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</div><h2>SL Paper 1</h2><div class="specification">
<p>Let \(f\,{\text{:}}\,\mathbb{R} \times \mathbb{R} \to \mathbb{R} \times \mathbb{R}\) be defined by \(f\left( {x,\,y} \right) = \left( {x + 3y,\,2x - y} \right)\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Given that <em>A</em> is the interval \(\left\{ {x\,{\text{:}}\,0 \leqslant x \leqslant 3} \right\}\) and <em>B</em> is the interval \(\left\{ {y\,{\text{:}}\,0 \leqslant x \leqslant 4} \right\}\) then describe <em>A</em> × <em>B</em> in geometric form.</p>
<div class="marks">[3]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that the function \(f\) is a bijection.</p>
<div class="marks">[8]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Hence find the inverse function \({f^{ - 1}}\).</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><em>A</em> × <em>B</em> is a rectangle <em><strong>A1</strong></em></p>
<p>vertices at (0, 0), (3, 0), (0, 4) and (3, 4) or equivalent description <em><strong>A1</strong></em></p>
<p>and its interior <em><strong>A1</strong></em></p>
<p><strong>Note:</strong> Accept diagrammatic answers.</p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>need to prove it is injective and surjective <em><strong>R1</strong></em></p>
<p>need to show if \(f\left( {x,\,y} \right) = f\left( {u,\,v} \right)\) then \(\left( {x,\,y} \right) = \left( {u,\,v} \right)\) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow x + 3y = u + 3v\)</p>
<p>\(2x - y = 2u - v\) <em><strong>A1</strong></em></p>
<p>Equation 2 – 2 Equation 1 \( \Rightarrow y = v\)</p>
<p>Equation 1 + 3 Equation 2 \( \Rightarrow x = u\) <em><strong>A1</strong></em></p>
<p>thus \(\left( {x,\,y} \right) = \left( {u,\,v} \right) \Rightarrow f\) is injective</p>
<p>let \(\left( {s,\,t} \right)\) be any value in the co-domain \(\mathbb{R} \times \mathbb{R}\)</p>
<p>we must find \(\left( {x,\,y} \right)\) such that \(f\left( {x,\,y} \right) = \left( {s,\,t} \right)\) <em><strong>M1</strong></em></p>
<p>\(s = x + 3y\) and \(t = 2x - y\) <em><strong>M1</strong></em></p>
<p>\( \Rightarrow y = \frac{{2s - t}}{7}\) <em><strong>A1</strong></em></p>
<p>and \(x = \frac{{s + 3t}}{7}\) <em><strong>A1</strong></em></p>
<p>hence \(f\left( {x,\,y} \right) = \left( {s,\,t} \right)\) and is therefore surjective</p>
<p><em><strong>[8 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\({f^{ - 1}}\left( {x,\,y} \right) = \left( {\frac{{x + 3y}}{7},\,\frac{{2x - y}}{7}} \right)\) <em><strong>A1A1</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The group \(\left\{ {G,\left. * \right\}} \right.\) is defined on the set \(G = \left\{ {1,2,3,4,5,\left. 6 \right\}} \right.\) where \( * \) denotes </span><span style="font-family: times new roman,times; font-size: medium;">multiplication modulo \(7\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Draw the Cayley table for \(\left\{ {G,\left. * \right\}} \right.\) .</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) Determine the order of each element of \(\left\{ {G,\left. * \right\}} \right.\) .</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">(ii) Find all the proper subgroups of \(\left\{ {G,\left. * \right\}} \right.\) .</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Solve the equation \(x * 6 * x = 3\) where \(x \in G\) .</span></p>
<div class="marks">[3]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><br><img src="images/dizzy.png" alt></span><em><strong><span style="font-family: times new roman,times; font-size: medium;"> A3</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A2</strong></em> for 1 error, <em><strong>A1</strong></em> for 2 errors, <em><strong>A0</strong></em> for 3 or more errors.</span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) We first identify \(1\) as the identity <strong><em>(A1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Order of \(1 = 1\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Order of \(2 = 3\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Order of \(3 = 6\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Order of \(4 = 3\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Order of \(5 = 6\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Order of \(6 = 2\) <strong><em>A3</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A2</strong></em> for 1 error, <em><strong>A1</strong> </em>for 2 errors, <em><strong>A0</strong></em> for more than 2 errors.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) \(\left\{ {1,\left. 6 \right\}} \right.\) ; \(\left\{ {1,\left. {2,4} \right\}} \right.\) <strong><em>A1A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The equation is equivalent to</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(6 * x * x = 3\) <strong><em> M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x * x = 4\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(x = 2\) or \(5\) <strong><em>A1A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">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>Let \(G\) denote the set of \(2 \times 2\) matrices whose elements belong to \(\mathbb{R}\) and whose determinant is equal to 1. Let \( * \) denote matrix multiplication which may be assumed to be associative.</p>
</div>
<div class="specification">
<p>Let \(H\) denote the set of \(2 \times 2\) matrices whose elements belong to \(\mathbb{Z}\) and whose determinant is equal to 1.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that \(\{ G,{\text{ }} * \} \) is a group.</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>Determine whether or not \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \).</p>
<div class="marks">[4]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>closure: let <strong><em>A</em></strong>, <strong><em>B </em></strong>\( \in G\)</p>
<p>(because <strong><em>AB </em></strong>is a \(2 \times 2\) matrix)</p>
<p>and det(<strong><em>AB</em></strong>) = det(<strong><em>A</em></strong>)det(<strong><em>B</em></strong>) \( = 1 \times 1 = 1\) <strong><em>M1A1</em></strong></p>
<p>identity: the \(2 \times 2\) identity matrix has determinant 1 <strong><em>R1</em></strong></p>
<p>inverse: let <strong><em>A</em></strong> \( \in G\). Then <strong><em>A </em></strong>has an inverse because it is non-singular <strong><em>(R1)</em></strong></p>
<p>since <strong><em>AA</em></strong>\(^{ - 1} = \) <strong><em>I</em></strong>, det(<strong><em>A</em></strong>)det(<strong><em>A</em></strong>\(^{ - 1}\)) = det(<strong><em>I</em></strong>) = 1 therefore <strong><em>A</em></strong>\(^{ - 1} \in G\) <strong><em>R1</em></strong></p>
<p>associativity is assumed</p>
<p>the four axioms are satisfied therefore \(\{ G,{\text{ }} * \} \) is a group <strong><em>AG</em></strong></p>
<p><strong><em>[5 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>closure: let <strong><em>A</em></strong>, <strong><em>B</em></strong> \( \in H\). Then <strong><em>AB</em></strong> \( \in H\) because the arithmetic involved produces elements that are integers <strong><em>R1</em></strong></p>
<p>inverse: <strong><em>A</em></strong>\(^{ - 1} \in H\) because the calculation of the inverse involves interchanging the elements and dividing by the determinant which is 1 <strong><em>R1</em></strong></p>
<p>the identity (and associativity) follow as above <strong><em>R1</em></strong></p>
<p>therefore \(\{ H,{\text{ }} * \} \) is a subgroup of \(\{ G,{\text{ }} * \} \) <strong><em>A1</em></strong></p>
<p> </p>
<p><strong>Note:</strong> Award the <strong><em>A1 </em></strong>only if the first two <strong><em>R1 </em></strong>marks are awarded but not necessarily the third <strong><em>R1</em></strong>.</p>
<p> </p>
<p><strong>Note:</strong> Accept subgroup test.</p>
<p> </p>
<p><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>The relations \({\rho _1}\) and \({\rho _2}\) are defined on the Cartesian plane as follows</p>
<p>\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Leftrightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2\)</p>
<p>\(({x_1},{\text{ }}{y_1}){\rho _2}({x_2},{\text{ }}{y_2}) \Leftrightarrow \sqrt {x_1^2 + x_2^2} \leqslant \sqrt {y_1^2 + y_2^2} \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">For \({\rho _1}\) and \({\rho _2}\) determine whether or not each is reflexive, symmetric and transitive.</p>
<div class="marks">[11]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">For each of \({\rho _1}\) and \({\rho _2}\) which is an equivalence relation, describe the equivalence classes.</p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\({\rho _1}\)</p>
<p>\(({x_1},{\text{ }}{y_1}){\rho _1}({x_1},{\text{ }}{y_1}) \Rightarrow 0 = 0\;\;\;\)hence reflexive. <strong><em>R1</em></strong></p>
<p>\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2\)</p>
<p>\( \Rightarrow (x_1^2 - x_2^2) = - (y_1^2 - y_2^2)\)</p>
