Advanced Higher Maths 2016 Paper — Questions & Marking Instructions

Product, quotient, chain rules

3 Marks

2016 Q1(a)

Differentiate $\displaystyle y = x \tan^{-1} 2x.$

Show answer

\displaystyle \tan^{-1} 2x + \frac{2x}{1 + 4x^2}

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	evidence of use of product rule	$\displaystyle (\ldots)\tan^{-1}2x + x(\ldots)$
(a) •²	one resultant term of the product correct	$\displaystyle 1 \cdot \tan^{-1}2x$ or $\displaystyle x \cdot \dfrac{1}{1 + (2x)^2} \cdot 2$
(a) •³	complete differentiation	$\displaystyle \tan^{-1}2x + \dfrac{2x}{1 + 4x^2}$

Notes

(a) 1. Evidence for the award of •¹ should take the form $\displaystyle f(x) \times (\ldots) + g(x) \times (\ldots).$
(a) 2. For a candidate who interprets $\displaystyle \tan^{-1}2x$ as $\displaystyle (\tan 2x)^{-1}$ •³ is not available.
(a) 3. Accept $\displaystyle (2x)^2$ when awarding •³.

Product, quotient, chain rules

3 Marks

2016 Q1(b)

Given $\displaystyle f(x) = \frac{1 - x^2}{1 + 4x^2}$ , find $\displaystyle f'(x)$ , simplifying your answer.

Show answer

\displaystyle -\frac{10x}{(1 + 4x^2)^2}

Show marking instructions

Mark	Awarded for	Illustrative answer
(b) •⁴	evidence of use of quotient or product rule and one term of numerator correct	$\displaystyle (-2x)(1 + 4x^2) - \ldots$
(b) •⁵	complete differentiation correctly	$\displaystyle \dfrac{\ldots(1 - x^2) \cdot 8x}{(1 + 4x^2)^2}$
(b) •⁶	simplify answer	$\displaystyle -\dfrac{10x}{(1 + 4x^2)^2}$ or $\displaystyle \dfrac{-10x}{(1 + 4x^2)^2}$

Notes

(b) 1. Where a candidate uses the product rule, simplification to $\displaystyle -\dfrac{10x}{(1 + 4x^2)^2}$ or $\displaystyle -10x(1 + 4x^2)^{-2}$ will be required in order to obtain •⁶.
(b) 2. Incorrect working subsequent to a correct answer should be penalised in this instance eg an incorrect expansion of the denominator.

Parametric differentiation

2 Marks

2016 Q1(c)

A curve is given by the parametric equations $\displaystyle x = 6t$ and $\displaystyle y = 1 - \cos t.$
Find $\displaystyle \frac{dy}{dx}$ in terms of $\displaystyle t.$

Show answer

\displaystyle \frac{1}{6}\sin t

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Mark	Awarded for	Illustrative answer
(c) •⁷	correct derivatives	$\displaystyle 6$ and $\displaystyle \sin t$
(c) •⁸	find $\displaystyle \dfrac{dy}{dx}$	$\displaystyle \dfrac{1}{6}\sin t$

Geometric sequences and series

(3, 1, 2)6 Marks

2016 Q2

A geometric sequence has second and fifth terms 108 and 4 respectively.

(a)Calculate the value of the common ratio.

(b)State why the associated geometric series has a sum to infinity.

(c)Find the value of this sum to infinity.

Show answer

(a)

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

(b) A sum to infinity exists because

\displaystyle -1 < \frac{1}{3} < 1

(c) 486

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	interpret geometric series	$\displaystyle ar = 108$ and $\displaystyle ar^4 = 4$
(a) •²	evidence of strategy	$\displaystyle \dfrac{ar^4}{ar}$ , $\displaystyle r^3 = \dfrac{1}{27}$
(a) •³	value	$\displaystyle r = \dfrac{1}{3}$
(b) •⁴	know condition	$\displaystyle -1 < \dfrac{1}{3} < 1$
(c) •⁵	calculate the first term	$\displaystyle a = 324$
(c) •⁶	value	$\displaystyle \dfrac{324}{1 - \frac{1}{3}}$ or equivalent leading to $\displaystyle 486$

Notes

(a) 1. For •² accept $\displaystyle r^3 = \dfrac{1}{27}.$
(a) 2. For a statement of the answer only, award •¹ and •³. To earn •² there must be evidence of a strategy eg $\displaystyle 108 \to 36 \to 12 \to 4$ gives $\displaystyle r = \dfrac{1}{3}.$
(b) 1. For •⁴ $\displaystyle \dfrac{1}{3}$ may be replaced with a letter consistent with their answer to (a). However, in the case where a candidate obtains a value in (a) outside the open interval $\displaystyle (-1, 1)$ •⁴ will only be available where they also acknowledge that there is no sum to infinity.
(b) 2. Only award •⁴ for a strict inequality, whether it is expressed algebraically or in words.
(c) 1. For an incorrect value in (a) •⁶ will only be available provided the value satisfies the condition for convergence.
(c) 2. Where a candidate has used $\displaystyle S_\infty = \dfrac{a(1 - r^\infty)}{1 - r}$ full credit is available.

