Advanced Higher Maths 2021 Paper 2 — Questions & Marking Instructions

Product, quotient, chain rules

2 Marks

2021 P2 Q1

Given $\displaystyle f(x) = 3\sec 2x$ find the exact value of $\displaystyle f'\left(\frac{\pi}{8}\right).$

Show answer

\displaystyle 6\sqrt{2}

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Mark	Awarded for	Illustrative answer
•¹	differentiate	$\displaystyle f'(x) = 3\sec 2x\tan 2x \times 2$
•²	evaluate	$\displaystyle 6\sqrt{2}$

Euclidean algorithm

(3, 1)4 Marks

2021 P2 Q2

(a)Use the Euclidean algorithm to find integers $\displaystyle a$ and $\displaystyle b$ such that $\displaystyle 105a + 72b = 3.$

(b)Hence find integers $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle 105x + 72y = 360.$

Show answer

(a)

\displaystyle a = 11

,

\displaystyle b = -16

(b)

\displaystyle x = 1320

,

\displaystyle y = -1920

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	complete algorithm	$\displaystyle 105 = 72 + 33$ , $\displaystyle 72 = 2 \times 33 + 6$ , $\displaystyle 33 = 5 \times 6 + 3$ , $\displaystyle 6 = 2 \times 3$
(a) •²	equates gcd and evidence of substitution	$\displaystyle 3 = 33 - 5 \times (72 - 2 \times 33)$
(a) •³	$\displaystyle a$ and $\displaystyle b$ obtained	$\displaystyle a = 11,\ b = -16$
(b) •⁴	find $\displaystyle x$ and $\displaystyle y$	$\displaystyle x = 1320,\ y = -1920$

Integration by parts

3 Marks

2021 P2 Q3

Use integration by parts to find $\displaystyle \int (2x + 3)\cos 4x \,dx.$

Show answer

\displaystyle \frac{1}{4}(2x + 3)\sin 4x + \frac{1}{8}\cos 4x + c

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Mark	Awarded for	Illustrative answer
•¹	use integration by parts and start process	$\displaystyle \dfrac{1}{4}(2x + 3)\sin 4x - \ldots$
•²	complete application	$\displaystyle \ldots\displaystyle\int \dfrac{2}{4}\sin 4x\,dx$
•³	complete integration	$\displaystyle \dfrac{1}{4}(2x + 3)\sin 4x + \dfrac{1}{8}\cos 4x + c$

Parametric differentiation

(3, 2)5 Marks

2021 P2 Q4

A curve is defined parametrically by $\displaystyle x = \sin^{-1} 2t$ and $\displaystyle y = \tan^{-1} t.$

(a)Find $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}.$

(b)When $\displaystyle t = 0$ find the equation of the tangent to the curve.

Show answer

(a)

\displaystyle \frac{dx}{dt} = \frac{2}{\sqrt{1 - 4t^2}}

\displaystyle \frac{dy}{dt} = \frac{1}{1 + t^2}

(b)

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

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	start to find $\displaystyle \dfrac{dx}{dt}$	$\displaystyle \dfrac{1}{\sqrt{1 - (2t)^2}}$
(a) •²	complete $\displaystyle \dfrac{dx}{dt}$	$\displaystyle \ldots \times 2$
(a) •³	state $\displaystyle \dfrac{dy}{dt}$	$\displaystyle \dfrac{1}{1 + t^2}$
(b) •⁴	begin process	$\displaystyle (0, 0)$ OR $\displaystyle \dfrac{dy}{dx} = \dfrac{1}{2}$
(b) •⁵	find the equation of the tangent	$\displaystyle y = \dfrac{1}{2}x$

Matrix operationsDeterminant and inverse

(2, 2)4 Marks

2021 P2 Q5

A non-singular matrix $\displaystyle A$ satisfies the equation $\displaystyle A^2 = 2A + 5I$ , where $\displaystyle I$ is the identity matrix.