<p>\( \Rightarrow x_2^2 - x_1^2 = y_2^2 - y_1^2 \Rightarrow ({x_2},{\text{ }}{y_2}){\rho _1}({x_1},{\text{ }}{y_1})\;\;\;\)hence symmetric <strong><em>M1A1</em></strong></p>
<p>\(({x_1},{\text{ }}{y_1}){\rho _1}({x_2},{\text{ }}{y_2}) \Rightarrow x_1^2 - x_2^2 = y_1^2 - y_2^2{\text{ - i}}\)</p>
<p>\(({x_2},{\text{ }}{y_2}){\rho _1}({x_3},{\text{ }}{y_3}) \Rightarrow x_2^2 - x_3^2 = y_2^2 - y_3^2{\text{ - ii}}\) <strong><em>M1</em></strong></p>
<p>\({\text{i}} + {\text{ii}} \Rightarrow x_1^2 - x_3^2 = y_1^2 - y_3^2 \Rightarrow ({x_1},{\text{ }}{y_1}){\rho _1}({x_3},{\text{ }}{y_3})\;\;\;\)hence transitive <strong><em>A1</em></strong></p>
<p>\({\rho _2}\)</p>
<p>\(({x_1},{\text{ }}{y_1}){\rho _2}({x_1},{\text{ }}{y_1}) \Rightarrow \sqrt {2x_1^2} \leqslant \sqrt {2y_1^2} \;\;\;\)This is not true in the case of (3,1)</p>
<p>hence not reflexive. <strong><em>R1</em></strong></p>
<p>\(({x_1},{\text{ }}{y_1}){\rho _2}({x_2},{\text{ }}{y_2}) \Rightarrow \sqrt {x_1^2 + x_2^2} \leqslant \sqrt {y_1^2 + y_2^2} \)</p>
<p>\( \Rightarrow \sqrt {x_2^2 + x_1^2} \leqslant \sqrt {y_2^2\_y_1^2} \Rightarrow ({x_2},{\text{ }}{x_2}){\rho _2}({x_1},{\text{ }}{y_1})\;\;\;\)hence symmetric. <strong><em>A1</em></strong></p>
<p>it is not transitive. <strong><em>A1</em></strong></p>
<p>attempt to find a counterexample (<strong><em>M1)</em></strong></p>
<p>for example \((1,{\text{ }}0){\rho _2}(0,{\text{ 1)}}\) and \((0,{\text{ }}1){\rho _2}(1,{\text{ 0)}}\) <strong><em>A1</em></strong></p>
<p>however, it is not true that \((1,{\text{ }}0){\rho _2}(1,{\text{ 0)}}\) <strong><em>A1</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">\({\rho _1}\) is an equivalence relation <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">the equivalence classes for \({\rho _1}\) form a family of curves of the form</p>
<p class="p1">\({y^2} - {x^2} = k\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</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 class="p1">Most candidates attempted this question with many showing correctly that \({\rho _1}\) is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that \({\rho _2}\) is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">Most candidates attempted this question with many showing correctly that \({\rho _1}\) is an equivalence relation. Most candidates, however, were unable to find a counterexample to show that \({\rho _2}\) is not transitive although many suspected that was the case. Most candidates were unable to describe the equivalence classes.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>The permutation \(P\) is given by</p>
<p>\[P = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 3&4&5&6&2&1 \end{array}} \right).\]</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Determine the order of \(P\), justifying your answer.</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>Find \({P^2}\).</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>The permutation group \(G\) is generated by \(P\). Determine the element of \(G\) that is of order 2, giving your answer in cycle notation.</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>the order is 6 <strong><em>A1</em></strong></p>
<p>tracking 1 through successive powers of \(P\) returns to 1 after 6 transitions (or equivalent) <strong><em>R1</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>\({P^2} = (1{\text{ }}5{\text{ }}4)(2{\text{ }}6{\text{ }}3){\text{ or }}\left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 5&6&2&1&4&3 \end{array}} \right)\) <strong><em>(M1)A1</em></strong></p>
<p><strong><em>[2 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>since \(P\) is of order 6, \({P^3}\) will be of order 2 <strong><em>R1</em></strong></p>
<p>\({P^3} = \left( {\begin{array}{*{20}{c}} 1&2&3&4&5&6 \\ 2&1&4&3&6&5 \end{array}} \right)\) <strong><em>(M1)(A1)</em></strong></p>
<p>\({P^3} = (1{\text{ }}2)(3{\text{ }}4)(5{\text{ }}6)\) <strong><em>A1</em></strong></p>
<p><strong><em>[4 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 class="p1">The set \(P\) contains all prime numbers less than 2500.</p>
<p class="p1">The set \(Q\) is the set of all subsets of \(P\).</p>
</div>
<div class="specification">
<p class="p1">The set \(S\) contains all positive integers less than <span class="s1">2500</span>.</p>
<p class="p1">The function \(f:{\text{ }}S \to Q\) is defined by \(f(s)\) as the set of primes exactly dividing \(s\), for \(s \in S\).</p>
<p class="p1">For example \(f(4) = \{ 2\} ,{\text{ }}f(45) = \{ 3,{\text{ }}5\} \).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Explain why only one of the following statements is true</p>
<p class="p2">(i) <span class="Apple-converted-space"> \(17 \subset P\)</span><span class="s1">;</span></p>
<p class="p2">(ii) <span class="Apple-converted-space"> \(\{ 7,{\text{ }}17,{\text{ }}37,{\text{ }}47,{\text{ }}57\} \in Q\)</span><span class="s1">;</span></p>
<p class="p2">(iii) <span class="Apple-converted-space"> \(\phi \subset Q\)</span> and \(\phi \in Q\), where \(\phi \) is the empty set.</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"><span class="s1">(i) <span class="Apple-converted-space"> </span>State the value of \(f(1)\), </span>giving a reason for your answer.</p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span>Find \(n\left( {f(2310)} \right)\).</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 class="p1">Determine whether or not \(f\) <span class="s1">is</span></p>
<p class="p2">(i) <span class="Apple-converted-space"> </span>injective;</p>
<p class="p3">(ii) <span class="Apple-converted-space"> </span>surjective.</p>
<div class="marks">[4]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="s1">(i) <span class="Apple-converted-space"> </span>17 </span>is an element not a subset of \(P\) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1"><span class="s1">(ii) <span class="Apple-converted-space"> </span>57 </span>is not a prime number <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">(iii) <span class="Apple-converted-space"> </span>any demonstration that this is the true statement <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">because every set contains the empty set as a subset <span class="Apple-converted-space"> </span><strong><em>R1</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 class="p1">(i) <span class="Apple-converted-space"> \(f(1) = \phi \)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="s1">because 1 </span>has no prime factors <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> \(f(2310) = f(2 \times 3 \times 5 \times 7 \times 11){\text{ }}\left( { = \{ 2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11\} } \right)\)</span> <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="Apple-converted-space">\(n\left( {f(2310)} \right) = 5\) </span><strong><em>A1</em></strong></p>
<p class="p1"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>not injective <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">because, for example, \(f(2) = f(4) = \{ 2\} \) <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>not surjective <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p2">\({f^{ - 1}}(2,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}11,{\text{ }}13)\) does not belong to S because</p>
<p class="p2"><span class="Apple-converted-space">\(2 \times 3 \times 5 \times 7 \times 11 \times 13 > 2500\) </span><span class="s1"><strong><em>R1</em></strong></span></p>
<p class="p3"> </p>
<p class="p1"><strong>Note: <span class="Apple-converted-space"> </span></strong>Accept any appropriate example.</p>
<p class="p3"> </p>
<p class="p1"><strong><em>[4 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;">
