}\) when finding approximations for angles between \(40^\circ\) and \(50^\circ\)<span class="s1">.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(f(x) = \sin x,{\text{ }}f'(x) = \cos x,{\text{ }}{f^{(2)}}(x) = - \sin x\) <strong><em>M1</em></strong></p>
<p>\(f\left( {\frac{{3\pi }}{4}} \right) = \frac{1}{{\sqrt 2 }},{\text{ }}f'\left( {\frac{{3\pi }}{4}} \right) = - \frac{1}{{\sqrt 2 }},{\text{ }}{f^{(2)}}\left( {\frac{{3\pi }}{4}} \right) = - \frac{1}{{\sqrt 2 }}\) <strong><em>A1</em></strong></p>
<p>hence the quadratic Taylor Polynomial is</p>
<p>\(\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}\left( {x - \frac{{3\pi }}{4}} \right) - \frac{1}{{\sqrt 2 }}\frac{{{{\left( {x - \frac{{3\pi }}{4}} \right)}^2}}}{{2!}}\) <strong><em>M1A1</em></strong></p>
<p>\(\left( {\frac{1}{{\sqrt 2 }}\left( {1 - \left( {x - \frac{{3\pi }}{4}} \right) - \frac{1}{2}{{\left( {x - \frac{{3\pi }}{4}} \right)}^2}} \right)} \right)\)</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(f(x) = \sin x,{\text{ }}{f^{(3)}}(x) = - \cos x\) <strong><em>(A1)</em></strong></p>
<p>the Lagrange form of the error term is: \(\left| {{R_n}(x)} \right| \leqslant \frac{{{{\left| {x - a} \right|}^{n + 1}}}}{{(n + 1)!}}\max \left| {{f^{n + 1}}(k)} \right|\)</p>
<p>\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\max \left| {{f^3}(k)} \right|\) <strong><em>(M1)</em></strong></p>
<p>\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\max \left| { - \cos k} \right|\) <strong><em>A1</em></strong></p>
<p>in this case \(\left| { - \cos k} \right| \leqslant \left| { - \cos 140} \right|\) <strong><em>(A1)</em></strong></p>
<p>\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\left| { - \cos 140} \right|\)</p>
<p>choosing \(140^\circ = \frac{{14\pi }}{{18}}\) <strong><em>M1</em></strong></p>
<p>\( \Rightarrow \left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {\frac{{14\pi }}{{18}} - \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\left| { - \cos \frac{{14\pi }}{{18}}} \right|\) <strong><em>A1</em></strong></p>
<p>therefore the maximum value of the error term is \(8.48 \times {10^{ - 5}}\) <strong><em>A1</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(\left| {{R_2}(x)} \right| \leqslant 8.48 \times {10^{ - 5}} = 0.0000848\) hence for angles between \(130^\circ\) and \(140^\circ\) the approximation will be accurate to 3 decimal places <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x - \frac{\pi }{4}} \right|}^3}}}{{3!}}\max \left| {\sin k} \right|\) <span class="Apple-converted-space"> </span>(<strong><em>M1)</em></strong></p>
<p class="p1">since the max value of \(\left| {{f^3}(k)} \right|\) is \(\sin 50^\circ \) which is the same as \(\left| {\cos 140^\circ } \right|\) <span class="Apple-converted-space"> </span><strong><em>A1R1</em></strong></p>
<p class="p1">then the error is the same <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part a) was answered successfully by most candidates. However, the majority of candidates struggled to gain full marks on the remainder of the question.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part b) candidates struggled to work out which angle to use to find the maximum value.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Part a) was answered successfully by most candidates. However, the majority of candidates struggled to gain full marks on the remainder of the question.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part d) most candidates understood that this was related to a translation of the sine graph but were unable to explain it convincingly.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. </span><span style="font-family: times new roman,times; font-size: medium;">However the weights vary with a standard deviation of \(0.54\) kg. A random sample of </span><span style="font-family: times new roman,times; font-size: medium;">\(24\)</span><span style="font-family: times new roman,times; font-size: medium;"> bags is taken to check that the mean weight is \(28\) kg.</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;">Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} {\rm{ .}}\]</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Use your answer to (a) to find an approximate expression for the cumulative distributive </span><span style="font-family: times new roman,times; font-size: medium;">function of \({\rm{N}}(0,1)\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>Hence</strong> find an approximate value for \({\rm{P}}( - 0.5 \le Z \le 0.5)\) , where </span><span style="font-family: times new roman,times; font-size: medium;">\(Z \sim {\rm{N}}(0,1)\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">A.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;">State and justify an appropriate test procedure giving the null and alternate </span><span style="font-family: times new roman,times; font-size: medium;">hypotheses.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">B.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;">What is the critical region for the sample mean if the probability of a Type I error </span><span style="font-family: times new roman,times; font-size: medium;">is to be \(3.5\%\)?</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">B.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;">If the mean weight of the bags is actually \(28\).1 kg, what would be the probability </span><span style="font-family: times new roman,times; font-size: medium;">of a Type II error?</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">B.c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = 1 + \left( { - \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} + \ldots \) <strong><em>M1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} - \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\) <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.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">(i) \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 - \frac{{{t^2}}}{2}} + \frac{{{t^4}}}{8} - \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\) <strong><em>M1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\) <strong><em>A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} - \ldots } \right)\) <strong><em>R1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">(ii) \({\rm{P}}( - 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 - \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} - \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} - \ldots } \right)\) <em><strong>M1</strong> </em></span></p>
<p style="margin-left: 30px;" align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = 0.38292 = 0.383\) <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;">[6 marks] </span></strong></em></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">this is a two tailed test of the sample mean \(\overline x \) </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">we use the central limit theorem to justify assuming that <em><strong>R1</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\overline X \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\) <strong><em>R1A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{H}}_0}:\mu = 28\) <strong><em>A1 </em></strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({{\rm{H}}_1}:\mu \ne 28\) <em><strong>A1 </strong></em></span></p>
<p><em><span style="font-family: times new roman,times; font-size: medium;"><strong>[5 marks]</strong> </span></em></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\) <strong><em>(M1)A1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">and (\(\overline x \le 28 - 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) ) <em><strong>(M1)(A1)(A1)</strong> </em></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\overline x \le 27.7676\) or \(\overline x \ge 28.2324\) </span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">so \(\overline x \le 27.8\) or \(\overline x \ge 28.2\) <strong><em>A1A1</em> </strong></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[7 marks] </span></strong></em></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">if \(\mu = 28.1\)</span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\(\overline X \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\) <strong><em>R1</em> </strong></span></p>
<p align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x < 28.2324)\)</span></p>
<p style="margin-left: 30px;" align="JUSTIFY"><span style="font-family: times new roman,times; font-size: medium;">\( = 0.884\) <em><strong>A1</strong> </em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Depending on the degree of accuracy used for the critical region the answer </span><span style="font-family: times new roman,times; font-size: medium;">for part (c) can be anywhere from \(0.8146\) to \(0.879\). </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></strong></em></p>
<div class="question_part_label">B.c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The derivation of a series from a given one by substitution seems not to be well known. This made finding the required series from \(({{\rm{e}}^x})\) in part (a) to be much more difficult than it need have been. The fact that this part was worth only 3 marks was a clear hint that an easy derivation was possible. </span></p>
<div class="question_part_label">A.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In part (b)(i) the \(0.5\) was usually missing which meant that this part came out incorrectly. </span></p>
<div class="question_part_label">A.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The conditions required in part (a) were rarely stated correctly and some candidates were unable to state the hypotheses precisely. There was some confusion with "less than" and "less than or equal to". </span></p>
<div class="question_part_label">B.a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">There was some confusion with "less than" and "less than or equal to". </span></p>
<div class="question_part_label">B.b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Levels of accuracy in the body of the question varied wildly leading to a wide range of answers to part (c). </span></p>
<div class="question_part_label">B.c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The diagram shows a sketch of the graph of \(y = {x^{ - 4}}\) for \(x > 0\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/bully.png" alt></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">By considering this sketch, show that, for \(n \in {\mathbb{Z}^ + }\) ,\[\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} < \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} .\]</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Let \(S = \sum\limits_{r = 1}^\infty {\frac{1}{{{r^4}}}} \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Use the result in (a) to show that, for \(n \ge 2\) , the value of \(S\) lies between</span></p>
