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HL Paper 2

Consider the identity $\frac{2 + 7 x}{(1 + 2 x) (1 - x)} \equiv \frac{A}{1 + 2 x} + \frac{B}{1 - x}$ , where $A, B \in ℤ$ .

Find the value of $A$ and the value of $B$ .

[3]

Hence, expand $\frac{2 + 7 x}{(1 + 2 x) (1 - x)}$ in ascending powers of $x$ , up to and including the term in $x^{2}$ .

[4]

Give a reason why the series expansion found in part (b) is not valid for $x = \frac{3}{4}$ .

[1]

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.

$2 + 7 x \equiv A (1 - x) + B (1 + 2 x)$

EITHER

substitutes $x = 1$ and attempts to solve for $B$ and substitutes $x = - \frac{1}{2}$ and attempts to solve for $A$ (M1)

$9 = 3 B \Rightarrow B = 3; \frac{3 A}{2} = - \frac{3}{2} \Rightarrow A = - 1$

forms $A + B = 2$ and $- A + 2 B = 7$ and attempts to solve for $A$ and $B$ (M1)

THEN

$A = - 1$ and $B = 3$ A1A1

[3 marks]

uses the binomial expansion on either $3 {(1 - x)}^{- 1}$ or ${(1 + 2 x)}^{- 1}$ M1

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

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

$3 + 3 x + 3 x^{2} - (1 - 2 x + 4 x^{2})$

so the expansion is $2 + 5 x - x^{2}$ (in ascending powers of $x$ ) A1

[4 marks]

${(1 + 2 x)}^{- 1}$ (is convergent) requires $|x| < \frac{1}{2}$ and $x = \frac{3}{4}$ is outside this so the expansion is not valid R1

[1 mark]

Examiners report

[N/A]

Use mathematical induction to prove that $\frac{d^{n}}{d x^{n}} (x e^{p x}) = p^{n - 1} (p x + n) e^{p x}$ for $n \in ℤ^{+}, p \in ℚ$ .

Markscheme

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

$n = 1 : LHS = \frac{d (x e^{p x})}{d x} = x p e^{p x} + e^{p x} = (p x + 1) e^{p x}, RHS = p^{0} (p x + 1) e^{p x}$

$LHS = RHS$ so true for $n = 1 :$ A1

Note: Award A1 if $n = 0$ is proved.

assume proposition true for $n = k$ , i.e. $\frac{d^{k}}{d x^{k}} (x e^{p x}) = p^{k - 1} (p x + k) e^{p x}$ M1

Notes: Do not award M1 if using $n$ instead of $k$ .
Assumption of truth must be present.
Subsequent marks are not dependent on this M1 mark.

$\frac{d^{k + 1}}{d x^{k + 1}} (x e^{p x}) = \frac{d}{d x} (\frac{d^{k}}{d x^{k}} (x e^{p x}))$ (M1)

$= \frac{d}{d x} (p^{k - 1} (p x + k) e^{p x})$ M1

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

$= p^{k} (p x + k) e^{p x} + e^{p x} (p^{k})$ A1

Note: Award A1 for correct derivative.

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

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

Note: The final A1 can be awarded for either of the two lines above.

hence true for $n = 1$ and $n = k$ true $\Rightarrow n = k + 1$ true R1

therefore true for all $n \in ℤ^{+}$

Note: Only award the final R1 if the three method marks have been awarded.

[7 marks]

Examiners report

[N/A]

At a gathering of $12$ teachers, seven are male and five are female. A group of five of these teachers go out for a meal together. Determine the possible number of groups in each of the following situations:

There are more males than females in the group.

[4]

Two of the teachers, Gary and Gerwyn, refuse to go out for a meal together.

[3]

Markscheme

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

identifying two or three possible cases (M1)

total number of possible groups is $(\begin{matrix} 7 \\ 5 \end{matrix}) + (\begin{matrix} 7 \\ 4 \end{matrix}) (\begin{matrix} 5 \\ 1 \end{matrix}) + (\begin{matrix} 7 \\ 3 \end{matrix}) (\begin{matrix} 5 \\ 2 \end{matrix})$ (A1)(A1)

Note: Award A1 for any two correct cases, A1 for the other one.

$= 21 + (35 \times 5) + (35 \times 10)$

$= 546$ A1

[4 marks]

METHOD 1

identifying at least two of the three possible cases- Gary goes, Gerwyn goes or neither goes (M1)

total number of possible groups is $(\begin{matrix} 10 \\ 5 \end{matrix}) + (\begin{matrix} 10 \\ 4 \end{matrix}) + (\begin{matrix} 10 \\ 4 \end{matrix})$ (A1)

$= 252 + 210 + 210$

$= 672$ A1

METHOD 2

identifying the overall number of groups and no. of cases where both Gary and Gerwyn go. (M1)

total number of possible groups is $(\begin{matrix} 12 \\ 5 \end{matrix}) - (\begin{matrix} 10 \\ 3 \end{matrix})$ (A1)

$= 792 - 120$

$= 672$ A1

[3 marks]

Examiners report

[N/A]

Consider the complex numbers $z = 2 (\cos \frac{π}{5} + i \sin \frac{π}{5})$ and $w = 8 (\cos \frac{2 k π}{5} - i \sin \frac{2 k π}{5})$ , where $k \in ℤ^{+}$ .

Suppose that $z w \in ℤ$ .

Find the modulus of $z w$ .

[1]

Find the argument of $z w$ in terms of $k$ .

[2]

Find the minimum value of $k$ .

[3]

c.i.

For the value of $k$ found in part (i), find the value of $z w$ .

[1]

c.ii.

Markscheme

$(|z w| =) 16$ A1

[1 mark]

attempt to find $arg (z) + arg (w)$ (M1)

$arg (z w) = arg (z) + arg (w)$

$= \frac{π}{5} - \frac{2 k π}{5} (= \frac{(1 - 2 k) π}{5})$ A1

[2 marks]

$z w \in ℤ \Rightarrow arg (z w)$ is a multiple of $π$ (M1)

$\Rightarrow 1 - 2 k$ is a multiple of $5$ (M1)

$k = 3$ A1

[3 marks]

c.i.

$z w = 16 (\cos (- π) + i \sin (- π))$

$- 16$ A1

[1 mark]

c.ii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

Solve the inequality ${x^2} > 2x + 1$ .

[2]

Use mathematical induction to prove that ${2^{n + 1}} > {n^2}$ for $n \in \mathbb{Z}$ , $n \geqslant 3$ .

[7]

Markscheme

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

$x < - 0.414,\,\,x > 2.41$ A1A1

$\left( {x < 1 - \sqrt 2 ,\,\,x > 1 + \sqrt 2 } \right)$

Note: Award A1 for −0.414, 2.41 and A1 for correct inequalities.

[2 marks]

check for $n = 3$ ,

16 > 9 so true when $n = 3$ A1

assume true for $n = k$

${2^{k + 1}} > {k^2}$ 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.

prove true for $n = k + 1$

${2^{k + 2}} = 2 \times {2^{k + 1}}$

$> 2{k^2}$ M1

$= {k^2} + {k^2}$ (M1)

$> {k^2} + 2k + 1$ (from part (a)) A1

which is true for $k$ ≥ 3 R1

Note: Only award the A1 or the R1 if it is clear why. Alternate methods are possible.

$= {\left( {K + 1} \right)^2}$

hence if true for $n = k$ true for $n = k +1$ , true for $n = 3$ so true for all $n$ ≥ 3 R1

Note: Only award the final R1 provided at least three of the previous marks are awarded.

[7 marks]

Examiners report

[N/A]

Consider the set of six-digit positive integers that can be formed from the digits $0, 1, 2, 3, 4, 5, 6, 7, 8$ and $9$ .

Find the total number of six-digit positive integers that can be formed such that

the digits are distinct.

[2]

the digits are distinct and are in increasing order.

[2]

Markscheme

$9 \times 9 \times 8 \times 7 \times 6 \times 5 (= 9 \times {}_{9}P_{5})$ (M1)

$= 136080 (= 9 \times \frac{9!}{4!})$ A1

Note: Award M1A0 for $10 \times 9 \times 8 \times 7 \times 6 \times 5 (= {}_{10}P_{6} = 151200 = \frac{10!}{4!})$ .

Note: Award M1A0 for ${}_{9}P_{6} = 60480$

[2 marks]

METHOD 1

EITHER

every unordered subset of $6$ digits from the set of $9$ non-zero digits can be arranged in exactly one way into a $6$ -digit number with the digits in increasing order. A1

${}^{9}C_{6} (\times 1)$ A1

THEN

$= 84$ A1

METHOD 2

EITHER

removes $3$ digits from the set of $9$ non-zero digits and these $6$ remaining digits can be arranged in exactly one way into a $6$ -digit number with the digits in increasing order. A1

${}^{9}C_{3} (\times 1)$ A1

THEN

$= 84$ A1

[2 marks]

Examiners report

Part (a) A number of candidates got the correct answer here with the valid approach and recognising that zero could not occupy the first position. Some lost a mark by including zero. Many candidates used an incorrect method with combinations or simplified permutations.

Part (b) Only very few candidates got the correct answer. Many left it blank or provided unreasonably enormous numbers as their answers.

Some candidates had the answer to part (b) showing in part (a).

A small number of candidates tried to list all possibilities but mostly unsuccessfully.

[N/A]

Consider the function $f (x) = \frac{x^{2} - x - 12}{2 x - 15}, x \in ℝ, x \neq \frac{15}{2}$ .

Find the coordinates where the graph of $f$ crosses the

$x$ -axis.

[2]

a.i.

$y$ -axis.

[1]

a.ii.

Write down the equation of the vertical asymptote of the graph of $f$ .

[1]

The oblique asymptote of the graph of $f$ can be written as $y = a x + b$ where $a, b \in ℚ$ .

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

[4]

Sketch the graph of $f$ for $- 30 \leq x \leq 30$ , clearly indicating the points of intersection with each axis and any asymptotes.

[3]

Express $\frac{1}{f (x)}$ in partial fractions.

[3]

e.i.

Hence find the exact value of $\int_{0}^{3} \frac{1}{f (x)} d x$ , expressing your answer as a single logarithm.

[4]

e.ii.

Markscheme

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

attempts to solve $x^{2} - x - 12 = 0$ (M1)

$(- 3, 0)$ and $(4, 0)$ A1

[2 marks]

a.i.

Note: In part (a), penalise once only, if correct values are given instead of correct coordinates.

$(0, \frac{4}{5})$ A1

[1 mark]

a.ii.

$x = \frac{15}{2}$ A1

Note: Award A0 for $x \neq \frac{15}{2}$ .
Award A1 in part (b), if $x = \frac{15}{2}$ is seen on their graph in part (d).

[1 mark]

METHOD 1

$(a x + b) (2 x - 15) \equiv x^{2} - x - 12$

attempts to expand $(a x + b) (2 x - 15)$ (M1)

$2 a x^{2} - 15 a x + 2 b x - 15 b \equiv x^{2} - x - 12$

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

equates coefficients of $x$ (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 2

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x}{2} + \frac{13}{4} + \dots$ M1

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

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 3

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

$\frac{x^{2} - x - 12}{2 x - 15} \equiv \frac{x}{2} + b + \frac{c}{2 x - 15}$ M1

$x^{2} - x - 12 \equiv \frac{(2 x - 15) x}{2} + (2 x - 15) b + c$

equates coefficients of $x$ : (M1)

$- 1 = - \frac{15}{2} + 2 b$

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

METHOD 4

attempts division on $\frac{x^{2} - x - 12}{2 x - 15}$ M1

$\frac{x^{2} - x - 12}{2 x - 15} = \frac{x}{2} + \frac{\frac{13 x}{2} - 12}{2 x - 15}$

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

$\frac{\frac{13 x}{2} - 12}{2 x - 15} = \frac{13}{4} + \dots$ M1

$b = \frac{13}{4}$ A1

$(y = \frac{x}{2} + \frac{13}{4})$

[4 marks]

two branches with approximately correct shape (for $- 30 \leq x \leq 30$ ) A1

their vertical and oblique asymptotes in approximately correct positions with both branches showing correct asymptotic behaviour to these asymptotes A1

their axes intercepts in approximately the correct positions A1

Note: Points of intersection with the axes and the equations of asymptotes are not required to be labelled.

