Advanced Higher Maths 2017 Paper — Questions & Marking Instructions

Binomial expansion

4 Marks

2017 Q1

Write down the binomial expansion of $\displaystyle \left(\frac{2}{y^2}-5y\right)^3$ and simplify your answer.

Show answer

\displaystyle \frac{8}{y^6} - \frac{60}{y^3} + 150 - 125y^3

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	write down binomial expansion	$\displaystyle \dbinom{3}{0}\left(\dfrac{2}{y^2}\right)^3 + \dbinom{3}{1}\left(\dfrac{2}{y^2}\right)^2(-5y) + \dbinom{3}{2}\left(\dfrac{2}{y^2}\right)(-5y)^2 + \dbinom{3}{3}(-5y)^3$
•²	resolve signs	(•², •³ and •⁴ together) $\displaystyle \dfrac{8}{y^6} - \dfrac{60}{y^3} + 150 - 125y^3$
•³	simplify coefficients or powers of $\displaystyle y$	(•², •³ and •⁴ together) $\displaystyle \dfrac{8}{y^6} - \dfrac{60}{y^3} + 150 - 125y^3$
•⁴	complete simplification and obtain expression	$\displaystyle \dfrac{8}{y^6} - \dfrac{60}{y^3} + 150 - 125y^3$

Notes

1. Accept any correct form for binomial coefficients.
2. Accept negative powers of $\displaystyle y.$
3. For candidates who expand $\displaystyle \left(\dfrac{2}{y^2} + 5y\right)^3$ using the Binomial Theorem •¹ and •² are not available.
4. Candidates who expand $\displaystyle \left(\dfrac{2}{y^2} - 5y\right)^3$ without using the Binomial Theorem may be awarded •², •³ and •⁴ but •¹ is not available.
5. •⁴ is not available for a final expression which contains the term ' $\displaystyle 150y^0$ '.
6. Do not award •⁴ where the candidate produces incorrect working subsequent to a correct simplification.

Partial fractions

4 Marks

2017 Q2

Express $\displaystyle \frac{x^2-6x+20}{(x+1)(x-2)^2}$ in partial fractions.

Show answer

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

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	state expression	$\displaystyle \dfrac{x^2 - 6x + 20}{(x + 1)(x - 2)^2} = \dfrac{A}{(x + 1)} + \dfrac{B}{(x - 2)} + \dfrac{C}{(x - 2)^2}$
•²	form equation	$\displaystyle x^2 - 6x + 20 = A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1)$
•³	obtain two of $\displaystyle A$ , $\displaystyle B$ and $\displaystyle C$	$\displaystyle A = 3,\ B = -2,\ C = 4$
•⁴	obtain final constant and state expression	$\displaystyle \dfrac{3}{(x + 1)} - \dfrac{2}{(x - 2)} + \dfrac{4}{(x - 2)^2}$

Notes

1. At •⁴ accept $\displaystyle \dfrac{3}{(x + 1)} + \dfrac{-2}{(x - 2)} + \dfrac{4}{(x - 2)^2}$ but do not accept $\displaystyle \dfrac{3}{(x + 1)} + -\dfrac{2}{(x - 2)} + \dfrac{4}{(x - 2)^2}.$
Commonly observed: 1. Alternative method: $\displaystyle \dfrac{A}{x + 1} + \dfrac{Bx + C}{(x - 2)^2}$ , award •¹; $\displaystyle x^2 - 6x + 20 = A(x - 2)^2 + (Bx + C)(x + 1)$ and one of $\displaystyle A$ , $\displaystyle B$ , $\displaystyle C$ , award •²; remaining two ( $\displaystyle A = 3$ , $\displaystyle B = -2$ , $\displaystyle C = 8$ ), award •³; $\displaystyle \dfrac{3}{x + 1} + \dfrac{8 - 2x}{(x - 2)^2} = \dfrac{3}{x + 1} + \dfrac{4 - 2(x - 2)}{(x - 2)^2} = \dfrac{3}{x + 1} - \dfrac{2}{(x - 2)} + \dfrac{4}{(x - 2)^2}$ , award •⁴.
Commonly observed: 2. Incorrect method: $\displaystyle \dfrac{A}{x + 1} + \dfrac{B}{(x - 2)^2}$ , do not award •¹; $\displaystyle x^2 - 6x + 20 = A(x - 2)^2 + B(x + 1)$ , do not award •²; $\displaystyle B = 4, A = 3$ or $\displaystyle B = 4, A = 1$ , award •³; $\displaystyle \dfrac{3}{x + 1} + \dfrac{4}{(x - 2)^2}$ or $\displaystyle \dfrac{1}{x + 1} + \dfrac{4}{(x - 2)^2}$ , do not award •⁴.

Product, quotient, chain rules

3 Marks

2017 Q3

On a suitable domain, a function is defined by $\displaystyle f(x)=\frac{e^{x^2-1}}{x^2-1}.$
Find $\displaystyle f'(x)$ simplifying your answer.

Show answer

\displaystyle \frac{2xe^{x^2-1}(x^2-2)}{(x^2-1)^2}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	evidence use of quotient rule with one term of numerator correct	$\displaystyle 2xe^{x^2 - 1}(x^2 - 1) - \ldots$
•²	complete differentiation	$\displaystyle \dfrac{\ldots 2xe^{x^2 - 1}}{(x^2 - 1)^2}$
•³	simplify	$\displaystyle \dfrac{2xe^{x^2 - 1}(x^2 - 2)}{(x^2 - 1)^2}$

Notes

1. At •³ accept $\displaystyle \dfrac{2x^3e^{x^2 - 1} - 4xe^{x^2 - 1}}{(x^2 - 1)^2}$ but not accept $\displaystyle \dfrac{2xe^{x^2 - 1}((x^2 - 1) - 1)}{(x^2 - 1)^2}$ (GM Principle (l)).
2. Do not award •³ where the candidate produces further incorrect simplification subsequent to a correct answer.
3. Where a candidate differentiates incorrectly •³ may be available provided like terms are collected in the numerator. Where this is not possible the expression should be fully factorised (this need not extend to exponential functions of differing powers). Where no simplification is possible •³ is not available.

