Advanced Higher Maths 2019 Paper — Questions & Marking Instructions

Product, quotient, chain rules

2 Marks

2019 Q1(a)

Differentiate $\displaystyle f(x) = x^6 \cot 5x.$

Show answer

\displaystyle 6x^5 \cot 5x - 5x^6 \text{cosec}^2 5x

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	evidence of product rule with one term correct	$\displaystyle 6x^5\cot 5x \pm x^6(\ldots)$ OR $\displaystyle -5x^6\operatorname{cosec}^2 5x + (\ldots)\cot 5x$
(a) •²	complete differentiation	$\displaystyle 6x^5\cot 5x - 5x^6\operatorname{cosec}^2 5x$

Notes

(a) 1. For candidates who produce a single term only, •¹ and •² are not available.
(a) 2. Award •² for final answers such as: $\displaystyle 6x^5\cot 5x + x^6(-5\operatorname{cosec}^2 5x)$ , $\displaystyle 6x^5\cot 5x - x^6 5\operatorname{cosec}^2 5x$ and $\displaystyle 6x^5\cot 5x - 5\operatorname{cosec}^2 5x \cdot x^6.$
(a) 3. Do not award •² for final answers such as: $\displaystyle 6x^5\cot 5x + -5x^6\operatorname{cosec}^2 5x$ , $\displaystyle 6x^5\cot 5x + x^6 - 5\operatorname{cosec}^2 5x$ and $\displaystyle 6x^5\cot 5x - 5\operatorname{cosec}^2 5x\, x^6.$
(a) 4. Where a candidate equates $\displaystyle f(x)$ to $\displaystyle f'(x)$ , •¹ is not available (see COR A).
Commonly observed: 1. (a) A: $\displaystyle f(x) = x^6\cot 5x = 6x^5\cot 5x - 5x^6\operatorname{cosec}^2 5x$ , award •² only.
Commonly observed: 2. (a) B: $\displaystyle x^6\cot 5x = x^6\tan^{-1}(5x)$ , $\displaystyle f'(x) = 6x^5\tan^{-1}(5x) + \dfrac{5x^6}{1 + (5x)^2}$ , award •² only.
Commonly observed: 3. (a) C: $\displaystyle f(x) = \dfrac{x^6}{\tan 5x}$ , $\displaystyle f'(x) = \dfrac{6x^5\tan 5x - 5x^6\sec^2 5x}{(\tan 5x)^2}$ , award •¹ and •².
Commonly observed: 4. (a) D: $\displaystyle f(x) = x^6(\tan 5x)^{-1}$ , $\displaystyle f'(x) = 6x^5(\tan 5x)^{-1} - x^6(\tan 5x)^{-2}5\sec^2 5x$ , award •¹ and •².
Commonly observed: 1. (b) A, candidates who rewrite the function as $\displaystyle y = 2 + \dfrac{9}{x^3 - 4}$ : •³ $\displaystyle y = 2 + 9(x^3 - 4)^{-1}$ stated (or implied at •⁴); •⁴ $\displaystyle -9(x^3 - 4)^{-2}\ldots$ ; •⁵ $\displaystyle -27x^2(x^3 - 4)^{-2}.$
Commonly observed: 2. (b) B, candidates who use product rule: •³ $\displaystyle 6x^2(x^3 - 4)^{-1} + (2x^3 + 1)\ldots$ or $\displaystyle \ldots(x^3 - 4)^{-1} - 3x^2(2x^3 + 1)(x^3 - 4)^{-2}$ ; •⁴ $\displaystyle 6x^2(x^3 - 4)^{-1} - 3x^2(2x^3 + 1)(x^3 - 4)^{-2}$ ; •⁵ $\displaystyle -27x^2(x^3 - 4)^{-2}.$

Product, quotient, chain rules

3 Marks

2019 Q1(b)

Given $\displaystyle y = \frac{2x^3+1}{x^3-4}$ , find $\displaystyle \frac{dy}{dx}.$ Simplify your answer.

Show answer

\displaystyle \frac{-27x^2}{(x^3 - 4)^2}

Show marking instructions

Mark	Awarded for	Illustrative answer
(b) •³	evidence use of quotient rule with denominator and one term of numerator correct	$\displaystyle \dfrac{6x^2(x^3 - 4) - \ldots}{(x^3 - 4)^2}$ OR $\displaystyle \dfrac{\ldots - (2x^3 + 1)(3x^2)}{(x^3 - 4)^2}$
(b) •⁴	complete differentiation	$\displaystyle \dfrac{6x^2(x^3 - 4) - (2x^3 + 1)(3x^2)}{(x^3 - 4)^2}$
(b) •⁵	simplify	$\displaystyle \dfrac{-27x^2}{(x^3 - 4)^2}$

Notes

Commonly observed: 1. (a) A: $\displaystyle f(x) = x^6\cot 5x = 6x^5\cot 5x - 5x^6\operatorname{cosec}^2 5x$ , award •² only.
Commonly observed: 2. (a) B: $\displaystyle x^6\cot 5x = x^6\tan^{-1}(5x)$ , $\displaystyle f'(x) = 6x^5\tan^{-1}(5x) + \dfrac{5x^6}{1 + (5x)^2}$ , award •² only.
Commonly observed: 3. (a) C: $\displaystyle f(x) = \dfrac{x^6}{\tan 5x}$ , $\displaystyle f'(x) = \dfrac{6x^5\tan 5x - 5x^6\sec^2 5x}{(\tan 5x)^2}$ , award •¹ and •².
Commonly observed: 4. (a) D: $\displaystyle f(x) = x^6(\tan 5x)^{-1}$ , $\displaystyle f'(x) = 6x^5(\tan 5x)^{-1} - x^6(\tan 5x)^{-2}5\sec^2 5x$ , award •¹ and •².
(b) 1. •⁵ is available only where candidates have multiplied out brackets and collected like terms in the numerator.
(b) 2. •⁵ is not available where a candidate produces further incorrect simplification subsequent to a correct answer.
Commonly observed: 1. (b) A, candidates who rewrite the function as $\displaystyle y = 2 + \dfrac{9}{x^3 - 4}$ : •³ $\displaystyle y = 2 + 9(x^3 - 4)^{-1}$ stated (or implied at •⁴); •⁴ $\displaystyle -9(x^3 - 4)^{-2}\ldots$ ; •⁵ $\displaystyle -27x^2(x^3 - 4)^{-2}.$
Commonly observed: 2. (b) B, candidates who use product rule: •³ $\displaystyle 6x^2(x^3 - 4)^{-1} + (2x^3 + 1)\ldots$ or $\displaystyle \ldots(x^3 - 4)^{-1} - 3x^2(2x^3 + 1)(x^3 - 4)^{-2}$ ; •⁴ $\displaystyle 6x^2(x^3 - 4)^{-1} - 3x^2(2x^3 + 1)(x^3 - 4)^{-2}$ ; •⁵ $\displaystyle -27x^2(x^3 - 4)^{-2}.$

Product, quotient, chain rules

3 Marks

2019 Q1(c)

For $\displaystyle f(x) = \cos^{-1} 2x$ , evaluate $\displaystyle f'\left(\frac{\sqrt{3}}{4}\right).$

Show answer

\displaystyle -4

Show marking instructions

Mark	Awarded for	Illustrative answer
(c) •⁶	start differentiation	$\displaystyle \dfrac{-1}{\sqrt{1 - (2x)^2}}$
(c) •⁷	complete differentiation	$\displaystyle \dfrac{-1}{\sqrt{1 - (2x)^2}} \times 2$
(c) •⁸	evaluate	$\displaystyle -4$

