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</div><h2>HL Paper 3</h2><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The function <em>f</em> is defined on the domain \(\left] { - \frac{\pi }{2},\frac{\pi }{2}} \right[{\text{ by }}f(x) = \ln (1 + \sin x)\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><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.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Explain briefly why your result shows that <em>f</em> is neither an even function nor an odd function.</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 style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Determine the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}}\).</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 style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f''(x) = \frac{{ - \sin x(1 + \sin x) - \cos x\cos x}}{{{{(1 + \sin x)}^2}}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{ - \sin x - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{(1 + \sin x)}^2}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \frac{1}{{1 + \sin x}}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) \(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({f^{(4)}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = - 1\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f'''(0) = 1,{\text{ }}{f^{(4)}}(0) = - 2\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f(x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} + \ldots \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px 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: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) the series contains even and odd powers of <em>x</em> <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} + \ldots - x}}{{{x^2}}}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 1}}{2} + \frac{x}{6} + \ldots }}{1}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \frac{1}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Use of l’Hopital’s Rule is also acceptable.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[3 marks]</em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The integral \({I_n}\) is defined by \({I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x,{\text{ for }}n \in \mathbb{N}} \) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \({I_0} = \frac{1}{2}(1 + {{\text{e}}^{ - \pi }})\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">By letting \(y = x - n\pi \) , show that \({I_n} = {{\text{e}}^{ - n\pi }}{I_0}\) .</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence determine the exact value of \(\int_0^\infty {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} \) .</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({I_0} = \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x} \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Award </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> for \({I_0} = \int_0^\pi {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Attempt at integration by parts, even if inappropriate modulus signs are present. </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>M1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\cos x{\text{d}}x} \) <strong>or</strong> \( = - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\cos x{\text{d}}x} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x} \) <strong>or</strong> \( = - \left[ {{{\text{e}}^{ - x}}\sin x + {{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \int_0^\pi {{{\text{e}}^{ - x}}\sin x{\text{d}}x} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = - \left[ {{{\text{e}}^{ - x}}\cos x} \right]_0^\pi - \left[ {{{\text{e}}^{ - x}}\sin x} \right]_0^\pi - {I_0}\) <strong>or</strong> \( - \left[ {{{\text{e}}^{ - x}}\sin x + {{\text{e}}^{ - x}}\cos x} \right]_0^\pi - {I_0}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Do not penalise absence of limits at this stage</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({I_0} = {{\text{e}}^{ - \pi }} + 1 - {I_0}\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({I_0} = \frac{1}{2}(1 + {{\text{e}}^{ - \pi }})\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> If modulus signs are used around cos </span><em style="font-family: 'times new roman', times; font-size: medium;">x</em><span style="font-family: 'times new roman', times; font-size: medium;"> , award no accuracy marks but do not penalise modulus signs around sin </span><em style="font-family: 'times new roman', times; font-size: medium;">x</em><span style="font-family: 'times new roman', times; font-size: medium;"> .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em> </em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 28.