Advanced Higher Maths 2018 Paper — Questions & Marking Instructions

Product, quotient, chain rules

2 Marks

2018 Q1(a)

Given $\displaystyle f(x) = \sin^{-1} 3x$ find $\displaystyle f'(x).$

Show answer

\displaystyle \frac{3}{\sqrt{1 - 9x^2}}

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	start differentiation	$\displaystyle \dfrac{1}{\sqrt{1 - (3x)^2}} \times \ldots$
(a) •²	apply chain rule and complete differentiation	$\displaystyle \dfrac{3}{\sqrt{1 - 9x^2}}$

Notes

(a) 1. For •¹ do not accept $\displaystyle \dfrac{1}{\sqrt{1 - 3x^2}} \times \ldots$ unless subsequently corrected.
(a) 2. For •² accept eg $\displaystyle \dfrac{3}{\sqrt{1 - (3x)^2}}$ or $\displaystyle \dfrac{1}{\sqrt{\frac{1}{9} - x^2}}.$
(a) 3. For candidates who interpret $\displaystyle \sin^{-1}3x$ as $\displaystyle (\sin 3x)^{-1}$ , •² is available for $\displaystyle -(\sin 3x)^{-2} \times 3\cos 3x.$
Commonly observed: 1. (b) Alternative method 1 (product rule): •³ $\displaystyle 5e^{5x}(7x + 1)^{-1}\ldots$ ; •⁴ $\displaystyle \ldots - 7e^{5x}(7x + 1)^{-2}.$ Alternative method 2 (logarithmic differentiation): •³ $\displaystyle \ln y = 5x - \ln|7x + 1|$ and $\displaystyle \dfrac{1}{y}\dfrac{dy}{dx} = \ldots$ ; •⁴ $\displaystyle \dfrac{dy}{dx} = y\left(5 - \dfrac{7}{7x + 1}\right).$

Product, quotient, chain rules

2 Marks

2018 Q1(b)

Differentiate $\displaystyle y = \frac{e^{5x}}{7x + 1}.$

Show answer

\displaystyle \frac{5(7x + 1)e^{5x} - 7e^{5x}}{(7x + 1)^2}

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Mark	Awarded for	Illustrative answer
(b) •³	evidence use of quotient rule with denominator and one term of numerator correct	$\displaystyle \dfrac{5(7x + 1)e^{5x}\ldots}{(7x + 1)^2}$ OR $\displaystyle \dfrac{\ldots - 7e^{5x}}{(7x + 1)^2}$
(b) •⁴	complete differentiation	$\displaystyle \dfrac{5(7x + 1)e^{5x} - 7e^{5x}}{(7x + 1)^2}$

Notes

Commonly observed: 1. (b) Alternative method 1 (product rule): •³ $\displaystyle 5e^{5x}(7x + 1)^{-1}\ldots$ ; •⁴ $\displaystyle \ldots - 7e^{5x}(7x + 1)^{-2}.$ Alternative method 2 (logarithmic differentiation): •³ $\displaystyle \ln y = 5x - \ln|7x + 1|$ and $\displaystyle \dfrac{1}{y}\dfrac{dy}{dx} = \ldots$ ; •⁴ $\displaystyle \dfrac{dy}{dx} = y\left(5 - \dfrac{7}{7x + 1}\right).$

Implicit differentiation

4 Marks

2018 Q1(c)

For $\displaystyle y \cos x + y^2 = 6x$ , use implicit differentiation to find $\displaystyle \frac{dy}{dx}.$

Show answer

\displaystyle \frac{6 + y \sin x}{\cos x + 2y}

Show marking instructions

Mark	Awarded for	Illustrative answer
(c) •⁵	start to differentiate product with one term correct	$\displaystyle \dfrac{dy}{dx}\cos x + \ldots$ OR $\displaystyle -y\sin x + \ldots$
(c) •⁶	complete differentiation of product	$\displaystyle \dfrac{dy}{dx}\cos x\ldots$ OR $\displaystyle -y\sin x$
(c) •⁷	differentiate remaining terms	$\displaystyle +2y\dfrac{dy}{dx} = 6$
(c) •⁸	express derivative explicitly in terms of $\displaystyle x$ and $\displaystyle y$	$\displaystyle \dfrac{dy}{dx} = \dfrac{6 + y\sin x}{\cos x + 2y}$

Notes

Commonly observed: 1. (b) Alternative method 1 (product rule): •³ $\displaystyle 5e^{5x}(7x + 1)^{-1}\ldots$ ; •⁴ $\displaystyle \ldots - 7e^{5x}(7x + 1)^{-2}.$ Alternative method 2 (logarithmic differentiation): •³ $\displaystyle \ln y = 5x - \ln|7x + 1|$ and $\displaystyle \dfrac{1}{y}\dfrac{dy}{dx} = \ldots$ ; •⁴ $\displaystyle \dfrac{dy}{dx} = y\left(5 - \dfrac{7}{7x + 1}\right).$
(c) 1. •⁵ and •⁶ are not available where the differentiation of $\displaystyle y\cos x$ leads to only one term.
(c) 2. •⁸ is available only where $\displaystyle \dfrac{dy}{dx}$ appears more than once after completing differentiation.

Partial fractionsIntegrating rational functions

4 Marks

2018 Q2

Use partial fractions to find $\displaystyle \int \frac{3x - 7}{x^2 - 2x - 15} \, dx.$

Show answer

\displaystyle 2 \ln|x + 3| + \ln|x - 5| + c

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	state expression	$\displaystyle \dfrac{3x - 7}{x^2 - 2x - 15} = \dfrac{A}{x + 3} + \dfrac{B}{x - 5}$
•²	form equation and find one unknown	$\displaystyle 3x - 7 = A(x - 5) + B(x + 3)$ AND eg $\displaystyle A = 2$
•³	find second unknown and write integral expression	$\displaystyle B = 1$ AND $\displaystyle \displaystyle\int\left(\dfrac{2}{x + 3} + \dfrac{1}{x - 5}\right)dx$ stated or implied by •⁴
•⁴	integrate	$\displaystyle 2\ln\|x + 3\| + \ln\|x - 5\| + c$

Notes

1. •² and •³ may be awarded where an attempt at factorisation leads to an incorrect linear denominator.
2. At •³ disregard the omission of $\displaystyle dx.$
3. At •⁴ disregard the omission of modulus signs.
4. •⁴ is not available where the constant of integration has either been omitted or first appears after an incorrect logarithmic term.
Commonly observed: 1. $\displaystyle \dfrac{3x - 7}{x^2 - 2x - 15} = \dfrac{A}{x - 3} + \dfrac{B}{x + 5}$ , do not award •¹; $\displaystyle 3x - 7 = A(x + 5) + B(x - 3)$ AND eg $\displaystyle A = \dfrac{1}{4}$ , award •²; $\displaystyle B = \dfrac{11}{4}$ AND $\displaystyle \displaystyle\int\left(\dfrac{1}{4(x - 3)} + \dfrac{11}{4(x + 5)}\right)dx$ stated or implied by •⁴, award •³; $\displaystyle \dfrac{1}{4}\ln|x - 3| + \dfrac{11}{4}\ln|x + 5| + c$ , award •⁴.