[3 marks]

attempts to split into partial fractions: (M1)

$\frac{2 x - 15}{(x + 3) (x - 4)} \equiv \frac{A}{x + 3} + \frac{B}{x - 4}$

$2 x - 15 \equiv A (x - 4) + B (x + 3)$

$A = 3$ A1

$B = - 1$ A1

$(\frac{3}{x + 3} - \frac{1}{x - 4})$

[3 marks]

e.i.

$\int_{0}^{3} (\frac{3}{x + 3} - \frac{1}{x - 4}) d x$

attempts to integrate and obtains two terms involving ‘ln’ (M1)

$= {[3 \ln |x + 3| - \ln |x - 4|]}_{0}^{3}$ A1

$= 3 \ln 6 - \ln 1 - 3 \ln 3 + \ln 4$ A1

$= 3 \ln 2 + \ln 4 (= \ln 8 + \ln 4)$

$= \ln 32 (= 5 \ln 2)$ A1

Note: The final A1 is dependent on the previous two A marks.

[4 marks]

e.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

e.i.

[N/A]

e.ii.

Given that ${\log _{10}}\left( {\frac{1}{{2\sqrt 2 }}\left( {p + 2q} \right)} \right) = \frac{1}{2}\left( {{{\log }_{10}}p + {{\log }_{10}}q} \right),{\text{ }}p > 0,{\text{ }}q > 0$ , find $p$ in terms of $q$ .

Markscheme

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

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = \frac{1}{2}({\log _{10}}p + {\log _{10}}q)$

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = \frac{1}{2}{\log _{10}}pq$ (M1)

${\log _{10}}\frac{1}{{2\sqrt 2 }}(p + 2q) = {\log _{10}}{(pq)^{\frac{1}{2}}}$ (M1)

$\frac{1}{{2\sqrt 2 }}(p + 2q) = {(pq)^{\frac{1}{2}}}$ (A1)

${(p + 2q)^2} = 8pq$

${p^2} + 4pq + 4{q^2} = 8pq$

${p^2} - 4pq + 4{q^2} = 0$

${(p - 2q)^2} = 0$ M1

hence $p = 2q$ A1

[5 marks]

Examiners report

[N/A]

Let $z = a + b{\text{i}}$ , $a$ , ${\text{b}} \in {\mathbb{R}^ + }$ and let ${\text{arg}}\,z = \theta$ .

Show the points represented by $z$ and $z - 2a$ on the following Argand diagram.

[1]

Find an expression in terms of θ for ${\text{arg}}\left( {z - 2a} \right)$ .

[1]

b.i.

Find an expression in terms of θ for ${\text{arg}}\left( {\frac{z}{{z - 2a}}} \right)$ .

[2]

b.ii.

Hence or otherwise find the value of θ for which ${\text{Re}}\left( {\frac{z}{{z - 2a}}} \right) = 0$ .

[3]

Markscheme

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

Note: Award A1 for $z$ in first quadrant and $z - 2a$ its reflection in the $y$ -axis.

[1 mark]

$\pi - \theta$ (or any equivalent) A1

[1 mark]

b.i.

${\text{arg}}\left( {\frac{z}{{z - 2a}}} \right) = {\text{arg}}\left( z \right) - {\text{arg}}\left( {z - 2a} \right)$ (M1)

$= 2\theta - \pi$ (or any equivalent) A1

[2 marks]

b.ii.

METHOD 1

if ${\text{Re}}\left( {\frac{z}{{z - 2a}}} \right) = 0$ then $2\theta - \pi = \frac{{n\pi }}{2}$ , ( $n$ odd) (M1)

$- \pi < 2\theta - \pi < 0 \Rightarrow n = - 1$

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

$\theta = \frac{\pi }{4}$ A1

METHOD 2

$\frac{{a + b{\text{i}}}}{{ - a + b{\text{i}}}} = \frac{{{b^2} - {a^2} - 2ab{\text{i}}}}{{{a^2} + {b^2}}}$ M1

${\text{Re}}\left( {\frac{z}{{z - 2a}}} \right) = 0 \Rightarrow {b^2} - {a^2} = 0$

$b = a$ A1

$\theta = \frac{\pi }{4}$ A1

Note: Accept any equivalent, eg $\theta = - \frac{{7\pi }}{4}$ .

[3 marks]

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

Solve ${z^2} = 4{{\text{e}}^{\frac{\pi }{2}{\text{i}}}}$ , giving your answers in the form

$r{{\text{e}}^{{\text{i}}\theta }}$ where $r$ , $\theta \in \mathbb{R}$ , $r > 0$ .

[3]

$a + {\text{i}}b$ where $a$ , $b \in \mathbb{R}$ .

[2]

Markscheme

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

$z = 2{{\text{e}}^{\frac{\pi }{4}{\text{i}}}}\,\left( { = 2{{\text{e}}^{0.785{\text{i}}}}} \right)$ A1

Note: Accept all answers in the form $2{{\text{e}}^{\left( {\frac{\pi }{4} + 2\pi n} \right){\text{i}}}}$ .

$z = 2{{\text{e}}^{\frac{{5\pi }}{4}{\text{i}}}}\,\left( { = 2{{\text{e}}^{3.93{\text{i}}}}} \right)$ OR $z = 2{{\text{e}}^{ - \frac{{3\pi }}{4}{\text{i}}}}\,\left( { = 2{{\text{e}}^{ - 2.36{\text{i}}}}} \right)$ (M1)A1

Note: Accept all answers in the form $2{{\text{e}}^{\left( { - \frac{{3\pi }}{4} + 2\pi n} \right){\text{i}}}}$ .

Note: Award M1A0 for correct answers in the incorrect form, eg $- 2{{\text{e}}^{\frac{\pi }{4}{\text{i}}}}$ .

[3 marks]

$z = 1.41 + 1.41{\text{i}}$ , $z = - 1.41 - 1.41{\text{i}}$ A1A1

$\left( {z = \sqrt 2 + \sqrt 2 {\text{i}},\,\,z = - \sqrt 2 - \sqrt 2 {\text{i}}} \right)$

[2 marks]

Examiners report

[N/A]

Consider the differential equation $x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2}$ for $x > 0$ and $y > 2 x$ . It is given that $y = 3$ when $x = 1$ .

Use Euler’s method, with a step length of $0.1$ , to find an approximate value of $y$ when $x = 1.5$ .

[4]

Use the substitution $y = v x$ to show that $x \frac{d v}{d x} = v^{2} - v - 2$ .

[3]

By solving the differential equation, show that $y = \frac{8 x + x^{4}}{4 - x^{3}}$ .

[10]

c.i.

Find the actual value of $y$ when $x = 1.5$ .

[1]

c.ii.

Using the graph of $y = \frac{8 x + x^{4}}{4 - x^{3}}$ , suggest a reason why the approximation given by Euler’s method in part (a) is not a good estimate to the actual value of $y$ at $x = 1.5$ .

[1]

c.iii.

Markscheme

attempt to use Euler’s method (M1)

$x_{n + 1} = x_{n} + 0.1; y_{n + 1} = y_{n} + 0.1 \times \frac{d y}{d x}$ , where $\frac{d y}{d x} = \frac{y^{2} - 2 x^{2}}{x^{2}}$

correct intermediate $y$ -values (A1)(A1)

$3.7, 4.63140 \dots, 5.92098, 7.79542 \dots$

Note: A1 for any two correct $y$ -values seen

$y = 10.6958 \dots$

$y = 10.7$ A1

Note: For the final A1, the value $10.7$ must be the last value in a table or a list, or be given as a final answer, not just embedded in a table which has further lines.

[4 marks]

$y = v x \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$ (A1)

replacing $y$ with $v x$ and $\frac{d y}{d x}$ with $v + x \frac{d v}{d x}$ M1

$x^{2} \frac{d y}{d x} = y^{2} - 2 x^{2} \Rightarrow x^{2} (v + x \frac{d v}{d x}) = v^{2} x^{2} - 2 x^{2}$ A1

$v + x \frac{d v}{d x} = v^{2} - 2$ (since $x > 0$ )

$x \frac{d v}{d x} = v^{2} - v - 2$ AG

[3 marks]

attempt to separate variables $v$ and $x$ (M1)

$\int \frac{d v}{v^{2} - v - 2} = \int \frac{d x}{x}$

$\int \frac{d v}{(v - 2) (v + 1)} = \int \frac{d x}{x}$ (A1)

attempt to express in partial fraction form M1

$\frac{1}{(v - 2) (v + 1)} \equiv \frac{A}{v - 2} + \frac{B}{v + 1}$

$\frac{1}{(v - 2) (v + 1)} = \frac{1}{3} (\frac{1}{v - 2} - \frac{1}{v + 1})$ A1

$\frac{1}{3} \int (\frac{1}{v - 2} - \frac{1}{v + 1}) d v = \int \frac{d x}{x}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| (+ c)$ A1

Note: Condone absence of modulus signs throughout.

EITHER

attempt to find $c$ using $x = 1, y = 3, v = 3$ M1

$c = \frac{1}{3} \ln \frac{1}{4}$

$\frac{1}{3} (\ln |v - 2| - \ln |v + 1|) = \ln |x| + \frac{1}{3} \ln \frac{1}{4}$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (\frac{{|x|}^{3}}{4})$

expressing both sides as a single logarithm (M1)

$\ln |\frac{v - 2}{v + 1}| = \ln (A {|x|}^{3})$

attempt to find $A$ using $x = 1, y = 3, v = 3$ M1

$A = \frac{1}{4}$

THEN

$|\frac{v - 2}{v + 1}| = \frac{1}{4} x^{3}$ (since $x > 0$ )

substitute $v = \frac{y}{x}$ (seen anywhere) M1

$\frac{\frac{y}{x} - 2}{\frac{y}{x} + 1} = \frac{1}{4} x^{3}$ (since $y > 2 x$ )

$(\Rightarrow \frac{y - 2 x}{y + x} = \frac{1}{4} x^{3})$

attempt to make $y$ the subject M1

$y - \frac{x^{3} y}{4} = 2 x + \frac{x^{4}}{4}$ A1

$y = \frac{8 x + x^{4}}{4 - x^{3}}$ AG

[10 marks]

c.i.

actual value at $y (1.5) = 27.3$ A1

[1 mark]

c.ii.

gradient changes rapidly (during the interval considered) OR

the curve has a vertical asymptote at $x = \sqrt[3]{4} (= 1.5874 \dots)$ R1

[1 mark]

c.iii.

Examiners report

Most candidates showed evidence of an attempt to use Euler's method in part a), although very few explicitly wrote down the formulae, they used in order to calculate successive y-value. In addition, many seemed to take a step-by-step approach rather than using the recursive capabilities of the graphical display calculator.

There were many good attempts at part b), but not all candidates recognised that this would help them to solve part c).

Part c) was done very well by many candidates, although there were a significant number who failed to recognise the need for partial fractions and could not make further progress. A common error was to integrate without a constant of integration, which meant that the initial condition could not be used. The reasoning given for the estimate being poor was often too vague and did not address the specific nature of the function given clearly enough.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Prove the identity ${(p + q)}^{3} - 3 p q (p + q) \equiv p^{3} + q^{3}$ .

[2]

The equation $2 x^{2} - 5 x + 1 = 0$ has two real roots, $α$ and $β$ .

Consider the equation $x^{2} + m x + n = 0$ , where $m, n \in ℤ$ and which has roots $\frac{1}{α^{3}}$ and $\frac{1}{β^{3}}$ .
Without solving $2 x^{2} - 5 x + 1 = 0$ , determine the values of $m$ and $n$ .