<p style="text-align: center;"><span style="font-family: times new roman,times; font-size: medium;">\(\sum\limits_{r = 1}^{n - 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) and \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that, by taking \(n = 8\) , the value of \(S\) can be deduced correct to three </span><span style="font-family: times new roman,times; font-size: medium;">decimal places and state this value.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) The exact value of \(S\) is known to be \(\frac{{{\pi ^4}}}{N}\)</span><span style="font-family: times new roman,times; font-size: medium;">where \(N \in {\mathbb{Z}^ + }\) . Determine the </span><span style="font-family: times new roman,times; font-size: medium;">value of \(N\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Now let \(T = \sum\limits_{r = 1}^\infty {\frac{{{{( - 1)}^{r + 1}}}}{{{r^4}}}} \) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Find the value of \(T\) correct to three decimal places.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><img src="images/bully2.png" alt></span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> (M1) </span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">total area of "upper" rectangles </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{{{n^4}}} \times 1 + \frac{1}{{{{(n + 1)}^4}}} \times 1 + \frac{1}{{{{(n + 2)}^4}}} \times 1 + \ldots = \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} \) <strong><em>M1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">total area of "lower" rectangles </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{1}{{{{(n + 1)}^4}}} \times 1 + \frac{1}{{{{(n + 2)}^4}}} \times 1 + \frac{1}{{{{(n + 3)}^4}}} \times 1 + \ldots = \sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} \) <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the total area under the curve from \(x = n\) to infinity lies between these two sums hence <span style="font-family: times new roman,times; font-size: medium;">\(\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} < \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} \)</span> <em><strong> R1AG</strong> </em></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[5 marks]</em> </span></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;">first evaluate the integral </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} = - \left[ {\frac{1}{{3{x^3}}}} \right]_n^\infty = \frac{1}{{3{n^3}}}\) <em><strong>M1A1</strong> </em></span></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,times; font-size: medium;">\(\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \frac{1}{{3{n^3}}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">adding \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} \) to both sides, <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S < \sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">similarly, </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} > \frac{1}{{3{n^3}}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">adding \(\sum\limits_{r = 1}^{n - 1} {\frac{1}{{{r^4}}}} \) to both sides, <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S > \sum\limits_{r = 1}^{n - 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) <strong><em> A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence the value of \(S\) lies between </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sum\limits_{r = 1}^{n - 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) and \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) <strong><em>AG </em></strong></span></p>
<p><strong><span style="font-family: times new roman,times; font-size: medium;"><em>[8 marks]</em> </span></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) putting \(n = 8\) , we find that </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S < 1.08243 \ldots \) and \(S > 1.08219 \ldots \) <strong><em>A1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it follows that \(S = 1.082\) to 3 decimal places <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) substituting this value of \(S\), </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(N \approx \frac{{{\pi ^4}}}{{1.082}} \approx 90.0268\) <strong><em>M1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(N = 90\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong><em> </em></strong></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 marks] </span></strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<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;">successive partial sums are </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">1 <em><strong> M1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">0.9375 </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">0.9498… </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">0.9459… </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">0.9475… </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">0.9467… </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">0.9471… <em><strong>A1</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it follows that correct to 3 decimal places \(T = 0.947\) <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;">\(T = S - \frac{2}{{16}}S\) <strong><em>M1A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">using part (c)(i) or \(0.94703…\) using the sum given in part (c)(ii) \(0.9471…\) </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">it follows that \(T = 0.947\) correct to 3 decimal places <strong><em>A1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks] </span></em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Most candidates attempted (a), although in many cases the explanations were poor and unconvincing. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">It was pleasing to see that some candidates who were unable to do part (a) moved on and made a reasonable attempt at (b) and (c). </span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">It was pleasing to see that some candidates who were unable to do part (a) moved on and made a reasonable attempt at (b) and (c). </span></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Attempts at (d) were often disappointing with candidates not realising that, in this case, the sum to infinity lies between any two successive partial sums. </span></p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the functions \({f_n}(x) = {\sec ^n}(x),{\text{ }}\left| x \right| < \frac{\pi }{2}\) and \({g_n}(x) = {f_n}(x)\tan x\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that</p>
<p>(i) \(\frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}} = n{g_n}(x)\);</p>
<p>(ii) \(\frac{{{\text{d}}{g_n}(x)}}{{{\text{d}}x}} = (n + 1){f_{n + 2}}(x) - n{f_n}(x)\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span></span>Use these results to show that the Maclaurin series for the function \({f_5}(x)\) up to and including the term in \({x^4}\) is \(1 + \frac{5}{2}{x^2} + \frac{{85}}{{24}}{x^4}\).</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>By considering the general form of its higher derivatives explain briefly why all coefficients in the Maclaurin series for the function \({f_5}(x)\) <span class="s1">are either positive or zero.</span></p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>Hence show that \({\sec ^5}(0.1) > 1.02535\).</p>
<div class="marks">[14]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \(\frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}} = n{\sec ^{n - 1}}(x)\sec (x)\tan (x)\)</span> <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p1"><span class="Apple-converted-space">\( = n{g_n}(x)\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p2"> </p>
<p class="p2">(ii) <span class="Apple-converted-space"> \(\frac{{{\text{d}}{g_n}(x)}}{{{\text{d}}x}} = \frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}}\tan (x) + {f_n}(x){\sec ^2}(x)\)</span> <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2">\(n{g_n}(x)\tan (x) + {f_{n + 2}}(x)\) or equivalent <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2">\(n{f_n}(x){\tan ^2}(x) + {f_{n + 2}}(x)\) or equivalent <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = (n + 1){f_{n + 2}}(x) - n{f_n}(x)\) </span><strong><em>AG</em></strong></p>
<p class="p3"> </p>
<p class="p2"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>M1A1 </em></strong>for the correct differentiation of a product and <strong><em>A1 </em></strong>for an intermediate result clearly leading to the <strong><em>AG</em></strong>.</p>
<p class="p3"> </p>
<p class="p2"><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \({f_5}(0) = 1\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\frac{{{\text{d}}{f_5}}}{{{\text{d}}x}}(0) = 5{g_5}(0) = 0\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}}(0) = 5\left( {6{f_7}(0) - 5{f_5}(0)} \right) = 5\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\frac{{{{\text{d}}^3}{f_5}}}{{{\text{d}}{x^3}}} = 30\frac{{{\text{d}}{f_7}}}{{{\text{d}}x}} - 25\frac{{{\text{d}}{f_5}}}{{{\text{d}}x}}\) </span><strong><em>M1</em></strong></p>
<p class="p1">hence \(\frac{{{{\text{d}}^3}{f_5}}}{{{\text{d}}{x^3}}}(0) = 30 \times 0 - 25 \times 0 = 0\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(\frac{{{{\text{d}}^4}{f_5}}}{{{\text{d}}{x^4}}} = 30\frac{{{{\text{d}}^2}{f_7}}}{{{\text{d}}{x^2}}} - 25\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}} = 210(8{f_9} - 7{f_7}) - 25\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}}\) </span><strong><em>M1A1</em></strong></p>
<p class="p1">hence \(\frac{{{{\text{d}}^4}{f_5}}}{{{\text{d}}{x^4}}}(0) = 210 - 125 = 85\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">hence \({f_5}(x) \approx 1 + \frac{5}{2}{x^2} + \frac{{85}}{{24}}{x^4}\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2"> </p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>each derivative of \({f_m}(x)\)<span class="s1"> </span>is a sum of terms of the form \({\text{A}}\,{\sec ^p}(x)\,\,{\tan ^q}(x)\)<span class="s1"> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="s2">where \(A \geqslant 0\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">when \(x = 0\) is substituted the result is the sum of positive and/or zero terms <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"> </p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>since the full series represents \({f_5}(x)\), the truncated series is a lower bound (or some equivalent statement) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">hence \({\sec ^5}(0.1) > 1 + \frac{5}{2}{0.1^2} + \frac{{85}}{{24}}{0.1^4}\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( = 1.025354\) </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\( > 1.02535\) </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[14 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">Part (a) was generally answered, albeit either with an excess of algebraic manipulation or with too little – candidates need to realise that when an answer is given in the question, they need to convincingly reach that answer.