Practice questions with worked answers

Complex numbers

Easy

Complex arithmetic · Year 12 · 5 marks

z = 3 + 4\mathrm{i}

and

w = 1 - 2\mathrm{i}

.

(a)
Find $zw$ in the form $a + b\mathrm{i}$ .
(2)
(b)
Find $|z|$ .
(1)
(c)
Find $\dfrac{z}{w}$ in the form $a + b\mathrm{i}$ .
(2)

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Part (a)2 marks

Expand the brackets exactly as you would with $x$ : every term times every term. $zw$ ${}= 3$ ${}- 6\mathrm{i}$ ${}+ 4\mathrm{i}$ ${}- 8\mathrm{i}^2$
Use $\mathrm{i}^2 = -1$ , so $-8\mathrm{i}^2 = +8$ , then collect real and imaginary parts. $= 11 - 2\mathrm{i}$

Part (b)1 mark

The modulus is the distance from the origin on an Argand diagram, so use Pythagoras. $|z| = \sqrt{3^2 + 4^2} = 5$

Part (c)2 marks

Multiply top and bottom by the conjugate of the denominator, which makes the denominator real. $\dfrac{(3 + 4\mathrm{i})(1 + 2\mathrm{i})}{(1 - 2\mathrm{i})(1 + 2\mathrm{i})}$ ${}= \dfrac{3 + 10\mathrm{i} + 8\mathrm{i}^2}{5}$
Replace $\mathrm{i}^2$ with $-1$ and divide each part by 5. $= -1 + 2\mathrm{i}$

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Complex numbers

Medium

Roots of a cubic · Year 12 · 5 marks

\mathrm{f}(z) = z^3 - 3z^2 + z + 5

(a)
Given that $2 - \mathrm{i}$ is a root of the equation $\mathrm{f}(z) = 0$ , solve $\mathrm{f}(z) = 0$ completely.
(5)

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Part (a)5 marks

A polynomial with real coefficients has complex roots in conjugate pairs. $2 + \mathrm{i}$ is also a root.
Two roots give two factors; multiplying them gives a real quadratic factor. $\bigl(z - (2 - \mathrm{i})\bigr)\bigl(z - (2 + \mathrm{i})\bigr)$ ${}= (z - 2)^2 - \mathrm{i}^2$
Expand $(z-2)^2$ and use $\mathrm{i}^2 = -1$ . $= z^2 - 4z + 5$
Divide, or compare coefficients: the constant term 5 = 5 × 1 gives the last factor. $z^3 - 3z^2 + z + 5$ ${}= (z^2 - 4z + 5)(z + 1)$
Set each factor to zero to list all three roots. $z$ ${}= -1,\ 2$ ${}+ \mathrm{i},\ 2$ ${}- \mathrm{i}$

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Matrices

Easy

Inverse of a 2 × 2 matrix · Year 12 · 4 marks

\mathbf{A}

{}= \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}

(a)
Find $\det \mathbf{A}$ .
(1)
(b)
Find $\mathbf{A}^{-1}$ .
(2)
(c)
A triangle of area 4 square units is transformed by $\mathbf{A}$ . State the area of its image.
(1)

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Part (a)1 mark

For $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ the determinant is $ad - bc$ . $\det\mathbf{A}$ ${}= 2 \times 3 - 1 \times 5$ ${}= 1$

Part (b)2 marks

Swap the numbers on the leading diagonal, change the signs of the other two, then divide by the determinant. $\mathbf{A}^{-1}$ ${}= \dfrac{1}{1}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$
Dividing by 1 changes nothing. $\mathbf{A}^{-1}$ ${}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$

Part (c)1 mark

The determinant is the area scale factor of the transformation.Area $= 4 \times |\det\mathbf{A}| = 4$ square units

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Proof

Medium

Induction with a sum · Year 12 · 5 marks

(a)
Prove by induction that, for all positive integers $n$ , $\sum_{r=1}^{n} r(r + 1) = \tfrac{1}{3}n(n + 1)(n + 2).$
(5)

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Part (a)5 marks

Basis: check the statement is true for the first value. $n = 1$ : left side $= 1 \times 2 = 2$ , right side $= \tfrac{1}{3}(1)(2)(3) = 2$
Assumption: suppose it is true for some $n = k$ .Assume $\sum_{r=1}^{k} r(r+1)$ ${}= \tfrac{1}{3}k(k+1)(k+2)$
The sum to $k + 1$ is the sum to $k$ plus the next term, where $r = k + 1$ . $\sum_{r=1}^{k+1} r(r+1)$ ${}= \tfrac{1}{3}k(k+1)(k+2)$ ${}+ (k+1)(k+2)$
Take out the common factor $\tfrac{1}{3}(k+1)(k+2)$ instead of expanding. $= \tfrac{1}{3}(k+1)(k+2)(k + 3)$
Show it is the original formula with $n$ replaced by $k + 1$ .This is $\tfrac{1}{3}n(n+1)(n+2)$ with $n = k + 1$ .
Conclusion: true for $n = 1$ , and true for $k$ means true for $k + 1$ .So it is true for all positive integers $n$ .

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Proof

Hard

Induction with divisibility · Year 12 · 6 marks

(a)
Prove by induction that $4^n + 6n - 1$ is divisible by 9 for all positive integers $n$ .
(6)

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Part (a)6 marks

Basis: 9 is divisible by 9. $\mathrm{f}(1) = 4 + 6 - 1 = 9$
Assumption step.Assume $\mathrm{f}(k) = 4^k + 6k - 1$ is divisible by 9.
Subtracting $4\mathrm{f}(k)$ cancels the $4^{k+1}$ terms. $\mathrm{f}(k+1)$ ${}- 4\mathrm{f}(k)$ ${}= 4^{k+1}$ ${}+ 6k$ ${}+ 5$ ${}- 4^{k+1}$ ${}- 24k$ ${}+ 4$
What is left is a multiple of 9. $= 9 - 18k = 9(1 - 2k)$
Rearrange to write $\mathrm{f}(k+1)$ in terms of $\mathrm{f}(k)$ . $\mathrm{f}(k+1)$ ${}= 4\mathrm{f}(k) + 9(1 - 2k)$
$\mathrm{f}(k)$ is a multiple of 9 by the assumption.Both terms are multiples of 9, so $\mathrm{f}(k+1)$ is too.
Conclusion: true for $n = 1$ , and true for $k$ means true for $k + 1$ .So it is true for all positive integers $n$ .

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Further algebra and functions

Medium

Roots of a cubic and their squares · Year 12 · 5 marks

The roots of

2x^3 - 3x^2 + 4x - 5 = 0

are

\alpha

,

\beta

and

\gamma

.

(a)
Write down $\alpha + \beta + \gamma$ , $\alpha\beta$ ${}+ \beta\gamma$ ${}+ \gamma\alpha$ and $\alpha\beta\gamma$ .
(2)
(b)
Find $\alpha^2 + \beta^2 + \gamma^2$ and explain what your answer tells you about the roots.
(3)

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Part (a)2 marks

For $ax^3 + bx^2 + cx + d$ : $\sum\alpha = -\tfrac{b}{a}$ , $\sum\alpha\beta = \tfrac{c}{a}$ , $\alpha\beta\gamma = -\tfrac{d}{a}$ . $\sum\alpha$ ${}= \tfrac{3}{2},$ $\quad \sum\alpha\beta$ ${}= 2,$ $\quad \alpha\beta\gamma$ ${}= \tfrac{5}{2}$

Part (b)3 marks

Squaring the sum of the roots gives the squares plus twice every pair. $\alpha^2 + \beta^2 + \gamma^2$ ${}= (\sum\alpha)^2$ ${}- 2\sum\alpha\beta$
Substitute the values from part (a). $= \tfrac{9}{4} - 4 = -\tfrac{7}{4}$
Two of the roots must be a complex conjugate pair.The squares of real numbers cannot add to a negative, so not all the roots are real.

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Further algebra and functions

Hard

Roots of a quartic · Year 12 · 6 marks

The roots of

x^4 + 2x^3 - x + 3 = 0

are

\alpha

,

\beta

,

\gamma

and

\delta

.

(a)
Find $\dfrac{1}{\alpha}$ ${}+ \dfrac{1}{\beta}$ ${}+ \dfrac{1}{\gamma}$ ${}+ \dfrac{1}{\delta}$ .
(3)
(b)
Find $\alpha^2$ ${}+ \beta^2$ ${}+ \gamma^2$ ${}+ \delta^2$ .
(3)

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Part (a)3 marks

For $ax^4 + bx^3 + cx^2 + dx + e$ : $\sum\alpha\beta\gamma$ ${}= -\tfrac{d}{a}$ and $\alpha\beta\gamma\delta$ ${}= \tfrac{e}{a}$ . $\sum\alpha\beta\gamma$ ${}= 1,$ $\quad \alpha\beta\gamma\delta$ ${}= 3$
Put the four fractions over the common denominator $\alpha\beta\gamma\delta$ . $\sum\dfrac{1}{\alpha}$ ${}= \dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}$
Substitute the values. $= \tfrac{1}{3}$

Part (b)3 marks

$\sum\alpha = -\tfrac{b}{a}$ and $\sum\alpha\beta = \tfrac{c}{a}$ ; there is no $x^2$ term, so $c = 0$ . $\sum\alpha$ ${}= -2,$ $\quad \sum\alpha\beta$ ${}= 0$
The same identity as for a cubic: it works for any number of roots. $\sum\alpha^2$ ${}= (\sum\alpha)^2$ ${}- 2\sum\alpha\beta$
Substitute the values. $= 4 - 0 = 4$

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Further algebra and functions

Medium

Method of differences · Year 12 · 4 marks

(a)
Show that $\dfrac{1}{r(r + 1)}$ ${}= \dfrac{1}{r}$ ${}- \dfrac{1}{r + 1}$ .
(1)
(b)
Hence find $\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r + 1)}$ , giving your answer as a single fraction.
(3)

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Part (a)1 mark

Combine the right-hand side over a common denominator. $\dfrac{1}{r} - \dfrac{1}{r+1}$ ${}= \dfrac{(r + 1) - r}{r(r + 1)}$ ${}= \dfrac{1}{r(r+1)}$

Part (b)3 marks

Write out the first few and last terms to see which parts cancel. $\left(1 - \tfrac{1}{2}\right)$ ${}+ \left(\tfrac{1}{2} - \tfrac{1}{3}\right)$ ${}+ \dots$ ${}+ \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right)$
Every term cancels except the first and the last. $= 1 - \dfrac{1}{n + 1}$
Write it as a single fraction, as asked. $= \dfrac{n}{n + 1}$

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Further vectors

Easy

Angle between two lines · Year 12 · 3 marks

Line

l_1

has direction

\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}

and line

l_2

has direction

\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}

.

(a)
Find the acute angle between the lines, to one decimal place.
(3)

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Part (a)3 marks

The angle between lines is the angle between their direction vectors, found with the scalar product. $\mathbf{a}\cdot\mathbf{b}$ ${}= 2 - 2 + 4$ ${}= 4$
Both vectors have magnitude $\sqrt{1 + 4 + 4} = 3$ . $\cos\theta$ ${}= \dfrac{4}{3 \times 3}$ ${}= \dfrac{4}{9}$
$\cos\theta$ is positive, so this angle is already acute. $\theta = 63.6^\circ$

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Polar coordinates

Medium

Area inside a polar curve · Year 13 · 5 marks

A curve has polar equation

r = 2(1 + \cos\theta)

,

0 \le \theta < 2\pi

.

(a)
Find the exact area enclosed by the curve.
(5)

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Part (a)5 marks

Area in polar coordinates is $\tfrac{1}{2}\int r^2\,\mathrm{d}\theta$ over one full turn. $A$ ${}= \tfrac{1}{2}\int_0^{2\pi} 4(1 + \cos\theta)^2\,\mathrm{d}\theta$
Expand the bracket before integrating. $= 2\int_0^{2\pi} \left(1 + 2\cos\theta + \cos^2\theta\right)\mathrm{d}\theta$
$\cos^2\theta$ cannot be integrated directly, so use the double angle formula. $\cos^2\theta$ ${}= \tfrac{1}{2}(1 + \cos 2\theta)$
Integrate each term. $= 2\left[\tfrac{3}{2}\theta + 2\sin\theta + \tfrac{1}{4}\sin 2\theta\right]_0^{2\pi}$
The sine terms are zero at both limits. $= 6\pi$

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Hyperbolic functions

Easy

Solving a hyperbolic equation · Year 13 · 4 marks

(a)
Solve $\cosh x = 2$ , giving your answers as natural logarithms.
(4)

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Part (a)4 marks

Replace $\cosh x$ with its definition in exponentials. $\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}$ ${}= 2$
Multiply through by $2\mathrm{e}^x$ to get a quadratic in $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}- 4\mathrm{e}^x$ ${}+ 1$ ${}= 0$
Solve the quadratic with the formula. $\mathrm{e}^x = 2 \pm \sqrt{3}$
Both values of $\mathrm{e}^x$ are positive, so both give solutions. $x = \ln(2 + \sqrt{3})$ or $x$ ${}= \ln(2 - \sqrt{3})$ ${}= -\ln(2 + \sqrt{3})$

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Differential equations

Medium

Integrating factor · Year 13 · 5 marks

(a)
Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}+ \dfrac{y}{x}$ ${}= x^2$ , $x > 0$ , given that $y = 1$ when $x = 1$ .
(5)

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Part (a)5 marks

The equation is in the form $y' + P(x)y = Q(x)$ , so the integrating factor is $\mathrm{e}^{\int P\,\mathrm{d}x}$ .Integrating factor $= \mathrm{e}^{\int \frac{1}{x}\,\mathrm{d}x}$ ${}= x$
Multiplying by $x$ makes the left side the derivative of a product. $\dfrac{\mathrm{d}}{\mathrm{d}x}(xy)$ ${}= x^3$
Integrate both sides; remember the constant. $xy = \tfrac{1}{4}x^4 + c$
Use the given point to find $c$ . $1$ ${}= \tfrac{1}{4} + c$ ${}\Rightarrow c$ ${}= \tfrac{3}{4}$
Divide by $x$ to make $y$ the subject. $y = \dfrac{x^3}{4} + \dfrac{3}{4x}$

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Differential equations

Hard

Modelling damped motion · Year 13 · 8 marks

A particle on a spring moves in a straight line. Its displacement

x

metres from equilibrium at time

t

seconds satisfies

\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 4x = 0.

At

t = 0

the particle is at equilibrium, moving with speed

2\ \mathrm{m\,s^{-1}}

.

(a)
Find $x$ in terms of $t$ .
(4)
(b)
Find the greatest displacement of the particle, and state the type of damping.
(4)

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Part (a)4 marks

Write the auxiliary equation by replacing each derivative with a power of $m$ . $m^2 + 4m + 4$ ${}= 0$ ${}\Rightarrow m$ ${}= -2$ (repeated)
A repeated root gives this form of general solution. $x = (A + Bt)\mathrm{e}^{-2t}$
Use $x = 0$ at $t = 0$ , then differentiate and use $\dot{x} = 2$ at $t = 0$ . $A = 0$ ; $\dot{x}$ ${}= B\mathrm{e}^{-2t}$ ${}- 2Bt\mathrm{e}^{-2t}$ , so $B = 2$
Substitute the constants. $x = 2t\mathrm{e}^{-2t}$

Part (b)4 marks

The greatest displacement is where the particle stops, so $\dot{x} = 0$ . $\dot{x}$ ${}= 2\mathrm{e}^{-2t}(1 - 2t)$ ${}= 0$ ${}\Rightarrow t$ ${}= \tfrac{1}{2}$
Substitute this time into $x$ . $x$ ${}= 2 \times \tfrac{1}{2} \times \mathrm{e}^{-1}$ ${}= \mathrm{e}^{-1}$
Give a sensible number of significant figures.Greatest displacement $= 0.368$ m
The auxiliary equation has a repeated root, so the particle returns to equilibrium without oscillating.Critical damping

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Proof

Easy

Proof by deduction · Year 12 · 3 marks

(a)
Prove that $n^2 - n$ is even for every integer $n$ .
(3)

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Part (a)3 marks

Factorise to see the structure of the expression. $n^2 - n = n(n - 1)$
Of any two numbers next to each other, one is a multiple of 2. $n$ and $n - 1$ are consecutive integers, so one of them is even.
Finish with a clear statement of what has been proved.An even number times any integer is even, so $n^2 - n$ is even.