(a)Express $\displaystyle A^4$ in the form $\displaystyle pA + qI$ , where $\displaystyle p, q \in \mathbb{Z}.$

(b)Express $\displaystyle A^{-1}$ in the form $\displaystyle rA + sI$ , where $\displaystyle r, s \in \mathbb{Q}.$

Show answer

(a)

\displaystyle 28A + 45I

(b)

\displaystyle \frac{1}{5}A - \frac{2}{5}I

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	find expression for $\displaystyle A^4$ in powers of two and less	$\displaystyle 4A^2 + 20A + 25I$ OR $\displaystyle 9A^2 + 10A$
(a) •²	substitute for $\displaystyle A^2$ and simplify	$\displaystyle 28A + 45I$
(b) •³	evidence of strategy	eg $\displaystyle A^{-1}A^2 = 2A^{-1}A + 5A^{-1}I$ OR $\displaystyle A(A - 2I) = 5I$
(b) •⁴	state expression	$\displaystyle \dfrac{1}{5}A - \dfrac{2}{5}I$

First order linear DE

4 Marks

2021 P2 Q6

Solve the differential equation

$\displaystyle \frac{dy}{dx} + 2xy = 14xe^{-x^2}$

given that when $\displaystyle x = 0$ , $\displaystyle y = 3.$ Express $\displaystyle y$ in terms of $\displaystyle x.$

Show answer

\displaystyle y = \frac{7x^2 + 3}{e^{x^2}}

(or

\displaystyle y = (7x^2 + 3)e^{-x^2}

)

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Mark	Awarded for	Illustrative answer
•¹	find integrating factor	$\displaystyle e^{x^2}$
•²	write in correct form	$\displaystyle e^{x^2}y = \displaystyle\int 14x\,dx$
•³	integrate right hand side including constant of integration	$\displaystyle e^{x^2}y = 7x^2 + c$
•⁴	find particular solution	$\displaystyle y = \dfrac{7x^2 + 3}{e^{x^2}}$

Binomial expansionComplex number operations

(3, 3)6 Marks

2021 P2 Q7

A complex number is defined by $\displaystyle z = a + 2i$ where $\displaystyle a$ is a positive real number.

(a)State and simplify the binomial expansion of $\displaystyle z^3.$

(b)Given that $\displaystyle z^3 + 3z = b + 148i$ where $\displaystyle b$ is a real number, find the values of $\displaystyle a$ and $\displaystyle b.$

Show answer

(a)

\displaystyle a^3 - 12a + (6a^2 - 8)i

(b)

\displaystyle a = 5

,

\displaystyle b = 80

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	binomial expansion	$\displaystyle \dbinom{3}{0}a^3 + \dbinom{3}{1}a^2(2i) + \dbinom{3}{2}a(2i)^2 + \dbinom{3}{3}(2i)^3$ stated or implied by •²
(a) •²	binomial coefficients and powers of 2	$\displaystyle a^3 + 6a^2i + 12ai^2 + 8i^3$
(a) •³	simplify	$\displaystyle a^3 - 12a + (6a^2 - 8)i$
(b) •⁴	substitute	$\displaystyle a^3 - 12a + (6a^2 - 8)i + 3a + 6i = b + 148i$
(b) •⁵	equate imaginary and real parts	$\displaystyle 6a^2 - 2 = 148$ and $\displaystyle a^3 - 9a = b$
(b) •⁶	find $\displaystyle a$ and $\displaystyle b$	$\displaystyle a = 5,\ b = 80$

Implicit differentiation

(4, 3)7 Marks

2021 P2 Q8

A curve is defined by $\displaystyle x^2y^3 + e^{2y} = 5.$

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

(b)Show that there is only one stationary point on the curve.

Show answer

(a)

\displaystyle \frac{-2xy^3}{3x^2y^2 + 2e^{2y}}

(b) Setting

\displaystyle \frac{dy}{dx} = 0

gives

\displaystyle -2xy^3 = 0

, so

\displaystyle x = 0

or

\displaystyle y = 0.

When

\displaystyle x = 0

,

\displaystyle e^{2y} = 5 \Rightarrow y = \frac{1}{2}\ln 5.

When

\displaystyle y = 0

,

\displaystyle 0 + 1 = 5

which has no solution.
Hence there is only one stationary point at

\displaystyle (0, \frac{1}{2}\ln 5).