[6]

Markscheme

METHOD 1

${(p + q)}^{3} - 3 p q (p + q) \equiv p^{3} + q^{3}$

attempts to expand ${(p + q)}^{3}$ M1

$p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3}$

${(p + q)}^{3} - 3 p q (p + q) \equiv p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3} - 3 p q (p + q)$

$\equiv p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3} - 3 p^{2} q - 3 p q^{2}$ A1

$\equiv p^{3} + q^{3}$ AG

Note: Condone the use of equals signs throughout.

METHOD 2

${(p + q)}^{3} - 3 p q (p + q) \equiv p^{3} + q^{3}$

attempts to factorise ${(p + q)}^{3} - 3 p q (p + q)$ M1

$\equiv (p + q) ({(p + q)}^{2} - 3 p q) (\equiv (p + q) (p^{2} - p q + q^{2}))$

$\equiv p^{3} - p^{2} q + p q^{2} + p^{2} q - p q^{2} + q^{3}$ A1

$\equiv p^{3} + q^{3}$ AG

Note: Condone the use of equals signs throughout.

METHOD 3

$p^{3} + q^{3} \equiv {(p + q)}^{3} - 3 p q (p + q)$

attempts to factorise $p^{3} + q^{3}$ M1

$\equiv (p + q) (p^{2} - p q + q^{2})$

$\equiv (p + q) ({(p + q)}^{2} - 3 p q)$ A1

$\equiv {(p + q)}^{3} - 3 p q (p + q)$ AG

Note: Condone the use of equals signs throughout.

[2 marks]

Note: Award a maximum of A1M0A0A1M0A0 for $m = - 95$ and $n = 8$ found by using $α, β = \frac{5 \pm \sqrt{17}}{4} (α, β = 0.219 \dots, 2.28 \dots)$ .
Condone, as appropriate, solutions that state but clearly do not use the values of $α$ and $β$ .
Special case: Award a maximum of A1M1A0A1M0A0 for $m = - 95$ and $n = 8$ obtained by solving simultaneously for $α$ and $β$ from product of roots and sum of roots equations.

product of roots of $x^{2} - \frac{5}{2} x + \frac{1}{2} = 0$

$α β = \frac{1}{2}$ (seen anywhere) A1

considers $(\frac{1}{α^{3}}) (\frac{1}{β^{3}})$ by stating $\frac{1}{{(α β)}^{3}} (= n)$ M1

Note: Award M1 for attempting to substitute their value of $α β$ into $\frac{1}{{(α β)}^{3}}$ .

$\frac{1}{{(α β)}^{3}} = \frac{1}{{(\frac{1}{2})}^{3}}$

$n = 8$ A1

sum of roots of $x^{2} - \frac{5}{2} x + \frac{1}{2} = 0$

$α + β = \frac{5}{2}$ (seen anywhere) A1

considers $\frac{1}{α^{3}}$ and $\frac{1}{β^{3}}$ by stating $\frac{{(α + β)}^{3} - 3 α β (α + β)}{{(α β)}^{3}} ({(\frac{α + β}{α β})}^{3} - \frac{3 (α + β)}{{(α β)}^{2}}) (= - m)$ M1

Note: Award M1 for attempting to substitute their values of $α + b$ and $α β$ into their expression. Award M0 for use of ${(α + β)}^{3} - 3 α β (α + β)$ only.

$= \frac{{(\frac{5}{2})}^{3} - (\frac{3}{2}) (\frac{5}{2})}{\frac{1}{8}} (= 125 - 30 = 95)$

$m = - 95$ A1

$(x^{2} - 95 x + 8 = 0)$

[6 marks]

Examiners report

[N/A]

Consider a geometric sequence with a first term of 4 and a fourth term of −2.916.

Find the common ratio of this sequence.

[3]

Find the sum to infinity of this sequence.

[2]

Markscheme

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

${u_4} = {u_1}{r^3} \Rightarrow - 2.916 = 4{r^3}$ (A1)

solving, $r = - 0.9$ (M1)A1

[3 marks]

${S_\infty } = \frac{4}{{1 - \left( { - 9} \right)}}$ (M1)

$= \frac{{40}}{{19}}\,\left( { = 2.11} \right)$ A1

[2 marks]

Examiners report

[N/A]

A particle $P$ moves in a straight line such that after time $t$ seconds, its velocity, $v$ in ${m s}^{- 1}$ , is given by $v = e^{- 3 t} \sin 6 t$ , where $0 < t < \frac{π}{2}$ .

At time $t$ , $P$ has displacement $s (t)$ ; at time $t = 0$ , $s (0) = 0$ .

At successive times when the acceleration of $P$ is $0 {m s}^{- 2}$ , the velocities of $P$ form a geometric sequence. The acceleration of $P$ is zero at times $t_{1}, t_{2}, t_{3}$ where $t_{1} < t_{2} < t_{3}$ and the respective velocities are $v_{1}, v_{2}, v_{3}$ .

Find the times when $P$ comes to instantaneous rest.

[2]

Find an expression for $s$ in terms of $t$ .

[7]

Find the maximum displacement of $P$ , in metres, from its initial position.

[2]

Find the total distance travelled by $P$ in the first $1.5$ seconds of its motion.

[2]

Show that, at these times, $\tan 6 t = 2$ .

[2]

e.i.

Hence show that $\frac{v_{2}}{v_{1}} = \frac{v_{3}}{v_{2}} = - e^{- \frac{π}{2}}$ .

[5]

e.ii.

Markscheme

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

$\frac{π}{6} (= 0.524)$ A1

$\frac{π}{3} (= 1.05)$ A1

[2 marks]

attempt to use integration by parts M1

$s = \int e^{- 3 t} \sin 6 t d t$

EITHER

$= - \frac{e^{- 3 t} \sin 6 t}{3} - \int - 2 e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} - \int - 4 e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} \sin 6 t}{3} - (\frac{2 e^{- 3 t} cos 6 t}{3} + 4 s)$

$5 s = \frac{-3 e^{- 3 t} \sin 6 t - 6 e^{- 3 t} cos 6 t}{9}$ M1

$= - \frac{e^{- 3 t} cos 6 t}{6} - \int \frac{1}{2} e^{- 3 t} cos 6 t d t$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \int \frac{1}{4} e^{- 3 t} sin 6 t d t)$ A1

$= - \frac{e^{- 3 t} cos 6 t}{6} - (\frac{e^{- 3 t} sin 6 t}{12} + \frac{1}{4} s)$

$\frac{5}{4} s = \frac{- 2 e^{- 3 t} cos 6 t - e^{- 3 t} sin 6 t}{12}$ M1

THEN

$s = - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15} (+ c)$ A1

at $t = 0, s = 0 \Rightarrow 0 = - \frac{2}{15} + c$ M1

$c = \frac{2}{15}$ A1

$s = \frac{2}{15} - \frac{e^{- 3 t} (sin 6 t + 2 cos 6 t)}{15}$

[7 marks]

EITHER

substituting $t = \frac{π}{6}$ into their equation for $s$ (M1)

$(s = \frac{2}{15} - \frac{e^{- \frac{π}{2}} (sin π + 2 cos π)}{15})$

using GDC to find maximum value (M1)

evaluating $\int_{0}^{\frac{π}{6}} v d t$ (M1)

THEN

$= 0.161 (= \frac{2}{15} (1 + e^{- \frac{π}{2}}))$ A1

[2 marks]

METHOD 1

EITHER

distance required $= \int_{0}^{1.5} |e^{- 3 t} \sin 6 t| d t$ (M1)

distance required $= \int_{0}^{\frac{π}{6}} e^{- 3 t} \sin 6 t d t + |\int_{\frac{π}{6}}^{\frac{π}{3}} e^{- 3 t} \sin 6 t d t| + \int_{\frac{π}{3}}^{1.5} e^{- 3 t} \sin 6 t d t$ (M1)

$(= 0.16105 \dots + 0.033479 \dots + 0.006806 \dots)$

THEN

$= 0.201 (m)$ A1

METHOD 2

using successive minimum and maximum values on the displacement graph (M1)

$0.16105 \dots + (0.16105 \dots - 0.12757 \dots) + (0.13453 \dots - 0.12757 \dots)$

$= 0.201 (m)$ A1

[2 marks]

valid attempt to find $\frac{d v}{d t}$ using product rule and set $\frac{d v}{d t} = 0$ M1

$\frac{d v}{d t} = e^{- 3 t} 6 \cos 6 t - 3 e^{- 3 t} \sin 6 t$ A1

$\frac{d v}{d t} = 0 \Rightarrow \tan 6 t = 2$ AG

[2 marks]

e.i.

attempt to evaluate $t_{1}, t_{2}, t_{3}$ in exact form M1

$6 t_{1} = arctan 2 (\Rightarrow t_{1} = \frac{1}{6} arctan 2)$

$6 t_{2} = π + arctan 2 (\Rightarrow t_{2} = \frac{π}{6} + \frac{1}{6} arctan 2)$

$6 t_{3} = 2 π + arctan 2 (\Rightarrow t_{3} = \frac{π}{3} + \frac{1}{6} arctan 2)$ A1

Note: The A1 is for any two consecutive correct, or showing that $6 t_{2} = π + 6 t_{1}$ or $6 t_{3} = π + 6 t_{2}$ .

showing that $\sin 6 t_{n + 1} = - \sin 6 t_{n}$

eg $\tan 6 t = 2 \Rightarrow \sin 6 t = \pm \frac{2}{\sqrt{5}}$ M1A1

showing that $\frac{e^{- 3 t_{n +1}}}{e^{- 3 t_{n}}} = e^{- \frac{π}{2}}$ M1

eg $e^{- 3 (\frac{π}{6} + k)} \div e^{- 3 k} = e^{- \frac{π}{2}}$

Note: Award the A1 for any two consecutive terms.

$\frac{v_{3}}{v_{2}} = \frac{v_{2}}{v_{1}} = - e^{- \frac{π}{2}}$ AG

[5 marks]

e.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

Find the term independent of $x$ in the expansion of $\frac{1}{x^{3}} {(\frac{1}{3 x^{2}} - \frac{x}{2})}^{9}$ .

Markscheme

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

use of Binomial expansion to find a term in either ${(\frac{1}{3 x^{2}} - \frac{x}{2})}^{9}, {(\frac{1}{3 x^{\frac{7}{3}}} - \frac{x^{\frac{2}{3}}}{2})}^{9}, {(\frac{1}{3} - \frac{x^{3}}{2})}^{9}, {(\frac{1}{3 x^{3}} - \frac{1}{2})}^{9}$ or ${(2 - 3 x^{3})}^{9}$ (M1)(A1)

Note: Award M1 for a product of three terms including a binomial coefficient and powers of the two terms, and A1 for a correct expression of a term in the expansion.

finding the powers required to be $2$ and $7$ (M1)(A1)

constant term is ${}_{9}C_{2} \times {(\frac{1}{3})}^{2} \times {(- \frac{1}{2})}^{7}$ (M1)

Note: Ignore all $x$ ’s in student’s expression.

therefore term independent of $x$ is $- \frac{1}{32} (= - 0.03125)$ A1

[6 marks]

Examiners report

[N/A]

Express the binomial coefficient $\left( \begin{gathered} 3n + 1 \hfill \\ 3n - 2 \hfill \\ \end{gathered} \right)$ as a polynomial in $n$ .

[3]

Hence find the least value of $n$ for which $\left( \begin{gathered} 3n + 1 \hfill \\ 3n - 2 \hfill \\ \end{gathered} \right) > {10^6}$ .

[3]

Markscheme

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

$\left( \begin{gathered} 3n + 1 \hfill \\ 3n - 2 \hfill \\ \end{gathered} \right) = \frac{{\left( {3n + 1} \right){\text{!}}}}{{\left( {3n - 2} \right){\text{!}}3{\text{!}}}}$ (M1)

$= \frac{{\left( {3n + 1} \right)3n\left( {3n - 1} \right)}}{{3{\text{!}}}}$ A1

$= \frac{9}{2}{n^3} - \frac{1}{2}n$ or equivalent A1

[3 marks]

attempt to solve $= \frac{9}{2}{n^3} - \frac{1}{2}n > {10^6}$ (M1)

$n > 60.57 \ldots$ (A1)

Note: Allow equality.