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Algebra and functions

Easy

Equal roots · Year 12 · 3 marks

(a)
The equation $x^2 + kx + 9 = 0$ has equal roots. Find the possible values of $k$ .
(3)

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Part (a)3 marks

Equal (repeated) roots happen when the discriminant is zero. $b^2 - 4ac = 0$
Here $a = 1$ , $b = k$ and $c = 9$ . $k^2 - 4 \times 1 \times 9$ ${}= 0$ ${}\Rightarrow k^2$ ${}= 36$
Square-rooting gives two answers: positive and negative. $k = 6$ or $k = -6$

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Sequences and series

Medium

Geometric series · Year 13 · 4 marks

A geometric series has first term 12 and sum to infinity 48.

(a)
Find the common ratio.
(2)
(b)
Find the sum of the first 10 terms, to 3 significant figures.
(2)

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Part (a)2 marks

Use $S_\infty = \dfrac{a}{1 - r}$ . $\dfrac{12}{1 - r} = 48$
Rearrange for $r$ . It is between −1 and 1, as it must be. $1 - r$ ${}= \tfrac{1}{4}$ ${}\Rightarrow r$ ${}= \tfrac{3}{4}$

Part (b)2 marks

Use $S_n = \dfrac{a(1 - r^n)}{1 - r}$ . $S_{10}$ ${}= \dfrac{12\left(1 - 0.75^{10}\right)}{1 - 0.75}$
Round to 3 significant figures as asked. $= 45.3$

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Trigonometry

Medium

A quadratic in cosine · Year 13 · 5 marks

(a)
Solve $2\sin^2 x = 3\cos x$ for $0^\circ \le x < 360^\circ$ .
(5)

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Part (a)5 marks

Use $\sin^2 x = 1 - \cos^2 x$ so there is only one trig function. $2(1 - \cos^2 x) = 3\cos x$
Rearrange into a quadratic in $\cos x$ . $2\cos^2 x + 3\cos x - 2 = 0$
Factorise as you would $2c^2 + 3c - 2$ . $(2\cos x - 1)(\cos x + 2) = 0$
Cosine is always between −1 and 1. $\cos x = \tfrac{1}{2}$ ( $\cos x = -2$ has no solutions)
Cosine is positive in the first and fourth quadrants. $x = 60^\circ,\ 300^\circ$

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Exponentials and logarithms

Hard

Modelling cooling · Year 12 · 7 marks

A cup of tea is left in a room. Its temperature

T

°C,

t

minutes later, is modelled by

T = 20 + 70\mathrm{e}^{-kt},

where

k

is a positive constant. After 10 minutes the temperature is 50 °C.

(a)
Find the value of $k$ , to 3 significant figures.
(3)
(b)
Find how long it takes for the tea to cool to 30 °C.
(3)
(c)
Give one limitation of the model.
(1)

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Part (a)3 marks

Substitute $t = 10$ and $T = 50$ , then isolate the exponential. $50$ ${}= 20 + 70\mathrm{e}^{-10k}$ ${}\Rightarrow \mathrm{e}^{-10k}$ ${}= \tfrac{3}{7}$
Take natural logs of both sides to bring down the power. $-10k$ ${}= \ln\tfrac{3}{7}$ ${}\Rightarrow k$ ${}= \tfrac{1}{10}\ln\tfrac{7}{3}$
Evaluate and round. $k = 0.0847$

Part (b)3 marks

Substitute $T = 30$ and isolate the exponential. $\mathrm{e}^{-kt} = \tfrac{1}{7}$
Take logs; $\ln\tfrac{1}{7} = -\ln 7$ . $t = \dfrac{\ln 7}{k}$
Use the unrounded value of $k$ to avoid rounding errors. $t = 23.0$ minutes

Part (c)1 mark

As $t$ grows, $T$ approaches 20, so 20 °C must be the room temperature.It assumes the room stays at exactly 20 °C.

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Differentiation

Easy

Gradient of a curve · Year 12 · 3 marks

A curve has equation

y = 2x^3 - 5x^2 + 4

.

(a)
Find the gradient of the curve at the point where $x = 2$ .
(3)

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Part (a)3 marks

Differentiate each term: multiply by the power, then reduce the power by 1. The constant disappears. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 6x^2 - 10x$
The gradient at a point is the derivative evaluated there.At $x = 2$ : $24 - 20 = 4$

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Differentiation

Medium

Stationary points · Year 12 · 6 marks

A curve has equation

y = x^3 - 6x^2 + 9x + 1

.

(a)
Find the coordinates of the stationary points of the curve and determine their nature.
(6)

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Part (a)6 marks

Stationary points are where the gradient is zero, so differentiate first. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 3x^2 - 12x + 9$
Set the derivative to zero and solve. $3(x - 1)(x - 3)$ ${}= 0$ ${}\Rightarrow x$ ${}= 1,\ x$ ${}= 3$
Substitute each $x$ into the original equation to get $y$ . $(1, 5)$ and $(3, 1)$
The second derivative tells you the shape at each point. $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}= 6x - 12$
Negative second derivative means the curve is bending down.At $x = 1$ : $-6 < 0$ , so $(1, 5)$ is a maximum.
Positive second derivative means the curve is bending up.At $x = 3$ : $6 > 0$ , so $(3, 1)$ is a minimum.

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Integration

Easy

A definite integral · Year 12 · 4 marks

(a)
Evaluate $\displaystyle\int_1^2 \left(3x^2 + \frac{2}{x^2}\right)\mathrm{d}x$ .
(4)

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Part (a)4 marks

Write $\frac{2}{x^2}$ as a power of $x$ so you can integrate it. $\int \left(3x^2 + 2x^{-2}\right)\mathrm{d}x$
Add 1 to each power and divide by the new power. $= \left[x^3 - 2x^{-1}\right]_1^2$
Substitute the top limit, then subtract the value at the bottom limit. $= (8 - 1) - (1 - 2)$
Take care with the double negative. $= 8$

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Statistical distributions

Medium

Binomial probabilities · Year 12 · 3 marks

X \sim \mathrm{B}(20, 0.3)

(a)
Find $\mathrm{P}(X \le 4)$ .
(1)
(b)
Find $\mathrm{P}(X \ge 8)$ .
(2)

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Part (a)1 mark

Use the cumulative binomial function on your calculator with $n = 20$ , $p = 0.3$ . $\mathrm{P}(X \le 4) = 0.2375$

Part (b)2 marks

The calculator gives 'less than or equal to', so use the complement. $\mathrm{P}(X \ge 8)$ ${}= 1 - \mathrm{P}(X \le 7)$
Subtract from 1. $= 1 - 0.7723 = 0.2277$

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Statistical hypothesis testing

Hard

Is the coin biased? · Year 12 · 5 marks

Sam thinks a coin is biased towards heads. He tosses it 20 times and gets 15 heads.

(a)
Test Sam's claim at the 5% significance level. State your hypotheses clearly.
(5)

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Part (a)5 marks

$p$ is the probability of heads. The claim is 'towards heads', so the test is one-tailed. $\mathrm{H_0}: p$ ${}= 0.5,$ $\quad \mathrm{H_1}: p > 0.5$
Assume the null hypothesis is true to find how likely the result is.Under $\mathrm{H_0}$ , $X \sim \mathrm{B}(20, 0.5)$
Find the probability of a result at least as extreme as the one seen. $\mathrm{P}(X \ge 15)$ ${}= 1 - \mathrm{P}(X \le 14)$ ${}= 0.0207$
Compare the probability with the significance level. $0.0207 < 0.05$ , so reject $\mathrm{H_0}$
Always give the conclusion in the context of the question, without certainty.There is evidence that the coin is biased towards heads.

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Kinematics

Easy

Constant acceleration · Year 12 · 4 marks

A car accelerates uniformly from

4\ \mathrm{m\,s^{-1}}

to

20\ \mathrm{m\,s^{-1}}

in 8 seconds.

(a)
Find the acceleration of the car.
(2)
(b)
Find the distance travelled in this time.
(2)

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Part (a)2 marks

Use $v = u + at$ with $u = 4$ , $v = 20$ , $t = 8$ . $20 = 4 + 8a$
Solve for $a$ . $a = 2\ \mathrm{m\,s^{-2}}$

Part (b)2 marks

Use $s = \tfrac{1}{2}(u + v)t$ , which needs no acceleration. $s = \tfrac{1}{2}(4 + 20) \times 8$
Include units. $s = 96$ m

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Forces and Newton's laws

Medium

Particle on an inclined plane · Year 12 · 4 marks

A particle of mass 2 kg is pulled up a smooth plane inclined at 30° to the horizontal by a force of 15 N acting parallel to the plane. Take

g = 9.8\ \mathrm{m\,s^{-2}}

.

(a)
Find the acceleration of the particle.
(4)

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Part (a)4 marks

Resolve along the plane and use $F = ma$ . The weight's component down the slope is $mg\sin 30^\circ$ . $15 - 2g\sin 30^\circ = 2a$
$2 \times 9.8 \times 0.5 = 9.8$ . $15 - 9.8 = 2a$
The plane is smooth, so there is no friction to include. $a = 2.6\ \mathrm{m\,s^{-2}}$ up the plane

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Electric circuits

Easy

Resistance of a wire · Year 12 · 2 marks

A wire is 0.80 m long with cross-sectional area

2.0 \times 10^{-7}\ \mathrm{m^2}

. Its material has resistivity

4.9 \times 10^{-7}\ \Omega\,\mathrm{m}

.

(a)
Calculate the resistance of the wire.
(2)

Worked answerHide answer

Part (a)2 marks

Resistivity links resistance to the length and area of a wire. $R$ ${}= \dfrac{\rho L}{A}$ ${}= \dfrac{4.9 \times 10^{-7} \times 0.80}{2.0 \times 10^{-7}}$
Give the unit, ohms. $R = 2.0\ \Omega$

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Electric circuits

Easy

Series and parallel resistors · Year 12 · 4 marks

A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel. This combination is in series with a 4.0 Ω resistor and a 12 V supply of negligible internal resistance.

(a)
Calculate the current from the supply.
(3)
(b)
Calculate the potential difference across the 3.0 Ω resistor.
(1)

Worked answerHide answer

Part (a)3 marks

Resistors in parallel combine with the reciprocal rule. $\dfrac{1}{R}$ ${}= \dfrac{1}{6.0}$ ${}+ \dfrac{1}{3.0}$ ${}\Rightarrow R$ ${}= 2.0\ \Omega$
Resistances in series add.Total $= 2.0 + 4.0 = 6.0\ \Omega$
Use $I = V/R$ for the whole circuit. $I$ ${}= \dfrac{12}{6.0}$ ${}= 2.0\ \mathrm{A}$

Part (b)1 mark

The full current passes through the 2.0 Ω parallel pair, and both parallel resistors share its p.d. $V$ ${}= 2.0 \times 2.0$ ${}= 4.0\ \mathrm{V}$

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Mechanics

Medium

Horizontal projection · Year 12 · 3 marks

A ball is thrown horizontally at

5.0\ \mathrm{m\,s^{-1}}

from a height of 1.8 m above level ground. Ignore air resistance.

(a)
Calculate the horizontal distance the ball travels before it hits the ground.
(3)

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Part (a)3 marks

Treat vertical motion on its own: it starts with zero vertical velocity and accelerates at $g$ . $1.8$ ${}= \tfrac{1}{2} \times 9.81 \times t^2$
Solve for the time of flight. $t = 0.606\ \mathrm{s}$
Horizontal velocity is constant, so distance = speed × time.Distance $= 5.0 \times 0.606$ ${}= 3.0\ \mathrm{m}$

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Mechanics

Medium

A collision · Year 12 · 4 marks

A 2.0 kg trolley moving at

3.0\ \mathrm{m\,s^{-1}}

collides with a stationary 1.0 kg trolley. They stick together.

(a)
Calculate their speed after the collision.
(2)
(b)
Show that the collision is inelastic.
(2)

Worked answerHide answer

Part (a)2 marks

Momentum is conserved: total before = total after. $2.0 \times 3.0 = 3.0 \times v$
The trolleys move together, so they share one velocity. $v = 2.0\ \mathrm{m\,s^{-1}}$

Part (b)2 marks

Use $E_k = \tfrac{1}{2}mv^2$ before and after.Kinetic energy before $= \tfrac{1}{2} \times 2.0 \times 3.0^2$ ${}= 9.0\ \mathrm{J}$
A collision that loses kinetic energy is inelastic.After $= \tfrac{1}{2} \times 3.0 \times 2.0^2$ ${}= 6.0\ \mathrm{J}$ , so kinetic energy is lost.

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Waves and the particle nature of light

Easy

The wave equation · Year 12 · 2 marks

A tuning fork produces a sound of frequency 440 Hz. The wavelength of the sound in air is 0.78 m.

(a)
Calculate the speed of sound in air.
(2)

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Part (a)2 marks

Wave speed = frequency × wavelength. $v = f\lambda = 440 \times 0.78$
Round to match the data. $v = 343\ \mathrm{m\,s^{-1}}$

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Waves and the particle nature of light

Medium

The photoelectric effect · Year 12 · 4 marks

Light of wavelength 400 nm falls on a metal surface with work function 2.3 eV.

(a)
Calculate the maximum kinetic energy of the emitted electrons, in eV.
(4)

Worked answerHide answer

Part (a)4 marks

Find the energy of one photon. Convert nm to m first. $E$ ${}= \dfrac{hc}{\lambda}$ ${}= \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{4.00 \times 10^{-7}}$ ${}= 4.97 \times 10^{-19}\ \mathrm{J}$
Convert the work function to joules so the units match. $\phi$ ${}= 2.3 \times 1.60 \times 10^{-19}$ ${}= 3.68 \times 10^{-19}\ \mathrm{J}$
$hf = \phi + E_{k(\max)}$ : the photon's energy pays the work function and the rest becomes kinetic energy. $E_{k(\max)}$ ${}= 4.97 \times 10^{-19}$ ${}- 3.68 \times 10^{-19}$ ${}= 1.29 \times 10^{-19}\ \mathrm{J}$
Divide by $1.60 \times 10^{-19}$ to convert back to electronvolts. $= 0.81\ \mathrm{eV}$

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Oscillations

Medium

A simple pendulum · Year 13 · 3 marks

A simple pendulum has length 0.90 m. Take

g = 9.81\ \mathrm{m\,s^{-2}}

.

(a)
Calculate its period.
(2)
(b)
State why the swings must be small.
(1)

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Part (a)2 marks

For small swings a pendulum moves with simple harmonic motion. $T$ ${}= 2\pi\sqrt{\dfrac{l}{g}}$ ${}= 2\pi\sqrt{\dfrac{0.90}{9.81}}$
Give 3 significant figures. $T = 1.90\ \mathrm{s}$

Part (b)1 mark

For large angles the motion is no longer simple harmonic.The formula uses $\sin\theta \approx \theta$ , which is only true for small angles.

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Gravitational fields

Hard

Geostationary orbit · Year 13 · 4 marks

A geostationary satellite orbits the Earth once every 24 hours. Mass of the Earth

= 5.97 \times 10^{24}

kg, radius of the Earth

= 6.37 \times 10^{6}

m,

G

{}= 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}

.

(a)
Show that the height of the satellite above the Earth's surface is about $3.6 \times 10^{7}$ m.
(4)

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Part (a)4 marks

Gravity provides the centripetal force that keeps the satellite in its circle. $\dfrac{GMm}{r^2}$ ${}= mr\left(\dfrac{2\pi}{T}\right)^2$
Cancel $m$ , rearrange for $r^3$ , and convert 24 hours to seconds. $r^3 = \dfrac{GMT^2}{4\pi^2}$ , with $T = 86\,400$ s
Take the cube root. This is measured from the centre of the Earth. $r = 4.22 \times 10^{7}\ \mathrm{m}$
Subtract the Earth's radius to get the height above the surface.Height $= 4.22 \times 10^{7}$ ${}- 6.37 \times 10^{6}$ ${}= 3.59 \times 10^{7}\ \mathrm{m}$

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Electric and magnetic fields

Hard

Capacitor discharge · Year 13 · 5 marks

A

470\ \mu\mathrm{F}

capacitor is charged to 12 V and then discharged through a

10\ \mathrm{k}\Omega

resistor.