<p class="p1">The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">The question caused a number of problems for candidates. In part (a) a number of candidates thought part (i) was correct as they did not realise it was an element and a number thought part (ii) was correct as they did not recognise 57 as a prime number. In both of these two cases, candidates then suggested part (iii) was false giving a variety of incorrect justifications. Part (b) was more successful for most candidates with many wholly correct answers seen. Part (c) again saw many correct answers, but some candidates tried to argue the opposite, incorrect viewpoint or in other cases gave no reason for their decisions, showing a complete misunderstanding of the command term “determine”.</p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The group \(\{ G,{\text{ }} * \} \) has a subgroup \(\{ H,{\text{ }} * \} \). The relation \(R\) is defined, for \(x,{\text{ }}y \in G\), by \(xRy\) if and only if \({x^{ - 1}} * y \in H\).</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) Show that \(R\) is an equivalence relation.</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) Given that \(G = \{ 0,{\text{ }} \pm 1,{\text{ }} \pm 2,{\text{ }} \ldots \} \), \(H = \{ 0,{\text{ }} \pm 4,{\text{ }} \pm 8,{\text{ }} \ldots \} \) and \( * \) denotes addition, find the equivalence class containing the number \(3\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) \(\underline {{\text{reflexive}}} \)</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;">\({x^{ - 1}}x = e \in H\) <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;">therefore \(xRx\) and \(R\) is reflexive <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\underline {{\text{symmetric}}} \)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Accept the word commutative.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">let \(xRy\) so that \({x^{ - 1}}y \in H\) <strong><em>M1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the inverse of \({x^{ - 1}}y\) is \({y^{ - 1}}x \in H\) <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;">therefore \(yRx\) and \(R\) is symmetric <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\underline {{\text{transitive}}} \)</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;">let \(xRy\) and \(yRz\) so \({x^{ - 1}}y \in H\) and \({y^{ - 1}}z \in H\) <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;">therefore \({x^{ - 1}}y\,{y^{ - 1}}z = {x^{ - 1}}z \in H\) <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;">therefore \(xRz\) and \(R\) is transitive <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">hence \(R\) is an equivalence relation <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>[8 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) the identity is \(0\) so the inverse of \(3\) is \(-3\) <strong><em>(R1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">the equivalence class of 3 contains \(x\) where \( - 3 + x \in H\) <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;">\( - 3 + x = 4n{\text{ }}(n \in \mathbb{Z})\) <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;">\(x = 3 + 4n{\text{ (n}} \in \mathbb{Z})\) <strong><em>A1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Accept \(\{ \ldots - 5,{\text{ }} - 1,{\text{ }}3,{\text{ }}7,{\text{ }} \ldots \} \) or \(x \equiv 3(\bmod 4)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>If no other relevant working seen award <strong><em>A3 </em></strong>for \(\{ 3 + 4n\} \) or \(\{ \ldots - 5,{\text{ }} - 1,{\text{ }}3,{\text{ }}7,{\text{ }} \ldots \} \) seen anywhere.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(G\) is a group. The elements \(a,b \in G\) , satisfy \({a^3} = {b^2} = e\) and \(ba = {a^2}b\) , where \(e\) is the </span><span style="font-family: times new roman,times; font-size: medium;">identity element of \(G\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \({(ba)^2} = e\) .</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;">Express \({(bab)^{ - 1}}\) in its simplest form.</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Given that \(a \ne e\) ,</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) show that \(b \ne e\) ;</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) show that \(G\) is not Abelian.</span></p>
<div class="marks">[6]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">EITHER</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(baba = ba{a^2}b\) <em><strong> M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = b{a^3}b\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {b^2}\) <em><strong> A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = e\) <em><strong>AG</strong></em></span></p>
<p align="LEFT"><strong><span style="font-family: times new roman,times; font-size: medium;">OR</span></strong></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(baba = {a^2}bba\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {a^2}{b^2}a\) <em><strong>(A1)</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {a^3}\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = e\) <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(bab = {a^2}bb\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = {a^2}\) <em><strong>(A1)</strong> </em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\({(bab)^{ - 1}} = a\) <em><strong>A1 </strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[3 marks] </span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) assume \(b = e\) <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then \(a = {a^2}\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow a = e\) which is a contradiction <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) if \(ab = ba\) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then \(ab = {a^2}b\) <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow a = e\) which is a contradiction <em><strong>R1</strong></em></span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;"> </span></strong></em></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></strong></em></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.</span></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was started by the majority of candidates, but only successfully completed by a few. Many candidates seemed to be aware of this style of question, but were either unable to make significant progress or manipulated the algebra in a contorted manner and hence lost valuable time. Also a number of candidates made assumptions about commutativity which were not justified. In part (c) the idea of a proof by contradiction was used by stronger candidates, but weaker candidates were often at a loss as how to start. Overall, the level and succinctness of meaningful algebraic manipulation shown by candidates was disappointing.</span></p>
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The set \({{\rm{S}}_1} = \left\{ {2,4,6,8} \right\}\) and \({ \times _{10}}\) denotes multiplication modulo \(10\).</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Write down the Cayley table for \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (ii) Show that \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) is a group.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (iii) Show that this group is cyclic.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Now consider the group \(\left\{ {{{\rm{S}}_1},{ \times _{20}}} \right\}\) where \({{\rm{S}}_2} = \left\{ {1,9,11,19} \right\}\) and \({{ \times _{20}}}\) denotes </span><span style="font-family: times new roman,times; font-size: medium;">multiplication modulo \(20\). Giving a reason, state whether or not \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) and </span><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ {{{\rm{S}}_1},{ \times _{20}}} \right\}\) are isomorphic.</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i)</span></p>
<p align="LEFT"><img src="data:image/png;base64,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" alt> <em><strong><span style="font-family: times new roman,times; font-size: medium;">A2</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A1</strong></em> for one error.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) closure: it is closed because no new elements are formed <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">identity: \(6\) is the identity element <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">inverses: \(4\) is self-inverse and (\(2\), \(8\)) form an inverse pair <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">associativity: multiplication is associative <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the four group axioms are satisfied</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(iii) any valid reason, <em>e.g.</em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(2\) (or \(8\)) has order \(4\), or \(2\) (or \(8\)) is a generator <strong><em>A2</em></strong></span></p>
<p align="LEFT"><strong><em><span style="font-family: times new roman,times; font-size: medium;"> </span></em></strong></p>
<p align="LEFT"><strong><em><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">the groups are not isomorphic <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">any valid reason, <em>e.g.</em> \({{\rm{S}}_2}\) is not cyclic or all its elements are self-inverse <strong><em>R2</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Parts (a) (i) and (a) (iii) were well answered in general. However, in (a) (ii), some candidates </span><span style="font-family: times new roman,times; font-size: medium;">lost marks by not showing convincingly that \(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) was a group. For example, in verifying </span><span style="font-family: times new roman,times; font-size: medium;">the group axioms, some candidates just made bald statements such as "\(\left\{ {{{\rm{S}}_1},{ \times _{10}}} \right\}\) is closed". This was not convincing because the question indicated that it was a group so that closure was implied by the question. It was necessary here to make some reference to the Cayley table which showed that no new elements were formed by the binary operation. To gain full marks on this style of question candidates need to clearly explain the reasoning used for deductions.</span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In (b), most candidates realised that the quickest way to establish isomorphism (or not) was to determine the order of each element. Candidates who knew that there are essentially only two different groups of order four had a slight advantage in this question. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Let <em><strong>S</strong></em> be the set of matrices given by</span></p>