Arithmetic sequences and series

(2, 3)5 Marks

2017 Q4

The fifth term of an arithmetic sequence is $\displaystyle -6$ and the twelfth term is $\displaystyle -34.$

(a)Determine the values of the first term and the common difference.

(b)Obtain algebraically the value of $\displaystyle n$ for which $\displaystyle S_n = -144.$

Show answer

(a)

\displaystyle a = 10

,

\displaystyle d = -4

(b)

\displaystyle n = 12

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	evidence use of valid strategy	eg $\displaystyle a + 4d = -6$ , $\displaystyle a + 11d = -34$
(a) •²	obtain values of $\displaystyle a$ and $\displaystyle d$	$\displaystyle a = 10,\ d = -4$
(b) •³	set up equation	$\displaystyle \dfrac{n}{2}\left[20 - 4(n - 1)\right] = -144$
(b) •⁴	rearrange to standard form	$\displaystyle 2n^2 - 12n - 144 = 0$
(b) •⁵	determine the value of $\displaystyle n$	$\displaystyle n > 0$ $\displaystyle \therefore n = 12$

Notes

(a) 1. Candidates who state correct values for both $\displaystyle a$ and $\displaystyle d$ without working may be awarded •¹ and •².
(b) 1. •⁴ may be awarded only where a quadratic equation has been expressed in standard form.
(b) 2. •⁵ may be awarded only where an invalid solution for $\displaystyle n$ has been discarded.

Systems of Equations

(4, 1, 1)6 Marks

2017 Q5

(a)(i) Use Gaussian elimination on the system of equations below to give an expression for $\displaystyle z$ in terms of $\displaystyle \lambda.$

$\displaystyle x + 2y - z = -3$
$\displaystyle 4x - 2y + 3z = 11$
$\displaystyle 3x + y + 2\lambda z = 8$

(ii) For what value of $\displaystyle \lambda$ is this system of equations inconsistent?

(b)Determine the solution of this system when $\displaystyle \lambda = -2.5.$

Show answer

(a)(i)

\displaystyle z = \frac{11}{4\lambda - 1}

(a)(ii)

\displaystyle \lambda = \frac{1}{4}

(b)

\displaystyle x = 2

,

\displaystyle y = -3

,

\displaystyle z = -1

Show marking instructions

Mark	Awarded for	Illustrative answer
(a)(i) •¹	set up augmented matrix	$\displaystyle \left(\begin{array}{ccc\|c}1 & 2 & -1 & -3\\4 & -2 & 3 & 11\\3 & 1 & 2\lambda & 8\end{array}\right)$
(a)(i) •²	obtain two zeros	$\displaystyle \left(\begin{array}{ccc\|c}1 & 2 & -1 & -3\\0 & -10 & 7 & 23\\0 & -5 & 2\lambda + 3 & 17\end{array}\right)$
(a)(i) •³	complete row operations	$\displaystyle \left(\begin{array}{ccc\|c}1 & 2 & -1 & -3\\0 & -10 & 7 & 23\\0 & 0 & 4\lambda - 1 & 11\end{array}\right)$
(a)(i) •⁴	obtain expression for $\displaystyle z$	$\displaystyle z = \dfrac{11}{4\lambda - 1}$
(a)(ii) •⁵	state value of $\displaystyle \lambda$	$\displaystyle \lambda = \dfrac{1}{4}$
(b) •⁶	find solution	$\displaystyle z = -1,\ y = -3,\ x = 2$

Notes

1. Only Gaussian Elimination (i.e. a systematic approach using EROs) is acceptable for the award of •² and •³.
2. Do not accept an answer of $\displaystyle (4\lambda - 1)z = 11$ when awarding •⁴.
3. At •⁴ accept an unsimplified expression for $\displaystyle z$ eg $\displaystyle z = \dfrac{5.5}{2\lambda - \frac{1}{2}}.$
4. Where decimal approximations are used •⁶ is available only where candidates work to 3sf or better.

Integration by substitution

6 Marks

2017 Q6

Use the substitution $\displaystyle u=5x^2$ to find the exact value of $\displaystyle \int_{0}^{\frac{1}{\sqrt{10}}} \frac{x}{\sqrt{1-25x^4}} dx.$

Show answer

\displaystyle \frac{\pi}{60}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	differentiate $\displaystyle 5x^2$	$\displaystyle \dfrac{du}{dx} = 10x$ or $\displaystyle du = 10x\,dx$
•²	find limits for $\displaystyle u$	$\displaystyle u = 0,\ u = \dfrac{1}{2}$
•³	replace ' $\displaystyle x\,dx$ '	$\displaystyle \dfrac{1}{10}\displaystyle\int\ldots du$
•⁴	obtain integrand	$\displaystyle \dfrac{1}{10}\displaystyle\int_0^{\frac{1}{2}}\dfrac{1}{\sqrt{1 - u^2}}\,du$
•⁵	integrate	$\displaystyle \dfrac{1}{10}\left[\sin^{-1}u\right]_0^{\frac{1}{2}}$
•⁶	evaluate	$\displaystyle \dfrac{\pi}{60}$