Binomial terms

(3, 2)5 Marks

2018 Q3

(a)Write down and simplify the general term in the binomial expansion of
$\displaystyle \left(2x + \frac{5}{x^2}\right)^9.$

(b)Hence, or otherwise, find the term independent of $\displaystyle x.$

Show answer

(a)

\displaystyle \binom{9}{r} 2^{9-r} 5^r x^{9-3r}

(b)

\displaystyle 672000

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	state general term	$\displaystyle \dbinom{9}{r}(2x)^{9-r}\left(\dfrac{5}{x^2}\right)^r$
(a) •²	simplify powers of $\displaystyle x$ OR coefficients	$\displaystyle 2^{9-r}5^r$ OR $\displaystyle x^{9-3r}$
(a) •³	state simplified general term (complete simplification)	$\displaystyle \dbinom{9}{r}2^{9-r}5^r x^{9-3r}$
(b) •⁴	determine value of $\displaystyle r$	$\displaystyle r = 3$
(b) •⁵	evaluate term	$\displaystyle 672000$

Notes

(a) 1. Where candidates write out a full binomial expansion, •¹, •² and •³ are not available unless the general term is identifiable in (b).
(a) 2. Candidates who write down $\displaystyle \dbinom{9}{r}2^{9-r}5^r x^{9-3r}$ with no working receive full marks.
(a) 3. •³ is unavailable to candidates who in (a) produce further incorrect simplification subsequent to a correct answer eg $\displaystyle 2^{9-r}5^r$ becomes $\displaystyle 10^9$ or $\displaystyle x^{9-3r}$ becomes $\displaystyle x^{6r}.$
Commonly observed: 1. (a) General term has not been isolated (the sum $\displaystyle \sum_{r=0}^{9}$ kept throughout): do not award •¹. Award •² and •³.
Commonly observed: 2. (a) General term has been isolated (the sum dropped at the last line): 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: disregard the incorrect use of the equals sign. Award •¹.
(b) 1. Where candidates write out a full expansion •⁴ may be awarded where this is complete and correct at least as far as the required term. •⁵ may be awarded only where the required term is clearly identified from the expansion.
(b) 2. •⁵ is not available to candidates who interpret the term independent of $\displaystyle x$ as substituting $\displaystyle x = 0.$
Commonly observed: 1. (b) Binomial expansion as far as required term: $\displaystyle 512x^9 + 11520x^6 + 115200x^3 + 672000\ldots$ or $\displaystyle \dfrac{1953125}{x^{18}} + \dfrac{7031250}{x^{15}} + \dfrac{11250000}{x^{12}} + \dfrac{10500000}{x^9} + \dfrac{6098400}{x^6} + \dfrac{2520000}{x^3} + 672000\ldots$

Complex number operations

(2, 1)3 Marks

2018 Q4

Given that $\displaystyle z_1 = 2 + 3i$ and $\displaystyle z_2 = p - 6i$ , $\displaystyle p \in \mathbb{R}$ , find:

(a) $\displaystyle z_1 \overline{z}_2$

(b)the value of $\displaystyle p$ such that $\displaystyle z_1 \overline{z}_2$ is a real number.

Show answer

(a)

\displaystyle (2p - 18) + (3p + 12)i

(b)

\displaystyle p = -4

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	state conjugate	$\displaystyle \overline{z}_2 = p + 6i$ stated or implied
(a) •²	substitute for $\displaystyle z_1$ , $\displaystyle \overline{z}_2$ , expand and apply $\displaystyle i^2 = -1$	$\displaystyle (2p - 18) + (3p + 12)i$
(b) •³	find value of $\displaystyle p$	$\displaystyle -4$

Notes

(a) 1. At •² accept $\displaystyle 2p + 12i + 3pi - 18.$
(a) 2. To award •² $\displaystyle (2 + 3i)$ must be multiplied by another complex number.
Commonly observed: 1. (a) Candidates find $\displaystyle z_1z_2$ : $\displaystyle 2p + 3pi - 12i + 18$ , do not award •¹, award •².

Euclidean algorithm

4 Marks

2018 Q5

Use the Euclidean algorithm to find integers $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle 306a + 119b = 17.$

Show answer

\displaystyle a = 2, b = -5

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	start process	$\displaystyle 306 = 2 \times 119 + 68$ ( $\displaystyle 119 = 1 \times 68 + 51$ )
•²	obtain remainder of 17	$\displaystyle 68 = 1 \times 51 + 17$ ( $\displaystyle 51 = 3 \times 17$ )
•³	express gcd in terms of 306 and 119	$\displaystyle 17 = -1 \times 119 + 2(306 - 2 \times 119)$
•⁴	obtain $\displaystyle a$ and $\displaystyle b$	$\displaystyle a = 2,\ b = -5$

Notes

1. At •² the gcd and the final line of working do not have to be stated explicitly.
2. The minimum requirement for •⁴ is $\displaystyle 306 \times 2 + 119 \times (-5) = 17.$
3. Do not accept $\displaystyle 306 \times 2 - 119 \times 5 = 17$ where the values of $\displaystyle a$ and $\displaystyle b$ have not been explicitly stated.

Parametric differentiation

5 Marks

2018 Q6

On a suitable domain, a curve is defined parametrically by $\displaystyle x = t^2 + 1$ and $\displaystyle y = \ln(3t + 2).$
Find the equation of the tangent to the curve where $\displaystyle t = -\frac{1}{3}.$

Show answer

\displaystyle y = -\frac{9}{2}x + 5

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	find $\displaystyle \dfrac{dy}{dt}$	$\displaystyle \dfrac{dy}{dt} = \dfrac{3}{3t + 2}$
•²	complete differentiation and relate derivatives	$\displaystyle \dfrac{dx}{dt} = 2t$ and $\displaystyle \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ stated or implied at •³
•³	evaluate gradient	$\displaystyle \dfrac{dy}{dx} = -\dfrac{9}{2}$
•⁴	find coordinates	$\displaystyle x = \dfrac{10}{9}$ , $\displaystyle y = 0$
•⁵	state equation of tangent	$\displaystyle y = -\dfrac{9}{2}x + 5$

Notes

1. Evidence for •² could include eg $\displaystyle \dfrac{dy}{dx} = \dfrac{\frac{3}{3t + 2}}{2t}.$
2. Where candidates evaluate $\displaystyle \dfrac{dx}{dt}$ and $\displaystyle \dfrac{dy}{dt}$ before finding $\displaystyle \dfrac{dy}{dx}$ , award •² for evaluating the individual derivatives and award •³ for evaluating $\displaystyle \dfrac{dy}{dx}.$
3. At •⁵ accept eg $\displaystyle y + \dfrac{9}{2}x - 5 = 0$ , $\displaystyle 2y + 9x = 10.$ Do not accept $\displaystyle y - 0 = \ldots$ or $\displaystyle y = -\dfrac{9}{2}\left(x - \dfrac{10}{9}\right).$
Commonly observed: 1. Incorrect expression for $\displaystyle \dfrac{dy}{dt}$ : $\displaystyle \dfrac{dy}{dt} = \dfrac{1}{3t + 2}$ leading to $\displaystyle y = -\dfrac{3}{2}x + \dfrac{5}{3}$ may be awarded a maximum of 4/5.