<p style="text-align: center;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(\left[ \begin{array}{l}<br>a\\<br>c<br>\end{array} \right.\left. \begin{array}{l}<br>b\\<br>d<br>\end{array} \right]\) ; \(a,b,c,d \in \mathbb{R}\), \(ad - bc = 1\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \(R\) is defined on \(S\) as follows. Given \(\boldsymbol{A}\) , </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{B} \in S\)</span> , \(\boldsymbol{ARB}\) if and only if there </span><span style="font-size: medium;"><span style="font-family: times new roman,times;">exists </span></span><span style="font-size: medium;"><span style="font-family: times new roman,times;"><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{X} \in S\)</span></span> </span></span><span style="font-size: medium;"><span style="font-family: times new roman,times;">such that </span></span><span style="font-family: times new roman,times; font-size: medium;">\(\boldsymbol{A} = \boldsymbol{BX}\)</span><span style="font-size: medium;"><span style="font-family: times new roman,times;"> .<br></span></span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(R\) is an equivalence relation.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relationship between \(a\) , \(b\) , \(c\) and \(d\) is changed to \(ad - bc = n\) . State, with </span><span style="font-family: times new roman,times; font-size: medium;">a reason, whether or not there are any non-zero values of \(n\) , other than \(1\), </span><span style="font-family: times new roman,times; font-size: medium;">for which \(R\) is an equivalence relation.</span></p>
<div class="marks">[2]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-size: medium; font-family: times new roman,times;">since \(\boldsymbol{A} = \boldsymbol{AI}\) where \(\boldsymbol{I}\) is the identity <strong><em>A1 </em></strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">and \(\det (\boldsymbol{I}) = 1\) , <em><strong>A1</strong> </em></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(R\) is reflexive </span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(\boldsymbol{ARB} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) where \(\det (\boldsymbol{X}) = 1\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">it follows that \(\boldsymbol{B} = \boldsymbol{A}{\boldsymbol{X}^{ - 1}}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">and \(\det ({\boldsymbol{X}^{ - 1}}) = \det{(\boldsymbol{X})^{ - 1}} = 1\) <em><strong>A1</strong> </em></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(R\) is symmetric </span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(\boldsymbol{ARB}\) and \(\boldsymbol{BRC} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) and \(\boldsymbol{B} = \boldsymbol{CY}\) where \(\det (\boldsymbol{X}) = \det (\boldsymbol{Y}) = 1\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">it follows that \(\boldsymbol{A} = \boldsymbol{CYX}\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(\det (\boldsymbol{YX}) = \det (\boldsymbol{Y})\det (\boldsymbol{X}) = 1\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">\(R\) is transitive </span></p>
<p><span style="font-size: medium; font-family: times new roman,times;">hence \(R\) is an equivalence relation <strong><em>AG </em></strong></span></p>
<p><span style="font-size: medium; font-family: times new roman,times;"><strong><em>[8 marks]</em> </strong></span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">for reflexivity, we require \(\boldsymbol{ARA}\) so that \(\boldsymbol{A} = \boldsymbol{AI}\) (for all \(\boldsymbol{A} \in S\) ) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">since \(\det (\boldsymbol{I}) = 1\) and we require \(\boldsymbol{I} \in S\) the only possibility is \(n = 1\) <strong><em>A1 </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[2 marks] </span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This question was not well done in general, again illustrating that questions involving both matrices and equivalence relations tend to cause problems for candidates. A common error was to assume, incorrectly, that ARB and BRC \( \Rightarrow A = BX\) and \(B = CX\) , not realizing that a different "\(x\)" is required each time. In proving that \(R\) is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">In proving that \(R\) is an equivalence relation, consideration of the determinant is necessary in this question although many candidates neglected to do this. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p class="p1">Consider the set \(S = \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\} \) <span class="s1">under the operation of addition modulo \(6\)</span>, denoted by \({ + _6}\).</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Construct the Cayley table for \(\{ S,{\text{ }}{ + _6}\} \).</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 \(\{ S,{\text{ }}{ + _6}\} \) forms an Abelian group.</p>
<div class="marks">[5]</div>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">State the order of each element.</p>
<div class="marks">[2]</div>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Explain whether or not the group is cyclic.</p>
<div class="marks">[2]</div>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space"><img src="images/Schermafbeelding_2015-12-15_om_06.42.41.png" alt> </span><strong><em>A2</em></strong></p>
<p class="p1"><strong>Note: <em>A1 </em></strong>for one or two errors in the table, <strong><em>A0 </em></strong>otherwise.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">closed no new elements <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1"><span class="s1">\(0\) </span>is identity (since \(0 + a = a + 0 = a,{\text{ }}a \in S\)) <strong><em>A1</em></strong></p>
<p class="p1"><span class="s1">\(0\)</span>, \(3\)<span class="s1"> </span>self inverse, \(1 \Leftrightarrow 5\) inverse pair, \(2 \Leftrightarrow 4\) inverse pair <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">all elements have an inverse</p>
<p class="p1">associativity is assumed over addition <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">since symmetry on leading diagonal in table or commutativity of addition <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">\( \Rightarrow \{ S,{\text{ }}{ + _6}\} \) is an Abelian group <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space"><img src="images/Schermafbeelding_2015-12-15_om_07.26.46.png" alt> </span><strong><em>A2</em></strong></p>
<p class="p1"><strong>Note: <em>A1 </em></strong>for one or two errors in the table, <strong><em>A0 </em></strong>otherwise.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">since there is an element with order \(6\) <strong>OR </strong>\(1\) or \(5\) are generators <span class="Apple-converted-space"> </span><strong><em>R1</em></strong></p>
<p class="p1">the group is cyclic <strong><em>A1</em></strong></p>
<div class="question_part_label">d.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.</p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.</p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This question was well answered in general although some candidates showed only commutativity, not realising that they also had to prove that it was a group.</p>
<div class="question_part_label">d.</div>
</div>
<br><hr><br><div class="question">
<p class="p1">Prove that the function \(f:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}\) defined by \(f(x,{\text{ }}y) = (2x + y,{\text{ }}x + y)\) is a bijection.</p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p>to be a bijection it must be injective and surjective <strong><em>R1</em></strong></p>
<p><strong>Note: </strong>This <strong><em>R1 </em></strong>may be awarded at any stage</p>
<p> </p>
<p>suppose \(f(x,{\text{ }}y) = f(u,{\text{ }}v)\) <strong><em>M1</em></strong></p>
<p>\(2x + y = 2u + v\;\;\;({\text{ - i}})\)</p>
<p>\(x + y = u + v\;\;\;({\text{ - ii}})\)</p>
<p>\({\text{i - ii}} \Rightarrow x = u\)</p>
<p>\({\text{i - 2(ii)}} \Rightarrow - y = - v\)</p>
<p>\( \Rightarrow x = u,{\text{ }}y = v\) <strong><em>A1</em></strong></p>
<p>thus \((x,{\text{ }}y) = (u,{\text{ }}v)\) hence injective <strong><em>A1</em></strong></p>
<p>let \(2x + y = s\;\;\;({\text{ - i}})\)</p>
<p>\(x + y = t\;\;\;({\text{ - ii}})\) <strong><em>M1</em></strong></p>
