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</div><h2>HL Paper 1</h2><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the values of <em>x </em>for which the vectors \(\left( {\begin{array}{*{20}{c}}<br> 1 \\ <br> {2\cos x} \\ <br> 0 <br>\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}<br> { - 1} \\ <br> {2\sin x} \\ <br> 1 <br>\end{array}} \right)\) are perpendicular, \(0 \leqslant x \leqslant \frac{\pi }{2}\).</span></p>
</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: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \(\arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right) = \arctan \left( {\frac{1}{p}} \right)\), where \(p \in {\mathbb{Z}^ + }\), find <em>p</em>.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">Hence find the value of \(\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right)\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</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: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Use the identity \(\cos 2\theta = 2{\cos ^2}\theta - 1\) to prove that \(\cos \frac{1}{2}x = \sqrt {\frac{{1 + \cos x}}{2}} ,{\text{ }}0 \leqslant x \leqslant \pi \).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find a similar expression for \(\sin \frac{1}{2}x,{\text{ }}0 \leqslant x \leqslant \pi \).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Hence find the value of \(\int_0^{\frac{\pi }{2}} {\left( {\sqrt {1 + \cos x} + \sqrt {1 - \cos x} } \right){\text{d}}x} \).</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that </span><span style="font-family: times new roman,times; font-size: medium;">\(\frac{{\sin 2\theta }}{{1 + \cos 2\theta }} = \tan \theta \)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Hence find the value of \(\cot \frac{\pi }{8}\) </span><span style="font-family: times new roman,times; font-size: medium;">in the form \(a + b\sqrt 2 \) , where \(a,b \in \mathbb{Z}\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Show that \(\cot \alpha = \tan \left( {\frac{\pi }{2} - \alpha } \right)\) for \(0 < \alpha < \frac{\pi }{2}\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Hence find \(\int_{\tan \alpha }^{\cot \alpha } {\frac{1}{{1 + {x^2}}}{\text{d}}x,{\text{ }}0 < \alpha < \frac{\pi }{2}} \).</p>
<div class="marks">[4]</div>
<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: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In the triangle ABC, \({\text{AB}} = 2\sqrt 3 \) , AC = 9 and \({\rm{B\hat AC}} = 150^\circ \) .</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: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Determine BC, giving your answer in the form \(k\sqrt 3 \), \(k \in {\mathbb{Z}^ + }\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The point D lies on (BC), and (AD) is perpendicular to (BC). Determine AD.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>Let \(z = 1 - \cos 2\theta - {\text{i}}\sin 2\theta ,{\text{ }}z \in \mathbb{C},{\text{ }}0 \leqslant \theta \leqslant \pi \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Solve \(2\sin (x + 60^\circ ) = \cos (x + 30^\circ ),{\text{ }}0^\circ \leqslant x \leqslant 180^\circ \).</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 \(\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the modulus and argument of \(z\) in terms of \(\theta \). Express each answer in its simplest form.</p>
<div class="marks">[9]</div>
<div class="question_part_label">c.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence find the cube roots of \(z\) in modulus-argument form.</p>
<div class="marks">[5]</div>
<div class="question_part_label">c.ii.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the value of \(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}\).</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">Show that \(\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi \) <span class="s1">where \(k \in \mathbb{Z}\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use the principle of mathematical induction to prove that</p>
<p class="p1">\(\sin x + \sin 3x + \ldots + \sin (2n - 1)x = \frac{{1 - \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi \) where \(k \in \mathbb{Z}\).</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">Hence or otherwise solve the equation \(\sin x + \sin 3x = \cos x\) <span class="s1">in the interval \(0 < x < \pi \).</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="specification">
<p style="font: 28px Helvetica; margin: 0px; text-align: justify;"><span style="font-family: 'times new roman', times; font-size: medium;">The diagram shows a tangent, (TP) , to the circle with centre O and radius <em>r</em> . The size of \({\rm{P\hat OA}}\) is \(\theta \) radians.</span></p>
<p style="font: normal normal normal 28px/normal Helvetica; text-align: center; margin: 0px;"> </p>