$\Rightarrow n = 61$ A1

[3 marks]

Examiners report

[N/A]

The following diagram shows part of the graph of $y = p + q \sin (r x)$ . The graph has a local maximum point at $(- \frac{9 π}{4}, 5)$ and a local minimum point at $(- \frac{3 π}{4}, - 1)$ .

Determine the values of $p$ , $q$ and $r$ .

[4]

Hence find the area of the shaded region.

[4]

Markscheme

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

the principal axis is $\frac{5 + (- 1)}{2} (= 2)$

so $p = 2$ A1

the amplitude is $\frac{5 - (- 1)}{2} (= 3)$

so $q = 3$ A1

EITHER

one period is $2 (- \frac{3 π}{4} - (- \frac{9 π}{4}))$ (M1)

$= 3 π$

$\Rightarrow \frac{2 π}{r} = 3 π$

Substituting a point eg $- 1 = 2 + \sin (- \frac{3 π}{4} r)$

$\sin (- \frac{3 π}{4} r) = - 1 \Rightarrow - \frac{3 π}{4} r = \dots - \frac{5 π}{2}, - \frac{π}{2}, \frac{3 π}{2}, \dots$

Choice of correct solution $- \frac{3 π}{4} r = - \frac{π}{2}$ (M1)

THEN

$\Rightarrow r = \frac{2}{3}$ A1

$(\Rightarrow y = 2 + 3 \sin (\frac{2 x}{3}))$

Note: $q$ and $r$ can be both given as negatives for full marks

[4 marks]

roots are $x = - 1.09459 \dots, x = - 3.617797 \dots$ (A1)

$\int_{- 3.617797 \dots}^{- 1.09459 \dots} (2 + 3 \sin (\frac{2 x}{3})) d x$ (M1)

$= - 1.66 (= - 1.66179 \dots)$ (A1)

so area $= 1.66 ({units}^{2})$ A1

[4 marks]

Examiners report

[N/A]

Mary, three female friends, and her brother, Peter, attend the theatre. In the theatre there is a row of $10$ empty seats. For the first half of the show, they decide to sit next to each other in this row.

For the second half of the show, they return to the same row of $10$ empty seats. The four girls decide to sit at least one seat apart from Peter. The four girls do not have to sit next to each other.

Find the number of ways these five people can be seated in this row.

[3]

Find the number of ways these five people can now be seated in this row.

[4]

Markscheme

$6 \times 5!$ (A1)(A1)

$= 720$ (accept $6!$ ) A1

[3 marks]

METHOD 1

(Peter apart from girls, in an end seat) ${}^{8}P_{4} (= 1680)$ OR

(Peter apart from girls, not in end seat) ${}^{7}P_{4} (= 840)$ (A1)

case 1: Peter at either end

$2 \times {}^{8}P_{4} (= 3360)$ OR $2 \times {}^{8}C_{4} \times 4! (= 3360)$ (A1)

case 2: Peter not at the end

$8 \times {}^{7}P_{4} (= 6720)$ OR $8 \times {}^{7}C_{4} \times 4! (= 6720)$ (A1)

Total number of ways $= 3360 + 6720$

$= 10080$ A1

METHOD 2

(Peter next to girl, in an end seat) $4 \times {}^{8}P_{3} (= 1344)$ OR

(Peter next to one girl, not in end seat) $2 \times 4 \times {}^{7}P_{3} (= 1680)$ OR

(Peter next to two girls, not in end seat) $4 \times 3 \times {}^{7}P_{2} (= 504)$ (A1)

case 1: Peter at either end

$2 \times 4 \times {}^{8}P_{3} (= 2688)$ (A1)

case 2: Peter not at the end

$8 (2 \times 4 \times {}^{7}P_{3} + 4 \times 3 \times {}^{7}P_{2}) (= 17472)$ (A1)

Total number of ways $= {}^{10}P_{5} - (2688 + 17472)$

$= 10080$ A1

[4 marks]

Examiners report

[N/A]

Write down and simplify the first three terms, in ascending powers of $x$ , in the Extended Binomial expansion of ${\left( {1 - x} \right)^{\frac{1}{3}}}$ .

[3]

By substituting $x = \frac{1}{9}$ find a rational approximation to $\sqrt[3]{9}$ .

[3]

Markscheme

${\left( {1 - x} \right)^{\frac{1}{3}}} = 1 + \frac{1}{3}\left( { - x} \right) + \frac{1}{3}\left( {\frac{{ - 2}}{3}} \right)\frac{{{{\left( { - x} \right)}^2}}}{{2{\text{!}}}} \ldots = 1 - \frac{x}{3} - \frac{{{x^2}}}{9} \ldots$ M1A1A1

[3 marks]

${\left( {\frac{8}{9}} \right)^{\frac{1}{3}}} \simeq 1 - \frac{1}{{27}} - \frac{1}{{729}} \Rightarrow \frac{2}{{\sqrt[3]{9}}} \simeq \frac{{701}}{{729}} \Rightarrow \sqrt[3]{9} \simeq \frac{{1458}}{{701}}$ M1A1A1

[3 marks]

Examiners report

[N/A]

Consider the equation ${x^5} - 3{x^4} + m{x^3} + n{x^2} + px + q = 0$ , where $m$ , $n$ , $p$ , $q \in \mathbb{R}$ .

The equation has three distinct real roots which can be written as ${\text{lo}}{{\text{g}}_2}\,a$ , ${\text{lo}}{{\text{g}}_2}\,b$ and ${\text{lo}}{{\text{g}}_2}\,c$ .

The equation also has two imaginary roots, one of which is $d{\text{i}}$ where $d \in \mathbb{R}$ .

The values $a$ , $b$ , and $c$ are consecutive terms in a geometric sequence.

Show that $abc = 8$ .

[5]

Show that one of the real roots is equal to 1.

[3]

Given that $q = 8{d^2}$ , find the other two real roots.

[9]

Markscheme

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

recognition of the other root $= - d{\text{i}}$ (A1)

${\text{lo}}{{\text{g}}_2}\,a + {\text{lo}}{{\text{g}}_2}\,b + {\text{lo}}{{\text{g}}_2}\,c + d{\text{i}} - d{\text{i}} = 3$ M1A1

Note: Award M1 for sum of the roots, A1 for 3. Award A0M1A0 for just ${\text{lo}}{{\text{g}}_2}\,a + {\text{lo}}{{\text{g}}_2}\,b + {\text{lo}}{{\text{g}}_2}\,c = 3$ .

${\text{lo}}{{\text{g}}_2}\,abc = 3$ (M1)

$\Rightarrow abc = {2^3}$ A1

$abc = 8$ AG

[5 marks]

METHOD 1

let the geometric series be ${u_1}$ , ${u_1}r$ , ${u_1}{r^2}$

${\left( {{u_1}r} \right)^3} = 8$ M1

${u_1}r = 2$ A1

hence one of the roots is ${\text{lo}}{{\text{g}}_2}2 = 1$ R1

METHOD 2

$\frac{b}{a} = \frac{c}{b}$

${b^2} = ac \Rightarrow {b^3} = abc = 8$ M1

$b = 2$ A1

hence one of the roots is ${\text{lo}}{{\text{g}}_2}2 = 1$ R1

[3 marks]

METHOD 1

product of the roots is ${r_1} \times {r_2} \times 1 \times d{\text{i}} \times - d{\text{i}} = - 8{d^2}$ (M1)(A1)

${r_1} \times {r_2} = - 8$ A1

sum of the roots is ${r_1} + {r_2} + 1 + d{\text{i}} + - d{\text{i}} = 3$ (M1)(A1)

${r_1} + {r_2} = 2$ A1

solving simultaneously (M1)

${r_1} = - 2$ , ${r_2} = 4$ A1A1

METHOD 2

product of the roots ${\text{lo}}{{\text{g}}_2}\,a \times {\text{lo}}{{\text{g}}_2}\,b \times {\text{lo}}{{\text{g}}_2}\,c \times d{\text{i}} \times - d{\text{i}} = - 8{d^2}$ M1A1

${\text{lo}}{{\text{g}}_2}\,a \times {\text{lo}}{{\text{g}}_2}\,b \times {\text{lo}}{{\text{g}}_2}\,c = - 8$ A1

EITHER

$a$ , $b$ , $c$ can be written as $\frac{2}{r}$ , $2$ , $2r$ M1

$\left( {{\text{lo}}{{\text{g}}_2}\frac{2}{r}} \right)\left( {{\text{lo}}{{\text{g}}_2}\,2} \right)\left( {{\text{lo}}{{\text{g}}_2}\,2r} \right) = - 8$

attempt to solve M1

$\left( {1 - {\text{lo}}{{\text{g}}_2}\,r} \right)\left( {1 + {\text{lo}}{{\text{g}}_2}\,r} \right) = - 8$

${\text{lo}}{{\text{g}}_2}\,r = \pm 3$

$r = \frac{1}{8}{\text{,}}\,\,8$ A1A1

$a$ , $b$ , $c$ can be written as $a$ , $2$ , $\frac{4}{a}$ M1

$\left( {{\text{lo}}{{\text{g}}_2}\,a} \right)\left( {{\text{lo}}{{\text{g}}_2}\,2} \right)\left( {{\text{lo}}{{\text{g}}_2}\,\frac{4}{a}} \right) = - 8$

attempt to solve M1

$a = \frac{1}{4}{\text{,}}\,\,16$ A1A1

THEN

$a$ and $c$ are $\frac{1}{4}{\text{,}}\,\,16$ (A1)

roots are −2, 4 A1

[9 marks]

Examiners report

[N/A]

The following diagram shows part of the graph of $2{x^2} = {\text{si}}{{\text{n}}^3}\,y$ for $0 \leqslant y \leqslant \pi$ .

The shaded region $R$ is the area bounded by the curve, the $y$ -axis and the lines $y = 0$ and $y = \pi$ .

Using implicit differentiation, find an expression for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .

[4]

a.i.

Find the equation of the tangent to the curve at the point $\left( {\frac{1}{4}{\text{, }}\frac{{5\pi }}{6}} \right)$ .

[4]

a.ii.

Find the area of $R$ .

[3]

The region $R$ is now rotated about the $y$ -axis, through $2\pi$ radians, to form a solid.

By writing ${{\text{si}}{{\text{n}}^3}\,y}$ as $\left( {1 - {\text{co}}{{\text{s}}^2}\,y} \right){\text{sin}}\,y$ , show that the volume of the solid formed is $\frac{{2\pi }}{3}$ .

[6]

Markscheme

valid attempt to differentiate implicitly (M1)

$4x = 3\,{\text{si}}{{\text{n}}^2}\,y\,{\text{cos}}\,y\frac{{{\text{d}}y}}{{{\text{d}}x}}$ A1A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4x}}{{3\,{\text{si}}{{\text{n}}^2}\,y\,{\text{cos}}\,y}}$ A1

[4 marks]

a.i.

at $\left( {\frac{1}{4}{\text{, }}\frac{{5\pi }}{6}} \right){\text{, }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4x}}{{3\,{\text{si}}{{\text{n}}^2}\,y\,{\text{cos}}\,y}} = \frac{1}{{3{{\left( {\frac{1}{2}} \right)}^2}\left( { - \frac{{\sqrt 3 }}{2}} \right)}}$ (M1)

$\Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{8}{{3\sqrt 3 }}\left( { = - 1.54} \right)$ A1

hence equation of tangent is

$y - \frac{{5\pi }}{6} = - 1.54\left( {x - \frac{1}{4}} \right)$ OR $y = - 1.54x + 3.00$ (M1)A1

Note: Accept $y = - 1.54x + 3$ .

[4 marks]

a.ii.