Notes

1. At •³ and •⁴ treat as bad form situations where candidates either omit limits or retain limits for $\displaystyle x.$
2. Where candidates attempt to integrate an expression containing both $\displaystyle u$ and $\displaystyle x$ , where $\displaystyle x$ 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 $\displaystyle \dfrac{1}{10}\left[\sin^{-1}(5x^2)\right]_0^{\frac{1}{\sqrt{10}}}$ , •² may be awarded.
4. Where candidates show no working but write down $\displaystyle \dfrac{1}{10}\left[\sin^{-1}(5x^2)\right]_0^{\frac{1}{\sqrt{10}}}$ , •¹ is not available.
5. •⁵ and •⁶ are unavailable to candidates who having been awarded •⁴ subsequently proceed to $\displaystyle \dfrac{1}{10}\left[\dfrac{(1 - u^2)^{\frac{1}{2}}}{-\frac{1}{2} \times 2u}\right].$
6. For candidates who integrate incorrectly, •⁶ may be available provided division by zero does not occur.
7. For candidates who, upon integrating, obtain a trigonometric expression and then work in degrees •⁶ is unavailable.
8. Disregard the appearance of a decimal approximation subsequent to a simplified exact value.

Determinant and inverseMatrix operations

(1, 1, 2, 2)6 Marks

2017 Q7

Matrices $\displaystyle P$ and $\displaystyle Q$ are defined by $\displaystyle P = \begin{pmatrix} x & 2 \\ -5 & -1 \end{pmatrix}$ and $\displaystyle Q = \begin{pmatrix} 2 & -3 \\ 4 & y \end{pmatrix}$ , where $\displaystyle x,y \in \mathbb{R}.$

(a)Given the determinant of $\displaystyle P$ is 2, obtain:

(i) The value of $\displaystyle x.$
(ii) $\displaystyle P^{-1}.$
(iii) $\displaystyle P^{-1}Q'$ , where $\displaystyle Q'$ is the transpose of $\displaystyle Q.$

(b)The matrix $\displaystyle R$ is defined by $\displaystyle R = \begin{pmatrix} 5 & -2 \\ z & -6 \end{pmatrix}$ , where $\displaystyle z \in \mathbb{R}.$

Determine the value of $\displaystyle z$ such that $\displaystyle R$ is singular.

Show answer

(a)(i)

\displaystyle x = 8

(a)(ii)

\displaystyle P^{-1} = \frac{1}{2}\begin{pmatrix} -1 & -2 \\ 5 & 8 \end{pmatrix}

(a)(iii)

\displaystyle \frac{1}{2}\begin{pmatrix} 4 & -4-2y \\ -14 & 20+8y \end{pmatrix}

(b)

\displaystyle z = 15

Show marking instructions

Mark	Awarded for	Illustrative answer
(a)(i) •¹	determine value of $\displaystyle x$	$\displaystyle x = 8$
(a)(ii) •²	find inverse	$\displaystyle P^{-1} = \dfrac{1}{2}\begin{pmatrix}-1 & -2\\5 & 8\end{pmatrix}$
(a)(iii) •³	state transpose	$\displaystyle Q' = \begin{pmatrix}2 & 4\\-3 & y\end{pmatrix}$
(a)(iii) •⁴	obtain product	$\displaystyle P^{-1}Q' = \begin{pmatrix}2 & -2 - y\\-7 & 10 + 4y\end{pmatrix}$
(b) •⁵	state condition for singularity	$\displaystyle \det R = 0$ or one row is a multiple of the other
(b) •⁶	obtain value for $\displaystyle z$	$\displaystyle z = 15$

Notes

(a) 1. At •² accept $\displaystyle P^{-1} = \dfrac{1}{2}\begin{pmatrix}-1 & -2\\5 & x\end{pmatrix}.$
(a) 2. For •⁴ accept $\displaystyle P^{-1}Q' = \dfrac{1}{2}\begin{pmatrix}4 & -4 - 2y\\-14 & 20 + 8y\end{pmatrix}$ but not $\displaystyle P^{-1}Q' = \dfrac{1}{2}\begin{pmatrix}-2 + 6 & -4 - 2y\\10 - 24 & 20 + 8y\end{pmatrix}.$
(a) 3. •⁴ may be awarded only where $\displaystyle y$ is present.
(b) 1. $\displaystyle \det R = 0$ may be stated or implied in the working for •⁶.
(b) 2. For an answer of $\displaystyle z = 15$ without justification, •⁵ is not available.

Euclidean algorithm

4 Marks

2017 Q8

Use the Euclidean algorithm to find integers $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle 1595a + 1218b = 29.$

Show answer

\displaystyle a = 13, b = -17

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	start process	$\displaystyle 1595 = 1 \times 1218 + 377$
•²	obtain remainder of 29	$\displaystyle 1218 = 3 \times 377 + 87$ , $\displaystyle 377 = 4 \times 87 + 29$ , $\displaystyle 87 = 3 \times 29 + 0$
•³	express gcd in terms of 377 and 1218	$\displaystyle 29 = 377 - 4(1218 - 3 \times 377)$
•⁴	state values of $\displaystyle a$ and $\displaystyle b$	$\displaystyle a = 13,\ b = -17$

Notes

1. At •² the gcd does not need to be explicitly stated.
2. The minimum requirement for •⁴ is $\displaystyle 1595 \times 13 + 1218 \times (-17) = 29.$

Separable DE

5 Marks

2017 Q9

Solve $\displaystyle \frac{dy}{dx} = e^{2x}(1+y^2)$ given that when $\displaystyle x=0$ , $\displaystyle y=1.$
Express $\displaystyle y$ in terms of $\displaystyle x.$

Show answer

\displaystyle y = \tan\left(\frac{1}{2}e^{2x} + \frac{\pi}{4} - \frac{1}{2}\right)

