Find the sum of the infinite geometric sequence 27, −9, 3, −1, ... .

Use mathematical induction to prove that for \(n \in {\mathbb{Z}^ + }\) ,

\[a + ar + a{r^2} + ... + a{r^{n - 1}} = \frac{{a(1 - {r^n})}}{{1 - r}}.\]

\(r = - \frac{1}{3}\)     (A1)

\({S_\infty } = \frac{{27}}{{1 + \frac{1}{3}}}\)     M1

\({S_\infty } = \frac{{81}}{4}\,\,\,\,\,( = 20.25)\)     A1     N1

[3 marks]

Attempting to show that the result is true for n = 1     M1

LHS = a and \({\text{RHS}} = \frac{{a(1 - r)}}{{1 - r}} = a\)     A1

Hence the result is true for n = 1

Assume it is true for n = k

\(a + ar + a{r^2} + ... + a{r^{k - 1}} = \frac{{a(1 - rk)}}{{1 - r}}\)     M1

Consider n = k + 1:

\(a + ar + a{r^2} + ... + a{r^{k - 1}} + a{r^k} = \frac{{a(1 - {r^k})}}{{1 - r}} + a{r^k}\)     M1

\( = \frac{{a(1 - {r^k}) + a{r^k}(1 - r)}}{{1 - r}}\)

\( = \frac{{a - a{r^k} + a{r^k} - a{r^{k + 1}}}}{{1 - r}}\)     A1

Note: Award A1 for an equivalent correct intermediate step.

\( = \frac{{a - a{r^{k + 1}}}}{{1 - r}}\)

\( = \frac{{a(1 - {r^{k + 1}})}}{{1 - r}}\)     A1

Note: Illogical attempted proofs that use the result to be proved would gain M1A0A0 for the last three above marks.

The result is true for \(n = k \Rightarrow \) it is true for \(n = k + 1\) and as it is true for \(n = 1\), the result is proved by mathematical induction.     R1     N0

Note: To obtain the final R1 mark a reasonable attempt must have been made to prove the k + 1 step.

[7 marks]

Part (a) was correctly answered by the majority of candidates, although a few found r = –3.

Part (b) was often started off well, but a number of candidates failed to initiate the n = k + 1 step in a satisfactory way. A number of candidates omitted the ‘P(1) is true’ part of the concluding statement.

Find integer values of \(m\) and \(n\) for which

\[m - n{\log _3}2 = 10{\log _9}6\]

METHOD 1

\(m - n{\log _3}2 = 10{\log _9}6\)

\(m - n{\log _3}2 = 5{\log _3}6\)    M1

\(m = {\log _3}\left( {{6^5}{2^n}} \right)\)    (M1)

\({3^m}{2^{ - n}} = {6^5} = {3^5} \times {2^5}\)    (M1)

\(m = 5,{\text{ }}n =  - 5\)    A1

Note:     First M1 is for any correct change of base, second M1 for writing as a single logarithm, third M1 is for writing 6 as \(2 \times 3\).

METHOD 2

\(m - n{\log _3}2 = 10{\log _9}6\)

\(m - n{\log _3}2 = 5{\log _3}6\)    M1

\(m - n{\log _3}2 = 5{\log _3}3 + 5{\log _3}2\)    (M1)

\(m - n{\log _3}2 = 5 + 5{\log _3}2\)    (M1)

\(m = 5,{\text{ }}n =  - 5\)    A1

Note:     First M1 is for any correct change of base, second M1 for writing 6 as \(2 \times 3\) and third M1 is for forming an expression without \({\log _3}3\).

[4 marks]

The first stage on this question was to change base, so each logarithm was written in the same base. Some candidates chose to move to base 10 or base e, rather than the more obvious base 3, but a few still successfully reached the correct answer having done this. A large majority though did not seem to know how to change the base of a logarithm.

Simplifying the expression further was a struggle for many candidates.

Expand \({(x + h)^3}\).

Hence find the derivative of \(f(x) = {x^3}\) from first principles.

\({(x + h)^3} = {x^3} + 3{x^2}h + 3x{h^2} + {h^3}\)     (M1)A1

[2 marks]

\(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{{(x + h)}^3} - {x^3}}}{h}\)     (M1)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{{x^3} + 3{x^2}h + 3x{h^2} + {h^3} - {x^3}}}{h}\)

\( = \mathop {\lim }\limits_{h \to 0} (3{x^2} + 3xh + {h^2})\)     A1

\( = 3{x^2}\)     A1

Note:     Do not award final A1 on FT if \( = 3{x^2}\) is not obtained

Note:     Final A1 can only be obtained if previous A1 is given

[3 marks]

Total [5 marks]

Well done although some did not use the binomial expansion.

Fine by those who knew what first principles meant, not by the others.

The fifth term of an arithmetic sequence is equal to 6 and the sum of the first 12 terms is 45.

Find the first term and the common difference.

use of either \({u_n} = {u_1} + (n - 1)d\) or \({S_n} = \frac{n}{2}\left( {2{u_1} + (n - 1)d} \right)\)     M1

\({u_1} + 4d = 6\)    (A1)

\(\frac{{12}}{2}(2{u_1} + 11d) = 45\)    (A1)

\( \Rightarrow 4{u_1} + 22d = 15\)

attempt to solve simultaneous equations     M1

\(4(6 - 4d) + 22d = 15\)

\(6d =  - 9 \Rightarrow d =  - 1.5\)    A1

\({u_1} = 12\)    A1

[6 marks]

Most candidates tackled this question through the use of the standard formula for arithmetic series. Others attempted a variety of trial and improvement approaches with varying degrees of success.

Find AB, giving your answer in the form \(a\sqrt {b - \sqrt 3 } \) , where a , \(b \in {\mathbb{Z}^ + }\) .

Calculate \({\rm{A\hat OB}}\) in terms of \(\pi \).

\({\text{AB}} = \sqrt {{1^2} + {{(2 - \sqrt 3 )}^2}} \)     M1

\( = \sqrt {8 - 4\sqrt 3 } \)     A1

\( = 2\sqrt {2 - \sqrt 3 } \)     A1

[3 marks]

METHOD 1

\(\arg {z_1} = - \frac{\pi }{4}{\text{ }}\arg {z_2} = - \frac{\pi }{3}\)     A1A1

Note: Allow \(\frac{\pi }{4}\) and \(\frac{\pi }{3}\) .

Note: Allow degrees at this stage.

\({\rm{A\hat OB}} = \frac{\pi }{3} - \frac{\pi }{4}\)

\( = \frac{\pi }{{12}}{\text{ (accept }} - \frac{\pi }{{12}})\)     A1

Note: Allow FT for final A1.

METHOD 2

attempt to use scalar product or cosine rule     M1

\(\cos {\rm{A\hat OB}} = \frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}\)     A1

\({\rm{A\hat OB}} = \frac{\pi }{{12}}\)     A1 

[3 marks]

It was disappointing to note the lack of diagram in many solutions. Most importantly the lack of understanding of the notation AB was apparent. Teachers need to make sure that students are aware of correct notation as given in the outline. A number used the cosine rule but then confused the required angle or sides.

It was disappointing to note the lack of diagram in many solutions. Most importantly the lack of understanding of the notation AB was apparent. Teachers need to make sure that students are aware of correct notation as given in the outline. A number used the cosine rule but then confused the required angle or sides.

Solve \(2\sin (x + 60^\circ ) = \cos (x + 30^\circ ),{\text{ }}0^\circ \leqslant x \leqslant 180^\circ \).

Show that \(\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}\).

Find the modulus and argument of \(z\) in terms of \(\theta \). Express each answer in its simplest form.

Hence find the cube roots of \(z\) in modulus-argument form.

\(2\sin (x + 60^\circ ) = \cos (x + 30^\circ )\)

\(2(\sin x\cos 60^\circ + \cos x\sin 60^\circ ) = \cos x\cos 30^\circ - \sin x\sin 30^\circ \)     (M1)(A1)

\(2\sin x \times \frac{1}{2} + 2\cos x \times \frac{{\sqrt 3 }}{2} = \cos x \times \frac{{\sqrt 3 }}{2} - \sin x \times \frac{1}{2}\)     A1

\( \Rightarrow \frac{3}{2}\sin x = - \frac{{\sqrt 3 }}{2}\cos x\)

\( \Rightarrow \tan x = - \frac{1}{{\sqrt 3 }}\)     M1

\( \Rightarrow x = 150^\circ \)     A1

[5 marks]

EITHER

choosing two appropriate angles, for example 60° and 45°     M1

\(\sin 105^\circ = \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \) and

\(\cos 105^\circ = \cos 60^\circ \cos 45^\circ - \sin 60^\circ \sin 45^\circ \)     (A1)

\(\sin 105^\circ + \cos 105^\circ = \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} + \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}\)     A1

\( = \frac{1}{{\sqrt 2 }}\)     AG

OR

attempt to square the expression     M1

\({(\sin 105^\circ + \cos 105^\circ )^2} = {\sin ^2}105^\circ + 2\sin 105^\circ \cos 105^\circ + {\cos ^2}105^\circ \)

\({(\sin 105^\circ + \cos 105^\circ )^2} = 1 + \sin 210^\circ \)     A1

\( = \frac{1}{2}\)     A1

\(\sin 105^\circ + \cos 105^\circ = \frac{1}{{\sqrt 2 }}\)   AG

[3 marks]

EITHER

\(z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta \)

\(\left| z \right| = \sqrt {{{(1 - \cos 2\theta )}^2} + {{(\sin 2\theta )}^2}} \)     M1

\(\left| z \right| = \sqrt {1 - 2\cos 2\theta + {{\cos }^2}2\theta + {{\sin }^2}2\theta } \)     A1

\( = \sqrt 2 \sqrt {(1 - \cos 2\theta )} \)     A1

\( = \sqrt {2(2{{\sin }^2}\theta )} \)

\( = 2\sin \theta \)     A1

let \(\arg (z) = \alpha \)

\(\tan \alpha = - \frac{{\sin 2\theta }}{{1 - \cos 2\theta }}\)     M1

\( = \frac{{ - 2\sin \theta \cos \theta }}{{2{{\sin }^2}\theta }}\)     (A1)

\( = - \cot \theta \)     A1

\(\arg (z) = \alpha = - \arctan \left( {\tan \left( {\frac{\pi }{2} - \theta } \right)} \right)\)     A1

\( = \theta - \frac{\pi }{2}\)     A1

OR

\(z = (1 - \cos 2\theta ) - {\text{i}}\sin 2\theta \)

\( = 2{\sin ^2}\theta - 2{\text{i}}\sin \theta \cos \theta \)     M1A1

\( = 2\sin \theta (\sin \theta - {\text{i}}\cos \theta )\)     (A1)

\( = - 2{\text{i}}\sin \theta (\cos \theta + {\text{i}}\sin \theta )\)     M1A1

\( = 2\sin \theta \left( {\cos \left( {\theta - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\theta - \frac{\pi }{2}} \right)} \right)\)     M1A1

\(\left| z \right| = 2\sin \theta \)     A1

\(\arg (z) = \theta - \frac{\pi }{2}\)     A1

[9 marks]

attempt to apply De Moivre’s theorem     M1

\({(1 - \cos 2\theta - {\text{i}}\sin 2\theta )^{\frac{1}{3}}} = {2^{\frac{1}{3}}}{(\sin \theta )^{\frac{1}{3}}}\left[ {\cos \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\theta - \frac{\pi }{2} + 2n\pi }}{3}} \right)} \right]\)     A1A1A1

Note:     A1 for modulus, A1 for dividing argument of \(z\) by 3 and A1 for \(2n\pi \).

Hence cube roots are the above expression when \(n = - 1,{\text{ }}0,{\text{ }}1\). Equivalent forms are acceptable.     A1

[5 marks]

Find the real part of \(\frac{{z + w}}{{z - w}}\).

Find the value of the real part of \(\frac{{z + w}}{{z - w}}\) when \(\left| z \right| = \left| w \right|\).

\(\frac{{z + w}}{{z - w}} = \frac{{\left( {a + c} \right) + {\text{i}}\left( {b + d} \right)}}{{\left( {a - c} \right) + {\text{i}}\left( {b - d} \right)}}\)

\( = \frac{{\left( {a + c} \right) + {\text{i}}\left( {b + d} \right)}}{{\left( {a - c} \right) + {\text{i}}\left( {b - d} \right)}} \times \frac{{\left( {a - c} \right) - {\text{i}}\left( {b - d} \right)}}{{\left( {a - c} \right) - {\text{i}}\left( {b - d} \right)}}\)     M1A1

real part \( = \frac{{\left( {a + c} \right)\left( {a - c} \right) + \left( {b + d} \right)\left( {b - d} \right)}}{{{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}} = \left( {\frac{{{a^2} - {c^2} + {b^2} - {d^2}}}{{{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}}} \right)\)     A1A1

Note: Award A1 for numerator, A1 for denominator.

[4 marks]

\(\left| z \right| = \left| w \right| \Rightarrow {a^2} + {b^2} = {c^2} + {d^2}\)     R1

hence real part = 0      A1

Note: Do not award R0A1.

[2 marks]

Consider the complex number \(\omega = \frac{{z + {\text{i}}}}{{z + 2}}\), where \(z = x + {\text{i}}y\) and \({\text{i}} = \sqrt { - 1} \).

(a)     If \(\omega = {\text{i}}\), determine z in the form \(z = r\,{\text{cis}}\,\theta \).

(b)     Prove that \(\omega = \frac{{({x^2} + 2x + {y^2} + y) + {\text{i}}(x + 2y + 2)}}{{{{(x + 2)}^2} + {y^2}}}\).

(c)     Hence show that when \(\operatorname{Re} (\omega) = 1\) the points \((x,{\text{ }}y)\) lie on a straight line, \({l_1}\), and write down its gradient.

(d)     Given \(\arg (z) = \arg (\omega) = \frac{\pi }{4}\), find \(\left| z \right|\).

(a)     METHOD 1

\(\frac{{z + {\text{i}}}}{{z + 2}} = {\text{i}}\)

\(z + {\text{i}} = {\text{i}}z + 2{\text{i}}\)     M1

\((1 - {\text{i}})z = {\text{i}}\)     A1

\(z = \frac{{\text{i}}}{{1 - {\text{i}}}}\)     A1

EITHER

\(z = \frac{{{\text{cis}}\left( {\frac{\pi }{2}} \right)}}{{\sqrt 2 \,{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)}}\)     M1

\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{4\pi }}{4}} \right)} \right)\)     A1A1

OR

\(z = \frac{{ - 1 + {\text{i}}}}{2}{\text{ }}\left( { = - \frac{1}{2} + \frac{1}{2}{\text{i}}} \right)\)     M1

\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)} \right)\)     A1A1

[6 marks]

METHOD 2

\({\text{i}} = \frac{{x + {\text{i}}(y + 1)}}{{x + 2 + {\text{i}}y}}\)     M1

\(x + {\text{i}}(y + 1) = - y + {\text{i}}(x + 2)\)     A1

\(x = - y;{\text{ }}x + 2 = y + 1\)     A1

solving, \(x = - \frac{1}{2};{\text{ }}y = \frac{1}{2}\)     A1

\(z = - \frac{1}{2} + \frac{1}{2}{\text{i}}\)

\(z = \frac{{\sqrt 2 }}{2}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right){\text{ }}\left( {{\text{or }}\frac{1}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{3\pi }}{4}} \right)} \right)\)     A1A1

Note: Award A1 fort the correct modulus and A1 for the correct argument, but the final answer must be in the form \({\text{r}}\,{\text{cis}}\,\theta \). Accept 135° for the argument.

[6 marks]

(b) substituting \(z = x + {\text{i}}y\) to obtain \(w = \frac{{x + (y + 1){\text{i}}}}{{(x + 2) + y{\text{i}}}}\)     (A1)

use of \((x + 2) - y{\text{i}}\) to rationalize the denominator     M1

\(\omega = \frac{{x(x + 2) + y(y + 1) + {\text{i}}\left( { - xy + (y + 1)(x + 2)} \right)}}{{{{(x + 2)}^2} + {y^2}}}\)     A1

\( = \frac{{({x^2} + 2x + {y^2} + y) + {\text{i}}(x + 2y + 2)}}{{{{(x + 2)}^2} + {y^2}}}\)     AG

[3 marks]

(c)     \(\operatorname{Re} \omega = \frac{{{x^2} + 2x + {y^2} + y}}{{{{(x + 2)}^2} + {y^2}}} = 1\)     M1

\( \Rightarrow {x^2} + 2x + {y^2} + y = {x^2} + 4x + 4 + {y^2}\)     A1

\( \Rightarrow y = 2x + 4\)     A1

which has gradient m = 2     A1

[4 marks]

(d)     EITHER

\(\arg (z) = \frac{\pi }{4} \Rightarrow x = y{\text{ (and }}x,{\text{ }}y > 0)\)     (A1)

\(\omega = \frac{{2{x^2} + 3x}}{{{{(x + 2)}^2} + {x^2}}} + \frac{{{\text{i}}(3x + 2)}}{{{{(x + 2)}^2} + {x^2}}}\)

if \(\arg (\omega) = \theta \Rightarrow \tan \theta = \frac{{3x + 2}}{{2{x^2} + 3x}}\)     (M1)

\(\frac{{3x + 2}}{{2{x^2} + 3x}} = 1\)     M1A1

OR

\(\arg (z) = \frac{\pi }{4} \Rightarrow x = y{\text{ (and }}x,{\text{ }}y > 0)\)     A1

\(\arg (w) = \frac{\pi }{4} \Rightarrow {x^2} + 2x + {y^2} + y = x + 2y + 2\)     M1

solve simultaneously     M1

\({x^2} + 2x + {x^2} + x = x + 2x + 2\) (or equivalent)     A1

THEN

\({x^2} = 1\)

\(x = 1{\text{ (as }}x > 0)\)     A1

Note: Award A0 for x = ±1.

\(\left| z \right| = \sqrt 2 \)     A1

Note: Allow FT from incorrect values of x.

[6 marks]

Total [19 marks]

Many candidates knew what had to be done in (a) but algebraic errors were fairly common. Parts (b) and (c) were well answered in general. Part (d), however, proved beyond many candidates who had no idea how to convert the given information into mathematical equations.

Find three distinct roots of the equation \(8{z^3} + 27 = 0,{\text{ }}z \in \mathbb{C}\) giving your answers in modulus-argument form.

The roots are represented by the vertices of a triangle in an Argand diagram.

Show that the area of the triangle is \(\frac{{27\sqrt 3 }}{{16}}\).

METHOD 1

\({z^3} =  - \frac{{27}}{8} = \frac{{27}}{8}(\cos \pi  + {\text{i}}\sin \pi )\)     M1(A1)

\( = \frac{{27}}{8}\left( {\cos (\pi  + 2n\pi ) + {\text{i}}\sin (\pi  + 2n\pi )} \right)\)     (A1)

\(z = \frac{3}{2}\left( {\cos \left( {\frac{{\pi  + 2n\pi }}{3}} \right) + {\text{i}}\sin \left( {\frac{{\pi  + 2n\pi }}{3}} \right)} \right)\)     M1

\({z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)\),

\({z_2} = \frac{3}{2}(\cos \pi  + {\text{i}}\sin \pi )\),

\({z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)\).     A2

Note:     Accept \( - \frac{\pi }{3}\) as the argument for \({z_3}\).

Note:     Award A1 for \(2\) correct roots.

Note:     Allow solutions expressed in Eulerian \((r{e^{{\text{i}}\theta }})\) form.

Note:     Allow use of degrees in mod-arg (r-cis) form only.

METHOD 2

\(8{z^3} + 27 = 0\)

\( \Rightarrow z =  - \frac{3}{2}\) so \((2z + 3)\) is a factor

Attempt to use long division or factor theorem:     M1

\( \Rightarrow 8{z^3} + 27 = (2z + 3)(4{z^2} - 6z + 9)\)

\( \Rightarrow 4{z^2} - 6z + 9 = 0\)     A1

Attempt to solve quadratic:     M1

\(z = \frac{{3 \pm 3\sqrt 3 {\text{i}}}}{4}\)     A1

\({z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)\),

\({z_2} = \frac{3}{2}(\cos \pi  + {\text{i}}\sin \pi )\),

\({z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)\).     A2

Note:     Accept \( - \frac{\pi }{3}\) as the argument for \({z_3}\).

Note:     Award A1 for \(2\) correct roots.

Note:     Allow solutions expressed in Eulerian \((r{e^{{\text{i}}\theta }})\) form.

Note:     Allow use of degrees in mod-arg (r-cis) form only.

METHOD 3

\(8{z^3} + 27 = 0\)

Substitute \(z = x + {\text{i}}y\)     M1

\(8({x^3} + 3{\text{i}}{x^2}y - 3x{y^2} - {\text{i}}{y^3}) + 27 = 0\)

\( \Rightarrow 8{x^3} - 24x{y^2} + 27 = 0\) and \(24{x^2}y - 8{y^3} = 0\)     A1

Attempt to solve simultaneously:     M1

\(8y(3{x^2} - {y^2}) = 0\)

\(y = 0,{\text{ }}y = x\sqrt 3 ,{\text{ }}y =  - x\sqrt 3 \)

\( \Rightarrow \left( {x =  - \frac{3}{2},{\text{ }}y = 0} \right),{\text{ }}x = \frac{3}{4},{\text{ }}y =  \pm \frac{{3\sqrt 3 }}{4}\)     A1

\({z_1} = \frac{3}{2}\left( {\cos \frac{\pi }{3} + {\text{i}}\sin \frac{\pi }{3}} \right)\),

\({z_2} = \frac{3}{2}(\cos \pi  + {\text{i}}\sin \pi )\),

\({z_3} = \frac{3}{2}\left( {\cos \frac{{5\pi }}{3} + {\text{i}}\sin \frac{{5\pi }}{3}} \right)\).     A2

Note:     Accept \( - \frac{\pi }{3}\) as the argument for \({z_3}\).

Note:     Award A1 for \(2\) correct roots.

Note:     Allow solutions expressed in Eulerian \((r{e^{{\text{i}}\theta }})\) form.

Note:     Allow use of degrees in mod-arg (r-cis) form only.

[6 marks]

EITHER

Valid attempt to use \({\text{area}} = 3\left( {\frac{1}{2}ab\sin C} \right)\)     M1

\( = 3 \times \frac{1}{2} \times \frac{3}{2} \times \frac{3}{2} \times \frac{{\sqrt 3 }}{2}\)     A1A1

Note:     Award A1 for correct sides, A1 for correct sin \(C\).