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	start to differentiate product with one term correct	$\displaystyle 2xy^3$ or $\displaystyle 3x^2y^2\dfrac{dy}{dx}$
(a) •²	complete differentiation of product	$\displaystyle 3x^2y^2\dfrac{dy}{dx}$ or $\displaystyle 2xy^3$
(a) •³	differentiate remaining terms	$\displaystyle \ldots 2e^{2y}\dfrac{dy}{dx} = 0$
(a) •⁴	write derivative explicitly in terms of $\displaystyle x$ and $\displaystyle y$	$\displaystyle \dfrac{-2xy^3}{3x^2y^2 + 2e^{2y}}$
(b) •⁵	express condition for stationary point	$\displaystyle \dfrac{-2xy^3}{3x^2y^2 + 2e^{2y}} = 0$
(b) •⁶	state corresponding values of both $\displaystyle x$ and $\displaystyle y$	$\displaystyle x = 0$ AND $\displaystyle y = 0$
(b) •⁷	show that there is one value of $\displaystyle y$ when $\displaystyle x = 0$ but no value of $\displaystyle x$ when $\displaystyle y = 0$	$\displaystyle \left(0, \dfrac{1}{2}\ln 5\right)$ AND eg $\displaystyle 1 = 5$ $\displaystyle \therefore$ no solution

Partial fractionsSeparable DE in context

(2, 8)10 Marks

2021 P2 Q9

(a)Express $\displaystyle \frac{1}{x(5 - x)}$ in partial fractions.

A small island is being populated by seals. The size of the seal population can be modelled by the differential equation

$\displaystyle \frac{dP}{dt} = \frac{1}{100}P(5 - P), \quad 0 < P < 5$

where $\displaystyle P$ (in hundreds) is the number of seals on the island $\displaystyle t$ years after the seals arrive.

(b)Given that there are 250 seals after 10 years, find an expression for $\displaystyle P$ in terms of $\displaystyle t.$

Show answer

(a)

\displaystyle \frac{1}{5x} + \frac{1}{5(5 - x)}

(b)

\displaystyle P = \frac{5e^{0.05t - 0.5}}{1 + e^{0.05t - 0.5}}

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	write template	$\displaystyle \dfrac{1}{x(5 - x)} = \dfrac{A}{x} + \dfrac{B}{5 - x}$
(a) •²	find constants and express in partial fractions	$\displaystyle \dfrac{1}{5x} + \dfrac{1}{5(5 - x)}$
(b) •³	write as integral equation	$\displaystyle \displaystyle\int \dfrac{1}{P(5 - P)}\,dP = \int \dfrac{1}{100}\,dt$
(b) •⁴	write LHS integral in partial fractions	$\displaystyle \dfrac{1}{5}\displaystyle\int\left(\dfrac{1}{P} + \dfrac{1}{5 - P}\right)dP$
(b) •⁵	integrate $\displaystyle \dfrac{1}{P}$	$\displaystyle \dfrac{1}{5}(\ln P\ldots$
(b) •⁶	complete integration	$\displaystyle \dfrac{1}{5}(\ln P - \ln(5 - P)) = \dfrac{1}{100}t + c$
(b) •⁷	substitute values	$\displaystyle \dfrac{1}{5}(\ln 2.5 - \ln(5 - 2.5)) = \dfrac{1}{10} + c$
(b) •⁸	evaluate constant of integration	$\displaystyle -\dfrac{1}{10}$
(b) •⁹	take exponentials	$\displaystyle \dfrac{P}{5 - P} = e^{0.05t - 0.5}$
(b) •¹⁰	write expression in terms of $\displaystyle t$	$\displaystyle P = \dfrac{5e^{0.05t - 0.5}}{1 + e^{0.05t - 0.5}}$

Proof by induction

5 Marks

2021 P2 Q10

Prove by induction that

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

for all positive integers $\displaystyle n \ge 2.$

Show answer

When

\displaystyle n=2

, LHS

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

and RHS

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

, so the statement is true.
Assume true for

\displaystyle n=k

:

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

Then

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

True for

\displaystyle n=2

, and true for

\displaystyle n=k

implies true for

\displaystyle n=k+1

, so by induction the statement holds for all integers

\displaystyle n\ge2.