(a)
Calculate the time constant.
(1)
(b)
Calculate the p.d. across the capacitor after 5.0 s.
(2)
(c)
Calculate the time taken for the p.d. to fall to 3.0 V.
(2)

Worked answerHide answer

Part (a)1 mark

Convert kΩ and μF to base units before multiplying. $RC$ ${}= 10 \times 10^{3} \times 470 \times 10^{-6}$ ${}= 4.7\ \mathrm{s}$

Part (b)2 marks

During discharge, $V = V_0\mathrm{e}^{-t/RC}$ . $V = 12\mathrm{e}^{-5.0/4.7}$
Evaluate and round sensibly. $V = 4.1\ \mathrm{V}$

Part (c)2 marks

Divide by 12, take natural logs, and rearrange for $t$ . $3.0$ ${}= 12\mathrm{e}^{-t/4.7}$ ${}\Rightarrow t$ ${}= 4.7\ln 4$
Falling to a quarter takes two half-lives' worth of time: $2 \times 0.69RC$ . $t = 6.5\ \mathrm{s}$

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Nuclear radiation

Medium

Half-life · Year 13 · 4 marks

A radioactive isotope has a half-life of 6.0 hours.

(a)
Calculate its decay constant in $\mathrm{s^{-1}}$ .
(2)
(b)
What fraction of the original nuclei remains after 24 hours?
(2)

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Part (a)2 marks

The decay constant and half-life are linked by $\lambda t_{1/2} = \ln 2$ . Convert hours to seconds. $\lambda$ ${}= \dfrac{\ln 2}{t_{1/2}}$ ${}= \dfrac{0.693}{6.0 \times 3600}$
Give the unit. $\lambda$ ${}= 3.2 \times 10^{-5}\ \mathrm{s^{-1}}$

Part (b)2 marks

Count how many half-lives fit into the time.24 hours = 4 half-lives
Each half-life halves the number left.Fraction $= \left(\tfrac{1}{2}\right)^4$ ${}= \tfrac{1}{16}$

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Proof

Easy

Sum of odd numbers · Year 12 · 4 marks

(a)
Prove by induction that, for all positive integers $n$ , $\sum_{r=1}^{n} (2r - 1) = n^2.$
(4)

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Part (a)4 marks

Start by checking the first case works. $n = 1$ : left side $= 1$ , right side $= 1^2 = 1$
Suppose the statement is true when $n = k$ .Assume $\sum_{r=1}^{k} (2r - 1) = k^2$
Adding one more term means adding the term with $r = k + 1$ . $\sum_{r=1}^{k+1} (2r - 1)$ ${}= k^2 + 2(k + 1) - 1$ ${}= k^2 + 2k + 1$
This is the formula with $n = k + 1$ , which is what we wanted. $= (k + 1)^2$
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Proof

Easy

Sum of the first n integers · Year 12 · 4 marks

(a)
Prove by induction that, for all positive integers $n$ , $\sum_{r=1}^{n} r = \tfrac{1}{2}n(n + 1).$
(4)

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Part (a)4 marks

Check the first case. $n = 1$ : left side $= 1$ , right side $= \tfrac{1}{2}(1)(2) = 1$
Suppose it is true when $n = k$ .Assume $\sum_{r=1}^{k} r$ ${}= \tfrac{1}{2}k(k + 1)$
Add the next term, $k + 1$ , then take out the common factor $\tfrac{1}{2}(k + 1)$ . $\sum_{r=1}^{k+1} r$ ${}= \tfrac{1}{2}k(k + 1)$ ${}+ (k + 1)$ ${}= \tfrac{1}{2}(k + 1)(k + 2)$
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Proof

Easy

Divisibility by 2 · Year 12 · 4 marks

(a)
Prove by induction that $3^n - 1$ is divisible by 2 for all positive integers $n$ .
(4)

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Part (a)4 marks

Check the first case. $n = 1$ : $3 - 1 = 2$ , which is divisible by 2
Suppose it is true when $n = k$ .Assume $3^k - 1$ is divisible by 2.
Write the next case using the one we assumed: $3 \times 3^k - 3 + 2$ . $3^{k+1} - 1 = 3(3^k - 1) + 2$
A sum of multiples of 2 is a multiple of 2.Both $3(3^k - 1)$ and 2 are multiples of 2, so $3^{k+1} - 1$ is too.
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Proof

Easy

Induction with a sequence · Year 12 · 4 marks

A sequence is defined by

u_1 = 1

and

u_{n+1} = 2u_n + 1

.

(a)
Prove by induction that $u_n = 2^n - 1$ for all positive integers $n$ .
(4)

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Part (a)4 marks

Check the formula gives the first term. $n = 1$ : $2^1 - 1 = 1 = u_1$
Suppose the formula is right for the $k$ th term.Assume $u_k = 2^k - 1$ .
Use the rule for the next term, then simplify. $u_{k+1}$ ${}= 2(2^k - 1) + 1$ ${}= 2^{k+1} - 1$
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Proof

Easy

Induction with matrices · Year 12 · 4 marks

\mathbf{M}

{}= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}

(a)
Prove by induction that $\mathbf{M}^n$ ${}= \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ for all positive integers $n$ .
(4)

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Part (a)4 marks

The formula with $n = 1$ gives $\mathbf{M}$ itself. $n = 1$ : $\mathbf{M}^1$ ${}= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$
Suppose it is true for $n = k$ .Assume $\mathbf{M}^k$ ${}= \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ .
$\mathbf{M}^{k+1}$ ${}= \mathbf{M}^k\mathbf{M}$ . Multiply rows by columns. $\mathbf{M}^{k+1}$ ${}= \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ ${}= \begin{pmatrix} 1 & k + 1 \\ 0 & 1 \end{pmatrix}$
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Proof

Medium

Induction with powers of 2 · Year 12 · 5 marks

(a)
Prove by induction that, for all positive integers $n$ , $\sum_{r=1}^{n} r \times 2^r = 2 + (n - 1)2^{n+1}.$
(5)

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Part (a)5 marks

Check the first case. $n = 1$ : left side $= 2$ , right side $= 2 + 0 = 2$
Suppose it is true when $n = k$ .Assume $\sum_{r=1}^{k} r \times 2^r$ ${}= 2 + (k - 1)2^{k+1}$ .
Add the next term, where $r = k + 1$ . $\sum_{r=1}^{k+1} r \times 2^r$ ${}= 2$ ${}+ (k - 1)2^{k+1}$ ${}+ (k + 1)2^{k+1}$
Collect the $2^{k+1}$ terms: $(k - 1) + (k + 1) = 2k$ . $= 2 + 2k \times 2^{k+1}$ ${}= 2 + k \times 2^{k+2}$
Rewrite it to match the formula exactly. $= 2$ ${}+ \bigl((k + 1) - 1\bigr)2^{(k+1)+1}$ , the formula with $n = k + 1$
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Proof

Medium

Divisibility by 4 · Year 12 · 5 marks

(a)
Prove by induction that $7^n - 3^n$ is divisible by 4 for all positive integers $n$ .
(5)

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Part (a)5 marks

Check the first case. $n = 1$ : $7 - 3 = 4$
Suppose it is true when $n = k$ .Assume $7^k - 3^k$ is divisible by 4.
Write the next case in terms of $7^k$ and $3^k$ . $7^{k+1} - 3^{k+1}$ ${}= 7 \times 7^k - 3 \times 3^k$
Split $7 \times 3^k$ off so the assumed expression appears: $-3 \times 3^k$ ${}= -7 \times 3^k + 4 \times 3^k$ . $= 7(7^k - 3^k) + 4 \times 3^k$
The first term by the assumption, the second because of the factor 4.Both terms are multiples of 4, so $7^{k+1} - 3^{k+1}$ is too.
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Proof

Hard

Induction with derivatives · Year 12 · 5 marks

(a)
Prove by induction that, for all positive integers $n$ , $\frac{\mathrm{d}^n}{\mathrm{d}x^n}\left(x\mathrm{e}^x\right) = (x + n)\mathrm{e}^x.$
(5)

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Part (a)5 marks

Use the product rule to check the first case. $n = 1$ : $\dfrac{\mathrm{d}}{\mathrm{d}x}(x\mathrm{e}^x)$ ${}= \mathrm{e}^x + x\mathrm{e}^x$ ${}= (x + 1)\mathrm{e}^x$
Suppose it is true for the $k$ th derivative.Assume $\dfrac{\mathrm{d}^k}{\mathrm{d}x^k}(x\mathrm{e}^x)$ ${}= (x + k)\mathrm{e}^x$ .
The next derivative is the derivative of the one we assumed. $\dfrac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(x\mathrm{e}^x)$ ${}= \dfrac{\mathrm{d}}{\mathrm{d}x}\bigl((x + k)\mathrm{e}^x\bigr)$
Product rule again, then factorise. This is the formula with $n = k + 1$ . $= \mathrm{e}^x$ ${}+ (x + k)\mathrm{e}^x$ ${}= \bigl(x + (k + 1)\bigr)\mathrm{e}^x$
It is true for $n = 1$ , and if it is true for one value it is true for the next, so it is true for every $n$ .So the statement is true for all positive integers $n$ .

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Complex numbers

Easy

Complex roots of a quadratic · Year 12 · 3 marks

(a)
Solve $z^2 + 6z + 13 = 0$ .
(3)

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Part (a)3 marks

Use the quadratic formula; the discriminant is negative, so the roots are complex. $z$ ${}= \dfrac{-6 \pm \sqrt{36 - 52}}{2}$
$\sqrt{-16} = 4\mathrm{i}$ . $= \dfrac{-6 \pm \sqrt{-16}}{2}$ ${}= \dfrac{-6 \pm 4\mathrm{i}}{2}$
The roots are a conjugate pair, as expected for real coefficients. $z = -3 + 2\mathrm{i}$ or $z = -3 - 2\mathrm{i}$

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Complex numbers

Easy

Modulus and argument · Year 12 · 3 marks

z = 1 + \sqrt{3}\,\mathrm{i}

(a)
Find $|z|$ .
(1)
(b)
Find $\arg z$ .
(1)
(c)
Write $z$ in modulus–argument form.
(1)

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Part (a)1 mark

The modulus is the distance from the origin. $|z|$ ${}= \sqrt{1^2 + (\sqrt{3})^2}$ ${}= 2$

Part (b)1 mark

$z$ is in the first quadrant, so no adjustment is needed. $\arg z$ ${}= \tan^{-1}\dfrac{\sqrt{3}}{1}$ ${}= \dfrac{\pi}{3}$

Part (c)1 mark

Modulus–argument form is $r(\cos\theta + \mathrm{i}\sin\theta)$ . $z$ ${}= 2\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right)$

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Complex numbers

Easy

The complex conjugate · Year 12 · 3 marks

z = 5 - 2\mathrm{i}

(a)
Write down $z^*$ .
(1)
(b)
Find $z + z^*$ and $zz^*$ .
(2)

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Part (a)1 mark

The conjugate changes the sign of the imaginary part. $z^* = 5 + 2\mathrm{i}$

Part (b)2 marks

The imaginary parts cancel, leaving twice the real part. $z + z^* = 10$
This is a difference of two squares, and equals $|z|^2$ . $zz^* = 25 - 4\mathrm{i}^2 = 29$

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Complex numbers

Easy

A circle in the Argand diagram · Year 12 · 2 marks

(a)
Describe the locus of points $z$ satisfying $|z - 2 - \mathrm{i}| = 3$ .
(2)

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Part (a)2 marks

$|z - a|$ is the distance from $z$ to the point $a$ . $|z - (2 + \mathrm{i})| = 3$
Every point exactly 3 away from $2 + \mathrm{i}$ .A circle with centre $(2, 1)$ and radius 3

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Complex numbers

Medium

De Moivre's theorem · Year 13 · 4 marks

(a)
Use de Moivre's theorem to find $(1 + \mathrm{i})^8$ .
(4)

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Part (a)4 marks

Modulus $\sqrt{2}$ , argument $\tfrac{\pi}{4}$ . $1 + \mathrm{i}$ ${}= \sqrt{2}\,\mathrm{e}^{\mathrm{i}\pi/4}$
Raise the modulus to the power 8 and multiply the argument by 8. $(1 + \mathrm{i})^8$ ${}= (\sqrt{2})^8\mathrm{e}^{2\pi\mathrm{i}}$
$\mathrm{e}^{2\pi\mathrm{i}} = 1$ . $= 16$

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Complex numbers

Medium

Cube roots · Year 13 · 4 marks

(a)
Solve $z^3 = 8\mathrm{i}$ , giving your answers in the form $r\mathrm{e}^{\mathrm{i}\theta}$ with $-\pi < \theta \le \pi$ .
(4)

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Part (a)4 marks

Write the right side in exponential form, including all the equivalent arguments. $8\mathrm{i}$ ${}= 8\mathrm{e}^{\mathrm{i}(\pi/2 + 2k\pi)}$
Take the cube root of the modulus and divide the argument by 3. $z$ ${}= 2\mathrm{e}^{\mathrm{i}(\pi/6 + 2k\pi/3)}$
Use $k = 0, 1, -1$ to keep the arguments in range. The three roots are equally spaced. $z$ ${}= 2\mathrm{e}^{\mathrm{i}\pi/6},\ 2\mathrm{e}^{5\mathrm{i}\pi/6},\ 2\mathrm{e}^{-\mathrm{i}\pi/2}$

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Complex numbers

Hard

Multiple angles · Year 13 · 5 marks

(a)
Use de Moivre's theorem to show that $\cos 3\theta$ ${}= 4\cos^3\theta - 3\cos\theta$ .
(5)

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Part (a)5 marks

De Moivre's theorem with $n = 3$ . $\cos 3\theta$ ${}+ \mathrm{i}\sin 3\theta$ ${}= (\cos\theta + \mathrm{i}\sin\theta)^3$
Expand with the binomial theorem, writing $c = \cos\theta$ and $s = \sin\theta$ . $= c^3$ ${}+ 3c^2(\mathrm{i}s)$ ${}+ 3c(\mathrm{i}s)^2$ ${}+ (\mathrm{i}s)^3$
$\mathrm{i}^2 = -1$ , so the real terms are $c^3$ and $-3cs^2$ .Real parts: $\cos 3\theta = c^3 - 3cs^2$
Use $s^2 = 1 - c^2$ to write everything in cosines. $= c^3 - 3c(1 - c^2)$
Collect the terms. $= 4\cos^3\theta - 3\cos\theta$

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Complex numbers

Hard

Two loci · Year 12 · 4 marks

(a)
Find the complex number $z$ that satisfies both $|z| = |z - 4|$ and $\arg z = \dfrac{\pi}{4}$ .
(4)

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Part (a)4 marks

Points the same distance from 0 and from 4. $|z| = |z - 4|$ : the perpendicular bisector of 0 and 4, the line $x = 2$
Points at angle $\tfrac{\pi}{4}$ from the positive real axis. $\arg z = \tfrac{\pi}{4}$ : the half-line $y = x$ , $x > 0$
Where the two loci cross: $x = 2$ , $y = 2$ . $z = 2 + 2\mathrm{i}$

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Matrices

Easy

Multiplying matrices · Year 12 · 2 marks

\mathbf{A}

{}= \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}

,

\mathbf{B}

{}= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

(a)
Find $\mathbf{AB}$ .
(2)

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Part (a)2 marks

Each entry is a row of $\mathbf{A}$ times a column of $\mathbf{B}$ . $\mathbf{AB}$ ${}= \begin{pmatrix} 1(0) + 2(1) & 1(1) + 2(0) \\ 3(0) + 4(1) & 3(1) + 4(0) \end{pmatrix}$
Multiplying by $\mathbf{B}$ on the right swaps the columns. $= \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$

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Matrices

Easy

Describing a transformation · Year 12 · 2 marks

(a)
Describe fully the transformation given by $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ .
(2)

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Part (a)2 marks

The columns show where the unit vectors go. $\begin{pmatrix} 1 \\ 0 \end{pmatrix} \to \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ , $\begin{pmatrix} 0 \\ 1 \end{pmatrix} \to \begin{pmatrix} -1 \\ 0 \end{pmatrix}$
Both unit vectors turn a quarter turn anticlockwise.A rotation of $90^\circ$ anticlockwise about the origin

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Matrices

Easy

Enlargement · Year 12 · 2 marks

(a)
Write down the matrix for an enlargement, scale factor 3, centre the origin.
(1)
(b)
State the area scale factor of this transformation.
(1)

