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HL Paper 1

The first term in an arithmetic sequence is $4$ and the fifth term is $\log_{2} 625$ .

Find the common difference of the sequence, expressing your answer in the form $\log_{2} p$ , where $p \in ℚ$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$u_{5} = 4 + 4 d = \log_{2} 625$ (A1)

$4 d = \log_{2} 625 - 4$

attempt to write an integer (eg $4$ or $1$ ) in terms of $\log_{2}$ M1

$4 d = \log_{2} 625 - \log_{2} 16$

attempt to combine two logs into one M1

$4 d = \log_{2} (\frac{625}{16})$

$d = \frac{1}{4} \log_{2} (\frac{625}{16})$

attempt to use power rule for logs M1

$d = \log_{2} {(\frac{625}{16})}^{\frac{1}{4}}$

$d = \log_{2} (\frac{5}{2})$ A1

[5 marks]

Note: Award method marks in any order.

Examiners report

[N/A]

Consider the integral $\int\limits_1^t {\frac{{ - 1}}{{x + {x^2}}}{\text{ }}} dx$ for $t > 1$ .

Very briefly, explain why the value of this integral must be negative.

[1]

Express the function $f\left( x \right) = \frac{{ - 1}}{{x + {x^2}}}$ in partial fractions.

[6]

Use parts (a) and (b) to show that ${\text{ln}}\left( {1 + t} \right) - {\text{ln}}\,t < {\text{ln}}\,2$ .

[4]

Markscheme

The numerator is negative but the denominator is positive. Thus the integrand is negative and so the value of the integral will be negative. R1AG

[1 mark]

$\frac{{ - 1}}{{x + {x^2}}} = \frac{{ - 1}}{{\left( {1 + x} \right)x}} \equiv \frac{A}{{1 + x}} + \frac{B}{x}$ M1M1A1

$\Rightarrow - 1 \equiv Ax + B(1 + x) \Rightarrow A = 1,\,B = - 1$ M1A1

$\frac{{ - 1}}{{x + {x^2}}} \equiv \frac{1}{{1 + x}} + \frac{{ - 1}}{x}$ A1

[6 marks]

$\int\limits_1^t {\frac{1}{{1 + x}} + \frac{{ - 1}}{x}dx} = \left[ {{\text{ln}}\left( {1 + x} \right) - {\text{ln}}\,x} \right]_1^t = {\text{ln}}\left( {1 + t} \right) - {\text{ln}}\,t - {\text{ln}}\,2$ M1A1A1

Hence ${\text{ln}}\left( {1 + t} \right) - {\text{ln}}\,t - {\text{ln}}\,2 < 0 \Rightarrow {\text{ln}}\left( {1 + t} \right) - {\text{ln}}\,t < {\text{ln}}\,2$ R1AG

[4 marks]

Examiners report

[N/A]

Three planes have equations:

$2x - y + z = 5$

$x + 3y - z = 4$ , where $a{\text{, }}b \in \mathbb{R}$ .

$3x - 5y + az = b$

Find the set of values of $a$ and $b$ such that the three planes have no points of intersection.

Markscheme

attempt to eliminate a variable (or attempt to find det $A$ ) M1

$\left( {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ 1&3&{ - 1} \\ 3&{ - 5}&a \end{array}\,} \right|\begin{array}{*{20}{c}} 5 \\ 4 \\ b \end{array}} \right) \to \left( {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ 0&7&{ - 3} \\ 0&{ - 14}&{a + 3} \end{array}\,} \right|\begin{array}{*{20}{c}} 5 \\ 3 \\ {b - 12} \end{array}} \right)$ (or det $A = 14\left( {a - 3} \right)$ )

(or two correct equations in two variables) A1

$\to \left( {\left. {\begin{array}{*{20}{c}} 2&{ - 1}&1 \\ 0&7&{ - 3} \\ 0&{ 0}&{a - 3} \end{array}\,} \right|\begin{array}{*{20}{c}} 5 \\ 3 \\ {b - 6} \end{array}} \right)$ (or solving det $A = 0$ )

(or attempting to reduce to one variable, e.g. $\left( {a - 3} \right)z = b - 6$ ) M1

$a = 3{\text{, }}b \ne 6$ A1A1

[5 marks]

Examiners report

[N/A]

Consider the equation ${(z - 1)}^{3} = i, z \in ℂ$ . The roots of this equation are $ω_{1}$ , $ω_{2}$ and $ω_{3}$ , where $Im (ω_{2}) > 0$ and $Im (ω_{3}) < 0$ .

The roots $ω_{1}$ , $ω_{2}$ and $ω_{3}$ are represented by the points $A$ , $B$ and $C$ respectively on an Argand diagram.

Consider the equation ${(z - 1)}^{3} = i z^{3}, z \in ℂ$ .

Verify that $ω_{1} = 1 + e^{i \frac{π}{6}}$ is a root of this equation.

[2]

a.i.

Find $ω_{2}$ and $ω_{3}$ , expressing these in the form $a + e^{i θ}$ , where $a \in ℝ$ and $θ > 0$ .

[4]

a.ii.

Plot the points $A$ , $B$ and $C$ on an Argand diagram.

[4]

Find $AC$ .

[3]

By using de Moivre’s theorem, show that $α = \frac{1}{1 - e^{i \frac{π}{6}}}$ is a root of this equation.

[3]

Determine the value of $Re (α)$ .

[6]

Markscheme

${(1 + e^{i \frac{π}{6}} - 1)}^{3}$

$= {(e^{i \frac{π}{6}})}^{3}$ A1

$= e^{i \frac{π}{2}}$ A1

$= \cos \frac{π}{2} + i \sin \frac{π}{2}$

$= i$ AG

Note: Candidates who solve the equation correctly can be awarded the above two marks. The working for part (i) may be seen in part (ii).

[2 marks]

a.i.

${(z - 1)}^{3} = e^{i (\frac{π}{2} + 2 πk)}$ (M1)

$z - 1 = e^{i (\frac{π}{6} + \frac{4 πk}{6})}$ (M1)

$(k = 1) \Rightarrow ω_{2} = 1 + e^{i \frac{5 π}{6}}$ A1

$(k = 2) \Rightarrow ω_{3} = 1 + e^{i \frac{9 π}{6}}$ A1

[4 marks]

a.ii.

EITHER

attempt to express $e^{i \frac{π}{6}}$ , $e^{i \frac{5 π}{6}}$ , $e^{i \frac{9 π}{6}}$ in Cartesian form and translate 1 unit in the positive direction of the real axis (M1)

attempt to express $w_{1}$ , $w_{2}$ and $w_{3}$ in Cartesian form (M1)

THEN

Note: To award A marks, it is not necessary to see $A$ , $B$ or $C$ , the $w_{1}$ , or the solid lines

A1A1A1

[4 marks]

valid attempt to find $ω_{1} - ω_{3} (or ω_{3} - ω_{1})$ M1

$ω_{1} - ω_{3} = (1 + \frac{\sqrt{3}}{2} + \frac{1}{2} i) - (1 - i) = \frac{\sqrt{3}}{2} + \frac{3}{2} i$ OR $\cos \frac{π}{6} + i \sin \frac{π}{6} + i \sin \frac{π}{2}$

valid attempt to find $|\frac{\sqrt{3}}{2} + \frac{3}{2} i|$ M1

$= \sqrt{\frac{3}{4} + \frac{9}{4}}$

$AC = \sqrt{3}$ A1

[3 marks]

METHOD 1

${(z - 1)}^{3} = i z^{3} \Rightarrow {(\frac{z - 1}{z})}^{3} = i$ M1

${(\frac{z - 1}{z})}^{3} = e^{i \frac{π}{2}}$ A1

$\frac{α - 1}{α} = e^{i \frac{π}{6}}$ A1

Note: This step to change from $z$ to $α$ may occur at any point in MS.

$α - 1 = α e^{i \frac{π}{6}}$

$α - α e^{i \frac{π}{6}} = 1$

$α (1 - e^{i \frac{π}{6}}) = 1$

$α = \frac{1}{1 - e^{i \frac{π}{6}}}$ AG

METHOD 2

${(z - 1)}^{3} = i z^{3} \Rightarrow {(\frac{z - 1}{z})}^{3} = i$ M1

${(1 - \frac{1}{z})}^{3} = e^{i \frac{π}{2}}$ A1

$1 - \frac{1}{z} = e^{i \frac{π}{6}}$ A1

Note: This step to change from $z$ to $α$ may occur at any point in MS.

$1 - e^{i \frac{π}{6}} = \frac{1}{α}$

$α = \frac{1}{1 - e^{i \frac{π}{6}}}$ AG

METHOD 3

LHS $= {(z - 1)}^{3} = {(\frac{1}{1 - e^{i \frac{π}{6}}} - 1)}^{3}$

$= {(\frac{e^{i \frac{π}{6}}}{1 - e^{i \frac{π}{6}}})}^{3}$

$= \frac{i}{{(1 - e^{i \frac{π}{6}})}^{3}} (= \frac{i}{\frac{5}{2} - \frac{3 \sqrt{3}}{2} + i (\frac{3 \sqrt{3}}{2} - \frac{5}{2})})$ M1A1

Note: Award M1 for applying de Moivre’s theorem (may be seen in modulus- argument form.)

RHS $= i z^{3} = i {(\frac{1}{1 - e^{i \frac{π}{6}}})}^{3}$

$= \frac{i}{{(1 - e^{i \frac{π}{6}})}^{3}}$ A1

${(z - 1)}^{3} = i z^{3}$ AG

METHOD 4

${(z - 1)}^{3} = i z^{3}$

$z^{3} - 3 z^{2} + 3 z - 1 = i z^{3}$

$(1 - i) z^{3} - 3 z^{2} + 3 z - 1 = 0$ (M1)

$(1 - i) {(\frac{1}{1 - e^{i \frac{π}{6}}})}^{3} - 3 {(\frac{1}{1 - e^{i \frac{π}{6}}})}^{2} + 3 (\frac{1}{1 - e^{i \frac{π}{6}}}) - 1$

$= (1 - i) - 3 (1 - e^{i \frac{π}{6}}) + 3 {(1 - e^{i \frac{π}{6}})}^{2} - {(1 - e^{i \frac{π}{6}})}^{3}$ (A1)

$= (1 - i) - 3 (1 - e^{i \frac{π}{6}}) + 3 (1 - 2 e^{i \frac{π}{6}} + e^{i \frac{π}{3}}) - (1 - 3 e^{i \frac{π}{6}} + 3 e^{i \frac{π}{3}} - e^{i \frac{π}{2}})$ A1

$= 0$ AG

Note: If the candidate does not interpret their conclusion, award (M1)(A1)A0 as appropriate.

[3 marks]

METHOD 1

$\frac{1}{1 - e^{i \frac{π}{6}}} = \frac{1}{1 - (\cos \frac{π}{6} + i \sin \frac{π}{6})}$ M1

$= \frac{2}{2 - \sqrt{3} - i}$ A1

attempt to use conjugate to rationalise M1

$= \frac{4 - 2 \sqrt{3} + 2 i}{{(2 - \sqrt{3})}^{2} + 1}$ A1

$= \frac{4 - 2 \sqrt{3} + 2 i}{8 - 4 \sqrt{3}}$ A1

$= \frac{1}{2} + \frac{1}{4 - 2 \sqrt{3}} i$

$\Rightarrow Re (α) = \frac{1}{2}$ A1

Note: Their final imaginary part does not have to be correct in order for the final three A marks to be awarded

METHOD 2

$\frac{1}{1 - e^{i \frac{π}{6}}} = \frac{1}{1 - (\cos \frac{π}{6} + i \sin \frac{π}{6})}$ M1

attempt to use conjugate to rationalise M1

$= \frac{1}{(1 - \cos \frac{π}{6}) - i \sin \frac{π}{6}} \times \frac{(1 - \cos \frac{π}{6}) + i \sin \frac{π}{6}}{(1 - \cos \frac{π}{6}) + i \sin \frac{π}{6}}$ A1

$= \frac{(1 - \cos \frac{π}{6}) + i \sin \frac{π}{6}}{{(1 - \cos \frac{π}{6})}^{2} + \sin^{2} \frac{π}{6}}$ A1

$= \frac{(1 - \cos \frac{π}{6}) + i \sin \frac{π}{6}}{1 - 2 \cos \frac{π}{6} + \cos^{2} \frac{π}{6} + \sin^{2} \frac{π}{6}}$

$= \frac{(1 - \cos \frac{π}{6}) + i \sin \frac{π}{6}}{2 - 2 \cos \frac{π}{6}}$ A1

$= \frac{1}{2} + \frac{i \sin \frac{π}{6}}{2 - 2 \cos \frac{π}{6}}$

$\Rightarrow Re (α) = \frac{1}{2}$ A1

Note: Their final imaginary part does not have to be correct in order for the final three A marks to be awarded

METHOD 3

attempt to multiply through by $- \frac{e^{- i \frac{π}{12}}}{e^{- i \frac{π}{12}}}$ M1

$\frac{1}{1 - e^{i \frac{π}{6}}} = - \frac{e^{- i \frac{π}{12}}}{e^{i \frac{π}{12}} - e^{- i \frac{π}{12}}}$ A1

attempting to re-write in r-cis form M1

$= - \frac{\cos (- \frac{π}{12}) + i \sin (- \frac{π}{12})}{\cos \frac{π}{12} + i \sin \frac{π}{12} - (\cos (- \frac{π}{12}) + i \sin (- \frac{π}{12}))}$ A1

$= - \frac{\cos \frac{π}{12} - i \sin \frac{π}{12}}{2 i \sin \frac{π}{12}}$ A1

$= \frac{1}{2} - \frac{1}{2 i} \cot \frac{π}{12} (= \frac{1}{2} + \frac{1}{2} i \cot \frac{π}{12})$

$\Rightarrow Re (α) = \frac{1}{2}$ A1

METHOD 4

attempt to multiply through by $\frac{1 - e^{- i \frac{π}{6}}}{1 - e^{- i \frac{π}{6}}}$ M1

$\frac{1}{1 - e^{i \frac{π}{6}}} = \frac{1 - e^{- i \frac{π}{6}}}{1 - e^{- i \frac{π}{6}} - e^{i \frac{π}{6}} + 1}$ A1

attempting to re-write in r-cis form M1

$= \frac{1 - \cos \frac{π}{6} - i \sin \frac{π}{6}}{2 - 2 \cos \frac{π}{6}}$ A1

attempt to re-write in Cartesian form M1

$= \frac{1 - \frac{\sqrt{3}}{2} - \frac{1}{2} i}{2 - \sqrt{3}} (= \frac{\frac{2 - \sqrt{3}}{2}}{2 - \sqrt{3}} + i \frac{\frac{1}{2}}{2 - \sqrt{3}})$

$\Rightarrow Re (α) = \frac{1}{2}$ A1

Note: Their final imaginary part does not have to be correct in order for the final A mark to be awarded

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

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

These four points form the vertices of a quadrilateral, Q.

Express w² and w³ in modulus-argument form.

[3]

a.i.

Sketch on an Argand diagram the points represented by w⁰ , w¹ , w² and w³.

[2]

a.ii.

Show that the area of the quadrilateral Q is $\frac{{21\sqrt 3 }}{2}$ .

[3]

Let $z = 2\left( {{\text{cos}}\frac{\pi }{n} + {\text{i}}\,{\text{sin}}\frac{\pi }{n}} \right),\,\,n \in {\mathbb{Z}^ + }$ . The points represented on an Argand diagram by ${z^0},\,\,{z^1},\,\,{z^2},\, \ldots \,,\,\,{z^n}$ form the vertices of a polygon ${P_n}$ .

Show that the area of the polygon ${P_n}$ can be expressed in the form $a\left( {{b^n} - 1} \right){\text{sin}}\frac{\pi }{n}$ , where $a,\,\,b\, \in \mathbb{R}$ .

[6]

Markscheme

${w^2} = 4\text{cis}\left( {\frac{{2\pi }}{3}} \right){\text{;}}\,\,{w^3} = 8{\text{cis}}\left( \pi \right)$ (M1)A1A1

Note: Accept Euler form.

Note: M1 can be awarded for either both correct moduli or both correct arguments.

Note: Allow multiplication of correct Cartesian form for M1, final answers must be in modulus-argument form.

[3 marks]

a.i.