Matrix operationsDeterminant and inverse

(2, 2, 1)5 Marks

2018 Q7

Matrices $\displaystyle C$ and $\displaystyle D$ are given by:
$\displaystyle C = \begin{pmatrix} -2 & 1 & 2 \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{pmatrix}$ and $\displaystyle D = \begin{pmatrix} 1 & 1 & 2 \\ k+3 & 0 & 2 \\ 1 & 1 & 1 \end{pmatrix}$ , where $\displaystyle k \in \mathbb{R}.$

(a)Obtain $\displaystyle 2C' - D$ where $\displaystyle C'$ is the transpose of $\displaystyle C.$

(b)(i) Find and simplify an expression for the determinant of $\displaystyle D.$
(ii) State the value of $\displaystyle k$ such that $\displaystyle D^{-1}$ does not exist.

Show answer

(a)

\displaystyle \begin{pmatrix} -5 & 1 & 0 \\ -k-1 & -2 & -2 \\ 3 & -1 & -3 \end{pmatrix}

(b)(i)

\displaystyle k + 3

(b)(ii)

\displaystyle -3

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	state transpose of $\displaystyle C$	$\displaystyle \begin{pmatrix}-2 & 1 & 1\\1 & -1 & 0\\2 & 0 & -1\end{pmatrix}$ stated or implied by •²
(a) •²	obtain matrix	$\displaystyle 2C' - D = \begin{pmatrix}-5 & 1 & 0\\-k - 1 & -2 & -2\\3 & -1 & -3\end{pmatrix}$
(b)(i) •³	begin to find determinant	$\displaystyle -(k + 3)\begin{vmatrix}1 & 2\\1 & 1\end{vmatrix} + 0\begin{vmatrix}1 & 2\\1 & 1\end{vmatrix} - 2\begin{vmatrix}1 & 1\\1 & 1\end{vmatrix}$
(b)(i) •⁴	simplify expression	$\displaystyle k + 3$
(b)(ii) •⁵	state value of $\displaystyle k$	$\displaystyle -3$

Notes

Commonly observed: 1. (a) $\displaystyle 2C - D = \begin{pmatrix}-5 & 1 & 2\\-k - 1 & -2 & -2\\1 & -1 & -3\end{pmatrix}$ , award •² only.
Commonly observed: 2. (b)(i) Expansion about the first row: $\displaystyle 1\begin{vmatrix}0 & 2\\1 & 1\end{vmatrix} - 1\begin{vmatrix}k + 3 & 2\\1 & 1\end{vmatrix} + 2\begin{vmatrix}k + 3 & 0\\1 & 1\end{vmatrix} = 1(0 - 2) - 1(k + 3 - 2) + 2(k + 3 - 0) = -2 - k - 1 + 2k + 6 = k + 3.$

Integration by substitution

4 Marks

2018 Q8

Using the substitution $\displaystyle u = \sin \theta$ , or otherwise, evaluate

$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 2 \sin^4 \theta \cos \theta \, d\theta.$

Show answer

\displaystyle \frac{31}{80}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	differentiate	$\displaystyle \dfrac{du}{d\theta} = \cos\theta$
•²	find limits for $\displaystyle u$	$\displaystyle u = \dfrac{1}{2}$ , $\displaystyle u = 1$
•³	rewrite integral	$\displaystyle 2\displaystyle\int_{1/2}^{1} u^4\,du$
•⁴	integrate and evaluate	$\displaystyle \dfrac{2}{5}\left[u^5\right]_{1/2}^{1}$ and $\displaystyle \dfrac{31}{80}$ (or $\displaystyle 0.3875$ )

Notes

1. •³ is available where candidates either omit limits or retain limits for $\displaystyle \theta.$
2. Where candidates attempt to integrate an expression containing both $\displaystyle u$ and $\displaystyle \theta$ , where $\displaystyle \theta$ is either inside the integrand or erroneously taken outside as a constant, only •¹ and •² may be available.
3. Where candidates do not change limits but who produce working leading to $\displaystyle \dfrac{2}{5}\left[\sin^5\theta\right]_{\pi/6}^{\pi/2}$ , •² may be awarded.
4. For candidates who arrive at $\displaystyle \dfrac{2}{5}\left[\sin^5\theta\right]_{\pi/6}^{\pi/2}$ by inspection full marks are still available.
5. For candidates who integrate incorrectly, •⁴ may be available provided division by zero does not occur.
6. •⁴ is not available to candidates who write limits using degrees.
7. At •⁴ accept decimal answers rounded to at least 3 significant figures.

Direct proof

(2, 1)3 Marks

2018 Q9

Prove directly that:

(a)the sum of any three consecutive integers is divisible by 3;

(b)any odd integer can be expressed as the sum of two consecutive integers.

Show answer

(a) Proof showing

\displaystyle (n - 1) + n + (n + 1) = 3n

, which is a multiple of 3 and thus divisible by 3.
(b) Proof showing any odd integer

\displaystyle 2k + 1

(where

\displaystyle k \in \mathbb{Z}

) can be written as

\displaystyle k + (k + 1)

, the sum of two consecutive integers.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	form the sum of three consecutive integers	$\displaystyle (n - 1) + n + (n + 1)$
(a) •²	communication	$\displaystyle 3n$ which is divisible by 3
(b) •³	appropriate form for odd number, decomposed into two consecutive integers	$\displaystyle 2k + 1 = k + (k + 1)$ , $\displaystyle k \in \mathbb{Z}$