Binomial terms

5 Marks

2016 Q3

Write down and simplify the general term in the binomial expansion of $\displaystyle \left(\frac{3}{x} - 2x\right)^{13}.$
Hence, or otherwise, find the term in $\displaystyle x^9.$

Show answer

General term:

\displaystyle \binom{13}{r}(3)^{13-r}(-2)^rx^{2r-13}

Term in

\displaystyle x^9

:

\displaystyle -1437696x^9

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	state general term	$\displaystyle ^{13}C_r\left(\dfrac{3}{x}\right)^{13-r}(-2x)^r$
•²	simplify powers of $\displaystyle x$ OR coefficients and signs	$\displaystyle (3)^{13-r}(-2)^r$ or $\displaystyle x^{2r-13}$
•³	state simplified general term (completes simplification)	$\displaystyle ^{13}C_r(3)^{13-r}(-2)^r x^{2r-13}$
•⁴	determine value of $\displaystyle r$	$\displaystyle 2r - 13 = 9 \Rightarrow r = 11$
•⁵	evaluate term	$\displaystyle -1437696x^9$

Notes

1. Accept $\displaystyle -1437696.$
2. For •¹ accept the initial appearance of $\displaystyle \displaystyle\sum_{r=0}^{13}{}^{13}C_r\left(\dfrac{3}{x}\right)^{13-r}(-2x)^r$ as bad form. •² and •³ are available only to candidates who simplify a general term correctly.
3. •⁴ and •⁵ are the only marks available to candidates who have not proceeded from a general term eg. an expansion using Pascal's Triangle. The required term must be explicitly identified in order for •⁵ to be awarded.
4. Starting with $\displaystyle ^{13}C_r\left(\dfrac{3}{x}\right)^r(-2x)^{13-r}$ leading to $\displaystyle r = 2$ can also gain full credit.
5. Accept $\displaystyle \dfrac{1}{x^{13-2r}}$ when awarding •² or •³.

Systems of Equations

4 Marks

2016 Q4

Below is a system of equations:

$\displaystyle x + 2y + 3z = 3$
$\displaystyle 2x - y + 4z = 5$
$\displaystyle x - 3y + 2\lambda z = 2$

Use Gaussian elimination to find the value of $\displaystyle \lambda$ which leads to redundancy.

Show answer

\displaystyle \lambda = \frac{1}{2}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	construct augmented matrix	$\displaystyle \left(\begin{array}{ccc\|c}1 & 2 & 3 & 3\\2 & -1 & 4 & 5\\1 & -3 & 2\lambda & 2\end{array}\right)$
•²	use row operations to establish first two zero elements	$\displaystyle \left(\begin{array}{ccc\|c}1 & 2 & 3 & 3\\0 & 5 & 2 & 1\\0 & -5 & 2\lambda - 3 & -1\end{array}\right)$
•³	establish third zero element OR recognise linear relationship between two rows	$\displaystyle \left(\begin{array}{ccc\|c}1 & 2 & 3 & 3\\0 & 5 & 2 & 1\\0 & 0 & 2\lambda - 1 & 0\end{array}\right)$ OR $\displaystyle 2\lambda - 3 = -2$
•⁴	state value of $\displaystyle \lambda$	$\displaystyle \lambda = \dfrac{1}{2}$

Notes

1. Elementary row operations must be carried out correctly for •² and •³ to be awarded.
2. •⁴ is only available where a candidate's final matrix exhibits redundancy.
3. Disregard any working/statement subsequent to $\displaystyle \lambda = \dfrac{1}{2}.$

Proof by induction

4 Marks

2016 Q5

Prove by induction that $\displaystyle \sum_{r=1}^{n}r(3r - 1) = n^2(n + 1)$ , $\displaystyle \forall n \in \mathbb{N}.$

Show answer

When

\displaystyle n=1

, LHS

\displaystyle =1(3-1)=2

and RHS

\displaystyle =1^{2}(1+1)=2

, so the statement is true.
Assume true for

\displaystyle n=k

:

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

Then

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

True for

\displaystyle n=1

, and true for

\displaystyle n=k

implies true for

\displaystyle n=k+1

, so by induction the statement holds

\displaystyle \forall n\in\mathbb{N}.

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	show true for $\displaystyle n = 1$	LHS: $\displaystyle 1(3 - 1) = 2$ , RHS: $\displaystyle 1^2(1 + 1) = 2$ , so true for $\displaystyle n = 1$
•²	assume true for $\displaystyle n = k$ AND consider $\displaystyle n = k + 1$	$\displaystyle \displaystyle\sum_{r=1}^{k}r(3r - 1) = k^2(k + 1)$ and $\displaystyle \displaystyle\sum_{r=1}^{k+1}r(3r - 1) = \ldots = \sum_{r=1}^{k}r(3r - 1) + (k + 1)(3(k + 1) - 1)$
•³	correct statement of sum to $\displaystyle (k + 1)$ terms using inductive hypothesis	$\displaystyle = k^2(k + 1) + (k + 1)(3k + 2) = (k + 1)[k^2 + 3k + 2] = (k + 1)(k + 1)(k + 2)$
•⁴	express explicitly in terms of $\displaystyle (k + 1)$ or achieve stated aim/goal AND communicate	$\displaystyle = (k + 1)^2((k + 1) + 1)$ , thus if true for $\displaystyle n = k$ then true for $\displaystyle n = k + 1$ but since true for $\displaystyle n = 1$ , then by induction true for all $\displaystyle n \in \mathbb{N}$

Notes

1. "RHS = 2, LHS = 2" 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 include: "If true for..."; "Suppose true for…"; "Assume true for…". However, *not* acceptable: "Consider $\displaystyle n = k$ ", "assume $\displaystyle n = k$ " and "True for $\displaystyle n = k$ ". Allow if appears at conclusion.
3. Full marks are available to candidates who state an aim/goal earlier in the proof and who subsequently achieve the stated aim/goal.
4. Minimum acceptable form for •⁴: "Then true for $\displaystyle n = k + 1$ , but since true for $\displaystyle n = 1$ , then true for all $\displaystyle n$ " or equivalent.