<p style="font: normal normal normal 28px/normal Helvetica; text-align: center; margin: 0px;"><img src="data:image/png;base64,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" alt></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the area of triangle AOP in terms of <em>r</em> and \(\theta \) .</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">Find the area of triangle POT in terms of <em>r</em> and \(\theta \) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 22.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Using your results from part (a) and part (b), show that \(\sin \theta < \theta < \tan \theta \) .</span></p>
<div class="marks">[2]</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;">The first three terms of a geometric sequence are \(\sin x,{\text{ }}\sin 2x\) and \(4\sin x{\cos ^2}x,{\text{ }} - \frac{\pi }{2} < x < \frac{\pi }{2}\).</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;">(a) Find the common ratio <em>r</em>.</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) Find the set of values of <em>x </em>for which the geometric series \(\sin x + \sin 2x + 4\sin x{\cos ^2}x + \ldots \) converges.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider \(x = \arccos \left( {\frac{1}{4}} \right),{\text{ }}x > 0\).</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;">(c) Show that the sum to infinity of this series is \(\frac{{\sqrt {15} }}{2}\).</span></p>
</div>
<br><hr><br><div class="question">
<p class="p1">Solve the equation \(\sin 2x - \cos 2x = 1 + \sin x - \cos x\) for \(x \in [ - \pi ,{\text{ }}\pi ]\).</p>
</div>
<br><hr><br><div class="specification">
<p class="p1">In triangle \({\text{ABC, BC}} = \sqrt 3 {\text{ cm}}\), \({\rm{A\hat BC}} = \theta \) and \({\rm{B\hat CA}} = \frac{\pi }{3}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that length \({\text{AB}} = \frac{3}{{\sqrt 3 \cos \theta + \sin \theta }}\).</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">Given that \(AB\) has a minimum value, determine the value of \(\theta \) <span class="s1">for which this occurs.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The following diagram shows the triangle ABC where \({\text{AB}} = 2,{\text{ AC}} = \sqrt 2 \) and \({\rm{B\hat AC}} = 15^\circ \).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2017-01-31_om_08.33.57.png" alt="M16/5/MATHL/HP1/ENG/TZ1/05.c"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Expand and simplify \({\left( {1 - \sqrt 3 } \right)^2}\).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">By writing \(15^\circ \) as \(60^\circ - 45^\circ \) find the value of \(\cos (15^\circ )\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1"><span class="s1">Find BC </span>in the form \(a + \sqrt b \) where \(a,{\text{ }}b \in \mathbb{Z}\).</p>
<div class="marks">[4]</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;">The triangle ABC is equilateral of side 3 cm. The point D lies on [BC] such that BD = 1 cm.</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;">Find \(\cos {\rm{D\hat AC}}\).</span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Let \(f(x) = \frac{{\sin 3x}}{{\sin x}} - \frac{{\cos 3x}}{{\cos x}}\)<em>.</em></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: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">For what values of <em>x </em>does \(f(x)\) not exist?</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Simplify the expression \(\frac{{\sin 3x}}{{\sin x}} - \frac{{\cos 3x}}{{\cos x}}\).</span></p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<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;">A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence.</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;">The angle of the largest sector is twice the angle of the smallest sector.</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;">Find the size of the angle of the smallest sector.</span></p>
</div>
<br><hr><br><div class="specification">
<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;">The diagram below shows a circular lake with centre O, diameter AB and radius 2 km.</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;"> </span></p>
<p style="font: normal normal normal 29px/normal Helvetica; text-align: center; margin: 0px;"><img src="data:image/png;base64,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" alt></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;"> </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;">Jorg needs to get from A to B as quickly as possible. He considers rowing to point P and then walking to point B. He can row at \(3{\text{ km}}\,{{\text{h}}^{ - 1}}\) and walk at \(6{\text{ km}}\,{{\text{h}}^{ - 1}}\). Let \({\rm{P\hat AB}} = \theta \) radians, and <em>t</em> be the time in hours taken by Jorg to travel from A to B.</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: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \(t = \frac{2}{3}(2\cos \theta + \theta )\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">Find the value of \(\theta \) for which \(\frac{{{\text{d}}t}}{{{\text{d}}\theta }} = 0\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">What route should Jorg take to travel from A to B in the least amount of time?