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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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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>
<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,iVBORw0KGgoAAAANSUhEUgAAATcAAAEsCAYAAACmM/RqAAAgAElEQVR4Ae2dDXBUVZr3n0TLUQR0EUY7woDiAk5lByQB2RcsImoiMzphQ+Eqo4lDwiwvH/GF2oRB31nHWoNiKFgTpVggSHA2Zcl2Kmy5A2E2UcdQI5jwoZTQKfEFAokWBAEjjsyk71vPDSd20t3pj3vPuV//U9XV3ffjPOf5ndv/Ph/3nidF0zSNkEAABEDAZQRSXeYP3AEBEAABnQDEDRcCCICAKwlA3FxZrXAKBEAA4oZrIAKBNqotuItSUlIoJeURWtty4eox56hxZSalFdbSmWCE07AJBGxEAOJmo8qwT1FGUV71Z9QdqKJsOkB/OPwF9WjZULo751Eaf/oCfWOfwqIkIBCRAMQtIhZsZAKpd06kh8Z20PHz31wVt+vo1pFj6M6H0ikNVw4uEpsTwCVq8wqytHipN9KwsT46/slJOssFCZ6kunWf0s+emESDLS0YjINAbAIQt9iMvHtE6o10822DrvofpK6Du6g5ezHNuf067zKB544hAHFzTFVZUdAb6Obbbiba+zm1tTXQK34fLZkzmnDRWFEXsJkoAVyniRLz1PHX000jhhBd3kuv/9thylo2m27HFeOpK8DJzl7r5MKj7LIJXEtDho0goiE0c+GvaJYP3VHZxJG/eQQgbuaxdGlOo2jV71bQ0xOGutQ/uOVWAuhkuLVmDfsVpK6WbVRFv6RnZ92OcTbDPJGBagJouakmbmt7LGiv0qOPnqSn3vhbCnw6hX6zPB23fdi6zlC4aATQcotGxqPbgxfPUqDjKAW+mEL/55mpEDaPXgducDsF67m5oRrhAwiAQH8CaLn1J4LvIAACriAAcXNFNcIJEACB/gQgbv2J4DsIgIArCEDcXFGNcpyora2l06dPy8kcuYKAZAIQN8mAnZr93r17ae7cuZSXlweBc2olerzcEDePXwCR3GdhmzFjhr7ro48+gsBFgoRttieAW0FsX0VqCxgqbKGWp0yZQrt376Zhw4aFbsZnELAtAbTcbFs16gvG42uixdbU1NRbgMrKSuIW3MqVK+nbb7/t3Y4PIGBnAnj8ys61o7BsLGw8vsaJhW369Om91pcuXap/XrZsmf5eUVFBN9xwQ+9+fAABOxKAuNmxVhSXSQgbt876C5soCgvc4cOHacuWLfomCJwgg3e7EoC42bVmLCgXdz9DW2z9i8CChgQCTiGACQWn1JTkcp4/fz5ssoDjlnLSNK3XuhhzQ7e0Fwk+2JQAxM2mFWOHYkUSNzuUC2UAgXgIYLY0Hko4BgRAwHEEIG6OqzIUGARAIB4CELd4KOEYEAABxxGAuDmuylBgEACBeAhA3OKhhGNAAAQcRwDi5rgqQ4FBAATiIQBxi4cSjgEBEHAcAYib46oMBQYBEIiHAMQtHko4BgRAwHEEIG6OqzIUGARAIB4CELd4KOEYEAABxxGAuDmuylBgEACBeAhA3OKhhGNAAAQcRwDi5rgqQ4FBAATiIQBxi4cSjgEBEHAcAYib46oMBQYBEIiHAMQtHko2PebQoUPEK+gaT1foTO1SSsvZSq3B6LmxPZUR6Nk3tqkq8SrDra2tlkT4Yruq2LItc64bVTWTnB2IW3LcBjyLY3/yS2biH+KvfvUrqqmpMWwmeOYd+pelr1NHjJx++9vf0gsvvBDjKPN2v/fee3TPPfco+yEGAgEaP348tbW1medEnDmVl5eTiDIW5ylJH8Y+Mlu3J0+IG/9TqUr877t8+XLii1VmOnDggB5LVMQZTdpW135a9+wfafhD6QNmwQx37txJ8+bNG/