Notes

Commonly observed: 1. (a) A: $\displaystyle f(x) = x^6\cot 5x = 6x^5\cot 5x - 5x^6\operatorname{cosec}^2 5x$ , award •² only.
Commonly observed: 2. (a) B: $\displaystyle x^6\cot 5x = x^6\tan^{-1}(5x)$ , $\displaystyle f'(x) = 6x^5\tan^{-1}(5x) + \dfrac{5x^6}{1 + (5x)^2}$ , award •² only.
Commonly observed: 3. (a) C: $\displaystyle f(x) = \dfrac{x^6}{\tan 5x}$ , $\displaystyle f'(x) = \dfrac{6x^5\tan 5x - 5x^6\sec^2 5x}{(\tan 5x)^2}$ , award •¹ and •².
Commonly observed: 4. (a) D: $\displaystyle f(x) = x^6(\tan 5x)^{-1}$ , $\displaystyle f'(x) = 6x^5(\tan 5x)^{-1} - x^6(\tan 5x)^{-2}5\sec^2 5x$ , award •¹ and •².
Commonly observed: 1. (b) A, candidates who rewrite the function as $\displaystyle y = 2 + \dfrac{9}{x^3 - 4}$ : •³ $\displaystyle y = 2 + 9(x^3 - 4)^{-1}$ stated (or implied at •⁴); •⁴ $\displaystyle -9(x^3 - 4)^{-2}\ldots$ ; •⁵ $\displaystyle -27x^2(x^3 - 4)^{-2}.$
Commonly observed: 2. (b) B, candidates who use product rule: •³ $\displaystyle 6x^2(x^3 - 4)^{-1} + (2x^3 + 1)\ldots$ or $\displaystyle \ldots(x^3 - 4)^{-1} - 3x^2(2x^3 + 1)(x^3 - 4)^{-2}$ ; •⁴ $\displaystyle 6x^2(x^3 - 4)^{-1} - 3x^2(2x^3 + 1)(x^3 - 4)^{-2}$ ; •⁵ $\displaystyle -27x^2(x^3 - 4)^{-2}.$
(c) 1. At •⁶ do not accept $\displaystyle \dfrac{-1}{\sqrt{1 - 2x^2}}$ unless either $\displaystyle \dfrac{\ldots}{\sqrt{1 - (2x)^2}}$ or $\displaystyle \dfrac{\ldots}{\sqrt{1 - 4x^2}}$ appears at •⁷.
(c) 2. •⁸ is available only where a candidate's answer is consistent with their stated derivative.
(c) 3. Where a candidate produces an incorrect, rounded answer; at least 2 significant figures are required for the award of •⁸.

Determinant and inverseMatrix operations

(3, 2, 1)6 Marks

2019 Q2

Matrix $\displaystyle A$ is defined by

$\displaystyle A = \begin{pmatrix} 2 & 1 & 4 \\ -3 & p & 2 \\ -1 & -2 & 5 \end{pmatrix}$

where $\displaystyle p \in \mathbb{R}.$
(a)Given that the determinant of $\displaystyle A$ is 3, find the value of $\displaystyle p.$

Matrix $\displaystyle B$ is defined by

$\displaystyle B = \begin{pmatrix} 0 & 1 \\ q & 3 \\ 4 & 0 \end{pmatrix}$

where $\displaystyle q \in \mathbb{R}.$
(b)Find $\displaystyle AB.$

(c)Explain why $\displaystyle AB$ does not have an inverse.

Show answer

(a)

\displaystyle -3

(b)

\displaystyle \begin{pmatrix} q+16 & 5 \\ -3q+8 & -12 \\ -2q+20 & -7 \end{pmatrix}

(c)

\displaystyle AB

is not a square matrix, and inverses are only defined for square matrices.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	begin process	eg $\displaystyle 2\begin{vmatrix}p & 2\\-2 & 5\end{vmatrix} - 1\begin{vmatrix}-3 & 2\\-1 & 5\end{vmatrix} + 4\begin{vmatrix}-3 & p\\-1 & -2\end{vmatrix}$
(a) •²	find determinant	$\displaystyle 14p + 45$
(a) •³	equate to 3 and find $\displaystyle p$	$\displaystyle -3$
(b) •⁴	any two simplified entries	(•⁴ and •⁵ together) $\displaystyle \begin{pmatrix}q + 16 & 5\\-3q + 8 & -12\\-2q + 20 & -7\end{pmatrix}$
(b) •⁵	complete multiplication	$\displaystyle \begin{pmatrix}q + 16 & 5\\-3q + 8 & -12\\-2q + 20 & -7\end{pmatrix}$
(c) •⁶	explain	$\displaystyle AB$ is not a square matrix AND a general statement about square matrices

Notes

(a) 1. Where a candidate interchanges any 2 rows, •¹ is available only where the determinant is equated to $\displaystyle -3.$ •² and •³ are still available.
(a) 2. At •² accept $\displaystyle 2(5p + 4) - 1(-13) + 4(6 + p).$
(b) 1. If the order of the resultant matrix is not $\displaystyle 3 \times 2$ award 0/2.
(b) 2. For the award of •⁴ and •⁵, accept $\displaystyle \begin{pmatrix}q + 16 & 5\\pq + 8 & -3 + 3p\\-2q + 20 & -7\end{pmatrix}.$
(c) 1. A general statement about square matrices could take the following form: only square matrices have an inverse; only square matrices have a determinant; only square matrices have an identity or unit matrix.
(c) 2. Where the answer contains incorrect information (before, between or after correct information), •⁶ is not available.
Commonly observed: 1. (c) A, acceptable explanations: "It's not a square matrix and inverses are only defined for square matrices"; "Since an identity matrix only exists for square matrices an inverse cannot be found. $\displaystyle AB$ is not a square matrix"; "You can only find an inverse if you can find a determinant. Only $\displaystyle 2 \times 2$ or $\displaystyle 3 \times 3$ matrices have a determinant. Since $\displaystyle AB$ is not $\displaystyle 2 \times 2$ or $\displaystyle 3 \times 3$ , you cannot find a determinant so it has no inverse".
Commonly observed: 2. (c) B, insufficient/unacceptable explanations: "It's not a square matrix so no inverse exists" (restates already given information); " $\displaystyle AB$ is not a square matrix. Only square matrices have an inverse. The determinant of $\displaystyle AB$ is 0"; "It's not a square matrix so it has no identity matrix to invert it with" (meaning of second part of the statement is unclear); "It's not a $\displaystyle 2 \times 2$ or a $\displaystyle 3 \times 3$ matrix so the determinant cannot be found" (no general comment linking determinant and square matrices).

Odd and even functionsSketching related functions

(1, 1)2 Marks

2019 Q3

The function $\displaystyle f(x)$ is defined by $\displaystyle f(x) = x^2 - a^2.$ The graph of $\displaystyle y = f(x)$ is shown in the diagram.

(a)State whether $\displaystyle f(x)$ is odd, even or neither. Give a reason for your answer.

(b)Sketch the graph of $\displaystyle y = |f(x)|.$

Show answer

(a) Even. The graph is symmetrical about the

\displaystyle y

-axis (or

\displaystyle f(-x) = f(x)

).
(b) A curve meeting the

\displaystyle x

-axis in sharp points at

\displaystyle -a

and

\displaystyle a

, with a local maximum at

\displaystyle (0, a^2)

on the

\displaystyle y

-axis, the parts for

\displaystyle |x| > a

unchanged, and line symmetry about the

\displaystyle y

-axis.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	state why function is even	graph is symmetrical about the $\displaystyle y$ -axis $\displaystyle \therefore$ even OR $\displaystyle f(-x) = (-x)^2 - a^2 = x^2 - a^2 = f(x)$ $\displaystyle \therefore$ even
(b) •²	sketch graph	a sketch of $\displaystyle y = \|f(x)\|$ : the parabola's middle section reflected above the $\displaystyle x$ -axis, so a maximum on the $\displaystyle y$ -axis between sharp points on the $\displaystyle x$ -axis at $\displaystyle -a$ and $\displaystyle a$ (both labelled), with the outer arms rising as before

Notes

(a) 1. Do not accept use of the word 'reflected'.
(a) 2. Accept phrases such as 'symmetrical in the $\displaystyle y$ -axis', 'symmetrical around the $\displaystyle y$ -axis' etc.
(a) 3. For justification using the graph, explicit mention of the $\displaystyle y$ -axis or the line $\displaystyle x = 0$ must be made.
(a) 4. •¹ is not available for only stating ' $\displaystyle f(-x) = f(x)$ $\displaystyle \therefore$ even' or ' $\displaystyle f(-a) = f(a)$ $\displaystyle \therefore$ even'.
(a) 5. •¹ is not available for ' $\displaystyle f(-x) = -x^2 - a^2 = x^2 - a^2 = f(x)$ $\displaystyle \therefore$ even'.
(a) 6. Where the answer contains incorrect information (before, between or after correct information), •¹ is not available.
(b) 1. The (local) maximum turning point must be on the $\displaystyle y$ -axis and the graph must exhibit line symmetry.
(b) 2. Do not award •² if the $\displaystyle x$ intercepts are not labelled.
(b) 3. Graph must not be 'smooth' at $\displaystyle x$ intercepts.
(b) 4. A candidate must make a reasonable attempt at reproduction when $\displaystyle x < -a$ and $\displaystyle x > a.$

Partial fractions (improper)

(1, 3)4 Marks

2019 Q4

(a)Express $\displaystyle \frac{3x^2+x-17}{x^2-x-12}$ in the form

$\displaystyle p + \frac{qx+r}{x^2-x-12}$

where $\displaystyle p$ , $\displaystyle q$ and $\displaystyle r$ are integers.

(b)Hence express $\displaystyle \frac{3x^2+x-17}{x^2-x-12}$ with partial fractions.