A1A1

[2 marks]

a.ii.

use of area = $\frac{1}{2}ab\,\,{\text{sin}}\,C$ M1

$\frac{1}{2} \times 1 \times 2 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 2 \times 4 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 4 \times 8 \times {\text{sin}}\frac{\pi }{3}$ A1A1

Note: Award A1 for $C = \frac{\pi }{3}$ , A1 for correct moduli.

$= \frac{{21\sqrt 3 }}{2}$ AG

Note: Other methods of splitting the area may receive full marks.

[3 marks]

$\frac{1}{2} \times {2^0} \times {2^1} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^1} \times {2^2} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^2} \times {2^3} \times {\text{sin}}\frac{\pi }{n} + \, \ldots \, + \frac{1}{2} \times {2^{n - 1}} \times {2^n} \times {\text{sin}}\frac{\pi }{n}$ M1A1

Note: Award M1 for powers of 2, A1 for any correct expression including both the first and last term.

$= {\text{sin}}\frac{\pi }{n} \times \left( {{2^0} + {2^2} + {2^4} + \, \ldots \, + {2^{n - 2}}} \right)$

identifying a geometric series with common ratio 2²(= 4) (M1)A1

$= \frac{{1 - {2^{2n}}}}{{1 - 4}} \times {\text{sin}}\frac{\pi }{n}$ M1

Note: Award M1 for use of formula for sum of geometric series.

$= \frac{1}{3}\left( {{4^n} - 1} \right){\text{sin}}\frac{\pi }{n}$ A1

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider the series $\ln x + p \ln x + \frac{1}{3} \ln x + \dots$ , where $x \in ℝ, x > 1$ and $p \in ℝ, p \neq 0$ .

Consider the case where the series is geometric.

Now consider the case where the series is arithmetic with common difference $d$ .

Show that $p = \pm \frac{1}{\sqrt{3}}$ .

[2]

a.i.

Hence or otherwise, show that the series is convergent.

[1]

a.ii.

Given that $p > 0$ and $S_{\infty} = 3 + \sqrt{3}$ , find the value of $x$ .

[3]

a.iii.

Show that $p = \frac{2}{3}$ .

[3]

b.i.

Write down $d$ in the form $k \ln x$ , where $k \in ℚ$ .

[1]

b.ii.

The sum of the first $n$ terms of the series is $\ln (\frac{1}{x^{3}})$ .

Find the value of $n$ .

[8]

b.iii.

Markscheme

EITHER

attempt to use a ratio from consecutive terms M1

$\frac{p \ln x}{\ln x} = \frac{\frac{1}{3} \ln x}{p \ln x}$ OR $\frac{1}{3} \ln x = (\ln x) r^{2}$ OR $p \ln x = \ln x (\frac{1}{3 p})$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in geometric sequence

Award M1 for $\frac{p}{1} = \frac{\frac{1}{3}}{p}$ .

$r = p$ and $r^{2} = \frac{1}{3}$ M1

THEN

$p^{2} = \frac{1}{3}$ OR $r = \pm \frac{1}{\sqrt{3}}$ A1

$p = \pm \frac{1}{\sqrt{3}}$ AG

Note: Award M0A0 for $r^{2} = \frac{1}{3}$ or $p^{2} = \frac{1}{3}$ with no other working seen.

[2 marks]

a.i.

EITHER

since, $|p| = \frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}} < 1$ R1

since, $|p| = \frac{1}{\sqrt{3}}$ and $- 1 < p < 1$ R1

THEN

$\Rightarrow$ the geometric series converges. AG

Note: Accept $r$ instead of $p$ .
Award R0 if both values of $p$ not considered.

[1 mark]

a.ii.

$\frac{\ln x}{1 - \frac{1}{\sqrt{3}}} (= 3 + \sqrt{3})$ (A1)

$\ln x = 3 - \frac{3}{\sqrt{3}} + \sqrt{3} - \frac{\sqrt{3}}{\sqrt{3}}$ OR $\ln x = 3 - \sqrt{3} + \sqrt{3} - 1 (\Rightarrow \ln x = 2)$ A1

$x = e^{2}$ A1

[3 marks]

a.iii.

METHOD 1

attempt to find a difference from consecutive terms or from $u_{2}$ M1

correct equation A1

$p \ln x - \ln x = \frac{1}{3} \ln x - p \ln x$ OR $\frac{1}{3} \ln x = \ln x + 2 (p \ln x - \ln x)$

Note: Candidates may use $\ln x^{1} + \ln x^{p} + \ln x^{\frac{1}{3}} + \dots$ and consider the powers of $x$ in arithmetic sequence.

Award M1A1 for $p - 1 = \frac{1}{3} - p$

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 2

attempt to use arithmetic mean $u_{2} = \frac{u_{1} + u_{3}}{2}$ M1

$p \ln x = \frac{\ln x + \frac{1}{3} \ln x}{2}$ A1

$2 p \ln x = \frac{4}{3} \ln x (\Rightarrow 2 p = \frac{4}{3})$ A1

$p = \frac{2}{3}$ AG

METHOD 3

attempt to find difference using $u_{3}$ M1

$\frac{1}{3} \ln x = \ln x + 2 d (\Rightarrow d = - \frac{1}{3} \ln x)$

$u_{2} = \ln x + \frac{1}{2} (\frac{1}{3} \ln x - \ln x)$ OR $p \ln x - \ln x = - \frac{1}{3} \ln x$ A1

$p \ln x = \frac{2}{3} \ln x$ A1

$p = \frac{2}{3}$ AG

[3 marks]

b.i.

$d = - \frac{1}{3} \ln x$ A1

[1 mark]

b.ii.

METHOD 1

$S_{n} = \frac{n}{2} ⌊2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)⌋$

attempt to substitute into $S_{n}$ and equate to $\ln (\frac{1}{x^{3}})$ (M1)

$\frac{n}{2} ⌊2 \ln x + (n - 1) \times (- \frac{1}{3} \ln x)⌋ = \ln (\frac{1}{x^{3}})$

$\ln (\frac{1}{x^{3}}) = - \ln x^{3} (= \ln x^{- 3})$ (A1)

$= - 3 \ln x$ (A1)

correct working with $S_{n}$ (seen anywhere) (A1)

$\frac{n}{2} ⌊2 \ln x - \frac{n}{3} \ln x + \frac{1}{3} \ln x⌋$ OR $n \ln x - \frac{n (n - 1)}{6} \ln x$ OR $\frac{n}{2} (\ln x + (\frac{4 - n}{3}) \ln x)$

correct equation without $\ln x$ A1

$\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ OR $n - \frac{n (n - 1)}{6} = - 3$ or equivalent

Note: Award as above if the series $1 + p + \frac{1}{3} + \dots$ is considered leading to $\frac{n}{2} (\frac{7}{3} - \frac{n}{3}) = - 3$ .

attempt to form a quadratic $= 0$ (M1)

$n^{2} - 7 n - 18 = 0$

attempt to solve their quadratic (M1)

$(n - 9) (n + 2) = 0$

$n = 9$ A1

METHOD 2

$\ln (\frac{1}{x^{3}}) = - \ln x^{3} (= \ln x^{- 3})$ (A1)

$= - 3 \ln x$ (A1)

listing the first $7$ terms of the sequence (A1)

$\ln x + \frac{2}{3} \ln x + \frac{1}{3} \ln x + 0 - \frac{1}{3} \ln x - \frac{2}{3} \ln x - \ln x + \dots$

recognizing first $7$ terms sum to $0$ M1

$8$ ^th term is $- \frac{4}{3} \ln x$ (A1)

$9$ ^th term is $- \frac{5}{3} \ln x$ (A1)

sum of $8$ ^th and $9$ ^th term $= - 3 \ln x$ (A1)

$n = 9$ A1

[8 marks]

b.iii.

Examiners report

Part (a)(i) was well done with few candidates incorrectly using the value of p to verify rather than to 'show' the given result. In part (a)(ii) most did not consider both values of r and some did know the condition for convergence of a geometric series. Part (a)(iii) was generally well done but some had difficulty in simplifying the surd. Part (b) (i) and (ii) was generally well done. Although many completely correct answers to part b (iii) were noted, weaker candidates often made errors in properties of logarithms or algebraic manipulation leading to an incorrect quadratic equation.

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

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 $a,{\text{ }}b \in \mathbb{Z}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

Consider $f\left( x \right) = \frac{{2x - 4}}{{{x^2} - 1}}{\text{, }} - 1 < x < 1$ .

For the graph of $y = f\left( x \right)$ ,

Find $f'\left( x \right)$ .

[2]

a.i.

Show that, if $f'\left( x \right) = 0$ , then $x = 2 - \sqrt 3$ .

[3]

a.ii.

find the coordinates of the $y$ -intercept.

[1]

b.i.

show that there are no $x$ -intercepts.

[2]

b.ii.

sketch the graph, showing clearly any asymptotic behaviour.

[2]

b.iii.

Show that $\frac{3}{{x + 1}} - \frac{1}{{x - 1}} = \frac{{2x - 4}}{{{x^2} - 1}}$ .

[2]

The area enclosed by the graph of $y = f\left( x \right)$ and the line $y = 4$ can be expressed as ${\text{ln}}\,v$ . Find the value of $v$ .

[7]

Markscheme

attempt to use quotient rule (or equivalent) (M1)

$f'\left( x \right) = \frac{{\left( {{x^2} - 1} \right)\left( 2 \right) - \left( {2x - 4} \right)\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}$ A1

$= \frac{{ - 2{x^2} + 8x - 2}}{{{{\left( {{x^2} - 1} \right)}^2}}}$

[2 marks]

a.i.

$f'\left( x \right) = 0$

simplifying numerator (may be seen in part (i)) (M1)

$\Rightarrow {x^2} - 4x + 1 = 0$ or equivalent quadratic equation A1

EITHER

use of quadratic formula

$\Rightarrow x = \frac{{4 \pm \sqrt {12} }}{2}$ A1

use of completing the square

${\left( {x - 2} \right)^2} = 3$ A1

THEN

$x = 2 - \sqrt 3$ (since $2 + \sqrt 3$ is outside the domain) AG

Note: Do not condone verification that $x = 2 - \sqrt 3 \Rightarrow f'\left( x \right) = 0$ .

Do not award the final A1 as follow through from part (i).

[3 marks]

a.ii.

(0, 4) A1

[1 mark]

b.i.

$2x - 4 = 0 \Rightarrow x = 2$ A1

outside the domain R1

[2 marks]

b.ii.

A1A1

award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching $x = \pm 1$ asymptotically

[2 marks]

b.iii.

valid attempt to combine fractions (using common denominator) M1

$\frac{{3\left( {x - 1} \right) - \left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}$ A1

$= \frac{{3x - 3 - x - 1}}{{{x^2} - 1}}$

$= \frac{{2x - 4}}{{{x^2} - 1}}$ AG

[2 marks]

$f\left( x \right) = 4 \Rightarrow 2x - 4 = 4{x^2} - 4$ M1

( $x = 0$ or) $x = \frac{1}{2}$ A1

area under the curve is $\int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ M1

$= \int_0^{\frac{1}{2}} {\frac{3}{{x + 1}} - \frac{1}{{x - 1}}{\text{d}}x}$

Note: Ignore absence of, or incorrect limits up to this point.

$= \left[ {3\,{\text{ln}}\,\left| {x + 1} \right| - {\text{ln}}\,\left| {x - 1} \right|} \right]_0^{\frac{1}{2}}$ A1

$= 3\,{\text{ln}}\frac{3}{2} - {\text{ln}}\frac{1}{2}\left( { - 0} \right)$

$= {\text{ln}}\frac{{27}}{4}$ A1

area is $2 - \int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ or $\int_0^{\frac{1}{2}} {4\,{\text{d}}x} - \int_0^{\frac{1}{2}} {f\left( x \right){\text{d}}x}$ M1

$= 2 - {\text{ln}}\frac{{27}}{4}$

$= {\text{ln}}\frac{{4\,{{\text{e}}^2}}}{{27}}$ A1

$\left( { \Rightarrow v = \frac{{4\,{{\text{e}}^2}}}{{27}}} \right)$

[7 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

A farmer has six sheep pens, arranged in a grid with three rows and two columns as shown in the following diagram.

Five sheep called Amber, Brownie, Curly, Daisy and Eden are to be placed in the pens. Each pen is large enough to hold all of the sheep. Amber and Brownie are known to fight.

Find the number of ways of placing the sheep in the pens in each of the following cases:

Each pen is large enough to contain five sheep. Amber and Brownie must not be placed in the same pen.

[4]

Each pen may only contain one sheep. Amber and Brownie must not be placed in pens which share a boundary.

[4]

Markscheme

METHOD 1

B has one less pen to select (M1)

EITHER

A and B can be placed in $6 \times 5$ ways (A1)

C, D, E have $6$ choices each (A1)

A (or B), C, D, E have $6$ choices each (A1)

B (or A) has only $5$ choices (A1)

THEN

$5 \times 6^{4} (= 6480)$ A1

METHOD 2

total number of ways $= 6^{5}$ (A1)

number of ways with Amber and Brownie together $= 6^{4}$ (A1)

attempt to subtract (may be seen in words) (M1)

$6^{5} - 6^{4}$

$= 5 \times 6^{4} (= 6480)$ A1

[4 marks]

METHOD 1

total number of ways $= 6! (= 720)$ (A1)

number of ways with Amber and Brownie sharing a boundary

$= 2 \times 7 \times 4! (= 336)$ (A1)

attempt to subtract (may be seen in words) (M1)

$720 - 336 = 384$ A1

METHOD 2

case 1: number of ways of placing A in corner pen

$3 \times 4 \times 3 \times 2 \times 1$

Four corners total no of ways is $4 \times (3 \times 4 \times 3 \times 2 \times 1) = 12 \times 4! (= 288)$ (A1)

case 2: number of ways of placing A in the middle pen

$2 \times 4 \times 3 \times 2 \times 1$

two middle pens so $2 \times (2 \times 4 \times 3 \times 2 \times 1) = 4 \times 4! (= 96)$ (A1)

attempt to add (may be seen in words) (M1)

total no of ways $= 288 + 96$

$= 16 \times 4! (= 384)$ A1

[4 marks]

Examiners report

[N/A]

Two distinct lines, ${l_1}$ and ${l_2}$ , intersect at a point ${\text{P}}$ . In addition to ${\text{P}}$ , four distinct points are marked out on ${l_1}$ and three distinct points on ${l_2}$ . A mathematician decides to join some of these eight points to form polygons.

The line ${l_1}$ has vector equation r₁ $= \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 1 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 1 \end{array}} \right)$ , $\lambda \in \mathbb{R}$ and the line ${l_2}$ has vector equation r₂ $= \left( {\begin{array}{*{20}{c}} { - 1} \\ 0 \\ 2 \end{array}} \right) + \mu \left( {\begin{array}{*{20}{c}} 5 \\ 6 \\ 2 \end{array}} \right)$ , $\mu \in \mathbb{R}$ .

The point ${\text{P}}$ has coordinates (4, 6, 4).

The point ${\text{A}}$ has coordinates (3, 4, 3) and lies on ${l_1}$ .

The point ${\text{B}}$ has coordinates (−1, 0, 2) and lies on ${l_2}$ .

Find how many sets of four points can be selected which can form the vertices of a quadrilateral.

[2]

a.i.

Find how many sets of three points can be selected which can form the vertices of a triangle.

[4]

a.ii.

Verify that ${\text{P}}$ is the point of intersection of the two lines.

[3]

Write down the value of $\lambda$ corresponding to the point ${\text{A}}$ .

[1]

Write down $\overrightarrow {{\text{PA}}}$ and $\overrightarrow {{\text{PB}}}$ .

[2]

Let ${\text{C}}$ be the point on ${l_1}$ with coordinates (1, 0, 1) and ${\text{D}}$ be the point on ${l_2}$ with parameter $\mu = - 2$ .

Find the area of the quadrilateral ${\text{CDBA}}$ .

[8]

Markscheme

appreciation that two points distinct from ${\text{P}}$ need to be chosen from each line M1

${}^4{C_2} \times {}^3{C_2}$

=18 A1

[2 marks]

a.i.

EITHER

consider cases for triangles including ${\text{P}}$ or triangles not including ${\text{P}}$ M1

$3 \times 4 + 4 \times {}^3{C_2} + 3 \times {}^4{C_2}$ (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

consider total number of ways to select 3 points and subtract those with 3 points on the same line M1

${}^8{C_3} - {}^5{C_3} - {}^4{C_3}$ (A1)(A1)

Note: Award A1 for 1st term, A1 for 2nd & 3rd term.