OR

Valid attempt to use \({\text{area}} = \frac{1}{2}{\text{base}} \times {\text{height}}\)     M1

\({\text{area}} = \frac{1}{2} \times \left( {\frac{3}{4} + \frac{3}{2}} \right) \times \frac{{6\sqrt 3 }}{4}\)     A1A1

Note:     A1 for correct height, A1 for correct base.

THEN

\( = \frac{{27\sqrt 3 }}{{16}}\)     AG

[3 marks]

Total [9 marks]

Consider the complex numbers \(z = 1 + 2{\text{i}}\) and \(w = 2 + a{\text{i}}\) , where \(a \in \mathbb{R}\) .

Find a when

(a)     \(\left| w \right| = 2\left| z \right|;\) ;

(b)     \({\text{Re}}(zw) = 2\operatorname{Im} (zw)\) .

(a)     \(\left| z \right| = \sqrt 5 \) and \(\left| w \right| = \sqrt {4 + {a^2}} \)

\(\left| w \right| = 2\left| z \right|\)

\(\sqrt {4 + {a^2}} = 2\sqrt 5 \)

attempt to solve equation     M1

Note: Award M0 if modulus is not used.

\(a = \pm 4\)     A1A1     N0

(b)     \(zw = (2 - 2a) + (4 + a){\text{i}}\)     A1

forming equation \(2 - 2a = 2(4 + a)\)     M1

\(a = - \frac{3}{2}\)     A1     N0

[6 marks]

Most candidates made good attempts to answer this question. Weaker candidates did not get full marks due to difficulties recognizing the notation and working with modulus of a complex number.

In the following Argand diagram the point A represents the complex number \( - 1 + 4{\text{i}}\) and the point B represents the complex number \( - 3 + 0{\text{i}}\). The shape of ABCD is a square. Determine the complex numbers represented by the points C and D.

C represents the complex number \(1 - 2{\text{i}}\)     A2

D represents the complex number \(3 + 2{\text{i}}\)     A2

[4 marks]

Use mathematical induction to prove that \(n({n^2} + 5)\) is divisible by 6 for \(n \in {\mathbb{Z}^ + }\).

let \({\text{P}}(n)\) be the proposition that \(n({n^2} + 5)\) is divisible by 6 for \(n \in {\mathbb{Z}^ + }\)

consider P(1):

when \(n = 1,{\text{ }}n({n^2} + 5) = 1 \times ({1^2} + 5) = 6\) and so P(1) is true     R1

assume \({\text{P}}(k)\) is true ie, \(k({k^2} + 5) = 6m\) where \(k,{\text{ }}m \in {\mathbb{Z}^ + }\)     M1

Note:     Do not award M1 for statements such as “let \(n = k\)”.

consider \({\text{P}}(k + 1)\):

\((k + 1)\left( {{{(k + 1)}^2} + 5} \right)\)    M1

\( = (k + 1)({k^2} + 2k + 6)\)

\( = {k^3} + 3{k^2} + 8k + 6\)    (A1)

\( = ({k^3} + 5k) + (3{k^2} + 3k + 6)\)    A1

\( = k({k^2} + 5) + 3k(k + 1) + 6\)    A1

\(k(k + 1)\) is even hence all three terms are divisible by 6     R1

\({\text{P}}(k + 1)\) is true whenever \({\text{P}}(k)\) is true and P(1) is true, so \({\text{P}}(n)\) is true for \(n \in {\mathbb{Z}^ + }\)     R1

Note:     To obtain the final R1, four of the previous marks must have been awarded.

[8 marks]

This proved to be a good discriminator. The average candidate seemed able to work towards \({\text{P}}(k + 1) = {k^3} + 3{k^2} + 8k + 6\), and a number made some further progress.

Unfortunately, even otherwise good candidates are still writing down incorrect or incomplete induction statements, such as ‘Let \(n = k\)’ rather than ‘Suppose true for \(n = k\)’ (or equivalent).

It was also noted than an increasing number of candidates this session assumed ‘\({\text{P}}(n)\) to be true’ before going to consider \({\text{P}}(n + 1)\). Showing a lack of understanding of the induction argument, these approaches scored very few marks.

Show that \({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x\) where \(r,\,x \in {\mathbb{R}^ + }\).

Express \(y\) in terms of \(x\). Give your answer in the form \(y = p{x^q}\), where p , q are constants.

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines \(x = 1\) and \(x = \alpha \) where \(\alpha  > 1\). The area of R is \(\sqrt 2 \).

Find the value of \(\alpha \).

METHOD 1

\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)\)     M1A1

\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\)     AG

[2 marks]

METHOD 2

\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}\)     M1

\( = \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}\)     A1

\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\)     AG

[2 marks]

METHOD 1

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y =  - \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}\)

\({\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)\)     M1A1

\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\)     A1

Note: For the final A mark, \(y\) must be expressed in the form \(p{x^q}\).

[5 marks]

METHOD 2

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)

\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0\)     M1

\(\sqrt 2 xy = 1\)     A1

\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\)     A1

Note: For the final A mark, \(y\) must be expressed in the form \(p{x^q}\).

[5 marks]

the area of R is \(\int\limits_1^\alpha  {\frac{1}{{\sqrt 2 }}} {x^{ - 1}}{\text{d}}x\)     M1

\( = \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha \)     A1

\( = \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha \)     A1

\(\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha  = \sqrt 2 \)     M1

\(\alpha  = {{\text{e}}^2}\)     A1

Note: Only follow through from part (b) if \(y\) is in the form \(y = p{x^q}\)

[5 marks]

An 81 metre rope is cut into n pieces of increasing lengths that form an arithmetic sequence with a common difference of d metres. Given that the lengths of the shortest and longest pieces are 1.5 metres and 7.5 metres respectively, find the values of n and d .

\(81 = \frac{n}{2}(1.5 + 7.5)\)     M1

\( \Rightarrow n = 18\)     A1

\(1.5 + 17d = 7.5\)     M1

\( \Rightarrow d = \frac{6}{{17}}\)     A1     N0

[4 marks]

There were many totally correct solutions to this question, but a number of candidates found two simultaneous equations and then spent a lot of time and working trying, often unsuccessfully, to solve these equations.

Given that \(\frac{z}{{z + 2}} = 2 - {\text{i}}\) , \(z \in \mathbb{C}\) , find z in the form \(a + {\text{i}}b\) .

METHOD 1
\(z = \left( {2 - {\text{i}}} \right)\left( {z + 2} \right)\)     M1
\( = 2z + 4 - {\text{i}}z - 2{\text{i}}\)
\(z\left( {1 - {\text{i}}} \right) = - 4 + 2{\text{i}}\)
\(z = \frac{{ - 4 + 2{\text{i}}}}{{1 - {\text{i}}}}\)     A1
\(z = \frac{{ - 4 + 2{\text{i}}}}{{1 - {\text{i}}}} \times \frac{{1 + {\text{i}}}}{{1 + {\text{i}}}}\)     M1
\( = - 3 - {\text{i}}\)     A1

METHOD 2

let \(z = a + {\text{i}}b\)
\(\frac{{a + {\text{i}}b}}{{a + {\text{i}}b + 2}} = 2 - {\text{i}}\)     M1
\(a + {\text{i}}b = \left( {2 - i} \right)\left( {\left( {a + 2} \right) + {\text{i}}b} \right)\)
\(a + {\text{i}}b = 2\left( {a + 2} \right) + 2b{\text{i}} - {\text{i}}\left( {a + 2} \right) + b\)
\(a + {\text{i}}b = 2a + b + 4 + \left( {2b - a - 2} \right){\text{i}}\)
attempt to equate real and imaginary parts     M1
\(a = 2a + b + 4\left( { \Rightarrow a + b + 4 = 0} \right)\)
and \(b = 2b - a - 2\left( { \Rightarrow - a + b - 2 = 0} \right)\)     A1

Note: Award Al for two correct equations.

\(b = - 1\); \(a = - 3\)     A1

\(z = - 3 - {\text{i}}\)

[4 marks]

A number of different methods were adopted in this question with some candidates working through their method to a correct answer. However many other candidates either stopped with \(z\) still expressed as a quotient of two complex numbers or made algebraic mistakes.

m and n are real numbers;

m and n are conjugate complex numbers.

attempt to equate real and imaginary parts     M1

equate real parts: \(4m + 4n = 16\); equate imaginary parts: \( -5m = 15\)     A1

\( \Rightarrow m = -3,{\text{ }}n = 7\)     A1

[3 marks]

let \(m = x + {\text{i}}y,{\text{ }}n = x - {\text{i}}y\)     M1

\( \Rightarrow (4 - 5{\text{i}})(x + {\text{i}}y) + 4(x - {\text{i}}y) = 16 + 15{\text{i}}\)

\( \Rightarrow 4x - 5{\text{i}}x + 4{\text{i}}y + 5y + 4x - 4{\text{i}}y = 16 + 15{\text{i}}\)

attempt to equate real and imaginary parts     M1

\(8x + 5y = 16,{\text{ }} -5x = 15\)     A1

\( \Rightarrow x = -3,{\text{ }}y = 8\)     A1

\(( \Rightarrow m = -3 + 8{\text{i}},{\text{ }}n = -3 - 8{\text{i}})\)

[4 marks]

Part (a) was generally well answered. In (b), however, some candidates put \(m = a + {\text{i}}b\) and \(n = c + {\text{i}}d\) which gave four equations for two unknowns so that no further progress could be made.

Part (a) was generally well answered. In (b), however, some candidates put \(m = a + {\text{i}}b\) and \(n = c + {\text{i}}d\) which gave four equations for two unknowns so that no further progress could be made.

Use De Moivre’s theorem to show that \({z^n} + {z^{ - n}} = 2\cos n\theta ,{\text{ }}n \in {\mathbb{Z}^ + }\).

Expand \({\left( {z + {z^{ - 1}}} \right)^4}\).

Hence show that \({\cos ^4}\theta  = p\cos 4\theta  + q\cos 2\theta  + r\), where \(p,{\text{ }}q\) and \(r\) are constants to be determined.

Show that \({\cos ^6}\theta  = \frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}\).

Hence find the value of \(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta } \).

S is rotated through \(2\pi \) radians about the x-axis. Find the value of the volume generated.

(i)     Write down an expression for the constant term in the expansion of \({\left( {z + {z^{ - 1}}} \right)^{2k}}\), \(k \in {\mathbb{Z}^ + }\).

(ii)     Hence determine an expression for \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta } \) in terms of k.

\({z^n} + {z^{ - n}} = \cos n\theta  + i\sin n\theta  + \cos ( - n\theta ) + i\sin ( - n\theta )\)     M1

\( = \cos n\theta  + \cos n\theta  + i\sin n\theta  - i\sin n\theta \)     A1

\( = 2\cos n\theta \)     AG

[2 marks]

(b)     \({\left( {z + {z^{ - 1}}} \right)^4} = {z^4} + 4{z^3}\left( {\frac{1}{z}} \right) + 6{z^2}\left( {\frac{1}{{{z^2}}}} \right) + 4z\left( {\frac{1}{{{z^3}}}} \right) + \frac{1}{{{z^4}}}\)     A1

Note:     Accept \({\left( {z + {z^{ - 1}}} \right)^4} = 16{\cos ^4}\theta \).

[1 mark]

METHOD 1

\({\left( {z + {z^{ - 1}}} \right)^4} = \left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 4\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 6\)     M1

\({(2\cos \theta )^4} = 2\cos 4\theta  + 8\cos 2\theta  + 6\)     A1A1

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

\({\cos ^4}\theta  = \frac{1}{8}\cos 4\theta  + \frac{1}{2}\cos 2\theta  + \frac{3}{8}\)     A1

\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)

METHOD 2

\({\cos ^4}\theta  = {\left( {\frac{{\cos 2\theta  + 1}}{2}} \right)^2}\)     M1

\( = \frac{1}{4}({\cos ^2}2\theta  + 2\cos 2\theta  + 1)\)     A1

\( = \frac{1}{4}\left( {\frac{{\cos 4\theta  + 1}}{2} + 2\cos 2\theta  + 1} \right)\)     A1

\({\cos ^4}\theta  = \frac{1}{8}\cos 4\theta  + \frac{1}{2}\cos 2\theta  + \frac{3}{8}\)     A1

\(\left( {{\text{or }}p = \frac{1}{8},{\text{ }}q = \frac{1}{2},{\text{ }}r = \frac{3}{8}} \right)\)

[4 marks]

\({\left( {z + {z^{ - 1}}} \right)^6} = {z^6} + 6{z^5}\left( {\frac{1}{z}} \right) + 15{z^4}\left( {\frac{1}{{{z^2}}}} \right) + 20{z^3}\left( {\frac{1}{{{z^3}}}} \right) + 15{z^2}\left( {\frac{1}{{{z^4}}}} \right) + 6z\left( {\frac{1}{{{z^5}}}} \right) + \frac{1}{{{z^6}}}\)     M1

\({\left( {z + {z^{ - 1}}} \right)^6} = \left( {{z^6} + \frac{1}{{{z^6}}}} \right) + 6\left( {{z^4} + \frac{1}{{{z^4}}}} \right) + 15\left( {{z^2} + \frac{1}{{{z^2}}}} \right) + 20\)

\({(2\cos \theta )^6} = 2\cos 6\theta  + 12\cos 4\theta  + 30\cos 2\theta  + 20\)     A1A1

Note:     Award A1 for RHS, A1 for LHS, independent of the M1.

\({\cos ^6}\theta  = \frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}\)     AG

Note:     Accept a purely trigonometric solution as for (c).

[3 marks]

\(\int_0^{\frac{\pi }{2}} {{{\cos }^6}\theta {\text{d}}\theta  = \int_0^{\frac{\pi }{2}} {\left( {\frac{1}{{32}}\cos 6\theta  + \frac{3}{{16}}\cos 4\theta  + \frac{{15}}{{32}}\cos 2\theta  + \frac{5}{{16}}} \right){\text{d}}\theta } } \)

\( = \left[ {\frac{1}{{192}}\sin 6\theta  + \frac{3}{{64}}\sin 4\theta  + \frac{{15}}{{64}}\sin 2\theta  + \frac{5}{{16}}\theta } \right]_0^{\frac{\pi }{2}}\)     M1A1

\( = \frac{{5\pi }}{{32}}\)     A1

[3 marks]

\({\text{V}} = \pi \int_0^{\frac{\pi }{2}} {{{\sin }^2}x{{\cos }^4}x{\text{d}}x} \)     M1

\( = \pi \int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x - \pi \int_0^{\frac{\pi }{2}} {{{\cos }^6}x{\text{d}}x} } \)     M1

\(\int_0^{\frac{\pi }{2}} {{{\cos }^4}x{\text{d}}x}  = \frac{{3\pi }}{{16}}\)     A1

\({\text{V}} = \frac{{3{\pi ^2}}}{{16}} - \frac{{5{\pi ^2}}}{{32}} = \frac{{{\pi ^2}}}{{32}}\)     A1

Note:     Follow through from an incorrect r in (c) provided the final answer is positive.

(i)     constant term = \(\left( \begin{array}{c}2k\\k\end{array} \right)\) \( = \frac{{(2k)!}}{{k!k!}} = \frac{{(2k)!}}{{{{(k!)}^2}}}{\text{ (accept }}C_k^{2k})\)     A1

(ii)     \({2^{2k}}\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta  = \frac{{(2k)!\pi }}{{{{(k!)}^2}}}\frac{\pi }{2}} \)     A1

          \(\int_0^{\frac{\pi }{2}} {{{\cos }^{2k}}\theta {\text{d}}\theta  = \frac{{(2k)!\pi }}{{{2^{2k + 1}}{{(k!)}^2}}}} \) \(\left( {{\rm{or}}\frac{{\left( \begin{array}{c}2k\\k\end{array} \right)\pi }}{{{2^{2k + 1}}}}} \right)\)     A1

[3 marks]

Part a) has appeared several times before, though with it again being a ‘show that’ question, some candidates still need to be more aware of the need to show every step in their working, including the result that \(\sin ( - n\theta ) =  - \sin (n\theta )\).

Part b) was usually answered correctly.

Part c) was again often answered correctly, though some candidates often less successfully utilised a trig-only approach rather than taking note of part b).

Part d) was a good source of marks for those who kept with the spirit of using complex numbers for this type of question. Some limited attempts at trig-only solutions were seen, and correct solutions using this approach were extremely rare.

Part e) was well answered, though numerical slips were often common. A small number integrated \(\sin n\theta \) as \(n\cos n\theta \).

A large number of candidates did not realise the help that part e) inevitably provided for part f). Some correctly expressed the volume as \(\pi \int {{{\cos }^4}x{\text{d}}x - \pi \int {{{\cos }^6}x{\text{d}}x} } \) and thus gained the first 2 marks but were able to progress no further. Only a small number of able candidates were able to obtain the correct answer of \(\frac{{{\pi ^2}}}{{32}}\).

Part g) proved to be a challenge for the vast majority, though it was pleasing to see some of the highest scoring candidates gain all 3 marks.

Show that A is an arithmetic sequence, stating its common difference d in terms of r.

A particular geometric sequence has u₁ = 3 and a sum to infinity of 4.

Find the value of d.

METHOD 1

state that \({u_n} = {u_1}{r^{n - 1}}\) (or equivalent)      A1

attempt to consider \({{a_n}}\) and use of at least one log rule       M1

\({\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right| = {\text{lo}}{{\text{g}}_2}\left| {{u_1}} \right| + \left( {n - 1} \right){\text{lo}}{{\text{g}}_2}\left| r \right|\)      A1

(which is an AP) with \(d = {\text{lo}}{{\text{g}}_2}\left| r \right|\) (and 1^st term \({\text{lo}}{{\text{g}}_2}\left| {{u_1}} \right|\))      A1

so A is an arithmetic sequence      AG

Note: Condone absence of modulus signs.

Note: The final A mark may be awarded independently.

Note: Consideration of the first two or three terms only will score M0.

[4 marks]

METHOD 2

consideration of \(\left( {d = } \right){a_{n + 1}} - {a_n}\)      M1

\(\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| {{u_{n + 1}}} \right| - {\text{lo}}{{\text{g}}_2}\left| {{u_n}} \right|\)

\(\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| {\frac{{{u_{n + 1}}}}{{{u_n}}}} \right|\)     M1

\(\left( d \right) = {\text{lo}}{{\text{g}}_2}\left| r \right|\)     A1

which is constant      R1

Note: Condone absence of modulus signs.

Note: The final A mark may be awarded independently.

Note: Consideration of the first two or three terms only will score M0.

attempting to solve \(\frac{3}{{1 - r}} = 4\)     M1

\(r = \frac{1}{4}\)     A1

\(d =  - \,2\)     A1

[3 marks]

If each of these equations defines a plane, show that, for any value of a , the planes do not intersect at a unique point.

Find the value of b for which the intersection of the planes is a straight line.

METHOD 1

\(\det \left( {\begin{array}{*{20}{c}}
  1&3&{a - 1} \\
  2&2&{a - 2} \\
  3&1&{a - 3}
\end{array}} \right)\)     M1

\( = 1\left( {2(a - 3) - (a - 2)} \right) - 3\left( {2(a - 3) - 3(a - 2)} \right) + (a - 1)(2 - 6)\)

(or equivalent)     A1

= 0 (therefore there is no unique solution)     A1

[3 marks] 

METHOD 2 

\(\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
2&2&{a - 2}\\
3&1&{a - 3}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
1\\
b
\end{array}} \right):\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&{ - 8}&{ - 2a}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 3}
\end{array}} \right)\)     M1A1

\(:\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&0&0
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 1}
\end{array}} \right)\) (and 3 zeros imply no unique solution)     A1

[3 marks]

METHOD 1

\(\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
2&2&{a - 2}\\
3&1&{a - 3}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
1\\
b
\end{array}} \right):\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&{ - 8}&{ - 2a}
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 3}
\end{array}} \right)\)     M1A1

\(:\left( {\begin{array}{*{20}{c}}
1&3&{a - 1}\\
0&{ - 4}&{ - a}\\
0&0&0
\end{array}} \right|\left. {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
{b - 1}
\end{array}} \right)\)     A1

b = 1     A1     N2

Note: Award M1 for an attempt to use row operations.

[4 marks]

METHOD 2

b = 1     A4

Note: Award A4 only if “ b −1 ” seen in (a).

[4 marks]

The best candidates used row reduction correctly in part a) and were hence able to deduce b = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).

The best candidates used row reduction correctly in part a) and were hence able to deduce b = 1 in part b) for an easy final 4 marks. The determinant method was often usefully employed in part a).

By expressing \({z_1}\) and \({z_2}\) in modulus-argument form write down the modulus of \(w\);

By expressing \({z_1}\) and \({z_2}\) in modulus-argument form write down the argument of \(w\).

Find the smallest positive integer value of \(n\), such that \({w^n}\) is a real number.

\({z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)\) and \({z_2} = \sqrt 2 {\text{cis}}\left( {\frac{\pi }{4}} \right)\)     A1A1

Note:     Award A1A0 for correct moduli and arguments found, but not written in mod-arg form.

\(\left| w \right| = \sqrt 2 \)     A1

[3 marks]

\({z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)\) and \({z_2} = \sqrt 2 {\text{cis}}\left( {\frac{\pi }{4}} \right)\)     A1A1

Note:     Award A1A0 for correct moduli and arguments found, but not written in mod-arg form.

\(\arg w = \frac{\pi }{{12}}\)     A1

Notes:     Allow FT from incorrect answers for \({z_1}\) and \({z_2}\) in modulus-argument form.

[1 mark]

EITHER

\(\sin \left( {\frac{{\pi n}}{{12}}} \right) = 0\)     (M1)

OR

\(\arg ({w^n}) = \pi \)     (M1)

\(\frac{{n\pi }}{{12}} = \pi \)

THEN

\(\therefore n = 12\)     A1

[2 marks]

(i)     Write down \({z_1}\) in Cartesian form.

(ii)     Hence determine \({({z_1} + {z_2})^ * }\) in Cartesian form.

(i)     Write \({z_2}\) in modulus-argument form.

(ii)     Hence solve the equation \({z^3} = {z_2}\) .