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Part (a)1 mark

Each unit vector is multiplied by 3. $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$

Part (b)1 mark

The area scale factor is the determinant: $3 \times 3 - 0 = 9$ . $\det = 9$

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Matrices

Easy

A singular matrix · Year 12 · 2 marks

(a)
Find the value of $k$ for which $\begin{pmatrix} k & 2 \\ 3 & 6 \end{pmatrix}$ is singular.
(2)

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Part (a)2 marks

A matrix is singular when its determinant is zero. $6k - 6 = 0$
Solve for $k$ . $k = 1$

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Matrices

Medium

Solving equations with an inverse · Year 12 · 3 marks

(a)
Use $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}^{-1}$ ${}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ to solve $2x + y = 5$ , $5x + 3y = 13$ .
(3)

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Part (a)3 marks

Write the equations as $\mathbf{A}\mathbf{x} = \mathbf{b}$ , so $\mathbf{x}$ ${}= \mathbf{A}^{-1}\mathbf{b}$ . $\begin{pmatrix} x \\ y \end{pmatrix}$ ${}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 13 \end{pmatrix}$
Multiply out. $= \begin{pmatrix} 15 - 13 \\ -25 + 26 \end{pmatrix}$
Check: $4 + 1 = 5$ and $10 + 3 = 13$ . $x = 2,\ y = 1$

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Matrices

Medium

A 3 × 3 determinant · Year 12 · 3 marks

(a)
Find the determinant of $\begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{pmatrix}$ .
(3)

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Part (a)3 marks

Expand along the top row, alternating signs $+, -, +$ . $2\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}$ ${}- 1\begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix}$ ${}+ 0$
Work out each 2 × 2 determinant. $= 2(5) - 1(2)$
$= 8$

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Matrices

Medium

Invariant lines · Year 12 · 4 marks

\mathbf{M}

{}= \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}

(a)
Find the invariant lines of the form $y = mx$ .
(4)

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Part (a)4 marks

Transform a general point on the line. $\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix}$ ${}= \begin{pmatrix} x + 2mx \\ 3mx \end{pmatrix}$
For the line to be invariant, the image must also satisfy $y = mx$ . $3mx = m(x + 2mx)$
Divide by $x$ and solve. $2m^2 - 2m = 0 \Rightarrow m = 0$ or $m = 1$
$y = 0$ and $y = x$

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Matrices

Hard

Three planes · Year 12 · 5 marks

(a)
Show that the planes $x + y + z = 6$ , $x + 2y + 3z = 14$ and $2x + 3y + 4z = 20$ do not meet at a single point.
(2)
(b)
Describe how the three planes are arranged.
(3)

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Part (a)2 marks

If the matrix of coefficients is singular, there is no unique solution. $\det\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{pmatrix}$ ${}= 1(-1) - 1(-2) + 1(-1)$ ${}= 0$
The matrix is singular, so there is no single point of intersection.

Part (b)3 marks

Look for a combination of equations that gives the third.The first equation plus the second gives $2x + 3y + 4z = 20$ .
The third plane contains every point the first two share.This is the third equation, so the equations are consistent.
Consistent but singular means infinitely many solutions on a line.The planes form a sheaf: they all meet in one common line.

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Matrices

Hard

Combining transformations · Year 12 · 4 marks

\mathbf{P}

is a reflection in

y = x

and

\mathbf{Q}

is a rotation of

90^\circ

anticlockwise about the origin.

(a)
Find the matrix of the transformation $\mathbf{P}$ followed by $\mathbf{Q}$ .
(3)
(b)
Describe this as a single transformation.
(1)

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Part (a)3 marks

Write each matrix from where the unit vectors go. $\mathbf{P}$ ${}= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\ \mathbf{Q}$ ${}= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
$\mathbf{P}$ first means $\mathbf{P}$ goes on the right: $\mathbf{QP}$ . $\mathbf{QP}$ ${}= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$

Part (b)1 mark

It sends $x$ to $-x$ and keeps $y$ .A reflection in the $y$ -axis

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Further algebra and functions

Easy

Roots of a quadratic · Year 12 · 3 marks

The roots of

x^2 - 5x + 7 = 0

are

\alpha

and

\beta

.

(a)
Write down $\alpha + \beta$ and $\alpha\beta$ .
(1)
(b)
Find $\alpha^2 + \beta^2$ .
(2)

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Part (a)1 mark

Sum $= -\tfrac{b}{a}$ , product $= \tfrac{c}{a}$ . $\alpha + \beta$ ${}= 5,\ \alpha\beta$ ${}= 7$

Part (b)2 marks

Squaring the sum gives the squares plus $2\alpha\beta$ . $\alpha^2 + \beta^2$ ${}= (\alpha + \beta)^2$ ${}- 2\alpha\beta$
$= 25 - 14 = 11$

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Further algebra and functions

Easy

Roots of a quartic · Year 12 · 4 marks

The roots of

2x^4 - 3x^3 + x - 6 = 0

are

\alpha, \beta, \gamma, \delta

.

(a)
Write down $\sum\alpha$ , $\sum\alpha\beta$ , $\sum\alpha\beta\gamma$ and $\alpha\beta\gamma\delta$ .
(4)

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Part (a)4 marks

$-\tfrac{b}{a}$ , with $a = 2$ , $b = -3$ . $\sum\alpha = \tfrac{3}{2}$
$\tfrac{c}{a}$ : there is no $x^2$ term, so $c = 0$ . $\sum\alpha\beta = 0$
$-\tfrac{d}{a}$ , with $d = 1$ . $\sum\alpha\beta\gamma$ ${}= -\tfrac{1}{2}$
$\tfrac{e}{a}$ , with $e = -6$ . $\alpha\beta\gamma\delta = -3$

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Further algebra and functions

Easy

Using standard sums · Year 12 · 4 marks

(a)
Show that $\displaystyle\sum_{r=1}^{n} r(r + 2)$ ${}= \tfrac{1}{6}n(n + 1)(2n + 7)$ .
(4)

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Part (a)4 marks

Expand, then split into standard sums. $\sum r(r + 2) = \sum r^2 + 2\sum r$
Use the formulae for $\sum r^2$ and $\sum r$ . $= \tfrac{1}{6}n(n+1)(2n+1)$ ${}+ n(n+1)$
Take out the common factor $\tfrac{1}{6}n(n+1)$ . $= \tfrac{1}{6}n(n+1)\bigl((2n + 1) + 6\bigr)$
$= \tfrac{1}{6}n(n+1)(2n + 7)$

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Further algebra and functions

Easy

Sum of squares · Year 12 · 2 marks

(a)
Find $\displaystyle\sum_{r=1}^{10} r^2$ .
(2)

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Part (a)2 marks

Use $\sum r^2$ ${}= \tfrac{1}{6}n(n+1)(2n+1)$ with $n = 10$ . $\tfrac{1}{6}(10)(11)(21)$
$= 385$

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Further algebra and functions

Easy

A Maclaurin series · Year 13 · 2 marks

(a)
Find the Maclaurin series of $\mathrm{e}^{2x}$ up to and including the term in $x^3$ .
(2)

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Part (a)2 marks

Replace $x$ with $2x$ in the series for $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}= 1$ ${}+ 2x$ ${}+ \dfrac{(2x)^2}{2!}$ ${}+ \dfrac{(2x)^3}{3!}$
Simplify each term. $= 1 + 2x + 2x^2 + \tfrac{4}{3}x^3$

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Further algebra and functions

Medium

Transformed roots · Year 12 · 3 marks

The roots of

x^3 - 2x + 5 = 0

are

\alpha, \beta, \gamma

.

(a)
Find a cubic equation with integer coefficients whose roots are $2\alpha, 2\beta, 2\gamma$ .
(3)

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Part (a)3 marks

Each new root is double an old root.Let $w = 2x$ , so $x = \tfrac{w}{2}$
Substitute into the original equation. $\dfrac{w^3}{8} - w + 5 = 0$
Multiply by 8 to clear the fraction. $w^3 - 8w + 40 = 0$

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Further algebra and functions

Hard

Series for ln(cos x) · Year 13 · 5 marks

(a)
Using the series for $\cos x$ and $\ln(1 + x)$ , find the series for $\ln(\cos x)$ up to the term in $x^4$ .
(5)

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Part (a)5 marks

Write $\cos x$ as 1 plus a small quantity $u$ . $\cos x = 1 + u$ , where $u$ ${}= -\dfrac{x^2}{2}$ ${}+ \dfrac{x^4}{24}$
Use the standard series; higher powers of $u$ give terms beyond $x^4$ . $\ln(1 + u)$ ${}= u - \dfrac{u^2}{2} + \dots$
Only the $x^4$ term of $u^2$ is needed. $u^2 = \dfrac{x^4}{4} + \dots$
Substitute. $\ln(\cos x)$ ${}= -\dfrac{x^2}{2}$ ${}+ \dfrac{x^4}{24}$ ${}- \dfrac{x^4}{8}$
$\tfrac{1}{24} - \tfrac{3}{24}$ ${}= -\tfrac{2}{24}$ . $= -\dfrac{x^2}{2}$ ${}- \dfrac{x^4}{12}$

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Further calculus

Easy

Volume of revolution · Year 12 · 3 marks

(a)
The region under $y = x^2$ for $0 \le x \le 2$ is rotated through $2\pi$ about the $x$ -axis. Find the exact volume.
(3)

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Part (a)3 marks

Volume about the $x$ -axis is $\pi\int y^2\,\mathrm{d}x$ , and $y^2 = x^4$ . $V = \pi\int_0^2 x^4\,\mathrm{d}x$
Integrate. $= \pi\left[\dfrac{x^5}{5}\right]_0^2$
$= \dfrac{32\pi}{5}$

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Further calculus

Easy

Volume with a square root · Year 12 · 3 marks

(a)
The region under $y = \sqrt{x}$ for $0 \le x \le 4$ is rotated through $2\pi$ about the $x$ -axis. Find the exact volume.
(3)

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Part (a)3 marks

Squaring $\sqrt{x}$ gives $x$ . $V = \pi\int_0^4 x\,\mathrm{d}x$
$= \pi\left[\dfrac{x^2}{2}\right]_0^4$
$= 8\pi$

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Further calculus

Easy

An improper integral · Year 13 · 3 marks

(a)
Evaluate $\displaystyle\int_1^\infty \frac{1}{x^2}\,\mathrm{d}x$ .
(3)

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Part (a)3 marks

Replace the infinite limit with $t$ and integrate. $\displaystyle\int_1^t x^{-2}\,\mathrm{d}x$ ${}= \left[-\frac{1}{x}\right]_1^t$ ${}= 1 - \frac{1}{t}$
The integral converges because the limit exists.As $t \to \infty$ , $\tfrac{1}{t} \to 0$ , so the integral is 1

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Further calculus

Easy

Mean value · Year 13 · 2 marks

(a)
Find the mean value of $f(x) = x^2$ over $0 \le x \le 3$ .
(2)

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Part (a)2 marks

Mean value $= \dfrac{1}{b - a}\int_a^b f(x)\,\mathrm{d}x$ . $\dfrac{1}{3 - 0}\displaystyle\int_0^3 x^2\,\mathrm{d}x$
$= \tfrac{1}{3} \times 9 = 3$

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Further calculus

Easy

Differentiating arctan · Year 13 · 2 marks

(a)
Differentiate $\arctan 2x$ .
(2)

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Part (a)2 marks

Use $\dfrac{\mathrm{d}}{\mathrm{d}x}\arctan u$ ${}= \dfrac{1}{1 + u^2}$ and the chain rule. $\dfrac{1}{1 + (2x)^2} \times 2$
$= \dfrac{2}{1 + 4x^2}$

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Further calculus

Medium

An integral giving π · Year 13 · 2 marks

(a)
Evaluate $\displaystyle\int_0^1 \frac{1}{1 + x^2}\,\mathrm{d}x$ .
(2)

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Part (a)2 marks

$\dfrac{1}{1 + x^2}$ integrates to $\arctan x$ . $\bigl[\arctan x\bigr]_0^1$
$\arctan 1 = \tfrac{\pi}{4}$ and $\arctan 0 = 0$ . $= \dfrac{\pi}{4}$

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Further calculus

Medium

Volume about the y-axis · Year 12 · 3 marks

(a)
The region between $y = x^2$ , the $y$ -axis and the line $y = 4$ is rotated through $2\pi$ about the $y$ -axis. Find the exact volume.
(3)

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Part (a)3 marks

About the $y$ -axis use $\pi\int x^2\,\mathrm{d}y$ ; here $x^2 = y$ . $V$ ${}= \pi\int_0^4 x^2\,\mathrm{d}y$ ${}= \pi\int_0^4 y\,\mathrm{d}y$
$= \pi\left[\dfrac{y^2}{2}\right]_0^4$
$= 8\pi$

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Further calculus

Medium

An exponential improper integral · Year 13 · 3 marks

(a)
Show that $\displaystyle\int_0^\infty \mathrm{e}^{-2x}\,\mathrm{d}x$ converges and find its value.
(3)

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Part (a)3 marks

Integrate up to $t$ first. $\displaystyle\int_0^t \mathrm{e}^{-2x}\,\mathrm{d}x$ ${}= \left[-\tfrac{1}{2}\mathrm{e}^{-2x}\right]_0^t$ ${}= \tfrac{1}{2}$ ${}- \tfrac{1}{2}\mathrm{e}^{-2t}$
As $t \to \infty$ , $\mathrm{e}^{-2t} \to 0$ , so the integral converges to $\tfrac{1}{2}$

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Further calculus

Hard

An inverse sine integral · Year 13 · 3 marks

(a)
Evaluate exactly $\displaystyle\int_0^{3/2} \frac{1}{\sqrt{9 - x^2}}\,\mathrm{d}x$ .
(3)

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Part (a)3 marks

$\int \dfrac{1}{\sqrt{a^2 - x^2}}\,\mathrm{d}x$ ${}= \arcsin\dfrac{x}{a}$ , with $a = 3$ . $\left[\arcsin\dfrac{x}{3}\right]_0^{3/2}$
$= \arcsin\tfrac{1}{2}$ ${}= \dfrac{\pi}{6}$

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Further calculus

Hard

Partial fractions with a quadratic factor · Year 13 · 6 marks

(a)
Express $\dfrac{2x + 3}{x(x^2 + 1)}$ in partial fractions.
(3)
(b)
Hence find $\displaystyle\int \frac{2x + 3}{x(x^2 + 1)}\,\mathrm{d}x$ .
(3)

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Part (a)3 marks

A quadratic factor that does not factorise needs a linear numerator. $\dfrac{2x + 3}{x(x^2 + 1)}$ ${}= \dfrac{A}{x}$ ${}+ \dfrac{Bx + C}{x^2 + 1}$
Multiply through by the denominator. $2x + 3 = A(x^2 + 1) + (Bx + C)x$
$x = 0$ gives $A$ ; compare $x^2$ and $x$ coefficients for $B$ and $C$ . $A = 3,\ B = -3,\ C = 2$

Part (b)3 marks

Split the second fraction into two simpler ones. $\displaystyle\int \left(\frac{3}{x} - \frac{3x}{x^2 + 1} + \frac{2}{x^2 + 1}\right)\mathrm{d}x$
The middle term is $\tfrac{f'(x)}{f(x)}$ form; the last is arctan. $= 3\ln|x|$ ${}- \tfrac{3}{2}\ln(x^2 + 1)$ ${}+ 2\arctan x$ ${}+ c$

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Further vectors

Easy

Equation of a line · Year 12 · 2 marks

(a)
Find a vector equation of the line through $A(1, 2, 3)$ and $B(4, 0, 5)$ .
(2)

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Part (a)2 marks

The direction is $\mathbf{b} - \mathbf{a}$ . $\overrightarrow{AB}$ ${}= \begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$
A point on the line plus any multiple of the direction. $\mathbf{r}$ ${}= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ ${}+ \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$

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Further vectors

Easy

Equation of a plane · Year 12 · 2 marks

(a)
Find the Cartesian equation of the plane through $(2, 1, -1)$ with normal vector $\begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}$ .
(2)

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Part (a)2 marks

The normal's components are the coefficients of $x$ , $y$ and $z$ . $x + 3y + 2z = d$
Substitute the point to find $d$ . $d = 2 + 3 - 2 = 3$ , so $x + 3y + 2z = 3$

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Further vectors

Easy

Perpendicular vectors · Year 12 · 2 marks

(a)
Show that $\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 5 \\ 1 \end{pmatrix}$ are perpendicular.
(2)

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Part (a)2 marks

Find the scalar product. $2(1) + (-1)(5) + 3(1) = 0$
$\mathbf{a}\cdot\mathbf{b}$ ${}= |\mathbf{a}||\mathbf{b}|\cos\theta$ ${}= 0$ means $\theta = 90^\circ$ .The scalar product is zero, so the vectors are perpendicular.