<p>\({\text{i - ii }}x = s - t\)</p>
<p>\( \Rightarrow y = 2t - s\)</p>
<p>both \(x\) and \(y\) are integer if \(s\) and \(t\) are integer <strong><em>R1</em></strong></p>
<p>hence it is surjective <strong><em>A1</em></strong></p>
<p>hence \(f\) is a bijection <strong><em>AG</em></strong></p>
<p><strong>Note: </strong>Accept a valid argument based on matrices</p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
<p class="p1">Most candidates were able to show that \(f\) was an injection although some candidates appear to believe that it is sufficient to show that \(f(x,{\text{ }}y)\) is unique. A significant minority failed to show that \(f\) is a surjection and most candidates failed to note that it had to be checked that all values were integers. Some candidates introduced a matrix to define the transformation which was often a successful alternative method.</p>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Show that the set \(S\) of numbers of the form \({2^m} \times {3^n}\) , where \(m,n \in \mathbb{Z}\) , forms a </span><span style="font-family: times new roman,times; font-size: medium;">group \(\left\{ {S, \times } \right\}\) under multiplication.</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Show that \(\left\{ {S, \times } \right\}\) is isomorphic to the group of complex numbers \(m + n{\rm{i}}\) under </span><span style="font-family: times new roman,times; font-size: medium;">addition, where \(m\), \(n \in \mathbb{Z}\) .</span></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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Closure: Consider the numbers \({2^{{m_1}}} \times {3^{{m_1}}}\) and \({2^{{m_2}}} \times {3^{{n_2}}}\) where <em><strong>M1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\({m_1},{m_2},{n_1},{n_2}, \in \mathbb{Z}\) . Then,</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Product \( = {2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}}\) which \( \in S\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Identity: \({2^0} \times {3^0} = 1 \in S\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Since \(({2^m} \times {3^n}) \times ({2^{ - m}} \times {3^{ - n}}) = 1\) and \({2^{ - m}} \times {3^{ - n}} \in S\) <strong><em>R1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then \({2^{ - m}} \times {3^{ - n}}\) is the inverse. <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Associativity: This follows from the associativity of multiplication. <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Consider the bijection</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f({2^m} \times {3^n}) = m + n{\rm{i}}\) <strong><em>(M1)</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Then</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f({2^{{m_1}}} \times {3^{{n_1}}}) \times ({2^{{m_2}}} \times {3^{{n_2}}}) = f({2^{{m_1} + {m_2}}} \times {3^{{n_1} + {n_2}}})\) <em><strong>M1A1</strong></em></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {m_1} + {m_2} + ({n_1} + {n_2}){\rm{i}}\) <strong><em>A1</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = ({m_1} + {n_1}{\rm{i}}) + ({m_2} + {n_2}{\rm{i}})\) <strong><em>(A1)</em></strong></span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = f({2^{{m_1}}} \times {3^{{n_1}}}) + f({2^{{m_2}}} \times {3^{{n_2}}})\) <strong><em>A1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[6 marks]</span></em></strong></p>
<div class="question_part_label">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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The relation \(R\) is defined on the set </span><span style="font-family: times new roman,times; font-size: medium;"><span style="font-family: times new roman,times; font-size: medium;">\(\mathbb{Z}\)</span> by \(aRb\) if and only if \(4a + b = 5n\) , </span><span style="font-family: times new roman,times; font-size: medium;">where \(a,b,n \in \mathbb{Z}\).</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that \(R\) is an equivalence relation.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">State the equivalence classes of \(R\) .</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><span style="font-family: times new roman,times; font-size: medium;">\(4a + b = 5n\) for \(a,b,n \in \mathbb{Z}\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">reflexive: </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4a + a = 5a\) so \(aRa\) , and \(R\) is reflexive <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">symmetric: </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4a + b = 5n\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4b + a = 5b - b + 5a - 4a\) <em><strong>M1</strong></em> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 5b + 5a - (4a + b)\) <strong><em>A1</em></strong> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\( = 5m\) so \(bRa\) , and \(R\) is symmetric <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">transitive: </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4a + b = 5n\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4b + c = 5k\) <strong><em>M1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4a + 5b + c = 5n + 5k\) <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(4a + c = 5(n + k - b)\) so \(aRc\) , and \(R\) is transitive <strong><em>A1</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">therefore \(R\) is an equivalence relation <strong><em>AG </em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[8 marks] </span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">equivalence classes are </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ { \ldots , - 10, - 5,0,5,10,\left. \ldots \right\}} \right.\) <strong><em>(M1)</em> </strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ { \ldots , - 9, - 4,1,6,11,\left. \ldots \right\}} \right.\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ { \ldots , - 8, - 3,2,7,12,\left. \ldots \right\}} \right.\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ { \ldots , - 7, - 2,3,8,13,\left. \ldots \right\}} \right.\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left\{ { \ldots , - 6, - 1,4,9,14,\left. \ldots \right\}} \right.\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">or \(\left\{ {\left\langle 0 \right\rangle ,\left\langle 1 \right\rangle ,\left\langle 2 \right\rangle ,\left\langle 3 \right\rangle ,\left. {\left\langle 4 \right\rangle } \right\}} \right.\) <em><strong>A2 </strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <strong><em>A2</em></strong> for all classes, <strong><em>A1</em></strong> for at least 2 correct classes. </span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks] </span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Part (a) was generally well done but not always in the most direct manner. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Too many missed the equivalence classes in part (b). </span></p>
<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: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">The function \(f:{\mathbb{R}^ + } \times {\mathbb{R}^ + } \to {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) is defined by \(f(x,{\text{ }}y) = \left( {xy,{\text{ }}\frac{x}{y}} \right)\).</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px 'Times New Roman';"><span style="font-family: 'times new roman', times; font-size: medium;">Prove that \(f\) is a bijection.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">we need to show that \(f\) is injective and surjective <strong><em>(R1)</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>R1 </em></strong>if seen anywhere in the solution.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\underline {{\text{injective}}} \)</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;">let \((a,{\text{ }}b)\) and \((c,{\text{ }}d) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\), and let \(f(a,{\text{ }}b) = f(c,{\text{ }}d)\) <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;">it follows that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(ab = cd\) and \(\frac{a}{b} = \frac{c}{d}\) <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;">multiplying these equations,</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;">\({a^2} = {c^2} \Rightarrow a = c\) and therefore \(b = d\) <strong>A1</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;">since \(f(a,{\text{ }}b) = f(c,{\text{ }}d) \Rightarrow (a,{\text{ }}b) = (c,{\text{ }}d),{\text{ }}f\) is injective <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note:</strong> Award <strong><em>R1 </em></strong>if stated anywhere as needing to be shown.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\underline {{\text{surjective}}} \)</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;">let \((p,{\text{ }}q) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\)</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">consider \(f(x,{\text{ }}y) = (p,{\text{ }}q)\) so \(xy = p\) and \(\frac{x}{y} = q\) <strong><em>M1A1</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;">multiplying these equations,</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;">\({x^2} = pq\) so \(x = \sqrt {pq} \) and therefore \(y = \sqrt {\frac{p}{q}} \) <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 given \((p,{\text{ }}q) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + },{\text{ }}\exists (x,{\text{ }}y) \in {\mathbb{R}^ + } \times {\mathbb{R}^ + }\) such that \(f(x,{\text{ }}y) = (p,{\text{ }}q)\) which shows that \(f\) is surjective <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>Award <strong><em>R1 </em></strong>if stated anywhere as needing to be shown.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(f\) is therefore a bijection</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[9 marks]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p>The transformations <em>T</em><sub>1</sub>, <em>T</em><sub>2</sub>, <em>T</em><sub>3</sub>, <em>T</em><sub>4</sub>, in the plane are defined as follows:</p>