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (b)(i), the results of part (a) were well used for up to the quadratic term. The obtaining of the cubic term, and more so the quartic term, was often not convincing. In part (ii), poor communication let down many candidates. In answering part (iii), many candidates failed to realise that in order to prove the stated inequality, they needed to actually write down the number 1.025354…, which is clearly greater than 1.02535.</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 differential equation\[\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x = x(\sec x - \tan x),{\text{ where }}y = 3{\text{ when }}x = 0.\]</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Use Euler’s method with a step length of \(0.1\) to find an approximate value for \(y\) </span><span style="font-family: times new roman,times; font-size: medium;">when \(x = 0.3\) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) By differentiating the above differential equation, obtain an expression </span><span style="font-family: times new roman,times; font-size: medium;">involving \(\frac{{{{\rm{d}}^{\rm{2}}}y}}{{{\rm{d}}{x^2}}}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Hence determine the Maclaurin series for \(y\) up to the term in \({{x^2}}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) Use the result in part (ii) to obtain an approximate value for \(y\) when </span><span style="font-family: times new roman,times; font-size: medium;">\(x = 0.3\) .</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Show that \(\sec x + \tan x\) is an integrating factor for solving this differential </span><span style="font-family: times new roman,times; font-size: medium;">equation.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) Solve the differential equation, giving your answer in the form \(y = f(x)\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) Hence determine which of the two approximate values for <strong><em>y</em></strong> when \(x = 0.3\) , </span><span style="font-family: times new roman,times; font-size: medium;">obtained in parts (a) and (b), is closer to the true value.</span></p>
<div class="marks">[11]</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;"><img src="images/org.png" alt></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: The <em><strong>A1</strong></em> marks above are for correct entries in the \(y\) column.<br></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;">\(y(0.3) \approx 2.21\) <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">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) use of product rule on either side <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{{\rm{d}}^{\rm{2}}}y}}{{{\rm{d}}{x^2}}} + \sec x\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y\sec x\tan x = \sec x - \tan x + x(\sec x\tan x - {\sec ^2}x)\) <strong><em> A1A1</em></strong></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;">(ii) \(y(0) = 3\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(y'(0) = - 3\), \(y''(0) = 4\) <strong><em>A1A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the quadratic approximation is</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(y = \left( {y(0) + xy'(0) + |\frac{{{x^2}y''(0)}}{2} = } \right)3 - 3x + 2{x^2}\) <em><strong> (M1)A1</strong></em></span></p>
<p align="LEFT"><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;">(iii) using this approximation, \(y(0.3) \approx 2.28\) <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;">[8 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) <strong>EITHER</strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\rm{d}}}{{{\rm{d}}x}}(\sec x + \tan x) = \sec x\tan x + {\sec ^2}x\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\sec x(\sec x + \tan x) = {\sec ^2}x + \sec x\tan x\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">as these two expressions are the same, this is an integrating factor <strong><em>R1AG</em></strong></span><strong><br><br><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;">\((\sec x + \tan x)\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y\sec x} \right) = (\sec x + \tan x)x(\sec x - \tan x)\) </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;"><strong>Note</strong>: RHS does not need to be shown.<br><br>\({\rm{LHS}} = \frac{{{\rm{d}}y}}{{{\rm{d}}x}}(\sec x + \tan x) + y(\sec x|\tan x + {\sec ^2}x)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{\rm{d}}}{{{\rm{d}}x}}y(\sec x + \tan x)\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">making LHS an exact derivative</span><strong><br><br><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;">integrating factor \( = {{\rm{e}}^{\int {\sec x{\rm{d}}x} }}\) <strong><em>M1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since \(\frac{{\rm{d}}}{{{\rm{d}}x}}\ln (\sec x + \tan x) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}} = \sec x\) </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,times; font-size: medium;">integrating factor \( = {{\rm{e}}^{\ln (\sec x + \tan x)}} = \sec x + \tan x\) <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) </span><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\rm{d}}}{{{\rm{d}}x}}(y\left[ {\sec x + \tan x} \right]) = x({\sec ^2}x - {\tan ^2}x) = x\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(y(\sec x + \tan x) = \frac{{{x^2}}}{2} + c\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> \(x = 0,y = 3 \Rightarrow c = 3\) <em><strong>M1A1</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(y = \frac{{{x^2} + 6}}{{2(\sec x + \tan x)}}\) </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;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) when \(x = 0.3,y = 2.245 \ldots \) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">the closer approximation is obtained by using the series in part (b) <strong><em>R1</em></strong></span></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;">[11 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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The function \(f\) is defined by \(f(x) = \frac{{{{\rm{e}}^x} + {{\rm{e}}^{ - x}}}}{2}\) </span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Obtain an expression for \({f^{(n)}}(x)\) , the <strong><em>n</em></strong>th derivative of \(f(x)\) with </span><span style="font-family: times new roman,times; font-size: medium;">respect to \(x\).</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Hence derive the Maclaurin series for \(f(x)\) up to and including the </span><span style="font-family: times new roman,times; font-size: medium;">term in \({x^4}\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iii) Use your result to find a rational approximation to \(f\left( {\frac{1}{2}} \right)\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (iv) Use the Lagrange error term to determine an upper bound to the error in </span><span style="font-family: times new roman,times; font-size: medium;">this approximation.</span></p>
<div class="marks">[13]</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;">Use the integral test to determine whether the series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} \) </span><span style="font-family: times new roman,times; font-size: medium;">is convergent </span><span style="font-family: times new roman,times; font-size: medium;">or divergent.</span></p>
<div class="marks">[9]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \({f^{(n)}}(x) = \frac{{{{\rm{e}}^x} + {{( - 1)}^n}{{\rm{e}}^{ - x}}}}{2}\) <strong><em>(M1)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) Coefficient of \({x^n} = \frac{{{f^{(n)}}(0)}}{{n!}}\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{1 + {{( - 1)}^n}}}{{2n!}}\) <strong><em>(A1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} + \ldots \) <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;">(iii) Putting \(x = \frac{1}{2}\) <strong><em> M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f(0.5) = 1 + \frac{1}{8} + \frac{1}{{16 \times 24}} = \frac{{433}}{{384}}\) <strong><em>(M1)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;">(iv) Lagrange error term \( = \frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{{{f^{(5)}}(c)}}{{120}} \times {\left( {\frac{1}{2}} \right)^5}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\({{f^{(5)}}(c)}\) </span><span style="font-family: times new roman,times; font-size: medium;">is an increasing function because – any valid reason, <em>e.g.</em> plotted a graph, positive derivative, increasing function minus a decreasing function, so this is maximized when \(x = 0.5\) . <strong><em>R1 </em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Therefore upper bound \( = \frac{{({{\rm{e}}^{0.5}} - {{\rm{e}}^{ - 0.5}})}}{{2 \times 120}} \times {\left( {\frac{1}{2}} \right)^5}\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 0.000136\) <em><strong>A1 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><em><strong> </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [13 marks] </span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">We consider \(\int_1^\infty {\frac{{\ln x}}{{{x^2}}}} {\rm{d}}x = \int_1^\infty {\ln x{\rm{d}}x} \left( { - \frac{1}{2}} \right)\) <strong><em>M1A1</em> </strong></span></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \left[ { - \frac{{\ln x}}{x}} \right]_1^\infty + \int_1^\infty {\frac{{1x}}{{{x^2}}}} {\rm{d}}x\) </span><strong><span style="font-family: times new roman,times; font-size: medium;"><em>A1A1</em> </span></strong></p>
<p style="margin-left: 30px;"><span style="font-family: Times New Roman; font-size: medium;">\( = \left[ { - \frac{{\ln x}}{x}} \right]_1^\infty - \left[ {\frac{1}{x}} \right]_1^\infty \) </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;">Now \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0\) <strong><em>R1</em> </strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\ln x}}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0\) </span><em><span style="font-family: times new roman,times; font-size: medium;"><strong>M1A1</strong> </span></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">The integral is convergent with value \(1\) and so therefore is the series. <em><strong>R1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> [9 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The random variable \(X\) has probability density function given by</p>
<p class="p1">\[f(x) = \left\{ {\begin{array}{*{20}{l}}<br> {x{{\text{e}}^{ - x}},}&{{\text{for }}x \geqslant 0,} \\ <br> {0,}&{{\text{otherwise}}} <br>\end{array}} \right..\]</p>
</div>
<div class="specification">
<p class="p1">A sample of size <span class="s1">50 </span>is taken from the distribution of \(X\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use l’Hôpital’s rule to show that \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{{\text{e}}^x}}} = 0\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>Find \({\text{E}}({X^2})\)<span class="s1">.</span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>Show that \({\text{Var}}(X) = 2\).</p>