</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;">Give reasons for your answer.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Write down the expansion of \({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3}\) in the form \(a + {\text{i}}b\) , where \(a\) and \(b\) </span><span style="font-family: times new roman,times; font-size: medium;">are in terms of \({\sin \theta }\) and \({\cos \theta }\) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Hence show that \(\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Similarly show that \(\cos 5\theta = 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta \) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Hence</strong> solve the equation \(\cos 5\theta + \cos 3\theta + \cos \theta = 0\) , where \(\theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">By considering the solutions of the equation \(\cos 5\theta = 0\) , show that </span><span style="font-family: times new roman,times; font-size: medium;">\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \)</span><span style="font-family: times new roman,times; font-size: medium;"> and state the value of \(\cos \frac{{7\pi }}{{10}}\)</span><span style="font-family: times new roman,times; font-size: medium;">.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">e.</div>
</div>
<br><hr><br><div class="question">
<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;">From a vertex of an equilateral triangle of side \(2x\), a circular arc is drawn to divide the triangle into two regions, as shown in the diagram below.</span></p>
<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;"> </span></p>
<p style="font: normal normal normal 25px/normal Helvetica; text-align: center; margin: 0px;"><img src="data:image/png;base64,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" alt></p>
<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;"> </span></p>
<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;">Given that the areas of the two regions are equal, find the radius of the arc in terms of <em>x</em>.</span></p>
</div>
<br><hr><br><div class="specification">
<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;">The angle \(\theta \) lies in the first quadrant and \(\cos \theta = \frac{1}{3}\).</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: 30.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Write down the value of \(\sin \theta \) .</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">Find the value of \(\tan 2\theta \) .</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 29.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the value of \(\cos \left( {\frac{\theta }{2}} \right)\) , giving your answer in the form \(\frac{{\sqrt a }}{b}\) where <em>a</em> , \(b \in {\mathbb{Z}^ + }\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1"><span class="s1">The following diagram shows the curve \(y = a\sin \left( {b(x + c)} \right) + d\), where \(a\)</span>, <span class="s1">\(b\)</span>, <span class="s1">\(c\) and \(d\) </span>are all positive constants. The curve has a maximum point at \((1,{\text{ }}3.5)\) and a minimum point at \((2,{\text{ }}0.5)\).</p>
<p class="p1" style="text-align: center;"><img src="images/Schermafbeelding_2017-01-31_om_07.33.57.png" alt="M16/5/MATHL/HP1/ENG/TZ1/03"></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Write down the value of \(a\) and the value of \(d\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the value of \(b\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find the smallest possible value of \(c\), given \(c > 0\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">The logo, for a company that makes chocolate, is a sector of a circle of radius \(2\) cm, shown as shaded in the diagram. The area of the logo is \(3\pi {\text{ c}}{{\text{m}}^2}\)<span class="s1">.</span></p>
<p class="p1" style="text-align: center;"><span class="s1"><img src="images/Schermafbeelding_2015-12-22_om_11.15.57.png" alt></span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find, in radians, the value of the angle \(\theta \), as indicated on the diagram.</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">Find the total length of the perimeter of the logo.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 36.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In the diagram below, AD is perpendicular to BC.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 36.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">CD = 4, BD = 2 and AD = 3. \({\rm{C}}\hat {\rm{A}}{\rm{D}} = \alpha \) and \({\rm{B}}\hat {\rm{A}}{\rm{D}} = \beta \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 36.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"> </span></p>