<p><em>T</em><sub>1</sub> : A rotation of 360° about the origin<br><em>T</em><sub>2</sub> : An anticlockwise rotation of 270° about the origin<br><em>T</em><sub>3</sub> : A rotation of 180° about the origin<br><em>T</em><sub>4</sub> : An anticlockwise rotation of 90° about the origin.</p>
</div>
<div class="specification">
<p>The transformation <em>T</em><sub>5</sub> is defined as a reflection in the \(x\)-axis.</p>
</div>
<div class="specification">
<p>The transformation <em>T</em> is defined as the composition of <em>T</em><sub>3</sub> followed by <em>T</em><sub>5</sub> followed by <em>T</em><sub>4</sub>.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Copy and complete the following Cayley table for the transformations of <em>T</em><sub>1</sub>, <em>T</em><sub>2</sub>, <em>T</em><sub>3</sub>, <em>T</em><sub>4</sub>, under the operation of composition of transformations.</p>
<p><img src="data:image/png;base64,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"></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><em>Show that T</em><sub>1</sub>, <em>T</em><sub>2</sub>, <em>T</em><sub>3</sub>, <em>T</em><sub>4 </sub>under the operation of composition of transformations form a group. Associativity may be assumed.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Show that this group is cyclic.</p>
<div class="marks">[1]</div>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Write down the 2 × 2 matrices representing <em>T</em><sub>3</sub>, <em>T</em><sub>4</sub> and <em>T</em><sub>5</sub>.</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>Find the 2 × 2 matrix representing <em>T</em>.</p>
<div class="marks">[2]</div>
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Give a geometric description of the transformation <em>T</em>.</p>
<div class="marks">[1]</div>
<div class="question_part_label">d.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><img src="data:image/png;base64,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"> <em><strong>A2</strong></em></p>
<p><em><strong>[2 marks]</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>A1</strong> </em>for 6, 7 or 8 correct.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the table is closed – no new elements <em><strong>A1</strong></em></p>
<p><em>T</em><sub>1</sub> is the identity <em><strong>A1</strong></em></p>
<p><em>T</em><sub>3</sub> (and <em>T</em><sub>1</sub>) are self-inverse; <em>T</em><sub>2</sub> and <em>T</em><sub>4</sub> are an inverse pair. Hence every element has an inverse <em><strong>A1</strong></em></p>
<p>hence it is a group <em><strong>AG</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>all elements in the group can be generated by <em>T</em><sub>2</sub> (or <em>T</em><sub>4</sub>) <em><strong>R1</strong></em></p>
<p>hence the group is cyclic <em><strong> AG</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<p> </p>
<div class="question_part_label">b.ii.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p> </p>
<p><em>T</em><sub>3</sub> is represented by \(\left( {\begin{array}{*{20}{c}}<br> { - 1}&0 \\ <br> 0&{ - 1} <br>\end{array}} \right)\) <em><strong>A1</strong></em></p>
<p><em>T</em><sub>4</sub> is represented by \(\left( {\begin{array}{*{20}{c}}<br> 0&{ - 1} \\ <br> 1&0 <br>\end{array}} \right)\) <em><strong>A1</strong></em></p>
<p><em>T</em><sub>5</sub> is represented by \(\left( {\begin{array}{*{20}{c}}<br> 1&0 \\ <br> 0&{ - 1} <br>\end{array}} \right)\) <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">c.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>\(\left( {\begin{array}{*{20}{c}}<br> 0&{ - 1} \\ <br> 1&0 <br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br> 1&0 \\ <br> 0&{ - 1} <br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br> { - 1}&0 \\ <br> 0&{ - 1} <br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br> 0&{ - 1} \\ <br> { - 1}&0 <br>\end{array}} \right)\) <em><strong>(M1)A1</strong></em></p>
<p><strong>Note:</strong> Award <em><strong>M1A0</strong> </em>for multiplying the matrices in the wrong order.</p>
<p><em><strong>[2 marks]</strong></em></p>
<div class="question_part_label">d.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>a reflection in the line \(y = - x\) <em><strong>A1</strong></em></p>
<p><em><strong>[1 mark]</strong></em></p>
<div class="question_part_label">d.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<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.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">d.ii.</div>
</div>
<br><hr><br><div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(\{ G,{\text{ }} * \} \) is a group of order \(N\) and \(\{ H,{\text{ }} * \} \) is a proper subgroup of \(\{ G,{\text{ }} * \} \) of order \(n\).</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;">(a) Define the right coset of \(\{ H,{\text{ }} * \} \) containing the element \(a \in G\).</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) Show that each right coset of \(\{ H,{\text{ }} * \} \) contains \(n\) elements.</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 the union of the right cosets of \(\{ H,{\text{ }} * \} \) is equal to \(G\).</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) Show that any two right cosets of \(\{ H,{\text{ }} * \} \) are either equal or disjoint.</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) Give a reason why the above results can be used to prove that \(N\) is a multiple of \(n\).</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(a) the right coset containing \(a\) has the form \(\{ ha|h \in H\} \) <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>[1 mark]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong>Note: </strong>From here on condone the use of left cosets.</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica; min-height: 25.0px;"> </p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(b) let \(b\), \(c\) be distinct elements of \(H\)<em>. </em>Then, given \(a \in G\), by the Latin square property of the Cayley table, \(ba\) and \(ca\) are distinct <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;">therefore each element of \(H\) corresponds to a unique element in the coset which must therefore contain \(n\) elements <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(c) let \(d\) be any element of \(G\). Then since \(H\) contains the identity \(e\), \(ed = d\) will be in a coset <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">therefore every element of \(G\) will be contained in a coset which proves that the union of all the cosets is \(G\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[2 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(d) let the cosets of \(b\) and \(c{\text{ }}(b,{\text{ }}c \in G)\) contain a common element so that</span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">\(pb = qc\) where \(p,{\text{ }}q \in H\). Let \(r\) denote any other element \( \in H\) <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;">then</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;">\(rb = r{p^{ - 1}}qc\) <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;">since \(r{p^{ - 1}}q \in H\), this shows that all the other elements are common and the cosets are equal <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">since not all cosets can be equal, there must be other cosets which are disjoint <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[4 marks]</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em> </em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;">(e) the above results show that \(G\) is partitioned into a number of disjoint subsets containing \(n\) elements so that \(N\) must be a multiple of \(n\) <strong><em>R1</em></strong></span></p>