A4M3du376dioqKwmI7mGkjNK+PP/5Y/zpu3LjQzdI/cyuKo4rl5+dLt+UlA64XN25B8T9VbW2t9HoVIfLY0GuvvSbVHgsNR4GfNGmSATsXqGVjDVHxYsq57QcD5rNnzx59/3333TfgcWbtZJaqxfTIkSO6mJrlQ7z5vPPOO/qhWVlZ8Z6C4+Ig4GpxY2Hjlg3/+8+ePTsOHMkfEipsLKQjR45MPrMYZ/I/vfFuTJC6WrZTBc2nRRnDYlgk4lZUWVmZsmDMdXV1eplUiSkbY6Yq7QnofL2oaqHydcrp1ltvFeZd++5aceNK5O5hbm4uyQ4grFLY+Epsbm7WL8hZs2Ylf2F2NdPGCqLiRZk0OEYu/CfBAZtnzpwZ40jzdqsWU/GjHzt2rHlOxJETT5hwC/Xpp5+O42jjh1y+fFnPZMSIEcYzs3kOrhS3ULHh7qHMGJs8sJ+Xl6dXs+wWm7iWNmzYoIt28q3DC9SyuZHuXL2IMgbHvgTef/99vQs8UMBmUTYz3q0Q008//VQv+ujRo81wIe48mpqa9GMnT54c9zk4MD4Cros4HypsssWGha24uFhv1Rw8eFBqV1RUJ/vH//T19fViU4Lv3B19i/yjn6AXb78urnOfe+454mj0qpJqMWW/eMKExzCT/8NInA5fP6pbqImX0sFnaCYnDlAez8tks3p2bW1t2pQpU/QXf5admpqadF/5XVWqrq7WbXZ2diZp8qzWUJoxQB35tOyqo1p3T5j53uMCgUCS9hI77fLly7rNysrKxE40eHRRUZFWUlJiMJfEThfXz8GDBxM70cDRXI/8+1RVnwaKavjU2H0SBwm3uE9IdotNIOFuWmdnJ6nqrrFdntErKSkxcHvEcJq1pplYu75/naWG0gyi7CoKdLdT/YIJFHph8Lilqtsj+F4zbkFlZ2cLzMrep02bpswWG/ryyy9NmPFOrMiDBg3ShzQSO8uZR6ewPDqz6OGl5i4bD5iq+iGGl8CpW85R48qH6YFDiymwawGNu6psKSkpukMuukScWkEodxIEXDXmpnK8JAnWOAUEQEAhAVe13BRy84QptNw8Uc2udTJ0aMW1TsIxEAAB7xGAuHmvzuExCHiCAMTNE9UMJ0HAewQgbt6rc3gMAp4gAHHzRDXDSRDwHgGIm/fqHB6DgCcIQNw8Uc1wEgS8RwDi5r06h8cg4AkCEDdPVDOcBAHvEYC4ea/O4TEIeIIAxM0T1QwnQcB7BCBu3qvzqx5foY79r1NBWgqlpKTR/StrqbVrgKClnuUEx51KAOLm1Jojojlz5iQZZStIXQd30R7KozfaNepu30E//+J5yvrX9+jSADxKS0tp9erVAxxh7i5ebnzq1KnKghXzOoDM1IrEXBcuXKjMNAcZ4oURmLFbE8RNUs1y4A9eRlpWEnFEk1q7LniSmi9Opqem+vRFKVN9f0+FBY8Svfk/1HwpcutNRNxSuawUL6fOSZXNDz/8kKwInMLXCS/lPnHiRFmXS1i+R48e1bdZ4W9YYSRtgLhJAMv/hhwpvaqqSkLuPVkaiiOaegdlZY0KWW33Cn1x4iSNXzGHpg6NfEmYEnErARpCTMXqygmcmvShHIX9jjvuSPr8ZE/kANucDAfYTqAAHPODVzxO6s8xATuWHmp4oXJk0IeAiOPAa/JzPABZiWNFlJWVGc/+64DWULVKy1+1R2vnwAkhScTC4E25ubka+6Qq1dfX62v9q4iFwT6J2A1+v1+Vi712mCvzVZlMu35UFjpBW7yOPpJJBISw8YUj80dpTmCRbu1iwyrNJwL6ZK3S/IGLfUgIcWNf+DMLjqrEP3aVP3irAqdwoB9my4F/VCUr6lOVb6F2IvdBLG1LOtM4j5uoil8qQt9NmjTJAKxUGjprNbVr31F783YqpW00N+tZqj1zJSzPxsZGfVtmZmbYPhkbRPjCxYsXy8g+Yp4ceIfT8OHDI+6XtZG7wpweeeQRWSbC8hUxWn/84x+H7XPTBlfFULCqYljYVMUvFYPP5sURvY58GU/RS//+Fzo0fgPtC1yivNv7/sA5sHVZWZmBiFuJ1YxqMeXStbe362NQw4YNS6ywBo/muKVFRUXK2