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	separate variables and write down integral equation	$\displaystyle \displaystyle\int \dfrac{dy}{1 + y^2} = \int e^{2x}\,dx$
•²	integrate LHS	$\displaystyle \tan^{-1}y$
•³	integrate RHS	$\displaystyle \dfrac{1}{2}e^{2x} + c$
•⁴	evaluate constant of integration	$\displaystyle c = \dfrac{\pi}{4} - \dfrac{1}{2}$
•⁵	express $\displaystyle y$ in terms of $\displaystyle x$	$\displaystyle y = \tan\left(\dfrac{1}{2}e^{2x} + \dfrac{\pi}{4} - \dfrac{1}{2}\right)$

Notes

1. Do not withhold •¹ where $\displaystyle dy$ and $\displaystyle dx$ have been omitted.
2. For candidates who integrate the LHS and obtain a logarithmic expression, •² and •⁴ are not available.
3. For candidates who omit a constant of integration, •³ may be awarded but •⁴ and •⁵ are unavailable.
4. At •⁴ accept a decimal value for the constant of integration correct to at least 3sf ( $\displaystyle 0.285$ ).
5. For candidates who work in degrees, •⁴ is unavailable but •⁵ may be awarded.
6. At •⁵ do not accept e.g. $\displaystyle y = \tan\left(\dfrac{1}{2}e^{2x}\right) + \dfrac{\pi}{4} - \dfrac{1}{2}$ , $\displaystyle y = \tan\dfrac{1}{2}e^{2x} + \dfrac{\pi}{4} - \dfrac{1}{2}.$
7. Candidates who use either Integration by Parts or the Integrating Factor Method receive 0/5.

Summation formulae

(2, 2)4 Marks

2017 Q10

$\displaystyle S_n$ is defined by $\displaystyle \sum_{r=1}^{n}\left(r^2 + \frac{1}{3}r\right).$

(a)Find an expression for $\displaystyle S_n$ , fully factorising your answer.

(b)Hence find an expression for $\displaystyle \sum_{r=10}^{2p}\left(r^2 + \frac{1}{3}r\right)$ where $\displaystyle p > 5.$

Show answer

(a)

\displaystyle \frac{n(n+1)^2}{3}

(b)

\displaystyle \frac{2p(2p+1)^2}{3} - 300

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	substitute formulae	$\displaystyle \displaystyle\sum_{r=1}^{n}\left(r^2 + \dfrac{1}{3}r\right) = \dfrac{n(n + 1)(2n + 1)}{6} + \dfrac{1}{3}\left(\dfrac{n(n + 1)}{2}\right) = \dfrac{n(n + 1)((2n + 1) + 1)}{6}$
(a) •²	factorise fully	$\displaystyle = \dfrac{n(n + 1)^2}{3}$
(b) •³	substitute $\displaystyle 2p$ and 9	$\displaystyle \dfrac{2p(2p + 1)^2}{3}$ and $\displaystyle \dfrac{9(9 + 1)^2}{3}$ , $\displaystyle \dfrac{2p(2p + 1)^2}{3} - \dfrac{9(9 + 1)^2}{3}$
(b) •⁴	obtain expression	$\displaystyle = \dfrac{2p(2p + 1)^2}{3} - 300$

Notes

(a) 1. At •² do not accept $\displaystyle \dfrac{n(n + 1)(n + 1)}{3}$ or $\displaystyle \dfrac{2n(n + 1)^2}{6}.$

Logarithmic differentiation

5 Marks

2017 Q11

Given $\displaystyle y = x^{2x^3+1}$ , use logarithmic differentiation to find $\displaystyle \frac{dy}{dx}.$
Write your answer in terms of $\displaystyle x.$

Show answer

\displaystyle x^{2x^3+1} \left(6x^2\ln x + \frac{2x^3+1}{x}\right)

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	take logarithms of both sides and apply rule	$\displaystyle \ln y = (2x^3 + 1)\ln x$
•²	differentiate LHS	$\displaystyle \dfrac{1}{y}\dfrac{dy}{dx}$
•³	evidence use of product rule and one term correct	$\displaystyle 6x^2\ln x$ or $\displaystyle \dfrac{2x^3 + 1}{x}$
•⁴	complete differentiation	$\displaystyle 6x^2\ln x + \dfrac{2x^3 + 1}{x}$
•⁵	write $\displaystyle \dfrac{dy}{dx}$ in terms of $\displaystyle x$	$\displaystyle \dfrac{dy}{dx} = x^{2x^3 + 1}\left(6x^2\ln x + \dfrac{2x^3 + 1}{x}\right)$

Notes

1. Accept 'log' in lieu of 'ln'.
2. For candidates who do not attempt to use the product rule, •³ and •⁴ are not available.
Commonly observed: 1. For candidates who express $\displaystyle y$ as $\displaystyle e^{(2x^3 + 1)\ln x}$ : •¹ is not available; •² writing in the form $\displaystyle y = e^{(2x^3 + 1)\ln x}$ ; •³ apply chain rule $\displaystyle \dfrac{dy}{dx} = e^{(2x^3 + 1)\ln x} \times \dfrac{d}{dx}((2x^3 + 1)\ln x)$ ; •⁴ evidence use of product rule and one term correct, $\displaystyle 6x^2\ln x$ or $\displaystyle \dfrac{2x^3 + 1}{x}$ ; •⁵ complete differentiation, $\displaystyle \dfrac{dy}{dx} = x^{2x^3 + 1}\left(6x^2\ln x + \dfrac{2x^3 + 1}{x}\right)$ or $\displaystyle e^{(2x^3 + 1)\ln x}\left(6x^2\ln x + \dfrac{2x^3 + 1}{x}\right).$ Note: if a candidate writes $\displaystyle y = e^{(2x^3 + 1)}\ln x$ , only marks •⁴ and •⁵ are available.