Maclaurin Series

6 Marks

2016 Q6

Find Maclaurin expansions for $\displaystyle \sin 3x$ and $\displaystyle e^{4x}$ up to and including the term in $\displaystyle x^3.$
Hence obtain an expansion for $\displaystyle e^{4x}\sin 3x$ up to and including the term in $\displaystyle x^3.$

Show answer

\displaystyle \sin 3x = 3x - \frac{9}{2}x^3 \dots

\displaystyle e^{4x} = 1 + 4x + 8x^2 + \frac{32}{3}x^3 \dots

\displaystyle e^{4x}\sin 3x = 3x + 12x^2 + \frac{39}{2}x^3 \dots

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	for either function: first derivative and two evaluations OR all three derivatives OR all four evaluations	$\displaystyle f(x) = \sin 3x$ , $\displaystyle f(0) = 0$ ; $\displaystyle f'(x) = 3\cos 3x$ , $\displaystyle f'(0) = 3$ ; $\displaystyle f''(x) = -9\sin 3x$ , $\displaystyle f''(0) = 0$ ; $\displaystyle f'''(x) = -27\cos 3x$ , $\displaystyle f'''(0) = -27$ ; $\displaystyle f(x) = f(0) + f'(0)x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3\ldots$
•²	complete derivatives and evaluations AND substitute	$\displaystyle f(x) = 3x - \dfrac{27}{3!}x^3 = 3x - \dfrac{9}{2}x^3$
•³	for second function: first derivative and two evaluations OR all three derivatives OR all four evaluations	$\displaystyle f(x) = e^{4x}$ , $\displaystyle f(0) = 1$ ; $\displaystyle f'(x) = 4e^{4x}$ , $\displaystyle f'(0) = 4$ ; $\displaystyle f''(x) = 16e^{4x}$ , $\displaystyle f''(0) = 16$ ; $\displaystyle f'''(x) = 64e^{4x}$ , $\displaystyle f'''(0) = 64$
•⁴	complete derivatives and evaluations AND substitute	$\displaystyle f(x) = 1 + 4x + \dfrac{16x^2}{2} + \dfrac{64x^3}{6} = 1 + 4x + 8x^2 + \dfrac{32}{3}x^3$
•⁵	multiply expressions	$\displaystyle e^{4x}\sin 3x = \left(3x - \dfrac{9}{2}x^3\ldots\right)\left(1 + 4x + 8x^2 + \dfrac{32}{3}x^3\ldots\right) = 24x^3 - \dfrac{9}{2}x^3 + 12x^2 + 3x\ldots$
•⁶	multiply out and simplify	$\displaystyle = 3x + 12x^2 + \dfrac{39}{2}x^3\ldots$

Notes

1. Method 1: if a candidate chooses to use the product rule to obtain the Maclaurin series for $\displaystyle e^{4x}\sin 3x$ without first obtaining series for $\displaystyle e^{4x}$ and $\displaystyle \sin 3x$ separately then only •⁵ and •⁶ are potentially available. In this instance for the award of •⁵ apply the same principle as that used to award •¹ and •³: $\displaystyle f(x) = e^{4x}\sin 3x$ , $\displaystyle f(0) = 0$ ; $\displaystyle f'(x) = 4e^{4x}\sin 3x + 3e^{4x}\cos 3x$ , $\displaystyle f'(0) = 3$ ; $\displaystyle f''(x) = 7e^{4x}\sin 3x + 24e^{4x}\cos 3x$ , $\displaystyle f''(0) = 24$ ; $\displaystyle f'''(x) = -44e^{4x}\sin 3x + 117e^{4x}\cos 3x$ , $\displaystyle f'''(0) = 117.$
2. At •⁶ the appearance of terms in $\displaystyle x^4$ or above should be disregarded.
3. Method 2: for a candidate who writes down $\displaystyle \sin 3x = 3x - \dfrac{(3x)^3}{3!}\ldots$ without first writing down the series for $\displaystyle \sin x$ then •¹ may be awarded. A similar principle may be applied to the awarding of •³ if required.

Matrix operationsDeterminant and inverse

(1, 3, 2)6 Marks

2016 Q7

A is the matrix $\displaystyle \begin{pmatrix} 2 & 0 \\ \lambda & -1 \end{pmatrix}.$

(a)Find the determinant of matrix A.