HJxefyUI5qpmqgxiCjp09FySxpdz4lC2LZs2UIcPdxYayp2YT744AP9ILNvWUi963/RP2b/IGIBPvroI5o5c2bEfTI2qhZT9uHw4cOUlZUlw52oeYoW6rx586IeI2MHz8w++OCDMrK2VZ4QN4PVsWvXLmJh4wjwKuKX7tixQ86/blc7BY7fQ/eOHxpGhGfVJk+eHLZdxga+hUa1mLIfXIfp6ekyXIqaZ11dnb7vvvvui3qM2Tv4FiJOPJvv+hQ6AIfPiRPgWTaVA+08AG00Wnj3ab+2wJetlW7b1zND2n1aa1g1T8uv+kT7OgSBmFBQGRGdzbM9mTPNIS72fuRBdmarMrGPRusymfKyTdV8kymn0XOkhPbjbgUn2fcoidBzsf6BEFS4H6GuI7R1yRNUuL1nEN2XX06vFf+C5mT03NQrjhZ8wU8QwbuTCEibUFi2bJnOQbbAOQm2bco6OJ0WVH9CC6ptUyIUBARMJyBF3ISgyRY4tChMvx6QIQi4hoAUcWM6qgTONTUBR0AABEwlIE3cuJQQOFPrCpmBAAgkQECquHE5IHAJ1AYOBQEQMI2AdHHjkoYK3NChQyk/P980B5ARCIAACEQioETc2DALHAvb2LFjI5UD20AABEDAVAJS7nMztYTIzDICuM/NMvQwbAIBPH5lAkRkAQIgYD8CEDf71QlKBAIgYAIBiJsJEJEFCICA/QhA3OxXJygRCICACQQgbiZARBYgAAL2IwBxs1+d2LZEvDAnR93idxWJ1x7jFWY4EpbKxDbFumeq7DJXjpuqKrEtN8csZY4QN1VXkwvs8MKcOTk51NbWpsQb/sHz4gs33HCDEntshBfLZJtnz55VZpMNbdiwgThuqqo0d+5c4vB+bk4QNzfX7kC+dbXSH9YWUFpKCqWkpNH9K9+klo7w4DAiC26tvfzyy1RSUqIs1iXHF+DYDSrFjZeK56Rq5WG2xYLKAag5roGKJFrCro5ZipabiktJvo2Eu4nBk7RzcyPRzyupXdOou30H/fyLVyhz/uvU0hU54jy32nj5bw5moiJxl4nt/fSnP1VhTrfBHL0gqKdOndL9HTNmjDK2VhhCy80K6iba5PEhjryViMAF/99puumxBfTQuJ54Cam+6fTMc8sp+73/oLf3h49v8T+96labOeELEwPNkd+tElSOZqaqhfr555/rYEaNGpUYIIcdDXFzWIWFFleMSU2cODGhH0bq2OmUdft1oVlR6m1jaJKvz6beLzU1NUpbbSymHKFJ9QILVgrqjBkzennL/nDs2DG9C6xKTGX7Ey1/iFs0Mjbfzt02HtznbmJhYaE5pR00LWL0Kx5g55aFqjEajqvJyezwhQNBskpQeayNo4vJDgkZ6vv+/fuV1WWoXeWfjUaYwfnqCXD0Io5MVVRUZFIUo27tYsNvtOzyfRGjX7EtlZGh2K/c3FylYDmCGfvJUbBUJWbKNqurq1WZ1O2wTb/fr9SmFcaULXmkXLVdapAD+T755JP6v31FRUVC3dGoSLqaaVP1cFr9eiYNjnAQt9pURWJn/ziGqN/vj1ASeZukxYMdoMiihTpr1qwBjjJ3l7h/78477zQ3YxvmBnGzYaVEKxL/8PPy8vTdfBOmOWMmF6hl8y4atmo5ZQyOPEoxf/78aEWSsp3FVHX0dw6MrHqtwR/+8If6rS4jR46UwjFSpsOHD6fq6mrid7cnaeu58b07fM+QWIVXJkhVcVJl+hArb54N5VnRw4cP63eym/ODuEJndm6l/xk/nwomhEeax3pusWoF++1MIPJftQklZmHjgWjRDDYhywGzYFtC5AY80OBO/sHH8zJoJurp69evJ7OEraNxK7095FF6qlfYLtGx6jfpvUuR73WLWijsAAE7EpA10CcGS8vKymSZ6JNvZWWlPjjL7zITD8bG85JZBuN5X9QC/lVaViRfsqu0QHePBeGncXvIAQTUEyCZJoXgqJppE/ZkC5xMZvLz/k477V+i+SIJG/m07Kqj2lVt6xVx+WWCBRAwn4C0MTdupfK9Q7fccot+j5SKsTe2yV1TcV+WKpt2bJGbUSaMuZlBEXlYRUDqbCnfPsAzXyw2POOm4nYCIWhsk5P4bhVg2AUBELCGgLQJBeGOuI2AH+FRlVjQhKjyg9BIIAAC3iMgtVsqcIquYmdnp5LWm7DLwsY3SJozuyhy9c47uqXeqWs3eqpE3KwYe3NjZan2CeKmmjjsmUlAibhxgfmm3vHjx5t0V72ZCJBXNAIQt2hksN0JBJSJmxNgoIx9CUDc+vLAN2cRkD6h4CwcKC0IgIBbCEDc3FKT8AMEQKAPAYhbHxz4AgIg4BYCEDe31CT8AAEQ6EMA4tYHB76AAAi4hQDEzS01mYwfXa3UWFtDawtyaGXjuWRyUHIO34ytYjmrUGf43sypU6dKX7KLFyAtLS0lfleVeBmy1atX689+q7JphR2ImxXU7WAzeIy2Pv5/qdr/EpVs70y4RPxj5B9IIiEFEzZy9QRedVjE2kw2j0TP41CGHOZP9oq1586do/Lycrp8+XKiRUz6eI5sz9HFVDzrnXQhTTgR4mYCREdmkTqBFrzzNr3x/HLKTtAB/ufn5c7r6uqora0twbMTO1x1NHYuHfvHgsPLccsWABFDVFVkMfbvgw8+oJKSksQqwoFHQ9wcWGlWFplDCvKTJpy4RSX7R/n73/9eD4YzefJkZW5zN5HD7c2bN0+6TW5F5ebmSrcjDHBLmwPwpKeni02ufYe4ubZqzXeMx744eDDHSuUAxrIXJOAfogjObE4wnNhMWLA5ligv567CJrcSeWxPVQoEArqpn/zkJ6pMWmYH4mYZeucYZpHhAf2CggI9WpNpIQVjIODuE6fs7EQ7zjEyjrKb/eSxNhbv6dOnRznKvM1sj4V0woQJ5mUaI6ePP/5YP0K0vmMc7ujdUherdDQZFL6XAEfd4q4Mr5GncvHPDRs26F022V1f4WhVVZU+ifC73/1ObJL6LsYrVcYQPXLkiC7eKlqlUuHFkTnELQ5IXj+EhY2jmalozQjWPBvLrZr6+nqxSeo7dw/F8vSqxPTEiRO6Tz/60Y+k+haauZgoCd3m1s8QN7fWrIl+HTx4kCZNmmRijrGz4plYTpmZmbEPNuEIFnCeRBArR5uQZcwsWFA5yZ6RFQUR9rww3sY+Sx1z4zEFvhlSVVJ5IyT7tmfPHumu8bJD8bxkFkS1sLEvPM7m9/uV/fB5xvLFF19UZk/4yC1iVYnv2WOmKluKqnyLZEequM2cOZM2btwYya7p23j6XuV4EI/P5OTkKL9z3nRwNs2Qu4Z8L52qxF1uVRMXwif2UWVXn1uIzFRVS1H4adm7+dECv8+RAzJPmTLl+w0SP/n9fj3OpqoYqZcvX9aKioqUBIKWg+2s1lCa0RubVA/AHBKQmW0iKLMc8shVDQGpK/HyDZ98XxTPCsm+J4q7pKNGjVI68M1dU6tmElX8G2IlXhWUYUMWAakTCnfffbde7v3790sXNxZPHhDmm0tVNfV5Op3v+eKEOKmyLlHkCwLJEZA65sZ9ex6o5UdMVKQ5c+bozzuqsCVsCIHjGz9Z4FSvXiHKgXcQAIG+BKSKG5t68MEH9YeQ+5qV840nMHglB5WzpuwJBE5OfSJXEDBCQLq48ZgbJ17dQXYK7QbLttU//1CB42cwVd4C078s+A4CICD5PjcGLJ5hE8+0yYSuuhvc3xchcLt37/bOdHt/CPgOAjYhIL3lxj94Ho8SD0HL9ltlNziSL+yvZ+4jigQA20DAJgSk3goifOTHPgYNGiR9xpTt8Xgbr2qq6vlA4aMb33EriBtr1Ts+KRE37+B0l6cQN3fVp9e8kd4t9RpQ+AsCIGAPAhA3e9QDSgECIGAyAYibyUCRHQiAgD0IQNzsUQ8oBQiAgMkEIG4mA0V2IAAC9iAAcbNHPaAUNiPAK77wC8m5BCBuzq07lFwiAV6M9IknnlAmcKtXr6aFCxdK9Cg8a34kksMKqn4WO7wkcrZA3ORwRa6SCPDS7rJbVPxcMK/wwj98fuJEdmJ/OD7rxIkTZZvqkz8HvOYke63FPkYVfoG4KYQNU8YI8HJSvLS77Ef5ampq9IIuWrTIWIHjPPvAgQP6kWKRiThPM3SYENT8/HxD+ZhxsqxFJiBuZtQO8pBKgH+ILGwi9J7MWAehIf5UPSO8bds2fd1DlYF