Show answer

(a)

\displaystyle 3 + \frac{4x+19}{x^2-x-12}

(b)

\displaystyle 3 - \frac{1}{x+3} + \frac{5}{x-4}

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	complete algebraic division and express in required form	$\displaystyle 3 + \dfrac{4x + 19}{x^2 - x - 12}$
(b) •²	state expression	$\displaystyle \dfrac{A}{x + 3} + \dfrac{B}{x - 4}$
(b) •³	form linear equation and obtain one constant	$\displaystyle 4x + 19 = B(x + 3) + A(x - 4)$ , $\displaystyle B = 5$ or $\displaystyle A = -1$
(b) •⁴	obtain final constant and state full expression	$\displaystyle 3 - \dfrac{1}{x + 3} + \dfrac{5}{x - 4}$

Notes

(b) 1. Where a candidate incorrectly factorises, •² is not available but •³ and •⁴ may still be awarded, including the situations illustrated in the Commonly Observed Responses.
(b) 2. Do not accept $\displaystyle 3 + -\dfrac{1}{x + 3} + \dfrac{5}{x - 4}$ at •⁴. Accept $\displaystyle 3 + \dfrac{-1}{x + 3} + \dfrac{5}{x - 4}.$
Commonly observed: 1. (b) $\displaystyle 3 + \dfrac{4x + 19}{x^2 - x - 12} = \dfrac{A}{x + 3} + \dfrac{B}{x - 4}$ , award •²; $\displaystyle 4x + 19 = A(x - 4) + B(x + 3)$ , $\displaystyle A = -1$ or $\displaystyle B = 5$ , award •³; leading to a final answer of $\displaystyle 3 - \dfrac{1}{x + 3} + \dfrac{5}{x - 4}$ , award •⁴.
Commonly observed: 2. (b) As 1, but leading to a final answer of $\displaystyle -\dfrac{1}{x + 3} + \dfrac{5}{x - 4}$ : award •² and •³, do not award •⁴.
Commonly observed: 3. (b) $\displaystyle \dfrac{4x + 19}{x^2 - x - 12} = \dfrac{A}{x + 3} + \dfrac{Bx + C}{x - 4}$ , award •²; $\displaystyle 4x + 19 = A(x - 4) + (Bx + C)(x + 3)$ , $\displaystyle A = -1$ or $\displaystyle B = 0$ or $\displaystyle C = 5$ , award •³; leading to $\displaystyle 3 + \dfrac{5}{x - 4} - \dfrac{1}{x + 3}$ , award •⁴ (award 2/3 if $\displaystyle B \neq 0$ ).
Commonly observed: 4. (b) $\displaystyle \dfrac{3x^2 + x - 17}{x^2 - x - 12} = \dfrac{A}{x + 3} + \dfrac{B}{x - 4}$ , do not award •²; $\displaystyle 3x^2 + x - 17 = A(x - 4) + B(x + 3)$ , $\displaystyle A = -1$ or $\displaystyle B = 5$ , award •³ but •⁴ is not available.
Commonly observed: 5. (b) $\displaystyle \dfrac{3x^2 + x - 17}{x^2 - x - 12} = \dfrac{A}{x + 3} + \dfrac{Bx + C}{x - 4}$ , do not award •²; $\displaystyle 3x^2 + x - 17 = A(x - 4) + (Bx + C)(x + 3)$ , $\displaystyle A = -1$ or $\displaystyle B = 3$ or $\displaystyle C = -7$ , award •³ but •⁴ is not available.
Commonly observed: 6. (b) As 5, then $\displaystyle 3x^2 + x - 17 = A(x - 4) + (Bx + C)(x + 3)$ with $\displaystyle A = -1$ or $\displaystyle B = 3$ or $\displaystyle C = -7$ , award •²; $\displaystyle \dfrac{3x - 7}{x - 4} = 3 + \dfrac{5}{x - 4}$ leading to $\displaystyle 3 - \dfrac{1}{x + 3} + \dfrac{5}{x - 4}$ , award •³ and •⁴.

Parametric differentiation

(2, 2)4 Marks

2019 Q5

For $\displaystyle x = \ln(2t+7)$ and $\displaystyle y = t^2$ , $\displaystyle t > 0$ , find

(a) $\displaystyle \frac{dy}{dx}$

(b) $\displaystyle \frac{d^2y}{dx^2}.$

Show answer

(a)

\displaystyle \frac{1}{2}(2t+7)(2t)

or

\displaystyle 2t^2+7t

(b)

\displaystyle \frac{1}{2}(2t+7)(4t+7)

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	find $\displaystyle \dfrac{dx}{dt}$	$\displaystyle \dfrac{2}{2t + 7}$
(a) •²	find $\displaystyle \dfrac{dy}{dx}$	$\displaystyle 2t^2 + 7t$
(b) •³	differentiate $\displaystyle \dfrac{dy}{dx}$ w.r.t. $\displaystyle t$ and evidence of strategy	$\displaystyle (4t + 7) \times \ldots$
(b) •⁴	find $\displaystyle \dfrac{d^2y}{dx^2}$	$\displaystyle \dfrac{1}{2}(2t + 7)(4t + 7)$

Notes

(a) 1. For •² do not accept $\displaystyle \dfrac{t}{\frac{1}{2t + 7}}.$
Commonly observed: 1. (a) Candidates who express $\displaystyle y$ explicitly as a function of $\displaystyle x$ : •¹ $\displaystyle y = \dfrac{1}{4}(e^x - 7)^2$ ; •² $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{2}(e^x - 7)e^x.$
(b) 1. •³ and •⁴ are not available to candidates who only differentiate $\displaystyle \dfrac{dy}{dx}$ w.r.t. $\displaystyle t.$ Evidence of multiplication or division by a function of $\displaystyle t$ - other than $\displaystyle \ln(2t + 7)$ or $\displaystyle t^2$ - must be present.
(b) 2. At •⁴, accept $\displaystyle \dfrac{1}{2}(8t^2 + 42t + 49).$
Commonly observed: 1. (b) Candidates who express $\displaystyle y$ explicitly as a function of $\displaystyle x$ : $\displaystyle \dfrac{1}{2}(e^x - 7)e^x + \dfrac{1}{2}e^x(e^x)$ , award •³ for one term correct; $\displaystyle e^{2x} - \dfrac{7}{2}e^x$ , award •⁴.
Commonly observed: 2. (b) Candidates who take a formula approach: $\displaystyle \dfrac{2\frac{2}{2t + 7} - \ldots}{\left(\frac{2}{2t + 7}\right)^3}$ or $\displaystyle \dfrac{\ldots - 2t \times (-4(2t + 7)^{-2})}{\left(\frac{2}{2t + 7}\right)^3}$ , award •³; $\displaystyle \dfrac{1}{2}(2t + 7)(4t + 7)$ , award •⁴.

Related rates of change

3 Marks

2019 Q6

A spherical balloon of radius $\displaystyle r$ cm, $\displaystyle r > 0$ , deflates at a constant rate of 60 cm $\displaystyle ^3$ s $\displaystyle ^{-1}.$
Calculate the rate of change of the radius with respect to time when $\displaystyle r = 3.$
[The volume of a sphere is given by $\displaystyle V = \frac{4}{3}\pi r^3.$ ]

Show answer

\displaystyle -\frac{5}{3\pi}

cm s

\displaystyle ^{-1}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	evidence of relationship	$\displaystyle \dfrac{dV}{dr} = 4\pi r^2$ AND $\displaystyle \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$ OR $\displaystyle \dfrac{dr}{dt} = \dfrac{dV}{dt} \times \dfrac{dr}{dV}$
•²	substitute	$\displaystyle -60 = 4\pi(3)^2\dfrac{dr}{dt}$ OR $\displaystyle \dfrac{dr}{dt} = \dfrac{-60}{4\pi(3)^2}$
•³	evaluate	$\displaystyle -\dfrac{5}{3\pi}\ \text{cms}^{-1}$

Notes

1. At •³ units are required. Accept decimal equivalent to at least 2 significant figures ( $\displaystyle -0.53\ \text{cms}^{-1}$ ).
2. •² may be implied at •³.
Commonly observed: 1. A, candidate attaches units to an exact value but omits them from a final answer (correctly rounded or otherwise): $\displaystyle -\dfrac{5}{3\pi}\ \text{cms}^{-1} = -0.5$ , award •³.
Commonly observed: 2. B, candidate attaches units to an incorrect decimal approximation and not to the exact value (or appropriately rounded decimal approximation): $\displaystyle -\dfrac{5}{3\pi}$ or $\displaystyle -0.53 = -0.5\ \text{cms}^{-1}$ , do not award •³.