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Further vectors

Easy

A line meeting a plane · Year 12 · 3 marks

(a)
Find where the line $\mathbf{r}$ ${}= \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ ${}+ t\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ meets the plane $x + y + z = 9$ .
(3)

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Part (a)3 marks

Substitute the line's coordinates into the plane's equation. $(1 + t) + t + (2 + t) = 9$
$3t + 3 = 9 \Rightarrow t = 2$
Put $t = 2$ back into the line. $(3, 2, 4)$

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Further vectors

Medium

Angle between a line and a plane · Year 12 · 4 marks

(a)
Find the acute angle between the line with direction $\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ and the plane $2x - y + 2z = 7$ .
(4)

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Part (a)4 marks

First find the angle with the plane's normal $(2, -1, 2)$ . $\cos\theta$ ${}= \dfrac{(1)(2) + (2)(-1) + (2)(2)}{3 \times 3}$ ${}= \dfrac{4}{9}$
$\theta = 63.6^\circ$
The normal is at right angles to the plane, so subtract from $90^\circ$ .Angle with the plane $= 90^\circ - 63.6^\circ$ ${}= 26.4^\circ$

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Further vectors

Medium

Distance from a point to a plane · Year 12 · 3 marks

(a)
Find the perpendicular distance from $(1, 2, 3)$ to the plane $2x - y + 2z = 10$ .
(3)

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Part (a)3 marks

Substitute the point into the plane's equation and divide by the normal's length. $\dfrac{|2(1) - 2 + 2(3) - 10|}{\sqrt{2^2 + 1^2 + 2^2}}$
$= \dfrac{|-4|}{3}$
$= \dfrac{4}{3}$

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Further vectors

Medium

Skew lines · Year 12 · 4 marks

(a)
Show that the lines $\mathbf{r}$ ${}= \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ ${}+ s\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ and $\mathbf{r}$ ${}= \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ ${}+ t\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ are skew.
(4)

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Part (a)4 marks

Skew means not parallel and not meeting.The directions are not multiples of each other, so the lines are not parallel.
Set the coordinates equal and solve two of the equations. $x$ : $1 + s = 2 \Rightarrow s = 1$ ; $y$ : $1 = t \Rightarrow t = 1$
Check the third equation. $z$ : $s = 1$ but $1 + t = 2$ , and $1 \ne 2$
The equations are inconsistent, so the lines do not meet: they are skew.

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Further vectors

Hard

Foot of a perpendicular · Year 12 · 5 marks

(a)
Find the foot of the perpendicular from $P(2, 3, 4)$ to the plane $x + 2y + 2z = 3$ .
(5)

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Part (a)5 marks

The perpendicular from $P$ goes in the direction of the plane's normal. $\mathbf{r}$ ${}= \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ ${}+ t\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$
Substitute into the plane to find where it lands. $(2 + t)$ ${}+ 2(3 + 2t)$ ${}+ 2(4 + 2t)$ ${}= 3$
$16 + 9t$ ${}= 3$ ${}\Rightarrow t$ ${}= -\tfrac{13}{9}$
Put $t$ back into the line.Foot $= \left(\tfrac{5}{9}, \tfrac{1}{9}, \tfrac{10}{9}\right)$

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Further vectors

Hard

Line of intersection of two planes · Year 12 · 5 marks

(a)
Find a vector equation of the line where the planes $x + y + z = 6$ and $x - y + 2z = 5$ meet.
(5)

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Part (a)5 marks

Use a parameter for one variable, then solve for the other two.Let $z = \lambda$ : $x + y = 6 - \lambda$ , $x - y = 5 - 2\lambda$
Add and subtract the two equations. $x$ ${}= \dfrac{11 - 3\lambda}{2},\ y$ ${}= \dfrac{1 + \lambda}{2}$
Separate the constants from the $\lambda$ terms. Any multiple of the direction is fine. $\mathbf{r}$ ${}= \begin{pmatrix} 5.5 \\ 0.5 \\ 0 \end{pmatrix}$ ${}+ \lambda\begin{pmatrix} -1.5 \\ 0.5 \\ 1 \end{pmatrix}$

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Polar coordinates

Easy

Polar to Cartesian · Year 13 · 2 marks

(a)
Convert the polar point $\left(4, \dfrac{\pi}{3}\right)$ to Cartesian coordinates.
(2)

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Part (a)2 marks

Use $x = r\cos\theta$ and $y = r\sin\theta$ . $x$ ${}= 4\cos\dfrac{\pi}{3}$ ${}= 2,\ y$ ${}= 4\sin\dfrac{\pi}{3}$ ${}= 2\sqrt{3}$

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Polar coordinates

Easy

Cartesian to polar · Year 13 · 2 marks

(a)
Convert the Cartesian point $(-1, 1)$ to polar coordinates with $0 \le \theta < 2\pi$ .
(2)

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Part (a)2 marks

$r$ is the distance from the origin. $r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$
The point is in the second quadrant, so $\theta = \pi - \tfrac{\pi}{4}$ . $\theta = \dfrac{3\pi}{4}$

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Polar coordinates

Easy

A circle in polar form · Year 13 · 3 marks

(a)
Find a Cartesian equation for $r = 4\cos\theta$ and describe the curve.
(3)

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Part (a)3 marks

Multiply by $r$ so you can use $r^2 = x^2 + y^2$ and $r\cos\theta = x$ . $r^2 = 4r\cos\theta$
$x^2 + y^2 = 4x$
Complete the square. $(x - 2)^2 + y^2 = 4$ : a circle, centre $(2, 0)$ , radius 2

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Polar coordinates

Easy

A line in polar form · Year 13 · 1 mark

(a)
Write the half-line $y = x$ , $x > 0$ , in polar form.
(1)

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Part (a)1 mark

Every point on it is at angle $\tfrac{\pi}{4}$ from the initial line, whatever $r$ is. $\theta = \dfrac{\pi}{4}$

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Polar coordinates

Easy

Area of a polar sector · Year 13 · 2 marks

(a)
Find the area enclosed by $r = 3$ and the half-lines $\theta = 0$ and $\theta = \dfrac{\pi}{3}$ .
(2)

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Part (a)2 marks

Area $= \tfrac{1}{2}\int r^2\,\mathrm{d}\theta$ . $\tfrac{1}{2}\displaystyle\int_0^{\pi/3} 9\,\mathrm{d}\theta$
Matches the sector formula $\tfrac{1}{2}r^2\theta$ . $= \dfrac{3\pi}{2}$

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Polar coordinates

Medium

One loop of a rose curve · Year 13 · 4 marks

(a)
Find the area of the loop of $r = \sin 2\theta$ for $0 \le \theta \le \dfrac{\pi}{2}$ .
(4)

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Part (a)4 marks

Use $\tfrac{1}{2}\int r^2\,\mathrm{d}\theta$ . $A$ ${}= \tfrac{1}{2}\displaystyle\int_0^{\pi/2} \sin^2 2\theta\,\mathrm{d}\theta$
$\sin^2 A$ ${}= \tfrac{1}{2}(1 - \cos 2A)$ . $= \tfrac{1}{4}\displaystyle\int_0^{\pi/2} (1 - \cos 4\theta)\,\mathrm{d}\theta$
$= \tfrac{1}{4}\left[\theta - \tfrac{1}{4}\sin 4\theta\right]_0^{\pi/2}$
$= \dfrac{\pi}{8}$

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Polar coordinates

Medium

Where two curves meet · Year 13 · 3 marks

(a)
Find the polar coordinates of the points where $r = 1 + \cos\theta$ and $r = 3\cos\theta$ meet, for $-\pi < \theta \le \pi$ , other than the pole.
(3)

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Part (a)3 marks

Set the two expressions for $r$ equal. $1 + \cos\theta$ ${}= 3\cos\theta$ ${}\Rightarrow \cos\theta$ ${}= \tfrac{1}{2}$
Find $r$ from either equation. $\left(\tfrac{3}{2}, \tfrac{\pi}{3}\right)$ and $\left(\tfrac{3}{2}, -\tfrac{\pi}{3}\right)$

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Polar coordinates

Hard

Tangents parallel to the initial line · Year 13 · 5 marks

(a)
Find the points on $r = 1 + \cos\theta$ , $0 < \theta < \pi$ , where the tangent is parallel to the initial line.
(5)

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Part (a)5 marks

A tangent parallel to the initial line is where $y$ is stationary. $y$ ${}= r\sin\theta$ ${}= \sin\theta$ ${}+ \sin\theta\cos\theta$
$\sin\theta\cos\theta$ ${}= \tfrac{1}{2}\sin 2\theta$ . $\dfrac{\mathrm{d}y}{\mathrm{d}\theta}$ ${}= \cos\theta + \cos 2\theta$
Set to zero and use $\cos 2\theta = 2\cos^2\theta - 1$ . $2\cos^2\theta + \cos\theta - 1$ ${}= 0$ ${}\Rightarrow (2\cos\theta - 1)(\cos\theta + 1)$ ${}= 0$
$\cos\theta = -1$ gives $\theta = \pi$ , outside the range. $\cos\theta$ ${}= \tfrac{1}{2}$ ${}\Rightarrow \theta$ ${}= \tfrac{\pi}{3}$
The point $\left(\tfrac{3}{2}, \tfrac{\pi}{3}\right)$

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Polar coordinates

Hard

Area of a limaçon · Year 13 · 5 marks

(a)
Find the exact area enclosed by $r = 2 + \sin\theta$ , $0 \le \theta < 2\pi$ .
(5)

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Part (a)5 marks

Square $r$ first. $A$ ${}= \tfrac{1}{2}\displaystyle\int_0^{2\pi} (4 + 4\sin\theta + \sin^2\theta)\,\mathrm{d}\theta$
Needed to integrate $\sin^2\theta$ . $\sin^2\theta$ ${}= \tfrac{1}{2}(1 - \cos 2\theta)$
A full turn of sine integrates to zero. $\displaystyle\int_0^{2\pi} 4\sin\theta\,\mathrm{d}\theta$ ${}= 0$
$4$ gives $8\pi$ , and $\sin^2\theta$ gives $\pi$ . $A = \tfrac{1}{2}(8\pi + \pi)$
$= \dfrac{9\pi}{2}$

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Hyperbolic functions

Easy

Evaluating sinh · Year 13 · 2 marks

(a)
Find the exact value of $\sinh(\ln 3)$ .
(2)

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Part (a)2 marks

Use $\sinh x$ ${}= \tfrac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})$ ; $\mathrm{e}^{-\ln 3} = \tfrac{1}{3}$ . $\dfrac{\mathrm{e}^{\ln 3} - \mathrm{e}^{-\ln 3}}{2}$ ${}= \dfrac{3 - \tfrac{1}{3}}{2}$
$= \dfrac{4}{3}$

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Hyperbolic functions

Easy

The hyperbolic identity · Year 13 · 3 marks

(a)
Prove that $\cosh^2 x - \sinh^2 x \equiv 1$ .
(3)

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Part (a)3 marks

Square the definition; $\mathrm{e}^x\mathrm{e}^{-x} = 1$ . $\cosh^2 x$ ${}= \tfrac{1}{4}(\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x})$
$\sinh^2 x$ ${}= \tfrac{1}{4}(\mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x})$
Everything cancels except the constants.Subtracting leaves $\tfrac{1}{4}(4) = 1$

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Hyperbolic functions

Easy

Differentiating hyperbolic functions · Year 13 · 3 marks

(a)
Differentiate $\cosh 3x$ .
(1)
(b)
Differentiate $x\sinh x$ .
(2)

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Part (a)1 mark

$\cosh$ differentiates to $\sinh$ (no minus sign), times 3 by the chain rule. $3\sinh 3x$

Part (b)2 marks

Product rule: differentiate $x$ , then $\sinh x$ . $\sinh x + x\cosh x$

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Hyperbolic functions

Easy

Inverse tanh as a logarithm · Year 13 · 2 marks

(a)
Express $\operatorname{artanh}\tfrac{1}{2}$ as a natural logarithm.
(2)

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Part (a)2 marks

Use $\operatorname{artanh} x$ ${}= \tfrac{1}{2}\ln\dfrac{1 + x}{1 - x}$ . $\tfrac{1}{2}\ln\dfrac{1 + \tfrac{1}{2}}{1 - \tfrac{1}{2}}$
$= \tfrac{1}{2}\ln 3$

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Hyperbolic functions

Medium

A hyperbolic equation · Year 13 · 5 marks

(a)
Solve $3\cosh x - \sinh x = 3$ .
(5)

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Part (a)5 marks

Write both in exponentials. $\tfrac{3}{2}(\mathrm{e}^x + \mathrm{e}^{-x})$ ${}- \tfrac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})$ ${}= 3$
Collect like terms and halve. $\mathrm{e}^x$ ${}+ 2\mathrm{e}^{-x}$ ${}= 3$
Multiply by $\mathrm{e}^x$ to get a quadratic in $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}- 3\mathrm{e}^x$ ${}+ 2$ ${}= 0$
$(\mathrm{e}^x - 1)(\mathrm{e}^x - 2)$ ${}= 0$
$x = 0$ or $x = \ln 2$

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Hyperbolic functions

Medium

Logarithmic form of arsinh · Year 13 · 4 marks

(a)
Show that $\operatorname{arsinh} x$ ${}= \ln\bigl(x + \sqrt{x^2 + 1}\bigr)$ .
(4)

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Part (a)4 marks

Rewrite the inverse as the original function.Let $y = \operatorname{arsinh} x$ , so $x$ ${}= \sinh y$ ${}= \tfrac{1}{2}(\mathrm{e}^y - \mathrm{e}^{-y})$
Multiply by $2\mathrm{e}^y$ . $\mathrm{e}^{2y}$ ${}- 2x\mathrm{e}^y$ ${}- 1$ ${}= 0$
Quadratic formula in $\mathrm{e}^y$ . $\mathrm{e}^y$ ${}= x \pm \sqrt{x^2 + 1}$
$x - \sqrt{x^2 + 1}$ is always negative, so reject it. $\mathrm{e}^y > 0$ , so $y$ ${}= \ln\bigl(x + \sqrt{x^2 + 1}\bigr)$

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Hyperbolic functions

Medium

An arsinh integral · Year 13 · 3 marks

(a)
Evaluate exactly $\displaystyle\int_0^2 \frac{1}{\sqrt{x^2 + 4}}\,\mathrm{d}x$ .
(3)

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Part (a)3 marks

$\int \dfrac{1}{\sqrt{x^2 + a^2}}\,\mathrm{d}x$ ${}= \operatorname{arsinh}\dfrac{x}{a}$ . $\left[\operatorname{arsinh}\dfrac{x}{2}\right]_0^2$
Use the logarithmic form. $= \operatorname{arsinh} 1$ ${}= \ln(1 + \sqrt{2})$

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Hyperbolic functions

Hard

An equation with cosh 2x · Year 13 · 5 marks

(a)
Solve $\cosh 2x - 3\sinh x = 0$ , giving answers as logarithms.
(5)

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Part (a)5 marks

Use $\cosh 2x = 1 + 2\sinh^2 x$ to get one function. $1 + 2\sinh^2 x - 3\sinh x = 0$
Factorise the quadratic in $\sinh x$ . $(2\sinh x - 1)(\sinh x - 1) = 0$
$x$ ${}= \operatorname{arsinh} 1$ ${}= \ln(1 + \sqrt{2})$
$\tfrac{1}{2}$ ${}+ \sqrt{\tfrac{1}{4} + 1}$ ${}= \tfrac{1 + \sqrt{5}}{2}$ . $x$ ${}= \operatorname{arsinh}\tfrac{1}{2}$ ${}= \ln\dfrac{1 + \sqrt{5}}{2}$

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Hyperbolic functions

Hard

A hyperbolic substitution · Year 13 · 6 marks

(a)
Use the substitution $x = \sinh u$ to find $\displaystyle\int_0^1 \sqrt{1 + x^2}\,\mathrm{d}x$ exactly.
(6)