AsymptotesOdd and even functions

(2, 2, 1)5 Marks

2017 Q12

In the diagram below part of the graph of $\displaystyle y=f(x)$ has been omitted. The point $\displaystyle (-1,-2)$ lies on the graph and the line $\displaystyle y=\frac{1}{2}x-3$ is an asymptote.

Graph showing an asymptote and part of a curve with point (-1,-2)

Given that $\displaystyle f(x)$ is an odd function:

(a)Copy and complete the diagram, including any asymptotes and any points you know to be on the graph.

(b) $\displaystyle g(x)=|f(x)|.$ On a separate diagram, sketch $\displaystyle g(x).$
Include known asymptotes and points.

(c)State the range of values of $\displaystyle f'(x)$ given that $\displaystyle f'(0)=2.$

Show answer

(a) Graph drawn with half-turn symmetry at the origin. New features include asymptote

\displaystyle y = \frac{1}{2}x + 3

and point

\displaystyle (1, 2)

clearly indicated.
(b) Graph showing the points

\displaystyle (-1, 2)

and

\displaystyle (1, 2)

, with asymptotes meeting on the positive y-axis forming a V-shape.
(c)

\displaystyle \frac{1}{2} < f'(x) \le 2

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	show half-turn symmetry and indicate $\displaystyle (1, 2)$	(•¹ and •² together) a sketch: the curve through $\displaystyle (-1, -2)$ , $\displaystyle O$ and $\displaystyle (1, 2)$ with half-turn symmetry about the origin, rising smoothly through $\displaystyle O$ , between the dashed asymptotes $\displaystyle y = \frac{1}{2}x - 3$ (through $\displaystyle -3$ on the $\displaystyle y$ -axis) and $\displaystyle y = \frac{1}{2}x + 3$ (through $\displaystyle 3$ ), approaching the first in quadrant 3 and the second in quadrant 1
(a) •²	demonstrate graph approaching parallel asymptote through $\displaystyle (0, 3)$	as •¹
(b) •³	apply modulus function to graph obtained in (a)	(•³ and •⁴ together) a sketch of $\displaystyle \|f(x)\|$ : a V-shaped curve with a cusp at $\displaystyle O$ , through $\displaystyle (-1, 2)$ and $\displaystyle (1, 2)$ , rising along the dashed asymptotes $\displaystyle y = \frac{1}{2}x + 3$ and $\displaystyle y = -\frac{1}{2}x + 3$ , which meet at $\displaystyle 3$ on the $\displaystyle y$ -axis
(b) •⁴	illustrate asymptotes meeting on the $\displaystyle y$ -axis	as •³
(c) •⁵	state range	$\displaystyle \dfrac{1}{2} < f'(x) \leq 2$

Notes

(a) 1. To award •¹ the candidate's graph should exhibit a smooth change in concavity at the origin.
(a) 2. Evidence of $\displaystyle (1, 2)$ may appear in (b).
(a) 3. At •² accept $\displaystyle y = \dfrac{1}{2}x + 3$ in lieu of $\displaystyle (0, 3).$
(a) 4. Where a candidate's graph diverges from the asymptote in quadrant 1, •² is not available.
(a) 5. For Graph 1 in the Commonly Observed Responses •¹, •², •³ and •⁴ are not available.
(a) 6. For Graph 2 in the Commonly Observed Responses •¹, •² and •³ are not available but •⁴ may be available where a second asymptote appears in (b).
Commonly observed: 1. (a) Graph 1: the given branch reflected in the $\displaystyle y$ -axis instead, through $\displaystyle (-1, 2)$ in quadrant 2. Graph 2 (one asymptote only): the given branch reflected in the $\displaystyle y$ -axis into quadrant 4, through $\displaystyle (1, -2)$ , meeting the other in a cusp at $\displaystyle O.$
(b) 1. To receive any credit, a candidate's graph from (a) must have a section lying in quadrant 1.
(b) 2. •⁴ is still available where a candidate's graph diverges from the asymptotes.
(b) 3. At •⁴ disregard the application of the modulus function to asymptotes.
(b) 4. Showing the image points is not required at •³.
(c) 1. Do not accept $\displaystyle \dfrac{1}{2} \leq f'(x) \leq 2$ or $\displaystyle \dfrac{1}{2} < f'(x) < 2.$
(c) 2. Accept ' $\displaystyle f'(x) > \dfrac{1}{2}$ and $\displaystyle f'(x) \leq 2$ ' but not ' $\displaystyle f'(x) > \dfrac{1}{2}$ or $\displaystyle f'(x) \leq 2$ '.
(c) 3. Accept ' $\displaystyle f'(x)$ is greater than $\displaystyle \dfrac{1}{2}$ and $\displaystyle f'(x)$ is less than or equal to 2'. Do not accept ' $\displaystyle f'(x)$ is between $\displaystyle \dfrac{1}{2}$ and 2'.

Proof by contrapositive

4 Marks

2017 Q13

Let $\displaystyle n$ be an integer.
Using proof by contrapositive, show that if $\displaystyle n^2$ is even, then $\displaystyle n$ is even.

Show answer

The contrapositive is: "If

\displaystyle n

is odd, then

\displaystyle n^2

is odd."
Let

\displaystyle n = 2k + 1

for some integer

\displaystyle k.

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

Since

\displaystyle 2k^2 + 2k

is an integer,

\displaystyle n^2

is odd.
Since the contrapositive statement is true, the original statement is true.