56−10−4

THEN

= 42 A1

[4 marks]

a.ii.

METHOD 1

substitution of (4, 6, 4) into both equations (M1)

$\lambda = 3$ and $\mu = 1$ A1A1

(4, 6, 4) AG

METHOD 2

attempting to solve two of the three parametric equations M1

$\lambda = 3$ and $\mu = 1$ A1

check both of the above give (4, 6, 4) M1AG

Note: If they have shown the curve intersects for all three coordinates they only need to check (4,6,4) with one of " $\lambda$ " or " $\mu$ ".

[3 marks]

$\lambda = 2$ A1

[1 mark]

$\overrightarrow {{\text{PA}}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ { - 2} \\ { - 1} \end{array}} \right)$ , $\overrightarrow {{\text{PB}}} = \left( {\begin{array}{*{20}{c}} { - 5} \\ { - 6} \\ { - 2} \end{array}} \right)$ A1A1

Note: Award A1A0 if both are given as coordinates.

[2 marks]

METHOD 1

area triangle ${\text{ABP}} = \frac{1}{2}\left| {\overrightarrow {{\text{PB}}} \times \overrightarrow {{\text{PA}}} } \right|$ M1

$\left( { = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} { - 5} \\ { - 6} \\ { - 2} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} { - 1} \\ { - 2} \\ { - 1} \end{array}} \right)} \right|} \right) = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} 2 \\ { - 3} \\ 4 \end{array}} \right)} \right|$ A1

$= \frac{{\sqrt {29} }}{2}$ A1

EITHER

$\overrightarrow {{\text{PC}}} = 3\overrightarrow {\,{\text{PA}}}$ , $\overrightarrow {{\text{PD}}} = 3\overrightarrow {\,{\text{PB}}}$ (M1)

area triangle ${\text{PCD}} = 9 \times$ area triangle ${\text{ABP}}$ (M1)A1

$= \frac{{9\sqrt {29} }}{2}$ A1

${\text{D}}$ has coordinates (−11, −12, −2) A1

area triangle ${\text{PCD}} = \frac{1}{2}\left| {\overrightarrow {{\text{PD}}} \times \overrightarrow {{\text{PC}}} } \right| = \frac{1}{2}\left| {\left( {\begin{array}{*{20}{c}} { - 15} \\ { - 18} \\ { - 6} \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 6} \\ { - 3} \end{array}} \right)} \right|$ M1A1

Note: A1 is for the correct vectors in the correct formula.

$= \frac{{9\sqrt {29} }}{2}$ A1

THEN

area of ${\text{CDBA}} = \frac{{9\sqrt {29} }}{2} - \frac{{\sqrt {29} }}{2}$

$= 4\sqrt {29}$ A1

METHOD 2

${\text{D}}$ has coordinates (−11, −12, −2) A1

area $= \frac{1}{2}\left| {\overrightarrow {{\text{CB}}} \times \overrightarrow {{\text{CA}}} } \right| + \frac{1}{2}\left| {\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{BD}}} } \right|$ M1

Note: Award M1 for use of correct formula on appropriate non-overlapping triangles.

Note: Different triangles or vectors could be used.

$\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} { - 2} \\ 0 \\ 1 \end{array}} \right)$ , $\overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ 2 \end{array}} \right)$ A1

$\overrightarrow {{\text{CB}}} \times \overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} { - 4} \\ 6 \\ { - 8} \end{array}} \right)$ A1

$\overrightarrow {{\text{BC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ 0 \\ { - 1} \end{array}} \right)$ , $\overrightarrow {{\text{BD}}} = \left( {\begin{array}{*{20}{c}} { - 10} \\ { - 12} \\ { - 4} \end{array}} \right)$ A1

$\overrightarrow {{\text{BC}}} \times \overrightarrow {{\text{BD}}} = \left( {\begin{array}{*{20}{c}} { - 12} \\ {18} \\ { - 24} \end{array}} \right)$ A1

Note: Other vectors which might be used are $\overrightarrow {{\text{DA}}} = \left( {\begin{array}{*{20}{c}} {14} \\ {16} \\ {5} \end{array}} \right)$ , $\overrightarrow {{\text{BA}}} = \left( {\begin{array}{*{20}{c}} {4} \\ {4} \\ {1} \end{array}} \right)$ , $\overrightarrow {{\text{DC}}} = \left( {\begin{array}{*{20}{c}} {12} \\ {12} \\ {3} \end{array}} \right)$ .

Note: Previous A1A1A1A1 are all dependent on the first M1.

valid attempt to find a value of $\frac{1}{2}\left| {a \times b} \right|$ M1

Note: M1 independent of triangle chosen.

area $= \frac{1}{2} \times 2 \times \sqrt {29} + \frac{1}{2} \times 6 \times \sqrt {29}$

$= 4\sqrt {29}$ A1

Note: accept $\frac{1}{2} \sqrt {116} + \frac{1}{2}\sqrt {1044}$ or equivalent.

[8 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

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

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

In the following Argand diagram, the points $Z_{1}$ , $O$ and $Z_{2}$ are the vertices of triangle $Z_{1} {OZ}_{2}$ described anticlockwise.

The point $Z_{1}$ represents the complex number $z_{1} = r_{1} e^{i α}$ , where $r_{1} > 0$ . The point $Z_{2}$ represents the complex number $z_{2} = r_{2} e^{i θ}$ , where $r_{2} > 0$ .

Angles $α, θ$ are measured anticlockwise from the positive direction of the real axis such that $0 \leq α, θ < 2 π$ and $0 < α - θ < π$ .

In parts (c), (d) and (e), consider the case where $Z_{1} {OZ}_{2}$ is an equilateral triangle.

Let $z_{1}$ and $z_{2}$ be the distinct roots of the equation $z^{2} + a z + b = 0$ where $z \in ℂ$ and $a, b \in ℝ$ .

Show that $z_{1} {z_{2}}^{*} = r_{1} r_{2} e^{i (α - θ)}$ where ${z_{2}}^{*}$ is the complex conjugate of $z_{2}$ .

[2]

Given that $Re (z_{1} {z_{2}}^{*}) = 0$ , show that $Z_{1} {OZ}_{2}$ is a right-angled triangle.

[2]

Express $z_{1}$ in terms of $z_{2}$ .

[2]

c.i.

Hence show that ${z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2}$ .

[4]

c.ii.

Use the result from part (c)(ii) to show that $a^{2} - 3 b = 0$ .

[5]

Consider the equation $z^{2} + a z + 12 = 0$ , where $z \in ℂ$ and $a \in ℝ$ .

Given that $0 < α - θ < π$ , deduce that only one equilateral triangle $Z_{1} {OZ}_{2}$ can be formed from the point $O$ and the roots of this equation.

[3]

Markscheme

${z_{2}}^{*} = r_{2} e^{-i θ}$ (A1)

$z_{1} {z_{2}}^{*} = r_{1} e^{i α} r_{2} e^{-i θ}$ A1

$z_{1} {z_{2}}^{*} = r_{1} r_{2} e^{i (α - θ)}$ AG

Note: Accept working in modulus-argument form

[2 marks]

$Re (z_{1} {z_{2}}^{*}) = r_{1} r_{2} \cos (α - θ) (= 0)$ A1

$α - θ = arcos 0 (r_{1}, r_{2} > 0)$

$α - θ = \frac{π}{2}$ (as $0 < α - θ < π$ ) A1

so $Z_{1} {OZ}_{2}$ is a right-angled triangle AG

[2 marks]

EITHER

$\frac{z_{1}}{z_{2}} (= \frac{r_{1}}{r_{2}} e^{i (α - θ)}) = e^{i \frac{π}{3}}$ (since $r_{1} = r_{2}$ ) (M1)

$z_{1} = r_{2} e^{i (θ + \frac{π}{3})} (= r_{2} e^{i θ} e^{i \frac{π}{3}})$ (M1)

THEN

$z_{1} = z_{2} e^{i \frac{π}{3}}$ A1

Note: Accept working in either modulus-argument form to obtain $z_{1} = z_{2} (\cos \frac{π}{3} + i \sin \frac{π}{3})$ or in Cartesian form to obtain $z_{1} = z_{2} (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ .

[2 marks]

c.i.

substitutes $z_{1} = z_{2} e^{i \frac{π}{3}}$ into ${z_{1}}^{2} + {z_{2}}^{2}$ M1

${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{2 π}{3}} + {z_{2}}^{2} (= {z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1))$ A1

EITHER

$e^{i \frac{2 π}{3}} + 1 = e^{i \frac{π}{3}}$ A1

${z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1) = {z_{2}}^{2} (- \frac{1}{2} + \frac{\sqrt{3}}{2} i + 1)$

$= {z_{2}}^{2} (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ A1

THEN

${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{π}{3}}$

$= z_{2} (z_{2} e^{i \frac{π}{3}})$ and $z_{2} e^{i \frac{π}{3}} = z_{1}$ A1

so ${z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2}$ AG

Note: For candidates who work on the LHS and RHS separately to show equality, award M1A1 for ${z_{1}}^{2} + {z_{2}}^{2} = {z_{2}}^{2} e^{i \frac{2 π}{3}} + {z_{2}}^{2} (= {z_{2}}^{2} (e^{i \frac{2 π}{3}} + 1))$ , A1 for $z_{1} z_{2} = {z_{2}}^{2} e^{i \frac{π}{3}}$ and A1 for $e^{i \frac{2 π}{3}} + 1 = e^{i \frac{π}{3}}$ . Accept working in either modulus-argument form or in Cartesian form.

[4 marks]

c.ii.

METHOD 1

$z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ (A1)

$a^{2} = {z_{1}}^{2} + {z_{2}}^{2} + 2 z_{1} z_{2}$ A1

$a^{2} = 2 z_{1} z_{2} + z_{1} z_{2} (= 3 z_{1} z_{2})$ A1

substitutes $b = z_{1} z_{2}$ into their expression M1

$a^{2} = 2 b + b$ OR $a^{2} = 3 b$ A1

Note: If $z_{1} + z_{2} = - a$ is not clearly recognized, award maximum (A0)A1A1M1A0.

so $a^{2} - 3 b = 0$ AG

METHOD 2

$z_{1} + z_{2} = - a$ and $z_{1} z_{2} = b$ (A1)

${(z_{1} + z_{2})}^{2} = {z_{1}}^{2} + {z_{2}}^{2} + 2 z_{1} z_{2}$ A1

${(z_{1} + z_{2})}^{2} = 2 z_{1} z_{2} + z_{1} z_{2} (= 3 z_{1} z_{2})$ A1

substitutes $b = z_{1} z_{2}$ and $z_{1} + z_{2} = - a$ into their expression M1

$a^{2} = 2 b + b$ OR $a^{2} = 3 b$ A1

Note: If $z_{1} + z_{2} = - a$ is not clearly recognized, award maximum (A0)A1A1M1A0.

so $a^{2} - 3 b = 0$ AG

[5 marks]

$a^{2} - 3 \times 12 = 0$

$a = \pm 6 (\Rightarrow z^{2} \pm 6 z + 12 = 0)$ A1

for $a = - 6 :$

$z_{1} = 3 + \sqrt{3} i, z_{2} = 3 - \sqrt{3} i$ and $α - θ = - \frac{5 π}{3}$ which does not satisfy $0 < α - θ < π$ R1

for $a = 6 :$

$z_{1} = - 3 - \sqrt{3} i, z_{2} = - 3 + \sqrt{3} i$ and $α - θ = \frac{π}{3}$ A1

so (for $0 < α - θ < π$ ), only one equilateral triangle can be formed from point $O$ and the two roots of this equation AG

[3 marks]

Examiners report

The vast majority of candidates scored full marks in parts (a) and (b). If they did not, it was normally due to the lack of rigour in setting out of the answer to a "show that" question. Part (c) was, though, more often than not poorly done. Many candidates could not use the given condition (equilateral triangle) to find $z_{1}$ in terms of $z_{2}$ . Part (d) was well answered by a rather high number of candidates.

Only a handful of students made good progress in (e), not even finding the possible values for $a$ .

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

Consider the following system of equations where $a \in \mathbb{R}$ .

$2x + 4y - z = 10$

$x + 2y + az = 5$

$5x + 12y = 2a$ .

Find the value of $a$ for which the system of equations does not have a unique solution.

[2]

Find the solution of the system of equations when $a = 2$ .

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

an attempt at a valid method eg by inspection or row reduction (M1)

$2 \times {R_2} = {R_1} \Rightarrow 2a = - 1$

$\Rightarrow a = - \frac{1}{2}$ A1

[2 marks]

using elimination or row reduction to eliminate one variable (M1)

correct pair of equations in 2 variables, such as

$\left. {\begin{array}{*{20}{c}} {5x + 10y = 25} \\ {5x + 12y = 4} \end{array}} \right\}$ A1

Note: Award A1 for $z$ = 0 and one other equation in two variables.

attempting to solve for these two variables (M1)

$x = 26$ , $y = - 10.5$ , $z = 0$ A1A1

Note: Award A1A0 for only two correct values, and A0A0 for only one.

Note: Award marks in part (b) for equivalent steps seen in part (a).

[5 marks]

Examiners report

[N/A]

A team of four is to be chosen from a group of four boys and four girls.

Find the number of different possible teams that could be chosen.

[3]

Find the number of different possible teams that could be chosen, given that the team must include at least one girl and at least one boy.

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

$\left( {\begin{array}{*{20}{c}} 8 \\ 4 \end{array}} \right)$ (A1)

$= \frac{{8{\text{!}}}}{{4{\text{!}}4{\text{!}}}} = \frac{{8 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2 \times 1}} = 7 \times 2 \times 5$ (M1)

= 70 A1

METHOD 2

recognition that they need to count the teams with 0 boys, 1 boy… 4 boys M1

$1 + \left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) + 1$

$= 1 + \left( {4 \times 4} \right) + \left( {6 \times 6} \right) + \left( {4 \times 4} \right) + 1$ (A1)

= 70 A1

[3 marks]

EITHER

recognition that the answer is the total number of teams minus the number of teams with all girls or all boys (M1)

70 − 2

recognition that the answer is the total of the number of teams with 1 boy,

2 boys, 3 boys (M1)

$\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right) \times \left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right) = \left( {4 \times 4} \right) + \left( {6 \times 6} \right) + \left( {4 \times 4} \right)$

THEN

= 68 A1

[2 marks]

Examiners report

[N/A]

Find the solution of ${\log _2}x - {\log _2}5 = 2 + {\log _2}3$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\log _2}x - {\log _2}5 = 2 + {\log _2}3$

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]

Examiners report

[N/A]

Let S be the sum of the roots found in part (a).

Find the roots of ${z^{24}} = 1$ which satisfy the condition $0 < {\text{arg}}\left( z \right) < \frac{\pi }{2}$ , expressing your answers in the form $r{e^{{\text{i}}\theta }}$ , where $r$ , $\theta \in {\mathbb{R}^ + }$ .

[5]

Show that Re S = Im S.

[4]

b.i.

By writing $\frac{\pi }{{12}}$ as $\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$ , find the value of cos $\frac{\pi }{{12}}$ in the form $\frac{{\sqrt a + \sqrt b }}{c}$ , where $a$ , $b$ and $c$ are integers to be determined.

[3]

b.ii.

Hence, or otherwise, show that S = $\frac{1}{2}\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)\left( {1 + {\text{i}}} \right)$ .

[4]

b.iii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\left( {r\left( {{\text{cos}}\,\theta + {\text{i}}\,{\text{sin}}\,\theta } \right)} \right)^{24}} = 1\left( {{\text{cos}}\,0 + {\text{i}}\,{\text{sin}}\,0} \right)$

use of De Moivre’s theorem (M1)

${r^{24}} = 1 \Rightarrow r = 1$ (A1)

$24\theta = 2\pi n \Rightarrow \theta = \frac{{\pi n}}{{12}}$ , $\left( {n \in \mathbb{Z}} \right)$ (A1)

$0 < {\text{arg}}\left( z \right) < \frac{\pi }{2} \Rightarrow n =$ 1, 2, 3, 4, 5

$z = {\text{e}}\frac{{\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{2\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{3\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{4\pi {\text{i}}}}{{12}}$ or ${\text{e}}\frac{{5\pi {\text{i}}}}{{12}}$ A2

Note: Award A1 if additional roots are given or if three correct roots are given with no incorrect (or additional) roots.