Let \(z = r\,{\text{cis}}\theta \) , where \(r \in {\mathbb{R}^ + }\)  and \(0 \leqslant \theta  < 2\pi \) . Find all possible values of r and \(\theta \) ,

(i)     if \({z^2} = {(1 + {z_2})^2}\);

(ii)     if \(z = - \frac{1}{{{z_2}}}\).

Find the smallest positive value of n for which \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} \in {\mathbb{R}^ + }\) .

(i)     \({z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2} \Rightarrow {z_1} = - 2\sqrt 3 {\text{i}}\)     A1

(ii)     \({z_1} + {z_2} = - 2\sqrt 3 {\text{i}} - 1 + \sqrt 3 {\text{i}} = - 1 - \sqrt 3 {\text{i}}\)     A1

\({({z_1} + {z_2})^ * } = - 1 + \sqrt 3 {\text{i}}\)     A1

[3 marks]

(i)     METHOD 1

\({z^2} = {\left( {1 - 1 + \sqrt 3 {\text{i}}} \right)^2} = - 3\left( { \Rightarrow z =  \pm \sqrt 3 {\text{i}}} \right)\)     M1

\(z = \sqrt 3 \,{\text{cis}}\frac{\pi }{2}{\text{ or }}{z_1} = \sqrt 3 \,{\text{cis}}\frac{{3\pi }}{2}\left( { = \sqrt 3 \,{\text{cis}}\left( {\frac{{ - \pi }}{2}} \right)} \right)\)     A1A1

so \(r = \sqrt 3 {\text{ and }}\theta  = \frac{\pi }{2}{\text{ or }}\theta  = \frac{{3\pi }}{2}\left( { = \frac{{ - \pi }}{2}} \right)\) 

Note: Accept \(r\,{\text{cis}}(\theta )\) form.

 METHOD 2

\({z^2} = {\left( {1 - 1 + \sqrt 3 {\text{i}}} \right)^2} = - 3 \Rightarrow {z^2} = 3{\text{cis}}\left( {(2n + 1)\pi } \right)\)     M1

 \({r^2} = 3 \Rightarrow r = \sqrt 3 \)     A1

\(2\theta  = (2n + 1)\pi  \Rightarrow \theta  = \frac{\pi }{2}{\text{ or }}\theta  = \frac{{3\pi }}{2}{\text{ (as }}0 \leqslant \theta  < 2\pi )\)     A1 

Note: Accept \(r\,{\text{cis}}(\theta )\) form.

(ii)     METHOD 1

\(z = - \frac{1}{{2{\text{cis}}\frac{{2\pi }}{3}}} \Rightarrow z = \frac{{{\text{cis}}\pi }}{{2{\text{cis}}\frac{{2\pi }}{3}}}\)     M1

\( \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)

so \(r = \frac{1}{2}{\text{ and }}\theta  = \frac{\pi }{3}\)     A1A1

METHOD 2

\({z_1} = - \frac{1}{{ - 1 + \sqrt 3 {\text{i}}}} \Rightarrow {z_1} = - \frac{{ - 1 - \sqrt 3 {\text{i}}}}{{\left( { - 1 + \sqrt 3 {\text{i}}} \right)\left( { - 1 - \sqrt 3 {\text{i}}} \right)}}\)     M1

\(z = \frac{{1 + \sqrt 3 {\text{i}}}}{4} \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)

so \(r = \frac{1}{2}{\text{ and }}\theta  = \frac{\pi }{3}\)     A1A1

[6 marks]

\(\frac{{{z_1}}}{{{z_2}}} = \sqrt 3 \,{\text{cis}}\frac{{5\pi }}{6}\)     (A1)

\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} = {\sqrt 3 ^n}{\text{cis}}\frac{{5n\pi }}{6}\)     A1

equating imaginary part to zero and attempting to solve     M1

obtain n = 12     A1

Note: Working which only includes the argument is valid.

[4 marks]

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

Many candidates were perhaps fortunate in this question due to there being several follow through marks available. Part a) was often done correctly. In part b), incorrect answers of \(2\,{\text{cis}}\left( { - \frac{\pi }{3}} \right)\) were common, though many of these candidates often applied De Moivre’s Theorem correctly to their answers. In c) the majority found \({z^2} = - 3\) but could then get no further. The second part was often poorly done, with those rationalising the Cartesian form of \({z_1}\) having the most success. Part d) posed problems for a great many, and correct solutions were rarely seen. \({\text{cis}}\left( {\frac{{5n\pi }}{6}} \right)\) was often seen, but then finding n = 12 proved to be a step too far for many. In general, the manipulation of complex numbers in polar form is not well understood.

(a)     The sum of the first six terms of an arithmetic series is 81. The sum of its first eleven terms is 231. Find the first term and the common difference.

(b)     The sum of the first two terms of a geometric series is 1 and the sum of its first four terms is 5. If all of its terms are positive, find the first term and the common ratio.

(c)     The \({r^{{\text{th}}}}\) term of a new series is defined as the product of the \({r^{{\text{th}}}}\) term of the arithmetic series and the \({r^{{\text{th}}}}\) term of the geometric series above. Show that the \({r^{{\text{th}}}}\) term of this new series is \((r + 1){2^{r - 1}}\) .

Using mathematical induction, prove that

\[\sum\limits_{r = 1}^n {(r + 1){2^{r - 1}} = n{2^n},{\text{ }}n \in {\mathbb{Z}^ + }.} \]

(a)     \({S_6} = 81 \Rightarrow 81 = \frac{6}{2}(2a + 5d)\)     M1A1

\( \Rightarrow 27 = 2a + 5d\)

\({S_{11}} = 231 \Rightarrow 231 = \frac{{11}}{2}(2a + 10d)\)     M1A1

\( \Rightarrow 21 = a + 5d\)

solving simultaneously, a = 6 , d = 3     A1A1

[6 marks]

(b)     \(a + ar = 1\)     A1

\(a + ar + a{r^2} + a{r^3} = 5\)     A1

\( \Rightarrow (a + ar) + a{r^2}(1 + r) = 5\)

\( \Rightarrow 1 + a{r^2} \times \frac{1}{a} = 5\)

obtaining \({r^2} - 4 = 0\)     M1

\( \Rightarrow r = \pm 2\)

\(r = 2\,\,\,\,\,\)(since all terms are positive)     A1

\(a = \frac{1}{3}\)     A1

[5 marks]

(c)     \({\text{AP }}{r^{{\text{th}}}}{\text{ term is }}3r + 3\)     A1

\({\text{GP }}{r^{{\text{th}}}}{\text{ term is }}\frac{1}{3}{2^{r - 1}}\)     A1

\(3(r + 1) \times \frac{1}{3}{2^{r - 1}} = (r + 1){2^{r - 1}}\)     M1AG

[3 marks]

Total [14 marks]

prove: \({P_n}:\sum\limits_{r = 1}^n {(r + 1){2^{r - 1}} = n{2^n},{\text{ }}n \in {\mathbb{Z}^ + }.} \)

show true for n = 1 , i.e.

\({\text{LHS}} = 2 \times {2^0} = 2 = {\text{RHS}}\)     A1

assume true for n = k , i.e.     M1

\(\sum\limits_{r = 1}^k {(r + 1){2^{r - 1}} = k{2^k},{\text{ }}k \in {\mathbb{Z}^ + }} \)

consider n = k +1

\(\sum\limits_{r = 1}^{k + 1} {(r + 1){2^{r - 1}} = k{2^k} + (k + 2){2^k}} \)     M1A1

\( = {2^k}(k + k + 2)\)

\( = 2(k + 1){2^k}\)     A1

\( = (k + 1){2^{k + 1}}\)     A1

hence true for n = k + 1

\({P_{k + 1}}\) is true whenever \({P_k}\) is true, and \({P_1}\)is true, therefore \({P_n}\) is true     R1

for \(n \in {\mathbb{Z}^ + }\)

[7 marks]

Parts (a), (b) and (c) were answered successfully by a large number of candidates. Some, however, had difficulty with the arithmetic.

In part (d) many candidates showed little understanding of sigma notation and proof by induction. There were cases of circular reasoning and using n, k and r randomly. A concluding sentence almost always appeared, even if the proof was done incorrectly, or not done at all.

the value of \(d\);

the value of \(r\);

EITHER

the first three terms of the geometric sequence are \(9\), \(9r\) and \(9{r^2}\)     (M1)

\(9 + 3d = 9r( \Rightarrow 3 + d = 3r)\) and \(9 + 7d = 9{r^2}\)     (A1)

attempt to solve simultaneously     (M1)

\(9 + 7d = 9{\left( {\frac{{3 + d}}{3}} \right)^2}\)

OR

the \({{\text{1}}^{{\text{st}}}}\), \({{\text{4}}^{{\text{th}}}}\) and \({{\text{8}}^{{\text{th}}}}\) terms of the arithmetic sequence are

\(9,{\text{ }}9 + 3d,{\text{ }}9 + 7d\)     (M1)

\(\frac{{9 + 7d}}{{9 + 3d}} = \frac{{9 + 3d}}{9}\)     (A1)

attempt to solve     (M1)

THEN

\(d = 1\)     A1

[4 marks]

\(r = \frac{4}{3}\)     A1

Note:     Accept answers where a candidate obtains \(d\) by finding \(r\) first. The first two marks in either method for part (a) are awarded for the same ideas and the third mark is awarded for attempting to solve an equation in \(r\).

[1 mark]

Show that \(\sin \left( {\theta  + \frac{\pi }{2}} \right) = \cos \theta \).

Consider \(f(x) = \sin (ax)\) where \(a\) is a constant. Prove by mathematical induction that \({f^{(n)}}(x) = {a^n}\sin \left( {ax + \frac{{n\pi }}{2}} \right)\) where \(n \in {\mathbb{Z}^ + }\) and \({f^{(n)}}(x)\) represents the \({{\text{n}}^{{\text{th}}}}\) derivative of \(f(x)\).

\(\sin \left( {\theta  + \frac{\pi }{2}} \right) = \sin \theta \cos \frac{\pi }{2} + \cos \theta \sin \frac{\pi }{2}\)     M1

\( = \cos \theta \)     AG

Note:     Accept a transformation/graphical based approach.

[1 mark]

consider \(n = 1,{\text{ }}f'(x) = a\cos (ax)\)     M1

since \(\sin \left( {ax + \frac{\pi }{2}} \right) = \cos ax\) then the proposition is true for \(n = 1\)     R1

assume that the proposition is true for \(n = k\) so \({f^{(k)}}(x) = {a^k}\sin \left( {ax + \frac{{k\pi }}{2}} \right)\)     M1

\({f^{(k + 1)}}(x) = \frac{{{\text{d}}\left( {{f^{(k)}}(x)} \right)}}{{{\text{d}}x}}\;\;\;\left( { = a\left( {{a^k}\cos \left( {ax + \frac{{k\pi }}{2}} \right)} \right)} \right)\)     M1

\( = {a^{k + 1}}\sin \left( {ax + \frac{{k\pi }}{2} + \frac{\pi }{2}} \right)\) (using part (a))     A1

\( = {a^{k + 1}}\sin \left( {ax + \frac{{(k + 1)\pi }}{2}} \right)\)     A1

given that the proposition is true for \(n = k\) then we have shown that the proposition is true for \(n = k + 1\). Since we have shown that the proposition is true for \(n = 1\) then the proposition is true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note:     Award final R1 only if all prior M and R marks have been awarded.

[7 marks]

Total [8 marks]

Solve the equation \(2 - {\log _3}(x + 7) = {\log _{\tfrac{1}{3}}}2x\) .

\({\log _3}\left( {\frac{9}{{x + 7}}} \right) = {\log _3}\frac{1}{{2x}}\)     M1M1A1

Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log and A1 for a correct equation with maximum one log expression each side.

\(x + 7 = 18x\)     M1

\(x = \frac{7}{{17}}\)     A1

[5 marks] 

Some good solutions to this question and few candidates failed to earn marks on the question. Many were able to change the base of the logs, and many were able to deal with the 2, but of those who managed both, poor algebraic skills were often evident. Many students attempted to change the base into base 10, resulting in some complicated algebra, few of which managed to complete successfully. 

Solve the equation \({4^x} + {2^{x + 2}} = 3\).

attempt to form a quadratic in \({2^x}\)     M1

\({({2^x})^2} + 4 \bullet {2^x} - 3 = 0\)    A1

\({2^x} = \frac{{ - 4 \pm \sqrt {16 + 12} }}{2}{\text{ }}\left( { =  - 2 \pm \sqrt 7 } \right)\)    M1

\({2^x} =  - 2 + \sqrt 7 {\text{ }}\left( {{\text{as }} - 2 - \sqrt 7  < 0} \right)\)    R1

\(x = {\log _2}\left( { - 2 + \sqrt 7 } \right){\text{ }}\left( {x = \frac{{\ln \left( { - 2 + \sqrt 7 } \right)}}{{\ln 2}}} \right)\)    A1

Note: Award R0 A1 if final answer is \(x = {\log _2}\left( { - 2 + \sqrt 7 } \right)\).

[5 marks]

Write down and simplify the expansion of \({(2 + x)^4}\) in ascending powers of \(x\).

Hence find the exact value of \({(2.1)^4}\).

\({(2 + x)^4} = {2^4} + 4 \bullet {2^3}x + 6 \bullet {2^2}{x^2} + 4 \bullet 2{x^3} + {x^4}\)     M1(A1)

Note:     Award M1 for an expansion, by whatever method, giving five terms in any order.

\( = 16 + 32x + 24{x^2} + 8{x^3} + {x^4}\)     A1

Note:     Award M1A1A0 for correct expansion not given in ascending powers of \(x\).

[3 marks]

let \(x = 0.1\) (in the binomial expansion)     (M1)

\({2.1^4} = 16 + 3.2 + 0.24 + 0.008 + 0.0001\)     (A1)

\( = 19.4481\)     A1

Note:     At most one of the marks can be implied.

[3 marks]

Total [6 marks]

A geometric sequence has first term a, common ratio r and sum to infinity 76. A second geometric sequence has first term a, common ratio \({r^3}\) and sum to infinity 36.

Find r.

for the first series \(\frac{a}{{1 - r}} = 76\)     A1

for the second series \(\frac{a}{{1 - {r^3}}} = 36\)     A1

attempt to eliminate a e.g. \(\frac{{76(1 - r)}}{{1 - {r^3}}} = 36\)     M1

simplify and obtain \(9{r^2} + 9r - 10 = 0\)     (M1)A1

Note: Only award the M1 if a quadratic is seen.

obtain \(r = \frac{{12}}{{18}}\) and \(- \frac{{30}}{{18}}\)     (A1)

\(r = \frac{{12}}{{18}}\left( { = \frac{2}{3} = 0.666 \ldots } \right)\)     A1

Note: Award A0 if the extra value of r is given in the final answer.

Total [7 marks]

Almost all candidates obtained the cubic equation satisfied by the common ratio of the first sequence, but few were able to find its roots. One of the roots was \(r = 1\).

The sum of the first two terms of a geometric series is 10 and the sum of the first four terms is 30.

(a)     Show that the common ratio \(r\) satisfies \({r^2} = 2\).

(b)     Given \(r = \sqrt 2 \)

          (i)     find the first term;

          (ii)     find the sum of the first ten terms.

(a)     METHOD 1

\(a + ar = 10\)     A1

\(a + ar + a{r^2} + a{r^3} = 30\)     A1

\(a + ar = 10 \Rightarrow a{r^2} + a{r^3} = 10{r^2}\)     or     \(a{r^2} + a{r^3} = 20\)     M1

\(10 + 10{r^2} = 30\)     or     \({r^2}(a + ar) = 20\)     A1

\( \Rightarrow {r^2} = 2\)     AG

METHOD 2

\(\frac{{a(1 - {r^2})}}{{1 - r}} = 10\) and \(\frac{{a(1 - {r^4})}}{{1 - r}} = 30\)     M1A1

\( \Rightarrow \frac{{1 - {r^4}}}{{1 - {r^2}}} = 3\)     M1

leading to either \(1 + {r^2} = 3{\text{   (or }}{r^4} - 3{r^2} + 2 = 0)\)   A1

\( \Rightarrow {r^2} = 2\)     AG

[4 marks]

(b)     (i)     \(a + a\sqrt 2  = 10\)

\( \Rightarrow a = \frac{{10}}{{1 + \sqrt 2 }}\)   or   \(a = 10\left( {\sqrt 2  - 1} \right)\)     A1

(ii)     \({S_{10}} = \frac{{10}}{{1 + \sqrt 2 }}\left( {\frac{{{{\sqrt 2 }^{10}} - 1}}{{\sqrt 2  - 1}}} \right){\text{ }}( = 10 \times 31)\)     M1

\( = 310\)     A1

[3 marks]

Total [7 marks]

This question was invariably answered very well. Candidates showed some skill in algebraic manipulation to derive the given answer in part a). Poor attempts at part b) were a rarity, though the final mark was sometimes lost after a correctly substituted equation was seen but with little follow-up work.

Use de Moivre’s theorem to find the value of \({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3}\).

Use mathematical induction to prove that

\[{(\cos \theta  - {\text{i}}\sin \theta )^n} = \cos n\theta  - {\text{i}}\sin n\theta {\text{ for }}n \in {\mathbb{Z}^ + }.\]

Find an expression in terms of \(\theta \) for \({(z)^n} + {(z{\text{*}})^n},{\text{ }}n \in {\mathbb{Z}^ + }\) where \(z{\text{*}}\) is the complex conjugate of \(z\).

(i)     Show that \(zz{\text{*}} = 1\).

(ii)     Write down the binomial expansion of \({(z + z{\text{*}})^3}\) in terms of \(z\) and \(z{\text{*}}\).

(iii)     Hence show that \(\cos 3\theta  = 4{\cos ^3}\theta  - 3\cos \theta \).

Hence solve \(4{\cos ^3}\theta  - 2{\cos ^2}\theta  - 3\cos \theta  + 1 = 0\) for \(0 \leqslant \theta  < \pi \).

\({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3} = \cos \pi  + {\text{i}}\sin \pi \)    M1

\( =  - 1\)    A1

[2 marks]

show the expression is true for \(n = 1\)     R1

assume true for \(n = k,{\text{ }}{(\cos \theta  - {\text{i}}\sin \theta )^k} = \cos k\theta  - {\text{i}}\sin k\theta \)     M1

Note:     Do not accept “let \(n = k\)” or “assume \(n = k\)”, assumption of truth must be present.

\({(\cos \theta  - {\text{i}}\sin \theta )^{k + 1}} = {(\cos \theta  - {\text{i}}\sin \theta )^k}(\cos \theta  - {\text{i}}\sin \theta )\)

\( = (\cos k\theta  - {\text{i}}\sin k\theta )(\cos \theta  - {\text{i}}\sin \theta )\)    M1

\( = \cos k\theta \cos \theta  - \sin k\theta \sin \theta  - {\text{i}}(\cos k\theta \sin \theta  + \sin k\theta \cos \theta )\)    A1

Note:     Award A1 for any correct expansion.

\( = \cos \left( {(k + 1)\theta } \right) - {\text{i}}\sin \left( {(k + 1)\theta } \right)\)    A1

therefore if true for \(n = k\) true for \(n = k + 1\), true for \(n = 1\), so true for all \(n( \in {\mathbb{Z}^ + })\)     R1

Note:     To award the final R mark the first 4 marks must be awarded.

[6 marks]

\({(z)^n} + {(z{\text{*}})^n} = {(\cos \theta  + {\text{i}}\sin \theta )^n} + {(\cos \theta  - {\text{i}}\sin \theta )^n}\)

\( = \cos n\theta  + {\text{i}}\sin n\theta  + \cos n\theta  - {\text{i}}\sin n\theta  = 2\cos (n\theta )\)    (M1)A1

[2 marks]

(i)     \(zz* = (\cos \theta  + {\text{i}}\sin \theta )(\cos \theta  - {\text{i}}\sin \theta )\)

\( = {\cos ^2}\theta  + {\sin ^2}\theta \)    A1

\( = 1\)    AG

Note:     Allow justification starting with \(|z| = 1\).

(ii)     \({(z + z{\text{*}})^3} = {z^3} + 3{z^2}z{\text{*}} + 3z{({z^*})^2} + (z{\text{*}})3\left( { = {z^3} + 3z + 3z{\text{*}} + {{(z{\text{*}})}^3}} \right)\)     A1

(iii)     \({(z + z{\text{*}})^3} = {(2\cos \theta )^3}\)     A1

\({z^3} + 3z + 3z{\text{*}} + {(z{\text{*}})^3} = 2\cos 3\theta  + 6\cos \theta \)    M1A1

\(\cos 3\theta  = 4{\cos ^3}\theta  - 3\cos \theta \)   AG

Note:     M1 is for using \(zz{\text{*}} = 1\), this might be seen in d(ii).

[5 marks]

\(4{\cos ^3}\theta  - 2{\cos ^2}\theta  - 3\cos \theta  + 1 = 0\)

\(4{\cos ^3}\theta  - 3\cos \theta  = 2{\cos ^2}\theta  - 1\)

\(\cos (3\theta ) = \cos (2\theta )\)    A1A1

Note:     A1 for \(\cos (3\theta )\) and A1 for \(\cos (2\theta )\).

\(\theta  = 0\)    A1

or \(3\theta  = 2\pi  - 2\theta {\text{ }}({\text{or }}3\theta  = 4\pi  - 2\theta )\)     M1

\(\theta  = \frac{{2\pi }}{5},{\text{ }}\frac{{4\pi }}{5}\)    A1A1

Note:     Do not accept solutions via factor theorem or other methods that do not follow “hence”.

[6 marks]

This was well done by most candidates who correctly applied de Moivre’s theorem.

This question was poorly done, which was surprising as it is very similar to the proof of de Moivre’s theorem which is stated as being required in the course guide. Many candidates spotted that they needed to use trigonometric identities but fell down through not being able to set out the proof in a logical form.

This was well done by the majority of candidates.

(d) parts (i) and (ii) were well done by the candidates, who were able to successfully use trigonometrical identities and the binomial theorem.

(d)(iii) This is a familiar technique that has appeared in several recent past papers and was successfully completed by many of the better candidates. Some candidates though neglected the instruction ‘hence’ and tried to derive the expression using trigonometric identities.

Again some candidates ignored ‘hence’ and tried to form a polynomial equation. Many candidates obtained the solution \(\cos (2\theta ) = \cos (3\theta )\) and hence the solution \(\theta  = 0\). Few were able to find the other solutions which can be obtained from consideration of the unit circle or similar methods.