<p style="font: normal normal normal 36px/normal Helvetica; text-align: center; margin: 0px;"><img src="data:image/png;base64,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" alt></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 36.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the exact value of \(\cos (\alpha - \beta )\) .</span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the curve defined by the equation \({x^2} + \sin y - xy = 0\) .</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: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the gradient of the tangent to the curve at the point \((\pi ,{\text{ }}\pi )\) .</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: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Hence, show that \(\tan \theta = \frac{1}{{1 + 2\pi }}\), where \(\theta \) is the acute angle between the tangent to the curve at \((\pi ,{\text{ }}\pi )\) and the line <em>y </em>= <em>x </em>.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that \(\frac{{\cos A + \sin A}}{{\cos A - \sin A}} = \sec 2A + \tan 2A\) .</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) Prove the trigonometric identity \(\sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\).</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) Given \(f(x) = \sin({x + \frac{\pi }{6}})\sin({x - \frac{\pi }{6}}),{\text{ }}x \in \left[ {0,{\text{ }}\pi } \right]\), find the range of \(f\).</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) Given \(g(x) = \csc( {x + \frac{\pi }{6}})\csc( {x - \frac{\pi }{6}}),{\text{ }}x \in \left[ {0,{\text{ }}\pi } \right],{\text{ }}x \ne \frac{\pi }{6},{\text{ }}x \ne \frac{{5\pi }}{6}\), find the range of \(g\).</span></p>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the equation \(\frac{{\sqrt 3 - 1}}{{\sin x}} + \frac{{\sqrt 3 + 1}}{{\cos x}} = 4\sqrt 2 ,{\text{ }}0 < x < \frac{\pi }{2}\). Given that \(\sin \left( {\frac{\pi }{{12}}} \right) = \frac{{\sqrt 6 - \sqrt 2 }}{4}\) and \(\cos \left( {\frac{\pi }{{12}}} \right) = \frac{{\sqrt 6 + \sqrt 2 }}{4}\)</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">verify that \(x = \frac{\pi }{{12}}\) is a solution to the equation;</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">hence find the other solution to the equation for \(0 < x < \frac{\pi }{2}\).</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the functions \(f(x) = \tan x,{\text{ }}0 \le \ x < \frac{\pi }{2}\) and \(g(x) = \frac{{x + 1}}{{x - 1}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne 1\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Find an expression for \(g \circ f(x)\), stating its domain.</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">Hence show that \(g \circ f(x) = \frac{{\sin x + \cos x}}{{\sin x - \cos x}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Let \(y = g \circ f(x)\)<span class="s1">, find an exact value for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) </span>at the point on the graph of \(y = g \circ f(x)\) where \(x = \frac{\pi }{6}\), expressing your answer in the form \(a + b\sqrt 3 ,{\text{ }}a,{\text{ }}b \in \mathbb{Z}\).</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>Show that the area bounded by the graph of \(y = g \circ f(x)\), the \(x\)-axis and the lines \(x = 0\) and \(x = \frac{\pi }{6}\) is \(\ln \left( {1 + \sqrt 3 } \right)\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<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 diagram below shows the boundary of the cross-section of a water channel.</span></p>
<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;"> </span></p>
<p style="font: normal normal normal 25px/normal Helvetica; text-align: center; margin: 0px;"><img 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" alt></p>
<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;"> </span></p>
<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 equation that represents this boundary is \(y = 16\sec \left( {\frac{{\pi x}}{{36}}} \right) - 32\) where <em>x</em> and <em>y</em> are both measured in cm.</span></p>
<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 top of the channel is level with the ground and has a width of 24 cm. The maximum depth of the channel is 16 cm.</span></p>
<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;">Find the width of the water surface in the channel when the water depth is 10 cm.</span></p>
<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;">Give your answer in the form \(a\arccos b\) where \(a,{\text{ }}b \in \mathbb{R}\) .</span></p>
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<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">In the triangle ABC, </span><span style="font-family: 'times new roman', times; font-size: medium;"><span style="font-family: 'times new roman', times; font-size: medium;">\({\rm{A\hat BC}} = 90^\circ\)</span> , \({\text{AC}} = \sqrt {\text{2}}\) and AB = BC + 1.</span></p>