[5 marks]

Re S = ${\text{cos}}\frac{\pi }{{12}} + {\text{cos}}\frac{{2\pi }}{{12}} + {\text{cos}}\frac{{3\pi }}{{12}} + {\text{cos}}\frac{{4\pi }}{{12}} + {\text{cos}}\frac{{5\pi }}{{12}}$

Im S = ${\text{sin}}\frac{\pi }{{12}} + {\text{sin}}\frac{{2\pi }}{{12}} + {\text{sin}}\frac{{3\pi }}{{12}} + {\text{sin}}\frac{{4\pi }}{{12}} + {\text{sin}}\frac{{5\pi }}{{12}}$ A1

Note: Award A1 for both parts correct.

but ${\text{sin}}\frac{{5\pi }}{{12}} = {\text{cos}}\frac{\pi }{{12}}$ , ${\text{sin}}\frac{{4\pi }}{{12}} = {\text{cos}}\frac{{2\pi }}{{12}}$ , ${\text{sin}}\frac{{3\pi }}{{12}} = {\text{cos}}\frac{{3\pi }}{{12}}$ , ${\text{sin}}\frac{{2\pi }}{{12}} = {\text{cos}}\frac{{4\pi }}{{12}}$ and ${\text{sin}}\frac{\pi }{{12}} = {\text{cos}}\frac{{5\pi }}{{12}}$ M1A1

⇒ Re S = Im S AG

Note: Accept a geometrical method.

[4 marks]

b.i.

${\text{cos}}\frac{\pi }{{12}} = {\text{cos}}\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) = {\text{cos}}\frac{\pi }{4}{\text{cos}}\frac{\pi }{6} + {\text{sin}}\frac{\pi }{4}{\text{sin}}\frac{\pi }{6}$ M1A1

$= \frac{{\sqrt 2 }}{2}\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 2 }}{2}\frac{1}{2}$

$= \frac{{\sqrt 6 + \sqrt 2 }}{4}$ A1

[3 marks]

b.ii.

${\text{cos}}\frac{{5\pi }}{{12}} = {\text{cos}}\left( {\frac{\pi }{6} + \frac{\pi }{4}} \right) = {\text{cos}}\frac{\pi }{6}{\text{cos}}\frac{\pi }{4} - {\text{sin}}\frac{\pi }{6}{\text{sin}}\frac{\pi }{4}$ (M1)

Note: Allow alternative methods eg ${\text{cos}}\frac{{5\pi }}{{12}} = {\text{sin}}\frac{\pi }{{12}} = {\text{sin}}\left( {\frac{\pi }{4} - \frac{\pi }{6}} \right)$ .

$= \frac{{\sqrt 3 }}{2}\frac{{\sqrt 2 }}{2} - \frac{1}{2}\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6 - \sqrt 2 }}{4}$ (A1)

Re S = ${\text{cos}}\frac{\pi }{{12}} + {\text{cos}}\frac{{2\pi }}{{12}} + {\text{cos}}\frac{{3\pi }}{{12}} + {\text{cos}}\frac{{4\pi }}{{12}} + {\text{cos}}\frac{{5\pi }}{{12}}$

Re S = $\frac{{\sqrt 2 + \sqrt 6 }}{4} + \frac{{\sqrt 3 }}{2} + \frac{{\sqrt 2 }}{2} + \frac{1}{2} + \frac{{\sqrt 6 - \sqrt 2 }}{4}$ A1

$= \frac{1}{2}\left( {\sqrt 6 + 1 + \sqrt 2 + \sqrt 3 } \right)$ A1

$= \frac{1}{2}\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)$

S = Re(S)(1 + i) since Re S = Im S, R1

S = $\frac{1}{2}\left( {1 + \sqrt 2 } \right)\left( {1 + \sqrt 3 } \right)\left( {1 + {\text{i}}} \right)$ AG

[4 marks]

b.iii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

Prove by mathematical induction that $\frac{d^{n}}{d x^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ for $n \in ℤ^{+}$ .

[7]

Hence or otherwise, determine the Maclaurin series of $f (x) = x^{2} e^{x}$ in ascending powers of $x$ , up to and including the term in $x^{4}$ .

[3]

Hence or otherwise, determine the value of $\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}]$ .

[4]

Markscheme

For $n = 1$

LHS: $\frac{d}{d x} (x^{2} e^{x}) = x^{2} e^{x} + 2 x e^{x} (= e^{x} (x^{2} + 2 x))$ A1

RHS: $(x^{2} + 2 (1) x + 1 (1 - 1)) e^{x} (= e^{x} (x^{2} + 2 x))$ A1

so true for $n = 1$

now assume true for $n = k$ ; i.e. $\frac{d^{k}}{d x^{k}} (x^{2} e^{x}) = [x^{2} + 2 k x + k (k - 1)] e^{x}$ M1

Note: Do not award M1 for statements such as "let $n = k$ ". Subsequent marks can still be awarded.

attempt to differentiate the RHS M1

$\frac{d^{k + 1}}{d x^{k + 1}} (x^{2} e^{x}) = \frac{d}{d x} ([x^{2} + 2 k x + k (k - 1)] e^{x})$

$= (2 x + 2 k) e^{x} + (x^{2} + 2 k x + k (k - 1)) e^{x}$ A1

$= [x^{2} + 2 (k + 1) x + k (k + 1)] e^{x}$ A1

so true for $n = k$ implies true for $n = k + 1$

therefore $n = 1$ true and $n = k$ true $\Rightarrow n = k + 1$ true

therefore, true for all $n \in ℤ^{+}$ R1

Note: Award R1 only if three of the previous four marks have been awarded

[7 marks]

METHOD 1

attempt to use $\frac{d^{n}}{d x^{n}} (x^{2} e^{x}) = [x^{2} + 2 n x + n (n - 1)] e^{x}$ (M1)

Note: For $x = 0$ , $\frac{d^{n}}{d x^{n}} {(x^{2} e^{x})}_{x = 0} = n (n - 1)$ may be seen.

$f (0) = 0, f' (0) = 0, f'' (0) = 2, f''' (0) = 6, f^{(4)} (0) = 12$

use of $f (x) = f (0) + x f' (0) + \frac{x^{2}}{2!} f'' (0) + \frac{x^{3}}{3!} f''' (0) + \frac{x^{4}}{4!} f^{(4)} (0) + \dots$ (M1)

$\Rightarrow f (x) \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ A1

METHOD 2

' $x^{2} \times$ Maclaurin series of $e^{x}$ ' (M1)

$x^{2} (1 + x + \frac{x^{2}}{2!} + \dots)$ (A1)

$\Rightarrow f (x) \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ A1

[3 marks]

METHOD 1

attempt to substitute $x^{2} e^{x} \approx x^{2} + x^{3} + \frac{1}{2} x^{4}$ into $\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}$ M1

$\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}} \approx \frac{{(x^{2} + x^{3} + \frac{1}{2} x^{4} (+ \dots) - x^{2})}^{3}}{x^{9}}$ (A1)

EITHER

$= \frac{{(x^{3} + \frac{1}{2} x^{4} + \dots)}^{3}}{x^{9}}$ A1

$= \frac{x^{9} (+ higher order terms)}{x^{9}}$

${(\frac{x^{3} + \frac{1}{2} x^{4} (+ \dots)}{x^{3}})}^{3}$ A1

${(1 + \frac{1}{2} x (+ \dots))}^{3}$

THEN

$= 1 (+ higher order terms)$

so $\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}] = 1$ A1

METHOD 2

$\lim_{x \to 0} [\frac{{(x^{2} e^{x} - x^{2})}^{3}}{x^{9}}] = \lim_{x \to 0} {(\frac{x^{2} e^{x} - x^{2}}{x^{3}})}^{3}$ M1

$= \lim_{x \to 0} {(\frac{e^{x} - 1}{x})}^{3}$ (A1)

attempt to use L'Hôpital's rule M1

$= \lim_{x \to 0} {(\frac{e^{x} - 0}{1})}^{3}$

$= {[\lim_{x \to 0} e^{x}]}^{3}$

$= 1$ A1

[4 marks]

Examiners report

[N/A]

Find the value of $\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4}$ .

[2]

Show that $\frac{{1 - \cos 2x}}{{2\sin x}} \equiv \sin x,{\text{ }}x \ne k\pi$ where $k \in \mathbb{Z}$ .

[2]

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 $k \in \mathbb{Z}$ .

[9]

Hence or otherwise solve the equation $\sin x + \sin 3x = \cos x$ in the interval $0 < x < \pi$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sin \frac{\pi }{4} + \sin \frac{{3\pi }}{4} + \sin \frac{{5\pi }}{4} + \sin \frac{{7\pi }}{4} + \sin \frac{{9\pi }}{4} = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{2}$ (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 $2x = \frac{{5\pi }}{6}$

$\sin x + \sin 3x = \cos x \Rightarrow 2\sin 2x\cos x = \cos x$ 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]

Examiners report

[N/A]

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.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

Prove by mathematical induction that $\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right)$ , where $n \in \mathbb{Z},n \geqslant 3$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {n - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right)$

show true for $n = 3$ (M1)

${\text{LHS}} = \left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) = 1$ $\,\,\,$ ${\text{RHS}} = \left( {\begin{array}{*{20}{c}} 3 \\ 3 \end{array}} \right) = 1$ A1

hence true for $n = 3$

assume true for $n = k:\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {k - 1} \\ 2 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} k \\ 3 \end{array}} \right)$ M1

consider for $n = k + 1:\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \ldots + \left( {\begin{array}{*{20}{c}} {k - 1} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right)$ (M1)

$= \left( {\begin{array}{*{20}{c}} k \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right)$ A1

$= \frac{{k!}}{{(k - 3)!3!}} + \frac{{k!}}{{(k - 2)!2!}}\,\,\,\left( { = \frac{{k!}}{{3!}}\left[ {\frac{1}{{(k - 3)!}} + \frac{3}{{(k - 2)!}}} \right]} \right)$ or any correct expression with a visible common factor (A1)

$= \frac{{k!}}{{3!}}\left[ {\frac{{k - 2 + 3}}{{(k - 2)!}}} \right]$ or any correct expression with a common denominator (A1)

$= \frac{{k!}}{{3!}}\left[ {\frac{{k + 1}}{{(k - 2)!}}} \right]$

Note: At least one of the above three lines or equivalent must be seen.

$= \frac{{(k + 1)!}}{{3!(k - 2)!}}$ or equivalent A1

$= \left( {\begin{array}{*{20}{c}} {k + 1} \\ 3 \end{array}} \right)$

Result is true for $k = 3$ . If result is true for $k$ it is true for $k + 1$ . Hence result is true for all $k \geqslant 3$ . Hence proved by induction. R1

Note: In order to award the R1 at least [5 marks] must have been awarded.

[9 marks]

Examiners report

[N/A]

Consider the equation ${z^4} = - 4$ , where $z \in \mathbb{C}$ .

Solve the equation, giving the solutions in the form $a + {\text{i}}b$ , where $a{\text{, }}b \in \mathbb{R}$ .

[5]

The solutions form the vertices of a polygon in the complex plane. Find the area of the polygon.

[2]

Markscheme

METHOD 1

$\left| z \right| = \sqrt[4]{4}\left( { = \sqrt 2 } \right)$ (A1)

${\text{arg}}\left( {{z_1}} \right) = \frac{\pi }{4}$ (A1)

first solution is $1 + {\text{i}}$ A1

valid attempt to find all roots (De Moivre or +/− their components) (M1)

other solutions are $- 1 + {\text{i}}$ , $- 1 - {\text{i}}$ , $1 - {\text{i}}$ A1

METHOD 2

${z^4} = - 4$

${\left( {a + {\text{i}}b} \right)^4} = - 4$

attempt to expand and equate both reals and imaginaries. (M1)

${a^4} + 4{a^3}b{\text{i}} - 6{a^2}{b^2} - 4a{b^3}{\text{i}} + {b^4} = - 4$

$\left( {{a^4} - 6{a^4} + {a^4} = - 4 \Rightarrow } \right)a = \pm 1$ and $\left( {4{a^3}b - 4a{b^3} = 0 \Rightarrow } \right)a = \pm b$ (A1)

first solution is $1 + {\text{i}}$ A1

valid attempt to find all roots (De Moivre or +/− their components) (M1)

other solutions are $- 1 + {\text{i}}$ , $- 1 - {\text{i}}$ , $1 - {\text{i}}$ A1

[5 marks]

complete method to find area of ‘rectangle' (M1)

$= 4$ A1

[2 marks]

Examiners report

[N/A]

Consider the function ${f_n}(x) = (\cos 2x)(\cos 4x) \ldots (\cos {2^n}x),{\text{ }}n \in {\mathbb{Z}^ + }$ .

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

[2]

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 $m \in \mathbb{Z}$ .

[8]

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

[3]

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

[8]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

even function A1

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

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]

Examiners report

[N/A]

Consider integers $a$ and $b$ such that $a^{2} + b^{2}$ is exactly divisible by $4$ . Prove by contradiction that $a$ and $b$ cannot both be odd.

Markscheme

Assume that $a$ and $b$ are both odd. M1

Note: Award M0 for statements such as “let $a$ and $b$ be both odd”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

Then $a = 2 m + 1$ and $b = 2 n + 1$ A1

$a^{2} + b^{2} \equiv {(2 m + 1)}^{2} + {(2 n + 1)}^{2}$

$= 4 m^{2} + 4 m + 1 + 4 n^{2} + 4 n + 1$ A1

$= 4 (m^{2} + m + n^{2} + n) + 2$ (A1)

( $4 (m^{2} + m + n^{2} + n)$ is always divisible by $4$ ) but $2$ is not divisible by $4$ . (or equivalent) R1

$\Rightarrow a^{2} + b^{2}$ is not divisible by $4$ , a contradiction. (or equivalent) R1

hence $a$ and $b$ cannot both be odd. AG

Note: Award a maximum of M1A0A0(A0)R1R1 for considering identical or two consecutive odd numbers for $a$ and $b$ .

[6 marks]

Examiners report

Most candidates did not present their proof in a formal manner and merely relied on an algebraic approach rendering the proof incomplete. Very few candidates earned the first mark for making a clear assumption that a and b are both odd. A significant number of candidates only considered consecutive or identical odd numbers. The required reasoning to complete the proof were often poorly expressed or missing altogether. Only a small number of candidates were awarded all the available marks for this question.

Let the roots of the equation ${z^3} = - 3 + \sqrt 3 {\text{i}}$ be $u$ , $v$ and $w$ .

On an Argand diagram, $u$ , $v$ and $w$ are represented by the points U, V and W respectively.

Express $- 3 + \sqrt 3 {\text{i}}$ in the form $r{{\text{e}}^{{\text{i}}\theta }}$ , where $r > 0$ and $- \pi < \theta \leqslant \pi$ .

[5]

Find $u$ , $v$ and $w$ expressing your answers in the form $r{{\text{e}}^{{\text{i}}\theta }}$ , where $r > 0$ and $- \pi < \theta \leqslant \pi$ .

[5]

Find the area of triangle UVW.

[4]

By considering the sum of the roots $u$ , $v$ and $w$ , show that

${\text{cos}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{7\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}} = 0$ .

[4]

Markscheme

attempt to find modulus (M1)

$r = 2\sqrt 3 \left( { = \sqrt {12} } \right)$ A1

attempt to find argument in the correct quadrant (M1)

$\theta = \pi + {\text{arctan}}\left( { - \frac{{\sqrt 3 }}{3}} \right)$ A1

$= \frac{{5\pi }}{6}$ A1

$- 3 + \sqrt 3 {\text{i}} = \sqrt {12} {{\text{e}}^{\frac{{5\pi {\text{i}}}}{6}}}\left( { = 2\sqrt 3 {{\text{e}}^{\frac{{5\pi {\text{i}}}}{6}}}} \right)$

[5 marks]

attempt to find a root using de Moivre’s theorem M1

${12^{\frac{1}{6}}}{{\text{e}}^{\frac{{5\pi {\text{i}}}}{{18}}}}$ A1

attempt to find further two roots by adding and subtracting $\frac{{2\pi }}{3}$ to the argument M1

${12^{\frac{1}{6}}}{{\text{e}}^{ - \frac{{7\pi {\text{i}}}}{{18}}}}$ A1

${12^{\frac{1}{6}}}{{\text{e}}^{\frac{{17\pi {\text{i}}}}{{18}}}}$ A1

Note: Ignore labels for $u$ , $v$ and $w$ at this stage.