Notes

(a) 1. Candidates may form the sum $\displaystyle n + (n + 1) + (n + 2)$ leading to $\displaystyle 3(n + 1)$ at •².
(a) 2. Withhold •¹ where candidates construct an expression of the form $\displaystyle an + (an + 1) + (an + 2)$ , where $\displaystyle a \neq \pm 1$ and $\displaystyle a \in \mathbb{Z}.$ •² may be available. eg $\displaystyle 4n + (4n + 1) + (4n + 2)$ leading to $\displaystyle 3(4n + 1)$ which is divisible by 3.
(a) 3. Where candidates equate an expression for the sum of 3 consecutive integers to a multiple of 3 see commonly observed responses.
(a) 4. At •¹ accept an expression such as $\displaystyle n + n + 1 + n + 2.$
(a) 5. Withhold •¹ and •² where candidates form one (or more) sum of 3 specific consecutive numbers eg $\displaystyle 2 + 3 + 4.$
Commonly observed: 1. (a) A: $\displaystyle x + (x + 1) + (x + 2) = 3k$ , $\displaystyle 3x + 3 = 3k$ , $\displaystyle 3(x + 1) = 3k$ , where $\displaystyle k = x + 1$ and $\displaystyle 3k$ is divisible by 3. Award 2/2.
Commonly observed: 2. (a) B: as A, but "where $\displaystyle k = x + 1$ " only. The candidate has defined $\displaystyle k$ but has not communicated divisibility. Award •¹ but not •².
Commonly observed: 3. (a) C: $\displaystyle (k + 1) + (k + 2) + (k + 3) = 3k + 6$ , $\displaystyle \dfrac{3k + 6}{3} = k + 2$ therefore statement is true. The candidate has explicitly divided and there is no requirement to state that $\displaystyle k + 2$ is an integer. Award 2/2.
(b) 1. Where candidates write down $\displaystyle 2k + 1 = k + (k + 1)$ and omit $\displaystyle k \in \mathbb{Z}$ , award •³.
(b) 2. Where candidates omit brackets and write down $\displaystyle k + k + 1$ do not award •³ unless the candidate demonstrates that $\displaystyle k$ and $\displaystyle k + 1$ are two consecutive integers eg writing $\displaystyle k, k + 1.$
(b) 3. Where candidates begin with consecutive integers, •³ may be awarded only where $\displaystyle 2k + 1$ is associated with any odd integer and not by a restatement of the assertion in the question.
Commonly observed: 1. (b) A: $\displaystyle k + (k + 1) = 2k + 1$ odd. Do not award •³. The candidate has not defined a general odd integer.
Commonly observed: 2. (b) B: $\displaystyle k + (k + 1) = 2k + 1$ and $\displaystyle 2k + 1$ represents **any** odd integer. Award •³. The candidate has defined a general odd integer.
Commonly observed: 3. (b) C: $\displaystyle 2k + 1$ , $\displaystyle k + (k + 1) = 2k + 1.$ Award •³. The candidate has begun with the general form for an odd integer.

Locus in the complex plane

3 Marks

2018 Q10

Given $\displaystyle z = x + iy$ sketch the locus in the complex plane given by $\displaystyle |z| = |z - 2 + 2i|.$

Show answer

A straight line representing the perpendicular bisector of the line segment joining

\displaystyle (0, 0)

and

\displaystyle (2, -2).

Equation of the line is

\displaystyle y = x - 2.

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	substitute, collect real and imaginary parts and equate moduli	$\displaystyle \|x + iy\| = \|(x - 2) + (y + 2)i\|$
•²	process to obtain a linear equation in $\displaystyle x$ and $\displaystyle y$	eg $\displaystyle y = x - 2$
•³	sketch consistent with equation	complete sketch: a straight line of positive gradient through $\displaystyle (0, -2)$

Notes

1. At •³ accept any line passing through an appropriate point, which has positive gradient.
2. Do not withhold •³ for axes which are unlabelled. Accept $\displaystyle x$ and $\displaystyle y$ in lieu of 'Re' and 'Im'. Disregard any appearance of $\displaystyle i$ in the diagram.

Transformation matrices

(1, 1, 2, 1)5 Marks

2018 Q11

(a)Obtain the matrix, $\displaystyle A$ , associated with an anticlockwise rotation of $\displaystyle \frac{\pi}{3}$ radians about the origin.

(b)Find the matrix, $\displaystyle B$ , associated with a reflection in the x-axis.

(c)Hence obtain the matrix, $\displaystyle P$ , associated with an anticlockwise rotation of $\displaystyle \frac{\pi}{3}$ radians about the origin followed by reflection in the x-axis, expressing your answer using exact values.

(d)Explain why matrix $\displaystyle P$ is not associated with rotation about the origin.

Show answer

(a)

\displaystyle \begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\[7pt] \dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix}

(b)

\displaystyle \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

(c)

\displaystyle \begin{pmatrix} \dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \\[7pt] -\dfrac{\sqrt{3}}{2} & -\dfrac{1}{2} \end{pmatrix}

(d)

\displaystyle P

is not associated with rotation about the origin because it is not in the general form of a rotation matrix

\displaystyle \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

(e.g. elements on the leading diagonal are not equal).

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	obtain $\displaystyle A$	$\displaystyle \begin{pmatrix}\cos\dfrac{\pi}{3} & -\sin\dfrac{\pi}{3}\\[7pt]\sin\dfrac{\pi}{3} & \cos\dfrac{\pi}{3}\end{pmatrix}$
(b) •²	obtain $\displaystyle B$	$\displaystyle \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$
(c) •³	correct order for multiplication ( $\displaystyle P = BA$ )	$\displaystyle \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}\dfrac{1}{2}\begin{pmatrix}1 & -\sqrt{3}\\\sqrt{3} & 1\end{pmatrix}$
(c) •⁴	multiplication completed and appearance of exact values	$\displaystyle \dfrac{1}{2}\begin{pmatrix}1 & -\sqrt{3}\\-\sqrt{3} & -1\end{pmatrix}$
(d) •⁵	valid explanation	eg compare the elements of $\displaystyle P$ with the general form of a rotation matrix

Notes

(a) 1. Accept $\displaystyle \begin{pmatrix}\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2}\\[7pt]\dfrac{\sqrt{3}}{2} & \dfrac{1}{2}\end{pmatrix}.$
(c) 1. Common factor not required for •⁴.
(c) 2. •⁴ is unavailable to candidates who have incorrectly identified $\displaystyle B = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}.$
Commonly observed: 1. (c) Incorrect order of multiplication: $\displaystyle \dfrac{1}{2}\begin{pmatrix}1 & -\sqrt{3}\\\sqrt{3} & 1\end{pmatrix}\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix} = \dfrac{1}{2}\begin{pmatrix}1 & \sqrt{3}\\\sqrt{3} & -1\end{pmatrix}.$ Do not award •³. •⁴ is still available on follow through.
(d) 1. •⁵ may be awarded where a candidate's explanation makes reference to the specific entries of the leading diagonal of $\displaystyle P$ ("they should be equal but are not") or trailing diagonal ("one must be negative of the other but is not").
(d) 2. Withhold •⁵ for a statement such as 'Matrix $\displaystyle P$ represents a reflection' unless the reflection is specified eg 'Matrix $\displaystyle P$ is a reflection in the line $\displaystyle y = -\dfrac{1}{\sqrt{3}}x$ '.
(d) 3. •⁵ may also be awarded where candidates investigate the images of at least two points, neither of which is O.
(d) 4. Candidates who respond with reference to the constituent transformations of $\displaystyle P$ may be awarded •⁵ only where there is reference to both transformations and communication demonstrates understanding that an even number of reflections are required in order to produce a rotation.
Commonly observed: 1. (d) Examples where •⁵ would be awarded: " $\displaystyle P$ is not associated with rotation about the origin as it is in the form $\displaystyle \begin{pmatrix}\cos\theta & -\sin\theta\\-\sin\theta & -\cos\theta\end{pmatrix}$ " (explanation makes explicit reference to their form of $\displaystyle P$ and implicitly refers to the form of a general rotation matrix); "Because $\displaystyle P$ can no longer be expressed as a certain angle of rotation. $\displaystyle \cos^{-1}\left(\frac{1}{2}\right) \neq \cos^{-1}\left(-\frac{1}{2}\right)$ $\displaystyle \therefore$ cannot be expressed as an angle"; the four entries turned into angles $\displaystyle \frac{\pi}{3}, \frac{\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}$ "and as the angles found are not all equal, this transformation is not a rotation about the origin".
Commonly observed: 2. (d) Examples where •⁵ would not be awarded: "As matrix $\displaystyle P$ doesn't exhibit $\displaystyle \begin{pmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{pmatrix}$ " (no reference made to form of $\displaystyle P$ ); "Because it gets reflected in the process …"; "Since it is followed by a reflection"; "cos values different so do not relate to same angle"; "Rotation is around a point on graph not the origin"; "It would put the point back to where it started"; "Rotations must have the matrix $\displaystyle \begin{pmatrix}a & -b\\b & a\end{pmatrix}$ ".