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<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that cos \(\hat A - \sin \hat A = \frac{1}{{\sqrt 2 }}\).</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;">By squaring both sides of the equation in part (a), solve the equation to find the angles in the triangle.</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><span style="font-family: 'times new roman', times; font-size: medium;">Apply Pythagoras’ theorem in the triangle ABC to find BC, and hence show that \(\sin \hat A = \frac{{\sqrt 6 - \sqrt 2 }}{4}\).</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;">Hence, or otherwise, calculate the length of the perpendicular from B to [AC].</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<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;">The diagram below shows a curve with equation \(y = 1 + k\sin x\) , defined for \(0 \leqslant x \leqslant 3\pi \) .</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;"> </span></p>
<p style="font: normal normal normal 29px/normal Helvetica; text-align: center; margin: 0px;"><img src="data:image/png;base64,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" alt></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;"> </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;">The point \({\text{A}}\left( {\frac{\pi }{6}, - 2} \right)\) lies on the curve and \({\text{B}}(a,{\text{ }}b)\) is the maximum point.</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;">(a) Show that <em>k</em> = – 6 .</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;">(b) Hence, find the values of <em>a</em> and <em>b</em> .</span></p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Show that \({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = \frac{{2\cos n\theta }}{{{{\cos }^n}\theta }},\;\;\;\cos \theta \ne 0\).</p>
<p>(ii) Hence verify that \({\text{i}}\tan \frac{{3\pi }}{8}\) is a root of the equation \({(1 + z)^4} + {(1 - z)^4} = 0,\;\;\;z \in \mathbb{C}\).</p>
<p>(iii) State another root of the equation \({(1 + z)^4} + {(1 - z)^4} = 0,\;\;\;z \in \mathbb{C}\).</p>
<div class="marks">[10]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>(i) Use the double angle identity \(\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\) to show that \(\tan \frac{\pi }{8} = \sqrt 2 - 1\).</p>
<p>(ii) Show that \(\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1\).</p>
<p>(iii) Hence find the value of \(\int_0^{\frac{\pi }{8}} {\frac{{2\cos 4x}}{{{{\cos }^2}x}}{\text{d}}x} \).</p>
<div class="marks">[13]</div>
<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: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">In the triangle PQR, PQ = 6 , PR = <em>k </em>and <span style="font-family: 'times new roman', times; font-size: medium;">\({\rm{P\hat QR}} = 30^\circ \) .</span></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: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">For the case <em>k </em>= 4 , find the two possible values of QR.</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: 12.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Determine the values of <em>k </em>for which the conditions above define a unique triangle.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">The following diagram shows a sector of a circle where \({\rm{A\hat OB}} = x\) radians and the length of the \({\text{arc AB}} = \frac{2}{x}{\text{ cm}}\).</p>
<p class="p1">Given that the area of the sector is \(16{\text{ c}}{{\text{m}}^2}\), find the length of the arc \(AB\)<span class="s1">.</span></p>
<p class="p1" style="text-align: center;"><span class="s1"><img src="images/Schermafbeelding_2016-01-28_om_15.11.09.png" alt></span></p>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \(\frac{\pi }{2} < \alpha < \pi \) and \(\cos \alpha = - \frac{3}{4}\), find the value of sin 2<span style="font: 12.5px Times;">α </span>.</span></p>
</div>
<br><hr><br><div class="question">
<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;">The vectors <strong><em>a</em></strong> , <strong><em>b</em></strong> , <strong><em>c</em></strong> satisfy the equation <strong><em>a</em></strong> + <strong><em>b</em></strong> + <strong><em>c</em></strong> = <strong>0</strong> . Show that <strong><em>a</em></strong> \( \times \) <strong><em>b</em></strong> = <strong><em>b</em></strong> \( \times \) <strong><em>c</em></strong> = <strong><em>c</em></strong> \( \times \) <strong><em>a</em></strong> .</span></p>
</div>