<div class="marks">[10]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">State the central limit theorem.</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the probability that the sample mean is less than <span class="s1">2.3</span><span class="s2">.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">attempt to apply l’Hôpital’s rule <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\(\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{{{\text{e}}^x}}}\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2">then \(\mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{{{\text{e}}^x}}}\)</p>
<p class="p2">then \(\mathop {\lim }\limits_{x \to \infty } \frac{6}{{{{\text{e}}^x}}}\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = 0\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> \({\text{E}}({X^2}) = \mathop {\lim }\limits_{R \to \infty } \int\limits_0^R {{x^3}{{\text{e}}^{ - x}}{\text{d}}x} \)</span> <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">attempt at integration by parts <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">the integral \( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + \mathop \smallint \limits_0^R 3{x^2}{{\text{e}}^{ - x}}{\text{d}}x\) <span class="Apple-converted-space"> </span><strong><em>A1A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6x{{\text{e}}^{ - x}}{\text{d}}x} \) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6{{\text{e}}^{ - x}}{\text{d}}x} \) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + [ - 6{{\text{e}}^{ - x}}]_0^R\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="s2">\( = 6\) </span>when \(R \to \infty \) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"> </p>
<p class="p1">(ii) <span class="Apple-converted-space"> \({\text{E}}(X) = 2\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\({\text{Var}}(X) = {\text{E}}({X^2}) - {\left( {{\text{E}}(X)} \right)^2} = 6 - {2^2}\) </span><span class="s1"><strong><em>M1</em></strong></span></p>
<p class="p2"><span class="Apple-converted-space">\( = 2\) </span><span class="s1"><strong><em>AG</em></strong></span></p>
<p class="p1"><strong><em>[10 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">if a random sample of size \(n\) is taken from <strong><em>any </em></strong>distribution \(X\), <span class="s1">with \({\text{E}}(X) = \mu \) and \({\text{Var}}(X) = {\sigma ^2}\), </span>then, for <span class="s1"><strong><em>large n</em></strong></span>, <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">the sample mean \(\bar X\) has approximate distribution \({\text{N}}\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}} \right)\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(\bar X \sim {\text{N}}\left( {2,{\text{ }}\frac{2}{{50}} = {{(0.2)}^2}} \right)\) </span><strong><em>(A1)</em></strong></p>
<p class="p2"><span class="Apple-converted-space">\({\text{P}}(\bar X < 2.3) = \left( {{\text{P}}(Z < 1.5)} \right) = 0.933\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In part (b) the infinite upper limit was rarely treated rigorously.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">In answering part (c) many failed to say that the Central Limit Theorem is valid for large samples and for any initial distribution. The parameters of the distribution were often not stated.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p>It is given that \(\left( {5x + y} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = \left( {x + 5y} \right)\) and that when \(x = 0,\,\,y = 2\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Use Euler’s method with step length 0.1 to find an approximate value of \(y\) when \(x = 0.4\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = - 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} - 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the Maclaurin expansion for \(y\) up to and including the term in \({{x^3}}\).</p>
<div class="marks">[5]</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>Euler’s method with step length \(h = 0.1\) to find \(y\) when \(x = 0.4\)</p>
<p><img 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A/akIUdK/ktiPg/HW66eQFu8h55fhZh1b13AsjBXwr7BMpO0kHUEdCqu6Rgt96Oe02rgA1fnfbtjSSpkDNiCHg2ccd9D2NlkvhgNg83L5qvksMbcOuqNTAB2PCXS1T3mFW5KGxeYmnClgFKmAgQASKgncB1jF65DOAyroxqmE1y3IJzI7QtkXa+kRoyRN19xfgTjnf/O0Z8x+SIWQLcp7gy9BkwdAWjmpu8Bd3nIn87KjLUYrbWUsGIQCwSmIvFKanQ4xxOnf0jAvXH3GArGmy3oQBtaLXwhh39RTcBbbqPl3EMg8daYFudDTSdgIWM9XE0serSLUTKagPQexpnnWMTl5K7iGMNNqwuuIam1g8j3pAnQ21iOeksESACEUVAh6Q1D6LE+CfU7/4+DvQ4wXGDeK/lOC7gAl7Jvwt5P30b9S84kV9RhsLtN+JM/8e4Pvgmvt94PqBhF1FFpMyoENCge72oLwfX+aN44eJ9qPheMbajD/1DLgya96PRflUlbvKKDQILsGbLIzA6D2L33sPocY6Bc76Pll/aALyEfMM21Av6j+B8QwMu5pfhe4UPAGf6MXT9IszfPwJ7oCe/cANi3j+A3wKS/mKdAOkc6wqPly+WtXY7ulhtwSrGlxH6IvbTl0vYcn0B27x6IcO95azFNswYc7Ph9nKmxypWUNvFHO5xNrHkimWdlToF0r26xSLo67bVMRP0zFjawmyjvOCXWHtpJoO+gNV2O1i0V4F40lqpvbbja8zR/SIr0EPoG/QFtezl0ruZvqCatVgc7FzdZgaYWGnLOTbKRzjczkr1YPqCF1m345q2JGYhlFJn2pQ93JbyLKev3Ox1lpOn5GaRAGk9i7DDmBTpHEb4s5w0aT3LwMOUnFJnevUZJiEoWSJABIgAESACRIAIBCNAhlowQnSeCBABIkAEiAARIAJhIkCGWpjAU7JEgAgQASJABIgAEQhGgAy1YIToPBEgAkSACBABIkAEwkSADLUwgadkiQARIAJEgAgQASIQjAAZasEI0XkiQASIABEgAkSACISJABlqYQJPyRIBIkAEiAARIAJEIBgBMtSCEaLzRIAIEAEiQASIABEIEwEy1MIEnpIlAkSACBABIkAEiEAwArKdCYIFpvNEgAgQASJABIgAESACM0uAMX4XLM/fDaKD/5WekPqTO3YIKLemiJ2SUUmUBEhrJZHYPCadY1NXtVKR1mpUYs+P11n6R68+pTTITQSIABEgAkSACBCBCCJAhloEiUFZIQJEgAgQASJABIiAlAAZalIa5CYCRIAIEAEiQASIQAQRIEMtgsSgrBABIkAEiAARIAJEQEogPg01zglrYxmyExKQYCjE/h4nOCkVcscAAQ4ueytqCjPAT8xMSMhDWWMPnCR0DGirVoSP0VGW5dWa19uAvPrz1K7VUEW636T7Zw4jHXthENo7XwcSkJBXDzu1+chU3GXH8ZpCr14GZJc1weoc05bXkQ6UGbwaC3pvQb39qrZrozBUHBpqV2CtfRx7bOtwxM3gtj6KS2WPo9Z6JQrloywHIsANvovDvwQ2HvwAjF2Do3sjhsofxLbaHrgCXUT+UUuAG3wPrzZZJflfi61rUxCHHZyEQTQ6p9A/c3/AiVffhtNXbD1MW+9GKlUCH5GIcXAX8ebhDmDjP8HBGNyO17FxqApZ2w7C6gpmWY9h8IQZTeNCA6b7sTZ1XsQUb9ozwrx/ntU5xKPY/XXb6phJX87ah93eQrrZcHs505vqmE30it3i8wuzxHDpxKJ9zvpOvs8GZHp+zmx1mxlk2ovhY/M3PrTmtfuEWaoLWGn7pdgUMkipYknnyffPbjZqqWWm0nY2HIRXNJ+OFa3dfV3s5MA1mRSC9sgM3o5Hu1m1qUJyD5dFExMHSp3j7FnjKvq63kVbRiqSE8Wi65CUtQ7be99FV1/sDp1Ou4Uf0RHOw3Lj17BElFjI61wsTkmFPqLzTZmbFIGR02iufQVVz1aj3twJe9An8kmlQhfNOIGp9M+X0dP8KtqqKvFcvRkd9pEZzy0lMHkCuuX3wLhkriwC3eIUrA7aQXMY6TmG2raf4dnn6mDusMfFGxLZrUxGLRYPuH50He2CfnUKFvuVvAtHu/ppTkss6i4p003rs5DuM9IlJ8gZpQSuwv7GIVTxr0E6q1Ccn430B56BmW7U0afnFPpnzv4WKqv4V99tqCrOR276ZpSZz8fFTTz6hJ4gxzd9DXelJwUOwNnxRmUDnHCis+rvkJ9rxANl5ph/OPMzVwITioEzLgdsvUBGugGJMVAcKkIoBC7D0jqInY/nKEbaQomDwkYegXlIK2r2zEO0vI26UhPQ+Tzyd9RpmOsSeaWJ6xxNoX/WpRWhVZjrZMGxulIYeYMtfw8O0dzjKKlSHEYsnTizswAmxUibrAC6FShqdYC5HbAcexmlRv75bDd2HLLEtFEeX4aaTHE6iB8CHFzWI2hctAs7M2+Jn2LHVUnnQp/5dRRVvg1H+w9g7HwVzT2X44oAFRbQ6TPxjaJKtDvaUG48jdrm06CXoFFQM1wWvNy4EPt2ZmkbRNHpkfmNb6Oy/TdoL89AZ+0x9IwE+wghCjgEyGJ8GWqJBqRnAL02R0xb3wG0jl9vlwWHm7+Acq2dQPySioGSz4U+ZxcqSoGm1g/pJh1Nik5j/6zT5+LpiseAphOwxPANPJrkDZzXK7AefgcLyh9DZqjTUnRLkPN0GUrRhlZL7D6YyTZlDwwyRs7oUrB261rAJi8PN9SPM076nF9OJUaOhM/Af4/1FY9gRaidQIwgiL9iJCI5PR0ZoCkOUaX9tPbPOiQmpyIjA5IPx6KKRpxkdgyDb76Gs+u/i6IVE8xNm4iGYOCn80JPFCqqz8XXiBrmIXXt/ciQPWVxcA30oTfW12GJ6mo6ycxzg+g48A7mP/zNcSPNdRaN9adopGWSSKPjMhcG+jJQ9lAaraMWHYJ5czmd/TPfr/8nMss2Ii3O7nLRI/kYnB31aJ7/ALb7jLQRnG9sQmcoo6AuB/oyi/FQWuyuoxZ3VViXtgn7Ss7h2R+1C6vUc852/OjZcyjZt4kadPS08OA5dZ2HubwIuSV/j1zDjeMr1s834Qj+XNs8iOCpUIhwE+BXsX/zV+MrmnNO9OyvRee67TAmxV33Fm41ppy+pv6ZuwhzcRaya8TFq8fgtP4Kb1rFHWbG4Ox5Gc91ZuI7xoVTzhNFMBMERmA3/wDbcv8eJbnJmOPbTeJmrDzyXzAIbz/GMGjeDUP2ft+HQZzTijfftPp2mOGcp7D/udNY9521mOR43EwUbtrjjMOe7BZklvwElYt+gcw5CZiz7QTSa36CEppkPu2VK2wR8h35k5uRX9WmkgV6xa0CJYq9OAx3H0CWYIxnoLD2Pbg2PI0f5iyh0bSoVHWS/fPwb/DjLINwwzcUvoBTrhxU/PA+6OPwDhf5svMG2F4Y859Hp19mg+0mcRndP34Ahjme7R9rT/0XNlSUIEcvX5PNL9oo90jgl/Hly8Dvi+Z1RnmRKPsTESCdJ6ITW+dI69jSM1BpSOdAZGLPn7SOPU3VSqTUmZ431CiRHxEgAkSACBABIkAEIoAAGWoRIAJlgQgQASJABIgAESACagTIUFOjQn5EgAgQASJABIgAEYgAArI5ahGQH8oCESACRIAIEAEiQATimoD0mwHZgrfSE3FNKIYLr5ykGMNFjfuikdbxUQVI5/jQmS8laR0fWvM6S//o1aeUBrmJABEgAkSACBABIhBBBMhQiyAxKCtEgAgQASJABIgAEZASIENNSoPcRIAIEAEiQASIABGIIAJkqEWQGJQVIkAEiAARIAJEgAhICcSuocbvAdhYhmx+DzFDIfb3iPvASYuv4nbZ0WE+gprCPJR1fKwSgLwihwAHl70VNYUZ3r0881DW2OPbB27ifHIY6dgLg2+PuQQk5NXDzk18FZ2NBALePQCD6jWV+hEJ5YyPPHDOHjSW5Qlt2FB4ED3OsdAK7t37M6/+PMabL2kfGsSZDj0C+/H9KDQkePrq7DI0+vZmnShtfh/XJpRlG7z1Yz+O20cUF0w2bkU0EXwYo4baFVhrH8ce2zoccTO4rY/iUtnjqLVemVgK7jzqH34GjS3P46lX/jRxWDobdgLc4Ls4/Etg48EPwNg1OLo3Yqj8QWyrFTdrniCL3B9w4tW34fQFCbbHnC8gOcJMgBv8V3x/90GJduoZmlL9UI+SfKebgKsHtdt+CFtePdzsGqyFH6Ns20HfJtzBkxvD4LFq7K53yIKS9jIcYT4Yw+CbTfglNuCgg4G5HejeOITyrG8HuSdzcFkP4rEXRpB/pB+MDaPj3rMoNO6FeVA05icbd5iRhJo8v9cn/wfwW33Gxp/bVsdM+nLWPuz2FsjNhtvLmd5Ux2yi1wRFFa5HJittvzRBqOg8FTs6f876Tr7PBmR6fs5sdZsZZNqr6eRmo5ZaZiptZ8Nqp2PEL3a0lggy2s2qC55kpQWrGCZsz1OpH5L0osAZvTp72qte1g4vsfbSu5mp7hyTNW1VHfh2fIAVlJSwAr1eck3sah+VWrs/YidP/odcT/c5VmfSM7n2CpGFMHfL78PuftZStGr8usnGrUgq0g6VOsfgiNpV9HW9i7aMVCQnisXTISlrHbb3vouuvquh2rIUPiIJzMNy49ewRJRYyONcLE5JhT5ofi+jp/lVtFVV4rl6Mzr8htKDRkABwkLgCqyHjgBPPI68xTcGycFU6keQqOn09BDg+tF19DQy0g1I9MW4AFl596L36K/RN/4e03dW5nBZcOgF4ImS+7FYdoK0l+EI94HuSzAab4Osq9YtRMpqw4Q54/p+jaNtS5CePF47oLsVq+5ZCWfTCVhGOGCScU+YcASelLGLwPyFniWh8XdBvzoFi/1K14WjXf2SeQyhR09XRD6Bm9ZnId1npPvnl7O/hcoqK4A2VBXnIzd9M8rM5+HyD0o+EUOAfw3yCl7ANuzMXDClXAWrH1OKnC7WTMBzI74Fq1MWym/ifAxtwR6qRaO9AJnz52hOk7TXjGoWAv451t+1XGKkS5Pk4BroQ6/US3B7H8adfegfEl9/+gUCMFHcauEj28/PlIns7GrIncsBWy8UT2karqMgMUDgMiytg9j5eI5ipE1eNF1aEVoZg9thwbG6Uhh5gy1/Dw4Fm8Moj4aOZpOAOHqyMytAx64lM9rqh5aYKMxUCYg34mXyERNN0UqN9ls0XQGQ9hpBzU6wkQ/ReuZ+PG5SjLT5UtchMTkVGQgwuKJPRcriub7QMkfQuGWho+Ig9gy1qMBOmZx+AnznfQSNi3ZhZ6a2zlunz8Q3iirR7mhDufE0aptPQ/k90fTnk2IMncAVWA93YNm+ncicYKR04nhDrx8Tx0dnw0bAZcHhlqXYV7JGo9FO2odNK9WEr8D68ltYtO9bE7ZnXdpGlJUa0PZMNRrOe3pmzmnBv7Ra4ZRNbZImoi1u6RXR4I49Qy3RgPQMoNfmoFdZ0VADpyuPfOfd/AWUT2LERafPxdMVjwHivIfpyhPFMw0E+Jvsa2hZ+jd4cEmAJ2gtqUyhfmiJnsKESkAcMfkItoFQJh14jPalu9ZPOGouyw1pL8MR3gNPe25e8LcaHqgXIqfiCNpKruOZlTcLy2zVnvoAv+u5ANPWu5HqZ72EEnd4KYSaul9RQ40g4sLrUrB261q/bHFD/TjjXIuta1P850P4hSaPqCLAXcSbh3+P9RWPYMWkRly8N42AT2lRRSPGMst/+PEzPJ+fgjm+Ne8WIZefY9hWjPQ5BsjXz1Ip/pTrh0qc5DVlArrUu7HVpPwoZAxD/X1wmu7H2tR5/mmMnEZzdTnyk2/0rp2YgISbc1HldKKteCXmJGxBvV3ywRhp788wjD7Csiln16KiaJW20dDENNy3pxEOxsAcjSj58hycsW1B2UNpfvfxkOMOI4dQk449Qw3zkLr2fmTIRke88yECNf5QqVH4yCHADaLjwDuY//A3x40011k01p8K4TUmX8YDk5EAACAASURBVD/+E5llG5EWgy0icsSaTE4WIqfSwq8dJPm/hPbSTMBUB5vbgdaiFX6dti+laakfvtjIMZ0EhIfqJWhq/VDSVl0YsA0GGDEBkJSDSn4tLml9GG5HqV4PU905uFkzitK8Bh5pP51qTTkuztmBA83z8PB20Ujj4Dr/Buo7tS0szzmP45kdb+O+t0uRkyTvqKca95QLN8MRyEs7w4nNVvS6tE3YV3IOz/6oXVilnnO240fPnkPJvk2+GzE3aEax4euooQnksyXL9KfjOg9zeRFyS/4euQbJE/Z8E47gz71PbN5V7LP3exfR5Fe6/hXe9K2KPQZnz8t4rjMT3zEunP48UoyzRsCvTWuqH7OWPUrIj8A8pG3Zg5KeWvyoYxAcxuDsOIhnex7Cvi3iiImy/fpFou5B2qtzCYsvv0uEGeXbHkVJSS4Mc7y7EyTMwfyVzYDBs/yGX/sV88rvFvRGDb7133+BRZU/QYlsDrK2uMWoovU3Jg014BZklvwElYt+gcw5CZiz7QTSa5QCq0n2MTrKsjAnvRhtsKIqdxFtK6SGKRL8+G1jntyM/Ko2ldwEecU9/Bv8OMsgvEozFL6AU64cVPzwPuhjtDWoAIp9r6nUj9inEzklTFyDkiPlWNR4P+YkpGBb6zLUHNk14STzoJkn7YMims0A3OAxPGnMR1Xn+D4wvvQness10oEyfsup+d9B68id+N6Zn+O7a/Sy0fNJx+3LQHQ4EvgVefmsJiQkCMPJ0ZFtyuVkCZDOkyUXfdeR1tGn2WRyTDpPhlp0XkNaR6duoeZaqTONIYRKkMITASJABIgAESACRGCWCJChNkugKRkiQASIABEgAkSACIRKgAy1UIlReCJABIgAESACRIAIzBIB2Ry1WUqTkiECRIAIEAEiQASIABEIQMD7+YBw9gZpGOkJqT+5Y4eAcpJi7JSMSqIkQForicTmMekcm7qqlYq0VqMSe368ztI/evUppUFuIkAEiAARIAJEgAhEEAEy1CJIDMoKESACRIAIEAEiQASkBMhQk9IgNxEgAkSACBABIkAEIogAGWoRJAZlhQgQASJABIgAESACUgKxa6hxTlgby5CdkIAEQyH29zjBSUuu5nbZcbymEAb+mgQDssuaYHWOqYUkv3AQmKw+0rqQkIHCmlbYXYraMNm4w8EhytPknKewvzBD2A0lIXsvzPaRICXiMNKx19suxX0Cxd8tqLdf9VwvbjkjtF/xvKct59Wfl7R/lfjy6mFXVAn4xSdJK0iO4/E05+xBY1meoKuh8CB6tPSdnBM9+8U+Nw9l5vNwyeCpaOXTV9RDSxgxUpWwftprCSPGF4e/QTVTYzIC+/H9KOS3hOL1yy5Do2+/5fHw0jrE37drjtsV9cGzzaMQh68eeONU6qip/SrjM0DeV4znLZyuGDXUrsBa+zj22NbhiJvBbX0Ul8oeR+1EG7BzF/Hm4Q5g4z/BwRjcjtexcagKWdsOejfzDqdMlDYmrQ9fF/bghcsbhbrARv8Z9/bugfHJYxgUb8yTjpt0CZmA6wMca5uDh3/eC+Z2oHvjIHYbf4SOEVEMtRgvw9LaBpWdAgHfXoEcRiwn0KQaSLH3K/cHnHj1bUl8epi23o1UWW84hsETZnl8vrTU8hjnfq4e1G77IWx59XCza7AWfoyyoH3nFfz22Cng4cNwsGtwdG/E0O5H8WzHxxKYWrTXEsYbpRbttYSR5DC+nFo0UxIZw+CbTfglNuCgg3nb/RDKs74tvyfzdeixl3A5vwFu5sZoRw56C/PxpPni+EPWyIdobbIqEwCgbMPa2i83+B5elcWn6CtUUgqLF7/XJ/8H8Ft9xsaf21bHTPpy1j7s9hbIzYbby5neVMdsopeiqO6+LnZy4JrMV4gHmay0/ZLMP5oPolXnyerjXxcYcw+0sCL9uK6TjTvS60Hkaf056zv5PhuQtsHhdlYq0UKV6fBJVl3RxhzS6xjfpivYhrpzzON9ibVX17J2h7wNMz7+DdJ272ajllpmKm1nw6qJeT1Hu1m1qULSh0wUOLznwq/z58xWt5npZUwvsfbSu5nJp48/I/92x1+TKY9Hi/ZawgjJa9FeSxj/ssyWT7i11qSZEob7I3by5H9426n3pPscqzPpJVqr1aFrbKBlF9P77uV8mz/AKtoH5HExvt7sYHW2z8dT1tR+P2GW6oKIvL8rdZY9Q4bFUpz2RK+ir+tdtGWkIjlRLJ4OSVnrsL33XXT1eV+TKNLVLb8HxiVzZb66xSlYrZd50UGYCExOH7W6AOj0t+OeDAeaWj8E/9JtcnGHCURUJzsPy41fwxKxWQLghvpxJv0RbFmzIGDJuE9vwzdLc6GXXAfOjjcqL2DT2hQI3hyH5G8WI0cvbcNXYX+jDmc2SUfLLqOn+VW0VVXiuXozOlRfu3IY6TmG2raf4dnn6mDuUL5+CZjV+DzB9aPr6GlkpBuQ6COwAFl596L36K/RF2Cw1K/dcR+j/8xtKNlyJ5K88WjRXksYT3RatNcSxlfIuHNo0cwPiu5LMBpv87RT8aRuIVJWG8QjQLUOzYV+1Z3IcLah1XIZwHV8mvwNlOYskcXF2d9C5Zk1WJs6zxufxvY7chrNta+g6tlq1Js7/afDjOcu7C5p1xf2zExLBgTBu6BfnYLFfqXrwtGu/vFhVC0J3vQ13JUudhtaLqAws0pgQn1cGLB95J8dbyfhPNOPoQA3EeGiCeP2j5Z8QiEwAntHPcqfc6LsyC5k+h6q/OPQ6ZdjueI81/drmJPzkSd2zrovIm25op3yfYF5EXbkLfN17EKnXsW/OmlDVXE+ctM3+8+LEozABjjhRGfV3yE/14gHyswR3ZH7U5s9H16Lo223YHXKQh9nX+ptgR+OfWHAwWXvQH35j9FX9iJKMm/xndKivZYwfIRatNcSxpe5uHYE1kw7lj/H+ruWe4x7lwO23s/8LvUMljhwpv9jcJgLfdpSycMAH5x/GD+F5B2541MXNLVf/iHuEKr4qRKdVSjOz0b6A89omC/rl8VZ8fAzZWYl1ZlMRBAciqe7ySTIz3npxJmdBTApRtomExtdM90EtOiTiOT0ZUCAm4W6Mc/nU0vc012eOIpPmOR7M9Jzi1H1yntoff8jxYThYCz4zrkHdzzyV7LROeVVgjF3x0ask7RfXVoRWoU5qBYcqyuFkTfY8vfgkHT+qm4Filodwlway7GXUWrk+/Ld2HHIEmI+lTmKxWMOroE+9GIZ0pPHx9O0l5SfzL0G89NzUVz1Ft5v7Uaf8kMfWWRatFcPo0V7LWFk2YnLg1A1U4HEzzU7cz8eN3lH2hINSM8A2lRHYA3qDwF8tMLD2BI8sk4yYqep/c5DWlEzGD830vI26kpNQOfzyN9RF5lz0sWXusp3oqJ/1P0Kc170/nMjAvkHKiD/jrvgALOMyibGBAodNf4xo7NWfQTdwfQFDeycoOU15rC8wkqNy/3riKii1rjF8BH6G+laux3drKHUxIBVrKilXzHvZAKo/PyWDcHmj/FzXnYEnX/idrSxcqN0roxKuu4B1l5uYvDNlVEJE0av8Orsnf+LzfI5QsIcwnKm9/MPAMrtYJaGUmYEmL6oRT6PUXqJFu21hGGMadFeSxhp9mbaHV6tFaXTqpniMsb4uWFPsmrLJ5IzYj1axQrqetkof8YXv7JujV/Gz0HeIJsbOX7O59LUfq8xR/sPmDFC5qQrdY69ETWvZd5rc0zh6fcKrIffwYLyxyZ8JaPynEBes0IgBH2SjKjofBclqMHK+XNgKHwBpz44i57OO7FVnN8ky3MIccuuo4NQCej0a1D4/AHUmf6Ed7ovaG6vnpEyI7KSJui+hCftLyIvK/DcNz6/On0unq54DGg6AUugL091S5DzdBlKIc6VCbWksRxeh8TkVGTgI9gG5AtrhFRqnR6Zhf+Al+o2w/mOBbYAo2patNcSRqv2mupHSAWNocAaNZOXmIPL+hqaF/wtdkpecQM6JOU8jc62IuCZDMxPyEBh7Ql88Lsz6Az4tbV35DTvy745jfK0vEea2u9c6HN2oaIUvrnLqnGFyXOCni5MOZpqsroUrN261i8WYdKyU8unt/ynxK/h7PrvomiFYs6LX6zkMfsEQtVHh8S0POxp7OU/a4ajcTe+jN/DVroTD6WJk0/FUoQat3gd/U6aQID2Gjg+bZ2z1ps1f4MQDA3Zx0cqqQsPgOmTfL2nEl8MeelS78ZW042KEo1hqL8PzoA3WUVw4XAeUtfeD5PaKcFPi/ZawogJaNFeSxgxvnj8DaaZnAk3+C4On12LiqJVirlmfLgkpN33XTTyS3iwXjSWfBU440Jp2UakqVkqGh/GhBxoar/8VJn0aZg2JS/zdBypFX864g1jHJ6KkyF7QvbOowjaaYzB2VGP5vkPYLvPSBvB+cYmdAZ62g5jSeMv6anqw19fiR1Nf4m3K4yKp7Cpxh1/akxPifkPPobHJxUHi1RT5xzKzZrvG/4TmYFuBmJ+XA70ZRarGPdigDj+FYztJYqRCF7XQZX16Sbi5OmnL6zPQrri4xHhKi3aawnjy4IW7bWE8UUYh44gmkmIcM4OHGieh4e3i0YaB9f5N1DfKV03z3sBN4iOZ/ag6b79qMhZKIll3Kn9YQyApvbrwkBfBsoeSvP/KGY82fC4xPe4yneion90/vLvwB9kxnLP2kueeQYPyt6Je9bS2iDxG2a2lnJhjgTPQvY/wfpr0cYnenXWpo+/rrxCbjZqO8lery5gywteZN3KtbaYtrhJ6ykScPezlqJMZix9hVkEDTzzQky++YOe+NU19J7TNCcl0Bw2fn7iL9kxi8M7H+4ac3T/lJUq1mhzOyzs2DGLb902t6OL1ZZW+6/RNkUc03V5RLRpfl6ncQMrF9a48s73MdbK5vjKdfWukWUsZQ1ePYR+2rSD1Z1TX+FOy3ykwGG0aK8lzHSpNrl4wqu1Rs3Edl7d7ZlrJvS/LazUqJffV4X7rHL+2TCztb/GqguMrKC2y9cG/WkFnoOqqf3y89+O/dLbD3nmw3XXlqus0eaf8mz4KHX2rXKrPDEbmZnRNNwO1l1bwPR8ZZB0BmKaqp2G0kATjlU+TBAjicLf6NTZ20Fo0EeuqzhB1VMH6tpt3o5DKpz2uKVXRYM78rQeZufqijxtktdSX8CqW8YNIpGpXEPRl//lO+dSycOV9Ny4221rYEW+m8S4P2PihGHPg5i+oJq9rlInxAnkPD8hj6+fZLYI/qgoUnQWDNqCVQzQS4zxcf5yXd1s9FwDK9CLD8WrWEF1y/iNc/wyr0uL9hOF0aK9ljB+GZtVj/BqrVEzhaHm0V3UWfHrGwTxLHbsqTt1rN2mbqz7YPMfjBSpf+ynqf2KHxgI9xS+7r0WPE1f4jPvUOqcwCfJj+Xxe2d5neEZ2qNUZ4UA6TwrmCMiEdI6ImSY8UyQzjOOOGISIK0jRooZzYhS5xicozaj/ChyIkAEiAARIAJEgAjMGgEy1GYNNSVEBIgAESACRIAIEIHQCJChFhovCk0EiAARIAJEgAgQgVkjIJujNmupUkJEgAgQASJABIgAESACqgSk3wzcIA0hPSH1J3fsEFBOUoydklFJlARIayWR2DwmnWNTV7VSkdZqVGLPj9dZ+kevPqU0yE0EiAARIAJEgAgQgQgiQIZaBIlBWSECRIAIEAEiQASIgJQAGWpSGuQmAkSACBABIkAEiEAEESBDLYLEoKwQASJABIgAESACREBKIHYNNc4Ja2MZshMSkGAoxP4eJzhpyVXdY3D2HEShIQEJCQZkl5lhdwW/SjUq8pyQAOc8hf2FGcKOGAnZe2G2j0wYXjjpsuN4TSEMvKaCPk2wOsf8ruOcPWgsy/PEbShEzXE7XNJQIx0oEzTm45H+G5BXf15eT/zCbkG9/ao0NkBLGPkV8XPEOdGzX9QsD2Xm83ItNJDgBs0oNqhwl7bxhAwU1rT6t1ctYTTVK+obZFJJuWruX4FQ2z1pL6MeGQdTbdN83XnzCGqE/j8LZR0qm7IDUNf+Y3SUZSn6bW8fnlcPu9rtmrsIc3GWom/nMNKx13svkd4DeLdKXxNu8uKuVcq9pUT/6PwNvim7f7ncbPT0MdbQ7dmwWdy3Tl/azoLsOuYfVQT7RITOo2dYS8P7ng13xT1Z9eWsfdgdmJy7nx2r/Slr8+4BJ+oDxcbPjN8c2lTEagUdxb3pVrGiln7vRtyS/T/99g5VbhCssg+ob286MatawohhZ/c3/Fp/wk63vM66xQ3Yu19kBfpMVtp+STsIYd9Afv9IpTZ8G982vnHzaC+rK1jF9EUtbMBXjTSE0VSvIrtvmH2dJ9O/MsZCbfekvV87mX2tlVmYWpsW++1A++z6Uguk/XA7K/XtDyvdNzTQntxi/6w8L+4tKo3D6/br4325mjWHUueY3JTdbatjJtmN33tznkgA90fs5Mn/8N7MeT2818jimTWdZiwhZQWYsYQCRvw56zv5vuRmyhgTGt/EN3B3Xxc7OXBNFqugM6TX8Zsyb2Zy49rbUH06XmLt1bWsXTAeJNHxedhQx2y+mzx/Y+GNvoqJDUgtYSTJzKYz3Fr7a+bpHOX6TESENwh2sZLSbUyvMNT82zhjns2fx+uDpjBa6lWE9w2zrbM/Vw39Kwu13ZP2ai1jtrVW5mFKbZrvK42Z4w9Xysh9x4G05+vZAVbRPiC5T/MX8f3KDlZn+9wXg8fhZqOWA6ygpIQV6BWG2vBJVl3R5hks8F3Fx1/BNtSdU8TvCzBrDqXOMfjq8yr6ut5FW0YqkhPF4umQlLUO23vfRVef4rWVOKSp+xKMxtsgXgGMYaj/ItJLHsSapHFfMTj9TpbAPCw3fg1LJEi5oX6cSX8EW9YsCBipbvk9MC6ZKzuvW5yC1XqJF9ePrqOnkZFuQKLPey70q+5EhrMNrZbLAMch+ZvFyNFL47oK+xt1OLPpbqT68sVhpOcYatt+hmefq4O5Q/H6VIhfSxhfRuLO4acZ9zH6z9yGki13IikoDQ4u6yt4AY+iJG+pIrRaGwd0+ttxT4YDTa0fYgRawgB+eQTgV6+ob5DwV+OqoX9FKO2etJcAjyinX3vR3KavwHroh6hdthf7nrwHel8/qyzeRNpfx6fJ30BpzhLJfRrg7G+h8swarE2dJ4/MZcGhF4AnSu7HYvkZcJ/ehm+W5srzwdnxRuUFbFqbIotfcWlYDgPiCktupiNR4WbdBf3qFCz2K10Xjnb1y+cgqaXpsqOj/od4ru8xHClZI7npqwUmv8kTGIG9ox7lzzlRdmQXMn2GdQgx3vQ13JXuve27HLD1fuZ3sefG68CZ/o/B6b6ItOUKM4GvM+ZF2JG3bLyBCo22AU440Vn1d8jPNeIB5ZxFLWH8chOPHhxc9g7Ul/8YfWUvoiTzluAQxE52Zxbm+4V2YcD2kZ8vdAuRstoA55l+DHFawvhH4fOR1iufJ4B47xumo39FkHZP2ktrXIS6Q2vTnN2MvU/9F7bf8yn++VueucmGwv04rpybPKH2c6FPW6q4H/MPDqeQvCNX8pDNI+MNwyPAEwXInD/Hj6FOvxzLFfcbru/XMCfnI09p8PldPfsefqbM7GdhmlMUbtZQjKpoTcM7wXB+OnKLn8cr75/E+30aJrlrjZ7CjRMQJuDfjPTcYlS98h5a3/8oxEnmHEYsnTizswAmcaQt0YD0DKDt6K/R5zep1IDVKQvHDbHxnEBooHdsxDoxHv6cbgWKWh1gbgcsx15GqRHorNqNHYcs4/nUEkaSTnw6+cm/azA/PRfFVW/h/dZu9AX9QEfSySo6Uw/DRCSnLwPa1EfIPQ9pWsKoKaJSr4Rg1DcIGKbUv8L74c1E7Z60V6uVkeUXapv2jsIaV+P2r6xDSWMv2Ggv/hH1MO2og9XXHwTTXoWC8JC9BI+sk74NE0fltmGnlodCIVo+jz2445G/kr3tUUkxLF6xZ6hNCaMOSTn74GDX4LC8glI0IN+4F+ZB/y8Lp5QMXQwk5aDSweB2dKOhFKjK34wnzReDj3aK7FwWvNy4EPt2Zo0/YenS8FDZY9C37cdzDWc9BhX/hdG/tKLHuQzpyeMvRMVoILwi68EdeV9Wfx2n0yPzG99GZftv0F6egc7aY+gZUViBWsKMJxhnroXIqbR4DN6G7bzQMD55DIMKhONQ+E72NbQs++4EI2/zkPbQTpTqX8czzx3FeaGjH4PT+g5ae654H9K0hBlP1edSq1fCSeobfIym4piw3ZP2U0E7e9eG2qY9o9v6NXn4Zqbe87CcuAqPfe+7MHVWY2+zHRy0aO9fQs9DthFZ0ulJLgsOtyzFvlDehgkG3xeRlxV4+o1/6rPnE3uGmndUpdfmGB/5CJnnXOgzt+P5l/4RJuf/h24bjaqFjFDjBTr9GhQ+fwB1pj/hne4LGjW7Auvhd7Cg/DHF61L+Zvo0OtuKgGcyMJ9fsqH2BD743Rl0mu73n8PA51FrA9UtQc7TZSiFd66bWvm0hFG7Lh78eGO28B/wUt1mON+xwOZ7ilYUXuhk9dj14FLV0U9f6CQjKjrfRQlqsHL+HBgKX8CpD86ip/NObBXnmGgJ44uQdwSqV9JAcd43TEv/ys8nVGn3pL20okW+W2ubFuaxOfzKo0u9G1tNgHCv1qq9LBbvKJjsIZtvwx1Yumt9SCNjqgafLK0wH4ifMSi/MhD9o+9X7cs/xjxfCCo/8Q9SOvc5Vme6O7TlBIJEGe7TkamzumbqrK6xgWM/ZQ3nNC6aEkRDvl5s0LoEixDXNpWviyQ51RJGEnwmnZGotf8Xg1IC3q8H/ZZNkXxCH/DLbS11aKIw01uvpKWaaffs6qzOcFL9K5PGRdprqSezq7WWHHnvrb6v6tWu8S6FoWy7Ql+5nJnq/n92ub2c6UNt9/z1GxRf5QdcvkPsQ9RsAL4e7oio+7xS59gbUcM8pK69HxlNJ2DxvaLi4BroQ2+gUZVAxjI/H+PCX4xPVg8UjvynSIAfGh/G+ruWj7/GVI1xDM6OejTPfwDbV4gfBIzgfGMTOn1aSy7kBtHxzB403bcfFTkLJSdEp9oTmXhO5dflQF9mMR5KU3xdJA2qJYw0fFy5Pe3wwvospKvOPRNfLzJ+2SDvvxvD7eXQYzPqbJ+DtRYhza/X4utFJXY0/SXerjCqv8LGRGFCrFe8ZnHbN0xj/wppuyfto7MrCNam+VItQFaeCfre0zgrXaBcmO/Ij4AvwxeEKUdim+d/g7d71VEw76v18f6DgQ23o1Svh6nuHNysGUXK/lvrW5VwCiTav0oLTvSPzt/gCzJ61lzawKotnwhF9BybWGlDt3ch1gHWXr6ZFdT1stHohKCa67DrLCxkmMmMpa8wi7gQavsPmKmggZ0blSxiJoar7vbyH2a2lnJmVHvqUj6psWFma3+NVRcYJ16zR+2JzEvN7bCwY8csvnV2hIUaS6tl669pCaMqwix5hldr7/p1xlLWYBEXkW5j5aYdrE42GiqGq2UWqf4+RuJIi9qTsJuN2k6y16sL2PKCF70L6/ou9DqChQleryK9b5h9nYP3r0zZfsXjYO1eJh9pL8PBWx/wLX2qPDULx2JbDdKmRa19fTe/LGk/aylaxfRiPy8sdF7ETNIwshJMpD0fMIRRMGGUTbGOmiStkN6qSK6bSadSZ5/qyhMzmYlZiVtc8Z6/sUtuFmLaSkONeVc25znw//zKyS3eG4x4TSz8hl/nYXaurmh8mFtfwKpbxg0iH2NZY/d2EGpGGqQNUFxtWs+MpXWs3buLgS9OhcNta2BFAToKt6ONlRv1Ql0An8fXTzKbwpDQEkaR5KwehldrcVcI8ZXDKlZQ3eI1zqUYxM4/FENN7MQ9bbuu3abyMKUljMZ6FeF9Q1h0DtK/+hlqTGO7l1YNcdFx2WLHWnTVEiY6tQ+L1j5NNLZpWd/tu5ixURtrqy7w9v2SQRFJkHGnqKHaAxpv+J1jdUUHAjzcjcciuCY01HiDr9Q3YKO4MmyHSp0T+JzwI3r8nodeZzgH+CjtGSZAOs8w4AiKnrSOIDFmMCuk8wzCjbCoSesIE2SGsqPU2W+2xwylS9ESASJABIgAESACRIAIhEiADLUQgVFwIkAEiAARIAJEgAjMFgEy1GaLNKVDBIgAESACRIAIEIEQCcjmqIV4LQUnAkSACBABIkAEiAARmGYC0m8GbpDGLT0h9Sd37BBQTlKMnZJRSZQESGslkdg8Jp1jU1e1UpHWalRiz4/XWfpHrz6lNMhNBIgAESACRIAIEIEIIkCGWgSJQVkhAkSACBABIkAEiICUABlqUhrkJgJEgAgQASJABIhABBEgQy2CxKCsEAEiQASIABEgAkRASiB2DTXOCWtjGbITEpBgKMT+Hic4acmDuscwaN4NQ1497KFdGDRmCgBwzlPYX5gh7IiRkL0XZvtISFg4pxVvvlGDQgOv7150qG3KDoAbNKPYsAX19qvj8Y90oIy/jq8bsn8D8urPq9cT7iLMxVmK8x+joyxLEYc3Tqo347zJRQSIABEgApMmEKOG2hVYax/HHts6HHEzuK2P4lLZ46i1XtEMihv8V3x/90E4NV9BATUTcH2AY21z8PDPe8HcDnRvHMRu448CGlvyeMfg7DmIb2U+BnP/bSjsHAZz7ENOkkpV5i7i2Pf/F+plInIYsZxAk8xPTGEttq5NgX9MYxg8Vo3d9Q4xoOd35EO0NlnlfsKRHqatdyPVPyKVsORFBIgAESACRCAwgZi8lXB2M/bWrkTF07nQ6wCdPhdPV6xE7V6zttExVw9q976HhfetCkyOzkySwFVcsHyOu7Z/TdAGOj3WFBdiO9rQarkcJE4OLutBbHuwB6uPteHne7YiJy0pwDW8sV6NUwu/DL0sxGVYTi/CLxzXhL1t+SVphP/hdpRuuB9rU+fJQgN8moew99SNuE8WEW/wncWNXrvP3AAABdxJREFUvxiAW4xD+L2E9tKNAQw+RdR0SASIABEgAkQgCIEYNNSuoq/rXbRlpCI5USyeDklZ67C991109UleganCuQLroSPAE48jb/GNqiHIcyoE5mG58WtYIkrDv54c6seZ9EewZc2CiSN2WXBoz6tY9uIP8eQavcrIl3g5b1y9ghfwKEryloqenl+OQ/I3i5Gjnyvxvwr7G3U4s0llFIxP8wXgiZL7sVhyBXAdnyZ/A6U5S2T54OxvofLMGhWDT3YxHRABIkAEiAAR0ERAcrvUFD7yA3H96DraBf3qFCz2K10Xjnb1q89BEkom3uC3YWdmEKMh8klEQQ5HYO+oR/lzTpQd2YVMn2GtlvWrsDfX4Cnk4h7udXxLmGOWgcKaVthdikmEonG1MwvzlVHpvoi05YpROL7OmBdhR94ymdEFiEZ7ATLnz1HENBf6tKVIlPnyDwmnkLwjl157yrjQAREgAkSACEyWgJ8pM9mIIuY6lwO2XiAj3aC4iWrIoeQGL78Ba7iWgoRGQJjQfzPSc4tR9cp7aH3/I7gmikEwwE/DuGYVvnLPE2h0uDF6bg9Q+y3sOGSRXCsxriY0/MYT4/p+DfMdG7FuiXSUTWq03zIeeCKXYPAtwSPrblMYfBNdROeIABEgAkSACAQmEHuGWuCyBjlzBdbDHVi2b2eQkZ0g0dBpbQSSclDpYHA7utFQClTlb8aT5ouBRzsFA/wWrMlbj0zhtaUOiSu24nv/uBadT9WgWfiqkzeuXkPLsu+iJFOjcQV+FKwHd+R9GbJxNpcFh1uWYl/JGs0Gv8fgMyJL7cMGbVQoFBEgAkSACBABGYHYM9QSDUjPAHptDskoi6zMKgfeG/zSv8GDslEVlaDkNa0EdPo1KHz+AOpMf8I73RcCaibMY/P7UnMeUtfeDxM+gm3ABQjGlR67HlyqfURLGAX7IvKypK+6PUb70l3rZXPpJi54AINv4ovoLBEgAkSACBCBCQnEnqGmS8HarWv9Cu250QdafuEyepp/hufzUzDHt67WIuRWWYG2YqTPmWB9Lb+UyCNkAgE0k8ajW5yC1XoHzvR/rDLqtgzpyTdhpOcYqp9/EMlzxPXR5uDm3OfhxOsoTv8zJKisbaY6CjZyGs3V5chPvnF8jbSbc1HldKKteCXmJCjWZeMzqmrwSUtAbiJABIgAESACoROIPUMNnlGWjKYTsPgWQeXgGuhDr0lt+QUe2kLkVFrkyzUwfpmFTMBUB5vbgdaiFdpHaULXIc6vcGHANoz1dy0P/Jox6cvI225A76nfwen7dkCq601IytkHh2ypDDeG28uhx2bU2T4Hay1CmqzGBxgF876W9S3dwcfJL9+h18NUdw5u1oyiNPkyHqoGX5yrSsUnAkSACBCBqROQ3bamHl1kxKBL24R9Jefw7I/ahZs652zHj549h5J9m3w3as+K9V9HTQiL4EZG6aI8F94V/rPLmmB1jgEYg7PjICqHtuEpk2QSvhiupsf7OnQhjE/sxfp3/hf2NpwV/Djn+6hrvCjTNSQ60zYKFsDgCykzFJgIEAEiQASIgD+BmDTUgFuQWfITVC76BTLnJGDOthNIr/lJCBPM/UGRzzQR0H0Bt9/zFdiqCpBluBEJhm/j1StfR8PPC7EiyFeauiUP4kBnDTL+7W8wP4HX9S0seKJm0rpyfd04Zfw61kx18j9v8J26Pfg6cNOEkKIhAkSACBCB+CGQwPj3O4AwF8frjJ/Sx2FJ+b0tSef4EJ60Jp3jg0D8lJLadHxordQ5RkfU4kNMKiURIAJEgAgQASIQ2wTIUIttfal0RIAIEAEiQASIQBQTIEMtisWjrBMBIkAEiAARIAKxTUA2Ry22i0qlIwJEgAgQASJABIhA5BOQziW/Qcyu1FP0o18iQASIABEgAkSACBCB8BGgV5/hY08pEwEiQASIABEgAkRgQgJkqE2Ih04SASJABIgAESACRCB8BMhQCx97SpkIEAEiQASIABEgAhMS+L/pdABwI43xDgAAAABJRU5ErkJggg=="></p>