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Part (a)6 marks

$\sqrt{1 + \sinh^2 u} = \cosh u$ and $\mathrm{d}x = \cosh u\,\mathrm{d}u$ . $\displaystyle\int \cosh u \cdot \cosh u\,\mathrm{d}u$ ${}= \int \cosh^2 u\,\mathrm{d}u$
Use $\cosh 2u = 2\cosh^2 u - 1$ . $= \tfrac{1}{2}\displaystyle\int (1 + \cosh 2u)\,\mathrm{d}u$
$\tfrac{1}{2}\sinh 2u$ ${}= \sinh u\cosh u$ . $= \tfrac{1}{2}(u + \sinh u\cosh u)$
Change the limits to $u$ .Limits: $u = 0$ to $u = \operatorname{arsinh} 1$ , where $\sinh u = 1$ , $\cosh u = \sqrt{2}$
$= \tfrac{1}{2}\bigl(\ln(1 + \sqrt{2}) + \sqrt{2}\bigr)$

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Differential equations

Easy

First-order linear equation · Year 13 · 4 marks

(a)
Find the general solution of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}+ 2y$ ${}= \mathrm{e}^x$ .
(4)

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Part (a)4 marks

The equation is $y' + P(x)y = Q(x)$ with $P = 2$ .Integrating factor $= \mathrm{e}^{\int 2\,\mathrm{d}x}$ ${}= \mathrm{e}^{2x}$
Multiply through by the integrating factor. $\dfrac{\mathrm{d}}{\mathrm{d}x}(y\mathrm{e}^{2x})$ ${}= \mathrm{e}^{3x}$
Integrate both sides. $y\mathrm{e}^{2x}$ ${}= \tfrac{1}{3}\mathrm{e}^{3x}$ ${}+ c$
Divide by $\mathrm{e}^{2x}$ . $y$ ${}= \tfrac{1}{3}\mathrm{e}^x$ ${}+ c\mathrm{e}^{-2x}$

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Easy

Two real roots · Year 13 · 3 marks

(a)
Find the general solution of $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}- 5\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}+ 6y$ ${}= 0$ .
(3)

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Part (a)3 marks

Auxiliary equation. $m^2 - 5m + 6 = 0$
$m = 2$ or $m = 3$
Two different real roots give two exponentials. $y$ ${}= A\mathrm{e}^{2x}$ ${}+ B\mathrm{e}^{3x}$

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Easy

Imaginary roots · Year 13 · 3 marks

(a)
Find the general solution of $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}+ 9y$ ${}= 0$ .
(3)

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Part (a)3 marks

Auxiliary equation. $m^2 + 9$ ${}= 0$ ${}\Rightarrow m$ ${}= \pm 3\mathrm{i}$
Roots $\pm q\mathrm{i}$ give $A\cos qx + B\sin qx$ . $y = A\cos 3x + B\sin 3x$

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Easy

A repeated root · Year 13 · 3 marks

(a)
Find the general solution of $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}- 4\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}+ 4y$ ${}= 0$ .
(3)

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Part (a)3 marks

Auxiliary equation. $m^2 - 4m + 4 = 0 \Rightarrow m = 2$ (repeated)
A repeated root needs the extra factor $x$ . $y = (A + Bx)\mathrm{e}^{2x}$

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Easy

A particular integral · Year 13 · 4 marks

(a)
Find the general solution of $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}+ y$ ${}= 3x + 1$ .
(4)

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Part (a)4 marks

Solve $m^2 + 1 = 0$ first.Complementary function: $A\cos x + B\sin x$
The right side is linear, so try a linear particular integral.Try $y = ax + b$ : $0 + ax + b = 3x + 1$ , so $a = 3,\ b = 1$
General solution = complementary function + particular integral. $y = A\cos x + B\sin x + 3x + 1$

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Medium

Complex roots with conditions · Year 13 · 5 marks

(a)
Solve $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}+ 2\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}+ 5y$ ${}= 0$ , given that $y = 1$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 1$ when $x = 0$ .
(5)

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Part (a)5 marks

Auxiliary equation, by the quadratic formula. $m^2 + 2m + 5$ ${}= 0$ ${}\Rightarrow m$ ${}= -1 \pm 2\mathrm{i}$
Roots $p \pm q\mathrm{i}$ give $\mathrm{e}^{px}(A\cos qx + B\sin qx)$ . $y$ ${}= \mathrm{e}^{-x}(A\cos 2x + B\sin 2x)$
Use $y(0)$ , then differentiate and use $y'(0)$ . $A = 1$ ; $y'(0)$ ${}= -A + 2B$ ${}= 1$ ${}\Rightarrow B$ ${}= 1$
$y$ ${}= \mathrm{e}^{-x}(\cos 2x + \sin 2x)$

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Simple harmonic motion · Year 13 · 4 marks

A particle moves so that

\ddot{x} = -4x

. At

t = 0

,

x = 3

and

\dot{x} = 0

.

(a)
Find $x$ in terms of $t$ .
(3)
(b)
State the period of the motion.
(1)

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Part (a)3 marks

Auxiliary equation $m^2 + 4 = 0$ gives $m = \pm 2\mathrm{i}$ . $x = A\cos 2t + B\sin 2t$
Use the starting conditions. $A = 3$ , and $\dot{x}(0)$ ${}= 2B$ ${}= 0$ ${}\Rightarrow B$ ${}= 0$
$x = 3\cos 2t$

Part (b)1 mark

The period is $\tfrac{2\pi}{\omega}$ with $\omega = 2$ . $\dfrac{2\pi}{2} = \pi$ seconds

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Differential equations

Hard

Modelling speed · Year 13 · 7 marks

A cyclist's speed

v

m s⁻¹ at time

t

seconds is modelled by

\frac{\mathrm{d}v}{\mathrm{d}t} + \frac{v}{t + 1} = 2,

with

v = 0

when

t = 0

.

(a)
Find $v$ in terms of $t$ .
(5)
(b)
Find the speed after 3 seconds and comment on the model for large $t$ .
(2)

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Part (a)5 marks

Standard first-order linear form.Integrating factor $= \mathrm{e}^{\int \frac{1}{t+1}\,\mathrm{d}t}$ ${}= t + 1$
Multiply through by the integrating factor. $\dfrac{\mathrm{d}}{\mathrm{d}t}\bigl(v(t + 1)\bigr)$ ${}= 2(t + 1)$
Integrate, then use the starting condition. $v(t + 1) = (t + 1)^2 + c$ ; $t = 0$ , $v = 0$ gives $c = -1$
$v = t + 1 - \dfrac{1}{t + 1}$

Part (b)2 marks

Substitute $t = 3$ . $v$ ${}= 4 - \tfrac{1}{4}$ ${}= 3.75\ \mathrm{m\,s^{-1}}$
A good model would level off at a top speed.The speed keeps increasing without limit, which is unrealistic for a cyclist.

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Proof

Easy

Consecutive integers · Year 12 · 2 marks

(a)
Prove that the sum of any three consecutive integers is a multiple of 3.
(2)

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Part (a)2 marks

Call the integers $n$ , $n + 1$ and $n + 2$ , then add them. $n + (n + 1) + (n + 2) = 3n + 3$
Taking out the factor 3 shows the sum is 3 times a whole number. $= 3(n + 1)$ , which is a multiple of 3

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Easy

Disproof by counter-example · Year 12 · 2 marks

(a)
Show that the statement “ $n^2 + n + 41$ is prime for every positive integer $n$ ” is false.
(2)

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Part (a)2 marks

One example where it fails is enough to disprove a statement. $n = 40$ : $40^2 + 40 + 41 = 1681$
Say clearly why the example breaks the statement. $1681 = 41^2$ , which is not prime, so the statement is false

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Proof by deduction · Year 12 · 2 marks

(a)
Prove that $(n + 3)^2 - (n - 3)^2$ is a multiple of 12 for every integer $n$ .
(2)

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Part (a)2 marks

Expand both brackets and simplify. $(n^2 + 6n + 9)$ ${}- (n^2 - 6n + 9)$ ${}= 12n$
$12n$ is 12 times an integer, so it is a multiple of 12

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Easy

Even products · Year 12 · 3 marks

(a)
Prove that $n^2 - n$ is even for every integer $n$ .
(3)

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Part (a)3 marks

Factorise: this is a product of two consecutive integers. $n^2 - n = n(n - 1)$
Of any two consecutive integers, one must be even.One of $n$ and $n - 1$ is even
An even number times any integer is even, so $n^2 - n$ is even

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Medium

Proof by exhaustion · Year 12 · 4 marks

(a)
Prove that the square of any integer is either a multiple of 3 or one more than a multiple of 3.
(4)

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Part (a)4 marks

Split into every possible case: the remainder on dividing by 3 is 0, 1 or 2.Every integer is $3k$ , $3k + 1$ or $3k + 2$
A multiple of 3. $(3k)^2 = 3(3k^2)$
One more than a multiple of 3. $(3k + 1)^2 = 3(3k^2 + 2k) + 1$
Also one more than a multiple of 3, so every case is covered. $(3k + 2)^2$ ${}= 9k^2 + 12k + 4$ ${}= 3(3k^2 + 4k + 1) + 1$

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Medium

√2 is irrational · Year 13 · 5 marks

(a)
Prove by contradiction that $\sqrt{2}$ is irrational.
(5)

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Part (a)5 marks

Proof by contradiction: assume the opposite and show it leads to something impossible.Assume $\sqrt{2} = \dfrac{a}{b}$ , with $a$ and $b$ integers that have no common factor
Square both sides. If $a$ were odd, $a^2$ would be odd. $a^2 = 2b^2$ , so $a^2$ is even, so $a$ is even
Substitute back in.Write $a = 2c$ : $4c^2 = 2b^2 \Rightarrow b^2 = 2c^2$
The same argument as for $a$ .So $b$ is even too
End by saying what the contradiction is.Then $a$ and $b$ share the factor 2, which contradicts our assumption, so $\sqrt{2}$ is irrational

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Medium

Contradiction with even numbers · Year 13 · 3 marks

(a)
Prove by contradiction that if $n^2$ is even, then $n$ is even.
(3)

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Part (a)3 marks

Assume the opposite of what you want to prove.Assume $n^2$ is even but $n$ is odd, so $n = 2k + 1$
Square the odd number. $n^2$ ${}= 4k^2 + 4k + 1$ ${}= 2(2k^2 + 2k) + 1$ , which is odd
This contradicts $n^2$ being even, so $n$ must be even

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Proof

Hard

Infinitely many primes · Year 13 · 5 marks

(a)
Prove by contradiction that there are infinitely many prime numbers.
(5)

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Part (a)5 marks

Assume the opposite.Assume there are finitely many primes, $p_1, p_2, \dots, p_n$
Build a number from every prime on the list.Let $N = p_1 p_2 \cdots p_n + 1$
So none of the listed primes divides $N$ .Dividing $N$ by any $p_i$ leaves remainder 1
Every integer greater than 1 has a prime factor.So $N$ is prime, or has a prime factor not on the list
Either way there is a prime missing from the list: a contradiction, so there are infinitely many primes

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Proof

Hard

An irrational logarithm · Year 13 · 4 marks

(a)
Prove by contradiction that $\log_2 3$ is irrational.
(4)

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Part (a)4 marks

$\log_2 3 > 0$ , so both can be taken positive.Assume $\log_2 3 = \dfrac{p}{q}$ , with $p, q$ positive integers
Undo the logarithm, then raise both sides to the power $q$ . $2^{p/q} = 3 \Rightarrow 2^p = 3^q$
A power of 2 is even; a power of 3 is odd. $2^p$ is even but $3^q$ is odd
An even number cannot equal an odd number, so $\log_2 3$ is irrational

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Algebra and functions

Easy

Surds · Year 12 · 4 marks

(a)
Simplify $\sqrt{48} + \sqrt{27}$ .
(2)
(b)
Rationalise the denominator of $\dfrac{4}{3 - \sqrt{5}}$ .
(2)

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Part (a)2 marks

Take out the largest square factor: $48 = 16 \times 3$ , $27 = 9 \times 3$ . $\sqrt{48} = 4\sqrt{3}$ and $\sqrt{27} = 3\sqrt{3}$
$4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}$

Part (b)2 marks

Multiply top and bottom by $3 + \sqrt{5}$ ; the bottom becomes $9 - 5$ . $\dfrac{4(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}$ ${}= \dfrac{4(3 + \sqrt{5})}{4}$
$= 3 + \sqrt{5}$

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Simultaneous equations · Year 12 · 4 marks

(a)
Solve the simultaneous equations $y = x + 1$ and $x^2 + y^2 = 13$ .
(4)

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Part (a)4 marks

Substitute the linear equation into the quadratic one. $x^2 + (x + 1)^2 = 13$
Expand and simplify. $2x^2 + 2x - 12$ ${}= 0$ ${}\Rightarrow x^2 + x - 6$ ${}= 0$
$(x + 3)(x - 2) = 0$
Find each $y$ from $y = x + 1$ . $x = 2, y = 3$ or $x = -3, y = -2$

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The factor theorem · Year 12 · 4 marks

\mathrm{f}(x) = x^3 - 4x^2 + x + 6

(a)
Show that $(x - 2)$ is a factor of $\mathrm{f}(x)$ .
(1)
(b)
Factorise $\mathrm{f}(x)$ completely.
(3)

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Part (a)1 mark

Factor theorem: if $\mathrm{f}(a) = 0$ then $(x - a)$ is a factor. $\mathrm{f}(2) = 8 - 16 + 2 + 6 = 0$ , so $(x - 2)$ is a factor

Part (b)3 marks

Divide by $(x - 2)$ , or compare coefficients. $\mathrm{f}(x)$ ${}= (x - 2)(x^2 - 2x - 3)$
Factorise the quadratic. $= (x - 2)(x - 3)(x + 1)$

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A quadratic inequality · Year 12 · 3 marks

(a)
Solve $x^2 - x - 12 < 0$ .
(3)

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Part (a)3 marks

Find the critical values first. $(x - 4)(x + 3)$ ${}= 0$ ${}\Rightarrow x$ ${}= 4, x$ ${}= -3$
Sketch it: we want where it is negative.The graph is a ∪ shape, below the axis between its roots
$-3 < x < 4$

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Partial fractions · Year 13 · 3 marks

(a)
Express $\dfrac{5x + 1}{(x - 1)(x + 2)}$ in partial fractions.
(3)

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Part (a)3 marks

Write it as $\dfrac{A}{x - 1}$ ${}+ \dfrac{B}{x + 2}$ and multiply through. $5x + 1 = A(x + 2) + B(x - 1)$
Choose values of $x$ that make one bracket zero. $x = 1$ : $6 = 3A \Rightarrow A = 2$ ; $x = -2$ : $-9 = -3B \Rightarrow B = 3$
$\dfrac{2}{x - 1}$ ${}+ \dfrac{3}{x + 2}$

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Algebra and functions

Medium

Composite and inverse functions · Year 13 · 5 marks

\mathrm{f}(x) = 2x + 3

and

\mathrm{g}(x) = x^2

, for all real

x

.

(a)
Find $\mathrm{f}^{-1}(x)$ .
(1)
(b)
Solve $\mathrm{fg}(x) = \mathrm{gf}(x)$ .
(4)

Worked answerHide answer

Part (a)1 mark

Swap: $x = 2y + 3$ , then make $y$ the subject. $\mathrm{f}^{-1}(x)$ ${}= \dfrac{x - 3}{2}$

Part (b)4 marks

$\mathrm{fg}$ means do $\mathrm{g}$ first, then $\mathrm{f}$ . $\mathrm{fg}(x) = 2x^2 + 3$ and $\mathrm{gf}(x) = (2x + 3)^2$
Set equal, expand and simplify. $2x^2 + 3$ ${}= 4x^2 + 12x + 9$ ${}\Rightarrow x^2 + 6x + 3$ ${}= 0$
Complete the square or use the formula. $x = -3 \pm \sqrt{6}$

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Algebra and functions

Medium

A modulus equation · Year 13 · 4 marks

(a)
Solve $|2x - 1| = x + 4$ .
(4)

Worked answerHide answer

Part (a)4 marks

Case 1: the inside is positive. $2x - 1 = x + 4 \Rightarrow x = 5$
Case 2: the inside is negative. Both answers give $x + 4 > 0$ , so both work. $-(2x - 1)$ ${}= x + 4$ ${}\Rightarrow x$ ${}= -1$

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Algebra and functions

Hard

Roots and the discriminant · Year 12 · 5 marks

(a)
Find the values of $k$ for which $kx^2 + 4x + (k - 3) = 0$ has two distinct real roots.
(5)

Worked answerHide answer

Part (a)5 marks

Two distinct real roots means the discriminant is positive. $b^2 - 4ac > 0$ : $16 - 4k(k - 3) > 0$
Divide by $-4$ , which reverses the inequality. $k^2 - 3k - 4 < 0$
Critical values $-1$ and $4$ ; a ∪ shape is negative between them. $(k - 4)(k + 1) < 0$ , so $-1 < k < 4$
If $k = 0$ the equation is not a quadratic and has only one root. $-1 < k < 4$ , $k \ne 0$

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Algebra and functions

Hard

Modelling a ball's height · Year 12 · 8 marks

The height of a ball,

h

metres,

t

seconds after it is thrown is modelled by

h = 1.5 + 14t - 4.9t^2

.