<p style="margin: 0.0px 0.0px 0.0px 0.0px; font: 21.0px Helvetica;"><span style="font-family: 'times new roman', times; font-size: medium;"><strong><em>[1 mark]</em></strong></span></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="specification">
<p><span style="font-family: times new roman,times; font-size: medium;">The set \(S\) contains the eight matrices of the form\[\left( {\begin{array}{*{20}{c}}<br>a&0&0\\<br>0&b&0\\<br>0&0&c<br>\end{array}} \right)\]where \(a\), \(b\), \(c\) can each take one of the values \( + 1\) or \( - 1\) .</span></p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">Show that any matrix of this form is its own inverse.</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;">Show that \(S\) forms an Abelian group under matrix multiplication.</span></p>
<div class="marks">[9]</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;">Giving a reason, state whether or not this group is cyclic.</span></p>
<div class="marks">[1]</div>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>a&0&0\\<br>0&b&0\\<br>0&0&c<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>a&0&0\\<br>0&b&0\\<br>0&0&c<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a^2}}&0&0\\<br>0&{{b^2}}&0\\<br>0&0&{{c^2}}<br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>A1M1</strong></em></span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;">\( = \left( {\begin{array}{*{20}{c}}<br>1&0&0\\<br>0&1&0\\<br>0&0&1<br>\end{array}} \right)\)</span><span style="font-family: times new roman,times; font-size: medium;"> <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this shows that each matrix is self-inverse</span></p>
<p><em><strong> <span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">closure:</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{{a_1}}&0&0\\<br>0&{{b_1}}&0\\<br>0&0&{{c_1}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{{a_2}}&0&0\\<br>0&{{b_2}}&0\\<br>0&0&{{c_2}}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a_1}{a_2}}&0&0\\<br>0&{{b_1}{b_2}}&0\\<br>0&0&{{c_1}{c_2}}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1A1</span></strong></em></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> \( = \left( {\begin{array}{*{20}{c}}<br>{{a_3}}&0&0\\<br>0&{{b_3}}&0\\<br>0&0&{{c_3}}<br>\end{array}} \right)\)</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">where each of \({a_3}\), \({b_3}\), \({c_3}\) can only be \( \pm 1\) <strong><em>A1</em></strong></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">this proves closure</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">identity: the identity matrix is the group identity <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">inverse: as shown above, every element is self-inverse <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">associativity: this follows because matrix multiplication is associative <em><strong>A1</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">\(S\) is therefore a group <em><strong>AG</strong></em></span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">Abelian:</span></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{{a_2}}&0&0\\<br>0&{{b_2}}&0\\<br>0&0&{{c_2}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{{a_1}}&0&0\\<br>0&{{b_1}}&0\\<br>0&0&{{c_1}}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a_2}{a_1}}&0&0\\<br>0&{{b_2}{b_1}}&0\\<br>0&0&{{c_2}{c_1}}<br>\end{array}} \right)\) </span><strong><em><span style="font-family: times new roman,times; font-size: medium;">A1</span></em></strong></p>
<p><span style="font-family: Times New Roman; font-size: medium;">\(\left( {\begin{array}{*{20}{c}}<br>{{a_1}}&0&0\\<br>0&{{b_1}}&0\\<br>0&0&{{c_1}}<br>\end{array}} \right)\left( {\begin{array}{*{20}{c}}<br>{{a_2}}&0&0\\<br>0&{{b_2}}&0\\<br>0&0&{{c_2}}<br>\end{array}} \right) = \left( {\begin{array}{*{20}{c}}<br>{{a_1}{a_2}}&0&0\\<br>0&{{b_1}{b_2}}&0\\<br>0&0&{{c_1}{c_2}}<br>\end{array}} \right)\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">A1</span></strong></em></p>
<p><span style="font-family: times new roman,times; font-size: medium;"><strong>Note:</strong> Second line may have been shown whilst proving closure, however a r</span><span style="font-family: times new roman,times; font-size: medium;">eference to it must be made here.</span></p>
<p><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><span style="font-family: times new roman,times; font-size: medium;">we see that the same result is obtained either way which proves commutativity </span><span style="font-family: times new roman,times; font-size: medium;">so that the group is Abelian <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;"> [9 marks]</span></em></strong></p>
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">since all elements (except the identity) are of order \(2\), the group is not cyclic </span><span style="font-family: times new roman,times; font-size: medium;">(since \(S\) contains \(8\) elements) <strong><em>R1</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[1 mark]</span></em></strong></p>
<div class="question_part_label">c.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">c.</div>
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Prove that the number \(14 641\) is the fourth power of an integer in any base greater </span><span style="font-family: times new roman,times; font-size: medium;">than \(6\).</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 align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">For \(a,b \in \mathbb{Z}\)</span><span style="font-family: times new roman,times; font-size: medium;"> the relation \(aRb\) is defined if and only if \(\frac{a}{b} = {2^k}\) </span><span style="font-family: times new roman,times; font-size: medium;">, \(k \in \mathbb{Z}\)</span><span style="font-family: times new roman,times; font-size: medium;"> .</span></p>
<p style="margin-left: 30px;" align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> (i) Prove that \(R\) is an equivalence relation.</span></p>
<p style="margin-left: 30px;"><span style="font-family: times new roman,times; font-size: medium;"> (ii) List the equivalence classes of \(R\) on the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.</span></p>
<div class="marks">[8]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(14641\) (base \(a > 6\) ) \( = {a^4} + 4{a^3} + 6{a^2} + 4a + 1\) , <em><strong>M1A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( = {(a + 1)^4}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">this is the fourth power of an integer <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[3 marks]</span></em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(i) \(aRa\) since \(\frac{a}{a} = 1 = {2^0}\)</span><span style="font-family: times new roman,times; font-size: medium;"> , hence \(R\) is reflexive <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(aRb \Rightarrow \frac{a}{b} = {2^k} \Rightarrow \frac{b}{a} = {2^{ - k}} \Rightarrow bRa\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">so R is symmetric <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(aRb\) and \(bRc \Rightarrow \frac{a}{b} = {2^m}\), \(m \in \mathbb{Z}\)</span><span style="font-family: times new roman,times; font-size: medium;"> and \(bRc \Rightarrow \frac{b}{c} = {2^n}\) , \(n \in \mathbb{Z}\) </span><em><strong><span style="font-family: times new roman,times; font-size: medium;">M1</span></strong></em></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow \frac{a}{b} \times \frac{b}{c} = \frac{a}{c} = {2^{m + n}}\) , \(m + n \in \mathbb{Z}\) </span><span style="font-family: times new roman,times; font-size: medium;"> <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\( \Rightarrow aRc\) so transitive <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">hence \(R\) is an equivalence relation <em><strong>AG</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">(ii) equivalence classes are {1, 2, 4, 8} , {3, 6} , {5, 10} , {7} , {9} <strong><em>A3</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"><strong>Note</strong>: Award <em><strong>A2</strong></em> if one class missing, </span><span style="font-family: times new roman,times; font-size: medium;"><em><strong>A1</strong></em> if two classes missing, </span><span style="font-family: times new roman,times; font-size: medium;"><em><strong>A0</strong></em> if three or more classes missing.</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;"> </span></p>