(b)Show that $\displaystyle A^2$ can be expressed in the form $\displaystyle pA + qI$ , stating the values of $\displaystyle p$ and $\displaystyle q.$

(c)Obtain a similar expression for $\displaystyle A^4.$

Show answer

(a)

\displaystyle -2

(b)

\displaystyle A^2 = A + 2I

, so

\displaystyle p = 1

and

\displaystyle q = 2

(c)

\displaystyle 5A + 6I

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	calculate determinant	$\displaystyle -2$
(b) •²	find $\displaystyle A^2$	$\displaystyle A^2 = \begin{pmatrix}4 & 0\\\lambda & 1\end{pmatrix}$
(b) •³	use an appropriate method	$\displaystyle A^2 = \begin{pmatrix}2 & 0\\\lambda & -1\end{pmatrix} + \begin{pmatrix}2 & 0\\0 & 2\end{pmatrix}$ , $\displaystyle A^2 = A + 2I$
(b) •⁴	write in required form and explicitly state values of $\displaystyle p$ and $\displaystyle q$	$\displaystyle p = 1$ and $\displaystyle q = 2$
(c) •⁵	square expression found in (b)	$\displaystyle A^4 = (A + 2I)^2 = A^2 + 4AI + 4I^2 = A + 2I + 4A + 4I$
(c) •⁶	substitute for $\displaystyle A^2$ and complete process	$\displaystyle = 5A + 6I$

Notes

(a) 1. If a candidate chooses to find $\displaystyle A^{-1}$ then •¹ is only available where ' $\displaystyle \det A$ ' is clearly identified.
Commonly observed: 1. (a) $\displaystyle A^{-1} = \dfrac{1}{\det A}(\ldots)$ , $\displaystyle A^{-1} = \dfrac{1}{-2}(\ldots)$ : award •¹. $\displaystyle A^{-1} = \dfrac{1}{-2}(\ldots)$ alone: do not award •¹.
(b) 1. $\displaystyle \begin{pmatrix}4 & 0\\\lambda & 1\end{pmatrix} = \begin{pmatrix}2 & 0\\\lambda & -1\end{pmatrix} + 2\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}$ is acceptable for •⁴ provided the values of $\displaystyle p$ and $\displaystyle q$ are explicitly stated.
(c) 1. •⁵ may be obtained by squaring $\displaystyle \begin{pmatrix}4 & 0\\\lambda & 1\end{pmatrix}$ to give $\displaystyle \begin{pmatrix}16 & 0\\5\lambda & 1\end{pmatrix}$ and identifying the coefficient of $\displaystyle A$ as 5. This leads to •⁶ using the same method as in (b).
(c) 2. Accept equivalent expressions eg $\displaystyle = A^2 + 4A + 4I.$
(c) 3. Candidates may calculate $\displaystyle A^3$ first so •⁵ can be awarded for $\displaystyle A^3 = 3A + 2I.$

Argand diagramde Moivre's theorem

(1, 2, 3)6 Marks

2016 Q8

Let $\displaystyle z = \sqrt{3} - i.$

(a)Plot $\displaystyle z$ on an Argand diagram.

(b)Let $\displaystyle w = az$ where $\displaystyle a > 0$ , $\displaystyle a \in \mathbb{R}.$
Express $\displaystyle w$ in polar form.

(c)Express $\displaystyle w^8$ in the form $\displaystyle ka^n(x + i\sqrt{y})$ where $\displaystyle k, x, y \in \mathbb{Z}.$

Show answer

(a) Point plotted at

\displaystyle (\sqrt{3}, -1)

in the 4th quadrant of the Argand diagram.
(b)

\displaystyle 2a\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)

(c)

\displaystyle 128a^8(-1 + i\sqrt{3})

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	correctly plot $\displaystyle z$ on Argand diagram	a point in quadrant 4 at $\displaystyle (\sqrt{3}, -1)$ , with $\displaystyle \sqrt{3}$ and $\displaystyle -1$ marked on the axes
(b) •²	find modulus or argument	$\displaystyle \|w\| = 2a$ or $\displaystyle \arg(w) = -\dfrac{\pi}{6}$
(b) •³	complete and express in polar form	$\displaystyle w = 2a\left(\cos\left(-\dfrac{\pi}{6}\right) + i\sin\left(-\dfrac{\pi}{6}\right)\right)$
(c) •⁴	process modulus	$\displaystyle 256a^8$
(c) •⁵	process argument	$\displaystyle \ldots\left(\cos\left(-\dfrac{8\pi}{6}\right) + i\sin\left(-\dfrac{8\pi}{6}\right)\right)$
(c) •⁶	evaluate and express in form $\displaystyle ka^n(x + i\sqrt{y})$	$\displaystyle w^8 = 128a^8(-1 + i\sqrt{3})$