(i)     Express each of the complex numbers \({z_1} = \sqrt 3  + {\text{i, }}{z_2} = - \sqrt 3  + {\text{i}}\) and \({z_3} = - 2{\text{i}}\) in modulus-argument form.

(ii)     Hence show that the points in the complex plane representing \({z_1}\), \({z_2}\) and \({z_3}\) form the vertices of an equilateral triangle.

(iii)     Show that \({\text{z}}_1^{3n} + z_2^{3n} = 2z_3^{3n}\) where \(n \in \mathbb{N}\).

(i)     State the solutions of the equation \({z^7} = 1\) for \(z \in \mathbb{C}\), giving them in modulus-argument form.

(ii)     If w is the solution to \({z^7} = 1\) with least positive argument, determine the argument of 1 + w. Express your answer in terms of \(\pi \).

(iii)     Show that \({z^2} - 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\) is a factor of the polynomial \({z^7} - 1\). State the two other quadratic factors with real coefficients.

(i)     \({z_1} = 2{\text{cis}}\left( {\frac{\pi }{6}} \right),{\text{ }}{z_2} = 2{\text{cis}}\left( {\frac{{5\pi }}{6}} \right),{\text{ }}{z_3} = 2{\text{cis}}\left( { - \frac{\pi }{2}} \right){\text{ or }}2{\text{cis}}\left( {\frac{{3\pi }}{2}} \right)\)     A1A1A1

Note: Accept modulus and argument given separately, or the use of exponential (Euler) form.

Note: Accept arguments given in rational degrees, except where exponential form is used.

(ii)     the points lie on a circle of radius 2 centre the origin     A1

differences are all \(\frac{{2\pi }}{3}(\bmod 2\pi )\)     A1

\( \Rightarrow \) points equally spaced \( \Rightarrow \) triangle is equilateral     R1AG

Note: Accept an approach based on a clearly marked diagram.

(iii)     \({\text{z}}_1^{3n} + z_2^{3n} = {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right) + {2^{3n}}{\text{cis}}\left( {\frac{{5n\pi }}{2}} \right)\)     M1

\( = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\)     A1

\(2z_3^{3n} = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{9n\pi }}{2}} \right) = 2 \times {2^{3n}}{\text{cis}}\left( {\frac{{n\pi }}{2}} \right)\)     A1AG

[9 marks]

(i)     attempt to obtain seven solutions in modulus argument form     M1

\(z = {\text{cis}}\left( {\frac{{2k\pi }}{7}} \right),{\text{ }}k = 0,{\text{ }}1 \ldots 6\)     A1

(ii)     w has argument \(\frac{{2\pi }}{7}\) and 1 + w has argument \(\phi \),

then \(\tan (\phi ) = \frac{{\sin \left( {\frac{{2\pi }}{7}} \right)}}{{1 + \cos \left( {\frac{{2\pi }}{7}} \right)}}\)     M1

\( = \frac{{2\sin \left( {\frac{\pi }{7}} \right)\cos \left( {\frac{\pi }{7}} \right)}}{{2{{\cos }^2}\left( {\frac{\pi }{7}} \right)}}\)     A1

\( = \tan \left( {\frac{\pi }{7}} \right) \Rightarrow \phi  = \frac{\pi }{7}\)     A1

Note: Accept alternative approaches.

(iii)     since roots occur in conjugate pairs,     (R1)

\({z^7} - 1\) has a quadratic factor \(\left( {z - {\text{cis}}\left( {\frac{{2\pi }}{7}} \right)} \right) \times \left( {z - {\text{cis}}\left( { - \frac{{2\pi }}{7}} \right)} \right)\)     A1

\( = {z^2} - 2z\cos \left( {\frac{{2\pi }}{7}} \right) + 1\)     AG

other quadratic factors are \({z^2} - 2z\cos \left( {\frac{{4\pi }}{7}} \right) + 1\)     A1

and \({z^2} - 2z\cos \left( {\frac{{6\pi }}{7}} \right) + 1\)     A1

[9 marks]

(i) A disappointingly large number of candidates were unable to give the correct arguments for the three complex numbers. Such errors undermined their efforts to tackle parts (ii) and (iii).

Many candidates were successful in part (i), but failed to capitalise on that – in particular, few used the fact that roots of \({z^7} - 1 = 0\) come in complex conjugate pairs.

Calculate \(\frac{{{z_1}}}{{{z_2}}}\) giving your answer both in modulus-argument form and Cartesian form.

Using your results, find the exact value of tan 75° , giving your answer in the form \(a + \sqrt b \) , a , \(b \in {\mathbb{Z}^ + }\) .

in Cartesian form

\({z_1} = 2 \times - \frac{{\sqrt 3 }}{2} + 2 \times \frac{1}{2}{\text{i}}\)     M1

\( = - \sqrt 3 + {\text{i}}\)     A1

\(\frac{{{z_1}}}{{{z_2}}} = \frac{{ - \sqrt 3 + {\text{i}}}}{{ - 1 + {\text{i}}}}\)

\( = \frac{{\left( { - \sqrt 3 + {\text{i}}} \right)}}{{( - 1 + {\text{i}})}} \times \frac{{( - 1 - {\text{i}})}}{{( - 1 - {\text{i}})}}\)     M1

\( = \frac{{1 + \sqrt 3 }}{2} + \frac{{\left( {\sqrt 3 - 1} \right)}}{2}{\text{i}}\)     A1

in modulus-argument form

\({{\text{z}}_2} = \sqrt 2 {\text{cis}}135^\circ \)     A1

\(\frac{{{z_1}}}{{{z_2}}} = \frac{{2{\text{cis}}150^\circ }}{{\sqrt 2 {\text{cis}}135^\circ }}\)

\( = \sqrt 2 {\text{cis}}15^\circ \)     A1A1

[7 marks]

equating the two expressions for \(\frac{{{z_1}}}{{{z_2}}}\)

\(\cos 15^\circ = \frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}\)     A1

\(\sin 15^\circ = \frac{{\sqrt 3 - 1}}{{2\sqrt 2 }}\)     A1

\(\tan 75^\circ = \frac{{\cos 15^\circ }}{{\sin 15^\circ }} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\)     M1

\( = \frac{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\)     A1

\( = 2 + \sqrt 3 \)     A1

[5 marks]

Show that \(\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \sqrt {n + 1}  - \sqrt n \) where \(n \ge 0,{\text{ }}n \in \mathbb{Z}\).

Hence show that \(\sqrt 2  - 1 < \frac{1}{{\sqrt 2 }}\).

Prove, by mathematical induction, that \(\sum\limits_{r = 1}^{r = n} {\frac{1}{{\sqrt r }} > \sqrt n } \) for \(n \ge 2,{\text{ }}n \in \mathbb{Z}\).

\(\frac{1}{{\sqrt n  + \sqrt {n + 1} }} = \frac{1}{{\sqrt n  + \sqrt {n + 1} }} \times \frac{{\sqrt {n + 1}  - \sqrt n }}{{\sqrt {n + 1}  - \sqrt n }}\)     M1

\( = \frac{{\sqrt {n + 1}  - \sqrt n }}{{(n + 1) - n}}\)     A1

\( = \sqrt {n + 1}  - \sqrt n \)     AG

[2 marks]

\(\sqrt 2  - 1 = \frac{1}{{\sqrt 2  + \sqrt 1 }}\)     A2

\( < \frac{1}{{\sqrt 2 }}\)     AG

[2 marks]

consider the case \(n = 2\): required to prove that \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \)     M1

from part (b) \(\frac{1}{{\sqrt 2 }} > \sqrt 2  - 1\)

hence \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \) is true for \(n = 2\)     A1

now assume true for \(n = k:\sum\limits_{r = 1}^{r = k} {\frac{1}{{\sqrt r }} > \sqrt k } \)     M1

\(\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} > \sqrt k \)

attempt to prove true for \(n = k + 1:\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \)     (M1)

from assumption, we have that \(\frac{1}{{\sqrt 1 }} +  \ldots  + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt k  + \frac{1}{{\sqrt {k + 1} }}\)     M1

so attempt to show that \(\sqrt k  + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \)     (M1)

EITHER

\(\frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1}  - \sqrt k \)     A1

\(\frac{1}{{\sqrt {k + 1} }} > \frac{1}{{\sqrt k  + \sqrt {k + 1} }}\), (from part a), which is true     A1

OR

\(\sqrt k  + \frac{1}{{\sqrt {k + 1} }} = \frac{{\sqrt {k + 1} \sqrt k  + 1}}{{\sqrt {k + 1} }}\)     A1

\( > \frac{{\sqrt k \sqrt k  + 1}}{{\sqrt {k + 1} }} = \sqrt {k + 1} \)     A1

THEN

so true for \(n = 2\) and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \ge 2\)     R1

Note:     Award R1 only if all previous M marks have been awarded.

[9 marks]

Total [13 marks]

Prove by mathematical induction \(\sum\limits_{r = 1}^n {r(r!) = (n + 1)! - 1} \), \(n \in {\mathbb{Z}^ + }\).

let \(n = 1\)

LHS \( = 1 \times 1! = 1\)

RHS \( = (1 + 1)! - 1 = 2 - 1 = 1\)

hence true for \(n = 1\)     R1

assume true for \(n = k\)

\(\sum\limits_{r = 1}^k {r(r!) = (k + 1)! - 1} \)     M1

\(\sum\limits_{r = 1}^{k + 1} {r(r!) = (k + 1)! - 1}  + (k + 1) \times (k + 1)!\)     M1A1

\( = (k + 1)!(1 + k + 1) - 1\)

\( = (k + 1)!(k + 2) - 1\)     A1

\( = (k + 2)! - 1\)     A1

hence if true for \(n = k\), true for \(n = k + 1\)     R1

since the result is true for \(n = 1\) and \({\text{P}}(k) \Rightarrow {\text{P}}(k + 1)\) the result is proved by mathematical induction \(\forall n \in {\mathbb{Z}^ + }\)     R1

[8 marks]

This question was done poorly on a number of levels. Many students knew the structure of induction but did not show that they understood what they were doing. The general notation was poor for both the induction itself and the sigma notation.

In noting the case for \(n = 1\) too many stated the equation rather than using the LHS and RHS separately and concluding with a statement. There were also too many who did not state the conclusion for this case.

Many did not state the assumption for \(n = k\) as an assumption.

Most stated the equation for \(n = k + 1\) and worked with the equation. Also common was the lack of sigma and inappropriate use of n and k in the statement. There were some very nice solutions however.

The final conclusion was often not complete or not considered which would lead to the conclusion that the student did not really understand what induction is about.

(i)     \(p =  - (\alpha  + \beta  + \gamma )\);

(ii)     \(q = \alpha \beta  + \beta \gamma  + \gamma \alpha \);

(iii)     \(c =  - \alpha \beta \gamma \).

It is now given that \(p =  - 6\) and \(q = 18\) for parts (b) and (c) below.

(i)     In the case that the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) form an arithmetic sequence, show that one of the roots is \(2\).

(ii)     Hence determine the value of \(c\).

In another case the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \) form a geometric sequence. Determine the value of \(c\).

(i)-(iii) given the three roots \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \), we have

\({x^3} + p{x^2} + qx + c = (x - \alpha )(x - \beta )(x - \gamma )\)     M1

\( = \left( {{x^2} - (\alpha  + \beta )x + \alpha \beta } \right)(x - \gamma )\)     A1

\( = {x^3} - (\alpha  + \beta  + \gamma ){x^2} + (\alpha \beta  + \beta \gamma  + \gamma \alpha )x - \alpha \beta \gamma \)     A1

comparing coefficients:

\(p =  - (\alpha  + \beta  + \gamma )\)     AG

\(q = (\alpha \beta  + \beta \gamma  + \gamma \alpha )\)     AG

\(c =  - \alpha \beta \gamma \)     AG

[3 marks]

METHOD 1

(i)     Given \( - \alpha  - \beta  - \gamma  =  - 6\)

And \(\alpha \beta  + \beta \gamma  + \gamma \alpha  = 18\)

Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \)

So \(\beta  - \alpha  = \gamma  - \beta \)     M1

or \(2\beta  = \alpha  + \gamma \)

Attempt to solve simultaneous equations:     M1

\(\beta  + 2\beta  = 6\)     A1

\(\beta  = 2\)     AG

(ii)     \(\alpha  + \gamma  = 4\)

\(2\alpha  + 2\gamma  + \alpha \gamma  = 18\)

\( \Rightarrow {\gamma ^2} - 4\gamma  + 10 = 0\)

\( \Rightarrow \gamma  = \frac{{4 \pm {\text{i}}\sqrt {24} }}{2}\)     (A1)

Therefore \(c =  - \alpha \beta \gamma  =  - \left( {\frac{{4 + {\text{i}}\sqrt {24} }}{2}} \right)\left( {\frac{{4 - {\text{i}}\sqrt {24} }}{2}} \right)2 =  - 20\)     A1

METHOD 2

(i)     let the three roots be \(\alpha ,{\text{ }}\alpha  - d,{\text{ }}\alpha  + d\)     M1

adding roots     M1

to give \(3\alpha  = 6\)     A1

\(\alpha  = 2\)     AG

(ii)     \(\alpha \) is a root, so \({2^3} - 6 \times {2^2} + 18 \times 2 + c = 0\)     M1

\(8 - 24 + 36 + c = 0\)

\(c =  - 20\)     A1

METHOD 3

(i)     let the three roots be \(\alpha ,{\text{ }}\alpha  - d,{\text{ }}\alpha  + d\)     M1

adding roots     M1

to give \(3\alpha  = 6\)     A1

\(\alpha  = 2\)     AG

(ii)     \(q = 18 = 2(2 - d) + (2 - d)(2 + d) + 2(2 + d)\)     M1

\({d^2} =  - 6 \Rightarrow d = \sqrt 6 {\text{i}}\)

\( \Rightarrow c =  - 20\)     A1

[5 marks]

METHOD 1

Given \( - \alpha  - \beta  - \gamma  =  - 6\)

And \(\alpha \beta  + \beta \gamma  + \gamma \alpha  = 18\)

Let the three roots be \(\alpha ,{\text{ }}\beta ,{\text{ }}\gamma \).

So \(\frac{\beta }{\alpha } = \frac{\gamma }{\beta }\)     M1

or \({\beta ^2} = \alpha \gamma \)

Attempt to solve simultaneous equations:     M1

\(\alpha \beta  + \gamma \beta  + {\beta ^2} = 18\)

\(\beta (\alpha  + \beta  + \gamma ) = 18\)

\(6\beta  = 18\)

\(\beta  = 3\)     A1

\(\alpha  + \gamma  = 3,{\text{ }}\alpha  = \frac{9}{\gamma }\)

\( \Rightarrow {\gamma ^2} - 3\gamma  + 9 = 0\)

\( \Rightarrow \gamma  = \frac{{3 \pm {\text{i}}\sqrt {27} }}{2}\)     (A1)(A1)

Therefore \(c =  - \alpha \beta \gamma  =  - \left( {\frac{{3 + {\text{i}}\sqrt {27} }}{2}} \right)\left( {\frac{{3 - {\text{i}}\sqrt {27} }}{2}} \right)3 =  - 27\)     A1

METHOD 2

let the three roots be \(a,{\text{ }}ar,{\text{ }}a{r^2}\)     M1

attempt at substitution of \(a,{\text{ }}ar,{\text{ }}a{r^2}\) and \(p\) and \(q\) into equations from (a)     M1

\(6 = a + ar + a{r^2}\left( { = a(1 + r + {r^2})} \right)\)     A1

\(18 = {a^2}r + {a^2}{r^3} + {a^2}{r^2}\left( { = {a^2}r(1 + r + {r^2})} \right)\)     A1

therefore \(3 = ar\)     A1

therefore \(c =  - {a^3}{r^3} =  - {3^3} =  - 27\)     A1

[6 marks]

Total [14 marks]

(i)     Show that \(\frac{{{v_{n + 1}}}}{{{v_n}}}\) is a constant.

(ii)     Write down the first term of the sequence \(\{ {v_n}\} \).

(iii)     Write down a formula for \({v_n}\) in terms of \(a\), \(d\) and \(n\).

Let \({S_n}\) be the sum of the first \(n\) terms of the sequence \(\{ {v_n}\} \).

(i)     Find \({S_n}\), in terms of \(a\), \(d\) and \(n\).

(ii)     Find the values of \(d\) for which \(\sum\limits_{i = 1}^\infty  {{v_i}} \) exists.

You are now told that \(\sum\limits_{i = 1}^\infty  {{v_i}} \) does exist and is denoted by \({S_\infty }\).

(iii)     Write down \({S_\infty }\) in terms of \(a\) and \(d\) .

(iv)     Given that \({S_\infty } = {2^{a + 1}}\) find the value of \(d\) .

Let \(\{ {w_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be a geometric sequence with first term equal to \(p\) and common ratio \(q\), where \(p\) and \(q\) are both greater than zero. Let another sequence \(\{ {z_n}\} \) be defined by \({z_n} = \ln {w_n}\).

Find \(\sum\limits_{i = 1}^n {{z_i}} \) giving your answer in the form \(\ln k\) with \(k\) in terms of \(n\), \(p\) and \(q\).

(i)     METHOD 1

\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{{u_{n + 1}}}}}}{{{2^{{u_n}}}}}\)     M1

\( = {2^{{u_{n + 1}} - {u_n}}} = {2^d}\)     A1

METHOD 2

\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{a + nd}}}}{{{2^{a + (n - 1)d}}}}\)     M1

\( = {2^d}\)     A1

(ii)     \( = {2^a}\)     A1

Note:     Accept \( = {2^{{u_1}}}\).

(iii)     EITHER

\({v_n}\) is a GP with first term \({2^a}\) and common ratio \({2^d}\)

\({v_n} = {2^a}{({2^d})^{(n - 1)}}\)

OR

\({u_n} = a + (n - 1)d\) as it is an AP

THEN

\({v_n} = {2^a}^{ + (n - 1)d}\)     A1

[4 marks]

(i)     \({S_n} = \frac{{{2^a}\left( {{{({2^d})}^n} - 1} \right)}}{{{2^d} - 1}} = \frac{{{2^a}({2^{dn}} - 1)}}{{{2^d} - 1}}\)     M1A1

Note:     Accept either expression.

(ii)     for sum to infinity to exist need \( - 1 < {2^d} < 1\)     R1

\( \Rightarrow \log {2^d} < 0 \Rightarrow d\log 2 < 0 \Rightarrow d < 0\)     (M1)A1

Note:     Also allow graph of \({2^d}\).

(iii)     \({S_\infty } = \frac{{{2^a}}}{{1 - {2^d}}}\)     A1

(iv)     \(\frac{{{2^a}}}{{1 - {2^d}}} = {2^{a + 1}} \Rightarrow \frac{1}{{1 - {2^d}}} = 2\)     M1

\( \Rightarrow 1 = 2 - {2^{d + 1}} \Rightarrow {2^{d + 1}} = 1\)

\( \Rightarrow d =  - 1\)     A1

[8 marks]

METHOD 1

\({w_n} = p{q^{n - 1}},{\text{ }}{z_n} = \ln p{q^{n - 1}}\)     (A1)

\({z_n} = \ln p + (n - 1)\ln q\)     M1A1

\({z_{n + 1}} - {z_n} = (\ln p + n\ln q) - (\ln p + (n - 1)\ln q) = \ln q\)

which is a constant so this is an AP

(with first term \(\ln p\) and common difference \(\ln q\))

\(\sum\limits_{i = 1}^n {{z_i} = \frac{n}{2}\left( {2\ln p + (n - 1)\ln q} \right)} \)     M1

\( = n\left( {\ln p + \ln {q^{\left( {\frac{{n - 1}}{2}} \right)}}} \right) = n\ln \left( {p{q^{\left( {\frac{{n - 1}}{2}} \right)}}} \right)\)     (M1)

\( = \ln \left( {{p^n}{q^{\frac{{n(n - 1)}}{2}}}} \right)\)     A1

METHOD 2

\(\sum\limits_{i = 1}^n {{z_i} = \ln p + \ln pq + \ln p{q^2} +  \ldots  + \ln p{q^{n - 1}}} \)     (M1)A1

\( = \ln \left( {{p^n}{q^{\left( {1 + 2 + 3 +  \ldots  + (n - 1)} \right)}}} \right)\)     (M1)A1

\( = \ln \left( {{p^n}{q^{\frac{{n(n - 1)}}{2}}}} \right)\)     (M1)A1

[6 marks]

Total [18 marks]

Method of first part was fine but then some algebra mistakes often happened. The next two parts were generally good.

Given that (a) indicated that there was a common ratio a disappointing number thought it was an AP. Although some good answers in the next parts, there was also some poor notational misunderstanding with the sum to infinity still involving \(n\).

Not enough candidates realised that this was an AP.

Express \(\frac{1}{{{{(1 - {\text{i}}\sqrt 3 )}^3}}}{\text{ in the form }}\frac{a}{b}{\text{ where }}a,{\text{ }}b \in \mathbb{Z}\) .

METHOD 1

\(r = 2,{\text{ }}\theta = - \frac{\pi }{3}\)     (A1)(A1)

\(\therefore {(1 - {\text{i}}\sqrt 3 )^{ - 3}} = {2^{ - 3}}{\left( {\cos \left( { - \frac{\pi }{3}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{3}} \right)} \right)^{ - 3}}\)     M1

\( = \frac{1}{8}(\cos \pi  + {\text{i}}\sin \pi )\)     (M1)

\( = - \frac{1}{8}\)     A1

[5 marks]

METHOD 2

\((1 - \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 1 - 2\sqrt 3 {\text{i}} - 3\,\,\,\,\,( = - 2 - 2\sqrt 3 {\text{i}})\)     (M1)A1

\(( - 2 - 2\sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = - 8\)     (M1)(A1)

\(\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}\)     A1

[5 marks]

METHOD 3

Attempt at Binomial expansion     M1

\({(1 - \sqrt 3 {\text{i}})^3} = 1 + 3( - \sqrt 3 {\text{i}}) + 3{( - \sqrt 3 {\text{i}})^2} + {( - \sqrt 3 {\text{i}})^3}\)     (A1)

\( = 1 - 3\sqrt 3 {\text{i}} - 9 + 3\sqrt 3 {\text{i}}\)     (A1)

\( = - 8\)     A1

\(\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}\)     M1

[5 marks]

Most candidates made a meaningful attempt at this question using a variety of different, but correct methods. Weaker candidates sometimes made errors with the manipulation of the square roots, but there were many fully correct solutions.