4hKDKZBnPRcH1+vDDD8vpGqtZzRxWnEjADjEUVMeqKCkp0eN+sF0ViWN+MOfq6moV5nptcPyP3Nzc3u9WfaisrNT951grbW1tUYrxF63d/09XY3qkawv8J7Ru7Tutfd9rWr6PNBpbrjX/JfxUCt+ELSDQQ8BqcQsVtqamJunVEggE9B8QBxtSlURgo+g/bPNLYpWgRvOkvr7+qnCRNmA9d5/Q/AvSNfKVav49r2mlVZ9oX0fLlAMcDbAPuzxGQIhZrHcVWPjHzi0LLsuAF7yJheHWDLcgVLXauOjsI9tVmYSgqooUF49vXMfiuote393axYZVmo9I8y3wa6e7B84Z4jYwH0/tFRdXrHfZUFjYWGT4dfDgQdnm9PzFj4tbEaoS+8msVdpk36wQ1HiYijpgJlEF7mKDVurzadlVR7UY2ha75RbrQsd+6v3HAQuwwDVg3jUQcRxSFzfSqF+M3UjiidnSeKZ0cAwIgIByApcuXeprM3iGGtfWEz2RTfTJZ3S6K9h3f79vMeOW9gQe73cWvnqCABar9EQ128ZJflIiLy9Pj2BXWVlJS5cuDSnbFTpTV0FvT1pMr6U30KF1G6i+uZDS2v5AgQcXUO7t14Uc2/NRbcsteJJqCzMpZ+sxGlhzw8oZ/4ZgB+1fX0BpKSmUkpJDK2uPUVf8ZydwZJC6WutpbcHfEYuAbqt6P3VIcyxS0bgM71Htf66lgruepcZLSo1HKpA527paqbG2htYW5NDKxnPm5BkxF+vqMNixl9aLa+f+Z6m2tV8rJWJ5zdkYPFNLhWmP0dbWP5uToQm5RBe2P1Pr1scoJeVRqqBf0tq80XRtWjo9lNVOr/xrFX369/MjCptepEh9VTnbvtNO+5doPopvMDC5MnylHfDv0Pa1f6dpvffBZGilDWeTy26As7pP/7e2bt1uLfA1D2uKe258Wlb5vgGnpwfIMuFd3YEq7Wfz87X5fK+Pb5XWcDHWEGtiJsT4UWJnGTy6+6hW9bN5Wv78dI1ITt2JElpWh18f0vzb/qS1c3V1t2v71uVrPgn1J/zs8y5up6B5WlXg2z67rPrCkwc8ecTXm5mTK4pmS7u1r5tf1fJXrNDy45zpSAZ092dN2runWdhEOqs1lGZovtIG7aLYZMr7t9pn7/6p31T0t1qgap4UkRm4yPLsWiJuV51l4c6WKm5W1WEEu/oguVwh78H6ldZcvkRbUTpf89lE3OKaIR34BxB1r5puaVczbawgKl7xMN1mQhM2WhapY6dTVmjfO3iOThwaRSsem0xDo52U1PbraWzWNLq9D73r6LYxd5EvqfxwknoCVtVhuN3gFyfo0Phf0GNTh0nEEKSulu1UQU/SipzREu0kljU/esapqamJpk+fntjJMY7u8/OMcWySuy9Qy8YaouJ8yhhyTZJ5JHoaj6U00tZV/0afrXyNVmTcnGgGSR8/aHYmjR+sAGvSJcSJsQioq8NL1Nq4lVaVddDKmiWUIfO6EQ2MRZk0JBYAhfsrKiro4MGDpgsbuyD5Vyj+LebTImUCc44aV06lIeMfoMJX/ov+VL+PPosxZWxOXZ6n5voztGjxrH4tOnNyRy4qCCisw0uNtDLtJhr/QCG9sv2PVP+nzyVNfDG3kAaGTAFNoop4SSlZCwbIFbeuZtrsH02rV0ylwUk4ntwpw2nWmmbSutupedtTRK/Mpaxn6uiM1IlEFvEaqh6xRKGIJ0cHZ0UjoLgOh86iNe0adbfvo22lfJnOo2dqT0q4i4D9eov8dy5X2oOJRlnl9sTFraOWCvRbH/j2h2iv+2ltyxlq2dxIo5fMNtCS+St11C4awM5V+3etpZa/9sOW6qOMghfo36vmUceuZgrE03qL27d+N5ewiL/9N7RqUaYxEU/Wfj/X8TUJAmbVYYKmU31TqeClV6kqu5N27TtufutNb2D4aMmc0bK7aQl6ruDwqFMNRneIxyQo2uMYaqai9Vk3mdPs3Se0unVvaUf1W0KMQkvmfMyWJkOtzzk2qUPzZ/W/f9BczHyHvcfxGFMfVg76knjLLV7BvdrsvrryCN9yQtrFBir1+Si76ih1a2/TgnHXx5tbkscFqev0Z3Rc1iA/Pw7y6i4a8vg/0AQxltF1hKq37iV1t2QmiQan9RCwRR120enARZp971hjLf+wOk2lobNWU3vP6j89v0Gtmy42rCIfzaOqwLek1S+gcfJUIKxEKje4yK0rdKZ2KaXdv5KqWzr0sYtgRwO9vOY8PVvyoIGucZTq6DpGtasW0AMr/jc9kPaD77vOQ7Kphm4x+SKNUgZsNkbAijq8+pTO/SvfpJaOK0R0hToaX6c1X8ynkuxR3us6GqvBgc9W2spMYLmSxMvVrX19dFvPypx6Vzhdyy/3a8360wqJ5zbgGb13eUfqcqvpbuvlC+v6m/v0h+jCDMjC9J09N14L2/q7jK6TZXV4UTtatUBfk0z3zZevlfube55WMJ1lpAxFV1XhdRqpGAq2pbCNgeUPe71KAA/Oe7Xm3eG3i7ql7qgQeAECKgjs3btXD7qjwpZVNtBys4q8A+yi5eaASkqyiHPmzNHPrKurSzIH+5+Glpv96wgltIgAt244CLSKxEv+cNR5WTE8Q31gv3bu3EklJSWhm133GeLmuiqFQ2YQ4PilHCR5x44dZmQXMw9uQXHUeRWpvLxcj5Nq9oPqKsqeiA10SxOh5bFjvdotFQsncnW///77xM8/ykzcWrvlllsofPVZ861yq41FW8YqHOaX1liOEDdj/Fx9thfFjaPbz5w5U6/X2tpaGjlypPQ65qjry5Yto87OTpId5d4LY22iwmLGUBAH4t3NBHiJqD/Sno+baeevz1PBgRdp1lDvjViwsBUXFxSr7vgAAA+2SURBVOtr+PMyPCqEjVttLGzcapMtbNzV5rE2v9/v5ou51zfvXcG9ruODIBBs3UaPv/AG+YtLaPtlsdV77+vXr6ctW7boXTZZy/D0p1pTU6Nvmj9/fv9dpn9n36ZMmUKzZ882PW9bZqjgRmGYcASB8AfwxVMCqoov7MXzbnaZKisr9TX8I8bKNNvY1fw4sj3HDigpKZFk4ftsRQBojjbvlSS15cbNYB634UFMFYnHSMQ4kQp7bGv79u0qTPXaKC0tJX4hmUeAJxBE1zA/P9+8jGPktGvXLr0LXFRUFONI47u5i81dbc+02ohI6pjbuHHj9Fr58ssvjdeODXPg+4SOHDmivGRfffWVUpv8JyXqUqZhq54E5B9+IBBQ4mMov5dfflm/10wFW7arqqsd6qOVn6W23FQ7duutt+om+cfo1jRt2jR9XEilf9xS5MF2NydVAhPKcPfu3foERug2fDaPgHRx4wHM9vZ280o8QE4jRowYYK+cXW4WUkGMZ9hU3cwqbHrhnWdHVczIeoFlJB+li1tWVhadOnUqkm3Hb+NWFP/wrUgDtqRMXq68rKyMCgoKyAtCbkVdwqYcAtLFTU6xI+c6atQofYcV42CRS2T+1vT0dD3Ttra26Jn78qi6z+qr2tVVWEPf36V/zogvbM/y5cv1Wwi80D2NDhV7nEZAurjdfPPNyv7xZT8m079yBw/uEQcVDzv3t63yO3PdtGmT3krle8GQQMAJBKSL24QJEyzrusmugDFjxugmzp07J9tUWP6XL6u925Zn2vguen64+9ChQ2HlwQYQsBsB6eKm2mG+Z+jDDz9UbVaZPTGr9/nnn5tnUw8QfAONL9xB1PESPXDTSMrZeiws/8LCQn01CVNth1nBBhAwh4D0B+f5Bl6+zy0vL8+cEsfIhdffuvHGG0nFci5882djYyM98sgj0p8LDHWbbxzmyQwhdKH7zPwsbogOvf+MJzJUd//N9Al5eYeAdHHzDkr3eRpJ3NznJTxyKwHXdUvdWlHwCwRAIDECELfEeOFoEAABhxCAuDmkolBMEACBxAhA3BLjhaNBAAQcQgDi5pCKQjFBAAQSIwBxS4wXjgYBEHAIAYibQyoKxQQBWQT4fk03LoogdbFKWZWBfEEABMwhIMIYpqWlkduiz6PlZs41glxAwDQCHHmew/3JTvz0kFhJR4U92f70zx/i1p+IZ75fotY/rKeCtBQ97kTK/SupuqWDgp7x356Ossjw4gRDhw6VWkB+TJGDM/Oz2LwisCsXzfRKJBz4GUrgO+103Wvauj0B7Wve3N2u7VuXr/noZ1p581