$x = \sqrt {\frac{1}{2}{\text{si}}{{\text{n}}^3}\,y}$ (M1)

$\int_0^\pi {\sqrt {\frac{1}{2}{\text{si}}{{\text{n}}^3}\,y\,{\text{d}}y} }$ (A1)

$= 1.24$ A1

[3 marks]

use of volume $= \int {\pi {x^2}} \,{\text{d}}y$ (M1)

$= \int_0^\pi {\frac{1}{2}} \pi \,{\text{si}}{{\text{n}}^3}\,y\,{\text{d}}y$ A1

$= \frac{1}{2}\pi \int_0^\pi {\left( {{\text{sin}}\,y - {\text{sin}}\,y\,{\text{co}}{{\text{s}}^2}\,y} \right){\text{d}}y}$

Note: Condone absence of limits up to this point.

reasonable attempt to integrate (M1)

$= \frac{1}{2}\pi \left[ { - {\text{cos}}\,y + \frac{1}{3}{\text{co}}{{\text{s}}^3}\,y} \right]_0^\pi$ A1A1

Note: Award A1 for correct limits (not to be awarded if previous M1 has not been awarded) and A1 for correct integrand.

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

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

Note: Do not accept decimal answer equivalent to $\frac{{2\pi }}{3}$ .

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

Phil takes out a bank loan of $150 000 to buy a house, at an annual interest rate of 3.5%. The interest is calculated at the end of each year and added to the amount outstanding.

To pay off the loan, Phil makes annual deposits of $P at the end of every year in a savings account, paying an annual interest rate of 2% . He makes his first deposit at the end of the first year after taking out the loan.

David visits a different bank and makes a single deposit of $Q , the annual interest rate being 2.8%.

Find the amount Phil would owe the bank after 20 years. Give your answer to the nearest dollar.

[3]

Show that the total value of Phil’s savings after 20 years is $\frac{{({{1.02}^{20}} - 1)P}}{{(1.02 - 1)}}$ .

[3]

Given that Phil’s aim is to own the house after 20 years, find the value for $P$ to the nearest dollar.

[3]

David wishes to withdraw $5000 at the end of each year for a period of $n$ years. Show that an expression for the minimum value of $Q$ is

$\frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} + \ldots + \frac{{5000}}{{{{1.028}^n}}}$ .

[3]

d.i.

Hence or otherwise, find the minimum value of $Q$ that would permit David to withdraw annual amounts of $5000 indefinitely. Give your answer to the nearest dollar.

[3]

d.ii.

Markscheme

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

$150000 \times {1.035^{20}}$ (M1)(A1)

$= \$ 298468$ A1

Note: Only accept answers to the nearest dollar. Accept $298469.

[3 marks]

attempt to look for a pattern by considering 1 year, 2 years etc (M1)

recognising a geometric series with first term $P$ and common ratio 1.02 (M1)

EITHER

$P + 1.02P + \ldots + {1.02^{19}}P{\text{ }}\left( { = P(1 + 1.02 + \ldots + {{1.02}^{19}})} \right)$ A1

explicitly identify ${u_1} = P,{\text{ }}r = 1.02$ and $n = 20$ (may be seen as ${S_{20}}$ ). A1

THEN

${s_{20}} = \frac{{({{1.02}^{20}} - 1)P}}{{(1.02 - 1)}}$ AG

[3 marks]

$24.297 \ldots P = 298468$ (M1)(A1)

$P = 12284$ A1

Note: Accept answers which round to 12284.

[3 marks]

METHOD 1

$Q({1.028^n}) = 5000(1 + 1.028 + {1.028^2} + {1.028^3} + \ldots + {1.028^{n - 1}})$ M1A1

$Q = \frac{{5000\left( {1 + 1.028 + {{1.028}^2} + {{1.028}^3} + ... + {{1.028}^{n - 1}}} \right)}}{{{{1.028}^n}}}$ A1

$= \frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} + \ldots + \frac{{5000}}{{{{1.028}^n}}}$ AG

METHOD 2

the initial value of the first withdrawal is $\frac{{5000}}{{1.028}}$ A1

the initial value of the second withdrawal is $\frac{{5000}}{{{{1.028}^2}}}$ R1

the investment required for these two withdrawals is $\frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}}$ R1

$Q = \frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} + \ldots + \frac{{5000}}{{{{1.028}^n}}}$ AG

[3 Marks]

d.i.

sum to infinity is $\frac{{\frac{{5000}}{{1.028}}}}{{1 - \frac{1}{{1.028}}}}$ (M1)(A1)

$= 178571.428 \ldots$

so minimum amount is $178572 A1

Note: Accept answers which round to $178571 or $178572.

[3 Marks]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

Eight boys and two girls sit on a bench. Determine the number of possible arrangements, given that

the girls do not sit together.

[3]

the girls do not sit on either end.

[2]

the girls do not sit on either end and do not sit together.

[3]

Markscheme

METHOD 1

$10{\text{!}} - 2 \times 9{\text{!}}\,\left( { = 2903040} \right)$ (A1)(A1)A1

Note: Award A1 for $10{\text{!}}$ , A1 for $2 \times 9{\text{!}}$ , A1 for final answer.

METHOD 2

$9 \times 8 \times 8{\text{!}}$ (A1)(A1)A1

Note: Award A1 for $9 \times 8$ or equivalent, A1 for $8{\text{!}}$ and A1 for answer.

[3 marks]

METHOD 1

$8 \times 7 \times 8{\text{!}}\,\left( { = 2257920} \right)$ (A1)A1

Note: Award (A1) for $8 \times 7$ , A1 for final answer.

METHOD 2

$10{\text{!}} - 2 \times 8{\text{!}} - 2 \times 2 \times 7 \times 8{\text{!}}$

Note: Award A1 for $10{\text{!}}$ minus EITHER subtracted terms and A1 for final correct answer.

[2 marks]

METHOD 1

$8 \times 7 \times \left( {8{\text{!}} - 2 \times 7{\text{!}}} \right)\,\left( { = 1693440} \right)$ (A1)(A1)A1

Note: Award (A1) for $8 \times 7$ , (A1) for ${2 \times 7{\text{!}}}$ , A1 for final answer. $\left( {8{\text{!}} - 2 \times 7{\text{!}}} \right)$ can be replaced by $6 \times 7{\text{!}}$ or ${}^7{P_2} \times 6{\text{!}}$ which may be awarded the second A1.

METHOD 2

their answer to (a) $- 2 \times 8{\text{!}} - 2 \times 2 \times 7 \times 8{\text{!}}$ (A1)(A1)A1

Note: Award A1 for subtracting each of the terms and A1 for final answer.

METHOD 3

their answer to (b) $- 2 \times 7 \times 8{\text{!}}$ or equivalent (A1)A2

Note: Award A1 for the subtraction and A2 for final answer.

[3 marks]

Examiners report

[N/A]

The complex numbers $w$ and $z$ satisfy the equations

$\frac{w}{z} = 2{\text{i}}$

${{\text{z}}^ * } - 3w = 5 + 5{\text{i}}$ .

Find $w$ and $z$ in the form $a + b{\text{i}}$ where $a$ , ${\text{b}} \in \mathbb{Z}$ .

Markscheme

substituting $w = 2{\text{i}}z$ into ${{\text{z}}^ * } - 3w = 5 + 5{\text{i}}$ M1

${{\text{z}}^ * } - 6{\text{i}}z = 5 + 5{\text{i}}$ A1

let $z = x + y{\text{i}}$

comparing real and imaginary parts of $\left( {x - y{\text{i}}} \right) - 6{\text{i}}\left( {x + y{\text{i}}} \right) = 5 + 5{\text{i}}$ M1

to obtain $x + 6y = 5$ and $- 6x - y = 5$ A1

attempting to solve for $x$ and $y$ ) M1

$x = - 1$ and $y = 1$ so $z = - 1 + {\text{i}}$ A1

hence $w = - 2 - 2{\text{i}}$ A1

[7 marks]

Examiners report

[N/A]

Boxes of mixed fruit are on sale at a local supermarket.

Box A contains 2 bananas, 3 kiwifruit and 4 melons, and costs $6.58.

Box B contains 5 bananas, 2 kiwifruit and 8 melons and costs $12.32.

Box C contains 5 bananas and 4 kiwifruit and costs $3.00.

Find the cost of each type of fruit.

Markscheme

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

let $b$ be the cost of one banana, $k$ the cost of one kiwifruit, and $m$ the cost of one melon

attempt to set up three linear equations (M1)

$2b + 3k + 4m = 658$

$5b + 2k + 8m = 1232$

$5b + 4k = 300$ (A1)

attempt to solve three simultaneous equations (M1)

$b = 36,{\text{ }}k = 30,{\text{ }}m = 124$

banana costs ($)0.36, kiwifruit costs ($)0.30, melon costs ($)1.24 A1

[4 marks]

Examiners report

[N/A]

A random variable $X$ has probability density function

$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {3a}&{\text{,}}&{0 \leqslant x < 2}&{} \\ {a\left( {x - 5} \right)\left( {1 - x} \right)}&{\text{,}}&{2 \leqslant x \leqslant b}&{a{\text{, }}b \in {\mathbb{R}^ + }{\text{, }}3 < b \leqslant 5.} \\ 0&{\text{,}}&{{\text{otherwise}}}&{} \end{array}} \right.$

Consider the case where $b = 5$ .

Find the value of

Find, in terms of $a$ , the probability that $X$ lies between 1 and 3.

[4]

Sketch the graph of $f$ . State the coordinates of the end points and any local maximum or minimum points, giving your answers in terms of $a$ .

[4]

$a$ .

[4]

c.i.

${\text{E}}\left( X \right)$ .

[3]

c.ii.

the median of $X$ .

[4]

c.iii.

Markscheme

$\left( {{\text{P}}\left( {1 < X < 3} \right) = } \right)\int_1^2 {3a\,{\text{d}}x + a} \int_2^3 { - {x^2} + 6x - 5} \,{\text{d}}x$ (M1)(A1)(A1)

$= 3a + \frac{{11}}{3}a$

$= \frac{{20}}{3}a\,\left( { = 6.67a} \right)$ A1

[4 marks]

award A1 for (0, 3 $a$ ), A1 for continuity at (2, 3 $a$ ), A1 for maximum at (3, 4 $a$ ), A1 for (5, 0)

Note: Award A3 if correct four points are not joined by a straight line and a quadratic curve.

[4 marks]

${\text{P}}\left( {0 \leqslant X \leqslant 5} \right) = 6a + a\int_2^5 { - {x^2} + 6x - 5} \,{\text{d}}x$ (M1)

$= 15a$ (A1)

$15a = 1$ (M1)

$\Rightarrow a = \frac{1}{{15}}\left( { = 0.0667} \right)$ A1

[4 marks]

c.i.

${\text{E}}\left( X \right) = \frac{1}{5}\int_0^2 {x\,{\text{d}}x} + \frac{1}{{15}}\int_2^5 { - {x^3} + 6{x^2} - 5} x\,{\text{d}}x$ (M1)(A1)

= 2.35 A1

[3 marks]

c.ii.

attempt to use $\int_0^m {f\left( x \right)} {\text{d}}x = 0.5$ (M1)

$0.4 + a\int_2^m { - {x^2} + 6x - 5} \,{\text{d}}x = 0.5$ (A1)

$a\int_2^m { - {x^2} + 6x - 5} \,{\text{d}}x = 0.1$

attempt to solve integral using GDC and/or analytically (M1)

$\frac{1}{{15}}\left[ { - \frac{1}{3}{x^3} + 3{x^2} - 5x} \right]_2^m = 0.1$

$m = 2.44$ A1

[4 marks]

c.iii.

Examiners report

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Eight runners compete in a race where there are no tied finishes. Andrea and Jack are two of the eight competitors in this race.

Find the total number of possible ways in which the eight runners can finish if Jack finishes

in the position immediately after Andrea.

[2]

in any position after Andrea.