Find the value of k if \({\sum\limits_{r = 1}^\infty{k\left( {\frac{1}{3}} \right)}^r} = 7\).

\({u_1} = \frac{1}{3}k{\text{ , }}r = \frac{1}{3}\)     (A1)     (A1)

\(7 = \frac{{\frac{1}{3}k}}{{1 - \frac{1}{3}}}\)     M1

\(k = 14\)     A1

[4 marks]

The question was well done generally. Those that did make mistakes on the question usually had the first term wrong, but did understand to use the formula for an infinite geometric series.

Given that \({z_1} = 2\) and \({z_2} = 1 + \sqrt 3 {\text{i}}\) are roots of the cubic equation \({z^3} + b{z^2} + cz + d = 0\)

where b, c, \(d \in \mathbb{R}\),

(a)     write down the third root, \({z_3}\), of the equation;

(b)     find the values of b, c and d ;

(c)     write \({z_2}\) and \({z_3}\) in the form \(r{{\text{e}}^{{\text{i}}\theta }}\).

(a)     \(1 - \sqrt 3 {\text{i}}\)     A1

(b)     EITHER

\(\left( {z - (1 + \sqrt 3 {\text{i)}}} \right)\left( {z - (1 - \sqrt 3 {\text{i)}}} \right) = {z^2} - 2z + 4\)     (M1)A1

\(p(z) = (z - 2)({z^2} - 2z + 4)\)     (M1)

\( = {z^3} - 4{z^2} + 8z - 8\)     A1

therefore \(b = - 4,{\text{ }}c = 8,{\text{ }}d = - 8\)

OR

relating coefficients of cubic equations to roots

\( - b = 2 + 1 + \sqrt 3 {\text{i}} + 1 - \sqrt 3 {\text{i}} = 4\)     M1

\(c = 2(1 + \sqrt 3 {\text{i}}) + 2(1 - \sqrt 3 {\text{i}}) + (1 + \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 8\)

\( - d = 2(1 + \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 8\)

\(b = - 4,{\text{ }}c = 8,{\text{ }}d = - 8\)     A1A1A1

(c)     \({z_2} = 2{{\text{e}}^{\frac{{{\text{i}}\pi }}{3}}},{\text{ }}{z_3} = 2{{\text{e}}^{ - \frac{{{\text{i}}\pi }}{3}}}\)     A1A1A1

Note: Award A1 for modulus,

A1 for each argument.

[8 marks]

Parts a) and c) were done quite well by many but the method used in b) often lead to tedious and long algebraic manipulations in which students got lost and so did not get to the correct solution. Many did not give the principal argument in c).

Determine whether \({f_n}\) is an odd or even function, justifying your answer.

By using mathematical induction, prove that

\({f_n}(x) = \frac{{\sin {2^{n + 1}}x}}{{{2^n}\sin 2x}},{\text{ }}x \ne \frac{{m\pi }}{2}\) where \(m \in \mathbb{Z}\).

Hence or otherwise, find an expression for the derivative of \({f_n}(x)\) with respect to \(x\).

Show that, for \(n > 1\), the equation of the tangent to the curve \(y = {f_n}(x)\) at \(x = \frac{\pi }{4}\) is \(4x - 2y - \pi  = 0\).

even function     A1

since \(\cos kx = \cos ( - kx)\) and \({f_n}(x)\) is a product of even functions     R1

OR

even function     A1

since \((\cos 2x)(\cos 4x) \ldots  = \left( {\cos ( - 2x)} \right)\left( {\cos ( - 4x)} \right) \ldots \)     R1

Note:     Do not award A0R1.

[2 marks]

consider the case \(n = 1\)

\(\frac{{\sin 4x}}{{2\sin 2x}} = \frac{{2\sin 2x\cos 2x}}{{2\sin 2x}} = \cos 2x\)     M1

hence true for \(n = 1\)     R1

assume true for \(n = k\), ie, \((\cos 2x)(\cos 4x) \ldots (\cos {2^k}x) = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\)     M1

Note:     Do not award M1 for “let \(n = k\)” or “assume \(n = k\)” or equivalent.

consider \(n = k + 1\):

\({f_{k + 1}}(x) = {f_k}(x)(\cos {2^{k + 1}}x)\)     (M1)

\( = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\cos {2^{k + 1}}x\)     A1

\( = \frac{{2\sin {2^{k + 1}}x\cos {2^{k + 1}}x}}{{{2^{k + 1}}\sin 2x}}\)     A1

\( = \frac{{\sin {2^{k + 2}}x}}{{{2^{k + 1}}\sin 2x}}\)     A1

so \(n = 1\) true and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note:     To obtain the final R1, all the previous M marks must have been awarded.

[8 marks]

attempt to use \(f’ = \frac{{vu' - uv'}}{{{v^2}}}\) (or correct product rule)     M1

\({f’_n}(x) = \frac{{({2^n}\sin 2x)({2^{n + 1}}\cos {2^{n + 1}}x) - (\sin {2^{n + 1}}x)({2^{n + 1}}\cos 2x)}}{{{{({2^n}\sin 2x)}^2}}}\)     A1A1

Note:     Award A1 for correct numerator and A1 for correct denominator.

[3 marks]

\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{\left( {{2^n}\sin \frac{\pi }{2}} \right)\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right) - \left( {\sin {2^{n + 1}}\frac{\pi }{4}} \right)\left( {{2^{n + 1}}\cos \frac{\pi }{2}} \right)}}{{{{\left( {{2^n}\sin \frac{\pi }{2}} \right)}^2}}}\)     (M1)(A1)

\({f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{({2^n})\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right)}}{{{{({2^n})}^2}}}\)     (A1)

\( = 2\cos {2^{n + 1}}\frac{\pi }{4}{\text{ }}( = 2\cos {2^{n - 1}}\pi )\)     A1

\({f’_n}\left( {\frac{\pi }{4}} \right) = 2\)     A1

\({f_n}\left( {\frac{\pi }{4}} \right) = 0\)     A1

Note:     This A mark is independent from the previous marks.

\(y = 2\left( {x - \frac{\pi }{4}} \right)\)     M1A1

\(4x - 2y - \pi  = 0\)     AG

[8 marks]

Use the method of mathematical induction to prove that \({4^n} + 15n - 1\) is divisible by 9 for \(n \in {\mathbb{Z}^ + }\).

let \(P(n)\) be the proposition that \({4^n} + 15n - 1\) is divisible by 9

showing true for \(n = 1\)     A1

ie\(\,\,\,\,\,\)for \(n = 1,{\text{ }}{4^1} + 15 \times 1 - 1 = 18\)

which is divisible by 9, therefore \(P(1)\) is true

assume \(P(k)\) is true so \({4^k} + 15k - 1 = 9A,{\text{ }}(A \in {\mathbb{Z}^ + })\)     M1

Note:     Only award M1 if “truth assumed” or equivalent.

consider \({4^{k + 1}} + 15(k + 1) - 1\)

\( = 4 \times {4^k} + 15k + 14\)

\( = 4(9A - 15k + 1) + 15k + 14\)     M1

\( = 4 \times 9A - 45k + 18\)     A1

\( = 9(4A - 5k + 2)\) which is divisible by 9     R1

Note:     Award R1 for either the expression or the statement above.

since \(P(1)\) is true and \(P(k)\) true implies \(P(k + 1)\) is true, therefore (by the principle of mathematical induction) \(P(n)\) is true for \(n \in {\mathbb{Z}^ + }\)     R1

Note:     Only award the final R1 if the 2 M1s have been awarded.

[6 marks]

Expand and simplify \({\left( {{x^2} - \frac{2}{x}} \right)^4}\).

\({\left( {{x^2} - \frac{2}{x}} \right)^4} = {({x^2})^4} + 4{({x^2})^3}\left( { - \frac{2}{x}} \right) + 6{({x^2})^2}{\left( { - \frac{2}{x}} \right)^2} + 4({x^2}){\left( { - \frac{2}{x}} \right)^3} + {\left( { - \frac{2}{x}} \right)^4}\)     (M1)

\( = {x^8} - 8{x^5} + 24{x^2} - \frac{{32}}{x} + \frac{{16}}{{{x^4}}}\)     A3

Note: Deduct one A mark for each incorrect or omitted term.

[4 marks]

Most candidates solved this question correctly with most candidates who explained how they obtained their coefficients using Pascal’s triangle rather than the combination formula.

Expand \({(3 - x)^4}\) in ascending powers of \(x\) and simplify your answer.

\({(3 - x)^4} = {1.3^4} + {4.3^3}( - x) + {6.3^2}{( - x)^2} + 4.3{( - x)^3} + 1{( - x)^4}\) or equivalent     (M1)(A1)

\( = 81 - 108x + 54{x^2} - 12{x^3} + {x^4}\)     A1A1

Note:     A1 for ascending powers, A1 for correct coefficients including signs.

[4 marks]

(a)     Consider the following sequence of equations.

     \(1 \times 2 = \frac{1}{3}(1 \times 2 \times 3),\)

     \(1 \times 2 + 2 \times 3 = \frac{1}{3}(2 \times 3 \times 4),\)

     \(1 \times 2 + 2 \times 3 + 3 \times 4 = \frac{1}{3}(3 \times 4 \times 5),\)

     \( \ldots {\text{ .}}\)

(i)     Formulate a conjecture for the \({n^{{\text{th}}}}\) equation in the sequence.

(ii)     Verify your conjecture for n = 4 .

(b)     A sequence of numbers has the \({n^{{\text{th}}}}\) term given by \({u_n} = {2^n} + 3,{\text{ }}n \in {\mathbb{Z}^ + }\). Bill conjectures that all members of the sequence are prime numbers. Show that Bill’s conjecture is false.

(c)     Use mathematical induction to prove that \(5 \times {7^n} + 1\) is divisible by 6 for all \(n \in {\mathbb{Z}^ + }\).

(a)     (i)     \(1 \times 2 + 2 \times 3 + \ldots + n(n + 1) = \frac{1}{3}n(n + 1)(n + 2)\)     R1

(ii)     LHS = 40; RHS = 40     A1

[2 marks]

(b)     the sequence of values are:

5, 7, 11, 19, 35 … or an example     A1

35 is not prime, so Bill’s conjecture is false     R1AG

[2 marks]

(c)     \({\text{P}}(n):5 \times {7^n} + 1\) is divisible by 6

\({\text{P}}(1):36\) is divisible by \(6 \Rightarrow {\text{P}}(1)\) true     A1

assume \({\text{P}}(k)\) is true \((5 \times {7^k} + 1 = 6r)\)     M1

Note: Do not award M1 for statement starting ‘let n = k’.

Subsequent marks are independent of this M1.

consider \(5 \times {7^{k + 1}} + 1\)     M1

\( = 7(6r - 1) + 1\)     (A1)

\( = 6(7r - 1) \Rightarrow {\text{P}}(k + 1)\) is true     A1

P(1) true and \({\text{P}}(k)\) true \( \Rightarrow {\text{P}}(k + 1)\) true, so by MI \({\text{P}}(n)\) is true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Only award R1 if there is consideration of P(1), \({\text{P}}(k)\) and \({\text{P}}(k + 1)\) in the final statement.

Only award R1 if at least one of the two preceding A marks has been awarded.

[6 marks]

Total [10 marks]

Although there were a good number of wholly correct solutions to this question, it was clear that a number of students had not been prepared for questions on conjectures. The proof by induction was relatively well done, but candidates often showed a lack of rigour in the proof. It was fairly common to see students who did not appreciate the idea that \({\text{P}}(k)\) is assumed not given and this was penalised. Also it appeared that a number of students had been taught to write down the final reasoning for a proof by induction, even if no attempt of a proof had taken place. In these cases, the final reasoning mark was not awarded.

(i)     Show that \({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = \frac{{2\cos n\theta }}{{{{\cos }^n}\theta }},\;\;\;\cos \theta  \ne 0\).

(ii)     Hence verify that \({\text{i}}\tan \frac{{3\pi }}{8}\) is a root of the equation \({(1 + z)^4} + {(1 - z)^4} = 0,\;\;\;z \in \mathbb{C}\).

(iii)     State another root of the equation \({(1 + z)^4} + {(1 - z)^4} = 0,\;\;\;z \in \mathbb{C}\).

(i)     Use the double angle identity \(\tan 2\theta  = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\) to show that \(\tan \frac{\pi }{8} = \sqrt 2  - 1\).

(ii)     Show that \(\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1\).

(iii)     Hence find the value of \(\int_0^{\frac{\pi }{8}} {\frac{{2\cos 4x}}{{{{\cos }^2}x}}{\text{d}}x} \).

(i)     METHOD 1

\({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = {\left( {1 + {\text{i}}\frac{{\sin \theta }}{{\cos \theta }}} \right)^n} + {\left( {1 - {\text{i}}\frac{{\sin \theta }}{{\cos \theta }}} \right)^n}\)     M1

\( = {\left( {\frac{{\cos \theta  + i\sin \theta }}{{\cos \theta }}} \right)^n} + {\left( {\frac{{\cos \theta  - i\sin \theta }}{{\cos \theta }}} \right)^n}\)     A1

by de Moivre’s theorem     (M1)

\({\left( {\frac{{\cos \theta  + i\sin \theta }}{{\cos \theta }}} \right)^n} = \frac{{\cos n\theta  + i\sin n\theta }}{{{{\cos }^n}\theta }}\)     A1

recognition that \(\cos \theta  - i\sin \theta \) is the complex conjugate of \(\cos \theta  + i\sin \theta \)     (R1)

use of the fact that the operation of complex conjugation commutes with the operation of raising to an integer power:

\({\left( {\frac{{\cos \theta  - i\sin \theta }}{{\cos \theta }}} \right)^n} = \frac{{\cos n\theta  - i\sin n\theta }}{{{{\cos }^n}\theta }}\)     A1

\({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = \frac{{2\cos n\theta }}{{{{\cos }^n}\theta }}\)     AG

METHOD 2

\({(1 + {\text{i}}\tan \theta )^n} + {(1 - {\text{i}}\tan \theta )^n} = {(1 + {\text{i}}\tan \theta )^n} + {\left( {1 + {\text{i}}\tan ( - \theta )} \right)^n}\)     (M1)

\( = \frac{{{{(\cos \theta  + i\sin \theta )}^n}}}{{{{\cos }^n}\theta }} + \frac{{{{\left( {\cos ( - \theta ) + i\sin ( - \theta )} \right)}^n}}}{{{{\cos }^n}\theta }}\)     M1A1

Note:     Award M1 for converting to cosine and sine terms.

use of de Moivre’s theorem     (M1)

\( = \frac{1}{{{{\cos }^n}\theta }}\left( {\cos n\theta  + {\text{i}}\sin n\theta  + \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta )} \right)\)     A1

\( = \frac{{2\cos n\theta }}{{{{\cos }^2}\theta }}\;\;\;{\text{as}}\;\;\;\cos ( - n\theta ) = \cos n\theta \;\;\;{\text{and}}\;\;\;\sin ( - n\theta ) =  - \sin n\theta \)     R1AG

(ii)     \({\left( {1 + {\text{i}}\tan \frac{{3\pi }}{8}} \right)^4} + {\left( {1 - {\text{i}}\tan \frac{{3\pi }}{8}} \right)^4} = \frac{{2\cos \left( {4 \times \frac{{3\pi }}{8}} \right)}}{{{{\cos }^4}\frac{{3\pi }}{8}}}\)     (A1)

\( = \frac{{2\cos \frac{{3\pi }}{2}}}{{{{\cos }^4}\frac{{3\pi }}{8}}}\)     A1

\( = 0\;\;\;{\text{as}}\;\;\;\cos \frac{{3\pi }}{2} = 0\)     R1

Note:     The above working could involve theta and the solution of \(\cos (4\theta ) = 0\).

so \({\text{i}}\tan \frac{{3\pi }}{8}\) is a root of the equation     AG

(iii)     either \( - {\text{i}}\tan \frac{{3\pi }}{8}\;\;\;{\text{or}}\;\;\; - {\text{i}}\tan \frac{\pi }{8}\;\;\;{\text{or}}\;\;\;{\text{i}}\tan \frac{\pi }{8}\)     A1

Note:     Accept \({\text{i}}\tan \frac{{5\pi }}{8}\;\;\;{\text{or}}\;\;\;{\text{i}}\tan \frac{{7\pi }}{8}\).

Accept \( - \left( {1 + \sqrt 2 } \right){\text{i}}\;\;\;{\text{or}}\;\;\;\left( {1 - \sqrt 2 } \right){\text{i}}\;\;\;{\text{or}}\;\;\;\left( { - 1 + \sqrt 2 } \right){\text{i}}\).

[10 marks]

(i)     \(\tan \frac{\pi }{4} = \frac{{2\tan \frac{\pi }{8}}}{{1 - {{\tan }^2}\frac{\pi }{8}}}\)     (M1)

\({\tan ^2}\frac{\pi }{8} + 2\tan \frac{\pi }{8} - 1 = 0\)     A1

let \(t = \tan \frac{\pi }{8}\)

attempting to solve \({t^2} + 2t - 1 = 0\;\;\;{\text{for}}\;\;\;t\)     M1

\(t =  - 1 \pm \sqrt 2 \)     A1

\(\frac{\pi }{8}\) is a first quadrant angle and tan is positive in this quadrant, so

\(\tan \frac{\pi }{8} > 0\)     R1

\(\tan \frac{\pi }{8} = \sqrt 2  - 1\)     AG

(ii)     \(\cos 4x = 2{\cos ^2}2x - 1\)     A1

\( = 2{\left( {2{{\cos }^2}x - 1} \right)^2} - 1\)     M1

\( = 2\left( {4{{\cos }^4}x - 4{{\cos }^2}x + 1} \right) - 1\)     A1

\( = 8{\cos ^4}x - 8{\cos ^2}x + 1\)     AG

Note:     Accept equivalent complex number derivation.

(iii)     \(\int_0^{\frac{\pi }{8}} {\frac{{2\cos 4x}}{{{{\cos }^2}x}}{\text{d}}x = 2} \int_0^{\frac{\pi }{8}} {\frac{{8{{\cos }^4}x - 8{{\cos }^2}x + 1}}{{{{\cos }^2}x}}{\text{d}}x} \)

\( = 2\int_0^{\frac{\pi }{8}} {8{{\cos }^2}x - 8 + {{\sec }^2}x{\text{d}}x} \)     M1

Note:     The M1 is for an integrand involving no fractions.

use of \({\cos ^2}x = \frac{1}{2}(\cos 2x + 1)\)     M1

\( = 2\int_0^{\frac{\pi }{8}} {4\cos 2x - 4 + {{\sec }^2}x{\text{d}}x} \)     A1

\( = [4\sin 2x - 8x + 2\tan x]_0^{\frac{\pi }{8}}\)     A1

\( = 4\sqrt 2  - \pi  - 2\;\;\;\)(or equivalent)     A1

[13 marks]

Total [23 marks]

Fairly successful.

(i)     Most candidates attempted to use the hint. Those who doubled the angle were usually successful – but many lost the final mark by not giving a convincing reason to reject the negative solution to the intermediate quadratic equation. Those who halved the angle got nowhere.

(ii)     The majority of candidates obtained full marks.

(iii)     This was poorly answered, few candidates realising that part of the integrand could be re-expressed using \(\frac{1}{{{{\cos }^2}x}} = {\sec ^2}x\), which can be immediately integrated.

Write down the numerical value of the sum and of the product of the roots of this equation.

The roots of this equation are three consecutive terms of an arithmetic sequence.

Taking the roots to be \(\alpha {\text{ , }}\alpha  \pm \beta \) , solve the equation.

\({\text{sum}} = \frac{{45}}{9},{\text{ product}} = \frac{{40}}{9}\)     A1

[1 mark]

it follows that \(3\alpha = \frac{{45}}{9}\) and \(\alpha ({\alpha ^2} - {\beta ^2}) = \frac{{40}}{9}\)     A1A1

solving, \(\alpha = \frac{5}{3}\)     A1

\(\frac{5}{3}\left( {\frac{{25}}{9} - {\beta ^2}} \right) = \frac{{40}}{9}\)     M1

\(\beta = ( \pm )\frac{1}{3}\)     A1

the other two roots are 2, \(\frac{4}{3}\)     A1

[6 marks]

(a)     Show that the following system of equations has an infinite number of solutions.

     \(x + y + 2z =  - 2\)

     \(3x - y + 14z = 6\)

     \(x + 2y =  - 5\)

The system of equations represents three planes in space.

(b)     Find the parametric equations of the line of intersection of the three planes.

(a)     EITHER

\(\left( {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}\left| \begin{array}{c} - 2\\6\\ - 5\end{array} \right.} \right) \to \left( {\begin{array}{*{20}{c}}1&1&2\\0&1&{ - 2}\\0&0&0\end{array}\left| \begin{array}{c} - 2\\ - 3\\0\end{array} \right.} \right)\)     M1

row of zeroes implies infinite solutions, (or equivalent).     R1

Note:     Award M1 for any attempt at row reduction.

OR

\(\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}} \right| = 0\)     M1

\(\left| {\begin{array}{*{20}{c}}1&1&2\\3&{ - 1}&{14}\\1&2&0\end{array}} \right| = 0\) with one valid point     R1

OR

\(x + y + 2z =  - 2\)

\(3x - y + 14z = 6\)

\(x + 2y =  - 5\)   \( \Rightarrow x =  - 5 - 2y\)

substitute \(x =  - 5 - 2y\) into the first two equations:

\( - 5 - 2y + y + 2z =  - 2\)

\(3( - 5 - 2y) - y + 14z = 6\)     M1

\( - y + 2z = 3\)

\( - 7y + 14z = 21\)

the latter two equations are equivalent (by multiplying by 7) therefore an infinite number of solutions.     R1

OR

for example, \(7 \times {{\text{R}}_1} - {{\text{R}}_2}\) gives \(4x + 8y =  - 20\)     M1

this equation is a multiple of the third equation, therefore an infinite

number of solutions.     R1

[2 marks]

(b)     let \(y = t\)     M1

then \(x =  - 5 - 2t\)     A1

\(z = \frac{{t + 3}}{2}\)     A1

OR

let \(x = t\)     M1

then \(y = \frac{{ - 5 - t}}{2}\)     A1

\(z = \frac{{1 - t}}{4}\)     A1

OR

let \(z = t\)     M1

then \(x = 1 - 4t\)     A1

\(y =  - 3 + 2t\)     A1

OR

attempt to find cross product of two normal vectors:

eg: \(\left| {\begin{array}{*{20}{c}}i&j&k\\1&1&2\\1&2&0\end{array}} \right| =  - 4i + 2j + k\)     M1A1

\(x = 1 - 4t\)

\(y =  - 3 + 2t\)

\(z = t\)     A1

(or equivalent)

[3 marks]

Total [5 marks]

(a)     Show that \(\sin 2nx = \sin \left( {(2n + 1)x} \right)\cos x - \cos \left( {(2n + 1)x} \right)\sin x\).