(a)
Write $h$ in the form $a - 4.9(t - b)^2$ .
(3)
(b)
State the greatest height and when it happens.
(2)
(c)
Find when the ball hits the ground, and give one limitation of the model.
(3)

Worked answerHide answer

Part (a)3 marks

Take $-4.9$ out of the $t$ terms: $14 \div 4.9 = \tfrac{20}{7}$ . $h$ ${}= -4.9\left(t^2 - \tfrac{20}{7}t\right)$ ${}+ 1.5$
Complete the square. $= -4.9\left(t - \tfrac{10}{7}\right)^2$ ${}+ 4.9 \times \tfrac{100}{49}$ ${}+ 1.5$
$h$ ${}= 11.5$ ${}- 4.9\left(t - \tfrac{10}{7}\right)^2$

Part (b)2 marks

The squared bracket is never negative, so $h$ is largest when it is zero.11.5 m, when $t = \tfrac{10}{7} \approx 1.43$ s

Part (c)3 marks

Set $h = 0$ and take the positive root. $4.9\left(t - \tfrac{10}{7}\right)^2$ ${}= 11.5$ ${}\Rightarrow t$ ${}= \tfrac{10}{7}$ ${}+ \sqrt{\tfrac{11.5}{4.9}}$ ${}= 2.96$ s
Any sensible simplification the model makes.It ignores air resistance (or: the ball is treated as a particle)

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Coordinate geometry in the (x, y) plane

Easy

Equation of a line · Year 12 · 3 marks

(a)
Find the equation of the line through $(2, 3)$ and $(6, 11)$ in the form $y = mx + c$ .
(3)

Worked answerHide answer

Part (a)3 marks

Gradient is change in $y$ over change in $x$ . $m = \dfrac{11 - 3}{6 - 2} = 2$
Use $y - y_1 = m(x - x_1)$ . $y - 3 = 2(x - 2)$
$y = 2x - 1$

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Coordinate geometry in the (x, y) plane

Easy

A perpendicular line · Year 12 · 3 marks

(a)
Find the equation of the line perpendicular to $y = 3x + 2$ through $(3, 1)$ , in the form $ax + by + c = 0$ .
(3)

Worked answerHide answer

Part (a)3 marks

Perpendicular gradients multiply to $-1$ .Gradient $= -\tfrac{1}{3}$
$y - 1 = -\tfrac{1}{3}(x - 3)$
Multiply by 3 and rearrange. $x + 3y - 6 = 0$

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Coordinate geometry in the (x, y) plane

Easy

Centre and radius · Year 12 · 3 marks

(a)
Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$ .
(3)

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Part (a)3 marks

Complete the square in $x$ and in $y$ . $(x - 3)^2$ ${}- 9$ ${}+ (y + 2)^2$ ${}- 4$ ${}- 12$ ${}= 0$
$(x - 3)^2 + (y + 2)^2 = 25$
Compare with $(x - a)^2 + (y - b)^2 = r^2$ .Centre $(3, -2)$ , radius 5

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Coordinate geometry in the (x, y) plane

Easy

Midpoint and length · Year 12 · 3 marks

A(-1, 4)

and

B(5, -4)

(a)
Find the midpoint of $AB$ and the length $AB$ .
(3)

Worked answerHide answer

Part (a)3 marks

Average the coordinates.Midpoint $= \left(\dfrac{-1 + 5}{2}, \dfrac{4 - 4}{2}\right)$ ${}= (2, 0)$
Pythagoras on the differences. $AB = \sqrt{6^2 + 8^2}$
$= 10$

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Coordinate geometry in the (x, y) plane

Easy

Parametric to Cartesian · Year 13 · 2 marks

(a)
A curve has parametric equations $x = t + 1$ , $y = t^2$ . Find its Cartesian equation.
(2)

Worked answerHide answer

Part (a)2 marks

Make $t$ the subject of one equation. $t = x - 1$
Substitute into the other. $y = (x - 1)^2$

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Coordinate geometry in the (x, y) plane

Medium

Tangent to a circle · Year 12 · 4 marks

The circle

C

has equation

(x - 3)^2 + (y + 2)^2 = 25

.

(a)
Find the equation of the tangent to $C$ at $(6, 2)$ .
(4)

Worked answerHide answer

Part (a)4 marks

The radius goes from the centre $(3, -2)$ to the point.Radius gradient $= \dfrac{2 - (-2)}{6 - 3}$ ${}= \dfrac{4}{3}$
A tangent is perpendicular to the radius.Tangent gradient $= -\tfrac{3}{4}$
$y - 2 = -\tfrac{3}{4}(x - 6)$
$3x + 4y - 26 = 0$

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Coordinate geometry in the (x, y) plane

Medium

A line touching a circle · Year 12 · 4 marks

(a)
Find the values of $k$ for which the line $y = x + k$ is a tangent to $x^2 + y^2 = 8$ .
(4)

Worked answerHide answer

Part (a)4 marks

Substitute the line into the circle. $x^2 + (x + k)^2$ ${}= 8$ ${}\Rightarrow 2x^2$ ${}+ 2kx$ ${}+ k^2$ ${}- 8$ ${}= 0$
A tangent meets the circle once, so the discriminant is zero. $4k^2 - 8(k^2 - 8) = 0$
$k = \pm 4$

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Coordinate geometry in the (x, y) plane

Medium

An ellipse in parametric form · Year 13 · 3 marks

(a)
A curve has parametric equations $x = 2\cos t$ , $y = 3\sin t$ . Find its Cartesian equation.
(3)

Worked answerHide answer

Part (a)3 marks

Make the trig functions the subjects. $\cos t = \dfrac{x}{2}$ , $\sin t = \dfrac{y}{3}$
Use the identity to remove $t$ . $\cos^2 t + \sin^2 t = 1$
$\dfrac{x^2}{4}$ ${}+ \dfrac{y^2}{9}$ ${}= 1$

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Coordinate geometry in the (x, y) plane

Hard

A circle through three points · Year 12 · 6 marks

A(0, 2)

,

B(4, 6)

and

C(4, 2)

(a)
Show that angle $ACB$ is a right angle.
(2)
(b)
Hence find the equation of the circle through $A$ , $B$ and $C$ .
(4)

Worked answerHide answer

Part (a)2 marks

Look at the gradients: 0 and undefined. $AC$ is horizontal ( $y = 2$ ) and $CB$ is vertical ( $x = 4$ )
So $AC$ and $CB$ are perpendicular

Part (b)4 marks

The angle in a semicircle is a right angle. $AB$ is a diameter
Centre = midpoint of $AB = (2, 4)$
Distance from the centre to $A$ . $r^2 = (2 - 0)^2 + (4 - 2)^2 = 8$
$(x - 2)^2 + (y - 4)^2 = 8$

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Coordinate geometry in the (x, y) plane

Hard

A chord of a circle · Year 12 · 6 marks

(a)
The line $2x + y = 10$ meets the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ at $A$ and $B$ . Find the exact length $AB$ .
(6)

Worked answerHide answer

Part (a)6 marks

Substitute the line into the circle. $y = 10 - 2x$ : $x^2$ ${}+ (10 - 2x)^2$ ${}- 4x$ ${}- 2(10 - 2x)$ ${}- 20$ ${}= 0$
$5x^2 - 40x + 60$ ${}= 0$ ${}\Rightarrow x^2 - 8x + 12$ ${}= 0$
$x = 2$ or $x = 6$
Find each $y$ from the line. $A(2, 6)$ , $B(6, -2)$
$AB$ ${}= \sqrt{4^2 + 8^2}$ ${}= \sqrt{80}$ ${}= 4\sqrt{5}$

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Sequences and series

Easy

Binomial expansion · Year 12 · 3 marks

(a)
Expand $(1 + 2x)^4$ fully.
(3)

Worked answerHide answer

Part (a)3 marks

Use row 4 of Pascal's triangle: 1, 4, 6, 4, 1. $1$ ${}+ 4(2x)$ ${}+ 6(2x)^2$ ${}+ 4(2x)^3$ ${}+ (2x)^4$
Simplify each term. $= 1 + 8x + 24x^2 + 32x^3 + 16x^4$

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Sequences and series

Easy

A binomial coefficient · Year 12 · 2 marks

(a)
Find the coefficient of $x^3$ in the expansion of $(2 - x)^6$ .
(2)

Worked answerHide answer

Part (a)2 marks

The $x^3$ term uses $\binom{6}{3}$ and $2^{6-3}$ . $\dbinom{6}{3} \times 2^3 \times (-x)^3$
$20 \times 8 \times (-1) = -160$

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Sequences and series

Easy

An arithmetic series · Year 13 · 3 marks

An arithmetic series has first term 5 and common difference 3.

(a)
Find the 20th term.
(1)
(b)
Find the sum of the first 20 terms.
(2)

Worked answerHide answer

Part (a)1 mark

$u_n = a + (n - 1)d$ . $5 + 19 \times 3 = 62$

Part (b)2 marks

$S_n$ ${}= \tfrac{n}{2}\bigl(2a + (n - 1)d\bigr)$ . $S_{20}$ ${}= \tfrac{20}{2}\bigl(2(5) + 19(3)\bigr)$
$= 670$

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Sequences and series

Easy

Sum to infinity · Year 13 · 2 marks

(a)
A geometric series has first term 12 and common ratio $\tfrac{1}{3}$ . Find its sum to infinity.
(2)

Worked answerHide answer

Part (a)2 marks

$S_\infty = \dfrac{a}{1 - r}$ , which works because $|r| < 1$ . $S_\infty$ ${}= \dfrac{12}{1 - \tfrac{1}{3}}$
$= 18$

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Sequences and series

Easy

A recurrence relation · Year 13 · 2 marks

(a)
A sequence has $u_1 = 2$ and $u_{n+1} = 3u_n - 1$ . Find $u_2$ , $u_3$ and $u_4$ .
(2)

Worked answerHide answer

Part (a)2 marks

Use each term to find the next. $u_2 = 5$ , $u_3 = 14$
$u_4 = 41$

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Sequences and series

Medium

A general binomial expansion · Year 13 · 4 marks

(a)
Expand $\sqrt{1 + 4x}$ up to the term in $x^2$ , and state the values of $x$ for which it is valid.
(4)

Worked answerHide answer

Part (a)4 marks

Use $(1 + x)^n$ ${}= 1$ ${}+ nx$ ${}+ \tfrac{n(n-1)}{2!}x^2$ with $4x$ in place of $x$ . $(1 + 4x)^{1/2}$ ${}= 1$ ${}+ \tfrac{1}{2}(4x)$ ${}+ \dfrac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2!}(4x)^2$
$-\tfrac{1}{8} \times 16 = -2$ . $= 1 + 2x - 2x^2$
It needs $|4x| < 1$ .Valid for $|x| < \tfrac{1}{4}$

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Sequences and series

Medium

When does a sum pass 1000? · Year 13 · 4 marks

(a)
Find the least number of terms of $3 + 7 + 11 + \dots$ needed for the sum to exceed 1000.
(4)

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Part (a)4 marks

Arithmetic series with $a = 3$ , $d = 4$ . $S_n$ ${}= \tfrac{n}{2}\bigl(6 + 4(n - 1)\bigr)$ ${}= n(2n + 1)$
Solve $n(2n + 1) > 1000$ , or test values either side. $n = 22$ : $990$ ; $n = 23$ : $1081$
$n = 23$

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Sequences and series

Hard

Modelling population growth · Year 13 · 6 marks

A town had 2000 people at the start of 2024. The population is modelled as growing by 4% each year.

(a)
Write down an expression for the population $n$ years after the start of 2024.
(1)
(b)
Find the first year in which the model predicts more than 3000 people.
(4)
(c)
Give one reason why the model may not be reliable in the long term.
(1)

Worked answerHide answer

Part (a)1 mark

Growing by 4% means multiplying by 1.04 each year. $2000 \times 1.04^n$

Part (b)4 marks

$2000 \times 1.04^n > 3000$ ${}\Rightarrow 1.04^n > 1.5$
Take logs of both sides. $n > \dfrac{\ln 1.5}{\ln 1.04}$ ${}= 10.34$
The first whole number of years after 10.34. $n = 11$ , so during 2035

Part (c)1 mark

Any sensible reason.Growth cannot carry on for ever: housing and space are limited

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Sequences and series

Hard

Finding a common ratio · Year 13 · 6 marks

(a)
A geometric series has sum to infinity 9, and the sum of its first two terms is 5. Find the possible values of the first term and the common ratio.
(6)

Worked answerHide answer

Part (a)6 marks

Write both facts as equations. $\dfrac{a}{1 - r} = 9$ and $a + ar = 5$
Substitute for $a$ in the second equation. $a = 9(1 - r)$ , so $9(1 - r)(1 + r) = 5$
$(1 - r)(1 + r) = 1 - r^2$ . $1 - r^2$ ${}= \tfrac{5}{9}$ ${}\Rightarrow r^2$ ${}= \tfrac{4}{9}$
Both have $|r| < 1$ , so both are valid. $r = \tfrac{2}{3}, a = 3$ or $r = -\tfrac{2}{3}, a = 15$

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Trigonometry

Easy

The sine rule · Year 12 · 3 marks

In triangle

ABC

,

A = 40^\circ

,

B = 65^\circ

and

a = 8

cm.

(a)
Find $b$ .
(3)

Worked answerHide answer

Part (a)3 marks

Sine rule: each side over the sine of the opposite angle. $\dfrac{b}{\sin 65^\circ}$ ${}= \dfrac{8}{\sin 40^\circ}$
$b$ ${}= \dfrac{8\sin 65^\circ}{\sin 40^\circ}$
$= 11.3$ cm

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Trigonometry

Easy

The cosine rule · Year 12 · 2 marks

(a)
Two sides of a triangle are 5 cm and 7 cm, with $60^\circ$ between them. Find the third side.
(2)

Worked answerHide answer

Part (a)2 marks

Cosine rule, because we know two sides and the angle between them. $c^2$ ${}= 5^2$ ${}+ 7^2$ ${}- 2(5)(7)\cos 60^\circ$ ${}= 39$
$c = \sqrt{39} = 6.24$ cm

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Trigonometry

Easy

Solving a sine equation · Year 12 · 2 marks

(a)
Solve $\sin x = 0.5$ for $0^\circ \le x < 360^\circ$ .
(2)

Worked answerHide answer

Part (a)2 marks

The calculator value. $x = 30^\circ$
Sine is also positive in the second quadrant. $x$ ${}= 180^\circ - 30^\circ$ ${}= 150^\circ$

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Trigonometry

Easy

Area of a triangle · Year 12 · 2 marks

(a)
Find the exact area of a triangle with sides 5 cm and 7 cm and an angle of $60^\circ$ between them.
(2)

Worked answerHide answer

Part (a)2 marks

Area $= \tfrac{1}{2}ab\sin C$ . $\tfrac{1}{2}(5)(7)\sin 60^\circ$
$\sin 60^\circ$ ${}= \tfrac{\sqrt{3}}{2}$ . $= \dfrac{35\sqrt{3}}{4}$ cm²

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Trigonometry

Easy

Arcs and sectors · Year 13 · 2 marks

A sector of a circle has radius 6 cm and angle 0.8 radians.