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;"><em>[6 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Attempt to use the substitution \(y = x - n\pi \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(putting \(y = x - n\pi \) , \({\text{d}}y = {\text{d}}x\) and \(\left[ {n\pi ,{\text{ }}(n + 1)\pi } \right] \to [0,{\text{ }}\pi ]\))</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so \({I_n} = \int_0^\pi {{{\text{e}}^{ - (y + n\pi )}}|\sin (y + n\pi )|{\text{d}}y} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{ - n\pi }}\int_0^\pi {{{\text{e}}^{ - y}}|\sin (y + n\pi )|{\text{d}}y} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{ - n\pi }}\int_0^\pi {{{\text{e}}^{ - y}}\sin y{\text{d}}y} \) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{ - n\pi }}{I_0}\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_0^\infty {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} = \sum\limits_{n = 0}^\infty {{I_n}} \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \sum\limits_{n = 0}^\infty {{{\text{e}}^{ - n\pi }}{I_0}} \) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the \(\sum \) term is an infinite geometric series with common ratio \({{\text{e}}^{ - \pi }}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_0^\infty {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} = \frac{{{I_0}}}{{1 - {{\text{e}}^{ - \pi }}}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \frac{{1 + {{\text{e}}^{ - \pi }}}}{{2(1 - {{\text{e}}^{ - \pi }})}}{\text{ }}\left( { = \frac{{{{\text{e}}^\pi } + 1}}{{2({{\text{e}}^\pi } - 1)}}} \right)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in \({I_0}\) which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 34.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \(y = \ln \left( {\frac{{1 + {{\text{e}}^{ - x}}}}{2}} \right)\), show that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ - y}}}}{2} - 1\).</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 style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence, by repeated differentiation of the above differential equation, find the Maclaurin series for <em>y</em> as far as the term in \({x^3}\), showing that two of the terms are zero.</span></p>
<div class="marks">[11]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = \ln \left( {\frac{{1 + {{\text{e}}^{ - x}}}}{2}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2{{\text{e}}^{ - x}}}}{{2(1 + {{\text{e}}^{ - x}})}} = \frac{{ - {{\text{e}}^{ - x}}}}{{1 + {{\text{e}}^{ - x}}}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">now \(\frac{{1 + {{\text{e}}^{ - x}}}}{2} = {{\text{e}}^y}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow 1 + {{\text{e}}^{ - x}} = 2{{\text{e}}^y}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {{\text{e}}^{ - x}} = 2{{\text{e}}^y} - 1\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - 2{{\text{e}}^y} + 1}}{{2{{\text{e}}^y}}}\) <strong><em>(A1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Only one of the two above </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>A1</em></strong><span style="font-family: 'times new roman', times; font-size: medium;"> marks may be implied.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ - y}}}}{2} = - 1\) </span><strong style="font-family: 'times new roman', times; font-size: medium;"><em>AG</em></strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">Note:</strong><span style="font-family: 'times new roman', times; font-size: medium;"> Candidates may find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) as a function of x and then work backwards from the given answer. Award full marks if done correctly.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><strong style="font-family: 'times new roman', times; font-size: medium;">METHOD 2</strong></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = \ln \left( {\frac{{1 + {{\text{e}}^{ - x}}}}{2}} \right)\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {{\text{e}}^y} = \frac{{1 + {{\text{e}}^{ - x}}}}{2}\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow {{\text{e}}^{ - x}} = 2{{\text{e}}^y} - 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow x = - \ln (2{{\text{e}}^y} - 1)\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}y}} = - \frac{1}{{2{{\text{e}}^y} - 1}} \times 2{{\text{e}}^y}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2{{\text{e}}^y} - 1}}{{ - 2{{\text{e}}^y}}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ - y}}}}{2} - 1\) <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 33.