Summation formulae

(1, 2)3 Marks

2019 Q7

(a)Find an expression for $\displaystyle \sum_{r=1}^{n}(6r+13)$ in terms of $\displaystyle n.$

(b)Hence, or otherwise, find $\displaystyle \sum_{r=p+1}^{20}(6r+13).$

Show answer

(a)

\displaystyle 3n^2 + 16n

(b)

\displaystyle 1520 - 3p^2 - 16p

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	find expression	$\displaystyle 3n^2 + 16n$
(b) •²	substitute 20 and evidence of subtraction from this term	$\displaystyle (3 \times 20^2 + 16 \times 20) - \ldots$
(b) •³	substitute for $\displaystyle p$ and find expression	$\displaystyle 1520 - 3p^2 - 16p$

Notes

(a) 1. At •¹ accept $\displaystyle 6 \times \dfrac{n(n + 1)}{2} + 13 \times n.$
(a) 2. At •¹ accept $\displaystyle \dfrac{1}{2}n[38 + 6(n - 1)]$ obtained via an arithmetic series.
(b) 1. Where a candidate produces further incorrect simplification, subsequent to •¹ being awarded, •² is not available.
(b) 2. Award •² for $\displaystyle \displaystyle\sum_1^{20}(6r + 13) - \sum_1^{p}(6r + 13)$ only where the substitution is not carried out. Disregard errors in sigma notation provided a candidate produces an answer consistent with their response to (a).
(b) 3. Do not award •³ for incorrect working subsequent to a correct answer.
Commonly observed: 1. (b) A: $\displaystyle 6 \times \dfrac{n(n + 1)}{2} + 13$ , an incorrect expression from (a), leading to $\displaystyle (3 \times 20^2 + 3 \times 20 + 13) - \ldots$ , award •²; $\displaystyle 1260 - 3p^2 - 3p$ , award •³.

Second order homogeneous DE

5 Marks

2019 Q8

Find the particular solution of the differential equation

$\displaystyle \frac{d^2y}{dx^2} + 11\frac{dy}{dx} + 28y = 0$

given that $\displaystyle y = 0$ and $\displaystyle \frac{dy}{dx} = 9$ , when $\displaystyle x = 0.$

Show answer

\displaystyle y = 3e^{-4x} - 3e^{-7x}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	solve auxiliary equation	$\displaystyle m = -4, -7$
•²	state general solution	$\displaystyle y = Ae^{-4x} + Be^{-7x}$
•³	differentiate	$\displaystyle \dfrac{dy}{dx} = -4Ae^{-4x} - 7Be^{-7x}$ stated or implied at •⁴
•⁴	form equations and solve for a constant	$\displaystyle A = 3$ or $\displaystyle B = -3$
•⁵	find second constant and state particular solution	$\displaystyle y = 3e^{-4x} - 3e^{-7x}$

Notes

1. Do not withhold •² for the omission of ' $\displaystyle y =$ '.
2. Do not withhold •³ for the omission of ' $\displaystyle \dfrac{dy}{dx} =$ '.
3. To award •⁵, ' $\displaystyle y =$ ' must be present.

Binomial terms

(3, 2)5 Marks

2019 Q9

(a)Write down and simplify the general term in the binomial expansion of

$\displaystyle \left(2x^2 - \frac{d}{x^3}\right)^7$ ,

where $\displaystyle d$ is a constant.

(b)Given that the coefficient of $\displaystyle \frac{1}{x}$ is $\displaystyle -70\ 000$ , find the value of $\displaystyle d.$

Show answer

(a)

\displaystyle \binom{7}{r} 2^{7-r} (-d)^r x^{14-5r}

(b)

\displaystyle d = 5

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	state general term	$\displaystyle \dbinom{7}{r}(2x^2)^{7-r}\left(\dfrac{-d}{x^3}\right)^r$
(a) •²	simplify powers of $\displaystyle x$ or coefficients	$\displaystyle x^{14-5r}$ or $\displaystyle 2^{7-r}(-d)^r$
(a) •³	state simplified general term (complete simplification)	$\displaystyle \dbinom{7}{r}2^{7-r}(-d)^r x^{14-5r}$
(b) •⁴	obtain value of $\displaystyle r$	$\displaystyle r = 3$
(b) •⁵	find value of $\displaystyle d$	$\displaystyle d = 5$

Notes

(a) 1. Candidates may also start with a general term of $\displaystyle \dbinom{7}{r}(2x^2)^r\left(\dfrac{-d}{x^3}\right)^{7-r}$ to obtain a simplified general term of $\displaystyle \dbinom{7}{r}2^r(-d)^{7-r}x^{-21+5r}.$
(a) 2. Where candidates write out a full binomial expansion, •¹, •² and •³ are not available unless the general term is identifiable in (b).
(a) 3. Candidates who write down $\displaystyle \dbinom{7}{r}2^{7-r}(-d)^r x^{14-5r}$ with no working receive full marks.
(a) 4. •³ is unavailable to candidates who, in (a), produce further incorrect simplification subsequent to a correct answer eg $\displaystyle (-2d)^{7-2r}.$
(a) 5. Where $\displaystyle 2^{7-r}$ and $\displaystyle x^{14-5r}$ do not appear within a single term, •³ is not available.
Commonly observed: 1. (a) General term has not been isolated: $\displaystyle \displaystyle\sum_{r=0}^{7}\dbinom{7}{r}(2x^2)^{7-r}\left(\dfrac{-d}{x^3}\right)^r = \sum_{r=0}^{7}\dbinom{7}{r}2^{7-r}(-d)^r x^{14-5r}.$ Do not award •¹. Award •² and •³.
Commonly observed: 2. (a) General term has been isolated: $\displaystyle \displaystyle\sum_{r=0}^{7}\dbinom{7}{r}(2x^2)^{7-r}\left(\dfrac{-d}{x^3}\right)^r = \dbinom{7}{r}2^{7-r}(-d)^r x^{14-5r}.$ Disregard the incorrect use of the final equals sign. Award •¹, •² and •³.
Commonly observed: 3. (a) Binomial expression has been equated to general term: $\displaystyle \left(2x^2 - \dfrac{d}{x^3}\right)^7 = \dbinom{7}{r}(2x^2)^{7-r}\left(\dfrac{-d}{x^3}\right)^r.$ Disregard the incorrect use of the equals sign. Award •¹.
Commonly observed: 4. (a) Negative sign omitted: $\displaystyle \dbinom{7}{r}(2x^2)^{7-r}\left(\dfrac{d}{x^3}\right)^r.$ Do not award •¹ but •² and •³ are still available.
Commonly observed: 5. (a) Brackets omitted around $\displaystyle -d$ : $\displaystyle \dbinom{7}{r}2^{7-r} - d^r x^{14-5r}.$ Do not award •³.
(b) 1. The alternative expansion leads to $\displaystyle r = 4.$
(b) 2. Where a candidate writes out a full expansion •⁴ may be awarded only where this is complete and correct at least as far as the required term (in either direction).
(b) 3. Where a candidate obtains an incorrect binomial expansion, •⁵ will be available only where the evaluation of a root is required.
Commonly observed: 1. (b) Binomial expansion: $\displaystyle 128x^{14} - 448dx^9 + 672d^2x^4 - 560d^3x^{-1} + 280d^4x^{-6} - 84d^5x^{-11} + 14d^6x^{-16} - d^7x^{-21}.$

Implicit differentiation

(3, 2)5 Marks

2019 Q10

A curve is defined implicitly by the equation $\displaystyle x^2 + y^2 = xy + 12.$

(a)Find an expression for $\displaystyle \frac{dy}{dx}$ in terms of $\displaystyle x$ and $\displaystyle y.$

(b)There are two points where the tangent to the curve has equation $\displaystyle x = k$ , $\displaystyle k \in \mathbb{R}.$
Find the values of $\displaystyle k.$

Show answer

(a)

\displaystyle \frac{y - 2x}{2y - x}

(b)

\displaystyle k = \pm 4

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	apply chain or product rule	$\displaystyle 2y\dfrac{dy}{dx}$ or $\displaystyle y + x\dfrac{dy}{dx}$
(a) •²	complete differentiation	$\displaystyle 2x + 2y\dfrac{dy}{dx} = y + x\dfrac{dy}{dx}$
(a) •³	express $\displaystyle \dfrac{dy}{dx}$ in terms of $\displaystyle x$ and $\displaystyle y$	$\displaystyle \dfrac{dy}{dx} = \dfrac{y - 2x}{2y - x}$
(b) •⁴	equate denominator of $\displaystyle \dfrac{dy}{dx}$ to zero	$\displaystyle 2y - x = 0$
(b) •⁵	calculate values of $\displaystyle k$	$\displaystyle k = \pm 4$

Notes

(a) 1. •³ is available only where $\displaystyle \dfrac{dy}{dx}$ appears more than once, after the candidate has completed their differentiation.
(b) 1. At •⁵, accept $\displaystyle x = \pm 4.$
(b) 2. Where a candidate equates the numerator to zero, •⁴ and •⁵ are not available.
Commonly observed: 1. (b) Intersection method: $\displaystyle y^2 - ky + (k^2 - 12) = 0$ (substitute for $\displaystyle x$ and express in general form); •⁴ $\displaystyle (-k)^2 - 4(k^2 - 12) = 0$ (communicate condition for equal roots); •⁵ $\displaystyle k = \pm 4.$

Disproof by counterexampleProof by contrapositive

(1, 1, 3)5 Marks

2019 Q11

Let $\displaystyle n$ be a positive integer.