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	write down contrapositive statement	The contrapositive of the original statement is: If $\displaystyle n$ is odd then $\displaystyle n^2$ is odd
•²	write down appropriate form for $\displaystyle n$	$\displaystyle n = 2k + 1$ , $\displaystyle k \in \mathbb{Z}$
•³	show $\displaystyle n^2$ is odd	$\displaystyle n^2 = 2(2k^2 + 2k) + 1$ which is odd
•⁴	communicate	contrapositive statement is true therefore original statement is true

Notes

1. A candidate who incorrectly states the contrapositive as $\displaystyle n^2$ is odd $\displaystyle \Rightarrow n$ is odd (or any other statement masquerading as the contrapositive) and subsequently demonstrates that when $\displaystyle n$ is odd then $\displaystyle n^2$ is odd may be awarded •³ only.
2. The minimum requirement for •¹ is a statement such as: $\displaystyle n$ is odd $\displaystyle \Rightarrow n^2$ is odd; $\displaystyle n$ is odd then $\displaystyle n^2$ is odd; $\displaystyle n$ is odd is a sufficient condition for $\displaystyle n^2$ is odd; $\displaystyle n$ is odd only if $\displaystyle n^2$ is odd; $\displaystyle n^2$ is odd when $\displaystyle n$ is odd. Do not accept " $\displaystyle n$ is odd, $\displaystyle n^2$ is odd" or " $\displaystyle n$ is odd when $\displaystyle n^2$ is odd".
3. At •² $\displaystyle k \in \mathbb{Z}$ is not required. Accept the form $\displaystyle n = 2k \pm a$ , where $\displaystyle a$ is a specified odd number.
4. For candidates who proceed from: eg $\displaystyle n = 2n + 1$ , •² and •⁴ are not available; eg $\displaystyle n = 2k$ , •², •³ and •⁴ are not available; eg $\displaystyle n = k + 1$ , •², •³ and •⁴ are not available ( $\displaystyle n$ is not always odd); eg $\displaystyle n = 4k + 1$ , •² is not available (not all odd numbers covered by this form).
5. At •³ accept $\displaystyle n^2 = 4(\ldots) + 1$ , $\displaystyle n^2 = 2k(\ldots) + 1$ or $\displaystyle n^2 = 4k(\ldots) + 1.$
6. At •³ candidates must state a conclusion eg "which is odd".
7. Candidates who carry out a proof by contradiction may be awarded •² and •³ only.
8. Candidates who write $\displaystyle \neg Q \Rightarrow \neg P$ may be awarded •¹ where they either identify $\displaystyle P$ and $\displaystyle Q$ or have written $\displaystyle P \Rightarrow Q.$

Second order non-homogeneous DE

10 Marks

2017 Q14

Find the particular solution of the differential equation

$\displaystyle \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 8\sin x + 19\cos x$

given that $\displaystyle y = 7$ and $\displaystyle \frac{dy}{dx} = \frac{1}{2}$ when $\displaystyle x = 0.$

Show answer

\displaystyle y = 5e^{3x} - 14xe^{3x} - \frac{1}{2}\sin x + 2\cos x

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	construct auxiliary equation	$\displaystyle m^2 - 6m + 9 = 0$
•²	solve auxiliary equation and state CF	$\displaystyle y = Ae^{3x} + Bxe^{3x}$
•³	state PI	$\displaystyle y = C\sin x + D\cos x$
•⁴	obtain first and second derivatives of PI	$\displaystyle \dfrac{dy}{dx} = C\cos x - D\sin x$ , $\displaystyle \dfrac{d^2y}{dx^2} = -C\sin x - D\cos x$
•⁵	substitute	$\displaystyle -C\sin x - D\cos x - 6(C\cos x - D\sin x) + 9(C\sin x + D\cos x) = 8\sin x + 19\cos x$
•⁶	derive equations	$\displaystyle 8C + 6D = 8$ , $\displaystyle -6C + 8D = 19$
•⁷	obtain both constants of PI	$\displaystyle C = -\dfrac{1}{2},\ D = 2$
•⁸	differentiate general solution	$\displaystyle \dfrac{dy}{dx} = 3Ae^{3x} + Be^{3x} + 3Bxe^{3x} - \dfrac{1}{2}\cos x - 2\sin x$
•⁹	determine first constant of general solution	$\displaystyle A = 5$ or $\displaystyle B = -14$
•¹⁰	determine second constant and state particular solution	$\displaystyle y = 5e^{3x} - 14xe^{3x} - \dfrac{1}{2}\sin x + 2\cos x$

Notes

1. •¹ is **not** available where ' $\displaystyle = 0$ ' has been omitted.
2. •² can be awarded if the Complementary Function appears later as part of the general solution, as opposed to being explicitly stated immediately after solving the Auxiliary Equation.
3. Do not penalise the omission of ' $\displaystyle y = \ldots$ ' provided it appears at •¹⁰.
4. For candidates who obtain a CF of $\displaystyle y = Ae^{-3x} + Bxe^{-3x}$ only •² is not available. In this case the particular solution is $\displaystyle y = 5e^{-3x} + 16xe^{-3x} - \dfrac{1}{2}\sin x + 2\cos x.$
5. For candidates who obtain two real and distinct roots •² and •⁸ are not available.
6. For candidates who obtain roots of the form $\displaystyle p \pm qi$ : if $\displaystyle p = 0$ and $\displaystyle q \neq 1$ •² and •⁸ are not available, otherwise only •² is not available.
7. For candidates who obtain a CF of $\displaystyle y = Ae^{3x} + Be^{3x}$ , •², •⁸, •⁹ and •¹⁰ are not available.
8. Where a candidate substitutes the given conditions into the CF to obtain values of $\displaystyle A$ and $\displaystyle B$ and then finds the PI correctly, •⁹ is not available.
9. Where a candidate does not find a PI only •¹, •², •⁸, •⁹ and •¹⁰ are available.
10. Where an error in the differentiation of the general solution results in the value of $\displaystyle B$ being unobtainable then •¹⁰ is not available.