(b)     Hence prove, by induction, that

\[\cos x + \cos 3x + \cos 5x +  \ldots  + \cos \left( {(2n - 1)x} \right) = \frac{{\sin 2nx}}{{2\sin x}},\]

for all \(n \in {\mathbb{Z}^ + }{\text{, }}\sin x \ne 0\).

(c)     Solve the equation \(\cos x + \cos 3x = \frac{1}{2},{\text{ }}0 < x < \pi \).

(a)     \(\sin (2n + 1)x\cos x - \cos (2n + 1)x\sin x = \sin (2n + 1)x - x\)     M1A1

\( = \sin 2nx\)     AG

[2 marks]

(b)     if n = 1     M1

\({\text{LHS}} = \cos x\)

\({\text{RHS}} = \frac{{\sin 2x}}{{2\sin x}} = \frac{{2\sin x\cos x}}{{2\sin x}} = \cos x\)     M1

so LHS = RHS and the statement is true for n = 1     R1

assume true for n = k     M1

Note: Only award M1 if the word true appears.

   Do not award M1 for ‘let n = k’ only.

   Subsequent marks are independent of this M1.

so \(\cos x + \cos 3x + \cos 5x +  \ldots  + \cos (2k - 1)x = \frac{{\sin 2kx}}{{2\sin x}}\)

if n = k + 1 then

\(\cos x + \cos 3x + \cos 5x +  \ldots  + \cos (2k - 1)x + \cos (2k + 1)x\)     M1

\( = \frac{{\sin 2kx}}{{2\sin x}} + \cos (2k + 1)x\)     A1

\( = \frac{{\sin 2kx + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\)     M1

\( = \frac{{\sin (2k + 1)x\cos x - \cos (2k + 1)x\sin x + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\)     M1

\( = \frac{{\sin (2k + 1)x\cos x + \cos (2k + 1)x\sin x}}{{2\sin x}}\)     A1

\( = \frac{{\sin (2k + 2)x}}{{2\sin x}}\)     M1

\( = \frac{{\sin 2(k + 1)x}}{{2\sin x}}\)     A1

so if true for n = k, then also true for n = k + 1

as true for n = 1 then true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Final R1 is independent of previous work.

[12 marks]

(c)     \(\frac{{\sin 4x}}{{2\sin x}} = \frac{1}{2}\)     M1A1

\(\sin 4x = \sin x\)

\(4x = x \Rightarrow x = 0\) but this is impossible

\(4x = \pi - x \Rightarrow x = \frac{\pi }{5}\)     A1

\(4x = 2\pi + x \Rightarrow x = \frac{{2\pi }}{3}\)     A1

\(4x = 3\pi - x \Rightarrow x = \frac{{3\pi }}{5}\)     A1

for not including any answers outside the domain     R1

Note: Award the first M1A1 for correctly obtaining \(8{\cos ^3}x - 4\cos x - 1 = 0\) or equivalent and subsequent marks as appropriate including the answers \(\left( { - \frac{1}{2},\frac{{1 \pm \sqrt 5 }}{4}} \right)\).

[6 marks]

Total [20 marks]

This question showed the weaknesses of many candidates in dealing with formal proofs and showing their reasoning in a logical manner. In part (a) just a few candidates clearly showed the result and part (b) showed that most candidates struggle with the formality of a proof by induction. The logic of many solutions was poor, though sometimes contained correct trigonometric work. Very few candidates were successful in answering part (c) using the unit circle. Most candidates attempted to manipulate the equation to obtain a cubic equation but made little progress. A few candidates guessed \(\frac{{2\pi }}{3}\) as a solution but were not able to determine the other solutions.

Find the cube roots of i in the form \(a + b{\text{i}}\), where \(a,{\text{ }}b \in \mathbb{R}\).

\({\text{i}} = \cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}\)     (A1)

\({z_1} = {{\text{i}}^{\frac{1}{3}}} = {\left( {\cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}} \right)^{\frac{1}{3}}} = \cos \frac{\pi }{6} + {\text{i}}\sin \frac{\pi }{6}\,\,\,\,\,\left( { = \frac{{\sqrt 3 }}{2} + \frac{1}{2}{\text{i}}} \right)\)     M1A1

\({z_2} = \cos \frac{{5\pi }}{6} + {\text{i}}\sin \frac{{5\pi }}{6}\,\,\,\,\,\left( { = - \frac{{\sqrt 3 }}{2} + \frac{1}{2}{\text{i}}} \right)\)     (M1)A1

\({z_3} = \cos \left( { - \frac{\pi }{2}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{2}} \right) = - {\text{i}}\)     A1

Note: Accept exponential and cis forms for intermediate results, but not the final roots.

Note: Accept the method based on expanding \({({\text{a}} + {\text{b}})^3}\). M1 for attempt, M1 for equating real and imaginary parts, A1 for finding a = 0 and \({\text{b}} = \frac{1}{2}\), then A1A1A1 for the roots.

[6 marks]

A varied response. Many knew how to solve this standard question in the most efficient way. A few candidates expanded \({(a + ib)^3}\) and solved the resulting fairly simple equations. A disappointing minority of candidates did not know how to start.

Factorize \({z^3} + 1\) into a linear and quadratic factor.

Let \(\gamma = \frac{{1 + {\text{i}}\sqrt 3 }}{2}\).

(i)     Show that \(\gamma \) is one of the cube roots of −1.

(ii)     Show that \({\gamma ^2} = \gamma - 1\).

(iii)     Hence find the value of \({(1 - \gamma )^6}\).

using the factor theorem z +1 is a factor     (M1)

\({z^3} + 1 = (z + 1)({z^2} - z + 1)\)     A1

[2 marks]

(i)     METHOD 1

\({z^3} = - 1 \Rightarrow {z^3} + 1 = (z + 1)({z^2} - z + 1) = 0\)     (M1)

solving \({z^2} - z + 1 = 0\)     M1

\(z = \frac{{1 \pm \sqrt {1 - 4} }}{2} = \frac{{1 \pm {\text{i}}\sqrt 3 }}{2}\)     A1

therefore one cube root of −1 is \(\gamma \)     AG

METHOD 2

\({\gamma ^2} = \left( {{{\frac{{1 + i\sqrt 3 }}{2}}^2}} \right) = \frac{{ - 1 + i\sqrt 3 }}{2}\)     M1A1

\({\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2} \times \frac{{1 + i\sqrt 3 }}{2} = \frac{{ - 1 - 3}}{4}\)     A1

= −1     AG

METHOD 3

\(\gamma  = \frac{{1 + i\sqrt 3 }}{2} = {e^{i\frac{\pi }{3}}}\)     M1A1

\({\gamma ^3} = {e^{i\pi }} = - 1\)     A1

(ii)     METHOD 1

as \(\gamma \) is a root of \({z^2} - z + 1 = 0\) then \({\gamma ^2} - \gamma + 1 = 0\)     M1R1

\(\therefore {\gamma ^2} = \gamma - 1\)     AG

Note: Award M1 for the use of \({z^2} - z + 1 = 0\) in any way.

Award R1 for a correct reasoned approach.

METHOD 2

\({\gamma ^2} = \frac{{ - 1 + i\sqrt 3 }}{2}\)     M1

\(\gamma - 1 = \frac{{1 + i\sqrt 3 }}{2} - 1 = \frac{{ - 1 + i\sqrt 3 }}{2}\)     A1

(iii)     METHOD 1

\({(1 - \gamma )^6} = {( - {\gamma ^2})^6}\)     (M1)

\( = {(\gamma )^{12}}\)     A1

\( = {({\gamma ^3})^4}\)     (M1)

\( = {( - 1)^4}\)

\( = 1\)     A1

METHOD 2

\({(1 - \gamma )^6}\)

\( = 1 - 6\gamma + 15{\gamma ^2} - 20{\gamma ^3} + 15{\gamma ^4} - 6{\gamma ^5} + {\gamma ^6}\)     M1A1

Note: Award M1 for attempt at binomial expansion.

use of any previous result e.g. \( = 1 - 6\gamma + 15{\gamma ^2} + 20 - 15\gamma  + 6{\gamma ^2} + 1\)     M1

= 1     A1

Note: As the question uses the word ‘hence’, other methods that do not use previous results are awarded no marks.

[9 marks]

In part a) the factorisation was, on the whole, well done.

Part (b) was done well by most although using a substitution method rather than the result above. This used much m retime than was necessary but was successful. A number of candidates did not use the previous results in part (iii) and so seemed to not understand the use of the ‘hence’.

find the value of \(d\).

determine the value of \(\sum\limits_{r = 1}^N {{u_r}} \).

use of \({u_n} = {u_1} + (n - 1)d\)     M1

\({(1 + 2d)^2} = (1 + d)(1 + 5d)\) (or equivalent)     M1A1

\(d = - 2\)     A1

[4 marks]

\(1 + (N - 1) \times - 2 = - 15\)

\(N = 9\)     (A1)

\(\sum\limits_{r = 1}^9 {{u_r}} = \frac{9}{2}(2 + 8 \times - 2)\)     (M1)

\( = - 63\)     A1

[3 marks]

If \({z_1} = a + a\sqrt 3 i\) and \({z_2} = 1 - i\), where a is a real constant, express \({z_1}\) and \({z_2}\) in the form \(r\,{\text{cis}}\,\theta \), and hence find an expression for \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6}\) in terms of a and i.

\({z_1} = 2a{\text{cis}}\left( {\frac{\pi }{3}} \right){\text{, }}{z_2} = \sqrt 2 {\text{ cis}}\left( { - \frac{\pi }{4}} \right)\)     M1     A1     A1

EITHER

\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = \frac{{{2^6}{a^6}{\text{cis(0)}}}}{{{{\sqrt 2 }^6}{\text{cis}}\left( {\frac{\pi }{2}} \right)}}\left( { = 8{a^6}{\text{cis}}\left( { - \frac{\pi }{2}} \right)} \right)\)     M1     A1     A1

OR

\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = {\left( {\frac{{2a}}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{7\pi }}{{12}}} \right)} \right)^6}\)     M1     A1

\( = 8{a^6}{\text{cis}}\left( { - \frac{\pi }{2}} \right)\)     A1

THEN

\( = - 8{a^6}{\text{i}}\)     A1

Note: Accept equivalent angles, in radians or degrees. 

Accept alternate answers without cis e.g. \({\text{ = }}\frac{{8{a^6}}}{{\text{i}}}\)

[7 marks]

Most students had an idea of what to do but were frequently let down in their calculations of the modulus and argument. The most common error was to give the argument of \({z_2}\) as \(\frac{{3\pi }}{4}\), failing to recognise that it should be in the fourth quadrant. There were also errors seen in the algebraic manipulation, in particular forgetting to raise the modulus to the power 6.

Verify that \(w\) is a root of the equation \({z^7} - 1 = 0,{\text{ }}z \in \mathbb{C}\).

(i)     Expand \((w - 1)(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6})\).

(ii)     Hence deduce that \(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6} = 0\).

Write down the roots of the equation \({z^7} - 1 = 0,{\text{ }}z \in \mathbb{C}\) in terms of \(w\) and plot these roots on an Argand diagram.

(i)     Given that \(\alpha  = w + {w^2} + {w^4}\), show that \(\alpha * = {w^6} + {w^5} + {w^3}\).

(ii)     Find the value of \(b\) and the value of \(c\).

Using the values for \(b\) and \(c\) obtained in part (d)(ii), find the imaginary part of \(\alpha \), giving your answer in surd form.

EITHER

\({w^7} = {\left( {\cos \frac{{2\pi }}{7} + {\text{i}}\sin \frac{{2\pi }}{7}} \right)^7}\)    (M1)

\( = \cos 2\pi  + {\text{i}}\sin 2\pi \)    A1

\( = 1\)    A1

so \(w\) is a root     AG

OR

\({z^7} = 1 = \cos (2\pi k) + {\text{i}}\sin (2\pi k)\)    (M1)

\(z = \cos \left( {\frac{{2\pi k}}{7}} \right) + {\text{i}}\sin \left( {\frac{{2\pi }}{7}} \right)\)    A1

\(k = 1 \Rightarrow z = \cos \left( {\frac{{2\pi }}{7}} \right) + {\text{i}}\sin \left( {\frac{{2\pi }}{7}} \right)\)    A1

so \(w\) is a root     AG

[3 marks]

(i)     \((w - 1)(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6})\)

\( = w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6} + {w^7} - 1 - w - {w^2} - {w^3} - {w^4} - {w^5} - {w^6}\)    M1

\( = {w^7} - 1{\text{ }}( = 0)\)    A1

(ii)     \({w^7} - 1 = 0\) and \(w - 1 \ne 0\)     R1

so \(1 + w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6} = 0\)     AG

[3 marks]

the roots are \(1,{\text{ }}w,{\text{ }}{w^2},{\text{ }}{w^3},{\text{ }}{w^4},{\text{ }}{w^5}\) and \({w^6}\)

7 points equidistant from the origin     A1

approximately correct angular positions for \(1,{\text{ }}w,{\text{ }}{w^2},{\text{ }}{w^3},{\text{ }}{w^4},{\text{ }}{w^5}\) and \({w^6}\)     A1

Note:     Condone use of cis notation for the final two A marks.

Note:     For the final A mark there should be one root in the first quadrant, two in the second, two in the third, one in the fourth, and one on the real axis.

[3 marks]

(i)     \(\alpha * = (w + {w^2} + {w^4})*\)

\( = w* + ({w^2})* + ({w^4})*\)    A1

since \( = w* = {w^6},{\text{ }}({w^2})* = {w^5}\) and \(({w^4})* = {w^3}\)     R1

\( \Rightarrow \alpha * = {w^6} + {w^5} + {w^3}\)    AG

(ii)     \(b =  - (\alpha  + \alpha *)\) (using sum of roots (or otherwise))     (M1)

\(b =  - (w + {w^2} + {w^3} + {w^4} + {w^5} + {w^6})\)    (A1)

\( =  - ( - 1)\)

\( = 1\)    A1

\(c = \alpha \alpha *\) (using product of roots (or otherwise))     (M1)

\(c = (w + {w^2} + {w^4})({w^6} + {w^5} + {w^3})\)

EITHER

\( = {w^{10}} + {w^9} + {w^8} + 3{w^7} + {w^6} + {w^5} + {w^4}\)    A1

\( = ({w^6} + {w^5} + {w^4} + {w^3} + {w^2} + w) + 3\)    M1

\( = 3 - 1\)    (A1)

OR

\( = {w^{10}} + {w^9} + {w^8} + 3{w^7} + {w^6} + {w^5} + {w^4}\left( { = {w^4}(1 + w + {w^3})({w^3} + {w^2} + 1)} \right)\)    A1

\( = {w^4}({w^6} + {w^5} + {w^4} + {w^2} + w + 1 + 3{w^3})\)    M1

\( = {w^4}({w^6} + {w^5} + {w^4} + {w^3} + {w^2} + w + 1 + 2{w^3})\)

\( = {w^4}(2{w^3})\)    (A1)

THEN

\( = 2\)    A1

[10 marks]

\({z^2} + z + 2 = 0 \Rightarrow z = \frac{{ - 1 \pm {\text{i}}\sqrt 7 }}{2}\)    M1A1

\(\operatorname{Im} (w + {w^2} + {w^4}) > 0\)    R1

\(\operatorname{Im} \alpha  = \frac{{\sqrt 7 }}{2}\)    A1

Note:     Final A mark is independent of previous R mark.

[4 marks]

The majority of candidates scored full marks in part (a).

The majority of candidates scored full marks in part (b)(i). It was expected to see \(w - 1 \ne 0\) stated for the (b)(ii) mark, though some did appreciate this.

In part (c), the roots were required to be stated in terms of \(w\). This was sometimes ignored, thankfully not too frequently. Clear Argand diagrams were not often seen, and candidates’ general presentation in this area could be improved. Having said this, most scripts were awarded at least 2 of 3 marks available.

Part (d) proved to be a good discriminator for the better candidates. The product and sum of roots formulae now seem to be better appreciated, and while only the best scored full marks, a good number were able to demonstrate the result \(b = 1\).

In part (e), of those candidates who reached this far in the paper, most were able to pick up two or three marks, albeit from sometimes following through incorrect work. A correct reason for choosing \({\text{i}}\sqrt 7 \) over \( - {\text{i}}\sqrt 7 \) was necessary, but rarely, if ever seen.

Determine the value of

(i)     \(1 + \omega  + {\omega ^2}\);

(ii)     \(1 + \omega {\text{*}} + {(\omega {\text{*}})^2}\).

Show that \((\omega  - 3{\omega ^2})({\omega ^2} - 3\omega ) = 13\).

Find the values of \(x\) that satisfy the equation \(\left| p \right| = \left| q \right|\).

Solve the inequality \(\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2}\).

(i)     METHOD 1

\(1 + \omega  + {\omega ^2} = \frac{{1 - {\omega ^3}}}{{1 - \omega }} = 0\)    A1

as \(\omega  \ne 1\)     R1

METHOD 2

solutions of \(1 - {\omega ^3} = 0\) are \(\omega  = 1,{\text{ }}\omega {\text{ = }}\frac{{ - 1 \pm \sqrt 3 {\text{i}}}}{2}\)     A1

verification that the sum of these roots is 0     R1

(ii)     \(1 + \omega {\text{*}} + {(\omega {\text{*}})^2} = 0\)     A2

[4 marks]

\((\omega  - 3{\omega ^2})({\omega ^2} - 3\omega ) =  - 3{\omega ^4} + 10{\omega ^3} - 3{\omega ^2}\)    M1A1

EITHER

\( =  - 3{\omega ^2}({\omega ^2} + \omega  + 1) + 13{\omega ^3}\)    M1

\( =  - 3{\omega ^2} \times 0 + 13 \times 1\)    A1

OR

\( =  - 3\omega  + 10 - 3{\omega ^2} =  - 3({\omega ^2} + \omega  + 1) + 13\)    M1

\( =  - 3 \times 0 + 13\)    A1

OR

substitution by \(\omega  = \frac{{ - 1 \pm \sqrt 3 {\text{i}}}}{2}\) in any form     M1

numerical values of each term seen     A1

THEN

\( = 13\)    AG

[4 marks]

\(\left| p \right| = \left| q \right| \Rightarrow \sqrt {{1^2} + {3^2}}  = \sqrt {{x^2} + {{(2x + 1)}^2}} \)    (M1)(A1)

\(5{x^2} + 4x - 9 = 0\)    A1

\((5x + 9)(x - 1) = 0\)    (M1)

\(x = 1,{\text{ }}x =  - \frac{9}{5}\)    A1

[5 marks]

\(pq = (1 - 3{\text{i}})\left( {x + (2x + 1){\text{i}}} \right) = (7x + 3) + (1 - x){\text{i}}\)    M1A1

\(\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2} \Rightarrow (7x + 3) + 8 < {(1 - x)^2}\)    M1

\( \Rightarrow {x^2} - 9x - 10 > 0\)    A1

\( \Rightarrow (x + 1)(x - 10) > 0\)    M1

\(x <  - 1,{\text{ }}x > 10\)    A1

[6 marks]

Expand and simplify \({\left( {x - \frac{2}{x}} \right)^4}\).

Hence determine the constant term in the expansion \((2{x^2} + 1){\left( {x - \frac{2}{x}} \right)^4}\).

\({\left( {x - \frac{2}{x}} \right)^4} = {x^4} + 4{x^3}\left( { - \frac{2}{x}} \right) + 6{x^2}{\left( { - \frac{2}{x}} \right)^2} + 4x{\left( { - \frac{2}{x}} \right)^3} + {\left( { - \frac{2}{x}} \right)^4}\)     (A2)

 Note: Award (A1) for 3 or 4 correct terms. 

Note: Accept combinatorial expressions, e.g. \(\left( {\begin{array}{*{20}{c}}
  4 \\
  2
\end{array}} \right)\) for 6.

\( = {x^4} - 8{x^2} + 24 - \frac{{32}}{{{x^2}}} + \frac{{16}}{{{x^4}}}\)     A1

[3 marks]

constant term from expansion of \((2{x^2} + 1){\left( {x - \frac{2}{x}} \right)^4} = -64 + 24 = -40\)     A2

Note: Award A1 for –64 or 24 seen.

[2 marks]

It was disappointing to see many candidates expanding \({\left( {x - \frac{2}{x}} \right)^4}\) by first expanding \({\left( {x - \frac{2}{x}} \right)^2}\) and then either squaring the result or multiplying twice by \(\left( {x - \frac{2}{x}} \right)\), processes which often resulted in arithmetic errors being made. Candidates at this level are expected to be sufficiently familiar with Pascal’s Triangle to use it in this kind of problem. In (b), some candidates appeared not to understand the phrase ‘constant term’.

It was disappointing to see many candidates expanding \({\left( {x - \frac{2}{x}} \right)^4}\) by first expanding \({\left( {x - \frac{2}{x}} \right)^2}\) and then either squaring the result or multiplying twice by \(\left( {x - \frac{2}{x}} \right)\), processes which often resulted in arithmetic errors being made. Candidates at this level are expected to be sufficiently familiar with Pascal's Triangle to use it in this kind of problem. In (b), some candidates appeared not to understand the phrase "constant term".

Find the value of \(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}\).

Show that \(\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi \) where \(k \in \mathbb{Z}\).

Use the principle of mathematical induction to prove that

\(\sin x + \sin 3x +  \ldots  + \sin (2n - 1)x = \frac{{1 - \cos 2nx}}{{2\sin x}},{\text{ }}n \in {\mathbb{Z}^ + },{\text{ }}x \ne k\pi \) where \(k \in \mathbb{Z}\).

Hence or otherwise solve the equation \(\sin x + \sin 3x = \cos x\) in the interval \(0 < x < \pi \).

\(\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}\)    (M1)A1

Note: Award M1 for 5 equal terms with \) + \) or \( - \) signs.