[3]

Markscheme

Jack and Andrea finish in that order (as a unit) so we are considering the arrangement of $7$ objects (M1)

$7! (= 5040)$ ways A1

[2 marks]

METHOD 1

the number of ways that Andrea finishes in front of Jack is equal to the number of ways that Jack finishes in front of Andrea (M1)

total number of ways is 8! (A1)

$\frac{8!}{2} (= 20160)$ ways A1

METHOD 2

the other six runners can finish in $6! (= 720)$ ways (A1)

when Andrea finishes first, Jack can finish in $7$ different positions

when Andrea finishes second, Jack can finish in $6$ different positions etc

$7 + 6 + 5 + 4 + 3 + 2 + 1 (= 28)$ ways (A1)

hence there are $(7 + 6 + 5 + 4 + 3 + 2 + 1) \times 6!$ ways

$28 \times 6! (= 20160)$ ways A1

[3 marks]

Examiners report

[N/A]

Prove by contradiction that $\log_{2} 5$ is an irrational number.

Markscheme

assume there exist $p, q \in ℕ$ where $q \geq 1$ such that $\log_{2} 5 = \frac{p}{q}$ M1A1

Note: Award M1 for attempting to write the negation of the statement as an assumption. Award A1 for a correctly stated assumption.

$\log_{2} 5 = \frac{p}{q} \Rightarrow 5 = 2^{\frac{p}{q}}$ A1

$5^{q} = 2^{p}$ A1

EITHER

$5$ is a factor of $5^{q}$ but not a factor of $2^{p}$ R1

$2$ is a factor of $2^{p}$ but not a factor of $5^{q}$ R1

$5^{q}$ is odd and $2^{p}$ is even R1

THEN

no $p, q \in ℕ$ (where $q \geq 1$ ) satisfy the equation $5^{q} = 2^{p}$ and this is a contradiction R1

so $\log_{2} 5$ is an irrational number AG

[6 marks]

Examiners report

[N/A]

A geometric sequence has ${u_4} = - 70$ and ${u_7} = 8.75$ . Find the second term of the sequence.

Markscheme

${u_1}{r^3} = - 70$ , ${u_1}{r^6} = 8.75$ (M1)

${r^3} = \frac{{8.75}}{{ - 70}} = - 0.125$ (A1)

$\Rightarrow r = - 0.5$ (A1)

valid attempt to find ${u_2}$ (M1)

for example: ${u_1} = \frac{{ - 70}}{{ - 0.125}} = 560$

${u_2} = 560 \times - 0.5$

$= - 280$ A1

[5 marks]

Examiners report

[N/A]

Consider the complex number $z = \frac{{2 + 7{\text{i}}}}{{6 + 2{\text{i}}}}$ .

Express $z$ in the form $a + {\text{i}}b$ , where $a,\,b \in \mathbb{Q}$ .

[2]

Find the exact value of the modulus of $z$ .

[2]

Find the argument of $z$ , giving your answer to 4 decimal places.

[2]

Markscheme

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

$z = \frac{{\left( {2 + 7{\text{i}}} \right)}}{{\left( {6 + 2{\text{i}}} \right)}} \times \frac{{\left( {6 - 2{\text{i}}} \right)}}{{\left( {6 - 2{\text{i}}} \right)}}$ (M1)

$= \frac{{26 + 38{\text{i}}}}{{40}} = \left( {\frac{{13 + 19{\text{i}}}}{{20}} = 0.65 + 0.95{\text{i}}} \right)$ A1

[2 marks]

attempt to use $\left| z \right| = \sqrt {{a^2} + {b^2}}$ (M1)

$\left| z \right| = \sqrt {\frac{{53}}{{40}}} \left( { = \frac{{\sqrt {530} }}{{20}}} \right)$ or equivalent A1

Note: A1 is only awarded for the correct exact value.

[2 marks]

EITHER

arg $z$ = arg(2 + 7i) − arg(6 + 2i) (M1)

arg $z$ = arctan $\left( {\frac{{19}}{{13}}} \right)$ (M1)

THEN

arg $z$ = 0.9707 (radians) (= 55.6197 degrees) A1

Note: Only award the last A1 if 4 decimal places are given.

[2 marks]

Examiners report

[N/A]

It is known that the number of fish in a given lake will decrease by 7% each year unless some new fish are added. At the end of each year, 250 new fish are added to the lake.

At the start of 2018, there are 2500 fish in the lake.

Show that there will be approximately 2645 fish in the lake at the start of 2020.

[3]

Find the approximate number of fish in the lake at the start of 2042.

[5]

Markscheme

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

EITHER

2019: 2500 × 0.93 + 250 = 2575 (M1)A1

2020: 2575 × 0.93 + 250 M1

2020: 2500 × 0.93² + 250(0.93 + 1) M1M1A1

Note: Award M1 for starting with 2500, M1 for multiplying by 0.93 and adding 250 twice. A1 for correct expression. Can be shown in recursive form.

THEN

(= 2644.75) = 2645 AG

[3 marks]

2020: 2500 × 0.93² + 250(0.93 + 1)
2042: 2500 × 0.93²⁴ + 250(0.93²³ + 0.93²² + … + 1) (M1)(A1)

$= 2500 \times {0.93^{24}} + 250\frac{{\left( {{{0.93}^{24}} - 1} \right)}}{{\left( {0.93 - 1} \right)}}$ (M1)(A1)

=3384 A1

Note: If recursive formula used, award M1 for u_n = 0.93 u_n₋₁ and u₀ or u₁ seen (can be awarded if seen in part (a)). Then award M1A1 for attempt to find u₂₄ or u₂₅ respectively (different term if other than 2500 used) (M1A0 if incorrect term is being found) and A2 for correct answer.

Note: Accept all answers that round to 3380.

[5 marks]

Examiners report

[N/A]

The coefficient of ${x^2}$ in the expansion of ${\left( {\frac{1}{x} + 5x} \right)^8}$ is equal to the coefficient of ${x^4}$ in the expansion of ${\left( {a + 5x} \right)^7},{\text{ }}a \in \mathbb{R}$ . Find the value of $a$ .

Markscheme

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

METHOD 1

$^8{C_r}{\left( {\frac{1}{x}} \right)^{8 - r}}{(5x)^r} = {}^8Cr{\left( 5 \right)^r}{x^{2r - 8}}$ (M1)

$r = 5$ (A1)

$^8{C_5} \times {5^5}{ = ^7}{C_4}{a^3} \times {5^4}$ M1A1

$56 \times 5 = 35{a^3}$

${a^3} = 8$ (A1)

$a = 2$ A1

METHOD 2

attempt to expand both binomials M1

$175000{x^2}$ A1

$21875{a^3}{x^4}$ A1

$175000 = 21875{a^3}$ M1

${a^3} = 8$ (A1)

$a = 2$ A1

[6 marks]

Examiners report

[N/A]

Write down the first three terms of the binomial expansion of $(1 + t)^{- 1}$ in ascending powers of $t$ .

[1]

By using the Maclaurin series for $\cos x$ and the result from part (a), show that the Maclaurin series for $sec x$ up to and including the term in $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ .

[4]

By using the Maclaurin series for $arctan x$ and the result from part (b), find $\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1})$ .

[3]

Markscheme

$1 - t + t^{2}$ A1

Note: Accept $1, - t$ and $t^{2}$ .

[1 mark]

$sec x = \frac{1}{1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} (- \dots)} (= {(1 - \frac{x^{2}}{2!} + (\frac{x^{4}}{4!} (- \dots)))}^{- 1})$ (M1)

$t = \cos x - 1$ or $sec x = 1 - (\cos x - 1) + {(\cos x - 1)}^{2}$ (M1)

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

$= 1 + \frac{x^{2}}{2} - \frac{x^{4}}{24} + \frac{x^{4}}{4}$ A1

so the Maclaurin series for $sec x$ up to and including the term in $x^{4}$ is $1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}$ AG

Note: Condone the absence of ‘…’

[4 marks]

$arctan 2 x = 2 x - \frac{{(2 x)}^{3}}{3} + \dots$

$\lim_{x \to 0} (\frac{x arctan 2 x}{sec x - 1}) = \lim_{x \to 0} (\frac{x (2 x - \frac{{(2 x)}^{3}}{3} + \dots)}{(1 + \frac{x^{2}}{2} + \frac{5 x^{4}}{24}) - 1})$ M1

$= \lim_{x \to 0} (\frac{2 x^{2} - \frac{8 x^{4}}{3} + \dots}{\frac{x^{2}}{2} + \frac{5 x^{4}}{24}})$ A1

$= \lim_{x \to 0} (\frac{2 x^{2} (1 - \frac{4 x^{2}}{3})}{\frac{x^{2}}{2} (1 + \frac{5 x^{2}}{12})})$

$= 4$ A1

Note: Condone missing ‘lim’ and errors in higher derivatives.
Do not award M1 unless $x$ is replaced by $2 x$ in $arctan$ .

[3 marks]

Examiners report

[N/A]

A biased coin is weighted such that the probability, $p$ , of obtaining a tail is $0.6$ . The coin is tossed repeatedly and independently until a tail is obtained.

Let $E$ be the event “obtaining the first tail on an even numbered toss”.

Find $P (E)$ .

Markscheme

METHOD 1

$E_{n}$ is the event “the first tail occurs on the $2$ nd, $4$ th, $6$ th, …, $2 n$ th toss”

$P (E) = Σ_{n = 1}^{\infty} P (E_{n})$ (A1)

Note: Award A1 for deducing that either $1$ head before a tail or $3$ heads before a tail or $5$ heads before a tail etc. is required. In other words, deduces $(2 n - 1)$ heads before a tail.

$P (E) = 0.4 \times 0.6 + {(0.4)}^{3} \times 0.6 + {(0.4)}^{5} \times 0.6 + \dots$ M1A1

Note: Award M1 for attempting to form an infinite geometric series.

Note: Award A1 for $P (E) = Σ_{n = 1}^{\infty} {(0.4)}^{2 n - 1} (0.6)$ .

uses $S_{\infty} = \frac{u_{1}}{1 - r}$ with $u_{1} = 0.6 \times 0.4$ and $r = {(0.4)}^{2}$ (M1)

Note: Award M1 for using $S_{\infty} = \frac{u_{1}}{1 - r}$ with $u_{1} = 0.4$ and $r = {(0.4)}^{2}$

$= \frac{0.6 \times 0.4}{1 - {(0.4)}^{2}}$ A1

$= 0.286 (= \frac{2}{7})$ A1

METHOD 2

let $T_{1}$ be the event “tail occurs on the first toss”

uses $P (E) = P (E T_{1}) P (T_{1}) + P (E T_{1}') P (T_{1}')$ M1

concludes that $P (E T_{1}) = 0$ and so $P (E) = P (E T_{1}') P (T_{1}')$ R1

$P (E T_{1}') = P (E') (= 1 - P (E))$ A1

Note: Award A1 for concluding: given that a tail is not obtained on the first toss, then $P (E T_{1}')$ is the probability that the first tail is obtained after a further odd number of tosses, $P (E')$ .

$P (T_{1}') = 0 .4$

$P (E) = 0 .4 (1 - P (E))$ A1

attempts to solve for $P (E)$ (M1)

$= 0.286 (= \frac{2}{7})$ A1

[6 marks]

Examiners report

[N/A]

Consider $z = \cos θ + i \sin θ$ where $z \in ℂ, z \neq 1$ .

Show that $Re (\frac{1 + z}{1 - z}) = 0$ .

Markscheme

$\frac{1 + z}{1 - z} = \frac{1 + \cos θ + i \sin θ}{1 - \cos θ - i \sin θ}$

attempt to use the complex conjugate of their denominator M1

$= \frac{(1 + \cos θ + i \sin θ) (1 - \cos θ + i \sin θ)}{(1 - \cos θ - i \sin θ) (1 - \cos θ + i \sin θ)}$ A1

$Re (\frac{1 + z}{1 - z}) = \frac{1 - \cos^{2} θ - \sin^{2} θ}{{(1 - \cos θ)}^{2} + \sin^{2} θ} (= \frac{1 - \cos^{2} θ - \sin^{2} θ}{2 - 2 \cos θ})$ M1A1

Note: Award M1 for expanding the numerator and A1 for a correct numerator. Condone either an incorrect denominator or the absence of a denominator.

using $\cos^{2} θ + \sin^{2} θ = 1$ to simplify the numerator (M1)

$Re (\frac{1 + z}{1 - z}) = 0$ AG

[5 marks]

Examiners report

[N/A]

Find the roots of the equation ${w^3} = 8{\text{i}}$ , $w \in \mathbb{C}$ . Give your answers in Cartesian form.