Proof by induction

5 Marks

2018 Q12

Prove by induction that, for all positive integers $\displaystyle n$ ,

$\displaystyle \sum_{r=1}^n 3^{r-1} = \frac{1}{2}(3^n - 1).$

Show answer

When

\displaystyle n=1

, LHS

\displaystyle =3^{0}=1

and RHS

\displaystyle =\frac{1}{2}(3-1)=1

, so the statement is true.
Assume true for

\displaystyle n=k

:

\displaystyle \sum_{r=1}^{k}3^{r-1}=\frac{1}{2}(3^{k}-1).

Then

\displaystyle \sum_{r=1}^{k+1}3^{r-1}=\frac{1}{2}(3^{k}-1)+3^{k}=\frac{3}{2}\times3^{k}-\frac{1}{2}=\frac{1}{2}(3^{k+1}-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 for $\displaystyle n = 1$	LHS: $\displaystyle 3^0 = 1$ , RHS: $\displaystyle \dfrac{1}{2}(3 - 1) = 1$ , so true for $\displaystyle n = 1$
•²	assume (statement) true for $\displaystyle n = k$ AND consider whether (statement) true for $\displaystyle n = k + 1$	Suitable statement and $\displaystyle \displaystyle\sum_{r=1}^{k}3^{r-1} = \dfrac{1}{2}(3^k - 1)$ AND $\displaystyle \displaystyle\sum_{r=1}^{k+1}3^{r-1} = \ldots$
•³	correct statement for sum to $\displaystyle (k + 1)$ terms using inductive hypothesis	$\displaystyle \ldots = \dfrac{1}{2}(3^k - 1) + 3^{(k+1)-1}$
•⁴	combine terms in $\displaystyle 3^k$	$\displaystyle \dfrac{3}{2} \times 3^k - \dfrac{1}{2}$
•⁵	express sum explicitly in terms of $\displaystyle (k + 1)$ or achieve stated aim/goal AND communicate	$\displaystyle \dfrac{1}{2}(3^{(k+1)} - 1).$ 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.
2. For •² acceptable phrases for $\displaystyle n = k$ contain: "If true for..."; "Suppose true for…"; "Assume true for…". For •² unacceptable phrases for $\displaystyle n = k$ contain: "Consider $\displaystyle n = k$ ", "assume $\displaystyle n = k$ " and "True for $\displaystyle n = k$ ". **An acceptable phrase may appear at •⁵.** For •², in addition to an acceptable phrase containing $\displaystyle n = k$ , accept: "Aim/goal: $\displaystyle \displaystyle\sum_{r=1}^{k+1}3^{r-1} = \dfrac{1}{2}(3^{k+1} - 1)$ ". For •² unacceptable phrases for $\displaystyle n = k + 1$ contain: "Consider true for $\displaystyle n = k + 1$ ", "true for $\displaystyle n = k + 1$ "; " $\displaystyle \displaystyle\sum_{r=1}^{k+1}3^{r-1} = \dfrac{1}{2}(3^{k+1} - 1)$ " (with no further working).
3. At •³ accept $\displaystyle \ldots = \dfrac{1}{2}(3^k - 1) + 3^k$ or $\displaystyle \ldots = \dfrac{1}{2}(3^k - 1) + 3^{k+1-1}.$
4. At •⁴ accept $\displaystyle \ldots = \dfrac{1}{2}(3 \times 3^k - 1).$
5. •⁵ is unavailable to candidates who write down the correct expression without algebraic justification.
6. Full marks are available to candidates who state an aim/goal earlier in the proof and who subsequently achieve the stated aim/goal.
7. Following the required algebra and statement of the inductive hypothesis, the minimum 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.

Related rates of change

(1, 5)6 Marks

2018 Q13

An engineer has designed a lifting device. The handle turns a screw which shortens the horizontal length and increases the vertical height.

The device is modelled by a rhombus, with each side 25 cm.

The horizontal length is $\displaystyle x$ cm, and the vertical height is $\displaystyle h$ cm as shown.

(a)Show that $\displaystyle h = \sqrt{2500 - x^2}.$

(b)The horizontal length decreases at a rate of 0.3 cm per second as the handle is turned.

Find the rate of change of the vertical height when $\displaystyle x = 30.$

Show answer

(a) Using Pythagoras' theorem on a quarter of the rhombus:

\displaystyle \left(\frac{x}{2}\right)^2 + \left(\frac{h}{2}\right)^2 = 25^2.

Rearranging gives

\displaystyle x^2 + h^2 = 2500

, hence

\displaystyle h = \sqrt{2500 - x^2}.