[5 marks]

METHOD 1
attempting to find the total area of (congruent) triangles UOV, VOW and UOW M1

Area $= 3\left( {\frac{1}{2}} \right)\left( {{{12}^{\frac{1}{6}}}} \right)\left( {{{12}^{\frac{1}{6}}}} \right){\text{sin}}\frac{{2\pi }}{3}$ A1A1

Note: Award A1 for $\left( {{{12}^{\frac{1}{6}}}} \right)\left( {{{12}^{\frac{1}{6}}}} \right)$ and A1 for ${\text{sin}}\frac{{2\pi }}{3}$

= $\frac{{3\sqrt 3 }}{4}\left( {{{12}^{\frac{1}{3}}}} \right)$ (or equivalent) A1

METHOD 2

UV² $= {\left( {{{12}^{\frac{1}{6}}}} \right)^2} + {\left( {{{12}^{\frac{1}{6}}}} \right)^2} - 2\left( {{{12}^{\frac{1}{6}}}} \right)\left( {{{12}^{\frac{1}{6}}}} \right){\text{cos}}\frac{{2\pi }}{3}$ (or equivalent) A1

UV $= \sqrt 3 \left( {{{12}^{\frac{1}{6}}}} \right)$ (or equivalent) A1

attempting to find the area of UVW using Area = ${\frac{1}{2}}$ × UV × VW × sin $\alpha$ for example M1

Area $= \frac{1}{2}\left( {\sqrt 3 \times {{12}^{\frac{1}{6}}}} \right)\left( {\sqrt 3 \times {{12}^{\frac{1}{6}}}} \right){\text{sin}}\frac{\pi }{3}$

= $\frac{{3\sqrt 3 }}{4}\left( {{{12}^{\frac{1}{3}}}} \right)$ (or equivalent) A1

[4 marks]

$u$ + $v$ + $w$ = 0 R1

${12^{\frac{1}{6}}}\left( {{\text{cos}}\left( { - \frac{{7\pi }}{{18}}} \right) + {\text{i}}\,{\text{sin}}\left( { - \frac{{7\pi }}{{18}}} \right) + {\text{cos}}\frac{{5\pi }}{{18}} + {\text{i}}\,{\text{sin}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}} + {\text{i}}\,{\text{sin}}\frac{{17\pi }}{{18}}} \right) = 0$ A1

consideration of real parts M1

${12^{\frac{1}{6}}}\left( {{\text{cos}}\left( { - \frac{{7\pi }}{{18}}} \right) + {\text{cos}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}}} \right) = 0$

${\text{cos}}\left( { - \frac{{7\pi }}{{18}}} \right) = {\text{cos}}\frac{{17\pi }}{{18}}$ explicitly stated A1

${\text{cos}}\frac{{5\pi }}{{18}} + {\text{cos}}\frac{{7\pi }}{{18}} + {\text{cos}}\frac{{17\pi }}{{18}} = 0$ AG

[4 marks]

Examiners report

[N/A]

Find the value of $\int_{1}^{9} (\frac{3 \sqrt{x} - 5}{\sqrt{x}}) d x$ .

Markscheme

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = \int (3 - 5 x^{- \frac{1}{2}}) d x$ (A1)

$\int \frac{3 \sqrt{x} - 5}{\sqrt{x}} d x = 3 x - 10 x^{\frac{1}{2}} (+ c)$ A1A1

substituting limits into their integrated function and subtracting (M1)

$3 (9) - 10 {(9)}^{\frac{1}{2}} - (3 (1) - 10 {(1)}^{\frac{1}{2}})$ OR $27 - 10 \times 3 - (3 - 10)$

$= 4$ A1

[5 marks]

Examiners report

A mixed response was noted for this question. Candidates who simplified the algebraic fraction before integrating were far more successful in gaining full marks in this question. Many candidates used other valid approaches such as integration by substitution and integration by parts with varying degrees of success. A small number of candidates substituted the limits without integrating.

Consider the three planes

$\prod_{1} : 2 x - y + z = 4$

$\prod_{2} : x - 2 y + 3 z = 5$

$\prod_{3} : - 9 x + 3 y - 2 z = 32$

Show that the three planes do not intersect.

[4]

Verify that the point $P (1, - 2, 0)$ lies on both $\prod_{1}$ and $\prod_{2}$ .

[1]

b.i.

Find a vector equation of $L$ , the line of intersection of $\prod_{1}$ and $\prod_{2}$ .

[4]

b.ii.

Find the distance between $L$ and $\prod_{3}$ .

[6]

Markscheme

METHOD 1

attempt to eliminate a variable M1

obtain a pair of equations in two variables

EITHER

$- 3 x + z = - 3$ and A1

$- 3 x + z = 44$ A1

$- 5 x + y = - 7$ and A1

$- 5 x + y = 40$ A1

$3 x - z = 3$ and A1

$3 x - z = - \frac{79}{5}$ A1

THEN

the two lines are parallel ( $- 3 \neq 44$ or $- 7 \neq 40$ or $3 \neq - \frac{79}{5}$ ) R1

Note: There are other possible pairs of equations in two variables.
To obtain the final R1, at least the initial M1 must have been awarded.

hence the three planes do not intersect AG

METHOD 2

vector product of the two normals $= (\begin{matrix} - 1 \\ - 5 \\ - 3 \end{matrix})$ (or equivalent) A1

$r = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ (or equivalent) A1

Note: Award A0 if “ $r =$ ” is missing. Subsequent marks may still be awarded.

Attempt to substitute $(1 + λ, - 2 + 5 λ, 3 λ)$ in $\prod_{3}$ M1

$- 9 (1 + λ) + 3 (- 2 + 5 λ) - 2 (3 λ) = 32$

$- 15 = 32$ , a contradiction R1

hence the three planes do not intersect AG

METHOD 3

attempt to eliminate a variable M1

$- 3 y + 5 z = 6$ A1

$- 3 y + 5 z = 100$ A1

$0 = 94$ , a contradiction R1

Note: Accept other equivalent alternatives. Accept other valid methods.
To obtain the final R1, at least the initial M1 must have been awarded.

hence the three planes do not intersect AG

[4 marks]

$\prod_{1} : 2 + 2 + 0 = 4$ and $\prod_{2} : 1 + 4 + 0 = 5$ A1

[1 mark]

b.i.

METHOD 1

attempt to find the vector product of the two normals M1

$(\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}) \times (\begin{matrix} 1 \\ - 2 \\ 3 \end{matrix})$

$= (\begin{matrix} - 1 \\ - 5 \\ - 3 \end{matrix})$ A1

$r = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ A1A1

Note: Award A1A0 if “ $r =$ ” is missing.
Accept any multiple of the direction vector.
Working for (b)(ii) may be seen in part (a) Method 2. In this case penalize lack of “ $r =$ ” only once.

METHOD 2

attempt to eliminate a variable from $\prod_{1}$ and $\prod_{2}$ M1

$3 x - z = 3$ OR $3 y - 5 z = - 6$ OR $5 x - y = 7$

Let $x = t$

substituting $x = t$ in $3 x - z = 3$ to obtain

$z = - 3 + 3 t$ and $y = 5 t - 7$ (for all three variables in parametric form) A1

$r = (\begin{matrix} 0 \\ - 7 \\ - 3 \end{matrix}) + λ (\begin{matrix} 1 \\ 5 \\ 3 \end{matrix})$ A1A1

Note: Award A1A0 if “ $r =$ ” is missing.
Accept any multiple of the direction vector. Accept other position vectors which satisfy both the planes $\prod_{1}$ and $\prod_{2}$ .

[4 marks]

b.ii.

METHOD 1

the line connecting $L$ and $\prod_{3}$ is given by $L_{1}$

attempt to substitute position and direction vector to form $L_{1}$ (M1)

$s = (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + t (\begin{matrix} - 9 \\ 3 \\ - 2 \end{matrix})$ A1

substitute $(1 - 9 t, - 2 + 3 t, - 2 t)$ in $\prod_{3}$ M1

$- 9 (1 - 9 t) + 3 (- 2 + 3 t) - 2 (- 2 t) = 32$

$94 t = 47 \Rightarrow t = \frac{1}{2}$ A1

attempt to find distance between $(1, - 2, 0)$ and their point $(- \frac{7}{2}, - \frac{1}{2}, - 1)$ (M1)

$= |(\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) + \frac{1}{2} (\begin{matrix} - 9 \\ 3 \\ - 2 \end{matrix}) - (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix})| = \frac{1}{2} \sqrt{{(- 9)}^{2} + 3^{2} + {(- 2)}^{2}}$

$= \frac{\sqrt{94}}{2}$ A1

METHOD 2

unit normal vector equation of $\prod_{3}$ is given by $\frac{(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix})}{\sqrt{81 + 9 + 4}}$ (M1)

$= \frac{32}{\sqrt{94}}$ A1

let $\prod_{4}$ be the plane parallel to $\prod_{3}$ and passing through $P$ ,
then the normal vector equation of $\prod_{4}$ is given by

$(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} 1 \\ - 2 \\ 0 \end{matrix}) = - 15$ M1

unit normal vector equation of $\prod_{4}$ is given by

$\frac{(\begin{matrix} - 9 \\ 3 \\ 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix})}{\sqrt{81 + 9 + 4}} = \frac{- 15}{\sqrt{94}}$ A1

distance between the planes is $\frac{32}{\sqrt{94}} - \frac{- 15}{\sqrt{94}}$ (M1)

$= \frac{47}{\sqrt{94}} (= \frac{\sqrt{94}}{2})$ A1

[6 marks]

Examiners report

Part (a) was well attempted using a variety of approaches. Most candidates were able to gain marks for part (a) through attempts to eliminate a variable with many subsequently making algebraic errors. Part (b)(i) was well done. For part (b)(ii) few successful attempts were noted, many candidates failed to use an appropriate notation "r =" while giving the vector equation of a line. Part (c) proved to be challenging for most candidates with very few correct answers seen. Many candidates did not attempt part (c).

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Prove by contradiction that the equation $2 x^{3} + 6 x + 1 = 0$ has no integer roots.

Markscheme

METHOD 1 (rearranging the equation)

assume there exists some $α \in ℤ$ such that $2 α^{3} + 6 α + 1 = 0$ M1

Note: Award M1 for equivalent statements such as ‘assume that $α$ is an integer root of $2 α^{3} + 6 α + 1 = 0$ ’. Condone the use of $x$ throughout the proof.

Award M1 for an assumption involving $α^{3} + 3 α + \frac{1}{2} = 0$ .

Note: Award M0 for statements such as “let’s consider the equation has integer roots…” ,“let $α \in ℤ$ be a root of $2 α^{3} + 6 α + 1 = 0$ …”

Note: Subsequent marks after this M1 are independent of this M1 and can be awarded.

attempts to rearrange their equation into a suitable form M1

EITHER

$2 α^{3} + 6 α = - 1$ A1

$α \in ℤ \Rightarrow 2 α^{3} + 6 α$ is even R1

$2 α^{3} + 6 α = - 1$ which is not even and so $α$ cannot be an integer R1

Note: Accept ‘ $2 α^{3} + 6 α = - 1$ which gives a contradiction’.

$1 = 2 (- α^{3} - 3 α)$ A1

$α \in ℤ \Rightarrow (- α^{3} - 3 α) \in ℤ$ R1

$\Rightarrow 1$ is even which is not true and so $α$ cannot be an integer R1

Note: Accept ‘ $\Rightarrow 1$ is even which gives a contradiction’.

$\frac{1}{2} = - α^{3} - 3 α$ A1

$α \in ℤ \Rightarrow (- α^{3} - 3 α) \in ℤ$ R1

$- α^{3} - 3 α$ is is not an integer $(= \frac{1}{2})$ and so $α$ cannot be an integer R1

Note: Accept ‘ $- α^{3} - 3 α$ is not an integer $(= \frac{1}{2})$ which gives a contradiction’.

$α = - \frac{1}{2 (α^{2} + 3)}$ A1

$α \in ℤ \Rightarrow - \frac{1}{2 (α^{2} + 3)} \in ℤ$ R1

$- \frac{1}{2 (α^{2} + 3)}$ is not an integer and so $α$ cannot be an integer R1

Note: Accept $- \frac{1}{2 (α^{2} + 3)}$ is not an integer which gives a contradiction’.

THEN

so the equation $2 x^{3} + 6 x + 1 = 0$ has no integer roots AG

METHOD 2

assume there exists some $α \in ℤ$ such that $2 α^{3} + 6 α + 1 = 0$ M1

Note: Award M1 for equivalent statements such as ‘assume that $α$ is an integer root of $2 α^{3} + 6 α + 1 = 0$ ’. Condone the use of $x$ throughout the proof. Award M1 for an assumption involving $α^{3} + 3 α + \frac{1}{2} = 0$ and award subsequent marks based on this.

Note: Award M0 for statements such as “let’s consider the equation has integer roots…” ,“let $α \in ℤ$ be a root of $2 α^{3} + 6 α + 1 = 0$ …”

Note: Subsequent marks after this M1 are independent of this M1 and can be awarded.

let $f (x) = 2 x^{3} + 6 x + 1$ (and $f (α) = 0$ )

$f' (x) = 6 x^{2} + 6 > 0$ for all $x \in ℝ \Rightarrow f$ is a (strictly) increasing function M1A1

$f (0) = 1$ and $f (- 1) = - 7$ R1

thus $f (x) = 0$ has only one real root between $- 1$ and $0$ , which gives a contradiction

(or therefore, contradicting the assumption that $f (α) = 0$ for some $α \in ℤ$ ), R1

so the equation $2 x^{3} + 6 x + 1 = 0$ has no integer roots AG

[5 marks]

Examiners report

[N/A]

Consider the quartic equation $z^{4} + 4 z^{3} + 8 z^{2} + 80 z + 400 = 0, z \in ℂ$ .

Two of the roots of this equation are $a + b i$ and $b + a i$ , where $a, b \in ℤ$ .

Find the possible values of $a$ .

Markscheme

METHOD 1

other two roots are $a - b i$ and $b - a i$ A1

sum of roots $= - 4$ and product of roots $= 400$ A1

attempt to set sum of four roots equal to $- 4$ or $4$ OR
attempt to set product of four roots equal to $400$ M1

$a + b i + a - b i + b + a i + b - a i = - 4$

$2 a + 2 b = - 4 (\Rightarrow a + b = - 2)$ A1

$(a + b i) (a - b i) (b + a i) (b - a i) = 400$

${(a^{2} + b^{2})}^{2} = 400$ A1

$a^{2} + b^{2} = 20$

attempt to solve simultaneous equations (M1)

$a = 2$ or $a = - 4$ A1A1

METHOD 2

other two roots are $a - b i$ and $b - a i$ A1

$(z - (a + b i)) (z - (a - b i)) (z - (b + a i)) (z - (b - a i)) (= 0)$ A1

$({(z - a)}^{2} + b^{2}) ({(z - b)}^{2} + a^{2}) (= 0)$

$(z^{2} - 2 a z + a^{2} + b^{2}) (z^{2} - 2 b z + b^{2} + a^{2}) (= 0)$ A1

Attempt to equate coefficient of $z^{3}$ and constant with the given quartic equation M1

$- 2 a - 2 b = 4$ and ${(a^{2} + b^{2})}^{2} = 400$ A1

attempt to solve simultaneous equations (M1)

$a = 2$ or $a = - 4$ A1A1

[8 marks]

Examiners report

[N/A]

Consider the expression $\frac{1}{\sqrt{1 + a x}} - \sqrt{1 - x}$ where $a \in ℚ, a \neq 0$ .

The binomial expansion of this expression, in ascending powers of $x$ , as far as the term in $x^{2}$ is $4 b x + b x^{2}$ , where $b \in ℚ$ .

Find the value of $a$ and the value of $b$ .

[6]

State the restriction which must be placed on $x$ for this expansion to be valid.

[1]

Markscheme

attempt to expand binomial with negative fractional power (M1)

$\frac{1}{\sqrt{1 + a x}} = {(1 + a x)}^{- \frac{1}{2}} = 1 - \frac{a x}{2} + \frac{3 a^{2} x^{2}}{8} + \dots$ A1

$\sqrt{1 - x} = {(1 - x)}^{\frac{1}{2}} = 1 - \frac{x}{2} - \frac{x^{2}}{8} + \dots$ A1

$\frac{1}{\sqrt{1 + a x}} - \sqrt{1 - x} = \frac{(1 - a)}{2} x + (\frac{3 a^{2} + 1}{8}) x^{2} + \dots$

attempt to equate coefficients of $x$ or $x^{2}$ (M1)

$x : \frac{1 - a}{2} = 4 b; x^{2} : \frac{3 a^{2} + 1}{8} = b$

attempt to solve simultaneously (M1)

$a = - \frac{1}{3}, b = \frac{1}{6}$ A1

[6 marks]

$|x| < 1$ A1

[1 mark]

Examiners report

[N/A]

Let $z = 1 - \cos 2\theta - {\text{i}}\sin 2\theta ,{\text{ }}z \in \mathbb{C},{\text{ }}0 \leqslant \theta \leqslant \pi$ .