(a)Find a counterexample to show that the following statement is false.

$\displaystyle n^2 + n + 1$ is always a prime number.

(b)(i) Write down the contrapositive of:
If $\displaystyle n^2 - 2n + 7$ is even then $\displaystyle n$ is odd.

(ii) Use the contrapositive to prove that if $\displaystyle n^2 - 2n + 7$ is even then $\displaystyle n$ is odd.

Show answer

(a) When

\displaystyle n=4

,

\displaystyle n^2+n+1 = 21

, which is not prime.
(b)(i) If

\displaystyle n

is even (not odd) then

\displaystyle n^2 - 2n + 7

is odd (not even).
(b)(ii) Let

\displaystyle n = 2k

,

\displaystyle k \in \mathbb{N}.

\displaystyle n^2 - 2n + 7 = (2k)^2 - 2(2k) + 7 = 4k^2 - 4k + 7 = 2(2k^2 - 2k + 3) + 1.

Since

\displaystyle 2k^2 - 2k + 3 \in \mathbb{N}

, this is odd. The contrapositive statement is true and therefore the original statement is true.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	state counterexample	eg when $\displaystyle n = 4$ , $\displaystyle n^2 + n + 1 = 21$ which is not prime
(b)(i) •²	write down contrapositive statement	If $\displaystyle n$ is even then $\displaystyle n^2 - 2n + 7$ is odd
(b)(ii) •³	write down appropriate form for $\displaystyle n$ AND substitute	$\displaystyle n = 2k$ , $\displaystyle k \in \mathbb{N}$ and $\displaystyle (2k)^2 - 2(2k) + 7$
(b)(ii) •⁴	show $\displaystyle n^2 - 2n + 7$ is odd	eg $\displaystyle 2(2k^2 - 2k + 3) + 1$ which is odd since $\displaystyle 2k^2 - 2k + 3 \in \mathbb{N}$
(b)(ii) •⁵	communicate	contrapositive statement is true AND therefore original statement is true

Notes

(a) 1. A candidate must demonstrate a value of $\displaystyle n$ , evaluate $\displaystyle n^2 + n + 1$ and communicate that this value is not prime.
(a) 2. Where the answer contains incorrect information (before, between or after correct information), •¹ is not available.
Commonly observed: 1. (a) $\displaystyle 4^2 + 4 + 1 = 21$ , which is not prime: award •¹ (value of $\displaystyle n$ has been demonstrated).
(b) 1. Marks •², •³, •⁴ and •⁵ are not available to a candidate whose statement of the contrapositive begins "If $\displaystyle n^2 - 2n + 7$ …".
(b) 2. Award •² for 'If $\displaystyle n$ is not odd then $\displaystyle n^2 - 2n + 7$ is not even'.
(b) 3. At •³ accept $\displaystyle k \in \mathbb{Z}^+$ but do not accept $\displaystyle k \in \mathbb{Z}.$
(b) 4. At •³ do not accept $\displaystyle n = 2n.$
(b) 5. At •³ the form of $\displaystyle n$ must be consistent with the candidate's response to b(i).
(b) 6. Do not withhold •⁴ for the omission of $\displaystyle 2k^2 - 2k + 3 \in \mathbb{N}.$
(b) 7. At •⁴ accept any valid expression of the form $\displaystyle ab + c$ , where $\displaystyle a$ is even, $\displaystyle b$ is an integer and $\displaystyle c$ is odd.
(b) 8. •⁵ is available only where a candidate's conclusion states that the contrapositive is true and links to the original statement.
(b) 9. Where a candidate's response mentions contradiction, •³, •⁴ and •⁵ are not available.
Commonly observed: 1. (b) Refer to note 3 when considering any of the responses below. Where a candidate uses $\displaystyle n = 2k + 1$ then $\displaystyle k$ must be suitably defined eg " $\displaystyle k$ is a whole number".
Commonly observed: 2. (b) A: "If $\displaystyle n$ is odd then $\displaystyle n^2 - 2n + 7$ is even", do not award •²; $\displaystyle n = 2k - 1$ , $\displaystyle k \in \mathbb{N}$ , $\displaystyle (2k - 1)^2 - 2(2k - 1) + 7$ , award •³; $\displaystyle 2(2k^2 - 4k + 5)$ which is even, award •⁴; "The contrapositive statement is true so the original statement is true", award •⁵.
Commonly observed: 3. (b) B: "If $\displaystyle n$ is odd then $\displaystyle n^2 - 2n + 7$ is odd", do not award •²; $\displaystyle n = 2k - 1$ , $\displaystyle (2k - 1)^2 - 2(2k - 1) + 7$ , award •³; $\displaystyle 2(2k^2 - 4k + 5)$ which is not odd, do not award •⁴; •⁵ is not available.
Commonly observed: 4. (b) C: "If $\displaystyle n$ is even then $\displaystyle n^2 - 2n + 7$ is even", do not award •²; $\displaystyle n = 2k$ , $\displaystyle (2k)^2 - 2(2k) + 7$ , award •³; $\displaystyle 2(2k^2 - 2k + 3) + 1$ which is odd, do not award •⁴; •⁵ is not available.

Number bases

3 Marks

2019 Q12

Express $\displaystyle 231_{11}$ in base 7.

Show answer

\displaystyle 543_7

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	convert to base 10	$\displaystyle 276$
•²	method leading to a quotient of 0 or equivalent	eg $\displaystyle 276 = 7 \times 39 + 3$ , $\displaystyle 39 = 7 \times 5 + 4$ , $\displaystyle 5 = 7 \times 0 + 5$
•³	express in base 7	$\displaystyle 543_7$

Notes

1. Where a candidate converts $\displaystyle 231_{10}$ into a number in base 7, •¹ is not available.
2. At •³, disregard the omission of base 7.
3. A candidate who finds three, or more, remainders and writes them in reverse order may be awarded •³.
Commonly observed: 1. A place-value table with $\displaystyle 7^3, 7^2, 7^1, 7^0$ over $\displaystyle 343, 49, 7, 1$ and $\displaystyle 5, 4, 3$ beneath: award •² for all entries in row 2 and the '5' in row 3; leading to 543 (identified), award •³.
Commonly observed: 2. Repeated division, $\displaystyle 276 \div 7$ giving 39 r 3, 5 r 4, 0 r 5: award •² (compare with illustrative scheme); leading to 543 (identified), award •³.
Commonly observed: 3. $\displaystyle 231 = 7 \times 33 + 0$ , $\displaystyle 33 = 7 \times 4 + 5$ , $\displaystyle 4 = 7 \times 0 + 4$ , leading to a final answer of 450: do not award •¹. Award •² and •³.

Separable DE in context

5 Marks

2019 Q13

An electronic device contains a timer circuit that switches off when the voltage, $\displaystyle V$ , reaches a set value.
The rate of change of the voltage is given by

$\displaystyle \frac{dV}{dt} = k(12 - V)$ ,

where $\displaystyle k$ is a constant, $\displaystyle t$ is the time in seconds, and $\displaystyle 0 \le V < 12.$
Given that $\displaystyle V = 2$ when $\displaystyle t = 0$ , express $\displaystyle V$ in terms of $\displaystyle k$ and $\displaystyle t.$

Show answer

\displaystyle V = 12 - 10e^{-kt}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	separate variables and write integral equation	$\displaystyle \displaystyle\int \dfrac{1}{12 - V}\,dV = \int k\,dt$
•²	integrate LHS	$\displaystyle -\ln(12 - V)$
•³	integrate RHS	$\displaystyle kt + c$
•⁴	evaluate constant of integration	$\displaystyle -\ln 10$
•⁵	express $\displaystyle V$ in terms of $\displaystyle k$ and $\displaystyle t$	$\displaystyle V = 12 - 10e^{-kt}$

Notes

1. Do not award •¹ where $\displaystyle \int\ldots dV$ and $\displaystyle \int\ldots dt$ do not appear.
2. For candidates who omit the constant of integration, •³ may be awarded but •⁴ and •⁵ are unavailable.
3. •⁵ is unavailable to candidates who omit the negative sign at •².
4. At •⁵, accept $\displaystyle V = 12 - \dfrac{10}{e^{kt}}$ or $\displaystyle V = \dfrac{12e^{kt} - 10}{e^{kt}}$ but do not accept the appearance of eg $\displaystyle e^{-kt + \ln 10}$ in the final answer.
Commonly observed: 1. Using integrating factor: $\displaystyle \dfrac{dV}{dt} + kV = 12k$ ; IF $\displaystyle = e^{kt}$ , award •¹; $\displaystyle \dfrac{d}{dt}(Ve^{kt}) = 12ke^{kt}$ , $\displaystyle Ve^{kt} = \displaystyle\int 12ke^{kt}\,dt$ , award •²; $\displaystyle Ve^{kt} = 12e^{kt} + c$ , award •³; $\displaystyle c = -10$ , award •⁴; $\displaystyle V = 12 - 10e^{-kt}$ , award •⁵.