<br><hr><br><div class="specification">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Consider the following functions:</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;"> \(h(x) = \arctan (x),{\text{ }}x \in \mathbb{R}\)</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;"> \(g(x) = \frac{1}{x}\), \(x\in \mathbb{R}\)</span><span style="font-family: 'times new roman', times; font-size: medium; background-color: #f7f7f7;">, \({\text{ }}x \ne 0\)</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Sketch the graph of \(y = h(x)\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Find an expression for the composite function \(h \circ g(x)\) and state its domain.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Given that \(f(x) = h(x) + h \circ g(x)\),</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;">(i) find \(f'(x)\) in simplified form;</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 \(f(x) = \frac{\pi }{2}\) for \(x > 0\).</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">c.</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;">Nigel states that \(f\) is an odd function and Tom argues that \(f\) is an even function.</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) State who is correct and justify your answer.</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 value of \(f(x)\) for \(x < 0\).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">d.</div>
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<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\sin \left( {\theta + \frac{\pi }{2}} \right) = \cos \theta \).</p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
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<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Consider \(f(x) = \sin (ax)\) where \(a\) is a constant. Prove by mathematical induction that \({f^{(n)}}(x) = {a^n}\sin \left( {ax + \frac{{n\pi }}{2}} \right)\) where \(n \in {\mathbb{Z}^ + }\) and \({f^{(n)}}(x)\) represents the \({{\text{n}}^{{\text{th}}}}\) derivative of \(f(x)\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
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<br><hr><br><div class="question">
<p>Solve the equation \({\sec ^2}x + 2\tan x = 0,{\text{ }}0 \leqslant x \leqslant 2\pi \).</p>
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<br><hr><br><div class="question">
<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) Show that \(\sin 2nx = \sin \left( {(2n + 1)x} \right)\cos x - \cos \left( {(2n + 1)x} \right)\sin x\).</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;">(b) <strong>Hence</strong> prove, by induction, that</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;">\[\cos x + \cos 3x + \cos 5x + \ldots + \cos \left( {(2n - 1)x} \right) = \frac{{\sin 2nx}}{{2\sin x}},\]</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;">for all \(n \in {\mathbb{Z}^ + }{\text{, }}\sin x \ne 0\).</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;">(c) Solve the equation \(\cos x + \cos 3x = \frac{1}{2},{\text{ }}0 < x < \pi \).</span></p>
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<br><hr><br><div class="question">
<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;">A triangle has sides of length \(({n^2} + n + 1)\), \((2n + 1)\) and \(({n^2} - 1)\) where \(n > 1\).</span></p>
<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;">(a) Explain why the side \(({n^2} + n + 1)\) must be the longest side of the triangle.</span></p>
<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;">(b) Show that the largest angle, \(\theta \), of the triangle is \(120^\circ \).</span></p>
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<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;">Given that \(\sin x + \cos x = \frac{2}{3}\), find \(\cos 4x\).</span></p>
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<br><hr><br><div class="question">
<p>Find all solutions to the equation \(\tan x + \tan 2x = 0\) where \(0^\circ \le x < 360^\circ\).</p>
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<br><hr><br><div class="question">
<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 diagram below shows two straight lines intersecting at O and two circles, each with centre O. The outer circle has radius <em>R</em> and the inner circle has radius <em>r</em> .</span></p>
<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;"> </span></p>
<p style="font: normal normal normal 25px/normal Helvetica; text-align: center; margin: 0px;"><img 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" alt></p>
<p style="font: normal normal normal 25px/normal Helvetica; text-align: center; margin: 0px;"> </p>
<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;">Consider the shaded regions with areas <em>A</em> and <em>B</em> . Given that \(A:B = 2:1\), find the <strong>exact</strong> value of the ratio \(R:r\) .</span></p>
</div>
<br><hr><br><div class="specification">