[2 marks]

\(\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \frac{{1 - (1 - 2{{\sin }^2}x)}}{{2\sin x}}\)    M1

\( \equiv \frac{{2{{\sin }^2}x}}{{2\sin x}}\)    A1

\( \equiv \sin x\)    AG

[2 marks]

let \({\text{P}}(n):\sin x + \sin 3x +  \ldots  + \sin (2n - 1)x \equiv \frac{{1 - \cos 2nx}}{{2\sin x}}\)

if \(n = 1\)

\({\text{P}}(1):\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x\) which is true (as proved in part (b))     R1

assume \({\text{P}}(k)\) true, \(\sin x + \sin 3x +  \ldots  + \sin (2k - 1)x \equiv \frac{{1 - \cos 2kx}}{{2\sin x}}\)     M1

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let \(n = k\)” only. Subsequent marks are independent of this M1.

consider \({\text{P}}(k + 1)\):

\({\text{P}}(k + 1):\sin x + \sin 3x +  \ldots  + \sin (2k - 1)x + \sin (2k + 1)x \equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}\)

\(LHS = \sin x + \sin 3x +  \ldots  + \sin (2k - 1)x + \sin (2k + 1)x\)    M1

\( \equiv \frac{{1 - \cos 2kx}}{{2\sin x}} + \sin (2k + 1)x\)    A1

\( \equiv \frac{{1 - \cos 2kx + 2\sin x\sin (2k + 1)x}}{{2\sin x}}\)

\( \equiv \frac{{1 - \cos 2kx + 2\sin x\cos x\sin 2kx + 2{{\sin }^2}x\cos 2kx}}{{2\sin x}}\)    M1

\( \equiv \frac{{1 - \left( {(1 - 2{{\sin }^2}x)\cos 2kx - \sin 2x\sin 2kx} \right)}}{{2\sin x}}\)    M1

\( \equiv \frac{{1 - (\cos 2x\cos 2kx - \sin 2x\sin 2kx)}}{{2\sin x}}\)    A1

\( \equiv \frac{{1 - \cos (2kx + 2x)}}{{2\sin x}}\)    A1

\( \equiv \frac{{1 - \cos 2(k + 1)x}}{{2\sin x}}\)

so if true for \(n = k\) , then also true for \(n = k + 1\)

as true for \(n = 1\) then true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

[9 marks]

EITHER

\(\sin x + \sin 3x = \cos x \Rightarrow \frac{{1 - \cos 4x}}{{2\sin x}} = \cos x\)    M1

\( \Rightarrow 1 - \cos 4x = 2\sin x\cos x,{\text{ }}(\sin x \ne 0)\)    A1

\( \Rightarrow 1 - (1 - 2{\sin ^2}2x) = \sin 2x\)    M1

\( \Rightarrow \sin 2x(2\sin 2x - 1) = 0\)    M1

\( \Rightarrow \sin 2x = 0\) or \(\sin 2x = \frac{1}{2}\)     A1

\(2x = \pi ,{\text{ }}2x = \frac{\pi }{6}\) and \(2x = \frac{{5\pi }}{6}\)

OR

\(\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x\)    M1A1

\( \Rightarrow (2\sin 2x - 1)\cos x = 0,{\text{ }}(\sin x \ne 0)\)    M1A1

\( \Rightarrow \sin 2x = \frac{1}{2}\) of \(\cos x = 0\)    A1

\(2x = \frac{\pi }{6},{\text{ }}2x = \frac{{5\pi }}{6}\) and \(x = \frac{\pi }{2}\)

THEN

\(\therefore x = \frac{\pi }{2},{\text{ }}x = \frac{\pi }{{12}}\) and \(x = \frac{{5\pi }}{{12}}\)     A1

Note: Do not award the final A1 if extra solutions are seen.

[6 marks]

Darren plays first, find the probability that he wins.

The game is now changed so that the ball chosen is replaced after each turn.

Darren still plays first.

Show that the probability of Darren winning has not changed.

probability that Darren wins \({\text{P}}(W) + {\text{P}}(RRW) + {\text{P}}(RRRRW)\)     (M1)

Note:     Only award M1 if three terms are seen or are implied by the following numerical equivalent.

Note:     Accept equivalent tree diagram for method mark.

\( = \frac{2}{6} + \frac{4}{6} \bullet \frac{3}{5} \bullet \frac{2}{4} + \frac{4}{6} \bullet \frac{3}{5} \bullet \frac{2}{4} \bullet \frac{1}{3} \bullet \frac{2}{2}\;\;\;\left( { = \frac{1}{3} + \frac{1}{5} + \frac{1}{{15}}} \right)\)     A2

Note:     A1 for two correct.

\( = \frac{3}{5}\)     A1

[4 marks]

METHOD 1

the probability that Darren wins is given by

\({\text{P}}(W) + {\text{P}}(RRW) + {\text{P}}(RRRRW) +  \ldots \)     (M1)

Note:     Accept equivalent tree diagram with correctly indicated path for method mark.

\({\text{P (Darren Win)}} = \frac{1}{3} + \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{1}{3} + \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{2}{3} \bullet \frac{1}{3} +  \ldots \)

or \( = \frac{1}{3}\left( {1 + \frac{4}{9} + {{\left( {\frac{4}{9}} \right)}^2} +  \ldots } \right)\)     A1

\( = \frac{1}{3}\left( {\frac{1}{{1 - \frac{4}{9}}}} \right)\)     A1

\( = \frac{3}{5}\)     AG

METHOD 2

\({\text{P (Darren wins)}} = {\text{P}}\)

\({\text{P}} = \frac{1}{3} + \frac{4}{9}{\text{P}}\)     M1A2

\(\frac{5}{9}{\text{P}} = \frac{1}{3}\)

\({\text{P}} = \frac{3}{5}\)     AG

[3 marks]

Total [7 marks]

Write down the value of \({u_1}\).

Find the value of \({u_6}\).

Prove that \(\{ {u_n}\} \) is an arithmetic sequence, stating clearly its common difference.

\({u_1} = 1\)    A1

[1 mark]

\({u_6} = {S_6} - {S_5} = 31\)    M1A1

[2 marks]

\({u_n} = {S_n} - {S_{n - 1}}\)    M1

\( = (3{n^2} - 2n) - \left( {3{{(n - 1)}^2} - 2(n - 1)} \right)\)

\( = (3{n^2} - 2n) - (3{n^2} - 6n + 3 - 2n + 2)\)

\( = 6n - 5\)    A1

\(d = {u_{n + 1}} - {u_n}\)    R1

\( = 6n + 6 - 5 - 6n + 5\)

\( = \left( {6(n + 1) - 5} \right) - (6n - 5)\)

\( = 6\) (constant)     A1

Notes: Award R1 only if candidate provides a clear argument that proves that the difference between ANY two consecutive terms of the sequence is constant. Do not accept examples involving particular terms of the sequence nor circular reasoning arguments (eg use of formulas of APs to prove that it is an AP). Last A1 is independent of R1.

[4 marks]

the degree of the polynomial;

the value of \({a_0}\).

the sum of the roots of the polynomial \( = \frac{{63}}{{16}}\)     (A1)

\(2\left( {\frac{{1 - {{\left( {\frac{1}{2}} \right)}^n}}}{{1 - \frac{1}{2}}}} \right) = \frac{{63}}{{16}}\)     M1A1

Note:     The formula for the sum of a geometric sequence must be equated to a value for the M1 to be awarded.

\(1 - {\left( {\frac{1}{2}} \right)^n} = \frac{{63}}{{64}} \Rightarrow {\left( {\frac{1}{2}} \right)^n} = \frac{1}{{64}}\)

\(n = 6\)     A1

[4 marks]

\(\frac{{{a_0}}}{{{a_n}}} = 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}},{\text{ (}}{{\text{a}}_n} = 16)\)     M1

\({a_0} = 16 \times 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}}\)

\({a_0} = {2^{ - 5}}\;\;\;\left( { = \frac{1}{{32}}} \right)\)     A1

[2 marks]

Total [6 marks]

Solve \({\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}\).

\({\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) - 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)\)

EITHER

\({\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}\)     M1

\( = \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}\)     A1

OR

\(\left( {{\text{ln}}\,x - 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)\)     M1A1

THEN

\({\text{ln}}\,x = 2\,{\text{ln}}\,2\) or \( - {\text{ln}}\,2\)     A1

\( \Rightarrow x = 4\) or \(x = \frac{1}{2}\)       (M1)A1   

Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.

solution is \(\frac{1}{2} < x < 4\)     A1

[6 marks]

Find the number of selections Grace could make if the largest integer drawn among the four cards is either a \(5\), a \(6\) or a \(7\).

Find the number of selections Grace could make if at least two of the four integers drawn are even.

use of the addition principle with \(3\) terms     (M1)

to obtain \(^4{C_3}{ + ^5}{C_3}{ + ^6}{C_3}{\text{ }}( = 4 + 10 + 20)\)     A1

number of possible selections is \(34\)     A1

[3 marks]

EITHER

recognition of three cases: (\(2\) odd and \(2\) even or \(1\) odd and \(3\) even or \(0\) odd and \(4\) even)     (M1)

\(\left( {^5{C_2}{ \times ^4}{C_2}} \right) + \left( {^5{C_1}{ \times ^4}{C_3}} \right) + \left( {^5{C_0}{ \times ^4}{C_4}} \right)\;\;\;( = 60 + 20 + 1)\)     (M1)A1

OR

recognition to subtract the sum of \(4\) odd and \(3\) odd and \(1\) even from the total     (M1)

\(^9{C_4}{ - ^5}{C_4} - \left( {^5{C_3}{ \times ^4}{C_1}} \right)\;\;\;( = 126 - 5 - 40)\)     (M1)A1

THEN

number of possible selections is \(81\)     A1

[4 marks]

Total [7 marks]

As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.

As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.

Solve the equation \({\log _2}(x + 3) + {\log _2}(x - 3) = 4\).

\({\log _2}(x + 3) + {\log _2}(x - 3) = 4\)

\({\log _2}({x^2} - 9) = 4\)     (M1)

\({x^2} - 9 = {2^4}{\text{ }}( = 16)\)     M1A1

\({x^2} = 25\)

\(x =  \pm 5\)     (A1)

\(x = 5\)     A1

[5 marks]

Find the solution of \({\log _2}x - {\log _2}5 = 2 + {\log _2}3\).

\({\log _2}x - {\log _2}5 = 2 + {\log _2}3\)

collecting at least two log terms     (M1)

eg\(\,\,\,\,\,\)\({\log _2}\frac{x}{5} = 2 + {\log _2}3{\text{ or }}{\log _2}\frac{x}{{15}} = 2\)

obtaining a correct equation without logs     (M1)

eg\(\,\,\,\,\,\)\(\frac{x}{5} = 12\)\(\,\,\,\)OR\(\,\,\,\)\(\frac{x}{{15}} = {2^2}\)     (A1)

\(x = 60\)     A1

[4 marks]

Show that this system does not have a unique solution for any value of \(\lambda \) .

(i)     Determine the value of \(\lambda \) for which the system is consistent.

(ii)     For this value of \(\lambda \) , find the general solution of the system.

using row operations,     M1

to obtain 2 equations in the same 2 variables     A1A1

for example \(y - z = 1\)

\(2y - 2z = \lambda  - 1\)

the fact that one of the left hand sides is a multiple of the other left hand side indicates that the equations do not have a unique solution, or equivalent     R1AG

[4 marks]

(i)     \(\lambda = 3\)     A1

(ii)     put \(z = \mu \)     M1

then \(y = 1 + \mu \)     A1

and \(x = - 2\mu \) or equivalent     A1

[4 marks]

Solve the equation \({z^3} = 8{\text{i}},{\text{ }}z \in \mathbb{C}\) giving your answers in the form \(z = r(\cos \theta  + {\text{i}}\sin \theta )\) and in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).

Consider the complex numbers \({z_1} = 1 + {\text{i}}\) and \({z_2} = 2\left( {\cos \left( {\frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right)\).

(i)     Write \({z_1}\) in the form \(r(\cos \theta  + {\text{i}}\sin \theta )\).

(ii)     Calculate \({z_1}{z_2}\) and write in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).

(iii)     Hence find the value of \(\tan \frac{{5\pi }}{{12}}\) in the form \(c + d\sqrt 3 \), where \(c,{\text{ }}d \in \mathbb{Z}\).

(iv)     Find the smallest value \(p > 0\) such that \({({z_2})^p}\) is a positive real number.

Note:     Accept answers and working in degrees, throughout.

\({z^3} = 8\left( {\cos \left( {\frac{\pi }{2} + 2\pi k} \right) + {\text{i}}\sin \left( {\frac{\pi }{2} + 2\pi k} \right)} \right)\)     (A1)

attempt the use of De Moivre’s Theorem in reverse     M1

\(z = 2\left( {\cos \left( {\frac{\pi }{6}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right);{\text{ }}2\left( {\cos \left( {\frac{{5\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{5\pi }}{6}} \right)} \right);\)

\(2\left( {\cos \left( {\frac{{9\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{9\pi }}{6}} \right)} \right)\)     A2

Note:     Accept cis form.

\(z =  \pm \sqrt 3  + {\text{i}},{\text{ }} - 2{\text{i}}\)     A2

Note:     Award A1 for two correct solutions in each of the two lines above.

[6 marks]

Note:     Accept answers and working in degrees, throughout.

(i)     \({z_1} = \sqrt 2 \left( {\cos \left( {\frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{4}} \right)} \right)\)     A1A1

(ii)     \(\left( {{z_2} = \left( {\sqrt 3  + {\text{i}}} \right)} \right)\)

\({z_1}{z_2} = (1 + {\text{i}})\left( {\sqrt 3  + {\text{i}}} \right)\)     M1

\( = \left( {\sqrt 3  - 1} \right) + {\text{i}}\left( {1 + \sqrt 3 } \right)\)     A1

(iii)     \({z_1}{z_2} = 2\sqrt 2 \left( {\cos \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right)} \right)\)     M1A1

Note:     Interpret “hence” as “hence or otherwise”.

\(\tan \frac{{5\pi }}{{12}} = \frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}}\)     A1

\( = 2 + \sqrt 3 \)     M1A1

Note:     Award final M1 for an attempt to rationalise the fraction.

(iv)     \({z_2}^p = {2^p}\left( {{\text{cis}}\left( {\frac{{p\pi }}{6}} \right)} \right)\)     (M1)

\({z_2}^p\) is a positive real number when \(p = 12\)     A1

Note:     Accept a solution based on part (a).

[11 marks]

Total [17 marks]

Let \(z = x + {\text{i}}y\) be any non-zero complex number.

(i)     Express \(\frac{1}{z}\) in the form \(u + {\text{i}}v\) .

(ii)     If \(z + \frac{1}{z} = k\) , \(k \in \mathbb{R}\) , show that either y = 0 or \({x^2} + {y^2} = 1\).

(iii)     Show that if \({x^2} + {y^2} = 1\) then \(\left| k \right| \leqslant 2\) .

Let \(w = \cos \theta + {\text{i}}\sin \theta \) .

(i)     Show that \({w^n} + {w^{ - n}} = 2\cos n\theta \) , \(n \in \mathbb{Z}\) .

(ii)     Solve the equation \(3{w^2} - w + 2 - {w^{ - 1}} + 3{w^{ - 2}} = 0\), giving the roots in the form \(x + {\text{i}}y\) .

(i)     \(\frac{1}{z} = \frac{1}{{x + {\text{i}}y}} \times \frac{{x - {\text{i}}y}}{{x - {\text{i}}y}} = \frac{x}{{{x^2} + {y^2}}} - {\text{i}}\frac{y}{{{x^2} + {y^2}}}\)     (M1)A1

(ii)     \(z + \frac{1}{z} = x + \frac{x}{{{x^2} + {y^2}}} + {\text{i}}\left( {y - \frac{y}{{{x^2} + {y^2}}}} \right) = k\)     (A1)

for k to be real, \(y - \frac{y}{{{x^2} + {y^2}}} = 0 \Rightarrow y({x^2} + {y^2} - 1) = 0\)     M1A1

hence, \(y = 0{\text{ or }}{x^2} + {y^2} - 1 = 0 \Rightarrow {x^2} + {y^2} = 1\)     AG

(iii)     when \({x^2} + {y^2} = 1,{\text{ }}z + \frac{1}{z} = 2x\)     (M1)A1

\(\left| x \right| \leqslant 1\)     R1

\( \Rightarrow \left| k \right| \leqslant 2\)     AG

[8 marks]

(i)     \({w^{ - n}} = \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta ) = \cos n\theta - {\text{i}}\sin n\theta \)     M1A1

\( \Rightarrow {w^n} + {w^{ - n}} = (\cos n\theta + {\text{i}}\sin n\theta ) + (\cos n\theta - {\text{i}}\sin n\theta ) = 2\cos n\theta \)     M1AG

(ii)     (rearranging)

\(3({w^2} + {w^{ - 2}}) - (w + {w^{ - 1}}) + 2 = 0\)     (M1)

\( \Rightarrow 3(2\cos 2\theta ) - 2\cos \theta + 2 = 0\)     A1

\( \Rightarrow 2(3\cos 2\theta - \cos \theta + 1) = 0\)

\( \Rightarrow 3(2{\cos ^2}\theta - 1) - \cos \theta + 1 = 0\)     M1

\( \Rightarrow 6{\cos ^2}\theta - \cos \theta - 2 = 0\)     A1

\( \Rightarrow (3\cos \theta - 2)(2\cos \theta + 1) = 0\)     M1

\(\therefore \cos \theta = \frac{2}{3},{\text{ }}\cos \theta = - \frac{1}{2}\)     A1A1

\(\cos \theta = \frac{2}{3} \Rightarrow \sin \theta = \pm \frac{{\sqrt 5 }}{3}\)     A1

\(\cos \theta = - \frac{1}{2} \Rightarrow \sin \theta = \pm \frac{{\sqrt 3 }}{2}\)     A1

\(\therefore w = \frac{2}{3} \pm \frac{{{\text{i}}\sqrt 5 }}{3}, - \frac{1}{2} \pm \frac{{{\text{i}}\sqrt 3 }}{2}\)     A1A1

Note: Allow FT from incorrect \(\cos \theta \) and/or \(\sin \theta \) .

[14 marks]

A large number of candidates did not attempt part (a), or did so unsuccessfully.

It was obvious that many candidates had been trained to answer questions of the type in part (b), and hence of those who attempted it, many did so successfully. Quite a few however failed to find all solutions.

Determine the roots of the equation \({(z + 2{\text{i}})^3} = 216{\text{i}}\), \(z \in \mathbb{C}\), giving the answers in the form \(z = a\sqrt 3  + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{Z}\).

METHOD 1

\(216{\text{i}} = 216\left( {\cos \frac{\pi }{2} + {\text{i}}\sin \frac{\pi }{2}} \right)\)     A1

\(z + 2{\text{i}} = \sqrt[3]{{216}}{\left( {\cos \left( {\frac{\pi }{2} + 2\pi k} \right) = {\text{i}}\sin \left( {\frac{\pi }{2} + 2\pi k} \right)} \right)^{\frac{1}{3}}}\)     (M1)

\(z + 2{\text{i}} = 6\left( {\cos \left( {\frac{\pi }{6} + \frac{{2\pi k}}{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6} + \frac{{2\pi k}}{3}} \right)} \right)\)     A1

\({z_1} + 2{\text{i}} = 6\left( {\cos \frac{\pi }{6} + {\text{i}}\sin \frac{\pi }{6}} \right) = 6\left( {\frac{{\sqrt 3 }}{2} + \frac{{\text{i}}}{2}} \right) = 3\sqrt 3  + 3{\text{i}}\)

\({z_2} + 2{\text{i}} = 6\left( {\cos \frac{{5\pi }}{6} + {\text{i}}\sin \frac{{5\pi }}{6}} \right) = 6\left( {\frac{{ - \sqrt 3 }}{2} + \frac{{\text{i}}}{2}} \right) =  - 3\sqrt 3  + 3{\text{i}}\)

\({z_3} + 2{\text{i}} = 6\left( {\cos \frac{{3\pi }}{2} + {\text{i}}\sin \frac{{3\pi }}{2}} \right) =  - 6{\text{i}}\)     A2

Note:     Award A1A0 for one correct root.

so roots are \({z_1} = 3\sqrt 3  + {\text{i, }}{z_2} =  - 3\sqrt 3  + {\text{i}}\) and \({z_3} =  - 8{\text{i}}\)     M1A1

Note:     Award M1 for subtracting 2i from their three roots.

METHOD 2

\({\left( {a\sqrt 3  + (b + 2){\text{i}}} \right)^3} = 216{\text{i}}\)

\({\left( {a\sqrt 3 } \right)^3} + 3{\left( {a\sqrt 3 } \right)^2}(b + 2){\text{i}} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} - {\text{i}}{(b + 2)^3} = 216{\text{i}}\)     M1A1

\({\left( {a\sqrt 3 } \right)^3} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} + {\text{i}}\left( {3{{\left( {a\sqrt 3 } \right)}^2}(b + 2) - {{(b + 2)}^3}} \right) = 216{\text{i}}\)

\({\left( {a\sqrt 3 } \right)^3} - 3\left( {a\sqrt 3 } \right){(b + 2)^2} = 0\) and \(3{\left( {a\sqrt 3 } \right)^2}(b + 2) - {(b + 2)^3} = 216\)     M1A1

\(a\left( {{a^2} - {{(b + 2)}^2}} \right) = 0\) and \(9{a^2}(b + 2) - {(b + 2)^3} = 216\)

\(a = 0\) or \({a^2} = {(b + 2)^2}\)

if \(a = 0,{\text{ }} - {(b + 2)^3} = 216 \Rightarrow b + 2 =  - 6\)

\(\therefore b =  - 8\)     A1

\((a,{\text{ }}b) = (0,{\text{ }} - 8)\)

if \({a^2} = {(b + 2)^2},{\text{ }}9{(b + 2)^2}(b + 2) - {(b + 2)^3} = 216\)

\(8{(b + 2)^3} = 216\)

\({(b + 2)^3} = 27\)

\(b + 2 = 3\)

\(b = 1\)

\(\therefore {a^2} = 9 \Rightarrow a =  \pm 3\)

\(\therefore (a,{\text{ }}b) = ( \pm 3,{\text{ }}1)\)     A1A1

so roots are \({z_1} = 3\sqrt 3  + {\text{i, }}{z_2} =  - 3\sqrt 3  + {\text{i}}\) and \({z_3} =  - 8{\text{i}}\)

METHOD 3

\({(z + 2{\text{i}})^3} - {( - 6{\text{i}})^3} = 0\)

attempt to factorise:     M1

\(\left( {(z + 2{\text{i}}) - ( - 6{\text{i}})} \right)\left( {{{(z + 2{\text{i}})}^2} + (z + 2{\text{i}})( - 6{\text{i}}) + {{( - 6{\text{i}})}^2}} \right) = 0\)     A1

\((z + 8{\text{i}})({z^2} - 2{\text{i}}z - 28) = 0\)     A1

\(z + 8{\text{i}} = 0 \Rightarrow z =  - 8{\text{i}}\)     A1

\({z^2} - 2{\text{i}}z - 28 = 0 \Rightarrow z = \frac{{2{\text{i}} \pm \sqrt { - 4 - (4 \times 1 \times  - 28)} }}{2}\)     M1

\(z = \frac{{2{\text{i}} \pm \sqrt {108} }}{2}\)

\(z = \frac{{2{\text{i}} \pm 6\sqrt 3 }}{2}\)

\(z = {\text{i}} \pm 3\sqrt 3 \)     A1A1

Special Case:

Note:     If a candidate recognises that \(\sqrt[3]{{216{\text{i}}}} =  - 6{\text{i}}\) (anywhere seen), and makes no valid progress in finding three roots, award A1 only.