Show marking instructions

Mark	Awarded for	Illustrative answer
•¹	show true for $\displaystyle n = 2$	LHS: $\displaystyle \dfrac{1}{2(2 - 1)} = \dfrac{1}{2}$ , RHS: $\displaystyle \dfrac{2 - 1}{2} = \dfrac{1}{2}$ so true for $\displaystyle n = 2$
•²	assume (statement) true for $\displaystyle n = k$ AND consider whether (statement) true for $\displaystyle n = k + 1$	suitable statement and $\displaystyle \displaystyle\sum_{r=2}^{k}\dfrac{1}{r(r - 1)} = \dfrac{k - 1}{k}$ AND $\displaystyle \displaystyle\sum_{r=2}^{k+1}\dfrac{1}{r(r - 1)} = \ldots$
•³	correct statement for sum to $\displaystyle (k + 1)$ terms using inductive hypothesis	$\displaystyle \ldots = \dfrac{k - 1}{k} + \dfrac{1}{(k + 1)((k + 1) - 1)}$
•⁴	express as a single fraction	$\displaystyle \dfrac{(k - 1)(k + 1) + 1}{k(k + 1)}$
•⁵	express explicitly in terms of $\displaystyle (k + 1)$ AND communicate	$\displaystyle \dfrac{k^2}{k(k + 1)}$ leading to $\displaystyle \dfrac{k + 1 - 1}{k + 1}.$ If true for $\displaystyle n = k$ then true for $\displaystyle n = k + 1.$ Also shown true for $\displaystyle n = 2$ therefore, by induction, true for all $\displaystyle n \geq 2.$

Arithmetic and geometric combined

(1, 1, 1, 1)4 Marks

2021 P2 Q11(a)(b)

Three consecutive terms of an arithmetic sequence are given by $\displaystyle x - 1, x - 7, 2x - 9.$

(a)(i) Find the common difference.
(ii) Hence find the value of $\displaystyle x.$

(b)Given that $\displaystyle x - 1$ is the $\displaystyle 21^{st}$ term, find

(i) the value of the first term
(ii) a simplified expression for the $\displaystyle n^{th}$ term of the sequence.

Show answer

(a)(i)

\displaystyle -6

(a)(ii)

\displaystyle x = -4

(b)(i)

\displaystyle 115

(b)(ii)

\displaystyle 121 - 6n

Show marking instructions

Mark	Awarded for	Illustrative answer
(a)(i) •¹	state common difference	$\displaystyle -6$
(a)(ii) •²	calculate $\displaystyle x$	$\displaystyle -4$
(b)(i) •³	calculate first term	$\displaystyle 115$
(b)(ii) •⁴	state simplified expression	$\displaystyle 121 - 6n$

Arithmetic and geometric combined

(3, 1, 2)6 Marks

2021 P2 Q11(c)(d)

Three consecutive terms of a geometric sequence are given by $\displaystyle y - 1, y - 7, 2y - 9.$

(c)Find the two possible values of $\displaystyle y$ and the corresponding common ratios.

One of the values of $\displaystyle y$ gives an associated geometric series which has a sum to infinity.

(d)(i) Identify the value of $\displaystyle y$ and justify your answer.
(ii) Determine whether $\displaystyle \frac{64}{3}$ is a possible value for this sum to infinity. Give a reason for your answer.

Show answer

(c)

\displaystyle y = 5

with common ratio

\displaystyle -\frac{1}{2}

\displaystyle y = -8

with common ratio

\displaystyle \frac{5}{3}

(d)(i)

\displaystyle y = 5

because

\displaystyle \left|-\frac{1}{2}\right| < 1.

(d)(ii) No. If the sum is

\displaystyle \frac{64}{3}

, then

\displaystyle a = 32

, which would mean the term

\displaystyle y-1

would be -4 (not 4 as required by

\displaystyle y=5

).