Notes

(a) 1. Do not penalise the omission of the diagonal line.
(a) 2. Treat alternative axis labels as bad form (to include the case where there are no labels).
(a) 3. Accept a point labelled using coordinates: $\displaystyle (\sqrt{3}, -1)$ and, in this instance, $\displaystyle (\sqrt{3}, -i).$
(a) 4. The minimum acceptable response for the award of •¹ is a point in quadrant 4 together with $\displaystyle \sqrt{3}$ and $\displaystyle -1$ (or $\displaystyle -i$ ).
(b) 1. For the award of •² and •³ accept any answer of the form $\displaystyle -\dfrac{\pi}{6} + 2k\pi$ , $\displaystyle k \in \mathbb{Z}.$
(b) 2. For the award of •² and •³ accept any answer of the form $\displaystyle (-30 + 360k)^\circ$ , $\displaystyle k \in \mathbb{Z}.$
(b) 3. A candidate who chooses to work in degrees can only be awarded •³ provided the degree symbol appears at some point within question 8.
(b) 4. Award •³ for $\displaystyle w = 2a\left(\cos\left(\dfrac{\pi}{6}\right) - i\sin\left(\dfrac{\pi}{6}\right)\right).$
(b) 5. At •³ do not accept $\displaystyle w = a\left[2\left(\cos\left(-\dfrac{\pi}{6}\right) + i\sin\left(-\dfrac{\pi}{6}\right)\right)\right].$
(b) 6. Working subsequent to the appearance of $\displaystyle -\dfrac{\pi}{6}$ should be penalised where it leads to the use of an incorrect argument.
(c) 1. Method 1: for the award of •⁵ accept any answer of the form $\displaystyle -\dfrac{4\pi}{3} + 2k\pi$ , $\displaystyle k \in \mathbb{Z}.$
(c) 2. Method 1: for the award of •⁵ accept any answer of the form $\displaystyle (-240 + 360k)^\circ$ , $\displaystyle k \in \mathbb{Z}.$
(c) 3. Method 1: a candidate who chooses to work in degrees can only be awarded •⁵ provided the degree symbol appears at some point within question 8.
(c) 4. Method 1: do not penalise unsimplified fractions.
(c) 5. Method 1: award •⁵ for $\displaystyle \ldots\left(\cos\dfrac{8\pi}{6} - i\sin\dfrac{8\pi}{6}\right).$
(c) 6. Methods 2 and 3: accept the omission of ' $\displaystyle a$ ' at •⁴ and •⁵ provided $\displaystyle a^8$ appears in the final answer.
(c) 7. Method 3: for the award of •⁴ a **full** expansion must be written out.

Integration by parts

6 Marks

2016 Q9

Obtain $\displaystyle \int x^7(\ln x)^2 \,dx.$

Show answer

\displaystyle \frac{1}{8}x^8(\ln x)^2 - \frac{1}{32}x^8(\ln x) + \frac{1}{256}x^8 + c

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	know to use integration by parts and start process	$\displaystyle \dfrac{1}{8}x^8(\ln x)^2 - \ldots$
•²	correct choice of functions to differentiate and integrate AND application thereof	$\displaystyle \ldots - \dfrac{1}{8}\displaystyle\int x^8 \times \dfrac{d}{dx}\left((\ln x)^2\right)dx$
•³	differentiate $\displaystyle (\ln x)^2$	$\displaystyle \dfrac{1}{8}x^8(\ln x)^2 - \dfrac{1}{4}\displaystyle\int x^7(\ln x)\,dx$
•⁴	know to use second application and begin process	$\displaystyle \ldots - \left[\dfrac{1}{32}x^8(\ln x) - \dfrac{1}{32}\displaystyle\int x^8\left(\dfrac{1}{x}\right)dx\right]$
•⁵	complete second application	$\displaystyle \ldots - \left[\dfrac{1}{32}x^8(\ln x) - \dfrac{1}{256}x^8\right]$
•⁶	simplify	$\displaystyle \dfrac{1}{8}x^8(\ln x)^2 - \dfrac{1}{32}x^8(\ln x) + \dfrac{1}{256}x^8 + c$

Notes

1. For candidates who attempt to integrate $\displaystyle (\ln x)^2$ and differentiate $\displaystyle x^7$ then •¹, •⁴ and •⁶ may be awarded but not •², •³ and •⁵.
2. Evidence of use of integration by parts would be the appearance of an attempt to integrate one term and differentiate the other.
3. For candidates who attempt to substitute for $\displaystyle \ln x$ eg $\displaystyle t = \ln x$ leading to $\displaystyle \int t^2e^{8t}\,dt$ then •¹ becomes available upon evidence of using integration by parts ie. $\displaystyle t^2 \cdot \dfrac{1}{8}e^{8t} - \ldots$ ; •⁶ is only available for a final answer expressed as a function of $\displaystyle x.$
4. For candidates who incorrectly differentiate $\displaystyle (\ln x)^2$ and do not require a second application of integration by parts, only •¹, •² and •⁶ are available.
5. Do not penalise the omission of " $\displaystyle +c$ ".

Disproof by counterexampleDirect proof

4 Marks

2016 Q10

For each of the following statements, decide whether it is true or false.
If true, give a proof; if false, give a counterexample.

A. If a positive integer $\displaystyle p$ is prime, then so is $\displaystyle 2p + 1.$

B. If a positive integer $\displaystyle n$ has remainder 1 when divided by 3, then $\displaystyle n^3$ also has remainder 1 when divided by 3.

Show answer

A: False. Counterexample: let

\displaystyle p=7

, then

\displaystyle 2p+1 = 15

, which is not prime.
B: True. Proof: let

\displaystyle n = 3a+1

, then

\displaystyle n^3 = (3a+1)^3 = 27a^3 + 27a^2 + 9a + 1 = 3(9a^3 + 9a^2 + 3a) + 1

, which leaves a remainder of 1 when divided by 3.

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	give counterexample	eg. choose $\displaystyle p = 7$ , $\displaystyle 2(7) + 1 = 15$ and since $\displaystyle 15 = 5 \times 3$ , hence not prime, statement is false
•²	set up $\displaystyle n$	$\displaystyle n = 3a + 1$ , $\displaystyle a \in \mathbb{N}_0$
•³	consider expansion of $\displaystyle n^3$	$\displaystyle n^3 = 27a^3 + 27a^2 + 9a + 1$
•⁴	complete proof with conclusion	$\displaystyle = 3(9a^3 + 9a^2 + 3a) + 1$ and statement such as "so $\displaystyle n^3$ has remainder 1 when divided by 3 $\displaystyle \therefore$ statement is true".