e9B4ooVL0b8EEqgfr6ej0CF0fikplEpK+ioiKNI3C5NfEihkheI9D9ufbuu6e07lC/u49qVdk+zVfaoF28uh3iFgpI7uempiZd2FhwZCUWMiFs/O5mYWOGmFCQ2vi3aeapd1BWVr+ypQ6nMZPS+m3EVxUEeFCfVzvmLmJFRYU0k8XFxXpwIV6Xb+nSpdLs2CVjjLnZpSZsUY5baPa9Yym+xcflFZiXVeIxIRWJVz6J9yWjPGK2kvN+/vnnpS4nNXHiRKqvr/eEsDFP6eLGa4+pClzMFwqv88/vKhLPNqkOkCzNx0sfU/2hh2lx9qiwi4KDXatMH3zwAeXk5CiZMVTpV39bLOIvvPCCvpkZyx7U59ZadnZ2/2K49rt0ceNgLaoCtvDS2+Xl5aRqCW5ehJPtqUxyfLxALZv+i0as/iVlDA6/JObOnavsD4NZ8g+Qu04cBV72HyMvxBnvy+x67uzspMOHDxPHpZAtbGaX3Qn5hV/JTii1h8sYXbj/Sh21i2J3se5aSy1/DQUYpK6Wt+jtYUW0KOPm0B29nzn2rGhh9G6U/IFbGSUlJXpIQW4huzGxoL3//vtKVo12I79YPmFCIRYhm+0X8QvClxi/lnx5G0nTNiZU4uCZ3bT5yAz6zYL0qGNtL774ot5NnD17trLl4tkJFtSvvvpKvx+rqanJlSKAJdsTulwTOhgtt4RwuevgYEcjvfr29fT4U0LYgtR17D9p63t9o3lxN5FbUaq7p/zD59lDnkXkG07d2oJz11VlH28gbvapC4UlCVJXay2tmv8krVjxAKVdI2YMr6Ehd79NlBY+X/rrX/9aD8ysunsqBC43N1e/XULVZJHCyoApSQQgbpLA2jnb4Jk6eiZrLr3yXkd4MbMfphl3XR+2fdiwYcTd0y1btpDq2VMWOPHsoxfuzwqDjw1JEcCYW1LYnH1S6u15VNWuUVWCbohZzPT09ATPNH44D76zqA4aNMh4ZsjBEwQgbp6oZvOctLLlhNslzKtHL+SEbqkXahk+goAHCSAoswcrPV6XEZQ5XlI4zo4E0HKzY62gTCAAAoYJQNwMI0QGIAACdiQAcbNjraBMIAAChglA3AwjRAYgAAJ2JABxs2OtoEwgAAKGCUDcDCNEBiAAAnYkAHGzY62gTCAAAoYJQNwMI0QGIOBsArzaiqpl3VWSgrippA1bIGAzArzKCi8ntWPHDpuVzHhxIG7GGTo2h2DHXlpf8HdXgzI/S7WtlxzrCwqeOAEWtry8PH0pqzVr1iSegc3PgLjZvIKkFa/rMNXtuYYef+MT0rrbad/Pz9DSrJep8RJizktjbqOMOTiNWASBV1vhJa3cliBubqvRuPz5Mx1v/pbufWoa+fgKSPXR1MICeor2UH3z+bhywEHOJcDCxjFMd+7cSZs2bXJtcBqIm3OvUQMlv57GZk2j20NqP/jFCTo0/hf02FT3/YMbAOXKUznaFi86ynEpJk2a5Eof2amQy1uOj4hbai5X8+OWXqLWxq20qqyDVtYsiRjaz1wP4suN/WxtbY3vYBOO4ut04cKFxK0aNyde0fi5557TQydOnz7dza4Sx2yUmkpKSjR+qUiBQEAjIo3fVSS/36/bU2FL2DDVx4sNWqmPdB+IsrVS/1Hta2GIg3lSzz72U3XKzc3VioqKlJltamrS/VV1rbJjgm+sd7MgdHZ26jYrKyvNytLW+Uhvubn7r0G9d9HjliZRlqGzaE27Rt3t+2hbKdErc+fRM7Unqf+UguqoV+zJ4sWL9a4Tt+BUJG7F+P1+Pci2iNegwq5KGzxpEAgEqLCwUKVZy2xB3CxDn5zh6HFLkw3KzPMJU6ngpVepKruTdu07Tl39imZFUObMzEy9FI2Njf1KI+8r3xYhIt2rEDgrIt1zvFuvxEqFuMn7rSjOWQRl1nioIfrrs3+mjEiRM1LH0Ix/nBGxzFZEveJWBscrVR1pi2+PEALH43BIziUQ6TJ3rjcouQECXXQ6cJFm3zs2LPJ8aFDmtrY2ZbcOPP300/rd8zyxwC0OVYkF7tKlS1RQUEBjx451ZaR7VSyttIOWm5X0rbIdPEm1hZl0/8o3qaXjChFdoY7G12