(b)

\displaystyle \frac{9}{40}

cm s

\displaystyle ^{-1}

(or

\displaystyle 0.225

cm s

\displaystyle ^{-1}

)

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	determine the relationship between $\displaystyle x$ and $\displaystyle h$	$\displaystyle x^2 + h^2 = 2500$ , $\displaystyle h = \sqrt{2500 - x^2}$
(b) •²	interpret rate of change of $\displaystyle x$	$\displaystyle \dfrac{dx}{dt} = -0.3$
(b) •³	find $\displaystyle \dfrac{dh}{dx}$	$\displaystyle \dfrac{dh}{dx} = -2x \cdot \dfrac{1}{2}(2500 - x^2)^{-\frac{1}{2}}$
(b) •⁴	form relationship	$\displaystyle \dfrac{dh}{dt} = \dfrac{dh}{dx} \cdot \dfrac{dx}{dt}$ stated or implied at •⁵
(b) •⁵	multiply by $\displaystyle \dfrac{dx}{dt}$	$\displaystyle \dfrac{dh}{dt} = \dfrac{0.3x}{\sqrt{2500 - x^2}}$
(b) •⁶	evaluate $\displaystyle \dfrac{dh}{dt}$	$\displaystyle \dfrac{dh}{dt} = \dfrac{9}{40}\ \text{cms}^{-1}$

Notes

(a) 1. Refer to general marking principle (m).
Commonly observed: 1. (a) $\displaystyle \left(\dfrac{1}{2}h\right)^2 = 25^2 - \left(\dfrac{1}{2}x\right)^2$ leading to correct answer, award •¹; $\displaystyle \dfrac{1}{2}h^2 = 25^2 - \dfrac{1}{2}x^2$ leading to $\displaystyle h^2 = 50^2 - x^2$ , do not award •¹; $\displaystyle \dfrac{1}{2}h^2 = 25^2 - \dfrac{1}{2}x^2$ leading to $\displaystyle h^2 = 4 \times 25^2 - x^2$ or equivalent, award •¹; $\displaystyle \dfrac{1}{2}h = \sqrt{25^2 - \dfrac{1}{4}x^2}$ moving directly to correct answer, award •¹.
(b) 1. At •⁵ candidates may evaluate the derivatives separately eg $\displaystyle \dfrac{dh}{dt} = -0.75 \times (-0.3).$
(b) 2. At •⁵ simplification is not required.
(b) 3. At •⁶ units are required. Accept decimal equivalent ( $\displaystyle 0.225\ \text{cms}^{-1}$ ).
(b) 4. Where candidates produce an incorrect answer, accept a decimal rounded to at least 2 significant figures.
(b) 5. Award •⁶ only where a candidate's final answer for $\displaystyle \dfrac{dh}{dt}$ is opposite in sign to that of $\displaystyle \dfrac{dx}{dt}.$
(b) 6. For candidates who do not show evidence of related rates, •⁴, •⁵ and •⁶ are not available.

Arithmetic and geometric combined

(2, 2, 2, 1, 3)10 Marks

2018 Q14

A geometric sequence has first term 80 and common ratio $\displaystyle \frac{1}{3}.$

(a)For this sequence, calculate:

(i) the $\displaystyle 7^{\text{th}}$ term;
(ii) the sum to infinity of the associated geometric series.

The first term of this geometric sequence is equal to the first term of an arithmetic sequence.
The sum of the first five terms of this arithmetic sequence is 240.

(b)(i) Find the common difference of this sequence.
(ii) Write down and simplify an expression for the $\displaystyle n^{\text{th}}$ term.

Let $\displaystyle S_n$ represent the sum of the first $\displaystyle n$ terms of this arithmetic sequence.

(c)Find the values of $\displaystyle n$ for which $\displaystyle S_n = 144.$

Show answer

(a)(i)

\displaystyle \frac{80}{729}

(a)(ii)

\displaystyle 120

(b)(i)

\displaystyle -16

(b)(ii)

\displaystyle 96 - 16n

(c)

\displaystyle n = 2, n = 9

Show marking instructions

Mark	Awarded for	Illustrative answer
(a)(i) •¹	multiply first term by a power of the common ratio	$\displaystyle 80\left(\dfrac{1}{3}\right)^{\ldots}$
(a)(i) •²	find term	$\displaystyle \dfrac{80}{729}$
(a)(ii) •³	substitute	$\displaystyle \dfrac{80}{1 - \frac{1}{3}}$
(a)(ii) •⁴	find sum to infinity	$\displaystyle 120$
(b)(i) •⁵	substitute	eg $\displaystyle \dfrac{5}{2}(2 \times 80 + (5 - 1)d) = 240$
(b)(i) •⁶	find common difference	$\displaystyle -16$
(b)(ii) •⁷	find simplified expression	$\displaystyle 96 - 16n$
(c) •⁸	set up equation	$\displaystyle \dfrac{n}{2}\left[160 + (n - 1)(-16)\right] = 144$
(c) •⁹	obtain quadratic equation in general form	$\displaystyle 16n^2 - 176n + 288 = 0$
(c) •¹⁰	find values of $\displaystyle n$	$\displaystyle n = 2,\ n = 9$

Notes

(a) 1. (i) At •¹ accept any integer index other than 0 or 1.
(a) 2. (i) Where candidates elect to repeatedly multiply, the minimum acceptable response for •¹ is $\displaystyle 80 \times \dfrac{1}{3} \times \dfrac{1}{3}\ldots.$
(a) 3. (i) At •² accept 0.11.
(a) 4. (i) Award full marks for a correct answer with no working.
(a) 5. (ii) A correct answer without working receives no credit.
(a) 6. (ii) •³ and •⁴ are available only where a formula has been used.
(a) 7. (ii) At •³ candidates may use either the formula for the sum to infinity or apply a limiting argument using the formula for the sum to $\displaystyle n$ terms (of a geometric progression).
(b) 1. (i) For a correct answer without working award 0/2.
(c) 1. At •⁹ ' $\displaystyle = 0$ ' must appear.
(c) 2. At •¹⁰ candidates must obtain 2 positive integer solutions.

Integration by partsFirst order linear DE

(3, 7)10 Marks

2018 Q15

(a)Use integration by parts to find $\displaystyle \int x \sin 3x \, dx.$

(b)Hence find the particular solution of

$\displaystyle \frac{dy}{dx} - \frac{2}{x}y = x^3 \sin 3x$ , $\displaystyle x \neq 0$

given that $\displaystyle x = \pi$ when $\displaystyle y = 0.$
Express your answer in the form $\displaystyle y = f(x).$

Show answer

(a)

\displaystyle -\frac{x}{3} \cos 3x + \frac{1}{9} \sin 3x + c

(b)

\displaystyle y = -\frac{x^3}{3}\cos 3x + \frac{x^2}{9}\sin 3x - \frac{\pi x^2}{3}

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	start integration by parts	$\displaystyle -\dfrac{x}{3}\cos 3x - \ldots$
(a) •²	complete integration by parts	$\displaystyle \ldots\displaystyle\int -\dfrac{1}{3}\cos 3x\,dx$
(a) •³	complete integration	$\displaystyle = -\dfrac{x}{3}\cos 3x + \dfrac{1}{9}\sin 3x + c$
(b) •⁴	identify integral form of integrating factor	$\displaystyle e^{\int -\frac{2}{x}dx}$
(b) •⁵	determine integrating factor	$\displaystyle \dfrac{1}{x^2}$
(b) •⁶	begin solution	$\displaystyle \dfrac{d}{dx}\left(\dfrac{1}{x^2}y\right) = \dfrac{1}{x^2}(x^3\sin 3x)$ stated or implied at •⁷
(b) •⁷	rewrite as integral equation	$\displaystyle \dfrac{1}{x^2}y = \displaystyle\int x\sin 3x\,dx$
(b) •⁸	integrate	$\displaystyle \dfrac{1}{x^2}y = -\dfrac{x}{3}\cos 3x + \dfrac{1}{9}\sin 3x + c$
(b) •⁹	evaluate constant	$\displaystyle c = -\dfrac{\pi}{3}$
(b) •¹⁰	form particular solution	$\displaystyle y = -\dfrac{x^3}{3}\cos 3x + \dfrac{x^2}{9}\sin 3x - \dfrac{\pi x^2}{3}$