Notes

1. Do not penalise the omission of $\displaystyle a \in \mathbb{N}_0$ in •².
2. Treat a statement such as $\displaystyle n = 3n + 1$ as bad form.
3. •³ can only be awarded for the correct expansion of $\displaystyle (3a + 1)^3.$
4. Minimum statement of conclusion in •⁴ is "true".
5. Where a candidate invokes an incorrect use of proof by contradiction full credit may still be available provided all relevant steps are included.

Related rates of change

4 Marks

2016 Q11

The height of a cube is increasing at the rate of 5 cm s $\displaystyle ^{-1}.$
Find the rate of increase of the volume when the height of the cube is 3 cm.

Show answer

\displaystyle 135 \text{ cm}^3\text{s}^{-1}

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	state differential equation	$\displaystyle \dfrac{dh}{dt} = 5$
•²	state relationship or apply chain rule	$\displaystyle \dfrac{dV}{dt} = \dfrac{dV}{dh} \cdot \dfrac{dh}{dt}$ , $\displaystyle V = h^3$
•³	find the rate of change of volume with respect to height	$\displaystyle \dfrac{dV}{dh} = 3h^2$
•⁴	evaluate	$\displaystyle \dfrac{dV}{dt} = 3h^2 \times 5 = 3(3)^2 \times 5 = 135\ \text{cm}^3\,\text{s}^{-1}$

Notes

1. A candidate who assumes that only the height changes – and that the length and breadth are constant – can be awarded •¹ and •² only.
2. Where a candidate uses the wrong formula for the volume of a cube only •¹ and •² are available.
3. A candidate using Method 1 who writes $\displaystyle \dfrac{dV}{dt} = 3h^2\dfrac{dh}{dt}$ can be awarded •² and •³.
4. To award •⁴ there must be evidence of substituting 3 and 5. Correct units must also be included.

Sketching related functions

(2, 2)4 Marks

2016 Q12

Below is a diagram showing the graph of a linear function, $\displaystyle y = f(x).$

On separate diagrams show:

(a) $\displaystyle y = |f(x) - c|$
(b) $\displaystyle y = |2f(x)|$

Show answer

(a) V-shaped graph with its vertex on the x-axis at

\displaystyle 2c

, passing through

\displaystyle 2c

on the positive y-axis.
(b) Symmetrical V-shaped graph with its vertex on the x-axis at

\displaystyle c

, passing through

\displaystyle 2c

on the positive y-axis.

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	correct shape	(•¹ and •² together) a V with its vertex on the $\displaystyle x$ -axis at $\displaystyle 2c$ , crossing the $\displaystyle y$ -axis at $\displaystyle 2c$
(a) •²	graph passes through $\displaystyle 2c$ on the positive $\displaystyle x$ - and $\displaystyle y$ -axes	as •¹
(b) •³	graph of $\displaystyle y = \|2f(x)\|$ passing through $\displaystyle 2c$ on the positive $\displaystyle y$ -axis	(•³ and •⁴ together) a symmetrical V with its vertex on the $\displaystyle x$ -axis at $\displaystyle c$ , crossing the $\displaystyle y$ -axis at $\displaystyle 2c$
(b) •⁴	correct shape (symmetrical V) meeting positive $\displaystyle x$ -axis at $\displaystyle c$	as •³

Notes

(a) 1. To award •², the second arm must be sketched to within 15° of the reflected angle.
(b) 1. For a candidate who sketches the graph of $\displaystyle y = 2f(x)$ award •³ for showing a straight line passing through $\displaystyle (0, -2c).$
(b) 2. To award •⁴, the second arm must be sketched to within 15° of the reflected angle.

Partial fractionsIntegrating rational functions

9 Marks

2016 Q13

Express $\displaystyle \frac{3x+32}{(x+4)(6-x)}$ in partial fractions and hence evaluate

$\displaystyle \int_{3}^{4}\frac{3x+32}{(x+4)(6-x)} \,dx.$

Give your answer in the form $\displaystyle \ln\left(\frac{p}{q}\right).$

Show answer

Partial fractions:

\displaystyle \frac{2}{x+4} + \frac{5}{6-x}

Integral evaluation:

\displaystyle \ln\left(\frac{486}{49}\right)

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	correct application of partial fractions	$\displaystyle \dfrac{3x + 32}{(x + 4)(6 - x)} = \dfrac{A}{x + 4} + \dfrac{B}{6 - x}$
•²	starts process	$\displaystyle 3x + 32 = A(6 - x) + B(x + 4)$
•³	calculate one value	$\displaystyle A = 2$
•⁴	calculate second value	$\displaystyle B = 5$
•⁵	re-state integral in partial fractions	$\displaystyle \displaystyle\int_3^4\left(\dfrac{2}{(x + 4)} + \dfrac{5}{(6 - x)}\right)dx$
•⁶	one term correctly integrated	$\displaystyle \left[2\ln\|x + 4\|\ldots\right.$
•⁷	integrate second term correctly	$\displaystyle \left.\ldots - 5\ln\|6 - x\|\right]_3^4$
•⁸	substitute limits	$\displaystyle (2\ln\|4 + 4\| - 5\ln\|6 - 4\|) - (2\ln\|3 + 4\| - 5\ln\|6 - 3\|)$
•⁹	evaluate to expected form	$\displaystyle = \ln\dfrac{486}{49}$

Notes

1. Do not penalise lack of modulus signs unless the candidate attempts to integrate $\displaystyle \dfrac{1}{x - 6}$ rather than $\displaystyle \dfrac{1}{6 - x}.$
2. Award maximum [8/9] for appropriate working leading to $\displaystyle \ln\dfrac{98}{243}$ (•⁹ lost) OR $\displaystyle \ln\dfrac{2048}{11907}$ (•⁷ lost).
3. Do not penalise unsimplified fractions in •⁹.