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 1</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}y = \ln 1 = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{2} - 1 = - \frac{1}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = - \frac{{{{\text{e}}^{ - y}}}}{2}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{{{{\text{e}}^{ - y}}}}{2}{\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2} - \frac{{{{\text{e}}^{ - y}}}}{2}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}\) <strong><em>M1A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{1}{2} \times \frac{1}{4} - \frac{1}{2} \times \frac{1}{4} = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = f(0) + f'(0)x + \frac{{f''(0)}}{{2!}}{x^2} + \frac{{f'''(0)}}{{3!}}{x^3}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow y = 0 - \frac{1}{2}x + \frac{1}{8}{x^2} + 0{x^3} + \ldots \) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">two of the above terms are zero <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>METHOD 2</strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}y = \ln 1 = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{2} - 1 = - \frac{1}{2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = \frac{{ - {{\text{e}}^{ - y}}}}{2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ - {{\text{e}}^{ - y}}}}{2}\left( {\frac{{{{\text{e}}^{ - y}}}}{2} - 1} \right) = \frac{{ - {{\text{e}}^{2y}}}}{4} + \frac{{{{\text{e}}^{ - y}}}}{2}\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = - \frac{1}{4} + \frac{1}{2} = \frac{1}{4}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \left( {\frac{{{{\text{e}}^{ - 2y}}}}{2} - \frac{{{{\text{e}}^{ - y}}}}{2}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}}\) <strong><em>M1A1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">when \(x = 0,{\text{ }}\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = - \frac{1}{2} \times \left( {\frac{1}{2} - \frac{1}{2}} \right) = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = f(0) + f'(0)x + \frac{{f''(0)}}{{2!}}{x^2} + \frac{{f'''(0)}}{{3!}}{x^3}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( \Rightarrow y = 0 - \frac{1}{2}x + \frac{1}{8}{x^2} + 0{x^3} + \ldots \) <strong><em>(M1)A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">two of the above terms are zero <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 35.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[11 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating \({{{\text{e}}^{ - y}}}\) with respect to <em>x</em>. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, although partial credit was given for this. In this case, they often found problems in simplifying the algebra.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Many candidates were successful in (a) with a variety of methods seen. In (b) the use of the chain rule was often omitted when differentiating \({{{\text{e}}^{ - y}}}\) with respect to <em>x</em>. A number of candidates tried to repeatedly differentiate the original expression, which was not what was asked for, although partial credit was given for this. In this case, they often found problems in simplifying the algebra.</span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">The function \(f\) is defined by \(f(x) = {{\text{e}}^{ - x}}\cos x + x - 1\).</p>
<p class="p1">By finding a suitable number of derivatives of \(f\), determine the first non-zero term in its Maclaurin series.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>\(f(0) = 0\) <strong><em>A1</em></strong></p>
<p>\(f'(x) = - {{\text{e}}^{ - x}}\cos x - {{\text{e}}^{ - x}}\sin x + 1\) <strong><em>M1A1</em></strong></p>
<p>\(f'(0) = 0\) <strong><em>(M1)</em></strong></p>
<p>\(f''(x) = 2{{\text{e}}^{ - x}}\sin x\) <strong><em>A1</em></strong></p>
<p>\(f''(0) = 0\)</p>
<p>\({f^{(3)}}(x) = - 2{{\text{e}}^{ - x}}\sin x + 2{{\text{e}}^{ - x}}\cos x\) <strong><em>A1</em></strong></p>
<p>\({f^{(3)}}(0) = 2\)</p>
<p>the first non-zero term is \(\frac{{2{x^3}}}{{3!}}\;\;\;\left( { = \frac{{{x^3}}}{3}} \right)\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award no marks for using known series.</p>
<p> </p>
<p><strong><em>[7 marks]</em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">Most students had a good understanding of the techniques involved with this question. A surprising number forgot to show \(f(0) = 0\). Some candidates did not simplify the second derivative which created extra work and increased the chance of errors being made.</p>
</div>
<br><hr><br><div class="specification">