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

[5]

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

[3]

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

[9]

c.i.

Hence find the cube roots of $z$ in modulus-argument form.

[5]

c.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

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

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

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

$= 2{\sin ^2}\theta - 2{\text{i}}\sin \theta \cos \theta$ 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]

c.i.

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]

c.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

Use the binomial theorem to expand ${(\cos θ + i \sin θ)}^{4}$ . Give your answer in the form $a + b i$ where $a$ and $b$ are expressed in terms of $\sin θ$ and $\cos θ$ .

[3]

Use de Moivre’s theorem and the result from part (a) to show that $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ .

[5]

Use the identity from part (b) to show that the quadratic equation $x^{2} - 6 x + 1 = 0$ has roots ${cot}^{2} \frac{π}{8}$ and ${cot}^{2} \frac{3 π}{8}$ .

[5]

Hence find the exact value of ${cot}^{2} \frac{3 π}{8}$ .

[4]

Deduce a quadratic equation with integer coefficients, having roots ${cosec}^{2} \frac{π}{8}$ and ${cosec}^{2} \frac{3 π}{8}$ .

[3]

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

uses the binomial theorem on ${(\cos θ + i \sin θ)}^{4}$ M1

$= {}_{4}C_{0} \cos^{4} θ + {}_{4}C_{1} \cos^{3} θ (i \sin θ) + {}_{4}C_{2} \cos^{2} θ (i^{2} \sin^{2} θ) + {}_{4}C_{3} \cos θ (i^{3} \sin^{3} θ) + {}_{4}C_{4} (i^{4} \sin^{4} θ)$ A1

$= (\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ A1

[3 marks]

(using de Moivre’s theorem with $n = 4$ gives) $\cos 4 θ + i \sin 4 θ$ (A1)

equates both the real and imaginary parts of $\cos 4 θ + i \sin 4 θ$ and $(\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ) + i (4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ)$ M1

$\cos 4 θ = \cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ$ and $\sin 4 θ = 4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ$

recognizes that $\cot 4 θ = \frac{\cos 4 θ}{\sin 4 θ}$ (A1)

substitutes for $\sin 4 θ$ and $\cos 4 θ$ into $\frac{\cos 4 θ}{\sin 4 θ}$ M1

$\cot 4 θ = \frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}$

divides the numerator and denominator by $\sin^{4} θ$ to obtain

$\cot 4 θ = \frac{\frac{\cos^{4} θ - 6 \cos^{2} θ \sin^{2} θ + \sin^{4} θ}{\sin^{4} θ}}{\frac{4 \cos^{3} θ \sin θ - 4 \cos θ \sin^{3} θ}{\sin^{4} θ}}$ A1

$\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ AG

[5 marks]

setting $\cot 4 θ = 0$ and putting $x = \cot^{2} θ$ in the numerator of $\cot 4 θ = \frac{\cot^{4} θ - 6 \cot^{2} θ + 1}{4 \cot^{3} θ - 4 \cot θ}$ gives $x^{2} - 6 x + 1 = 0$ M1

attempts to solve $\cot 4 θ = 0$ for $θ$ M1

$4 θ = \frac{π}{2}, \frac{3 π}{2}, \dots (4 θ = \frac{1}{2} (2 n + 1) π, n = 0, 1, \dots)$ (A1)

$θ = \frac{π}{8}, \frac{3 π}{8}$ A1

Note: Do not award the final A1 if solutions other than $θ = \frac{π}{8}, \frac{3 π}{8}$ are listed.

finding the roots of $\cot 4 θ = 0 (θ = \frac{π}{8}, \frac{3 π}{8})$ corresponds to finding the roots of $x^{2} - 6 x + 1 = 0$ where $x = \cot^{2} θ$ R1

so the equation $x^{2} - 6 x + 1 = 0$ as roots ${cot}^{2} \frac{π}{8}$ and ${cot}^{2} \frac{3 π}{8}$ AG

[5 marks]

attempts to solve $x^{2} - 6 x + 1 = 0$ for $x$ M1

$x = 3 \pm 2 \sqrt{2}$ A1

since ${cot}^{2} \frac{π}{8} > {cot}^{2} \frac{3 π}{8}, {cot}^{2} \frac{3 π}{8}$ has the smaller value of the two roots R1

Note: Award R1 for an alternative convincing valid reason.

so ${cot}^{2} \frac{3 π}{8} = 3 -2 \sqrt{2}$ A1

[4 marks]

let $y = {cosec}^{2} θ$

uses $\cot^{2} θ = {cosec}^{2} θ - 1$ where $x = \cot^{2} θ$ (M1)

$x^{2} - 6 x + 1 = 0 \Rightarrow {(y - 1)}^{2} - 6 (y - 1) + 1 = 0$ M1

$y^{2} - 8 y + 8 = 0$ A1

[3 marks]

Examiners report

[N/A]

Use the principle of mathematical induction to prove that

$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + n{\left( {\frac{1}{2}} \right)^{n - 1}} = 4 - \frac{{n + 2}}{{{2^{n - 1}}}}$ , where $n \in {\mathbb{Z}^ + }$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

if $n = 1$

${\text{LHS}} = 1\,{\text{;}}\,\,{\text{RHS}} = 4 - \frac{3}{{{2^0}}} = 4 - 3 = 1$ M1

hence true for $n = 1$

assume true for $n = k$ M1

Note: Assumption of truth must be present. Following marks are not dependent on the first two M1 marks.

so $1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + k{\left( {\frac{1}{2}} \right)^{k - 1}} = 4 - \frac{{k + 2}}{{{2^{k - 1}}}}$

if $n = k + 1$

$1 + 2\left( {\frac{1}{2}} \right) + 3{\left( {\frac{1}{2}} \right)^2} + 4{\left( {\frac{1}{2}} \right)^3} + \, \ldots \, + k{\left( {\frac{1}{2}} \right)^{k - 1}} + \left( {k + 1} \right){\left( {\frac{1}{2}} \right)^k}$

$= 4 - \frac{{k + 2}}{{{2^{k - 1}}}} + \left( {k + 1} \right){\left( {\frac{1}{2}} \right)^k}$ M1A1

finding a common denominator for the two fractions M1

$= 4 - \frac{{2\left( {k + 2} \right)}}{{{2^k}}} + \frac{{k + 1}}{{{2^k}}}$

$= 4 - \frac{{2\left( {k + 2} \right) - \left( {k + 1} \right)}}{{{2^k}}} = 4 - \frac{{k + 3}}{{{2^k}}}\left( { = 4 - \frac{{\left( {k + 1} \right) + 2}}{{{2^{\left( {k + 1} \right) - 1}}}}} \right)$ A1

hence if true for $n = k$ then also true for $n = k + 1$ , as true for $n = 1$ , so true (for all $n \in {\mathbb{Z}^ + }$ ) R1

Note: Award the final R1 only if the first four marks have been awarded.

[7 marks]

Examiners report

[N/A]

Consider the complex numbers ${z_1} = 1 + \sqrt 3 {\text{i, }}{z_2} = 1 + {\text{i}}$ and $w = \frac{{{z_1}}}{{{z_2}}}$ .

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

[3]

a.i.

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

[1]

a.ii.

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

[2]

Markscheme

${z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)$ 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]

a.i.

${z_1} = 2{\text{cis}}\left( {\frac{\pi }{3}} \right)$ 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]

a.ii.

EITHER

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

$\arg ({w^n}) = \pi$ (M1)

$\frac{{n\pi }}{{12}} = \pi$

THEN

$\therefore n = 12$ A1

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Consider the function $f\left( x \right) = x\,{{\text{e}}^{2x}}$ , where $x \in \mathbb{R}$ . The ${n^{{\text{th}}}}$ derivative of $f\left( x \right)$ is denoted by ${f^{\left( n \right)}}\left( x \right)$ .

Prove, by mathematical induction, that ${f^{\left( n \right)}}\left( x \right) = \left( {{2^n}x + n{2^{n - 1}}} \right){{\text{e}}^{2x}}$ , $n \in {\mathbb{Z}^ + }$ .

Markscheme

$f'\left( x \right) = {{\text{e}}^{2x}} + 2x{{\text{e}}^{2x}}$ A1

Note: This must be obtained from the candidate differentiating $f\left( x \right)$ .

$= \left( {{2^1}x + 1 \times {2^{1 - 1}}} \right){{\text{e}}^{2x}}$ A1

(hence true for $n = 1$ )

assume true for $n = k$ : M1

${f^{\left( k \right)}}\left( x \right) = \left( {{2^k}x + k{2^{k - 1}}} \right){{\text{e}}^{2x}}$

Note: Award M1 if truth is assumed. Do not allow “let $n = k$ ”.

consider $n = k + 1$ :

${f^{\left( {k + 1} \right)}}\left( x \right) = \frac{{\text{d}}}{{{\text{d}}x}}\left( {\left( {{2^k}x + k{2^{k - 1}}} \right){{\text{e}}^{2x}}} \right)$

attempt to differentiate ${f^{\left( k \right)}}\left( x \right)$ M1

${f^{\left( {k + 1} \right)}}\left( x \right) = {2^k}{{\text{e}}^{2x}} + 2\left( {{2^k}x + k{2^{k - 1}}} \right){{\text{e}}^{2x}}$ A1

${f^{\left( {k + 1} \right)}}\left( x \right) = \left( {{2^k} + {2^{k + 1}}x + k{2^k}} \right){{\text{e}}^{2x}}$

${f^{\left( {k + 1} \right)}}\left( x \right) = \left( {{2^{k + 1}}x + \left( {k + 1} \right){2^k}} \right){{\text{e}}^{2x}}$ A1

$= \left( {{2^{k + 1}}x + \left( {k + 1} \right){2^{\left( {k + 1} \right) - 1}}} \right){{\text{e}}^{2x}}$

True for $n = 1$ and $n = k$ true implies true for $n = k + 1$ .

Therefore the statement is true for all $n\left( { \in {\mathbb{Z}^ + }} \right)$ R1

Note: Do not award final R1 if the two previous M1s are not awarded. Allow full marks for candidates who use the base case $n = 0$ .

[7 marks]

Examiners report

[N/A]

Let $f\left( x \right) = \frac{{4x - 5}}{{{x^2} - 3x + 2}}$ $x \ne 1{\text{,}}\,x \ne 2$ .

Express $f(x)$ in partial fractions.

[6]

Use part (a) to show that $f(x)$ is always decreasing.

[3]

Use part (a) to find the exact value of $\int\limits_{ - 1}^0 {f(x)dx}$ , giving the answer in the form ${\text{ln}}\,q$ , $q \in \mathbb{Q}$ .

[4]

Markscheme

$f\left( x \right) = \frac{{4x - 5}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} \equiv \frac{A}{{x - 1}} + \frac{B}{{x - 2}}$ M1A1

$\Rightarrow 4x - 5 \equiv A\left( {x - 2} \right) + B\left( {x - 1} \right)$ M1A1

$x = 1 \Rightarrow A = 1$ $x = 2 \Rightarrow B = 3$ A1A1

$f\left( x \right) = \frac{1}{{x - 1}} + \frac{3}{{x - 2}}$

[6 marks]

$f'\left( x \right) = - {\left( {x - 1} \right)^{ - 2}} - 3{\left( {x - 2} \right)^{ - 2}}$ M1A1

This is always negative so function is always decreasing. R1AG

[3 marks]

$\int\limits_{ - 1}^0 {\frac{1}{{x - 1}} + \frac{3}{{x - 2}}{\text{ }}} dx = \left[ {{\text{ln}}\left| {x - 1} \right| + 3\,{\text{ln}}\left| {x - 2} \right|} \right]_{ - 1}^0$ M1A1

$= \left( {3\,{\text{ln}}\,2} \right) - \left( {\,{\text{ln}}\,2 + 3\,{\text{ln}}\,3} \right) = 2\,{\text{ln}}\,2 - 3\,{\text{ln}}\,3 = \,{\text{ln}}\frac{4}{{27}}$ A1A1

[4 marks]

Examiners report

[N/A]

Consider the complex numbers $z_{1} = 1 + b i$ and $z_{2} = (1 - b^{2}) - 2 b i$ , where $b \in ℝ, b \neq 0$ .

Find an expression for $z_{1} z_{2}$ in terms of $b$ .

[3]

Hence, given that $arg (z_{1} z_{2}) = \frac{π}{4}$ , find the value of $b$ .

[3]

Markscheme

$z_{1} z_{2} = (1 + b i) ((1 - b^{2}) - (2 b) i)$

$= (1 - b^{2} - 2 i^{2} b^{2}) + i (- 2 b + b - b^{3})$ M1

$= (1 + b^{2}) + i (- b - b^{3})$ A1A1

Note: Award A1 for $1 + b^{2}$ and A1 for $- b i - b^{3} i$ .

[3 marks]

$arg (z_{1} z_{2}) = arctan (\frac{- b - b^{3}}{1 + b^{2}}) = \frac{π}{4}$ (M1)

EITHER
$arctan (- b) = \frac{π}{4}$ (since $1 + b^{2} \neq 0$ , for $b \in ℝ$ ) A1

$- b - b^{3} = 1 + b^{2}$ (or equivalent) A1

THEN

$b = - 1$ A1

[3 marks]

Examiners report

Part (a) was generally well done with many completely correct answers seen. Part (b) proved to be challenging with many candidates incorrectly equating the ratio of their imaginary and real parts to $\frac{π}{4}$ instead of $\tan \frac{π}{4}$ . Stronger candidates realized that when $θ = \frac{π}{4}$ , it forms an isosceles right-angled triangle and equated the real and imaginary parts to obtain the value of b .

[N/A]

Chloe and Selena play a game where each have four cards showing capital letters A, B, C and D.
Chloe lays her cards face up on the table in order A, B, C, D as shown in the following diagram.

N17/5/MATHL/HP1/ENG/TZ0/10

Selena shuffles her cards and lays them face down on the table. She then turns them over one by one to see if her card matches with Chloe’s card directly above.
Chloe wins if no matches occur; otherwise Selena wins.

Chloe and Selena repeat their game so that they play a total of 50 times.
Suppose the discrete random variable X represents the number of times Chloe wins.

Show that the probability that Chloe wins the game is $\frac{3}{8}$ .

[6]

Determine the mean of X.

[3]

b.i.

Determine the variance of X.

[2]

b.ii.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

b.i.

${\sigma ^2} = np(1 - p) = 50 \times \frac{3}{8} \times \frac{5}{8} = \frac{{750}}{{64}}{\text{ }}\left( { = \frac{{375}}{{32}}} \right){\text{ }}( = 11.7)$ (M1)A1

[2 marks]

b.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

Consider the function defined by $f (x) = \frac{k x - 5}{x - k}$ , where $x \in ℝ \ \{k\}$ and $k^{2} \neq 5$ .

Consider the case where $k = 3$ .

State the equation of the vertical asymptote on the graph of $y = f (x)$ .

[1]

State the equation of the horizontal asymptote on the graph of $y = f (x)$ .

[1]

Use an algebraic method to determine whether $f$ is a self-inverse function.

[4]

Sketch the graph of $y = f (x)$ , stating clearly the equations of any asymptotes and the coordinates of any points of intersections with the coordinate axes.

[3]

The region bounded by the $x$ -axis, the curve $y = f (x)$ , and the lines $x = 5$ and $x = 7$ is rotated through $2 π$ about the $x$ -axis. Find the volume of the solid generated, giving your answer in the form $π (a + b \ln 2)$ , where $a, b \in ℤ$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$x = k$ A1

[1 mark]

$y = k$ A1

[1 mark]

METHOD 1

$(f \circ f) (x) = \frac{k (\frac{k x - 5}{x - k}) - 5}{(\frac{k x - 5}{x - k}) - k}$ M1

$= \frac{k (k x - 5) - 5 (x - k)}{k x - 5 - k (x - k)}$ A1

$= \frac{k^{2} x - 5 k - 5 x + 5 k}{k x - 5 - k x + k^{2}}$

$= \frac{k^{2} x - 5 x}{k^{2} - 5}$ A1

$= \frac{x (k^{2} - 5)}{k^{2} - 5}$

$= x$

$(f \circ f) (x) = x$ , (hence $f$ is self-inverse) R1

Note: The statement $f (f (x)) = x$ could be seen anywhere in the candidate’s working to award R1.