[4]

One of the roots ${w_1}$ satisfies the condition ${\text{Re}}\left( {{w_1}} \right) = 0$ .

Given that ${w_1} = \frac{z}{{z - {\text{i}}}}$ , express $z$ in the form $a + b{\text{i}}$ , where $a$ , $b \in \mathbb{Q}$ .

[3]

Markscheme

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

METHOD 1

${w^3} = 8{\text{i}}$

writing $8{\text{i}} = 8\left( {{\text{cos}}\left( {\frac{\pi }{2} + 2\pi k} \right) + {\text{i}}\,{\text{sin}}\left( {\frac{\pi }{2} + 2\pi k} \right)} \right)$ (M1)

Note: Award M1 for an attempt to find cube roots of $w$ using modulus-argument form.

cube roots $w = 2\left( {{\text{cos}}\left( {\frac{{\frac{\pi }{2} + 2\pi k}}{3}} \right) + {\text{i}}\,{\text{sin}}\left( {\frac{{\frac{\pi }{2} + 2\pi k}}{3}} \right)} \right)$ (M1)

i.e. $w = \sqrt 3 + {\text{i,}}\,\, - \sqrt 3 + {\text{i,}}\,\, - 2{\text{i}}$ A2

Note: Award A2 for all 3 correct, A1 for 2 correct.

Note: Accept $w = 1.73 + {\text{i}}$ and $w = - 1.73 + {\text{i}}$ .

METHOD 2

${w^3} + {\left( {2{\text{i}}} \right)^3} = 0$

$\left( {w + 2{\text{i}}} \right)\left( {{w^2} - 2w{\text{i}} - 4} \right) = 0$ M1

$w = \frac{{2{\text{i}} \pm \sqrt {12} }}{2}$ M1

$w = \sqrt 3 + {\text{i,}}\,\, - \sqrt 3 + {\text{i,}}\,\, - 2{\text{i}}$ A2

Note: Award A2 for all 3 correct, A1 for 2 correct.

Note: Accept $w = 1.73 + {\text{i}}$ and $w = - 1.73 + {\text{i}}$ .

[4 marks]

${w_1} = - 2{\text{i}}$

$\frac{z}{{z - {\text{i}}}} = - 2{\text{i}}$ M1

$z = - 2{\text{i}}\left( {z - {\text{i}}} \right)$

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

$z = \frac{{ - 2}}{{1 + 2{\text{i}}}}$ A1

$z = - \frac{2}{5} + \frac{4}{5}{\text{i}}$ A1

Note: Accept $a = - \frac{2}{5}{\text{,}}\,\,b = \frac{4}{5}$ .

[3 marks]

Examiners report

[N/A]

The function $f$ has a derivative given by $f' (x) = \frac{1}{x (k - x)}, x \in ℝ, x \neq o, x \neq k$ where $k$ is a positive constant.

Consider $P$ , the population of a colony of ants, which has an initial value of $1200$ .

The rate of change of the population can be modelled by the differential equation $\frac{d P}{d t} = \frac{P (k - P)}{5 k}$ , where $t$ is the time measured in days, $t \geq 0$ , and $k$ is the upper bound for the population.

At $t = 10$ the population of the colony has doubled in size from its initial value.

The expression for $f' (x)$ can be written in the form $\frac{a}{x} + \frac{b}{k - x}$ , where $a, b \in ℝ$ . Find $a$ and $b$ in terms of $k$ .

[3]

Hence, find an expression for $f (x)$ .

[3]

By solving the differential equation, show that $P = \frac{1200 k}{(k - 1200) e^{- \frac{t}{5}} + 1200}$ .

[8]

Find the value of $k$ , giving your answer correct to four significant figures.

[3]

Find the value of $t$ when the rate of change of the population is at its maximum.

[3]

Markscheme

$\frac{1}{x (k - x)} \equiv \frac{a}{x} + \frac{b}{k - x}$

$a (k - x) + b x = 1$ (A1)

attempt to compare coefficients OR substitute $x = k$ and $x = 0$ and solve (M1)

$a = \frac{1}{k}$ and $b = \frac{1}{k}$ A1

$f' (x) = \frac{1}{k x} + \frac{1}{k (k - x)}$

[3 marks]

attempt to integrate their $\frac{a}{x} + \frac{b}{k - x}$ (M1)

$f (x) \frac{1}{k} \int (\frac{1}{x} + \frac{1}{k - x}) d x$

$= \frac{1}{k} (\ln |x| - \ln |k - x|) (+ c) (= \frac{1}{k} \ln |\frac{x}{k - x}| (+ c))$ A1A1

Note: Award A1 for each correct term. Award A1A0 for a correct answer without modulus signs. Condone the absence of $+ c$ .

[3 marks]

attempt to separate variables and integrate both sides M1

$5 k \int \frac{1}{P (k - P)} d P = \int 1 d t$

$5 (\ln P - \ln (k - P)) = t + c$ A1

Note: There are variations on this which should be accepted, such as $\frac{1}{k} (\ln P - \ln (k - P)) = \frac{1}{5 k} t + c$ . Subsequent marks for these variations should be awarded as appropriate.

EITHER

attempt to substitute $t = 0, P = 1200$ into an equation involving $c$ M1

$c = 5 (\ln 1200 - \ln (k - 1200)) (= 5 \ln (\frac{1200}{k - 1200}))$ A1

$5 (\ln P - \ln (k - P)) = t + 5 (\ln 1200 - \ln (k - 1200))$ A1

$\ln (\frac{P (k - 1200)}{1200 (k - P)}) = \frac{t}{5}$

$\frac{P (k - 1200)}{1200 (k - P)} = e^{\frac{t}{5}}$ A1

$\ln (\frac{P}{k - P}) = \frac{t + c}{5}$

$\frac{P}{k - P} = A e^{\frac{t}{5}}$ A1

attempt to substitute $t = 0, P = 1200$ M1

$\frac{1200}{k - 1200} = A$ A1

$\frac{P}{k - P} = \frac{1200 e^{\frac{t}{5}}}{k - 1200}$ A1

THEN

attempt to rearrange and isolate $P$ M1

$P k - 1200 P = 1200 k e^{\frac{t}{5}} - 1200 P e^{\frac{t}{5}}$ OR $P k e^{- \frac{t}{5}} - 1200 P e^{- \frac{t}{5}} = 1200 k - 1200 P$ OR $\frac{k}{P} - 1 = \frac{k - 1200}{1200 e^{\frac{t}{5}}}$

$P (k - 1200 + 1200 e^{\frac{t}{5}}) = 1200 k e^{\frac{t}{5}}$ OR $P (k e^{- \frac{t}{5}} - 1200 e^{- \frac{t}{5}} + 1200) = 1200 k$ A1

$P = \frac{1200 k}{(k - 1200) e^{- \frac{t}{5}} + 1200}$ AG

[8 marks]

attempt to substitute $t = 10, P = 2400$ (M1)

$2400 = \frac{1200 k}{(k - 1200) e^{- 2} + 1200}$ (A1)

$k = 2845.34 \dots$

$k = 2845$ A1

Note: Award (M1)(A1)A0 for any other value of $k$ which rounds to $2850$

[3 marks]

attempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that $P = \frac{k}{2} (= 1422.67 \dots)$ (M1)

$t = 1.57814 \dots$

$= 1.58$ (days) A2

Note: Accept any value which rounds to $1.6$ .

[3 marks]

Examiners report

[N/A]

Let $P\left( z \right) = a{z^3} - 37{z^2} + 66z - 10$ , where $z \in \mathbb{C}$ and $a \in \mathbb{Z}$ .

One of the roots of $P\left( z \right) = 0$ is $3 + {\text{i}}$ . Find the value of $a$ .

Markscheme

METHOD 1

one other root is $3 - {\text{i}}$ A1

let third root be $\alpha$ (M1)

considering sum or product of roots (M1)

sum of roots $= 6 + \alpha = \frac{{37}}{a}$ A1

product of roots $= 10\alpha = \frac{{10}}{a}$ A1

hence $a = 6$ A1

METHOD 2

one other root is $3 - {\text{i}}$ A1

quadratic factor will be ${z^2} - 6z + 10$ (M1)A1

$P\left( z \right) = a{z^3} - 37{z^2} + 66z - 10 = \left( {{z^2} - 6z + 10} \right)\left( {az - 1} \right)$ M1

comparing coefficients (M1)

hence $a = 6$ A1

METHOD 3

substitute $3 + {\text{i}}$ into $P\left( z \right)$ (M1)

$a\left( {18 + 26{\text{i}}} \right) - 37\left( {8 + 6{\text{i}}} \right) + 66\left( {3 + {\text{i}}} \right) - 10 = 0$ (M1)A1

equating real or imaginary parts or dividing M1

$18a - 296 + 198 - 10 = 0$ or $26a - 222 + 66 = 0$ or $\frac{{10 - 66\left( {3 + {\text{i}}} \right) + 37\left( {8 + 6{\text{i}}} \right)}}{{18 + 26{\text{i}}}}$ A1

hence $a = 6$ A1

[6 marks]

Examiners report

[N/A]

The population, $P$ , of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$

where $t$ is the time measured in years and $k, N$ are positive constants.

The constant $N$ represents the maximum population of this species of marsupial that the island can sustain indefinitely.

Let $P_{0}$ be the initial population of marsupials.

In the context of the population model, interpret the meaning of $\frac{d P}{d t}$ .

[1]

Show that $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ .

[4]

Hence show that the population of marsupials will increase at its maximum rate when $P = \frac{N}{2}$ . Justify your answer.

[5]

Hence determine the maximum value of $\frac{d P}{d t}$ in terms of $k$ and $N$ .

[2]

By solving the logistic differential equation, show that its solution can be expressed in the form

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ .

[7]

After $10$ years, the population of marsupials is $3 P_{0}$ . It is known that $N = 4 P_{0}$ .

Find the value of $k$ for this population model.

[2]

Markscheme

rate of growth (change) of the (marsupial) population (with respect to time) A1

[1 mark]

Note: Do not accept growth (change) in the (marsupials) population per year.