Notes

(a) 1. When integrating, candidates who repeatedly multiply by 3 cannot be awarded •¹ but •² and •³ may still be available.
(a) 2. Candidates who communicate an intention to integrate $\displaystyle \sin 3x$ and differentiate $\displaystyle x$ but who inadvertently produce the derivatives of both $\displaystyle \sin 3x$ and $\displaystyle x$ cannot be awarded •¹ but •² and •³ are still available.
(a) 3. For candidates who choose $\displaystyle \sin 3x$ as the function to differentiate and $\displaystyle x$ as the function to integrate, •¹ is available only where this is processed correctly. •² and •³ are not available.
(a) 4. An error in sign when differentiating or integrating a trigonometric function should not be treated as a repeated error.
(a) 5. Do not withhold •³ for the omission of the constant of integration.
(a) 6. The evidence for •¹, •² and •³ may appear in (b).
(b) 1. Candidates who attempt to solve the equation using eg separation of variables or second order method receive 0/7.
(b) 2. Candidates who attempt to apply integration by parts to the entire differential equation receive 0/7.
(b) 3. At •⁵ accept an unsimplified integrating factor eg $\displaystyle e^{-2\ln x}.$
(b) 4. For candidates who omit the constant of integration, •⁸, •⁹ and •¹⁰ are not available.
(b) 5. •⁸ is available only where a candidate integrates correctly based on their RHS at •⁷.
(b) 6. Where candidates obtain an integrating factor which is a constant, •⁵, •⁹ and •¹⁰ are not available.
(b) 7. For candidates who proceed from •⁸ by multiplying through by $\displaystyle x^2$ : award •⁹ for multiplying through by $\displaystyle x^2$ and award •¹⁰ for evaluating the constant of integration **then** stating the particular solution.
(b) 8. For candidates who proceed from •⁸ by multiplying through by $\displaystyle x^2$ but who fail to multiply the constant of integration, •⁹ and •¹⁰ are not available.

Intersection of planes

(4, 2, 3, 1)10 Marks

2018 Q16

Planes $\displaystyle \pi_1$ , $\displaystyle \pi_2$ and $\displaystyle \pi_3$ have equations:

$\displaystyle \pi_1$ : $\displaystyle x - 2y + z = -4$
$\displaystyle \pi_2$ : $\displaystyle 3x - 5y - 2z = 1$
$\displaystyle \pi_3$ : $\displaystyle -7x + 11y + az = -11$

where $\displaystyle a \in \mathbb{R}.$

(a)Use Gaussian elimination to find the value of $\displaystyle a$ such that the intersection of the planes $\displaystyle \pi_1$ , $\displaystyle \pi_2$ and $\displaystyle \pi_3$ is a line.

(b)Find the equation of the line of intersection of the planes when $\displaystyle a$ takes this value.

The plane $\displaystyle \pi_4$ has equation $\displaystyle -9x + 15y + 6z = 20.$

(c)Find the acute angle between $\displaystyle \pi_1$ and $\displaystyle \pi_4.$

(d)Describe the geometrical relationship between $\displaystyle \pi_2$ and $\displaystyle \pi_4.$ Justify your answer.

Show answer

(a)

\displaystyle a = 8

(b)

\displaystyle x = 22 + 9t, y = 13 + 5t, z = t

(or equivalent form)
(c)

\displaystyle 43^{\circ}

(or

\displaystyle 0.75

rad)
(d) Planes

\displaystyle \pi_2

and

\displaystyle \pi_4

are parallel because the normal of

\displaystyle \pi_4

is a scalar multiple of the normal of

\displaystyle \pi_2.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	set up augmented matrix	$\displaystyle \begin{bmatrix}1 & -2 & 1 & -4\\3 & -5 & -2 & 1\\-7 & 11 & a & -11\end{bmatrix}$
(a) •²	obtain two zeros	$\displaystyle \begin{bmatrix}1 & -2 & 1 & -4\\0 & 1 & -5 & 13\\0 & -3 & a + 7 & -39\end{bmatrix}$
(a) •³	complete row operations	$\displaystyle \begin{bmatrix}1 & -2 & 1 & -4\\0 & 1 & -5 & 13\\0 & 0 & a - 8 & 0\end{bmatrix}$
(a) •⁴	obtain value for $\displaystyle a$	$\displaystyle a = 8$
(b) •⁵	introduce parameter and substitute	$\displaystyle z = t$ , $\displaystyle y - 5t = 13$
(b) •⁶	equation of line	$\displaystyle x = 22 + 9t$ , $\displaystyle y = 13 + 5t$ , $\displaystyle z = t$
(c) •⁷	write down normals	$\displaystyle \begin{pmatrix}1\\-2\\1\end{pmatrix}$ , $\displaystyle \begin{pmatrix}-3\\5\\2\end{pmatrix}$ stated or implied
(c) •⁸	start to find angle	$\displaystyle \cos\theta = \dfrac{-11}{\sqrt{38}\sqrt{6}}$ OR $\displaystyle \cos\theta = \dfrac{11}{\sqrt{38}\sqrt{6}}$
(c) •⁹	find acute angle	$\displaystyle 0.75$
(d) •¹⁰	explanation	Planes $\displaystyle \pi_2$ and $\displaystyle \pi_4$ are parallel because the normal of $\displaystyle \pi_4$ is a multiple of the normal of $\displaystyle \pi_2.$