<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;">Consider the following system of equations:</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;">\[x + y + z = 1\]</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;">\[2x + 3y + z = 3\]</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;">\[x + 3y - z = \lambda \]</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;">where \(\lambda \in \mathbb{R}\) .</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: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Show that this system does not have a unique solution for any value of \(\lambda \) .</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: 28.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Determine the value of \(\lambda \) for which the system is consistent.</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;">(ii) For this value of \(\lambda \) , find the general solution of the system.</span></p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Sketch the graph of \(y = \left| {\cos \left( {\frac{x}{4}} \right)} \right|\) for \(0 \leqslant x \leqslant 8\pi \).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Solve \(\left| {\cos \left( {\frac{x}{4}} \right)} \right| = \frac{1}{2}\) for \(0 \leqslant x \leqslant 8\pi \).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<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;">In triangle ABC, AB = 9 cm , AC = 12 cm , and \(\hat B\) is twice the size of \({\hat C}\) .</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;">Find the cosine of \({\hat C}\) .</span></p>
</div>
<br><hr><br><div class="specification">
<p>Consider \(w = 2\left( {{\text{cos}}\frac{\pi }{3} + {\text{i}}\,{\text{sin}}\frac{\pi }{3}} \right)\)</p>
</div>
<div class="specification">
<p>These four points form the vertices of a quadrilateral, <em>Q</em>.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Express <em>w</em><sup>2</sup> and <em>w</em><sup>3</sup> in modulus-argument form.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Sketch on an Argand diagram the points represented by <em>w</em><sup>0</sup> , <em>w</em><sup>1</sup> , <em>w</em><sup>2</sup> and <em>w</em><sup>3</sup>.</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>Show that the area of the quadrilateral <em>Q</em> is \(\frac{{21\sqrt 3 }}{2}\).</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Let \(z = 2\left( {{\text{cos}}\frac{\pi }{n} + {\text{i}}\,{\text{sin}}\frac{\pi }{n}} \right),\,\,n \in {\mathbb{Z}^ + }\). The points represented on an Argand diagram by \({z^0},\,\,{z^1},\,\,{z^2},\, \ldots \,,\,\,{z^n}\) form the vertices of a polygon \({P_n}\).</p>
<p>Show that the area of the polygon \({P_n}\) can be expressed in the form \(a\left( {{b^n} - 1} \right){\text{sin}}\frac{\pi }{n}\), where \(a,\,\,b\, \in \mathbb{R}\).</p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p>Let \(a = {\text{sin}}\,b,\,\,0 < b < \frac{\pi }{2}\).</p>
<p>Find, in terms of <em>b</em>, the solutions of \({\text{sin}}\,2x = - a,\,\,0 \leqslant x \leqslant \pi \).</p>
</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: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find all values of <em>x</em> for \(0.1 \leqslant x \leqslant 1\) such that \(\sin (\pi {x^{ - 1}}) = 0\).</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 style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find \(\int_{\frac{1}{{n + 1}}}^{\frac{1}{n}} {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}}){\text{d}}x} \), showing that it takes different integer values when <em>n</em> is even and when <em>n</em> is odd.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">Evaluate \(\int_{0.1}^1 {\left| {\pi {x^{ - 2}}\sin (\pi {x^{ - 1}})} \right|{\text{d}}x} \).</span></p>
<div class="marks">[2]</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: 34.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">If <em>x</em> satisfies the equation \(\sin \left( {x + \frac{\pi }{3}} \right) = 2\sin x\sin \left( {\frac{\pi }{3}} \right)\), show that \(11\tan x = a + b\sqrt 3 \), where <em>a</em>, <em>b</em> \( \in {\mathbb{Z}^ + }\).</span></p>
</div>
<br><hr><br><div class="specification">
<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;">Given that \(f(x) = 1 + \sin x,{\text{ }}0 \leqslant x \leqslant \frac{{3\pi }}{2}\),</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: 31.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">sketch the graph of \(f\);</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;"> </span></p>
<p style="font: normal normal normal 31px/normal Helvetica; text-align: center; margin: 0px;"><img src="data:image/png;base64,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" alt></p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">show that \({\left( {f(x)} \right)^2} = \frac{3}{2} + 2\sin x - \frac{1}{2}\cos 2x\);</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">find the volume of the solid formed when the graph of <em>f</em> is rotated through \(2\pi \) radians about the <em>x</em>-axis.</span></p>
<div class="marks">[4]</div>