<p><strong>Note:</strong> Accept 3 significant figures in the table.</p>
<p>first line of table <em><strong>(M1)(A1)</strong></em></p>
<p>line 2 <em><strong>(A1)</strong></em></p>
<p>line 3 <em><strong>(A1)</strong></em></p>
<p>hence \(y\) = 3.65 <em><strong>A1</strong></em></p>
<p><strong>Note:</strong> Accept any answer that rounds to 3.65.</p>
<p><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left( {5x + y} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 5y\)</p>
<p>\(\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}}\) <em><strong>M1A1A1</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>M1</strong> </em>for a valid attempt to differentiate, <em><strong>A1</strong> </em>for LHS, <em><strong>A1</strong> </em>for RHS.</p>
<p>\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}} - 5\frac{{{\text{d}}y}}{{{\text{d}}x}} - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\)</p>
<p>\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\) <em><strong>AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}1 - {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\)</p>
<p>\(\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = - 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\) <em><strong>M1A1A1A1</strong></em></p>
<p>\(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = - 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right) - 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} - \left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\)</p>
<p>\(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = - 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} - 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\) <em><strong>AG</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>when \(x = 0\,\,\,y = 2\)</p>
<p>when \(x = 0\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = 5\) <em><strong>A1</strong></em></p>
<p>when \(x = 0\,\,\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = - 12\) <em><strong>A1</strong></em></p>
<p>when \(x = 0\,\,\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = 120\) <em><strong>A1</strong></em></p>
<p><strong>Note:</strong> Allow follow through from incorrect values of derivatives.</p>
<p>\(y = 2 + 5x - 6{x^2} + 20{x^3}\) <em><strong>M1A1</strong></em></p>
<p><em><strong>[5 marks]</strong></em></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.</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 value of \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \cot x} \right)\) .</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;">Find the interval of convergence of the infinite series\[\frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \]</span></p>
<div class="marks">[10]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Find the Maclaurin series for \(\ln (1 + \sin x)\) up to and including the term </span><span style="font-family: times new roman,times; font-size: medium;">in \({x^3}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) <strong>Hence</strong> find a series for \(\ln (1 - \sin x)\) up to and including the term in \({x^3}\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) Deduce, by considering the difference of the two series, that \(\ln 3 \simeq \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\) .</span></p>
<div class="marks">[12]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><strong><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;"> \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \cot x} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan x - x}}{{x\tan x}}} \right)\) <em><strong>M1A1</strong></em></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{{\sec }^2}x - 1}}{{x{{\sec }^2}x + \tan x}}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">, using l’Hopital <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2{{\sec }^2}x\tan x}}{{2{{\sec }^2}x + 2x{{\sec }^2}x\tan x}}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 0\) <em><strong>A1</strong></em></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;"> \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \cot x} \right)\)</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x - x\cos x}}{{x\sin x}}} \right)\) </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;"> \( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x\sin x}}{{\sin x + x\cos x}}} \right)\) </span><span style="font-family: times new roman,times; font-size: medium;">, using l’Hopital <strong><em>A1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x + x\cos x}}{{2\cos x - x\sin x}}} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1A1</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><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><span style="font-family: times new roman,times; font-size: medium;">\({u_n} = \frac{{{{(x + 2)}^n}}}{{{3^n} \times n}}\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}} \times (n + 1)}}}}{{\frac{{{{(x + n)}^n}}}{{{3^n} \times n}}}} = \frac{{(x + 2)n}}{{3(n + 1)}}\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\mathop {\lim }\limits_{n \to \infty } \frac{{(x + 2)n}}{{3(n + 1)}} = \frac{{(x + 2)}}{3}\) </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;">\(\left| {\frac{{(x + 2)}}{3}} \right| < 1 \Rightarrow - 5 < x < 1\) </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;">if \(x = 1\) series is \(1 + \frac{1}{2} + \frac{1}{3} + \ldots \) </span><span style="font-family: times new roman,times; font-size: medium;">which diverges <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">if \(x = - 5\) series is \( - 1 + \frac{1}{2} - \frac{1}{3} + \ldots + \frac{{{{( - 1)}^n}}}{n}\) </span><span style="font-family: times new roman,times; font-size: medium;">which converges <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">hence interval is <span lang="EN-US">\( - 5 \le x < 1\) </span> <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [10 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(i) \(f(x) = \ln (1 + \sin x)\) , \(f(0) = 0\) <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) , \(f'(0) = 1\) </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; font-size: medium;">\(f''(x) = \frac{{ - \sin x(1 + \sin x) - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}} = \frac{{ - (1 + \sin x)}}{{{{(1 + \sin x)}^2}}} = \frac{{ - 1}}{{1 + \sin x}}\) , \(f''(0) = - 1\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) , \(f'''(0) = 1\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\ln (1 + \sin x) \approx x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \ldots \) <span lang="EN-US"> </span><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) <span lang="EN-US">\( - \sin x = \sin ( - x)\) </span> <em><strong>M1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">so, \(\ln (1 - \sin x) \approx - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} - \ldots \) <em><strong>A1</strong></em></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;">(iii) \(\ln (1 + \sin x) - ln(1 - \sin x)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \ln \left( {\frac{{1 + \sin x}}{{1 - \sin x}}} \right) \approx 2x + \frac{{{x^3}}}{3}\) <strong><em>M1A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">let </span><span style="font-family: times new roman,times; font-size: medium;">\(x = \frac{\pi }{6}\) then, \(\ln \left( {\frac{{1 + \frac{1}{2}}}{{1 - \frac{1}{2}}}} \right) = \ln 3 \approx 2\left( {\frac{\pi }{6}} \right) + \frac{{{{\left( {\frac{\pi }{6}} \right)}^3}}}{3}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">AG</span></strong></em></p>
<p> </p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[12 marks]</span></strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">There was some confusion in differentiating twice using l’Hopital’s Rule but the confusion was made worse by not taking care to write legibly. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This was in general well done but some students did not bother to test the end points. </span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">(c)(i) This was generally well done with various approaches being used. </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) This part was often done by using the differentiation all over again instead of using part (i) again demonstrating a lack of appreciation of where time and effort can be saved in answering questions and ignoring the word "Hence". </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(iii) Candidates often managed to work their way through this question but with lack of clarity as to where \(\frac{p}{6}\) came from. </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 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\cos ^4}x\) given that \(y = 1\) when \(x = 0\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Solve the differential equation, giving your answer in the form \(y = f(x)\).</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) (i) By differentiating both sides of the differential equation, show that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = - 10\sin x{\cos ^3}x\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Hence find the first four terms of the Maclaurin series for \(y\).</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) integrating factor \( = {e^{\int {\tan x{\text{d}}x} }}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{\ln \sec x}}\) <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;">\( = \sec x\) <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;">\(\sec x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x\tan x = 2{\cos ^3}x\) <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;">integrating,</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y\sec x = 2\int {{{\cos }^3}x{\text{d}}x} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2\int {\cos x(1 - {{\sin }^2}x){\text{d}}x} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 2\left( {\sin x - \frac{{{{\sin }^3}x}}{3}} \right) + C\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Condone the absence of \(C\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(substituting \(x = 0,{\text{ }}y = 1\))</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(1 = C\) <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;">the solution is</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = 2\cos x\left( {\sin x - \frac{{{{\sin }^3}x}}{3}} \right) + \cos x\) <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>[9 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) differentiating the equation,</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;">\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\frac{{{\text{d}}y}}{{{\text{d}}x}} = - 8{\cos ^3}x\sin x\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: <em>A1</em></strong> for each side.