Notes

(a) 1. Only Gaussian elimination (ie a systematic approach using EROs) is acceptable for the award of •² and •³.
(a) 2. For •³ accept any equivalent form.
(a) 3. •⁴ is not available unless the candidate's augmented matrix exhibits redundancy.
(b) 1. •⁵ and •⁶ are not available for substituting in either a numerical value or any expression in terms of $\displaystyle a.$
(b) 2. •⁵ is not available where the candidate substitutes into a row containing two other variables.
(b) 3. For •⁶ accept symmetric or vector form.
Commonly observed: 1. (b) For the line, $\displaystyle \mathbf{d} = \mathbf{n}_{\pi_1} \times \mathbf{n}_{\pi_4} = \begin{pmatrix}1\\-2\\1\end{pmatrix} \times \begin{pmatrix}3\\-5\\-2\end{pmatrix} = \begin{pmatrix}9\\5\\1\end{pmatrix}$ ; eg let $\displaystyle z = 0$ so that $\displaystyle x - 2y = -4$ , award •⁵; form second equation eg $\displaystyle 3x - 5y = 1$ and solve to give $\displaystyle x = 22$ , $\displaystyle y = 13$ leading to $\displaystyle \dfrac{x - 22}{9} = \dfrac{y - 13}{5} = \dfrac{z}{1}\ (= \lambda)$ , award •⁶.
(c) 1. At •⁷ accept the use of $\displaystyle \begin{pmatrix}-9\\15\\6\end{pmatrix}.$
(c) 2. Accept an answer in degrees which rounds to $\displaystyle 43^\circ.$
(c) 3. •⁹ is not available for incorrect working subsequent to a correct answer eg $\displaystyle 90^\circ - 43^\circ.$
(c) 4. At •⁷ accept eg $\displaystyle \pi_1 = \begin{pmatrix}1\\-2\\1\end{pmatrix}$ or $\displaystyle \pi_4 = \begin{pmatrix}-3\\5\\2\end{pmatrix}$ but not at •¹⁰.
(c) 5. For candidates who express an answer in degrees, the degree symbol must appear.
Commonly observed: 1. (c) Solution obtained by rearrangement of the vector product formula: $\displaystyle \sin\theta = \dfrac{\sqrt{107}}{\sqrt{6}\sqrt{38}}$ , award •⁸.
(d) 1. For the award of •¹⁰ a statement must compare normal vectors or coefficients of $\displaystyle x$ , $\displaystyle y$ and $\displaystyle z.$ Accept eg $\displaystyle \begin{pmatrix}\ldots\\\ldots\\\ldots\end{pmatrix} = -3\begin{pmatrix}\ldots\\\ldots\\\ldots\end{pmatrix}$ or 'The normals are multiples of one another' as justification for the planes being parallel.
(d) 2. Do not accept a plane equating to a vector eg $\displaystyle \pi_2 = \begin{pmatrix}-3\\5\\2\end{pmatrix}.$
(d) 3. Withhold •¹⁰ from candidates who provide a correct description but who subsequently write eg $\displaystyle \pi_4 = -3\pi_2$ or make reference to "direction vectors".
Commonly observed: 1. (d) The planes are parallel because: their normals are multiples of each other, award •¹⁰; $\displaystyle \pi_4 = -3\pi_2$ , do not award •¹⁰; their direction vectors are multiples of each other, do not award •¹⁰.

Product, quotient, chain rules

(2, 3, 2, 2, 1)10 Marks

2018 Q17

(a)Given $\displaystyle f(x) = e^{2x}$ obtain the Maclaurin expansion for $\displaystyle f(x)$ up to, and including, the term in $\displaystyle x^3.$

(b)On a suitable domain, let $\displaystyle g(x) = \tan x.$
(i) Show that the third derivative of $\displaystyle g(x)$ is given by
$\displaystyle g'''(x) = 2 \sec^4 x + 4 \tan^2 x \sec^2 x.$
(ii) Hence obtain the Maclaurin expansion for $\displaystyle g(x)$ up to and including the term in $\displaystyle x^3.$

(c)Hence, or otherwise, obtain the Maclaurin expansion for $\displaystyle e^{2x} \tan x$ up to, and including, the term in $\displaystyle x^3.$

(d)Write down the first three non-zero terms in the Maclaurin expansion for
$\displaystyle 2e^{2x} \tan x + e^{2x} \sec^2 x.$

Show answer

(a)

\displaystyle 1 + 2x + 2x^2 + \frac{4}{3}x^3

(b)(i)

\displaystyle g'(x) = \sec^2 x

, so

\displaystyle g''(x) = 2\sec x \times \sec x\tan x = 2\sec^2 x\tan x.

By the product rule,

\displaystyle g'''(x) = 2\sec^2 x \times \sec^2 x + 4\sec^2 x\tan x \times \tan x = 2\sec^4 x + 4\tan^2 x\sec^2 x.

(b)(ii)

\displaystyle x + \frac{1}{3}x^3

(c)

\displaystyle x + 2x^2 + \frac{7}{3}x^3

(d)

\displaystyle 1 + 4x + 7x^2

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	first derivative and two evaluations OR all three derivatives OR all four evaluations	$\displaystyle f(x) = e^{2x}$ , $\displaystyle f(0) = 1$ ; $\displaystyle f'(x) = 2e^{2x}$ , $\displaystyle f'(0) = 2$ ; $\displaystyle f''(x) = 4e^{2x}$ , $\displaystyle f''(0) = 4$ ; $\displaystyle f'''(x) = 8e^{2x}$ , $\displaystyle f'''(0) = 8$
(a) •²	obtain expression	$\displaystyle f(x) = 1 + 2x + 2x^2 + \dfrac{4}{3}x^3\ldots$
(b)(i) •³	find $\displaystyle g''(x)$	$\displaystyle g''(x) = 2\sec x\sec x\tan x$
(b)(i) •⁴	evidence of product rule	$\displaystyle g'''(x) = 2\sec^2 x(\ldots) + (\ldots)\tan x$
(b)(i) •⁵	complete proof	$\displaystyle g'''(x) = 2\sec^2 x(\sec^2 x) + (4\sec^2 x\tan x)\tan x$
(b)(ii) •⁶	completes ALL evaluations	$\displaystyle g(0) = 0$ , $\displaystyle g'(0) = 1$ , $\displaystyle g''(0) = 0$ , $\displaystyle g'''(0) = 2$
(b)(ii) •⁷	substitute	$\displaystyle g(x) = x + \dfrac{1}{3}x^3\ldots$
(c) •⁸	multiply expressions	$\displaystyle (1 + 2x + 2x^2 + \ldots)\left(x + \dfrac{1}{3}x^3\ldots\right)$
(c) •⁹	multiply out and simplify	$\displaystyle x + 2x^2 + \dfrac{7}{3}x^3\ldots$
(d) •¹⁰	write down terms	$\displaystyle 1 + 4x + 7x^2$

Notes

(a) 1. Simplification might not appear until (c).
(b) 1. (i) Candidates can be awarded •⁴ only where the product or quotient rule is required to differentiate their expression for $\displaystyle g''(x).$
(b) 2. (i) At •⁴ there must be clear evidence of the product rule (or quotient rule).
(b) 3. (i) •⁵ is not available to candidates who obtain an incorrect answer at •³.
(b) 4. (i) •⁵ can be awarded only where the candidate completes the differentiation correctly and shows clearly that the result is equivalent to the expression asked for in the question.
(b) 5. (ii) •⁷ is available only for powers of $\displaystyle x$ with numerical coefficients.
(c) 1. For candidates who proceed via differentiation •⁸ is available for obtaining all three derivatives correctly.
(c) 2. •⁹ is available only for powers of $\displaystyle x$ with numerical coefficients.

2018 Paper

Question 1(a)

Notes

Question 1(b)

Notes

Question 1(c)

Notes

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

Notes

Question 10

Notes

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

Notes

Question 16

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Question 17

Notes