<p class="p1">Let \(f(x) = {{\text{e}}^x}\sin x\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \(f''(x) = 2\left( {f'(x) - f(x)} \right)\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">By further differentiation of the result in part (a) , find the Maclaurin expansion of \(f(x)\), as far as the term in \({x^5}\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">\(f'(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x\) <span class="Apple-converted-space"> </span><strong><em>M1A1</em></strong></p>
<p class="p1">\(f''(x) = {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x = 2{{\text{e}}^x}\cos x\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( = 2\left( {{{\text{e}}^x}\sin x + {{\text{e}}^x}\cos x - {{\text{e}}^x}\sin x} \right)\) <span class="Apple-converted-space"> </span><strong><em>M1</em></strong></p>
<p class="p1">\( = 2\left( {f'(x) - f(x)} \right)\) <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(f(0) = 0,{\text{ }}f'(0) = 1,{\text{ }}f''(0) = 2(1 - 0) = 2\) <strong><em>(M1)A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>M1 </em></strong>for attempt to find \(f(0)\), \(f'(0)\) and \(f''(0)\).</p>
<p> </p>
<p>\(f'''(x) = 2\left( {f''(x) - f'(x)} \right)\) <strong><em>(M1)</em></strong></p>
<p>\(f'''(0) = 2(2 - 1) = 2,{\text{ }}{f^{IV}}(0) = 2(2 - 2) = 0,{\text{ }}{f^V}(0) = 2(0 - 2) = - 4\) <strong><em>A1</em></strong></p>
<p>so \(f(x) = x + \frac{2}{{2!}}{x^2} + \frac{2}{{3!}}{x^3} - \frac{4}{{5!}} + \ldots \) <strong><em>(M1)A1</em></strong></p>
<p>\( = x + {x^2} + \frac{1}{3}{x^3} - \frac{1}{{30}}{x^5} + \ldots \)</p>
<p><em><strong>[6 marks]</strong></em></p>
<p><em><strong>Total [10 marks]</strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(f(x) = 2x + \left| x \right|\) , \(x \in \mathbb{R}\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Prove that <em>f</em> is continuous but not differentiable at the point (0, 0) .</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Determine the value of \(\int_{ - a}^a {f(x){\text{d}}x} \) where \(a > 0\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">we note that \(f(0) = 0,{\text{ }}f(x) = 3x\) for \(x > 0\) and \(f(x) = x{\text{ for }}x < 0\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} x = 0\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} 3x = 0\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">since \(f(0) = 0\) , the function is continuous when <em>x</em> = 0 <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{h}{h} = 1\) <strong><em>M1A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3h}}{h} = 3\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">these limits are unequal <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">so <em>f</em> is not differentiable when <em>x</em> = 0 <strong><em>AG</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[7 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\int_{ - a}^a {f(x){\text{d}}x = \int_{ - a}^0 {x{\text{d}}x + \int_0^a {3x{\text{d}}x} } } \) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = \left[ {\frac{{{x^2}}}{2}} \right]_{ - a}^0 + \left[ {\frac{{3{x^2}}}{2}} \right]_0^a\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {a^2}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 27.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[3 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">The curves \(y = f(x)\) and \(y = g(x)\) </span>both pass through the point \((1,{\text{ }}0)\) and are defined by the differential equations \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x - {y^2}\) and \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = y - {x^2}\) <span class="s1">respectively.</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that the tangent to the curve \(y = f(x)\) <span class="s1">at the point \((1,{\text{ }}0)\) </span>is normal to the curve \(y = g(x)\) <span class="s1">at the point \((1,{\text{ }}0)\)</span>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find \(g(x)\).</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 class="p1">Use Euler’s method with steps of \(0.2\) <span class="s1">to estimate \(f(2)\) </span>to \(5\) <span class="s1">decimal places.</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Explain why \(y = f(x)\) cannot cross the isocline \(x - {y^2} = 0\), for \(x > 1\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Sketch the isoclines \(x - {y^2} = - 2,{\text{ }}0,{\text{ }}1\).</p>
<p>(ii) On the same set of axes, sketch the graph of \(f\).</p>
<div class="marks">[4]</div>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>gradient of \(f\) at \((1,{\text{ }}0)\) is \(1 - {0^2} = 1\) and the gradient of \(g\) at \((1,{\text{ }}0)\) is \(0 - {1^2} = - 1\) <strong><em>A1</em></strong></p>
<p>so gradient of normal is \(1\) <strong><em>A1</em></strong></p>