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">substituting for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\left( {2{{\cos }^4}x - y\tan x} \right) = - 8{\cos ^3}x\sin x\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y({\sec ^2}x - {\tan ^2}x) = - 8{\cos ^3}x\sin x - 2\tan x{\cos ^4}x\) (or equivalent) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = - 10\sin x{\cos ^3}x\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) differentiating again,</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;">\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}} = - 10{\cos ^4}x + {\text{term involving }} \sin x\) <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;">it follows that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y(0) = 1,{\text{ }}y'(0) = 2\) <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;">\(y''(0) = - 1,{\text{ }}y'''(0) = - 12\) <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;">attempting to use \(y = y(0) + xy'(0) + \frac{{{x^2}}}{2}y''(0) + \frac{{{x^3}}}{6}y'''(0) + \ldots \) <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;">\(y = 1 + 2x - \frac{{{x^2}}}{2} - 2{x^3}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[9 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) (i) Using l’Hôpital’s rule, show that</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;">\[\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^{\lambda x}}}} = 0;{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}\lambda \in {\mathbb{R}^ + }\]</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Using mathematical induction on \(n\), prove that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\[\int_0^\infty {{x^n}{{\text{e}}^{ - \lambda x}}{\text{d}}x = \frac{{n!}}{{{\lambda ^{n + 1}}}};{\text{ }}} n \in \mathbb{N},{\text{ }}\lambda \in {\mathbb{R}^ + }\]</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 probability density function</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;">\[f(x) = \left\{ \begin{array}{l}\frac{{{\lambda ^{n + 1}}{x^n}{{\rm{e}}^{ - \lambda x}}}}{{n!}}x \ge 0,n \in {\mathbb{Z}^ + },\lambda \in {\mathbb{R}^ + }\\{\rm{otherwise}}\end{array} \right.\]</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;">Giving your answers in terms of \(n\) and \(\lambda \), determine</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;">(i) \({\text{E}}(X)\);</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the mode of \(X\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) Customers arrive at a shop such that the number of arrivals in any interval of duration \(d\) hours follows a Poisson distribution with mean \(8d\). The third customer on a particular day arrives \(T\) hours after the shop opens.</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;">(i) Show that \({\text{P}}(T > t) = {{\text{e}}^{ - 8t}}\left( {1 + 8t + 32{t^2}} \right)\).</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;">(ii) Find an expression for the probability density function of \(T\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Deduce the mean and the mode of \(T\).</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) (i) using l’Hopital’s rule once,</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;">\(\mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^{\lambda x}}}} = \mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{n{x^{n - 1}}}}{{\lambda {{\text{e}}^{\lambda x}}}}\) <strong><em>(A1)(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>A1 </em></strong>for numerator, <strong><em>A1 </em></strong>for denominator.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">if \(n > 1\), this still gives \(\frac{\infty }{\infty }\) so differentiate again giving</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;">\(\mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{n(n - 1){x^{n - 2}}}}{{{\lambda ^2}{{\text{e}}^{\lambda x}}}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">if \(n > 2\), this still gives \(\frac{\infty }{\infty }\) so differentiate a further \(n - 2\) times giving <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mathop {{\text{lim}}}\limits_{x \to \infty } \frac{{n!}}{{{\lambda ^n}{{\text{e}}^{\lambda x}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = 0\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) first prove the result true for \(n = 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_0^\infty {{{\text{e}}^{ - \lambda x}}{\text{d}}x = - \frac{1}{\lambda }\left[ {{{\text{e}}^{ - \lambda x}}} \right]_0^\infty = \frac{1}{\lambda }} \) as required <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">assume the result is true for \(n = k\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_0^\infty {{x^k}{{\text{e}}^{ - \lambda x}}{\text{d}}x = \frac{{k!}}{{{\lambda ^{k + 1}}}}} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">consider, for \(n = k + 1\),</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_0^\infty {{x^{k + 1}}{{\text{e}}^{ - \lambda x}}{\text{d}}x = - \frac{1}{\lambda }\left[ {{x^{k + 1}}{{\text{e}}^{ - \lambda x}}} \right]_0^\infty + \frac{{k + 1}}{\lambda }\int_0^\infty {{x^k}{{\text{e}}^{ - \lambda x}}{\text{d}}x} } \) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = (0 + )\frac{{k + 1}}{\lambda } \times \frac{{k!}}{{{\lambda ^{k + 1}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{(k + 1)!}}{{{\lambda ^{k + 2}}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">therefore true for \(n = k \Rightarrow \) true for \(n = k + 1\) and since true for \(n = 0\), the result is proved by induction <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Only award the <strong><em>R1 </em></strong>if at least 4 of the previous marks have been awarded.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> If a candidate starts at \(n = 1\), do not award the first 2 marks but follow through thereafter.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[13 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) (i) \({\text{E}}(X) = \frac{{{\lambda ^{n + 1}}}}{{n!}}\int_0^\infty {{x^{n + 1}}} {{\text{e}}^{ - \lambda x}}{\text{d}}x\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{{\lambda ^{n + 1}}}}{{n!}} \times \frac{{(n + 1)!}}{{{\lambda ^{n + 2}}}}\) <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;">\( = \frac{{(n + 1)}}{\lambda }\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the mode satisfies \(f'(x) = 0\) <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;">\(f'(x) = \frac{{{\lambda ^{n + 1}}}}{{n!}}\left( {n{x^{n - 1}}{{\text{e}}^{ - \lambda x}} - \lambda {x^n}{{\text{e}}^{ - \lambda x}}} \right)\) <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;">mode \( = \frac{n}{\lambda }\) <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>[6 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) (i) \({\text{P}}(T > t) = {\text{P}}(0,{\text{ }}1{\text{ or 2 arrivals in }}[0,{\text{ }}t])\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{ - 8t}} + {{\text{e}}^{ - 8t}} \times 8t + {{\text{e}}^{ - 8t}} \times \frac{{{{(8t)}^2}}}{2}\) <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;">\( = {{\text{e}}^{ - 8t}}\left( {1 + 8t + 32{t^2}} \right)\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) differentiating,</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;">\( - f(t) = - 8{{\text{e}}^{ - 8t}}\left( {1 + 8t + 32{t^2}} \right) + {{\text{e}}^{ - 8t}}(8 + 64t)\) <strong><em>A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>A1 </em></strong>for LHS, <strong><em>A1 </em></strong>for RHS.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(t) = 256{t^2}{{\text{e}}^{ - 8t}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) with the previous notation, \(n = 2,{\text{ }}\lambda = 8\). <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;">mean \( = \frac{3}{8}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">mode \( = \frac{1}{4}\) <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>[8 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Do not follow through if they use a negative probability density function.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman'; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[27 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p>Draw slope fields for the following cases for \( - 2 \leqslant x \leqslant 2,\,\, - 2 \leqslant y \leqslant 2\)</p>
</div>
<div class="specification">
<p>Explain what isoclines tell you about the slope field in the following case:</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 1\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x - 1\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \) constant.</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>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( x \right)\).</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>The slope field for the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + y\) for \( - 4 \leqslant x \leqslant 4,\,\, - 4 \leqslant y \leqslant 4\) is shown in the following diagram.</p>
<p><img src="data:image/png;base64,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"></p>
<p>Explain why the slope field indicates that the only linear solution is \(y = - x - 1\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Given that all the isoclines from a slope field of a differential equation are straight lines through the origin, find two examples of the differential equation.</p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><img 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"> <em><strong>A2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.i.</div>
</div>