METHOD 1

attempts implicit differentiation on $\frac{d P}{d t} = k P - \frac{k P^{2}}{N}$ be expanding $k P (1 - \frac{P}{N})$ (M1)

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} - 2 \frac{k P}{N} \frac{d P}{d t}$ A1A1

$= k \frac{d P}{d t} (1 - \frac{2 P}{N})$ A1

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$ and so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

METHOD 2

attempts implicit differentiation (product rule) on $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ M1

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) \frac{d P}{d t})$ A1

substitutes $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ into their $\frac{d^{2} P}{d t^{2}}$ M1

$\frac{d^{2} P}{d t^{2}} = k (k P (1 - \frac{P}{N})) (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) k P (1 - \frac{P}{N}))$

$= k^{2} P {(1 - \frac{P}{N})}^{2} - k^{2} P (1 - \frac{P}{N}) (\frac{P}{N})$

$= k^{2} P (1 - \frac{P}{N}) (1 - \frac{P}{N} - \frac{P}{N})$ A1

so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

[4 marks]

$\frac{d^{2} P}{d t^{2}} = 0 \Rightarrow k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N}) = 0$ (M1)

$P = 0, \frac{N}{2}, N$ A2

Note: Award A1 for $P = \frac{N}{2}$ only.

uses the second derivative to show that concavity changes at $P = \frac{N}{2}$ or the first derivative to show a local maximum at $P = \frac{N}{2}$ M1

EITHER

a clearly labelled correct sketch of $\frac{d^{2} P}{d t^{2}}$ versus $P$ showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

a correct and clearly labelled sign diagram (table) showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

for example, $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (> 0)$ with $P = \frac{N}{4}$ and $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (< 0)$ with $P = \frac{3 N}{4}$ showing $P = \frac{N}{2}$ corresponds to a local maximum point for $\frac{d P}{d t}$ R1

so the population is increasing at its maximum rate when $P = \frac{N}{2}$ AG

[5 marks]

substitutes $P = \frac{N}{2}$ into $\frac{d P}{d t}$ (M1)

$\frac{d P}{d t} = k (\frac{N}{2}) (1 - \frac{\frac{N}{2}}{N})$

the maximum value of $\frac{d P}{d t}$ is $\frac{k N}{4}$ A1

[2 marks]

METHOD 1

attempts to separate variables M1

$\int \frac{N}{P (N - P)} d P = \int k d t$

attempts to write $\frac{N}{P (N - P)}$ in partial fractions form M1

$\frac{N}{P (N - P)} \equiv \frac{A}{P} + \frac{B}{(N - P)} \Rightarrow N \equiv A (N - P) + B P$

$A = 1, B = 1$ A1

$\frac{N}{P (N - P)} \equiv \frac{1}{P} + \frac{1}{(N - P)}$

$\int (\frac{1}{P} + \frac{1}{(N - P)}) d P = \int k d t$

$\Rightarrow \ln P - \ln (N - P) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (N - P)$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln P_{0} - \ln (N - P_{0})$

$k t = \ln (\frac{P}{N - P}) - \ln (\frac{P_{0}}{N - P_{o}}) (= \ln (\frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}}))$ A1

so $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 2

attempts to separate variables M1

$\int \frac{1}{P (1 - \frac{P}{N})} d P = \int k d t$

attempts to write $\frac{1}{P (1 - \frac{P}{N})}$ in partial fractions form M1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{A}{P} + \frac{B}{1 - \frac{P}{N}} \Rightarrow 1 \equiv A (1 - \frac{P}{N}) + B P$

$A = 1, B = \frac{1}{N}$ A1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})}$

$\int \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})} d P = \int k d t$

$\Rightarrow \ln P - \ln (1 - \frac{P}{N}) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (1 - \frac{P}{N})$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

$\ln (\frac{P}{1 - \frac{P}{N}}) = k t + C \Rightarrow \ln (\frac{N P}{N - P}) = k t + C$

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln (\frac{N P_{0}}{N - P_{0}})$

$k t = \ln (\frac{N P}{N - P}) - \ln (\frac{N P_{0}}{N - P_{0}}) (= \ln \frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 3

lets $u = \frac{1}{P}$ and forms $\frac{d u}{d t} = - \frac{1}{P^{2}} \frac{d P}{d t}$ M1

multiplies both sides of the differential equation by $- \frac{1}{P^{2}}$ and makes the above substitutions M1

$- \frac{1}{P^{2}} \frac{d P}{d t} = k (\frac{1}{N} - \frac{1}{P}) \Rightarrow \frac{d u}{d t} = k (\frac{1}{N} - u)$

$\frac{d u}{d t} + k u = \frac{k}{N}$ (linear first-order DE) A1

$IF = e^{\int k d t} = e^{k t} \Rightarrow e^{k t} \frac{d u}{d t} + k e^{k t} u = \frac{k}{N} e^{k t}$ (M1)

$\frac{d}{d t} (u e^{k t}) = \frac{k}{N} e^{k t}$

$u e^{k t} = \frac{1}{N} e^{k t} (+ C) (\frac{1}{P} e^{k t} = \frac{1}{N} e^{k t} (+ C))$ A1

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}, u = \frac{1}{P_{0}}$ and so $C = \frac{1}{P_{0}} - \frac{1}{N} (= \frac{N - P_{0}}{N P_{0}})$

$e^{k t} (\frac{N - P}{N P}) = \frac{N - P_{0}}{N P_{0}}$

$e^{k t} = (\frac{P}{N - P}) (\frac{N - P_{0}}{P_{0}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

[7 marks]

substitutes $t = 10, P = 3 P_{0}$ and $N = 4 P_{0}$ into $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ M1

$10 k = \ln 3 (\frac{4 P_{0} - P_{0}}{4 P_{0} - 3 P_{0}}) (= \ln 9)$

$k = 0.220 (= \frac{1}{10} \ln 9, = \frac{1}{5} \ln 3)$ A1

[2 marks]

Examiners report

An extremely tricky question even for the strong candidates. Many struggled to understand what was expected in parts (b) and (c). As the question was set all with pronumerals instead of numbers many candidates found it challenging, thrown at deep water for parts (b), (c) and (e). It definitely was the question to show their skills for the Level 7 candidates provided that they did not run out of time.

Part (a) Very well answered, mostly correctly referring to the rate of change. Some candidates did not gain this mark because their sentence did not include the reference to the rate of change. Worded explanations continue being problematic to many candidates.

Part (b) Many candidates were confused how to approach this question and did not realise that they
needed to differentiate implicitly. Some tried but with errors, some did not fully show what was required.

Part (c) Most candidates started with equating the second derivative to zero. Most gave the answer $P = \frac{N}{2}$ omitting the other two possibilities. Most stopped here. Only a small number of candidates provided the correct mathematical argument to show it is a local maximum.

Part (d) Well done by those candidates who got that far. Most got the correct answer, sometimes not fully simplified.

Part (e) Most candidates separated the variables, but some were not able to do much more. Some candidates knew to resolve into partial fractions and attempted to do so, mainly successfully. Then they integrated, again, mainly successfully and continued to substitute the initial condition and manipulate the equation accordingly.

Part (f) Algebraic manipulation of the logarithmic expression was too much for some candidates with a common error of 0.33 given as the answer. The strong candidates provided the correct exact or rounded value.

[N/A]

In a trial examination session a candidate at a school has to take 18 examination papers including the physics paper, the chemistry paper and the biology paper. No two of these three papers may be taken consecutively. There is no restriction on the order in which the other examination papers may be taken.

Find the number of different orders in which these 18 examination papers may be taken.

Markscheme

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

METHOD 1

consideration of all papers

all papers may be sat in 18! ways A1

number of ways of positioning “pairs” of science subjects $= 3! \times 17!$ A1

but this includes two copies of each “triple” (R1)

number of ways of positioning “triplets” of science subjects $= 3! \times 16!$ A1

hence number of arrangements is $18! - 3! \times 17! + 3! \times 16!$ M1A1

$( = 4.39 \times {10^{15}})$

METHOD 2

consideration of all the non-science papers (M1)

hence all non-science papers can be sat in 15! ways A1

there are $16 \times 15 \times 14{\text{ }}( = 3360)$ ways of positioning the three science papers (M1)A1

hence the number of arrangements is $16 \times 15 \times 14 \times 15!{\text{ (}} = 4.39 \times {10^{15}})$ (M1)A1

METHOD 3

consideration of all papers

all papers may be sat in 18! ways A1

number of ways of positioning exactly two science subjects $= 3! \times 15! \times 16 \times 15$ M1A1

number of ways of positioning “triplets” of science subjects $= 3! \times 16!$ A1

hence number of arrangements is $18! - 3! \times 16! - 3! \times 15! \times 16 \times 15$ M1A1

$( = 4.39 \times {10^{15}})$

[6 marks]

Examiners report

[N/A]

Consider the polynomial $P\left( z \right) \equiv {z^4} - 6{z^3} - 2{z^2} + 58z - 51,\,\,z \in \mathbb{C}$ .

Sketch the graph of $y = {x^4} - 6{x^3} - 2{x^2} + 58x - 51$ , stating clearly the coordinates of any maximum and minimum points and intersections with axes.

[6]

Hence, or otherwise, state the condition on $k \in \mathbb{R}$ such that all roots of the equation $P\left( z \right) = k$ are real.

[2]

Markscheme

shape A1

$x$ -axis intercepts at (−3, 0), (1, 0) and $y$ -axis intercept at (0, −51) A1A1

minimum points at (−1.62, −118) and (3.72, 19.7) A1A1

maximum point at (2.40, 26.9) A1

Note: Coordinates may be seen on the graph or elsewhere.

Note: Accept −3, 1 and −51 marked on the axes.

[6 marks]

from graph, 19.7 ≤ $k$ ≤ 26.9 A1A1

Note: Award A1 for correct endpoints and A1 for correct inequalities.

[2 marks]

Examiners report

[N/A]

Use mathematical induction to prove that ${\left( {1 - a} \right)^n} > 1 - na$ for $\left\{ {n\,{\text{:}}\,n \in {\mathbb{Z}^ + },\,n \geqslant 2} \right\}$ where $0 < a < 1$ .

Markscheme

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

Let ${{\text{P}}_n}$ be the statement: ${\left( {1 - a} \right)^n} > 1 - na$ for some ${n \in {\mathbb{Z}^ + },\,n \geqslant 2}$ where $0 < a < 1$ consider the case $n = 2{\text{:}}\,\,\,{\left( {1 - a} \right)^2} = 1 - 2a + {a^2}$ M1

$> 1 - 2a$ because ${a^2} < 0$ . Therefore ${{\text{P}}_2}$ is true R1

assume ${{\text{P}}_n}$ is true for some $n = k$

${\left( {1 - a} \right)^k} > 1 - ka$ M1

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

EITHER

consider ${\left( {1 - a} \right)^{k + 1}} = \left( {1 - a} \right){\left( {1 - a} \right)^k}$ M1

$> 1 - \left( {k + 1} \right)a + k{a^2}$ A1

$> 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}}$ is true (as $k{a^2} > 0$ ) R1

multiply both sides by $\left( {1 - a} \right)$ (which is positive) M1

${\left( {1 - a} \right)^{k + 1}} > \left( {1 - ka} \right)\left( {1 - a} \right)$

${\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a + k{a^2}$ A1

${\left( {1 - a} \right)^{k + 1}} > 1 - \left( {k + 1} \right)a \Rightarrow {{\text{P}}_{k + 1}}$ is true (as $k{a^2} > 0$ ) R1

THEN

${{\text{P}}_2}$ is true ${{\text{P}}_k}$ is true $\Rightarrow {{\text{P}}_{k + 1}}$ is true so ${{\text{P}}_n}$ true for all $n > 2$ (or equivalent) R1

Note: Only award the last R1 if at least four of the previous marks are gained including the A1.

[7 marks]

Examiners report

[N/A]

The 3rd term of an arithmetic sequence is 1407 and the 10th term is 1183.

Calculate the number of positive terms in the sequence.

Markscheme

1471 + (n − 1)(−32) > 0 (M1)

⇒ n < $\frac{{1471}}{{32}} + 1$

n < 46.96… (A1)

so 46 positive terms A1

[3 marks]

Examiners report

[N/A]

Consider the expansion of ${\left( {2 + x} \right)^n}$ , where $n \geqslant 3$ and $n \in \mathbb{Z}$ .

The coefficient of ${x^3}$ is four times the coefficient of ${x^2}$ . Find the value of $n$ .

Markscheme

attempt to find coefficients in binomial expansion (M1)

coefficient of ${x^2}\,{\text{:}}\,\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) \times {2^{n - 2}}$ ; coefficient of ${x^3}\,{\text{:}}\,\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right) \times {2^{n - 3}}$ A1A1

Note: Condone terms given rather than coefficients. Terms may be seen in an equation such as that below.

$\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right) \times {2^{n - 3}} = 4\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right) \times {2^{n - 2}}$ (A1)

attempt to solve equation using GDC or algebraically (M1)

$\left( {\begin{array}{*{20}{c}} n \\ 3 \end{array}} \right) = 8\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right)$

$\frac{{n{\text{!}}}}{{3{\text{!}}\left( {n - 3} \right){\text{!}}}} = \frac{{8n{\text{!}}}}{{2{\text{!}}\left( {n - 2} \right){\text{!}}}}$

$\frac{1}{3} = \frac{8}{{n - 2}}$

$n = 26$ A1

[6 marks]

Examiners report

[N/A]