<p>= Gradient of the tangent of \(f\) at \((1,{\text{ }}0)\) <strong><em>AG</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y = - {x^2}\)</p>
<p>integrating factor is \({{\text{e}}^{\int { - 1{\text{d}}x} }} = {{\text{e}}^{ - x}}\) <strong><em>M1</em></strong></p>
<p>\(y{{\text{e}}^{ - x}} = \int { - {x^2}{{\text{e}}^{ - x}}{\text{d}}x} \) <strong><em>A1</em></strong></p>
<p>\( = {x^2}{{\text{e}}^{ - x}} - \int {2x{{\text{e}}^{ - x}}{\text{d}}x} \) <strong><em>M1</em></strong></p>
<p>\( = {x^2}{{\text{e}}^{ - x}} + 2x{{\text{e}}^{ - x}} - \int {2{{\text{e}}^{ - x}}{\text{d}}x} \)</p>
<p>\( = {x^2}{{\text{e}}^{ - x}} + 2x{{\text{e}}^{ - x}} + 2{{\text{e}}^{ - x}} + c\) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note: </strong>Condone missing \( + c\) at this stage.</p>
<p> </p>
<p>\( \Rightarrow g(x) = {x^2} + 2x + 2 + c{{\text{e}}^x}\)</p>
<p>\(g(1) = 0 \Rightarrow c = - \frac{5}{{\text{e}}}\) <strong><em>M1</em></strong></p>
<p>\( \Rightarrow g(x) = {x^2} + 2x + 2 - 5{{\text{e}}^{x - 1}}\) <strong><em>A1</em></strong></p>
<p><strong><em>[6 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">use of \({y_{n + 1}} = {y_n} + hf'({x_n},{\text{ }}{y_n})\) <span class="Apple-converted-space"> </span><strong><em>(M1)</em></strong></p>
<p class="p1">\({x_0} = 1,{\text{ }}{y_0} = 0\)</p>
<p class="p1">\({x_1} = 1.2,{\text{ }}{y_1} = 0.2\) <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\({x_2} = 1.4,{\text{ }}{y_2} = 0.432\) <span class="Apple-converted-space"> </span><strong><em>(M1)(A1)</em></strong></p>
<p class="p1">\({x_3} = 1.6,{\text{ }}{y_3} = 0.67467 \ldots \)</p>
<p class="p1">\({x_4} = 1.8,{\text{ }}{y_4} = 0.90363 \ldots \)</p>
<p class="p1">\({x_5} = 2,{\text{ }}{y_5} = 1.1003255 \ldots \)</p>
<p class="p1">answer \( = 1.10033\) <span class="Apple-converted-space"> </span><strong><em>A1 <span class="Apple-converted-space"> </span>N3</em></strong></p>
<p class="p2"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Award <strong><em>A0 </em></strong>or <strong><em>N1 </em></strong><span class="s1">if \(1.10\) </span>given as answer.</p>
<p class="p1"><em><strong>[5 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">at the point \((1,{\text{ }}0)\), the gradient of \(f\) is positive so the graph of <span class="s1">\(f\) </span>passes into the first quadrant for \(x > 1\)</p>
<p class="p1">in the first quadrant below the curve \(x - {y^2} = 0\) the gradient of \(f\) is positive <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">the curve \(x - {y^2} = 0\) has positive gradient in the first quadrant <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">if \(f\) were to reach \(x - {y^2} = 0\) it would have gradient of zero, and therefore would not cross <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><strong><em>[3 marks]</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) and (ii)</p>
<p class="p1"><strong><em><img src="images/Schermafbeelding_2016-01-21_om_12.34.00.png" alt> A4</em></strong></p>
<p class="p2"> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for 3 correct isoclines.</p>
<p>Award <strong><em>A1 </em></strong>for \(f\) not reaching \(x - {y^2} = 0\).</p>
<p>Award <strong><em>A1 </em></strong>for turning point of \(f\) on \(x - {y^2} = 0\).</p>
<p>Award <strong><em>A1 </em></strong>for negative gradient to the left of the turning point.</p>
<p> </p>
<p><strong>Note: </strong>Award <strong><em>A1 </em></strong>for correct shape and position if curve drawn without any isoclines.</p>
<p><em><strong>[4 marks]</strong></em></p>
<p><em><strong>Total [20 marks]</strong></em></p>
<div class="question_part_label">e.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the functions \(f(x) = {(\ln x)^2},{\text{ }}x > 1\) and \(g(x) = \ln \left( {f(x)} \right),{\text{ }}x > 1\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Find \(f'(x)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Find \(g'(x)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 16.0px 'Times New Roman'; color: #3f3f3f;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Hence, show that \(g(x)\) is increasing on \(\left] {1,{\text{ }}\infty } \right[\).</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 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</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;">\[(\ln x)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{2}{x}y = \frac{{2x - 1}}{{(\ln x)}},{\text{ }}x > 1.\]</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;">(i) Find the general solution of the differential equation in the form \(y = h(x)\).</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) Show that the particular solution passing through the point with coordinates \(\left( {{\text{e, }}{{\text{e}}^2}} \right)\) is given by \(y = \frac{{{x^2} - x + {\text{e}}}}{{{{(\ln x)}^2}}}\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Sketch the graph of your solution for \(x > 1\), clearly indicating any asymptotes and any maximum or minimum points.</span></p>