<p><em><strong><span style="font-family: times new roman,times; font-size: medium;">[8 marks]</span></strong></em></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">This was not difficult but a surprising number of candidates were unable to do it. Care with notation and logic were lacking. </span></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p><span style="font-family: times new roman,times; font-size: medium;">The question was at first straightforward but some candidates mixed up the properties of an equivalence relation with those of a group. The idea of an equivalence class is still not clearly understood by many candidates so that some were missing. </span></p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The group \(\left\{ {G, + } \right\}\) is defined by the operation of addition on the set \(G = \left\{ {2n|n \in \mathbb{Z}} \right\}\) .</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">The group \(\left\{ {H, + } \right\}\) is defined by the operation of addition on the set \(H = \left\{ {4n|n \in \mathbb{Z}} \right\}\) </span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">Prove that \(\left\{ {G, + } \right\}\) and \(\left\{ {H, + } \right\}\) are isomorphic.</span></p>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question">
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">consider the function \(f:G \to H\) defined by \(f(g) = 2g\) where \(g \in G\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">given \({g_1}\), \({g_2} \in G,f({g_1}) = f({g_2}) \Rightarrow 2{g_1} = 2{g_2} \Rightarrow {g_1} = {g_2}\) (injective) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">given \(h \in H\) then \(h = 4n\) , so \(f(2n) = h\) and \(2n \in G\) (surjective) <strong><em>M1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">hence <em><strong>f</strong></em> is a bijection <em><strong>A1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">then, for \({g_1}\), \({g_2} \in G\)</span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f({g_1} + {g_2}) = 2({g_1} + {g_2})\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">\(f({g_1}) + f({g_2}) = 2{g_1} + 2{g_2}\) <strong><em>A1</em></strong></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">it follows that \(f({g_1} + {g_2}) = f({g_1}) + f({g_2})\) <em><strong>R1</strong></em></span></p>
<p align="LEFT"><span style="font-family: times new roman,times; font-size: medium;">which completes the proof that \(\left\{ {G, + } \right\}\) and \(\left\{ {H, + } \right\}\) are isomorphic <strong><em>AG</em></strong></span></p>
<p><strong><em><span style="font-family: times new roman,times; font-size: medium;">[7 marks]</span></em></strong></p>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question">
[N/A]
</div>
<br><hr><br><div class="question" style="padding-left: 20px; padding-right: 20px;">
<p class="p1">Use the Euclidean algorithm to find \(\gcd (162,{\text{ }}5982)\).</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"><span class="s1">The relation \(R\) </span>is defined on \({\mathbb{Z}^ + }\) by \(nRm\) if and only if \(\gcd (n,{\text{ }}m) = 2\).</p>
<p class="p1">(i) <span class="Apple-converted-space"> </span>By finding counterexamples show that \(R\) is neither reflexive nor transitive.</p>
<p class="p1">(ii) <span class="Apple-converted-space"> </span>Write down the set of solutions of \(nR6\).</p>
<div class="marks">[7]</div>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1"><span class="Apple-converted-space">\(5982 = 162 \times 36 + 150\) </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p1"><span class="Apple-converted-space">\(162 = 150 \times 1 + 12\) </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1">\(150 = 12 \times 12 + 6\)</p>
<p class="p1">\(12 = 6 \times 2 + 0 \Rightarrow \gcd \) is 6 <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><strong><em>[4 marks]</em></strong></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">(i) <span class="Apple-converted-space"> </span>for example, \(\gcd (4,{\text{ }}4) = 4\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p1"><span class="Apple-converted-space">\(4 \ne 2\) </span><span class="s1"><strong><em>R1</em></strong></span></p>
<p class="p2">so \(R\) is not reflexive <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p1">for example</p>
<p class="p1">\(\gcd (4,{\text{ }}2) = 2\) and \(\gcd (2,{\text{ }}8) = 2\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>M1A1</em></strong></span></p>
<p class="p1">but \(\gcd (4,{\text{ }}8) = 4{\text{ }}( \ne 2)\) <span class="Apple-converted-space"> </span><span class="s1"><strong><em>R1</em></strong></span></p>
<p class="p2">so \(R\) is not transitive <span class="Apple-converted-space"> </span><strong><em>AG</em></strong></p>
<p class="p2">(ii) <span class="Apple-converted-space"> </span><strong>EITHER</strong></p>
<p class="p2">even numbers <span class="Apple-converted-space"> </span><strong><em>A1</em></strong></p>
<p class="p1">not divisible by 6 <span class="Apple-converted-space"> </span><span class="s1"><strong><em>A1</em></strong></span></p>
<p class="p2"><strong>OR</strong></p>
<p class="p3"><span class="Apple-converted-space">\(\{ 2 + 6n:n \in \mathbb{N}\} {\text{ }} \cup \{ 4 + 6n:n \in \mathbb{N}\} \) </span><span class="s1"><strong><em>A1A1</em></strong></span></p>
<p class="p2"><strong>OR</strong></p>
<p class="p1"><span class="Apple-converted-space">\(2,{\text{ }}4,{\text{ }}8,{\text{ }}10,{\text{ }} \ldots \) </span><span class="s1"><strong><em>A2</em></strong></span></p>
<p class="p2"><strong><em>[7 marks]</em></strong></p>
<div class="question_part_label">b.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was a successful question for many students with many wholly correct answers seen. Part (a) was successfully answered by most candidates and those candidates usually had a reasonable understanding of how to complete part (b). A number were not fully successful in knowing how to explain their results.</p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p class="p1">This was a successful question for many students with many wholly correct answers seen. Part (a) was successfully answered by most candidates and those candidates usually had a reasonable understanding of how to complete part (b). A number were not fully successful in knowing how to explain their results.</p>
<div class="question_part_label">b.</div>
</div>
<br><hr><br><div class="specification">
<p>A sample of size 100 is taken from a normal population with unknown mean <em>μ</em> and known variance 36.</p>
</div>
<div class="specification">
<p>Another investigator decides to use the same data to test the hypotheses <em>H</em><sub>0</sub> : <em>μ</em> = 65 , <em>H<span style="font-size: 11.6667px;">1</span></em> : <em>μ</em> = 67.9.</p>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>An investigator wishes to test the hypotheses <em>H</em><sub>0</sub> : <em>μ</em> = 65, <em>H</em><sub>1</sub> : <em>μ</em> > 65.</p>
<p>He decides on the following acceptance criteria:</p>
<p>Accept <em>H</em><sub>0</sub> if the sample mean \(\bar x\) ≤ 66.5</p>
<p>Accept <em>H</em><sub>1</sub> if \(\bar x\) > 66.5</p>
<p>Find the probability of a Type I error.</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>She decides to use the same acceptance criteria as the previous investigator. Find the probability of a Type II error.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px; padding-right: 20px;">
<p>Find the critical value for \({\bar x}\) if she wants the probabilities of a Type I error and a Type II error to be equal.</p>
<div class="marks">[3]</div>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Markscheme</h2>
<div class="question" style="padding-left: 20px;">
<p>\(\bar X \sim {\text{N}}\left( {\mu ,\,\frac{{{\sigma ^2}}}{n}} \right)\)</p>
<p>\(\bar X \sim {\text{N}}\left( {65,\,\frac{{36}}{{100}}} \right)\) <em><strong>(A1)</strong></em></p>
<p>P(Type I Error) \( = {\text{P}}\left( {\bar X > 66.5} \right)\) <em><strong> (M1)</strong></em></p>
<p>= 0.00621 <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>P(Type II Error) = P(accept <em>H</em><sub>0 </sub>| <em>H<sub>1</sub> </em>is true)</p>
<p>\( = {\text{P}}\left( {\bar X \leqslant 66.5\left| {\mu = 67.9} \right.} \right)\) <em><strong>(M1)</strong></em></p>
<p>\( = {\text{P}}\left( {\bar X \leqslant 66.5} \right)\) when \(\bar X \sim {\text{N}}\left( {67.9,\,\frac{{36}}{{100}}} \right)\) <em><strong>(M1)</strong></em></p>
<p>= 0.00982 <em><strong>A1</strong></em></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
<p>the variances of the distributions given by <em>H</em><sub>0</sub> and <em>H<sub>1</sub></em> are equal, <strong> (R1)</strong></p>
<p>by symmetry the value of \({\bar x}\) lies midway between 65 and 67.9 <em><strong>(M1)</strong></em></p>
<p>\( \Rightarrow \bar x = \frac{1}{2}\left( {65 + 67.9} \right) = 66.45\) <em><strong>A1</strong></em></p>
<p><img 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"></p>
<p><em><strong>[3 marks]</strong></em></p>
<div class="question_part_label">b.ii.</div>
</div>
<h2 style="margin-top: 1em">Examiners report</h2>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">a.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.i.</div>
</div>
<div class="question" style="padding-left: 20px;">
[N/A]
<div class="question_part_label">b.ii.</div>
</div>
<br><hr><br>