Show marking instructions

Mark	Awarded for	Illustrative answer
(c) •⁵	equate ratios of two pairs of consecutive terms	eg $\displaystyle \dfrac{y - 7}{y - 1} = \dfrac{2y - 9}{y - 7}$
(c) •⁶	rearrange into quadratic equation in standard form	$\displaystyle y^2 + 3y - 40 = 0$
(c) •⁷	state values of $\displaystyle y$ and corresponding common ratios	$\displaystyle 5$ , $\displaystyle -\dfrac{1}{2}$ and $\displaystyle -8$ , $\displaystyle \dfrac{5}{3}$
(d)(i) •⁸	state value of $\displaystyle y$ , with justification	$\displaystyle 5$ and $\displaystyle \left\|-\dfrac{1}{2}\right\| < 1$
(d)(ii) •⁹	form equation	$\displaystyle \dfrac{a}{1 - \left(-\frac{1}{2}\right)} = \dfrac{64}{3}$
(d)(ii) •¹⁰	calculate $\displaystyle a$ and state reason	$\displaystyle a = 32$ and eg No. This value would lead to $\displaystyle -4$ , not 4, for the $\displaystyle y - 1$ term

Equations of lines or planes

(4, 1, 1, 2)8 Marks

2021 P2 Q12

The points $\displaystyle A(4, 0, 8)$ , $\displaystyle B(6, -5, 4)$ and $\displaystyle C(3, 4, 11)$ all lie on the plane $\displaystyle \pi_1.$

(a)Find the Cartesian equation of $\displaystyle \pi_1.$

The plane $\displaystyle \pi_2$ is parallel to $\displaystyle \pi_1$ and passes through the origin.

(b)State the equation of $\displaystyle \pi_2.$

A sphere touches $\displaystyle \pi_1$ , where $\displaystyle A$ is the point of contact. The sphere also has a single point of contact, $\displaystyle Q$ , with $\displaystyle \pi_2.$

(c)(i) Find parametric equations for the line $\displaystyle AQ.$
(ii) Hence find the coordinates for $\displaystyle Q.$

Show answer

(a)

\displaystyle x - 2y + 3z = 28

(b)

\displaystyle x - 2y + 3z = 0

(c)(i)

\displaystyle x = 4 + t

,

\displaystyle y = -2t

,

\displaystyle z = 8 + 3t

(c)(ii)

\displaystyle Q(2, 4, 2)

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	find two directed line segments	eg $\displaystyle \overrightarrow{AB} = \begin{pmatrix}2\\-5\\-4\end{pmatrix}$ , $\displaystyle \overrightarrow{AC} = \begin{pmatrix}-1\\4\\3\end{pmatrix}$
(a) •²	begin to find vector product	eg $\displaystyle \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\2 & -5 & -4\\-1 & 4 & 3\end{vmatrix}$
(a) •³	calculate a normal vector	eg $\displaystyle \mathbf{n} = \begin{pmatrix}1\\-2\\3\end{pmatrix}$ stated or implied at •⁴
(a) •⁴	obtain equation	$\displaystyle x - 2y + 3z = 28$
(b) •⁵	state equation of $\displaystyle \pi_2$	$\displaystyle x - 2y + 3z = 0$
(c)(i) •⁶	find parametric equations	$\displaystyle x = 4 + t$ , $\displaystyle y = -2t$ , $\displaystyle z = 8 + 3t$
(c)(ii) •⁷	substitute into equation of $\displaystyle \pi_2$	$\displaystyle 4 + t - 2(-2t) + 3(8 + 3t) = 0$
(c)(ii) •⁸	find coordinates of Q	$\displaystyle (2, 4, 2)$

de Moivre's theoremEquations with complex roots

(1, 1, 1, 2, 2)7 Marks

2021 P2 Q13

(a)Express $\displaystyle -1$ in the form $\displaystyle \cos\theta + i\sin\theta.$

The complex number $\displaystyle z_1$ is defined by $\displaystyle z_1 = \cos\frac{\pi}{5} + i\sin\frac{\pi}{5}.$

(b)Use de Moivre's theorem to show that $\displaystyle z_1$ is a root of the equation $\displaystyle z^5 + 1 = 0.$

The complex number $\displaystyle z_2$ is also a root of the equation $\displaystyle z^5 + 1 = 0.$ Roots $\displaystyle z_1$ and $\displaystyle z_2$ have been plotted on an Argand diagram, as shown.