METHOD 2

$f (x) = \frac{k x - 5}{x - k}$

$x = \frac{k y - 5}{y - k}$ M1

Note: Interchanging $x$ and $y$ can be done at any stage.

$x (y - k) = k y - 5$ A1

$x y - x k = k y - 5$

$x y - k y = x k - 5$

$y (x - k) = k x - 5$ A1

$y = f^{- 1} (x) = \frac{k x - 5}{x - k}$ (hence $f$ is self-inverse) R1

[4 marks]

attempt to draw both branches of a rectangular hyperbola M1

$x = 3$ and $y = 3$ A1

$(0, \frac{5}{3})$ and $(\frac{5}{3}, 0)$ A1

[3 marks]

METHOD 1

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

EITHER

attempt to express $\frac{3 x - 5}{x - 3}$ in the form $p + \frac{q}{x - 3}$ M1

$\frac{3 x - 5}{x - 3} = 3 + \frac{4}{x - 3}$ A1

attempt to expand ${(\frac{3 x - 5}{x - 3})}^{2}$ or ${(3 x - 5)}^{2}$ and divide out M1

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24 x - 56}{{(x - 3)}^{2}}$ A1

THEN

${(\frac{3 x - 5}{x - 3})}^{2} = 9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}$ A1

$volume = π \int_{5}^{7} (9 + \frac{24}{x - 3} + \frac{16}{{(x - 3)}^{2}}) d x$

$= π {[9 x + 24 \ln (x - 3) - \frac{16}{x - 3}]}_{5}^{7}$ A1

$= π ⌊(63 + 24 \ln 4 - 4) - (45 + 24 \ln 2 - 8)⌋$

$= π (22 + 24 \ln 2)$ A1

METHOD 2

$volume = π \int_{5}^{7} {(\frac{3 x - 5}{x - 3})}^{2} d x$ (M1)

substituting $u = x - 3 \Rightarrow \frac{d u}{d x} = 1$ A1

$3 x - 5 = 3 (u + 3) - 5 = 3 u + 4$

$volume = π \int_{2}^{4} {(\frac{3 u + 4}{u})}^{2} d u$ M1

$= π \int_{2}^{4} 9 + \frac{16}{u^{2}} + \frac{24}{u} d u$ A1

$= π {[9 u - \frac{16}{u} + 24 \ln u]}_{2}^{4}$ A1

Note: Ignore absence of or incorrect limits seen up to this point.

$= π (22 + 24 \ln 2)$ A1

[6 marks]

Examiners report

[N/A]

Let $\omega$ be one of the non-real solutions of the equation ${z^3} = 1$ .

Consider the complex numbers $p = 1 - 3{\text{i}}$ and $q = x + (2x + 1){\text{i}}$ , where $x \in \mathbb{R}$ .

Determine the value of

(i) $1 + \omega + {\omega ^2}$ ;

(ii) $1 + \omega {\text{*}} + {(\omega {\text{*}})^2}$ .

[4]

Show that $(\omega - 3{\omega ^2})({\omega ^2} - 3\omega ) = 13$ .

[4]

Find the values of $x$ that satisfy the equation $\left| p \right| = \left| q \right|$ .

[5]

Solve the inequality $\operatorname{Re} (pq) + 8 < {\left( {\operatorname{Im} (pq)} \right)^2}$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(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

$= - 3\omega + 10 - 3{\omega ^2} = - 3({\omega ^2} + \omega + 1) + 13$ M1

$= - 3 \times 0 + 13$ A1

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]

Examiners report

[N/A]

An arithmetic sequence ${u_1}{\text{, }}{u_2}{\text{, }}{u_3} \ldots$ has ${u_1} = 1$ and common difference $d \ne 0$ . Given that ${u_2}{\text{, }}{u_3}$ and ${u_6}$ are the first three terms of a geometric sequence

Given that ${u_N} = - 15$

find the value of $d$ .

[4]

determine the value of $\sum\limits_{r = 1}^N {{u_r}}$ .

[3]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

Solve the equation ${4^x} + {2^{x + 2}} = 3$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

Solve the simultaneous equations

${\text{lo}}{{\text{g}}_2}6x = 1 + 2\,{\text{lo}}{{\text{g}}_2}y$

$1 + {\text{lo}}{{\text{g}}_6}x = {\text{lo}}{{\text{g}}_6}\left( {15y - 25} \right)$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

use of at least one “log rule” applied correctly for the first equation M1

${\text{lo}}{{\text{g}}_2}6x = {\text{lo}}{{\text{g}}_2}2 + 2\,{\text{lo}}{{\text{g}}_2}y$

$= {\text{lo}}{{\text{g}}_2}2 + \,{\text{lo}}{{\text{g}}_2}{y^2}$

$= {\text{lo}}{{\text{g}}_2}\left( {2{y^2}} \right)$

$\Rightarrow 6x = 2{y^2}$ A1

use of at least one “log rule” applied correctly for the second equation M1

${\text{lo}}{{\text{g}}_6}\left( {15y - 25} \right) = 1 + {\text{lo}}{{\text{g}}_6}x$

$= {\text{lo}}{{\text{g}}_6}6 + {\text{lo}}{{\text{g}}_6}x$

$= {\text{lo}}{{\text{g}}_6}6x$

$\Rightarrow 15y - 25 = 6x$ A1

attempt to eliminate $x$ (or $y$ ) from their two equations M1

$2{y^2} = 15y - 25$

$2{y^2} - 15y + 25 = 0$

$\left( {2y - 5} \right)\left( {y - 5} \right) = 0$

$x = \frac{{25}}{{12}}{\text{,}}\,\,y = \frac{5}{2}{\text{,}}$ A1

or $x = \frac{{25}}{3}{\text{,}}\,\,y = 5$ A1

Note: $x$ , $y$ values do not have to be “paired” to gain either of the final two A marks.

[7 marks]

Examiners report

[N/A]

Solve the equation ${\log _2}(x + 3) + {\log _2}(x - 3) = 4$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\log _2}(x + 3) + {\log _2}(x - 3) = 4$

${\log _2}({x^2} - 9) = 4$ (M1)

${x^2} - 9 = {2^4}{\text{ }}( = 16)$ M1A1

${x^2} = 25$

$x = \pm 5$ (A1)

$x = 5$ A1

[5 marks]

Examiners report

[N/A]

The 1st, 4th and 8th terms of an arithmetic sequence, with common difference $d$ , $d \ne 0$ , are the first three terms of a geometric sequence, with common ratio $r$ . Given that the 1st term of both sequences is 9 find

the value of $d$ ;

[4]

the value of $r$ ;

[1]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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)

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

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]

Examiners report

[N/A]

Solve ${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) - 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)$

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

$\left( {{\text{ln}}\,x - 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)$ 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]

Examiners report

[N/A]

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.

M17/5/MATHL/HP1/ENG/TZ2/05

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

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

[4 marks]

Examiners report

[N/A]

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

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners report

[N/A]

Let $f(x) = \frac{1}{{1 - {x^2}}}$ for $- 1 < x < 1$ . Use partial fractions to find $\int {f\left( x \right)} {\text{ }}dx$ .

Markscheme

$\frac{1}{{1 - {x^2}}} = \frac{1}{{\left( {1 - x} \right)\left( {1 + x} \right)}} \equiv \frac{A}{{1 - x}} + \frac{B}{{1 + x}}$ M1M1A1

$\Rightarrow 1 \equiv A\left( {1 + x} \right) + B\left( {1 - x} \right) \Rightarrow A = B = \frac{1}{2}$ M1A1A1

$\int {\frac{{\tfrac{1}{2}}}{{1 - x}}} + \frac{{\tfrac{1}{2}}}{{1 + x}}dx = \frac{{ - 1}}{2}\ln \left( {1 - x} \right) + \frac{1}{2}\ln \left( {1 + x} \right) + c$ $\left( { = \ln k\sqrt {\frac{{1 + x}}{{1 - x}}} } \right)$ M1A1

[8 marks]

Examiners report

[N/A]

It is given that $2 \cos A \sin B \equiv \sin (A + B) - \sin (A - B)$ . (Do not prove this identity.)

Using mathematical induction and the above identity, prove that $Σ_{r = 1}^{n} \cos (2 r - 1) θ = \frac{\sin 2 n θ}{2 \sin θ}$ for $n \in ℤ^{+}$ .

Markscheme

let $P (n)$ be the proposition that $Σ_{r = 1}^{n} \cos (2 r - 1) θ = \frac{\sin 2 n θ}{2 \sin θ}$ for $n \in ℤ^{+}$

considering $P (1)$ :

$LHS = \cos (1) θ = \cos θ$ and $RHS = \frac{\sin 2 θ}{2 \sin θ} = \frac{2 \sin θ \cos θ}{2 \sin θ} = \cos θ = LHS$

so $P (1)$ is true R1

assume $P (k)$ is true, i.e. $Σ_{r = 1}^{k} \cos (2 r - 1) θ = \frac{\sin 2 k θ}{2 \sin θ} (k \in ℤ^{+})$ M1

Note: Award M0 for statements such as “let $n = k$ ”.

Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

considering $P (k + 1)$

$Σ_{r = 1}^{k + 1} \cos (2 r - 1) θ = Σ_{r = 1}^{k} \cos (2 r - 1) θ + \cos (2 (k + 1) - 1) θ$ M1

$= \frac{\sin 2 k θ}{2 \sin θ} + \cos (2 (k + 1) - 1) θ$ A1

$= \frac{\sin 2 k θ + 2 \cos ((2 k + 1) θ) \sin θ}{2 \sin θ}$

$= \frac{\sin 2 k θ + \sin ((2 k + 1) θ + θ) - \sin ((2 k + 1) θ - θ)}{2 \sin θ}$ M1

Note: Award M1 for use of $2 \cos A \sin B = \sin (A + B) - \sin (A - B)$ with $A = (2 k + 1) θ$ and $B = θ$ .

$= \frac{\sin 2 k θ + \sin (2 k + 2) θ - \sin 2 k θ}{2 \sin θ}$ A1

$= \frac{\sin 2 (k + 1) θ}{2 \sin θ}$ A1

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

Note: Award the final R1 mark provided at least five of the previous marks have been awarded.

[8 marks]

Examiners report

[N/A]

Consider the equation $\frac{2 z}{3 - z *} = i$ , where $z = x + i y$ and $x, y \in ℝ$ .

Find the value of $x$ and the value of $y$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

substituting $z = x + i y$ and $z * = x - i y$ M1

$\frac{2 (x + i y)}{3 - (x - i y)} = i$

$2 x + 2 i y = - y + i (3 - x)$

equate real and imaginary: M1

$y = - 2 x$ AND $2 y = 3 - x$ A1

Note: If they multiply top and bottom by the conjugate, the equations $6 x - 2 x^{2} + 2 y^{2} = 0$ and $6 y - 4 x y = {(3 - x)}^{2} + y^{2}$ may be seen. Allow for A1.

solving simultaneously:

$x = - 1, y = 2 (z = - 1 + 2 i)$ A1A1

[5 marks]

Examiners report

[N/A]

The following diagram shows the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ , with asymptotes at $y = - \frac{π}{4}$ and $y = \frac{3 π}{4}$ .

Describe a sequence of transformations that transforms the graph of $y = arctan x$ to the graph of $y = arctan (2 x + 1) + \frac{π}{4}$ for $x \in ℝ$ .

[3]

Show that $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ .

[4]

Verify that $arctan (2 x + 1) = arctan (\frac{x}{x + 1}) + \frac{π}{4}$ for $x \in ℝ, x > 0$ .

[3]

Using mathematical induction and the result from part (b), prove that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$ .

[9]

Markscheme

EITHER
horizontal stretch/scaling with scale factor $\frac{1}{2}$

Note: Do not allow ‘shrink’ or ‘compression’

followed by a horizontal translation/shift $\frac{1}{2}$ units to the left A2

Note: Do not allow ‘move’

horizontal translation/shift $1$ unit to the left

followed by horizontal stretch/scaling with scale factor $\frac{1}{2}$ A2

THEN

vertical translation/shift up by $\frac{π}{4}$ (or translation through $(\begin{matrix} 0 \\ \frac{π}{4} \end{matrix})$ A1
(may be seen anywhere)

[3 marks]

let $α = arctan p$ and $β = arctan q$ M1

$p = tan α$ and $q = tan β$ (A1)

$\tan (α + β) = \frac{p + q}{1 - p q}$ A1

$α + β = arctan (\frac{p + q}{1 - p q})$ A1

so $arctan p + arctan q \equiv arctan (\frac{p + q}{1 - p q})$ where $p, q > 0$ and $p q < 1$ . AG

[4 marks]

METHOD 1

$\frac{π}{4} = arctan 1$ (or equivalent) A1

$arctan (\frac{x}{x + 1}) + arctan 1 = arctan (\frac{\frac{x}{x + 1} + 1}{1 - \frac{x}{x + 1} (1)})$ A1

$= arctan (\frac{\frac{x + x + 1}{x + 1}}{\frac{x + 1 - x}{x + 1}})$ A1

$= arctan (2 x + 1)$ AG

METHOD 2

$\tan \frac{π}{4} = 1$ (or equivalent) A1

Consider $arctan (2 x + 1) - arctan (\frac{x}{x + 1}) = \frac{π}{4}$

$\tan (arctan (2 x + 1) - arctan (\frac{x}{x + 1}))$

$= arctan (\frac{2 x +1 - \frac{x}{x + 1}}{1 + \frac{x (2 x + 1)}{x + 1}})$ A1

$= arctan (\frac{(2 x +1) (x + 1) - x}{x + 1 + x (2 x + 1)})$ A1

$= arctan 1$ AG

METHOD 3

$tan (arctan (2 x + 1)) = \tan (arctan (\frac{x}{x + 1}) + \frac{π}{4})$

$\tan \frac{π}{4} = 1$ (or equivalent) A1

$LHS = 2 x + 1$ A1

$RHS = \frac{\frac{x}{x + 1} + 1}{1 - \frac{x}{x + 1}} (= 2 x + 1)$ A1

[3 marks]

let $P (n)$ be the proposition that $Σ_{r = 1}^{n} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{n}{n + 1})$ for $n \in ℤ^{+}$

consider $P (1)$

when $n = 1, Σ_{r = 1}^{1} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{1}{2}) = RHS$ and so $P (1)$ is true R1

assume $P (k)$ is true, ie. $Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) = arctan (\frac{k}{k + 1}) (k \in ℤ^{+})$ M1

Note: Award M0 for statements such as “let $n = k$ ”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

consider $P (k + 1)$ :

$Σ_{r = 1}^{k + 1} arctan (\frac{1}{2 r^{2}}) = Σ_{r = 1}^{k} arctan (\frac{1}{2 r^{2}}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ (M1)

$= arctan (\frac{k}{k + 1}) + arctan (\frac{1}{2 {(k + 1)}^{2}})$ A1

$= arctan (\frac{\frac{k}{k + 1} + \frac{1}{2 {(k + 1)}^{2}}}{1 - (\frac{k}{k + 1}) (\frac{1}{2 {(k + 1)}^{2}})})$ M1

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{2 {(k + 1)}^{3} - k})$ A1

Note: Award A1 for correct numerator, with $(k + 1)$ factored. Denominator does not need to be simplified

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

Note: Award A1 for denominator correctly expanded. Numerator does not need to be simplified. These two A marks may be awarded in any order

$= arctan (\frac{(k + 1) (2 k^{2} + 2 k + 1)}{(k + 2) (2 k^{2} + 2 k + 1)}) = arctan (\frac{k + 1}{k + 2})$ A1

Note: The word ‘arctan’ must be present to be able to award the last three A marks

$P (k + 1)$ is true whenever $P (k)$ is true and $P (1)$ is true, so

$P (n)$ is true for for $n \in ℤ^{+}$ R1

Note: Award the final R1 mark provided at least four of the previous marks have been awarded.
Note: To award the final R1, the truth of $P (k)$ must be mentioned. ‘ $P (k)$ implies $P (k + 1)$ ’ is insufficient to award the mark.