Proof by induction

5 Marks

2019 Q14

Prove by induction that

$\displaystyle \sum_{r=1}^{n}r!r = (n+1)! - 1$

for all positive integers $\displaystyle n.$

Show answer

When

\displaystyle n=1

, LHS

\displaystyle =1!\times1=1

and RHS

\displaystyle =2!-1=1

, so the statement is true.
Assume true for

\displaystyle n=k

:

\displaystyle \sum_{r=1}^{k}r!\,r=(k+1)!-1.

Then

\displaystyle \sum_{r=1}^{k+1}r!\,r=(k+1)!-1+(k+1)!(k+1)=(k+1)!(k+2)-1=\left((k+1)+1\right)!-1.

True for

\displaystyle n=1

, and true for

\displaystyle n=k

implies true for

\displaystyle n=k+1

, so by induction the statement holds for all positive integers

\displaystyle n.

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	show true when $\displaystyle n = 1$	when $\displaystyle n = 1$ , LHS $\displaystyle = 1! \times 1 = 1$ , RHS $\displaystyle = (1 + 1)! - 1 = 1$ so result is true when $\displaystyle n = 1.$
•²	assume (statement) true for $\displaystyle n = k$ AND consider whether (statement) true for $\displaystyle n = k + 1$	suitable statement AND $\displaystyle \displaystyle\sum_{r=1}^{k}r!\,r = (k + 1)! - 1$ AND $\displaystyle \displaystyle\sum_{r=1}^{k+1}r!\,r = \ldots$
•³	state sum to $\displaystyle (k + 1)$ terms using inductive hypothesis	$\displaystyle (k + 1)! - 1 + (k + 1)!(k + 1)$
•⁴	extract $\displaystyle (k + 1)!$ as common factor	$\displaystyle (k + 1)!(k + 2) - 1$
•⁵	express sum explicitly in terms of $\displaystyle (k + 1)$ or achieve stated aim/goal AND communicate	$\displaystyle ((k + 1) + 1)! - 1$ AND If true for $\displaystyle n = k$ then true for $\displaystyle n = k + 1.$ Also shown true for $\displaystyle n = 1$ therefore, by induction, true for all positive integers $\displaystyle n.$

Notes

1. "RHS = 1, LHS = 1" and/or "True for $\displaystyle n = 1$ " are insufficient for the award of •¹. A candidate must demonstrate evidence of substitution into both expressions. Accept $\displaystyle 2! - 1$ for RHS. Where a candidate does not independently evaluate the LHS and RHS, •¹ may still be awarded.
2. For •² acceptable phrases for $\displaystyle n = k$ contain: "If true for..."; "Suppose true for…"; "Assume true for…". For •² insufficient phrases for $\displaystyle n = k$ contain: "Consider $\displaystyle n = k$ ", "assume $\displaystyle n = k$ ", "let $\displaystyle n = k$ ". For an insufficient phrase, do not award •² unless an acceptable statement subsequently appears as part of the conclusion at •⁵. For •² unacceptable phrases for $\displaystyle n = k$ contain: "True for $\displaystyle n = k$ ", "Consider true for $\displaystyle n = k$ ". For an unacceptable phrase, do not award •² but •⁵ may still be available. For •² unacceptable phrases for $\displaystyle n = k + 1$ contain: "Consider true for $\displaystyle n = k + 1$ ", "true for $\displaystyle n = k + 1$ "; " $\displaystyle \displaystyle\sum_{r=1}^{k+1}r!\,r = (k + 2)! - 1$ " (with no further working).
3. At •⁴ accept $\displaystyle (k + 1)!(1 + k + 1) - 1.$
4. •⁵ is unavailable to candidates who have not been awarded •⁴.
5. Full marks are available to candidates who state an aim/goal earlier in the proof and who subsequently achieve the stated aim/goal, provided $\displaystyle ((k + 1) + 1)! - 1$ appears at some point.
6. Following the required algebra and statement of the inductive hypothesis, the minimal acceptable response for •⁵ is: "Then true for $\displaystyle n = k + 1$ , but since true for $\displaystyle n = 1$ , then true for all $\displaystyle n$ " or equivalent.

Intersection of planesIntersection of lines

(2, 3, 4)9 Marks

2019 Q15

The equations of two planes are given below.
$\displaystyle \pi_1: 2x - 3y - z = 9$
$\displaystyle \pi_2: x + y - 3z = 2$

(a)Verify that the line of intersection, $\displaystyle L_1$ , of these two planes has parametric equations
$\displaystyle x = 2\lambda + 3$
$\displaystyle y = \lambda - 1$
$\displaystyle z = \lambda$

(b)Let $\displaystyle \pi_3$ be the plane with equation $\displaystyle -2x + 4y + 3z = 4.$
Calculate the acute angle between the line $\displaystyle L_1$ and the plane $\displaystyle \pi_3.$

(c) $\displaystyle L_2$ is the line perpendicular to $\displaystyle \pi_3$ passing through $\displaystyle P(1, 3, -2).$
Determine whether or not $\displaystyle L_1$ and $\displaystyle L_2$ intersect.

Show answer

(a) Substitute

\displaystyle x, y, z

into both plane equations: e.g.

\displaystyle 2(2\lambda+3) - 3(\lambda-1) - \lambda = 4\lambda + 6 - 3\lambda + 3 - \lambda = 9

(matches

\displaystyle \pi_1

).

\displaystyle (2\lambda+3) + (\lambda-1) - 3(\lambda) = 2

(matches

\displaystyle \pi_2

). Both verify successfully.
(b)

\displaystyle 13^\circ

or

\displaystyle 0.229

radians
(c) The lines do not intersect (equating parameters leads to an inconsistency, e.g.

\displaystyle 0 = -5

).

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	verify that the line lies on one plane (or: substitute parameter for $\displaystyle x$ , $\displaystyle y$ or $\displaystyle z$ into both equations; or: use vector product to find direction vector OR substitute eg $\displaystyle z = 0$ to find common point)	eg $\displaystyle 2(2\lambda + 3) - 3(\lambda - 1) - \lambda = 9$ (or $\displaystyle 2x - 3y - \lambda = 9$ , $\displaystyle x + y - 3\lambda = 2$ ; or $\displaystyle 10\mathbf{i} + 5\mathbf{j} + 5\mathbf{k}$ OR $\displaystyle (3, -1, 0)$ )
(a) •²	verify for other plane and state conclusion (or: solve simultaneous equations leading to parametric equations; or: find parametric equations)	eg $\displaystyle 2\lambda + 3 + \lambda - 1 - 3\lambda = 2$ ; therefore the line lies on both planes (or $\displaystyle x = 2\lambda + 3;\ y = \lambda - 1;\ z = \lambda$ )
(b) •³	identify vectors	$\displaystyle \begin{pmatrix}2\\1\\1\end{pmatrix}$ , $\displaystyle \begin{pmatrix}-2\\4\\3\end{pmatrix}$
(b) •⁴	start to calculate angle	$\displaystyle \cos\theta = \left(\dfrac{3}{\sqrt{6}\sqrt{29}}\right)$
(b) •⁵	calculate complement	any answer which rounds to $\displaystyle 0.229$ or $\displaystyle 13^\circ$
(c) •⁶	parametric equations for $\displaystyle L_2$	$\displaystyle x = -2\mu + 1;\ y = 4\mu + 3;\ z = 3\mu - 2$
(c) •⁷	two equations for two parameters	any two from $\displaystyle 2\lambda + 3 = -2\mu + 1$ ; $\displaystyle \lambda - 1 = 4\mu + 3$ ; $\displaystyle \lambda = 3\mu - 2$
(c) •⁸	solve for two possible parameters	eg $\displaystyle \mu = -1;\ \lambda = 0$
(c) •⁹	substitute into remaining equation and state conclusion	eg LHS $\displaystyle = 0$ , RHS $\displaystyle = -5$ so lines do not intersect.