(c)Express $\displaystyle z_2$ in the form $\displaystyle \cos\theta + i\sin\theta.$

The remaining roots of the equation $\displaystyle z^5 + 1 = 0$ are $\displaystyle z_3$ , $\displaystyle z_4$ and $\displaystyle z_5.$

(d)Express $\displaystyle z_3$ , $\displaystyle z_4$ and $\displaystyle z_5$ in the form $\displaystyle \cos\theta + i\sin\theta$ , where $\displaystyle -\pi < \theta \le \pi.$

(e)Given $\displaystyle z_1 + z_2 + z_3 + z_4 + z_5 = 0$ , show algebraically that

$\displaystyle \cos\frac{\pi}{5} + \cos\frac{3\pi}{5} = \frac{1}{2}.$

Show answer

(a)

\displaystyle \cos\pi + i\sin\pi

(b)

\displaystyle z_1^5 = \cos\left(5 \times \frac{\pi}{5}\right) + i\sin\left(5 \times \frac{\pi}{5}\right) = \cos\pi + i\sin\pi = -1.

Therefore,

\displaystyle z_1^5 + 1 = -1 + 1 = 0.

(c)

\displaystyle \cos\frac{3\pi}{5} + i\sin\frac{3\pi}{5}

(d)

\displaystyle \cos\left(-\frac{\pi}{5}\right) + i\sin\left(-\frac{\pi}{5}\right)

,

\displaystyle \cos\left(-\frac{3\pi}{5}\right) + i\sin\left(-\frac{3\pi}{5}\right)

and

\displaystyle \cos\pi + i\sin\pi

(e) Equating the real parts to zero gives:

\displaystyle \cos\frac{\pi}{5} + \cos\frac{3\pi}{5} + \cos\left(-\frac{\pi}{5}\right) + \cos\left(-\frac{3\pi}{5}\right) + \cos\pi = 0

Since

\displaystyle \cos(-\theta) = \cos(\theta)

and

\displaystyle \cos\pi = -1

:

\displaystyle 2\cos\frac{\pi}{5} + 2\cos\frac{3\pi}{5} - 1 = 0

\displaystyle 2\left(\cos\frac{\pi}{5} + \cos\frac{3\pi}{5}\right) = 1

\displaystyle \cos\frac{\pi}{5} + \cos\frac{3\pi}{5} = \frac{1}{2}

Show marking instructions

Mark	Awarded for	Illustrative answer
(a) •¹	express in appropriate form	$\displaystyle -1 = \cos\pi + i\sin\pi$
(b) •²	verify root	$\displaystyle \cos\pi + i\sin\pi$
(c) •³	polar form	$\displaystyle z_2 = \cos\dfrac{3\pi}{5} + i\sin\dfrac{3\pi}{5}$
(d) •⁴	any one from	(•⁴ and •⁵ together) $\displaystyle \cos\left(-\dfrac{\pi}{5}\right) + i\sin\left(-\dfrac{\pi}{5}\right)$ or $\displaystyle \cos\left(-\dfrac{3\pi}{5}\right) + i\sin\left(-\dfrac{3\pi}{5}\right)$ or $\displaystyle \cos\pi + i\sin\pi$
(d) •⁵	remaining roots	$\displaystyle \cos\left(-\dfrac{\pi}{5}\right) + i\sin\left(-\dfrac{\pi}{5}\right)$ , $\displaystyle \cos\left(-\dfrac{3\pi}{5}\right) + i\sin\left(-\dfrac{3\pi}{5}\right)$ , $\displaystyle \cos\pi + i\sin\pi$
(e) •⁶	equate real part to zero	$\displaystyle \cos\dfrac{\pi}{5} + \cos\dfrac{3\pi}{5} + \cos\left(-\dfrac{\pi}{5}\right) + \cos\left(-\dfrac{3\pi}{5}\right) - 1 = 0$ , $\displaystyle 2\cos\dfrac{\pi}{5} + 2\cos\dfrac{3\pi}{5} = 1$
(e) •⁷	complete proof	leading to $\displaystyle \cos\dfrac{\pi}{5} + \cos\dfrac{3\pi}{5} = \dfrac{1}{2}$

2021 Paper 2

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11(a)(b)

Question 11(c)(d)

Question 12

Question 13