Intersection of lines

(5, 4)9 Marks

2016 Q14

Two lines $\displaystyle L_1$ and $\displaystyle L_2$ are given by the equations:

$\displaystyle L_1: x = 4 + 3\lambda, y = 2 + 4\lambda, z = -7\lambda$
$\displaystyle L_2: \frac{x - 3}{-2} = \frac{y - 8}{1} = \frac{z + 1}{3}$

(a)Show that the lines $\displaystyle L_1$ and $\displaystyle L_2$ intersect and find the point of intersection.

(b)Calculate the obtuse angle between the lines $\displaystyle L_1$ and $\displaystyle L_2.$

Show answer

(a) Point of intersection is

\displaystyle (7, 6, -7)

(b)

\displaystyle 135.6^\circ

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	convert any two components of $\displaystyle L_2$ to parametric form	two from $\displaystyle x = 3 - 2\mu$ , $\displaystyle y = 8 + \mu$ , $\displaystyle z = -1 + 3\mu$
(a) •²	two linear equations involving two distinct parameters	two from $\displaystyle 4 + 3\lambda = 3 - 2\mu$ , $\displaystyle 2 + 4\lambda = 8 + \mu$ , $\displaystyle -7\lambda = -1 + 3\mu$
(a) •³	find parameter values	$\displaystyle \lambda = 1,\ \mu = -2$
(a) •⁴	verify third component in both equations or equivalent	eg $\displaystyle z_1 = -7 \times 1$ and $\displaystyle z_2 = 3(-2) - 1$ therefore the lines intersect
(a) •⁵	find point of intersection	$\displaystyle (7, 6, -7)$
(b) •⁶	identify first direction vector	$\displaystyle \mathbf{d}_1 = 3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$
(b) •⁷	identify second direction vector	$\displaystyle \mathbf{d}_2 = -2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$
(b) •⁸	calculate magnitudes and scalar product	$\displaystyle \|\mathbf{d}_1\| = \sqrt{74}$ , $\displaystyle \|\mathbf{d}_2\| = \sqrt{14}$ and $\displaystyle \mathbf{d}_1 \cdot \mathbf{d}_2 = -6 + 4 - 21 = -23$
(b) •⁹	calculate obtuse angle	$\displaystyle \cos^{-1}\left(\dfrac{-23}{\sqrt{74}\sqrt{14}}\right) \approx 135.6^\circ$

Notes

(a) 1. A candidate who uses $\displaystyle \lambda$ as the second parameter can only be awarded •¹ unless this is rectified later in the question.
(a) 2. Do not penalise the omission of the statement 'therefore the lines intersect'.
(b) 1. For $\displaystyle L_1 = 3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$ and $\displaystyle L_2 = -2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ or equivalent, lose •⁶ but •⁷ is available (repeated error).
(b) 2. Do not penalise $\displaystyle \mathbf{L_1} = 3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$ and $\displaystyle \mathbf{L_2} = -2\mathbf{i} + \mathbf{j} + 3\mathbf{k}.$
(b) 3. For $\displaystyle L_1: 3\mathbf{i} + 4\mathbf{j} - 7\mathbf{k}$ and $\displaystyle L_2: -2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$ or equivalent, •⁶ and •⁷ are both available.
(b) 4. For the award of •⁹ accept $\displaystyle 136^\circ.$
(b) 5. •⁹ is not available to candidates who calculate an obtuse angle correctly but who subsequently calculate an acute angle.

Second order non-homogeneous DE

10 Marks

2016 Q15

Solve the differential equation

$\displaystyle \frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = 12x^2 + 2x - 5$

given $\displaystyle y = -6$ and $\displaystyle \frac{dy}{dx} = 3$ , when $\displaystyle x = 0.$

Show answer

\displaystyle y = 8e^{-3x} - 15e^{-2x} + 2x^2 - 3x + 1

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	state auxiliary equation	$\displaystyle m^2 + 5m + 6 = 0$ , $\displaystyle m = -3,\ m = -2$
•²	solve auxiliary equation and state complementary function	$\displaystyle y = Ae^{-3x} + Be^{-2x}$
•³	construct particular integral	$\displaystyle y = Cx^2 + Dx + E$
•⁴	differentiate particular integral	$\displaystyle \dfrac{dy}{dx} = 2Cx + D$ and $\displaystyle \dfrac{d^2y}{dx^2} = 2C$
•⁵	calculate one coefficient of the particular integral	$\displaystyle C = 2$
•⁶	calculate remaining coefficients	$\displaystyle D = -3,\ E = 1$ , $\displaystyle y = Ae^{-3x} + Be^{-2x} + 2x^2 - 3x + 1$
•⁷	differentiate general solution	$\displaystyle \dfrac{dy}{dx} = -3Ae^{-3x} - 2Be^{-2x} + 4x - 3$
•⁸	construct equations using given conditions	$\displaystyle A + B = -7$ and $\displaystyle 3A + 2B = -6$ or equivalent
•⁹	find one coefficient	$\displaystyle A = 8$ or $\displaystyle B = -15$
•¹⁰	find other coefficient and state particular solution	$\displaystyle y = 8e^{-3x} - 15e^{-2x} + 2x^2 - 3x + 1$

Notes

1. For •¹ do not penalise the omission of ' $\displaystyle = 0$ '.
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. A candidate who obtains $\displaystyle m = 2$ and $\displaystyle m = 3$ from a correct auxiliary equation, leading to $\displaystyle y = 20e^{3x} - 27e^{2x} + 2x^2 - 3x + 1$ cannot gain •² but all other marks are available.
4. Where a candidate substitutes the given conditions into the Complementary Function to obtain values of $\displaystyle A$ and $\displaystyle B$ and then finds the particular integral correctly •⁸ and •⁹ are unavailable.