[9 marks]

Examiners report

[N/A]

Use mathematical induction to prove that $\sum\limits_{r = 1}^n {r\left( {r{\text{!}}} \right)} = \left( {n + 1} \right){\text{!}} - 1$ , for $n \in {\mathbb{Z}^ + }$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

consider $n = 1$ . $1\left( {1{\text{!}}} \right) = 1$ and $2{\text{!}} - 1 = 1$ therefore true for $n = 1$ R1

Note: There must be evidence that $n = 1$ has been substituted into both expressions, or an expression such LHS=RHS=1 is used. “therefore true for $n = 1$ ” or an equivalent statement must be seen.

assume true for $n = k$ , (so that $\sum\limits_{r = 1}^k {r\left( {r{\text{!}}} \right)} = \left( {k + 1} \right){\text{!}} - 1$ ) M1

Note: Assumption of truth must be present.

consider $n = k + 1$

$\sum\limits_{r = 1}^{k + 1} {r\left( {r{\text{!}}} \right)} = \sum\limits_{r = 1}^k {r\left( {r{\text{!}}} \right)} + \left( {k + 1} \right)\left( {k + 1} \right){\text{!}}$ (M1)

${\text{ = }}\left( {k + 1} \right){\text{!}} - 1 + \left( {k + 1} \right)\left( {k + 1} \right){\text{!}}$ A1

${\text{ = }}\left( {k + 2} \right)\left( {k + 1} \right){\text{!}} - 1$ M1

Note: M1 is for factorising $\left( {k + 1} \right){\text{!}}$

${\text{ = }}\left( {k + 2} \right){\text{!}} - 1$

$= \left( {\left( {k + 1} \right) + 1} \right){\text{!}} - 1$

so if true for $n = k$ , then also true for $n = k + 1$ , and as true for $n = 1$ then true for all $n$ $\left( { \in {\mathbb{Z}^ + }} \right)$ R1

Note: Only award final R1 if all three method marks have been awarded.
Award R0 if the proof is developed from both LHS and RHS.

[6 marks]

Examiners report

[N/A]

Consider the equation ${z^4} + a{z^3} + b{z^2} + cz + d = 0$ , where $a$ , $b$ , $c$ , $d \in \mathbb{R}$ and $z \in \mathbb{C}$ .

Two of the roots of the equation are log₂6 and $i\sqrt 3$ and the sum of all the roots is 3 + log₂3.

Show that 6 $a$ + $d$ + 12 = 0.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$- i\sqrt 3$ is a root (A1)

$3 + {\text{lo}}{{\text{g}}_2}3 - {\text{lo}}{{\text{g}}_2}6\left( { = 3 + {\text{lo}}{{\text{g}}_2}\frac{1}{2} = 3 - 1 = 2} \right)$ is a root (A1)

sum of roots: $- a = 3 + {\text{lo}}{{\text{g}}_2}3 \Rightarrow a = - 3 - {\text{lo}}{{\text{g}}_2}3$ M1

Note: Award M1 for use of $- a$ is equal to the sum of the roots, do not award if minus is missing.

Note: If expanding the factored form of the equation, award M1 for equating $a$ to the coefficient of ${z^3}$ .

product of roots: ${\left( { - 1} \right)^4}d$ $= 2\left( {{\text{lo}}{{\text{g}}_2}6} \right)\left( {i\sqrt 3 } \right)\left( { - i\sqrt 3 } \right)$ M1

$= 6\,{\text{lo}}{{\text{g}}_2}6$ A1

Note: Award M1A0 for $d = - 6\,{\text{lo}}{{\text{g}}_2}6$

$6a + d + 12 = - 18 - 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}6 + 12$

EITHER

$= - 6 + 6\,{\text{lo}}{{\text{g}}_2}2 = 0$ M1A1AG

Note: M1 is for a correct use of one of the log laws.

$= - 6 - 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}3 + 6\,{\text{lo}}{{\text{g}}_2}2 = 0$ M1A1AG

Note: M1 is for a correct use of one of the log laws.

[7 marks]

Examiners report

[N/A]

Consider two events $A$ and $A$ defined in the same sample space.

Given that ${\text{P}}(A \cup B) = \frac{4}{9},{\text{ P}}(B|A) = \frac{1}{3}$ and ${\text{P}}(B|A') = \frac{1}{6}$ ,

Show that ${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A' \cap B)$ .

[3]

(i) show that ${\text{P}}(A) = \frac{1}{3}$ ;

(ii) hence find ${\text{P}}(B)$ .

[6]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)$ M1

$= {\text{P}}(A) + {\text{P}}(A \cap B) + {\text{P}}(A' \cap B) - {\text{P}}(A \cap B)$ M1A1

$= {\text{P}}(A) + {\text{P}}(A' \cap B)$ AG

METHOD 2

${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)$ M1

$= {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A|B) \times {\text{P}}(B)$ M1

$= {\text{P}}(A) + \left( {1 - {\text{P}}(A|B)} \right) \times {\text{P}}(B)$

$= {\text{P}}(A) + {\text{P}}(A'|B) \times {\text{P}}(B)$ A1

$= {\text{P}}(A) + {\text{P}}(A' \cap B)$ AG

[3 marks]

(i) use ${\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A' \cap B)$ and ${\text{P}}(A' \cap B) = {\text{P}}(B|A'){\text{P}}(A')$ (M1)

$\frac{4}{9} = {\text{P}}(A) + \frac{1}{6}\left( {1 - {\text{P}}(A)} \right)$ A1

$8 = 18{\text{P}}(A) + 3\left( {1 - {\text{P}}(A)} \right)$ M1

${\text{P}}(A) = \frac{1}{3}$ AG

(ii) METHOD 1

${\text{P}}(B) = {\text{P}}(A \cap B) + {\text{P}}(A' \cap B)$ M1

$= {\text{P}}(B|A){\text{P}}(A) + {\text{P}}(B|A'){\text{P}}(A')$ M1

$= \frac{1}{3} \times \frac{1}{3} + \frac{1}{6} \times \frac{2}{3} = \frac{2}{9}$ A1

METHOD 2

${\text{P}}(A \cap B) = {\text{P}}(B|A){\text{P}}(A) \Rightarrow {\text{P}}(A \cap B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ M1

${\text{P}}(B) = {\text{P}}(A \cup B) + {\text{P}}(A \cap B) - {\text{P}}(A)$ M1

${\text{P}}(B) = \frac{4}{9} + \frac{1}{9} - \frac{1}{3} = \frac{2}{9}$ A1

[6 marks]

Examiners report

[N/A]

The function $f$ is defined by $f\left( x \right) = \frac{{ax + b}}{{cx + d}}$ , for $x \in \mathbb{R},\,\,x \ne - \frac{d}{c}$ .

The function $g$ is defined by $g\left( x \right) = \frac{{2x - 3}}{{x - 2}},\,\,x \in \mathbb{R},\,\,x \ne 2$

Express $g\left( x \right)$ in the form $A + \frac{B}{{x - 2}}$ where A, B are constants.

Markscheme

$g\left( x \right) = 2 + \frac{1}{{x - 2}}$ A1A1

[2 marks]

Examiners report

[N/A]

Solve the equation $\log_{3} \sqrt{x} = \frac{1}{2 \log_{2} 3} + \log_{3} (4 x^{3})$ , where $x > 0$ .

Markscheme

attempt to use change the base (M1)

$\log_{3} \sqrt{x} = \frac{\log_{3} 2}{2} + \log_{3} (4 x^{3})$

attempt to use the power rule (M1)

$\log_{3} \sqrt{x} = \log_{3} \sqrt{2} + \log_{3} (4 x^{3})$

attempt to use product or quotient rule for logs, $\ln a + \ln b = \ln a b$ (M1)

$\log_{3} \sqrt{x} = \log_{3} (4 \sqrt{2} x^{3})$

Note: The M marks are for attempting to use the relevant log rule and may be applied in any order and at any time during the attempt seen.

$\sqrt{x} = 4 \sqrt{2} x^{3}$

$x = 32 x^{6}$

$x^{5} = \frac{1}{32}$ (A1)

$x = \frac{1}{2}$ A1

[5 marks]

Examiners report

[N/A]

Consider the expansion of ${(8 x^{3} - \frac{1}{2 x})}^{n}$ where $n \in ℤ^{+}$ . Determine all possible values of $n$ for which the expansion has a non-zero constant term.

Markscheme

EITHER

attempt to obtain the general term of the expansion

$T_{r + 1} = {}_{n}C_{r} {(8 x^{3})}^{n - r} {(- \frac{1}{2 x})}^{r}$ OR $T_{r + 1} = {}_{n}C_{n - r} {(8 x^{3})}^{r} {(- \frac{1}{2 x})}^{n - r}$ (M1)

recognize power of $x$ starts at $3 n$ and goes down by $4$ each time (M1)

THEN

recognizing the constant term when the power of $x$ is zero (or equivalent) (M1)

$r = \frac{3 n}{4}$ or $n = \frac{4}{3} r$ or $3 n - 4 r = 0$ OR $3 r - (n - r) = 0$ (or equivalent) A1

$r$ is a multiple of $3 (r = 3, 6, 9, \dots)$ or one correct value of $n$ (seen anywhere) (A1)

$n = 4 k, k \in ℤ^{+}$ A1

Note: Accept $n$ is a (positive) multiple of $4$ or $n = 4, 8, 12, \dots$
Do not accept $n = 4, 8, 12$

Note: Award full marks for a correct answer using trial and error approach
showing $n = 4, 8, 12, \dots$ and for recognizing that this pattern continues.

[5 marks]

Examiners report

There was a mixed response to this question. Candidates who used a trial and error approach were more successful in obtaining completely correct answers than those who tried to solve algebraically by finding the general term to form an equation relating n and r . Poor explanations were often noted in the trial and error approach. Candidates often failed to make progress after obtaining $n = \frac{4}{3} r$ in the algebraic approach. Some candidates did not attempt this question.

Let $f (x) = \sqrt{1 + x}$ for $x > - 1$ .

Show that $f'' (x) = - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ .

[3]

Use mathematical induction to prove that $f^{(n)} (x) = {(- \frac{1}{4})}^{n - 1} \frac{(2 n - 3)!}{(n - 2)!} {(1 + x)}^{\frac{1}{2} - n}$ for $n \in ℤ, n \geq 2$ .

[9]

Let $g (x) = e^{m x}, m \in ℚ$ .

Consider the function $h$ defined by $h (x) = f (x) \times g (x)$ for $x > - 1$ .

It is given that the $x^{2}$ term in the Maclaurin series for $h (x)$ has a coefficient of $\frac{7}{4}$ .

Find the possible values of $m$ .

[8]

Markscheme

attempt to use the chain rule M1

$f' (x) = \frac{1}{2} {(1 + x)}^{- \frac{1}{2}}$ A1

$f'' (x) = - \frac{1}{4} {(1 + x)}^{- \frac{3}{2}}$ A1

$= - \frac{1}{4 \sqrt{{(1 + x)}^{3}}}$ AG

Note: Award M1A0A0 for $f' (x) = \frac{1}{\sqrt{1 + x}}$ or equivalent seen

[3 marks]

let $n = 2$

$f^{''} (x) = (- \frac{1}{4 \sqrt{{(1 + x)}^{3}}} =) {(- \frac{1}{4})}^{1} \frac{1!}{0!} {(1 + x)}^{\frac{1}{2} - 2}$ R1

Note: Award R0 for not starting at $n = 2$ . Award subsequent marks as appropriate.

assume true for $n = k$ , (so $f^{(k)} (x) = {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k}$ ) M1

Note: Do not award M1 for statements such as “let $n = k$ ” or “ $n = k$ is true”. Subsequent marks can still be awarded.

consider $n = k + 1$

$LHS = f^{(k + 1)} (x) = \frac{d (f^{(k)} (x))}{d x}$ M1

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1}{2} - k) {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

EITHER

$RHS = f^{(k + 1)} (x) = {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ (or equivalent) A1

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for $\frac{(2 k - 1)!}{(k - 1)!} = \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} (= \frac{2 (2 k - 1) (2 k - 3)!}{(k - 2)!})$

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 2) (2 k - 3)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$(= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1})$

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= (\frac{1}{2} - k) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: The following A marks can be awarded in any order.

$= {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 3)!}{(k - 2)!} (\frac{1 - 2 k}{2}) {(1 + x)}^{\frac{1}{2} - k - 1}$

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1) (2 k - 3)!}{(k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for isolating $(2 k - 1)$ correctly.

$= (- \frac{1}{2}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(2 k - 2) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for multiplying top and bottom by $(k - 1)$ or $2 (k - 1)$ .

$= (- \frac{1}{4}) {(- \frac{1}{4})}^{k - 1} \frac{(2 k - 1)!}{(k - 1) (k - 2)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

Note: Award A1 for leading coefficient of $- \frac{1}{4}$ .

$= {(- \frac{1}{4})}^{k} \frac{(2 k - 1)!}{(k - 1)!} {(1 + x)}^{\frac{1}{2} - k - 1}$ A1

$= {(- \frac{1}{4})}^{(k + 1) - 1} \frac{(2 (k + 1) - 3)!}{((k + 1) - 2)!} {(1 + x)}^{\frac{1}{2} - (k + 1)} = RHS$

THEN

since true for $n = 2$ , and true for $n = k + 1$ if true for $n = k$ , the statement is true for all, $n \in ℤ, n \geq 2$ by mathematical induction R1

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

[9 marks]

METHOD 1

$h (x) = \sqrt{1 + x} e^{m x}$

using product rule to find $h' (x)$ (M1)

$h' (x) = \sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}$ A1

$h'' (x) = m (\sqrt{1 + x} m e^{m x} + \frac{1}{2 \sqrt{1 + x}} e^{m x}) + \frac{1}{2 \sqrt{1 + x}} m e^{m x} - \frac{1}{4 \sqrt{{(1 + x)}^{3}}} e^{m x}$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = m^{2} + \frac{1}{2} m + \frac{1}{2} m - \frac{1}{4} (= m^{2} + m - \frac{1}{4})$ A1

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 2

EITHER

attempt to find $f (0), f' (0), f'' (0)$ (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} f (0) = 1$

$f' (x) = \frac{1}{2} {(1 + x)}^{- \frac{1}{2}} f' (0) = \frac{1}{2}$

$f'' (x) = - \frac{1}{4} {(1 + x)}^{- \frac{3}{2}} f'' (0) = - \frac{1}{4}$

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

attempt to apply binomial theorem for rational exponents (M1)

$f (x) = {(1 + x)}^{\frac{1}{2}} = 1 + \frac{1}{2} x + \frac{(\frac{1}{2}) (- \frac{1}{2})}{2!} x^{2} \dots$

$f (x) = 1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots$ A1

THEN

$g (x) = 1 + m x + \frac{m^{2}}{2} x^{2} + \dots$ (A1)

$h (x) = (1 + \frac{1}{2} x - \frac{1}{8} x^{2} + \dots) (1 + m x + \frac{m^{2}}{2} x^{2} + \dots)$ (M1)

coefficient of $x^{2}$ is $\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8}$ A1

attempt to set equal to $\frac{7}{4}$ and solve M1

$\frac{m^{2}}{2} + \frac{m}{2} - \frac{1}{8} = \frac{7}{4}$

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

METHOD 3

$g' (x) = m e^{m x}$ and $g'' (x) = m^{2} e^{m x}$ (A1)

$h (x) = h (0) + x h' (0) + \frac{x^{2}}{2!} h'' (0) + \dots$

equating $x^{2}$ coefficient to $\frac{7}{4}$ M1

$\frac{h'' (0)}{2!} = \frac{7}{4} (\Rightarrow h'' (0) = \frac{7}{2})$

using product rule to find $h' (x)$ and $h'' (x)$ (M1)

$h' (x) = f (x) g' (x) + f' (x) g (x)$

$h'' (x) = f (x) g'' (x) + 2 f' (x) g' (x) + f'' (x) g (x)$ A1

substituting $x = 0$ into $h'' (x)$ M1

$h'' (0) = f (0) g'' (0) + 2 g' (0) f' (0) + g (0) f'' (0)$

$= 1 \times m^{2} + 2 m \times \frac{1}{2} + 1 \times (- \frac{1}{4}) (= m^{2} + m - \frac{1}{4})$ A1

$4 m^{2} + 4 m - 15 = 0$ A1

$(2 m + 5) (2 m - 3) = 0$

$m = - \frac{5}{2}$ or $m = \frac{3}{2}$ A1

[8 marks]

Examiners report

[N/A]