Equations of lines or planesIntersection of line and plane

(2, 4, 3)9 Marks

2017 Q15

(a)A beam of light passes through the points $\displaystyle B(7,8,1)$ and $\displaystyle T(-3,-22,6).$
Obtain parametric equations of the line representing the beam of light.

(b)A sheet of metal is represented by a plane containing the points $\displaystyle P(2,1,9)$ , $\displaystyle Q(1,2,7)$ and $\displaystyle R(-3,7,1).$
Find the Cartesian equation of the plane.

(c)The beam of light passes through a hole in the metal at point $\displaystyle H.$
Find the coordinates of $\displaystyle H.$

Show answer

(a)

\displaystyle x = 2\lambda + 7

,

\displaystyle y = 6\lambda + 8

,

\displaystyle z = -\lambda + 1

(or equivalent)
(b)

\displaystyle 4x + 2y - z = 1

(c)

\displaystyle H(3, -4, 3)

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	obtain direction vector	$\displaystyle \mathbf{d} = \begin{pmatrix}2\\6\\-1\end{pmatrix}$ or multiple thereof
(a) •²	state parametric equations	$\displaystyle x = 2\lambda + 7$ , $\displaystyle y = 6\lambda + 8$ , $\displaystyle z = -\lambda + 1$ or $\displaystyle x = 2\lambda - 3$ , $\displaystyle y = 6\lambda - 22$ , $\displaystyle z = -\lambda + 6$ , or equivalent
(b) •³	identify vectors	any two from $\displaystyle \overrightarrow{PQ} = \begin{pmatrix}-1\\1\\-2\end{pmatrix}$ , $\displaystyle \overrightarrow{PR} = \begin{pmatrix}-5\\6\\-8\end{pmatrix}$ , $\displaystyle \overrightarrow{QR} = \begin{pmatrix}-4\\5\\-6\end{pmatrix}$ or equivalent
(b) •⁴	evidence of strategy for finding normal	$\displaystyle \overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\-1 & 1 & -2\\-5 & 6 & -8\end{vmatrix}$ or equivalent
(b) •⁵	calculate normal	$\displaystyle \mathbf{n} = \begin{pmatrix}4\\2\\-1\end{pmatrix}$
(b) •⁶	obtain equation	$\displaystyle 4x + 2y - z = 1$
(c) •⁷	substitute into equation of plane	$\displaystyle 4(2\lambda + 7) + 2(6\lambda + 8) - (-\lambda + 1) = 1$
(c) •⁸	find $\displaystyle \lambda$	$\displaystyle \lambda = -2$
(c) •⁹	determine coordinates of H	$\displaystyle H(3, -4, 3)$

Notes

(a) 1. For candidates who express the equation in either symmetric or vector form •¹ is available for evidence of a correct direction vector; •² is unavailable unless parametric equations appear at (c).
(a) 2. Throughout the question accept horizontal vector notation eg $\displaystyle (2, 6, -1).$
(a) 3. A correct answer with no working receives full marks.
(a) 4. For an incorrect answer containing the correct direction vector but with no working, •¹ is available.
(a) 5. For an answer with an incorrect direction vector and no working neither •¹ nor •² are available.
Commonly observed: 1. (a) Unsimplified direction vector: $\displaystyle \mathbf{d} = \begin{pmatrix}-10\\-30\\5\end{pmatrix}$ ; parametric equations $\displaystyle x = -10\lambda + 7$ , $\displaystyle y = -30\lambda + 8$ , $\displaystyle z = 5\lambda + 1.$
(b) 1. Do not award •⁴ where the position vectors of $\displaystyle P$ , $\displaystyle Q$ or $\displaystyle R$ are used.
(c) 1. Do not accept a position vector at •⁹.
Commonly observed: 1. (c) For candidates who use the unsimplified direction vector from Commonly Observed Responses in (a), $\displaystyle \lambda = \dfrac{2}{5}.$

Volume of revolution

5 Marks

2017 Q16

On a suitable domain, a curve is defined by the equation $\displaystyle 4x^2 + 9y^2 = 36.$ A section of the curve in the first quadrant, illustrated in the diagram below, is rotated $\displaystyle 360^\circ$ about the $\displaystyle y$ -axis.

Curve of 4x^2 + 9y^2 = 36 rotated about y-axis

Calculate the exact value of the volume generated.

Show answer

\displaystyle 12\pi

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	state form of integral	$\displaystyle V = \pi\displaystyle\int x^2\,dy$ or $\displaystyle V = \pi\displaystyle\int (f(y))^2\,dy$
•²	rearrange and substitute for $\displaystyle x^2$	$\displaystyle V = \pi\displaystyle\int\left(9 - \dfrac{9}{4}y^2\right)dy$
•³	calculate limits to match variable	$\displaystyle \displaystyle\int_0^2\ldots dy$ or $\displaystyle y = 0,\ y = 2$
•⁴	integrate	$\displaystyle V = \pi\left[9y - \dfrac{3y^3}{4}\right]_0^2$
•⁵	evaluate	$\displaystyle V = 12\pi$ (cubic units)

Notes

1. $\displaystyle dy$ must appear for •¹ to be awarded.
2. •¹ may be awarded at •².
3. For candidates who write $\displaystyle V = \pi\int x^2\,dx$ , $\displaystyle V = \pi\int y^2\,dy$ or $\displaystyle V = \pi\int y^2\,dx$ and proceed to: (a) $\displaystyle V = \pi\displaystyle\int\left(9 - \dfrac{9}{4}y^2\right)dy$ , full credit may still be available; (b) $\displaystyle V = \pi\displaystyle\int\left(4 - \dfrac{4}{9}x^2\right)dx$ , •², •³, •⁴ and •⁵ may still be available; (c) $\displaystyle \pi\left[\dfrac{x^3}{3}\right]$ or $\displaystyle \pi\left[\dfrac{y^3}{3}\right]$ , only •³ is available.
4. •³ may be awarded at •⁴.
5. •⁵ is not available where a candidate's evaluation necessarily leads to a negative answer.
6. At •⁵ units are not required.