Notes

(a) 1. •² is available only where there is supporting algebraic evidence.
(a) 2. Where a candidate elects to substitute the parametric equations for $\displaystyle L_1$ into the equations of $\displaystyle \pi_1$ and $\displaystyle \pi_2$ and concludes that " $\displaystyle L_1$ intersects $\displaystyle \pi_1$ and $\displaystyle \pi_2$ ", do not award •².
(b) 1. At •³, accept the appearance of the vectors within an attempt to find a scalar or vector product.
(b) 2. For a candidate who uses $\displaystyle \sin^{-1}\left(\dfrac{3}{\sqrt{6}\sqrt{29}}\right)$ as a means of obtaining the complement directly (with no further processing) •⁴ and •⁵ may be awarded.
(b) 3. For a candidate who finds $\displaystyle \sin^{-1}\left(\dfrac{3}{\sqrt{6}\sqrt{29}}\right)$ and proceeds to find its complement, •⁴ is unavailable.
(b) 4. Do not award •⁵ where the degree symbol has been omitted.
Commonly observed: 1. (b) Use of definition of vector product: $\displaystyle \sin\theta = \dfrac{\sqrt{165}}{\sqrt{6}\sqrt{29}}$ , award •⁴.
(c) 1. Alternative responses: equating $\displaystyle x$ and $\displaystyle z$ , $\displaystyle 2\lambda + 3 = -2\mu + 1$ , $\displaystyle \lambda = 3\mu - 2$ , leading to $\displaystyle \lambda = -\dfrac{5}{4}$ , $\displaystyle \mu = \dfrac{1}{4}$ , LHS $\displaystyle = -\dfrac{9}{4}$ , RHS $\displaystyle = 4$ ; equating $\displaystyle y$ and $\displaystyle z$ , $\displaystyle \lambda - 1 = 4\mu + 3$ , $\displaystyle \lambda = 3\mu - 2$ , leading to $\displaystyle \lambda = -20$ , $\displaystyle \mu = -6$ , LHS $\displaystyle = -37$ , RHS $\displaystyle = 13.$
(c) 2. Where candidates employ the same parameter twice leading to $\displaystyle x = -2\lambda + 1;\ y = 4\lambda + 3;\ z = 3\lambda - 2$ only •⁶ may be awarded.
(c) 3. For a final response of " $\displaystyle 0 = -5$ so the lines do not intersect" do not award •⁹ unless the candidate subsequently communicates the inconsistency of $\displaystyle 0 = -5.$
Commonly observed: 1. (c) A: $\displaystyle z = 0$ , $\displaystyle z = -3 - 2$ , lines do not intersect, award •⁹.

Integration by partsVolume of revolution

(5, 3)8 Marks

2019 Q16

(a)Use integration by parts to find the exact value of

$\displaystyle \int_{0}^{1}(x^2 - 2x + 1)e^{4x}dx.$

(b)A solid is formed by rotating the curve with equation $\displaystyle y = 4(x - 1)e^{2x}$ between $\displaystyle x = 0$ and $\displaystyle x = 1$ through $\displaystyle 2\pi$ radians about the $\displaystyle x$ -axis.
Find the exact value of the volume of this solid.

Show answer

(a)

\displaystyle \frac{1}{32}(e^4 - 13)

(b)

\displaystyle \frac{\pi}{2}(e^4 - 13)

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	evidence of integration by parts	$\displaystyle \dfrac{e^{4x}}{4}(x^2 - 2x + 1) - \ldots$
(a) •²	complete first application	$\displaystyle \ldots\displaystyle\int (2x - 2)\dfrac{e^{4x}}{4}\,dx$
(a) •³	second application of integration by parts	$\displaystyle \ldots\left[\dfrac{e^{4x}}{16}(2x - 2) - \dfrac{1}{8}\displaystyle\int e^{4x}\,dx\right]$
(a) •⁴	complete integration and include limits	$\displaystyle \left[\dfrac{e^{4x}}{4}(x^2 - 2x + 1)\right]_0^1 - \left[\dfrac{1}{16}(2x - 2)e^{4x} - \dfrac{1}{32}e^{4x}\right]_0^1$
(a) •⁵	evaluate	$\displaystyle \dfrac{1}{32}(e^4 - 13)$
(b) •⁶	correct form of integral	$\displaystyle \pi\displaystyle\int_0^1 y^2\,dx$
(b) •⁷	find expression to integrate	$\displaystyle 16\pi\displaystyle\int_0^1 (x^2 - 2x + 1)e^{4x}\,dx$
(b) •⁸	integrate and evaluate	$\displaystyle \dfrac{\pi}{2}(e^4 - 13)$

Notes

(a) 1. Disregard the omission of ' $\displaystyle dx$ '.
(a) 2. Evidence of limits may not appear until •⁵.
(a) 3. Do not award •⁵ for a decimal approximation, unless preceded by the exact value.
(b) 1. For the award of •⁶, limits must appear at some point.
(b) 2. •⁶ is not available unless " $\displaystyle dx$ " appears at some point.
(b) 3. At •⁶, accept $\displaystyle \pi\displaystyle\int_0^1 [f(x)]^2\,dx.$
(b) 4. Evidence for the award of •⁷ must include all of the following: 16; $\displaystyle (x^2 - 2x + 1)$ or $\displaystyle (x - 1)^2$ ; $\displaystyle e^{4x}$ ; unless an exact value appears at •⁸.
(b) 5. Do not award •⁸ for a decimal approximation unless: preceded by an exact value OR •⁵ has been withheld for the same reason AND there is sufficient evidence for •⁷.
(b) 6. Do not award •⁸ for a negative volume (including eg $\displaystyle \dfrac{\pi}{2}(e^2 - 13)$ ).

Geometric sequences and series

(2, 1, 2, 2, 2, 1)10 Marks

2019 Q17

The first three terms of a sequence are given by
$\displaystyle 5x + 8,\ -2x + 1,\ x - 4$

(a)When $\displaystyle x = 11$ , show that the first three terms form the start of a geometric sequence, and state the value of the common ratio.

(b)Given that the entire sequence is geometric for $\displaystyle x = 11$
(i) state why the associated series has a sum to infinity
(ii) calculate this sum to infinity.

(c)There is a second value for $\displaystyle x$ that also gives a geometric sequence.
For this second sequence
(i) show that $\displaystyle x^2 - 8x - 33 = 0$
(ii) find the first three terms
(iii) state the value of $\displaystyle S_{2n}$ and justify your answer.

Show answer

(a) Terms are 63, -21, 7. Ratios are

\displaystyle \frac{-21}{63} = -\frac{1}{3}

and

\displaystyle \frac{7}{-21} = -\frac{1}{3}.

Common ratio is

\displaystyle r = -\frac{1}{3}.

(b)(i) The sum to infinity exists because

\displaystyle \left| -\frac{1}{3} \right| < 1.

(b)(ii)

\displaystyle \frac{189}{4}

or

\displaystyle 47.25

(c)(i) Equating ratios

\displaystyle \frac{-2x+1}{5x+8} = \frac{x-4}{-2x+1}

leads to

\displaystyle x^2-8x-33=0.

(c)(ii)

\displaystyle x = -3

, giving terms -7, 7, -7.
(c)(iii)

\displaystyle S_{2n} = 0

because

\displaystyle 2n

is an even number of terms, and the pairs cancel each other out.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	substitute and calculate one ratio	$\displaystyle \dfrac{-21}{63} = -\dfrac{1}{3}$ or $\displaystyle \dfrac{7}{-21} = -\dfrac{1}{3}$
(a) •²	calculate second ratio and state common ratio	$\displaystyle \dfrac{7}{-21} = -\dfrac{1}{3}$ or $\displaystyle \dfrac{-21}{63} = -\dfrac{1}{3}$ , so $\displaystyle r = -\dfrac{1}{3}$
(b)(i) •³	state condition	$\displaystyle \left\|-\dfrac{1}{3}\right\| < 1$
(b)(ii) •⁴	begin to substitute	$\displaystyle \dfrac{\ldots}{1 - \left(-\frac{1}{3}\right)}$
(b)(ii) •⁵	calculate sum	$\displaystyle \dfrac{189}{4}$ or $\displaystyle 47.25$
(c)(i) •⁶	equate ratios	$\displaystyle \dfrac{-2x + 1}{5x + 8} = \dfrac{x - 4}{-2x + 1}$
(c)(i) •⁷	perform algebraic manipulation leading to formation of quadratic equation	$\displaystyle x^2 - 8x - 33 = 0$
(c)(ii) •⁸	calculate second value of $\displaystyle x$	$\displaystyle x = -3$
(c)(ii) •⁹	find first three terms	$\displaystyle -7, 7, -7$
(c)(iii) •¹⁰	state $\displaystyle S_{2n}$ and justify	$\displaystyle 0$ since eg $\displaystyle 2n$ is even and so pairs of terms cancel each other out