[7 marks]

The complex number z is defined as \(z = \cos \theta + {\text{i}}\sin \theta \) .

(a)     State de Moivre’s theorem.

(b)     Show that \({z^n} - \frac{1}{{{z^n}}} = 2{\text{i}}\sin (n\theta )\) .

(c)     Use the binomial theorem to expand \({\left( {z - \frac{1}{z}} \right)^5}\) giving your answer in simplified form.

(d)     Hence show that \(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \) .

(e)     Check that your result in part (d) is true for \(\theta = \frac{\pi }{4}\) .

(f)     Find \(\int_0^{\frac{\pi }{2}} {{{\sin }^5}\theta {\text{d}}\theta } \) .

(g)     Hence, with reference to graphs of circular functions, find \(\int_0^{\frac{\pi }{2}} {{{\cos }^5}\theta {\text{d}}\theta } \) , explaining your reasoning.

(a)     any appropriate form, e.g. \({(\cos \theta + {\text{i}}\sin \theta )^n} = \cos (n\theta) + {\text{i}}\sin (n\theta)\)     A1

[1 mark]

(b)     \({z^n} = \cos n\theta + {\text{i}}\sin n\theta \)     A1

\(\frac{1}{{{z^n}}} = \cos ( - n\theta ) + {\text{i}}\sin ( - n\theta )\)     (M1)

\( = \cos n\theta  - {\text{i}}\sin (n\theta )\)     A1

therefore \({z^n} - \frac{1}{{{z^n}}} = 2{\text{i}}\sin (n\theta )\)     AG

[3 marks]

(c)     \({\left( {z - \frac{1}{z}} \right)^5} = {z^5} + \left( {\begin{array}{*{20}{c}}
  5 \\
  1
\end{array}} \right){z^4}\left( { - \frac{1}{z}} \right) + \left( {\begin{array}{*{20}{c}}
  5 \\
  2
\end{array}} \right){z^3}{\left( { - \frac{1}{z}} \right)^2} + \left( {\begin{array}{*{20}{c}}
  5 \\
  3
\end{array}} \right){z^2}{\left( { - \frac{1}{z}} \right)^3} + \left( {\begin{array}{*{20}{c}}
  5 \\
  4
\end{array}} \right)z{\left( { - \frac{1}{z}} \right)^4} + {\left( { - \frac{1}{z}} \right)^5}\)     (M1)(A1)

\( = {z^5} - 5{z^3} + 10z - \frac{{10}}{z} + \frac{5}{{{z^3}}} - \frac{1}{{{z^5}}}\)     A1

[3 marks]

(d)     \({\left( {z - \frac{1}{z}} \right)^5} = {z^5} - \frac{1}{{{z^5}}} - 5\left( {{z^3} - \frac{1}{{{z^3}}}} \right) + 10\left( {z - \frac{1}{z}} \right)\)     M1A1

\({(2{\text{i}}\sin \theta )^5} = 2{\text{i}}\sin 5\theta - 10{\text{i}}\sin 3\theta + 20{\text{i}}\sin \theta \)     M1A1

\(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \)     AG

[4 marks]

(e)     \(16{\sin ^5}\theta = \sin 5\theta - 5\sin 3\theta + 10\sin \theta \)

\({\text{LHS}} = 16{\left( {\sin \frac{\pi }{4}} \right)^5}\)

\( = 16{\left( {\frac{{\sqrt 2 }}{2}} \right)^5}\)

\( = 2\sqrt 2 \,\,\,\,\,\left( { = \frac{4}{{\sqrt 2 }}} \right)\)     A1

\({\text{RHS}} = \sin \left( {\frac{{5\pi }}{4}} \right) - 5\sin \left( {\frac{{3\pi }}{4}} \right) + 10\sin \left( {\frac{\pi }{4}} \right)\)

\( = - \frac{{\sqrt 2 }}{2} - 5\left( {\frac{{\sqrt 2 }}{2}} \right) + 10\left( {\frac{{\sqrt 2 }}{2}} \right)\)     M1A1

Note: Award M1 for attempted substitution.

\( = 2\sqrt 2 \,\,\,\,\,\left( { = \frac{4}{{\sqrt 2 }}} \right)\)     A1

hence this is true for \(\theta = \frac{\pi }{4}\)     AG

[4 marks]

(f)     \(\int_0^{\frac{\pi }{2}} {{{\sin }^5}\theta {\text{d}}\theta = \frac{1}{{16}}\int_0^{\frac{\pi }{2}} {(\sin 5\theta - 5\sin 3\theta + 10\sin \theta ){\text{d}}\theta } } \)     M1

\( = \frac{1}{{16}}\left[ { - \frac{{\cos 5\theta }}{5} + \frac{{5\cos 3\theta }}{3} - 10\cos \theta } \right]_0^{\frac{\pi }{2}}\)     A1

\( = \frac{1}{{16}}\left[ {0 - \left( { - \frac{1}{5} + \frac{5}{3} - 10} \right)} \right]\)     A1

\( = \frac{8}{{15}}\)     A1

[4 marks]

(g)     \(\int_0^{\frac{\pi }{2}} {{{\cos }^5}\theta {\text{d}}\theta = \frac{8}{{15}}} \) , with appropriate reference to symmetry and graphs.     A1R1R1

Note: Award first R1 for partially correct reasoning e.g. sketches of graphs of sin and cos.

Award second R1 for fully correct reasoning involving \({\sin ^5}\) and \({\cos ^5}\).

[3 marks]

Total [22 marks]

Many students in b) substituted for the second term (again not making the connection to part a)) on the LHS and multiplied by the conjugate, which some managed well but it is inefficient. The binomial expansion was done well even if students did not do the earlier part. The connection between d) and f) was missed by many which lead to some creative attempts at the integral. Very few attempted the last part and of those many attempted another integral, ignoring the hence, while others related to the graph of sin and cos but not to the particular graphs here.

Three girls and four boys are seated randomly on a straight bench. Find the probability that the girls sit together and the boys sit together.

METHOD 1

total number of arrangements 7!     (A1)

number of ways for girls and boys to sit together \( = 3! \times 4! \times 2\)     (M1)(A1)

Note:    Award M1A0 if the 2 is missing.

probability \(\frac{{3! \times 4! \times 2}}{{7!}}\)     M1

Note:     Award M1 for attempting to write as a probability.

\(\frac{{2 \times 3 \times 4! \times 2}}{{7 \times 6 \times 5 \times 4!}}\)

\( = \frac{2}{{35}}\)     A1

Note:     Award A0 if not fully simplified.

METHOD 2

\(\frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} + \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4}\)     (M1)A1A1

Note:     Accept \(\frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} \times 2\) or \(\frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times 2\).

\( = \frac{2}{{35}}\)     (M1)A1

Note:     Award A0 if not fully simplified.

[5 marks]

The following system of equations represents three planes in space.

\[x + 3y + z =  - 1\]

\[x + 2y - 2z = 15\]

\[2x + y - z = 6\]

Find the coordinates of the point of intersection of the three planes.

EITHER

eliminating a variable, \(x\), for example to obtain \(y + 3z =  - 16\) and \( - 5y - 3z = 8\)     M1A1

attempting to find the value of one variable     M1

point of intersection is \(( - 1,{\text{ }}2,{\text{ }} - 6)\)     A1A1A1

OR

attempting row reduction of relevant matrix, eg.      M1

correct matrix with two zeroes in a column, eg.      A1

further attempt at reduction     M1

point of intersection is \(( - 1,{\text{ }}2,{\text{ }} - 6)\)     A1A1A1

Note:     Allow solution expressed as \(x =  - 1,{\text{ }}y = 2,{\text{ }}z =  - 6\) for final A marks.

[6 marks]

This provided a generally easy start for many candidates. Most successful candidates obtained their answer through row reduction of a suitable matrix. Those choosing an alternative method often made slips in their algebra.

Write down the expansion of \({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3}\) in the form \(a + {\text{i}}b\) , where \(a\) and \(b\) are in terms of \({\sin \theta }\) and \({\cos \theta }\) .

Hence show that \(\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta \) .

Similarly show that \(\cos 5\theta = 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta \) .

Hence solve the equation \(\cos 5\theta + \cos 3\theta + \cos \theta = 0\) , where \(\theta \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\) .

By considering the solutions of the equation \(\cos 5\theta = 0\) , show that \(\cos \frac{\pi }{{10}} = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \) and state the value of \(\cos \frac{{7\pi }}{{10}}\).

\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = {\cos ^3}\theta + 3{\cos ^2}\theta \left( {{\text{i}}\sin \theta } \right) + 3\cos \theta {\left( {{\text{i}}\sin \theta } \right)^2} + {\left( {{\text{i}}\sin \theta } \right)^3}\)     (M1)

\( = {\cos ^3}\theta - 3\cos \theta {\sin ^2}\theta + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta - {{\sin }^3}\theta } \right)\)     A1

[2 marks]

from De Moivre’s theorem

\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^3} = \cos 3\theta + {\text{i}}\sin 3\theta \)     (M1)

\(\cos 3\theta + {\text{i}}\sin 3\theta = \left( {{{\cos }^3}\theta - 3\cos \theta {{\sin }^2}\theta } \right) + {\text{i}}\left( {3{{\cos }^2}\theta \sin \theta - {{\sin }^3}\theta } \right)\)

equating real parts     M1

\(\cos 3\theta = {\cos ^3}\theta - 3\cos \theta {\sin ^2}\theta \)

\( = {\cos ^3}\theta - 3\cos \theta \left( {1 - {{\cos }^2}\theta } \right)\)     A1

\( = {\cos ^3}\theta - 3\cos \theta + 3{\cos ^3}\theta \)

\( = 4{\cos ^3}\theta - 3\cos \theta \)     AG

Note: Do not award marks if part (a) is not used.

[3 marks]

\({\left( {\cos \theta + {\text{i}}\sin \theta } \right)^5} = \)

\({\cos ^5}\theta + 5{\cos ^4}\theta \left( {{\text{i}}\sin \theta } \right) + 10{\cos ^3}\theta {\left( {{\text{i}}\sin \theta } \right)^2} + 10{\cos ^2}\theta {\left( {{\text{i}}\sin \theta } \right)^3} + 5\cos \theta {\left( {{\text{i}}\sin \theta } \right)^4} + {\left( {{\text{i}}\sin \theta } \right)^5}\)     (A1)

from De Moivre’s theorem

\(\cos 5\theta = {\cos ^5}\theta - 10{\cos ^3}\theta {\sin ^2}\theta  + 5\cos \theta {\sin ^4}\theta \)     M1

\( = {\cos ^5}\theta - 10{\cos ^3}\theta \left( {1 - {{\cos }^2}\theta } \right) + 5\cos \theta {\left( {1 - {{\cos }^2}\theta } \right)^2}\)     A1

\( = {\cos ^5}\theta - 10{\cos ^3}\theta + 10{\cos ^5}\theta + 5\cos \theta - 10{\cos ^3}\theta + 5{\cos ^5}\theta \)

\(\therefore \cos 5\theta = 16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta \)     AG

Note: If compound angles used in (b) and (c), then marks can be allocated in (c) only.

[3 marks]

\(\cos 5\theta + \cos 3\theta + \cos \theta \)

\( = \left( {16{{\cos }^5}\theta - 20{{\cos }^3}\theta + 5\cos \theta } \right) + \left( {4{{\cos }^3}\theta - 3\cos \theta } \right) + \cos \theta = 0\)     M1

\(16{\cos ^5}\theta - 16{\cos ^3}\theta + 3\cos \theta = 0\)     A1

\(\cos \theta \left( {16{{\cos }^4}\theta - 16{{\cos }^2}\theta + 3} \right) = 0\)

\(\cos \theta \left( {4{{\cos }^2}\theta - 3} \right)\left( {4{{\cos }^2}\theta - 1} \right) = 0\)     A1

\(\therefore \cos \theta = 0\); \( \pm \frac{{\sqrt 3 }}{2}\); \( \pm \frac{1}{2}\)     A1

\(\therefore \theta = \pm \frac{\pi }{6}\); \(\pm \frac{\pi }{3}\); \( \pm \frac{\pi }{2}\)     A2

[6 marks]

\(\cos 5\theta = 0\)

\(5\theta = ...\frac{\pi }{2}\); \(\left( {\frac{{3\pi }}{2};\frac{{5\pi }}{2}} \right)\); \(\frac{{7\pi }}{2}\); \(...\)     (M1)

\(\theta = ...\frac{\pi }{{10}}\); \(\left( {\frac{{3\pi }}{{10}};\frac{{5\pi }}{{10}}} \right)\); \(\frac{{7\pi }}{10}\); \(...\)     (M1)

Note: These marks can be awarded for verifications later in the question.

now consider \(16{\cos ^5}\theta - 20{\cos ^3}\theta + 5\cos \theta = 0\)     M1

\(\cos \theta \left( {16{{\cos }^4}\theta - 20{{\cos }^2}\theta + 5} \right) = 0\)

\({\cos ^2}\theta = \frac{{20 \pm \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}\); \(\cos \theta = 0\)     A1

\(\cos \theta = \pm \sqrt {\frac{{20 \pm \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}} \)

\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{20 + \sqrt {400 - 4\left( {16} \right)\left( 5 \right)} }}{{32}}} \) since max value of cosine
\( \Rightarrow \) angle closest to zero     R1

\(\cos \frac{\pi }{{10}} = \sqrt {\frac{{4.5 + 4\sqrt {25 - 4\left( 5 \right)} }}{{4.8}}}  = \sqrt {\frac{{5 + \sqrt 5 }}{8}} \)     A1

\(\cos \frac{{7\pi }}{{10}} = - \sqrt {\frac{{5 - \sqrt 5 }}{8}} \)     A1A1

[8 marks]

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

This question proved to be very difficult for most candidates. Many had difficulties in following the instructions and attempted to use addition formulae rather than binomial expansions. A small number of candidates used the results given and made a good attempt to part (d) but very few answered part (e).

Given that \(y = \frac{1}{{1 - x}}\), use mathematical induction to prove that \(\frac{{{{\text{d}}^n}y}}{{{\text{d}}{x^n}}} = \frac{{n!}}{{{{(1 - x)}^{n + 1}}}},{\text{ }}n \in {\mathbb{Z}^ + }\).

proposition is true for n = 1 since \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{{{{(1 - x)}^2}}}\)     M1

\( = \frac{{1!}}{{{{(1 - x)}^2}}}\)     A1

Note: Must see the 1! for the A1.

assume true for n = k, \(k \in {\mathbb{Z}^ + }\), i.e. \(\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}} = \frac{{k!}}{{{{(1 - x)}^{k + 1}}}}\)     M1

consider \(\frac{{{{\text{d}}^{k + 1}}y}}{{{\text{d}}{x^{k + 1}}}} = \frac{{{\text{d}}\left( {\frac{{{{\text{d}}^k}y}}{{{\text{d}}{x^k}}}} \right)}}{{{\text{d}}x}}\)     (M1)

\( = (k + 1)k!{(1 - x)^{ - (k + 1) - 1}}\)     A1

\( = \frac{{(k + 1)!}}{{{{(1 - x)}^{k + 2}}}}\)     A1

hence, \({{\text{P}}_{k + 1}}\) is true whenever \({{\text{P}}_{k}}\) is true, and \({{\text{P}}_1}\) is true, and therefore the proposition is true for all positive integers     R1

Note: The final R1 is only available if at least 4 of the previous marks have been awarded.

[7 marks]

Most candidates were awarded good marks for this question. A disappointing minority thought that the \((k + 1)\)th derivative was the \((k)\)th derivative multiplied by the first derivative. Providing an acceptable final statement remains a perennial issue. 

Write down the first four terms of the expansion.

(i)     Show that \({n^3} - 9{n^2} + 14n = 0\).

(ii)     Hence find the value of \(n\).

\(1,{\text{ }}nx,{\text{ }}\frac{{n(n - 1)}}{2}{x^2},{\text{ }}\frac{{n(n - 1)(n - 2)}}{6}{x^3}\)    A1A1

Note:     Award A1 for the first two terms and A1 for the next two terms.

Note:     Accept \(\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right)\) notation.

Note:     Allow the terms seen in the context of an arithmetic sum.

Note:     Allow unsimplified terms, eg, those including powers of 1 if seen.

[2 marks]

(i)     EITHER

using \({u_3} - {u_2} = {u_4} - {u_3}\)     (M1)

\(\frac{{n(n - 1)}}{2} - n = \frac{{n(n - 1)(n - 2)}}{6} - \frac{{n(n - 1)}}{2}\)    A1

attempting to remove denominators and expanding (or vice versa)     M1

\(3{n^2} - 9n = {n^3} - 6{n^2} + 5n\) (or equivalent, eg, \(6{n^2} - 12n = {n^3} - 3{n^2} + 2n\))     A1

OR

using \({u_2} + {u_4} = 2{u_3}\)     (M1)

\(n + \frac{{n(n - 1)(n - 2)}}{6} = n(n - 1)\)    (A1)

attempting to remove denominators and expanding (or vice versa)     M1

\(6n + {n^3} - 3{n^2} + 2n = 6{n^2} - 6n\) (or equivalent)     (A1)

THEN

\({n^3} - 9{n^2} + 14n = 0\)    AG

(ii)     \(n(n - 2)(n - 7) = 0\) or \((n - 2)(n - 7) = 0\)     (A1)

\(n = 7\) only (as \(n \geqslant 3\))     A1

[6 marks]

This was another question that was very well answered by most candidates.

This was another question that was very well answered by most candidates. Some struggled in part (b) by attempting to find an expression for \(d\), a common difference, then substituting this in to further equations, where algebra tended to falter. The most fruitful technique was to apply \({u_3} - {u_2} = {u_4} - {u_3}\). Good presentation often helped candidates reach the final result. Correct factorisation was more often seen than not in the final section, though a small number thought it judicious to guess the correct answer(s) here.

Show that the probability that Chloe wins the game is \(\frac{3}{8}\).

Determine the mean of X.

Determine the variance of X.

METHOD 1

number of possible “deals” \( = 4! = 24\)     A1

consider ways of achieving “no matches” (Chloe winning):

Selena could deal B, C, D (ie, 3 possibilities)

as her first card     R1

for each of these matches, there are only 3 possible combinations for the remaining 3 cards     R1

so no. ways achieving no matches \( = 3 \times 3 = 9\)     M1A1

so probability Chloe wins \( = \frac{9}{{23}} = \frac{3}{8}\)     A1AG

METHOD 2

number of possible “deals” \( = 4! = 24\)     A1

consider ways of achieving a match (Selena winning)

Selena card A can match with Chloe card A, giving 6 possibilities for this happening     R1

if Selena deals B as her first card, there are only 3 possible combinations for the remaining 3 cards. Similarly for dealing C and dealing D     R1

so no. ways achieving one match is \( = 6 + 3 + 3 + 3 = 15\)     M1A1

so probability Chloe wins \( = 1 - \frac{{15}}{{24}} = \frac{3}{8}\)     A1AG

METHOD 3

systematic attempt to find number of outcomes where Chloe wins (no matches)

(using tree diag. or otherwise)     M1

9 found     A1

each has probability \(\frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times 1\)     M1

\( = \frac{1}{{24}}\)     A1

their 9 multiplied by their \(\frac{1}{{24}}\)     M1A1

\( = \frac{3}{8}\)     AG

[6 marks]

\(X \sim {\text{B}}\left( {50,{\text{ }}\frac{3}{8}} \right)\)     (M1)

\(\mu  = np = 50 \times \frac{3}{8} = \frac{{150}}{8}{\text{ }}\left( { = \frac{{75}}{4}} \right){\text{ }}( = 18.75)\)     (M1)A1

[3 marks]

\({\sigma ^2} = np(1 - p) = 50 \times \frac{3}{8} \times \frac{5}{8} = \frac{{750}}{{64}}{\text{ }}\left( { = \frac{{375}}{{32}}} \right){\text{ }}( = 11.7)\)     (M1)A1

[2 marks]

Consider the complex numbers \(u = 2 + 3{\text{i}}\) and \(v = 3 + 2{\text{i}}\).

(a)     Given that \(\frac{1}{u} + \frac{{1}}{v} = \frac{{10}}{w}\), express w in the form \(a + b{\text{i, }}a,{\text{ }}b \in \mathbb{R}\).

(b)     Find \(w\)* and express it in the form \(r{e^{{\text{i}}\theta }}\).

(a)     METHOD 1

\(\frac{1}{{2 + 3{\text{i}}}} + \frac{1}{{3 + 2{\text{i}}}} = \frac{{2 - 3{\text{i}}}}{{4 + 9}} + \frac{{3 - 2{\text{i}}}}{{9 + 4}}\)     M1A1

\(\frac{{10}}{w} = \frac{{5 - 5{\text{i}}}}{{13}}\)     A1

\(w = \frac{{130}}{{5 - 5{\text{i}}}}\)

\( = \frac{{130 \times 5 \times (1 + {\text{i}})}}{{50}}\)

\(w = 13 + 13{\text{i}}\)     A1

[4 marks]