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<p><img 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5pMl6WSoFcaTVJf709I6w09jzyc0Oj7/Px8cXx8vLIL6pEKQhc74ZWovLICAlU4X+xIhCbnxbSU/G+0fVApCjvfaJa1Pn0JhE87CEh6mZ7Oa72nJmACJjAlgWqcL5e6Hz58eGLryOsJxWQz6dvAJuvUHZlApgSqcb7wPzg4aEY8EPky/Cwdk5qpfUKJlb4NLJRiVsYEtiBQlfPVKAKlHRz9brHH7LAJ/M18B4DeNBSBqpxvl+Uc/XZRGWeZh5qNw9WtlkmgmtEObfMQ/XLnnZEPPALrF760CQ3/G9aMOuEJtxq+9js8QbcYiUC1kS+Xvxp69unTp8YRRzJszrqkY5JzltOymcCYBKp1vkS+PF/i/O+Yu9evbTvn+ysTL6mTQLXOV+bmSTAugyt40E8qzzblRMfVBqMeXEygdgJhX6zTx7DOP/ahtX1dHC9/frpwe4beMg6B6iPfOKa0JiZgAiURsPMtyVpBZFWePYg6VsMEtiJg57sVNm+0CwGNMtmlDW9rAqUTsPMt3YIFyu/It0CjWeTBCdj5Do7UDa4i4KFmq+h4XU0E7HxrsnYGuqYPt2QgjkUwgdkI2PnOhr6+jhnfy7hqj/Otz/bW+FcCHuf7KxMvGYnA9+/fm2+9+YGWkQC72aIIOPItylzlC+ucb/k2tAbDELDzHYajW9mAAC+zpzjtsAGsDKswRJCP0foEOoxxnHYYhqNb2YCAnC7pB5fyCPDua43RtgPe3X6OfHdn6BZ6EPBB2wNWplWdsx/GMHa+w3B0KxsSsPPdEFRm1dKoNzPRihXHzrdY05UneE1fL1Z+9I8//ijPUCsk9hsAV8DpucrOtycwV9+egHK+27dQzpY3NzcLvtgRJb+tXK8+QluOJfKV1M43X9uEk4wDt5Z8IY6Xd1jwov4IRc43gi656FCk83316pWHvOSyB/WQgwM4es6XiFcfZo340viIOvXYhQetWuRQM3KHyh+yo19cXAwKxY2NQ4DIlxs3Ue1FtHt9fd3AQ8cob2/DZipRdJI+c06LdL739/eLvb29hpsiKXYKn5Xn3JXW9x39q8VE9oruI+2LOsbWW9g1+hAoMu3QTvqzczgn1cfs+dXlCqb0opML+6MjxNKtOb78RTpfsBD98oYsPbLKDk8u2CVfAu2TJpJymY7j5cZUyREWJ38uz3G6kW4qpkHN6elpvjtXgZIV63w5kNnR//3vfz/t7OSBOZDv7u4KNEUdIqcOFmf18PDQOF3Gj6brSqKBHq9fv25EJiCIWrpOnlF1nUKvYp2v4OCAcbjpgZuerVXP0/kJcPDqZhs2Y54oUcvml3A7CbTvOdWwHb9at+p1w01ndw4i8lu6qcBvhtiw83379m3x9evXZh0pgU+fPjV1mWeQPdEpA8854D5+/Lh49+5dM2zszZs3ixcvXjTpBNZTl50aRypnSvvteSImDmAtx5BcytLPhw8fmr51B7ptZNrXgUO0jD7pAcQ865mm87TDb9pVno/fFFi0+6MN6qXLqZ/KrPaIBPf39xv9WaZ227J3/aYP+l9VNqmzanvWdbWBLutkxSbYHPtSl4hR/Nt9dvXRrpPD7z///LPRBf3Pz89zEKmXDOs4p/uoUny9OnDlpQR6OV8MRdGBpoMHh8HlFs6ZOvxxcPFoJc4RhyAjppdlODw5bQxLNKTLTx3IrMdp8VuOkH6pyx+F36znj34kA+vpj35Yx3JkYR6HrXZYLvnYRv2wHXUozKe/qYNcak91VJ/2WEfbtCmZ6JvtWCa5qUvbkkFTtdUIsFg0eVFdpmsZU9qkaDumcOSPeenNvGRCfunJMv74rfbSuixTXaasgx/2Zp5CX8gLb147qPZYzjz1OLlRh99izpR8L22xPN1OdViP3uQcWQ+7VCbZgWXUYZ/h5I6sqvvy5ctGLuRRu8zzBz/+kDFlyTx1qYO+2EjL6JP6BBdiwPx//vOf5jfr07Ybgf/3jzb0x7Y6dliNfmwnWdCB3yxHJ5Zrv0E2lqcFxuhM+8jAH21IfuqKEXUku/qmTfqhD6borIJ+x8fHT+u0XPLQFttQsEHaluoiL8vpG5mYUuhH+4fqakod7Me21GOe9imSkTbTnLTkRyZ4qG3aUl3qqLBcNmMb+uJP20kv1R9k+tijHB0dPfJ3eXn5eHFx8bhYLJ7maUbLmbJO61l3e3vb/GZ7rdN6pqxnOwptq6gfba9tWc+8pmzbrsN6ba/2tE26neqof9aly9RuulztaZrKz7K0b+ZTvZmn0J/6adfnd98/9UHbmqcPtcPyVE7V0ZR60lXbaJrWYRkllZ314teuqzZ2nbbbRQa1qb5TmbSOqbZN9dMy2tE8ddtttHXV+pStttMy1ZEMWq7faX8sk/xav+lUDGgv1UPL2/3SLrJpfTovOdR317as07aq19ZFyzVNWVCXdvmTLdL1bKN+u9plXbu++lk3XbddV3+0iZxjlGf/AzmII5+jEc5SSodw5uKsxplehTMWv3VmZXk6r3rplPXpWTxd195ebdEv0ZnOyjpTan27ja7ftKHxy13r1y1ThKSzO+0hR3r2J2LgjM4ySlqX30QP6MEZnzrInxbqs47SXt/WWZEF/NmOwlQRBvPixTrkVRQsOZqN/vePfpEnlUG2V9uKWNQPbTIvfRXJEG1KPrZlO/ghK3XZJi2sR6Z0G+q2C9upP7UBF7ZDV9iyXP0wTfdb9a0p6xUZc65h/5CcqcztPtUH8jEvvZmniAfz9MEfhfUw429VoR5tsB110aGrpG13rW8vk/1S+dp12r9TWaRHWieVQXVZzzxFfdE3uqi+2mI9dfnTvp+2v+188c6XHOLJyUmjP3DYUSm63GSenZWDupTCkDnyossOAHYSiqYc1OjLVDtMDroiP3/svDgN5GOeFBOXr6UWbMP9CfY3Hbjo8k8gXqpW3XKnx1FE/bq1nmZp8aMduInTVeSEWcdZOY1YuurntOzz58+NzBzcnDhU+E3RMg58nC36cXLJyfHCmyiPEwQHMFEj8jO/zGbSM/dpKr9OgLnLvI18usrYZltvswGBMXIZU7aZ5nHI2bRLmgdqryvht/Ji/wRWJUj8Q0Zkxj5p6VqWri9hPt3nSrbPOtbKveoYWlff6/sRKD7yTc8vXVFIuiynyDCVe9V8Kv+qermsI+JltAMF3iUyX8dSV1FE8roaiahnyiFNr6TLPb89gV5DzbbvZrwtU+dETrFduBznYFmWP23Xz+03+pFry33nh68468abpilTnFWptkAPdGQMOnlfpX+i5kLTfa7r2Ert6vn+BIqPfFPn26W+ohMOFHJY6Q2Ervq5LlOElat82IG/9IDtkpX15IJLLUS4etiAPHbkktpy3XEWmcNYuhUf+QqMnKx+p1MiMJyuHBhOeMghI2lfNc4TDYrnuiiw5Mtz9KQQua/TM8p+sOq4iqLjXHoU73y1c3CWTkc4tIFysHDQ8EddfssZt+v69+YEGGfLyY3L8K40Q9pS6U4L/TjJpBFhql+keV0hcpyss2skvafUpfhxvlPCcl+/EiAa7HNwclBzkvSJ71eWOS2R88VOq4KanGQuTZbic76lAY8mbx/HS+RL0eV7NBZR9JGdouiTqx52vrlaJqBc3LQh5+ubN3kb1853GvvY+U7D2b0sFk1eGOdby82q0o1OysFDzMazonO+47F1yx0ENOrEecQOOBktUs7XJ8rxjFL8aIfx0LjloQnoctZph6HJDt+ene7wTNstOu3QJuLfoxPgSx0uJlA7ATvf2veACfVX5MuYYBcTqJ2AnW/te8CE+mu0Q8mPF0+Iy10FJ+AbbsENnKN6PBXnG245WsYyTUnAke+UtN1X82hu9BfS2MwmsAkBO99NKLnOYAR4qZGLCZjAYmHn671gcgJ2wJMjd4cZErDzzdAokUXyE26RrWvd+hCw8+1Dy3UHIeCHLAbB6EYKJ2DnW7gBLb4JmECZBOx8y7RbsVL7sdViTWfBByZg5zswUDe3mgAvbPENt9WMvLYOAna+ddh5Yy31CPDGG7iiCZjAVgTsfLfCFncj3wyLa1trlhcBO9+87BFeGnK+dvDhzWwFNyBg57sBJFcZjgDvdSjZ+dbw5eLhrO2WVhGw811Fx+sGJ8CnaUrNK/PhT04eZ2dng3Nxg/UR8FvN6rP5rBqX/BkhfVrHn1OfdRcK07kj3zCmLEMRXiX5/PnzMoRNpLy+vk5+xZglhRJRr1KsY+dbiqWCyMlle4nOV1EvZig5Z53uRqRR9Jcu9/w0BOx8p+HsXhICJeZ80wgxyufUiXyxBS87cpmegJ3v9Myr75GcaWlFoxyQvUT51/H2U4frCA2/3s53eKZucQ0BObI11bJZTcohTZVEcL7taPfw8DAb3rUI4tEOtVg6Ez1LG+3AiYI8tUqUFwOlOWx0i6KX7FTC1JFvCVYKJmNJN6zS/HSEiFe7Ujvy1XJPpyNg5zsda/e0WCwuLi4W+/v7xbBgNIAKskcpSv1wInTUO49V7Xzn4V5tr0Rcnz59+kV/bvjIIfyycqYF6U2oSDfa4CzWaWQ/E+Zqu7Xzrdb0eSnODR85hFwkQx45J9+QysUqceSw841jyyI04TI+zTfKubGcS+D0Mn9uhSLleOdm6f5/JWDn+ysTLxmRABGkHlggsjw5OWkiXpblln+UnOA4Pj4ekcq0TafD5kq6+TktpfF7+238LtyDCfwgQDSp6Jb5N2/eNEO5dNMnl2izPRTr4ODghxKFz33//v1JAzvfJxSTz4SKfBmPyUGjg3tymu7wFwJpWgG7YB+WyU4PDw/Z3W2XzFImTZNoWcnTtn4l61Ky7GEiXy5h9cw9B8vnz58X7969C/koaEk7HAf63t7eTyKzjGFbTK+urn5al8OPNDLkEj3SEDP4ps7XNxLn2+PCRL5crr59+/aJJMOZiIRzu4P+JGAlM11pBJZxgkxzqjnhSJ1vpHSDGHfZROs8nY5AGOcLMh3U6WViepafDqt7SgmQz+UkiH2Y18GvPG9aN4f59IQdMSea6pcD71plCOV8MSKXiPrjN598cQ54vt375uam6RyHy5UJtsj94E8frlAqaz6Cw/fs42F4ptu0GPbFOkS8aa6Rg5/8YkmRDDqQt+YPJ8bIgC9fvjQ6KKJHHy6NWc7lsi6Z+f333383+8T5+fnkQ6VwsMiMHOTfmceRceAjI8tyLMgd8UU6Yp0eFxwTfFnEZSYCj8ELL2zS38XFRVHaSu5dp0dHR5PqLc7I3dU3y3Mu4p2zjNvKdnl5+XQ8MO8yH4FwaYf2OSy9qZNeTrbr5fibKEw3EZUn3VRORfhsN/Xdeg35I/fejqyIfCXbprpMXY9cdK756KlZuL8RCczn96frWZGMptP1PGxPihjb0Yv0IuJknvW3t7fN/LASbNYa0a5kbW8h+drL/XsaAu19Z5pe3UsXgfCRL+et+/v7nyKwNKc34nlt8Ka/ffvW3KwikmWeqFLz6EiUzDJyq0SXU0ZvROnpjZxlfSNfejUyOCQ3uJaAxvaenp6uresK4xEIe8OtC1k67hdHleMA/y65S1gGW1IcpBo4ESxLLfBkm2/0zGtRbn5y423qdNS8WufXe1XOl+iMAx8HQFkWneVnpvwlwukS+a5jauebvy0t4TQEqkg7CGX7aSWiNV8Ci85u066ba10trnPOXdt4mQlEJFCV89Wr9HAUFCJhHsJwGYYAVxWblE3rbdKW65hAqQSqSjukRsLpKup1/jclM/48qQdHwONzdg95E6gq8k1NkT42qqfF0vWeH4eAcr7jtO5WTaAcAtVGvjziyqOv6bAzR2Pj77hK9TAiwsUEaiYQ5n2+fY1I/pfcI86APwpPwGkMZN/2XH8zAjBeNgxtsxZcywRiEKg28o1hvvK0IO3ADU+PMS3PdpZ4WALV5nyHxejWNiVAakdXGptu43omEJGAnW9Eq1onEzCB7AnY+WZvIgtoAiYQkYCdb0SrZq6Tb7hlbiCLNwkBO99JMLuTlIDHVac0PF8rATvfWi1vvU3ABLcB5gcAAApJSURBVGYlYOc7K/76OmeomYsJmMBiYefrvWBSAgw1c9phUuTuLFMCdr6ZGiayWL7hFtm61m1TAn7CbVNSrjcYAb/VbDCUbqhgAo58CzZeqaLri8ylym+5TWAIAna+Q1B0G70I+PHiXrhcOSgBpx2CGjZntZx2yNk6lm0qAo58pyLtfhoCOF5/Rsg7gwl4qJn3gYkJ+IX1EwN3d9kScOSbrWniCsZXRFxMoHYCdr617wEz6K+vSM/Qtbs0gWwI2PlmY4o6BHn//n0dilpLE1hDoNpvuK3h4tUjEeDzQU47jAS3R7Ma7uebnz2gDVzVzndgoG5uPQG+Gu0yHwEcr77a7Rug89nBaYf52Ffbs6OtfEx/dnaWjzCVSWLnW5nBra4JpARevnyZ/vT8hATsfCeE7a7+IXB9fV0MCl5/yWX6q1evFlFuFqKTrj4eHh6KsUU0QZ3zjWbRAvQp6X2+nCjkdJWr5qZhqQVdLi8vn8QvyRZPQgeZceQbxJAlqaGoqySZJWvJjhcd2uz9bmVZdvqpne/0zKvvsaRoS0OyMFoaMZZqxFQfdCjJFqUyXya3ne8yMl4+GoGSoq27u7snDqVHvU+KJDOHh4fJL89OScDOd0ra7qshUEq0RX5Ul+malm7CduRbuj4ly+/3+ZZsvUJlL+V9vsipQtQbIe2Q6oRufshCFp5+6sh3eubV91haFIm8pcnctZPxVFuqR0npny59Sl/moWalW9DyD06AtMje3t5Tu1yq397ePv0udaadcvC39Oa1pCPfeflX2XufnC9515ubm0k5teXzpfmk+KvpzM63GlPno+jp6elGwuiBgKkfgU0jxPQyfSOhM62U6iQRI47ekG4lTO18S7BSMBnbkWVbPRzFu3fvmsU4P5zwXGVqxz+Wnno6T+3b8YrEfFM73/nYV9vzsjQCThlHy9jar1+/NqMLcL5T5lvpO33fQUnvoVi1Q7XfoRxh5MYqfUtY5xtuJVgpmIzLLuXT9yjgcKd0ukJM1C2HG9VBOeqVteed2vnOy7/K3vf393/SGydHtHl+ft4MhSL6PDg4+KnOVD/keDlBRHr6K73aaEfBU7F1Pz8TsPP9mYd/TUDg77//fuqFNAMD/0k5fPr0aUFUtiwyftpopBlkUT4aGeaSY2j1OJlJL9r2B0yHJrxde3a+23HzVjsQ0OB+nC6pBTm5uS+H0zTD3LLsgPeXTbtGOvxSyQsmJ2DnOznyujok4koveaW9RjDw1BVOTw5Y66eeppHh3LKMrXt6khm7L7e/nEA1zpezP38cWEReir6Wo/GaXQng0PgCBDlGeKcOjrZZxiVwDlFmGh1GGV4m+6W6eb8Xlfmn1Qw1I+9FtEWkxUcDu6Kx+c0RSwIOdMbr8oTY/f39k5NlOcu4ufX58+cslGb/oHByjvbYbRrJR9Mti51nSyGqeqtZ+kYndsg5hjJtaacQm+nKgxNgbuyRSRFitMeJ0/2elEMOVxohdugdlagm8oVTGn1xoCnfuCNDb76CAJEvnPn7448/GgeX42W9ngCLGBnibDXCIdLwuRW7XRGrqop8yTmmb6uShRwFi8TwU3jj0Ii4FFHiiHOLfDkZk3pgX4jooHTVMbyF3eK2BKpyvoJE7hdHoDvuLJdjUB1PdyOgS12cGQd+ernLOvPeja+3Lp9AVWkHmQuHoBssWoZDaC/TupynSp0gv/44qTCvG4zU0W8iUa3X01xD68cNTQoOF9ap4x26L7dnAsUSeKy4HB0dPf7++++PBL76u7q6yprI7e1tI/ObN28amdvyS490ip7p73R+SGXv7+8f6UsMkbWrvHv3rmuxl5lAVQS4/Ku6XF5eNg4jdUgsy71IXuTUfDrF8fEbXTSfrmd+mXPcVnf6Up+r2qCOiwnUTqCahyyWXZpwJ5i77+QlVbhM5qYLl8y5FmSkkCohfaCxs0qd6Dd1uNHYlWNtP/Swi67K69LGuhtWEUcU7MLO29ZJoMobbstMjUPTTTiG5nz48KFxajgyl/UEdELAua5iRk746upqfYOuYQKBCdj5toyLYyAiVCSHQ+mKGlub+WcPAjDlBqAH+/eA5qrhCNj5dpgUx5AWO9+Uxu7zOF9Fybu35hZMoEwCVQ41W2cqot4038tQLaUj1m3r9esJ+L0a6xm5RnwCdr4dNuaGUfqeWZwxf0PeoOrotppFq/LB1UCwotUTsPNdsQvggLk8JjeJ83XEtgKWV5mACfQiYOe7BheOV5EaDti5yjXAvNoETGAjAna+G2BKna8ezd1gM1cxARMwgaUE7HyXovmxgptvjHjQm7h4NaXL9gR0Mtu+BW9pAuUT8FCz8m1YnAYM5VMuvTjhLbAJDETAzncgkG5mcwLcuOQpOI+f3pyZa8YjYOcbz6bZa/TixYvFwcHBUxone4EtoAmMQMA53xGgusnVBPi0kPLnq2t6rQnEJWDnG9e21swETCBjAna+GRvHopmACcQlYOcb17bZavbx48dsZbNgJjAVATvfqUi7nycC3GxrvznuaaVnTKASAna+lRg6JzVPT089zCwng1iWWQjY+c6Cve5O+ewR78lwMYGaCdj51mz9mXR/eHj46X3JM4nhbk1gVgJ2vrPir7Nz3u3gnG+dtrfWPwjY+f5g4bmJCJDzdTGB2gnY+da+B8yg/5cvX2bo1V2aQF4E7HzzskcV0nz9+rUKPa2kCawiYOe7io7XjULAQ81GwepGCyNg51uYwSKIe3d3t+CL0C4mUDMBO9+arT+T7ox28DjfmeC722wI2PlmY4p6BHn58mU9ylpTE1hCwM53CRgvHo8AaQcXE6idgJ1v7XvADPoT+fpl6jOAd5dZEbDzzcocdQjDN9wc/dZha2u5nICd73I2XjMSgaOjo+brxSM172ZNoAgCdr5FmCmWkB7pEMue1mY7Ar9tt5m3MoHtCRwfHy8ODw+3b8BbmkAAAv50fAAjlqYCkS8PWTw+PpYmuuU1gcEIOO0wGEo3tCkB3ufrYgK1E7DzrX0PmEH//f39xcXFxQw9u0sTyIeA0w752KIaSUg5eJxvNea2oksIOPJdAsaLxyPAULP379+P14FbNoECCHi0QwFGiiYiL9bx1yyiWdX69CXgyLcvMdffmcC//vUvf8NtZ4puoHQCzvmWbkHLbwImUCQBR75Fmq1soZ3vLdt+ln4YAna+w3B0Kz0IMNTMjxj3AOaqIQnY+YY0a95K8fVihps5As7bTpZuXALO+Y7L1613EPjrr78We3t7fry4g40X1UPAkW89ts5GU5wv73Vw5JuNSSzIDATsfGeAXnuXpB0ofsS49j2hbv3tfOu2/yzaf//+3eN8ZyHvTnMiYOebkzUqkYUn3Pw6yUqMbTWXErDzXYrGK8YiwDfcXEygdgIe7VD7HmD9TcAEZiHgyHcW7O7UBEygdgJ2vrXvAdbfBExgFgJ2vrNgd6cmYAK1E7DzrX0PsP4mYAKzELDznQW7OzUBE6idgJ1v7XuA9TcBE5iFgJ3vLNjdqQmYQO0E/h9JzE/Fz0rpMgAAAABJRU5ErkJggg=="> <em><strong>A2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><img 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"> <em><strong>A2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">a.iii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the slope is the same everywhere <em><strong>A1</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>all points that have the same \(x\) coordinate have the same slope <em><strong>A1</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>this is where a straight line appears on the slope field<em><strong> A1</strong></em></p>
<p>There is no other straight line, all the other solutions are curves<em><strong> A1</strong></em></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>given \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {x,\,y} \right)\), the isoclines are \(f\left( {x,\,y} \right) = k\) <em><strong> (M1)</strong></em></p>
<p>here the isoclines are \(y = kx\) (or \(x = ky\)) <em><strong>(A1)</strong></em></p>
<p>any two differential equations of the correct form, for example</p>
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ky}}{x},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{kx}}{y},\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{y}{x}} \right),\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = {\text{sin}}\left( {\frac{x}{y}} \right)\) <em><strong>A1A1</strong></em></p>
<p><em><strong>[4 marks]</strong></em></p>
<div class="question_part_label">d.</div>
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<h2 style="margin-top: 1em">Examiners report</h2>
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<div class="question_part_label">a.i.</div>
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<div class="question_part_label">a.ii.</div>
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<div class="question_part_label">a.iii.</div>
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<div class="question_part_label">b.i.</div>
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<div class="question_part_label">b.ii.</div>
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<div class="question_part_label">c.</div>
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<div class="question_part_label">d.</div>
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