Notes

(a) 1. Where a candidate calculates the first three terms only, •¹ and •² are not available.
(a) 2. Where a candidate calculates the first three terms and simply states $\displaystyle r = -\dfrac{1}{3}$ , award •¹.
(a) 3. Where a candidate finds the first three terms followed by eg " $\displaystyle r = \dfrac{-21}{7}$ , so $\displaystyle r = -\dfrac{1}{3}$ ", do not award •¹.
(a) 4. Where a candidate calculates the first three terms and then substitutes one pair of numbers into the $\displaystyle n^{th}$ term formula to calculate $\displaystyle r$ , award •¹ only.
(a) 5. For the award of •², there must be evidence that the candidate has considered a second pair of terms.
Commonly observed: 1. (a) A, first three terms found followed by: $\displaystyle \dfrac{-21}{63} = -\dfrac{1}{3}$ , award •¹; $\displaystyle -21 \times \left(-\dfrac{1}{3}\right) = 7$ so $\displaystyle r = -\dfrac{1}{3}$ , award •².
(b) 1. (i) At •³, $\displaystyle -\dfrac{1}{3}$ may be replaced by a letter consistent with the candidate's answer in (a). However, in the case where a candidate obtains a value in (a) outside the open interval $\displaystyle (-1, 1)$ , •³ will be available only where they also acknowledge that there is no sum to infinity.
(b) 2. (i) Award •³ only for a strict inequality, whether expressed algebraically or in words.
(b) 3. (ii) Where a candidate calculates a common ratio outwith the open interval $\displaystyle (-1, 1)$ , •⁴ and •⁵ are not available.
(b) 4. (ii) Where a candidate writes $\displaystyle S_n = \dfrac{63\left(1 - \left(-\frac{1}{3}\right)^n\right)}{1 - \left(-\frac{1}{3}\right)}$ , •⁴ will be available only where a candidate states that as $\displaystyle n \to \infty$ $\displaystyle \left(-\dfrac{1}{3}\right)^n \to 0.$ •⁵ is still available.
(b) 5. (ii) For a correct answer with no working, •⁴ and •⁵ are not available.
(c) 1. (i) Evidence for the award of •⁷ must include the expansion of the products of two pairs of brackets.
(c) 2. (iii) For a descriptive justification, reference must be made either to an even number of terms or to the fact that $\displaystyle 2n$ is even (and the consequence thereof).
(c) 3. (iii) At •¹⁰ accept $\displaystyle S_{2n} = 0$ since $\displaystyle \dfrac{-7(1 - (-1)^{2n})}{1 - (-1)} = 0.$

Argand diagramde Moivre's theorem

(1, 3, 4, 2)10 Marks

2019 Q18

The complex number $\displaystyle w$ has been plotted on an Argand diagram, as shown below.

(a)Express $\displaystyle w$ in
(i) Cartesian form
(ii) polar form.

The complex number $\displaystyle z_1$ is a root of $\displaystyle z^3 = w$ , where
$\displaystyle z_1 = k\left(\cos \frac{\pi}{m} + i\sin \frac{\pi}{m}\right)$
for integers $\displaystyle k$ and $\displaystyle m.$

Given that $\displaystyle a = 4$ ,
(b)(i) use de Moivre's theorem to obtain the values of $\displaystyle k$ and $\displaystyle m$ , and
(ii) find the remaining roots.

Show answer

(a)(i)

\displaystyle a - a\sqrt{3}i

(a)(ii)

\displaystyle 2a\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)

(b)(i)

\displaystyle k = 2

and

\displaystyle m = -9

(b)(ii)

\displaystyle z_2 = 2\left(\cos\left(\frac{5\pi}{9}\right) + i\sin\left(\frac{5\pi}{9}\right)\right)

\displaystyle z_3 = 2\left(\cos\left(-\frac{7\pi}{9}\right) + i\sin\left(-\frac{7\pi}{9}\right)\right)

Show marking instructions

Mark	Awarded for	Illustrative answer
(a)(i) •¹	write in Cartesian form	$\displaystyle a - a\sqrt{3}i$
(a)(ii) •²	calculate modulus	$\displaystyle 2a$
(a)(ii) •³	calculate argument	$\displaystyle -\dfrac{\pi}{3}$
(a)(ii) •⁴	write in polar form	$\displaystyle 2a\left(\cos\left(-\dfrac{\pi}{3}\right) + i\sin\left(-\dfrac{\pi}{3}\right)\right)$
(b)(i) •⁵	begin process	$\displaystyle z_1 = 8^{\frac{1}{3}}\left(\cos\left(-\dfrac{\pi}{3}\right) + i\sin\left(-\dfrac{\pi}{3}\right)\right)^{\frac{1}{3}}$ stated or implied by •⁶
(b)(i) •⁶	complete process	$\displaystyle z_1 = 8^{\frac{1}{3}}\left(\cos\left(-\dfrac{\pi}{9}\right) + i\sin\left(-\dfrac{\pi}{9}\right)\right)$
(b)(i) •⁷	state value of $\displaystyle k$	$\displaystyle k = 2$
(b)(i) •⁸	state value of $\displaystyle m$	$\displaystyle m = -9$
(b)(ii) •⁹	begin to add or subtract $\displaystyle \dfrac{2\pi}{3}$ to or from argument of $\displaystyle z_1$	$\displaystyle \ldots \pm \dfrac{2\pi}{3}$ stated or implied by •¹⁰
(b)(ii) •¹⁰	state roots	$\displaystyle z_2 = 2\left(\cos\dfrac{5\pi}{9} + i\sin\dfrac{5\pi}{9}\right)$ , $\displaystyle z_3 = 2\left(\cos\left(-\dfrac{7\pi}{9}\right) + i\sin\left(-\dfrac{7\pi}{9}\right)\right)$

Notes

(a) 1. (ii) At •² accept $\displaystyle \sqrt{4a^2}$ , but it must be simplified at •⁴.
(a) 2. (ii) For •³, accept any answer of the form $\displaystyle -\dfrac{\pi}{3} + 2k\pi$ , $\displaystyle k \in \mathbb{Z}.$
(a) 3. (ii) Accept an argument expressed in degrees, with or without a degree symbol.
(a) 4. (ii) Evidence for •³ may not appear until b(i). In this case, •⁴ is not available.
(a) 5. (ii) At •⁴, accept $\displaystyle w = 2a\left(\cos\dfrac{\pi}{3} - i\sin\dfrac{\pi}{3}\right).$
(a) 6. (ii) Do not withhold •⁴ for an unsimplified modulus if •² has already been withheld for the same reason.
(b) 1. (i) Where the operations carried out on the modulus and argument are incompatible eg cubing the modulus and dividing the argument by three, do not award •⁵ or •⁶; however, •⁷ and •⁸ are still available.
(b) 2. (i) Where a candidate obtains a non-integer value for $\displaystyle k$ or $\displaystyle m$ , •⁷ or •⁸ is not available.
(b) 3. (ii) The addition of other multiples of $\displaystyle \dfrac{2\pi}{3}$ , leading to other forms of roots, is acceptable.
(b) 4. (ii) Where a candidate finds one further root, consistent with adding or subtracting $\displaystyle \dfrac{2\pi}{3}$ to their response to b(i) and without working, •⁹ may be awarded.
(b) 5. (ii) •¹⁰ is available only where a candidate produces exactly two roots, with consistent spacing, distinct from one another and also from $\displaystyle z_1.$
Commonly observed: 1. (b)(i) A: $\displaystyle z_1^3 = k^3\left(\cos\dfrac{\pi}{m} + i\sin\dfrac{\pi}{m}\right)^3$ , award •⁵ (stated or implied by •⁶); $\displaystyle z_1^3 = k^3\left(\cos\dfrac{3\pi}{m} + i\sin\dfrac{3\pi}{m}\right)$ , award •⁶.
Commonly observed: 2. (b)(i) B: $\displaystyle w^3 = 8^3\left(\cos\left(-\dfrac{\pi}{3}\right) + i\sin\left(-\dfrac{\pi}{3}\right)\right)^3$ , do not award •⁵; $\displaystyle w^3 = 8^3(\cos(-\pi) + i\sin(-\pi))$ , award •⁶; $\displaystyle k = 512$ , award •⁷; $\displaystyle m = -1$ , award •⁸.
Commonly observed: 3. (b)(i) C, answers without working: $\displaystyle k = 2$ and $\displaystyle m = -9$ , award full marks; $\displaystyle k = 2$ and $\displaystyle m \neq -9$ , award •⁷ only; $\displaystyle k \neq 2$ and $\displaystyle m = -9$ , award •⁸ only.

2019 Paper

Question 1(a)

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Question 1(b)

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Question 1(c)

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Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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Question 16

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Question 17

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Question 18

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