<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: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">The function <em>f </em>is defined on the domain \(\left[ {0,\,\frac{{3\pi }}{2}} \right]\) by \(f(x) = {e^{ - x}}\cos x\) .</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: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">State the two zeros of <em>f </em>.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">Sketch the graph of <em>f </em>.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">The region bounded by the graph, the <em>x</em>-axis and the <em>y</em>-axis is denoted by <em>A </em>and the region bounded by the graph and the <em>x</em>-axis is denoted by <em>B </em>. Show that the ratio of the area of <em>A </em>to the area of <em>B </em>is</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px Times;"><span style="font-family: 'times new roman', times; font-size: medium;">\[\frac{{{e^\pi }\left( {{e^{\frac{\pi }{2}}} + 1} \right)}}{{{e^\pi } + 1}}.\]</span></p>
<div class="marks">[7]</div>
<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: 25.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The function <em>f</em> is defined by \(f(x) = \frac{1}{{4{x^2} - 4x + 5}}\).</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: 26.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Express \(4{x^2} - 4x + 5\) in the form \(a{(x - h)^2} + k\) where <em>a</em>, <em>h</em>, \(k \in \mathbb{Q}\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The graph of \(y = {x^2}\) is transformed onto the graph of \(y = 4{x^2} - 4x + 5\). Describe a sequence of transformations that does this, making the order of transformations clear.</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">Sketch the graph of \(y = f(x)\).</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p 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;">Find the range of <em>f</em>.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">By using a suitable substitution show that \(\int {f(x){\text{d}}x = \frac{1}{4}\int {\frac{1}{{{u^2} + 1}}{\text{d}}u} } \).</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">e.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 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;">Prove that \(\int_1^{3.5} {\frac{1}{{4{x^2} - 4x + 5}}{\text{d}}x = \frac{\pi }{{16}}} \).</span></p>
<div class="marks">[7]</div>
<div class="question_part_label">f.</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: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) Sketch the graphs of \(y = \sin x\) and \(y = \sin 2x\) , on the same set of axes, for \(0 \leqslant x \leqslant \frac{\pi }{2}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) Find the x-coordinates of the points of intersection of the graphs in the domain \(0 \leqslant x \leqslant \frac{\pi }{2}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(iii) Find the area enclosed by the graphs.</span></p>
<div class="marks">[9]</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: 24.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">Find the value of \(\int_0^1 {\sqrt {\frac{x}{{4 - x}}} }{{\text{d}}x} \) using the substitution \(x = 4{\sin ^2}\theta \) .</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 style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">The increasing function <em>f</em> satisfies \(f(0) = 0\) and \(f(a) = b\) , where \(a > 0\) and \(b > 0\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(i) By reference to a sketch, show that \(\int_0^a {f(x){\text{d}}x = ab - \int_0^b {{f^{ - 1}}(x){\text{d}}x} } \) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 20.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(ii) <strong>Hence</strong> find the value of \(\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x} \) .</span></p>
<div class="marks">[8]</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: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Show that \(\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{3}} \right) = \frac{\pi }{4}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 23.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) Hence, or otherwise, find the value of \(\arctan (2) + \arctan (3)\) .</span></p>
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<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) Sketch the curve \(f(x) = \sin 2x\) , \(0 \leqslant x \leqslant \pi \) .</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) Hence sketch on a separate diagram the graph of \(g(x) = \csc 2x\) , \(0 \leqslant x \leqslant \pi \) , clearly stating the coordinates of any local maximum or minimum points and the equations of any asymptotes.</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) Show that tan \(x + \cot x \equiv 2\csc 2x\) .</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;">(d) Hence or otherwise, find the coordinates of the local maximum and local minimum points on the graph of \(y = \tan 2x + \cot 2x\) , \(0 \leqslant x \leqslant \frac{\pi }{2}\) .</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(e) Find the solution of the equation \(\csc 2x = 1.5\tan x - 0.5\) , \(0 \leqslant x \leqslant \frac{\pi }{2}\) .</span></p>
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