nNF/OpJHtUxCn00KDMqmYUJ0+erBOy4rnH5cuXI9K9VdenSXYhbiaBdFQ2qX9DP54+kQKv5FNm2g8oJW0h/ceFR2jbGwU0YXDkSyI0KLOq5xB5bKisrIys6B6KSPfcNUZyJgF0S51ZbwZLPZQmLKii9gWJhWXm7ml1dTVNmzbNoP34T585c6Z+X9ahQ4eU33DKArd58+b4C4sjbUUg8t+0rYqIwtiJQH5+vtLxL75Fg2drxSyxnVigLPYmgLil9q4fS0uHuKWW4odxgwTQcjMIEKeDAAjYkwDEzZ71glKBAAgYJABxMwgQp4MACNiTAMTNnvWCUoEACBgkAHEzCBCngwAI2JMAxM2e9YJSgQAIGCQAcTMIEKeDAAjYkwDEzZ71glKBAAgYJABxMwgQp4MACNiTAMTNnvWCUoEACBgkAHEzCBCngwAI2JMAxM2e9aK8VMEztVSY9hhtbf2zctswCAIyCEDcZFB1Wp7Bk1T3L8/T1g6nFRzlBYHoBKSL2969e4lfKtL58+f1Nff5XUXi5a9Vr/Fvvo8XqGVdOe0d/hPyqYAGG70EVEX26jXosQ/SxW3nzp3ELxXp3LlzxGHo+F1FOnLkiG5PhS1hw1wfg9TVsp0q6ElakTNamLDFO//wVf0pWuEw+zdq1ChSEWVL+MdLWMX7Euc4+V26uDkZjuvL3tVMGyuIihdl0hCbOVtRUaHsT5FX+S0tLVUabX7kyJG9UbasiBFhs+qWUhyImxSs8jLl1iIn/tc3li5Qy8YaouJ8yogSN8FY/sbOTk9P1wMkG8slvrO/+eYb3VZxcXF8J5h0FAdH5hgNOTk5SlqpA4Z87BcO0iQXLc0G4mYp/uSNGwusy93Rt8h/53JakXFzzEKoHlfkAg0ePFgvl4pIWyLa/JYtW5R2E0OD0MyYMUOJwMWsbBcdAHFzTWUmEHH+QjNt9vtoyZzREcP49UfC45g8eaIyccuNE8dJVZFUR5sXPvUXONWcRTnc+A5xc02txhtxfgX97YE6Kn9pDo28RgwwX0M3PfASddAOKhx/A6XkbKXW4PdgOECL6jEpYf3s2bPio/T30GjzKgf6WeDWrFmjB8J58sknCbOo5lQ1xM0cjg7KJZWGzlpN7X3GWLrpYsMq8tE8qgp8S1r9AhoXcmVs2rRJH9xXFa+UYYoI819++aVStixwHCt12bJlSruJHBdWdP+5FanqdialcBUbC7mEFVuGOccQmDRpkv6DLygoUNo95RZje3u7ck5WRZvnGVQWOBZYFjskYwQgbsb4eeZs/sGr7p5mZWXRqVOnlDMOHQc7fvy4UvsscBwbFsk4AcQtNc7QtTn0j1vK94Pdc889etR5FT9AHvfavn077d+/37WM4Zg8AhA3eWwdn3N/cWOHuNs0depU4haG7MTjTvxEhhh/k20P+buLAMTNXfVpqjeRxM1UA8gMBCQSwJibRLjIGgRAwDoCEDfr2MMyCICARAIQN4lwkTUIgIB1BCBu1rGHZRAAAYkEIG4S4SJrEAAB6whA3KxjD8sgAAISCUDcJMJF1iAAAtYRgLhZxx6WQQAEJBKAuEmEi6xBAASsIwBxs449LIMACEgkAHGTCBdZgwAIWEcA4mYde1gGARCQSADiJhEusgYBELCOAMTNOvawDAIgIJEAxE0iXGQNAiBgHQGIm3XsYRkEQEAiAYibRLjIGgRAwDoCEDfr2MMyCICARAIQN4lwkTUIgIB1BCBu1rGHZRAAAYkEIG4S4SJrEAAB6whA3KxjD8sgAAISCUDcJMJF1iAAAtYRgLhZxx6WQQAEJBKAuEmEi6xBAASsIwBxs449LIMACEgkAHGTCBdZgwAIWEfgWutMw7LdCWiaZvcionwgEJUAWm5R0WAHCICAkwlA3Jxceyg7CIBAVAIQt6hosAMEQMDJBCBuTq49lB0EQCAqAYhbVDTYAQIg4GQC/x+84Rnbn7iV/gAAAABJRU5ErkJggg=="></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>
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