Separable DE in context

9 Marks

2016 Q16

A beaker of liquid was placed in a fridge.

The rate of cooling is given by

$\displaystyle \frac{dT}{dt} = -k(T - T_F)$ , $\displaystyle k > 0$ ,

where $\displaystyle T_F$ is the constant temperature in the fridge and $\displaystyle T$ is the temperature of the liquid at time $\displaystyle t.$

The constant temperature in the fridge is $\displaystyle 4^\circ\text{C}.$
When first placed in the fridge, the temperature of the liquid was $\displaystyle 25^\circ\text{C}.$
At 12 noon, the temperature of the liquid was $\displaystyle 9.8^\circ\text{C}.$
At 12:15 pm, the temperature of the liquid had dropped to $\displaystyle 6.5^\circ\text{C}.$

At what time, to the nearest minute, was the liquid placed in the fridge?

Show answer

11:37 am

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	construct integral equation	$\displaystyle \displaystyle\int \dfrac{1}{(T - T_F)}\,dT = \int -k\,dt$
•²	integrate	$\displaystyle \ln(T - T_F) = -kt + c$
•³	find constant, $\displaystyle c$	$\displaystyle \ln(9.8 - 4) = -k(0) + c$ , $\displaystyle c = \ln 5.8$
•⁴	substitute using given information	$\displaystyle \ln(6.5 - 4) = -15k + \ln 5.8$
•⁵	find constant, $\displaystyle k$	$\displaystyle k = \dfrac{\ln 2.5 - \ln 5.8}{-15} = 0.05610\ldots$
•⁶	substitute given condition	$\displaystyle \ln(25 - 4) = -0.05610\ldots t + \ln 5.8$
•⁷	know how to find time	$\displaystyle t = \dfrac{\ln 21 - \ln 5.8}{-0.05610\ldots}$
•⁸	calculate time	$\displaystyle t = -22.93\ldots$
•⁹	state the time to the nearest minute	The liquid was placed in the fridge at 11:37 (am)

Notes

1. General note: many candidates may use a combination of the given methods. For all methods the evidence for •¹, •², •⁸ and •⁹ is the same. To award •³ up to •⁷ note that: two marks are awarded for using two different values of $\displaystyle T$ ; one mark is awarded for finding the constant of integration; one mark is awarded for finding or eliminating $\displaystyle k$ (refer to Note 6); one mark is awarded for dealing with the elapsed time (noon until 12:15).
2. Do not penalise the omission of integral symbols at •¹. (All Methods)
3. Do not penalise omission of " $\displaystyle +c$ " at •². However, it is necessary to access some later marks. (All Methods)
4. Where a candidate obtains an incorrect final answer because of earlier rounding, only •⁹ is unavailable. (All Methods)
5. For Method 1, if the candidate works in hours: •⁴ $\displaystyle \ln(6.5 - 4) = -0.25k + \ln(5.8)$ ; •⁵ $\displaystyle k = -4(\ln 2.5 - \ln 5.8) = 3.366\ldots$ ; •⁶ $\displaystyle \ln(25 - 4) = -3.366\ldots t + \ln 5.8$ ; •⁷ $\displaystyle t = \dfrac{\ln 21 - \ln 5.8}{-3.366\ldots}$ ; •⁸ $\displaystyle t = -0.3822\ldots$
6. For Method 2, if the candidate works in hours: •⁵ appearance of $\displaystyle (t + 0.25)$ ; •⁶ $\displaystyle \ln(6.5 - 4) = -k(t + 0.25) + \ln 21$ ; •⁷ $\displaystyle k = -\dfrac{1}{0.25}\ln\left(\dfrac{2.5}{5.8}\right) = 3.366\ldots$ ; •⁸ $\displaystyle t = \ln\left(\dfrac{21}{5.8}\right) \div 3.366\ldots = 0.3822\ldots$ (the PDF prints $\displaystyle \div 0.366\ldots$ , a slip).
7. In Method 2 •⁷ can be awarded for eliminating $\displaystyle k.$
8. For Method 4, if the candidate works in hours: •⁵ appearance of $\displaystyle (t + 0.25)$ ; •⁶ $\displaystyle 6.5 = 21e^{-k(t + 0.25)} + 4$ ; •⁷ $\displaystyle k = \dfrac{\ln\left(\frac{5.8}{21}\right) - \ln\left(\frac{2.5}{21}\right)}{0.25} = 3.366\ldots$ ; •⁸ $\displaystyle t = \ln\left(\dfrac{21}{5.8}\right) \div 3.366\ldots = 0.3822\ldots$

2016 Paper

Question 1(a)

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Question 1(b)

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Question 1(c)

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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