<div class="marks">[12]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) attempt at chain rule <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f'(x) = \frac{{2\ln x}}{x}\) <em><strong>A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) attempt at chain rule <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(g'(x) = \frac{2}{{x\ln x}}\) <em><strong>A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) \(g'(x)\) is positive on \(\left] {1,{\text{ }}\infty } \right[\) <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;">so \(g(x)\) is increasing on \(\left] {1,{\text{ }}\infty } \right[\) <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;"><strong><em>[5 marks]</em></strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) rearrange in standard form:</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;">\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{2}{{x\ln x}}y = \frac{{2x - 1}}{{{{(\ln x)}^2}}},{\text{ }}x > 1\) <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;">integrating factor:</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;">\({{\text{e}}^{\int {\frac{2}{{x\ln x}}{\text{d}}x} }}\) <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {{\text{e}}^{\ln \left( {{{(\ln x)}^2}} \right)}}\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\( = {(\ln x)^2}\) <em><strong>(A1)</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">multiply by integrating factor <strong><em>(M1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(\ln x)^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{{2\ln x}}{x}y = 2x - 1\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {y{{(\ln x)}^2}} \right) = 2x - 1{\text{ }}\left( {{\text{or }}y{{(\ln x)}^2} = \int {2x - 1{\text{d}}x} } \right)\) <em><strong>M1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">attempt to integrate: <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\({(\ln x)^2}y = {x^2} - x + c\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = \frac{{{x^2} - x + c}}{{{{(\ln x)}^2}}}\) <em><strong>A1</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) attempt to use the point \(\left( {{\text{e, }}{{\text{e}}^2}} \right)\) to determine 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;"><em>eg</em>, \({(\ln {\text{e}})^2}{{\text{e}}^2} = {{\text{e}}^2} - {\text{e}} + {\text{c}}\) or \({{\text{e}}^2} = \frac{{{{\text{e}}^2} - {\text{e}} + {\text{c}}}}{{{{(\ln {\text{e}})}^2}}}\) or \({{\text{e}}^2} = {{\text{e}}^2} - {\text{e}} + {\text{c}}\)</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;">\({\text{c}} = {\text{e}}\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(y = \frac{{{x^2} - x + {\text{e}}}}{{{{(\ln x)}^2}}}\) <em><strong>AG</strong></em></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.5px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) <br><img src="images/Schermafbeelding_2014-09-10_om_10.21.28.png" alt></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;">graph with correct shape <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;">minimum at \(x = 3.1\) (accept answers to a minimum of 2 s.f) <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;">asymptote shown at \(x = 1\) <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> </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>Note: </strong><em>y</em>-coor<span style="font-family: 'times new roman', times; font-size: medium;">dinate of minimum not required for <strong><em>A1</em></strong>;</span></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;"> Equation of asymptote not required for <strong><em>A1 </em></strong>if VA appears on the sketch.</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;"> Award <strong><em>A0 </em></strong>for asymptotes if more than one asymptote are shown</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 30.0px Helvetica; min-height: 36.0px;"> </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>[12 marks]</em></strong></span></p>
<div class="question_part_label">b.</div>
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<h2 style="margin-top: 1em">Examiners report</h2>
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[N/A]
<div class="question_part_label">a.</div>
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<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
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