Equations with complex rootsArgand diagram

(1, 6, 1)8 Marks

2017 Q17

The complex number $\displaystyle z = 2+i$ is a root of the polynomial equation $\displaystyle z^4 - 6z^3 + 16z^2 - 22z + q = 0$ , where $\displaystyle q \in \mathbb{Z}.$

(a)State a second root of the equation.

(b)Find the value of $\displaystyle q$ and the remaining roots.

(c)Show the solutions to $\displaystyle z^4 - 6z^3 + 16z^2 - 22z + q = 0$ on an Argand diagram.

Show answer

(a)

\displaystyle 2 - i

(b)

\displaystyle q = 15

Remaining roots are

\displaystyle 1 + \sqrt{2}i

and

\displaystyle 1 - \sqrt{2}i.

(c) An Argand diagram showing the points

\displaystyle (2, 1)

,

\displaystyle (2, -1)

,

\displaystyle (1, \sqrt{2})

, and

\displaystyle (1, -\sqrt{2})

plotted correctly relative to each other.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	state second root	$\displaystyle 2 - i$
(b) •²	obtain two linear factors	$\displaystyle z - (2 + i),\ z - (2 - i)$
(b) •³	obtain quadratic factor	$\displaystyle z^2 - 4z + 5$
(b) •⁴	set up algebraic division or equivalent	$\displaystyle z^2 - 4z + 5 \,\big)\, z^4 - 6z^3 + 16z^2 - 22z + q$
(b) •⁵	complete algebraic division	quotient $\displaystyle z^2 - 2z + 3$ , remainder $\displaystyle q - 15$
(b) •⁶	state value of $\displaystyle q$	$\displaystyle q = 15$
(b) •⁷	obtain remaining two roots	$\displaystyle 1 \pm \sqrt{2}i$
(c) •⁸	show all four solutions on an Argand diagram	four points: $\displaystyle 2 + i$ , $\displaystyle 2 - i$ , $\displaystyle 1 + \sqrt{2}i$ , $\displaystyle 1 - \sqrt{2}i$ in correct relative positions

Notes

(b) 1. For candidates who substitute either $\displaystyle 2 + i$ or $\displaystyle 2 - i$ into the equation, obtain a correct value of $\displaystyle q$ but who do not exhibit any other working, only •⁶ may be awarded.
(b) 2. •⁶ not available for a non-integer value of $\displaystyle q.$
(c) 1. Do not penalise the omission of the labels on the axes.
(c) 2. •⁸ is available only where 4 roots are illustrated.
(c) 3. Positional information is required for •⁸. In the illustrative scheme this is provided by the relative positions of the points. Where points are plotted inaccurately, positional information may be provided by coordinates eg $\displaystyle (2, 1)$ or the numbers 2 and 1 indicated on the appropriate axes. Accept $\displaystyle (2, i).$ The label $\displaystyle 2 + i$ is not of itself sufficient to award •⁸.
(c) 4. Accept separate labelled Argand diagrams.

Parametric differentiationRectilinear motion

(5, 2)7 Marks

2017 Q18

The position of a particle at time $\displaystyle t$ is given by the parametric equations $\displaystyle x=t\cos t$ , $\displaystyle y=t\sin t$ ( $\displaystyle t \ge 0$ ).

(a)Find an expression for the instantaneous speed of the particle.

The diagram below shows the path that the particle takes.

Spiral path of a particle with point A marked

(b)Calculate the instantaneous speed of the particle at point A.

Show answer

(a)

\displaystyle \sqrt{1 + t^2}

(b)

\displaystyle \sqrt{1 + 9\pi^2}

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	evidence of use of product rule to find either $\displaystyle \dfrac{dx}{dt}$ or $\displaystyle \dfrac{dy}{dt}$ with one term correct	eg $\displaystyle \dfrac{dx}{dt} = \cos t + \ldots$
(a) •²	obtain $\displaystyle \dfrac{dx}{dt}$ or $\displaystyle \dfrac{dy}{dt}$	$\displaystyle \dfrac{dx}{dt} = \cos t - t\sin t$
(a) •³	obtain remaining derivative	$\displaystyle \dfrac{dy}{dt} = \sin t + t\cos t$
(a) •⁴	state formula for instantaneous speed	speed $\displaystyle = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}$ stated or implied at •⁵
(a) •⁵	obtain expression	$\displaystyle \sqrt{(\cos t - t\sin t)^2 + (\sin t + t\cos t)^2} = \sqrt{1 + t^2}$
(b) •⁶	evidence of valid strategy to find value of $\displaystyle t$ and obtain at least one non-zero solution	$\displaystyle 0 = t\sin t$ and eg $\displaystyle t = \pi$
(b) •⁷	choose correct value for $\displaystyle t$ and calculate speed	$\displaystyle t = 3\pi$ , speed $\displaystyle = \sqrt{1 + 9\pi^2}$

Notes

(a) 1. At •⁵ the simplification to $\displaystyle \sqrt{1 + t^2}$ is not required.
(a) 2. •⁵ may only be awarded for substitution into an expression of the form $\displaystyle \sqrt{(\ldots)^2 + (\ldots)^2}.$
(b) 1. For candidates who obtain an expression for $\displaystyle \dfrac{dy}{dx}$ rather than instantaneous speed, •⁶ and •⁷ are still available.
(b) 2. At •⁷ accept a decimal answer provided it is accurate to at least 3sf ( $\displaystyle 9.48$ ).

2017 Paper

Question 1

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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

Notes