(a)
Find the arc length and the area of the sector.
(2)

Worked answerHide answer

Part (a)2 marks

$s = r\theta$ , with $\theta$ in radians.Arc $= 6 \times 0.8 = 4.8$ cm
$A = \tfrac{1}{2}r^2\theta$ .Area $= \tfrac{1}{2}(36)(0.8) = 14.4$ cm²

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Trigonometry

Medium

A double angle equation · Year 13 · 5 marks

(a)
Solve $\sin 2x = \cos x$ for $0 \le x < 2\pi$ .
(5)

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Part (a)5 marks

Use $\sin 2x = 2\sin x\cos x$ . $2\sin x\cos x - \cos x = 0$
Factorise. Do not divide by $\cos x$ , or you lose solutions. $\cos x(2\sin x - 1) = 0$
$\cos x = 0$ : $x$ ${}= \tfrac{\pi}{2}, \tfrac{3\pi}{2}$
$\sin x = \tfrac{1}{2}$ : $x$ ${}= \tfrac{\pi}{6}, \tfrac{5\pi}{6}$

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Trigonometry

Medium

The R form · Year 13 · 4 marks

(a)
Write $3\sin x + 4\cos x$ in the form $R\sin(x + \alpha)$ , with $R > 0$ and $0 < \alpha < \tfrac{\pi}{2}$ .
(3)
(b)
State the maximum value of $3\sin x + 4\cos x$ .
(1)

Worked answerHide answer

Part (a)3 marks

Expand $R\sin(x + \alpha)$ ${}= R\sin x\cos\alpha$ ${}+ R\cos x\sin\alpha$ and compare. $R\cos\alpha = 3$ , $R\sin\alpha = 4$
$R = \sqrt{3^2 + 4^2} = 5$
$\alpha$ ${}= \tan^{-1}\tfrac{4}{3}$ ${}= 0.927$

Part (b)1 mark

The largest value of $\sin$ is 1.5

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Trigonometry

Hard

Solving with the R form · Year 13 · 5 marks

(a)
Solve $3\sin x + 4\cos x = 2$ for $0^\circ \le x < 360^\circ$ .
(5)

Worked answerHide answer

Part (a)5 marks

Use $R = 5$ , $\alpha = 53.13^\circ$ . $5\sin(x + 53.13^\circ) = 2$
$\sin(x + 53.13^\circ) = 0.4$
The range for $x + 53.13^\circ$ is $53.13^\circ$ to $413.13^\circ$ ; $23.58^\circ$ is too small. $x + 53.13^\circ = 156.42^\circ$ or $383.58^\circ$
$x = 103.3^\circ$ or $x = 330.4^\circ$

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Trigonometry

Hard

Modelling tides · Year 13 · 6 marks

The depth of water in a harbour,

D

metres,

t

hours after midnight is modelled by

D = 5 + 2\sin(30t)^\circ

.

(a)
Find the greatest depth and the first time it happens.
(2)
(b)
A boat needs at least 6 m of water. Find the times between midnight and noon when the depth is exactly 6 m.
(4)

Worked answerHide answer

Part (a)2 marks

$\sin$ is at most 1.Greatest depth $= 5 + 2 = 7$ m
$30t = 90 \Rightarrow t = 3$ : 3 am

Part (b)4 marks

Set $D = 6$ . $\sin(30t)^\circ = 0.5$
Both solutions in the range $0 \le 30t \le 360$ . $30t = 30$ or $150$
The boat can enter between 1 am and 5 am. $t = 1$ and $t = 5$ : 1 am and 5 am

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Exponentials and logarithms

Easy

An exponential equation · Year 12 · 2 marks

(a)
Solve $3^x = 20$ , giving $x$ to 3 significant figures.
(2)

Worked answerHide answer

Part (a)2 marks

Take logs of both sides: $x\ln 3 = \ln 20$ . $x = \dfrac{\ln 20}{\ln 3}$
$x = 2.73$

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Exponentials and logarithms

Easy

Laws of logarithms · Year 12 · 3 marks

(a)
Write $2\log 3 + \log 4 - \log 6$ as a single logarithm.
(3)

Worked answerHide answer

Part (a)3 marks

$n\log a = \log a^n$ . $2\log 3 = \log 9$
Adding logs multiplies; subtracting divides. $\log\dfrac{9 \times 4}{6}$
$= \log 6$

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Exponentials and logarithms

Easy

Natural logarithms · Year 12 · 2 marks

(a)
Solve $\mathrm{e}^{2x} = 5$ , giving the exact answer.
(2)

Worked answerHide answer

Part (a)2 marks

$\ln$ undoes $\mathrm{e}$ . $2x = \ln 5$
$x = \tfrac{1}{2}\ln 5$

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Exponentials and logarithms

Easy

A logarithm equation · Year 12 · 2 marks

(a)
Solve $\log_2(x + 3) = 4$ .
(2)

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Part (a)2 marks

$\log_a y = b$ means $y = a^b$ . $x + 3 = 2^4 = 16$
$x = 13$

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Exponentials and logarithms

Easy

Logs in terms of p and q · Year 12 · 2 marks

\log_a 2 = p

and

\log_a 3 = q

(a)
Write $\log_a 12$ in terms of $p$ and $q$ .
(2)

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Part (a)2 marks

Write 12 using only 2s and 3s: $12 = 2^2 \times 3$ . $\log_a 12$ ${}= \log_a(2^2 \times 3)$ ${}= 2\log_a 2 + \log_a 3$
$= 2p + q$

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Exponentials and logarithms

Medium

A hidden quadratic · Year 12 · 4 marks

(a)
Solve $2^{2x} - 5 \times 2^x + 4 = 0$ .
(4)

Worked answerHide answer

Part (a)4 marks

$2^{2x} = (2^x)^2$ .Let $y = 2^x$ : $y^2 - 5y + 4 = 0$
$(y - 1)(y - 4) = 0$
$2^x = 1 \Rightarrow x = 0$ ; $2^x = 4 \Rightarrow x = 2$

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Exponentials and logarithms

Medium

Adding logarithms · Year 12 · 4 marks

(a)
Solve $\log_3 x + \log_3(x - 6) = 3$ .
(4)

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Part (a)4 marks

Combine the logs, then undo the log. $\log_3\bigl(x(x - 6)\bigr)$ ${}= 3$ ${}\Rightarrow x(x - 6)$ ${}= 27$
$x^2 - 6x - 27$ ${}= 0$ ${}\Rightarrow (x - 9)(x + 3)$ ${}= 0$
Reject $x = -3$ : you cannot take the log of a negative number. $x = 9$

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Exponentials and logarithms

Medium

Using a straight-line graph · Year 12 · 4 marks

Two variables are thought to follow

y = ab^x

. A graph of

\ln y

against

x

is a straight line with gradient 0.5 and intercept 1.2.

(a)
Find $a$ and $b$ .
(4)

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Part (a)4 marks

Take logs of $y = ab^x$ : this has the form $Y = c + mx$ . $\ln y = \ln a + x\ln b$
The intercept is $\ln a$ . $\ln a$ ${}= 1.2$ ${}\Rightarrow a$ ${}= \mathrm{e}^{1.2}$ ${}= 3.32$
The gradient is $\ln b$ . $\ln b$ ${}= 0.5$ ${}\Rightarrow b$ ${}= \mathrm{e}^{0.5}$ ${}= 1.65$

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Exponentials and logarithms

Hard

The value of a car · Year 12 · 7 marks

The value of a car, £

V

,

t

years after it is bought is modelled by

V = 18000\mathrm{e}^{-0.15t}

.

(a)
Find the value after 4 years.
(2)
(b)
Find when the value first falls below £5000.
(3)
(c)
Find the rate at which the value is falling when $t = 4$ .
(2)

Worked answerHide answer

Part (a)2 marks

Substitute $t = 4$ . $18000\mathrm{e}^{-0.6} = 9878.6$
About £9880

Part (b)3 marks

$\mathrm{e}^{-0.15t} < \dfrac{5000}{18000}$
Take logs. Dividing by $-0.15$ reverses the inequality. $-0.15t < \ln\tfrac{5}{18}$ ${}\Rightarrow t > \dfrac{\ln 3.6}{0.15}$
$t > 8.54$ years

Part (c)2 marks

$\dfrac{\mathrm{d}}{\mathrm{d}t}\mathrm{e}^{kt}$ ${}= k\mathrm{e}^{kt}$ . $\dfrac{\mathrm{d}V}{\mathrm{d}t}$ ${}= -2700\mathrm{e}^{-0.15t}$
At $t = 4$ : falling by about £1480 per year

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Differentiation

Easy

Differentiating powers · Year 12 · 3 marks

(a)
Differentiate $y = 4x^3 - 5x + \dfrac{7}{x}$ .
(3)

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Part (a)3 marks

Write fractions as negative powers first. $y = 4x^3 - 5x + 7x^{-1}$
Multiply by the power, then reduce the power by one. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 12x^2 - 5 - 7x^{-2}$

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Differentiation

Easy

Gradient with a square root · Year 12 · 3 marks

(a)
Find the gradient of $y = \sqrt{x} + 3x^2$ at $x = 4$ .
(3)

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Part (a)3 marks

$\sqrt{x} = x^{1/2}$ . $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= \tfrac{1}{2}x^{-1/2} + 6x$
$4^{-1/2} = \tfrac{1}{2}$ .At $x = 4$ : $\tfrac{1}{4} + 24 = 24.25$

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Differentiation

Easy

Equation of a tangent · Year 12 · 3 marks

(a)
Find the equation of the tangent to $y = x^2 - 3x + 2$ at the point where $x = 4$ .
(3)

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Part (a)3 marks

Find the point and the gradient there. $y = 16 - 12 + 2 = 6$ ; $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 2x - 3$ ${}= 5$
$y - 6 = 5(x - 4)$
$y = 5x - 14$

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Differentiation

Easy

The chain rule · Year 13 · 2 marks

(a)
Differentiate $y = (3x - 1)^5$ .
(2)

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Part (a)2 marks

Differentiate the outside, then multiply by the derivative of the inside. $5(3x - 1)^4 \times 3$
$= 15(3x - 1)^4$

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Differentiation

Medium

The product rule · Year 13 · 5 marks

(a)
Find the coordinates of the stationary points of $y = x^2\mathrm{e}^{2x}$ .
(5)

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Part (a)5 marks

Product rule: $u'v + uv'$ with $u = x^2$ , $v = \mathrm{e}^{2x}$ . $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 2x\mathrm{e}^{2x}$ ${}+ 2x^2\mathrm{e}^{2x}$
Set to zero and factorise. $\mathrm{e}^{2x}$ is never zero. $2x\mathrm{e}^{2x}(1 + x) = 0$
$(0, 0)$ and $\left(-1, \mathrm{e}^{-2}\right)$

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Differentiation

Medium

The quotient rule · Year 13 · 4 marks

(a)
Show that $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{x}{x^2 + 1}\right)$ ${}= \dfrac{1 - x^2}{(x^2 + 1)^2}$ , and find the $x$ -coordinates of the stationary points.
(4)

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Part (a)4 marks

Quotient rule: $\dfrac{vu' - uv'}{v^2}$ . $\dfrac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2}$
$= \dfrac{1 - x^2}{(x^2 + 1)^2}$
A fraction is zero when its top is zero. $1 - x^2 = 0 \Rightarrow x = \pm 1$

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Differentiation

Hard

Making the biggest box · Year 12 · 7 marks

An open box is made from a 24 cm square of card by cutting a square of side

x

cm from each corner and folding up the sides.

(a)
Show that the volume is $V = 4x^3 - 96x^2 + 576x$ .
(2)
(b)
Find the maximum volume, showing that it is a maximum.
(5)

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Part (a)2 marks

The base is $24 - 2x$ each way and the height is $x$ . $V = x(24 - 2x)^2$
$= x(576 - 96x + 4x^2)$ ${}= 4x^3 - 96x^2 + 576x$

Part (b)5 marks

$\dfrac{\mathrm{d}V}{\mathrm{d}x}$ ${}= 12x^2 - 192x + 576$ ${}= 12(x - 4)(x - 12)$
When $x = 12$ the base has width 0. $x = 4$ ( $x = 12$ gives no box)
Negative second derivative means a maximum. $\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2}$ ${}= 24x - 192$ ${}= -96 < 0$ , so a maximum
$V = 4(16)^2 = 1024$ cm³

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Differentiation

Hard

Implicit differentiation · Year 13 · 6 marks

A curve has equation

x^2 + xy + y^2 = 7

.

(a)
Find the equation of the tangent at $(1, 2)$ in the form $ax + by + c = 0$ .
(6)

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Part (a)6 marks

Differentiate each term with respect to $x$ ; $xy$ needs the product rule. $2x$ ${}+ y$ ${}+ x\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}+ 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 0$
Collect the $\tfrac{\mathrm{d}y}{\mathrm{d}x}$ terms. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= -\dfrac{2x + y}{x + 2y}$
At $(1, 2)$ : $-\tfrac{4}{5}$
$y - 2$ ${}= -\tfrac{4}{5}(x - 1)$ ${}\Rightarrow 4x + 5y - 14$ ${}= 0$

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Integration

Easy

Integrating powers · Year 12 · 2 marks

(a)
Find $\displaystyle\int (6x^2 - 4x + 3)\,\mathrm{d}x$ .
(2)

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Part (a)2 marks

Add one to the power and divide by the new power. Remember $+c$ . $2x^3 - 2x^2 + 3x + c$

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Integration

Easy

Integrating roots and fractions · Year 12 · 3 marks

(a)
Find $\displaystyle\int \left(x^{1/2} + \frac{2}{x^2}\right)\mathrm{d}x$ .
(3)

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Part (a)3 marks

Write the fraction as a negative power. $\displaystyle\int (x^{1/2} + 2x^{-2})\,\mathrm{d}x$
Dividing by $\tfrac{3}{2}$ means multiplying by $\tfrac{2}{3}$ . $\tfrac{2}{3}x^{3/2} - 2x^{-1} + c$

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Integration

Easy

Finding the equation of a curve · Year 12 · 3 marks

(a)
A curve has $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 3x^2 - 2$ and passes through $(1, 4)$ . Find its equation.
(3)

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Part (a)3 marks

Integrate the gradient. $y = x^3 - 2x + c$
Use the point to find $c$ . $4 = 1 - 2 + c \Rightarrow c = 5$
$y = x^3 - 2x + 5$

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Integration

Easy

Area under a curve · Year 12 · 3 marks

(a)
Find the area between $y = x^2 + 1$ , the $x$ -axis and the lines $x = 0$ and $x = 3$ .
(3)

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Part (a)3 marks

$\displaystyle\int_0^3 (x^2 + 1)\,\mathrm{d}x$ ${}= \left[\frac{x^3}{3} + x\right]_0^3$
$= 9 + 3 = 12$

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Integration

Medium

Area between a curve and a line · Year 12 · 5 marks

(a)
Find the area enclosed by $y = 4x - x^2$ and $y = x$ .
(5)

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Part (a)5 marks

Find where they meet. $4x - x^2$ ${}= x$ ${}\Rightarrow x$ ${}= 0, x$ ${}= 3$
Top minus bottom: the curve is above the line here. $\displaystyle\int_0^3 (4x - x^2 - x)\,\mathrm{d}x$ ${}= \int_0^3 (3x - x^2)\,\mathrm{d}x$
$\left[\tfrac{3}{2}x^2 - \tfrac{1}{3}x^3\right]_0^3$ ${}= 13.5 - 9$
$= 4.5$

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Integration

Medium

Integration by substitution · Year 13 · 3 marks

(a)
Use the substitution $u = x^2 + 1$ to find $\displaystyle\int x(x^2 + 1)^3\,\mathrm{d}x$ .
(3)

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Part (a)3 marks

Replace everything with $u$ , including $\mathrm{d}x$ . $\mathrm{d}u = 2x\,\mathrm{d}x$ , so $\displaystyle\int \tfrac{1}{2}u^3\,\mathrm{d}u$
$= \tfrac{1}{8}u^4$
Put $x$ back in. $= \tfrac{1}{8}(x^2 + 1)^4 + c$

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Integration

Medium

Integration by parts · Year 13 · 4 marks

(a)
Evaluate $\displaystyle\int_0^1 x\mathrm{e}^x\,\mathrm{d}x$ .
(4)

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Part (a)4 marks

Parts: $\int uv' = uv - \int u'v$ . Choose $u = x$ so it disappears when differentiated. $u = x$ , $v' = \mathrm{e}^x$ : $x\mathrm{e}^x$ ${}- \displaystyle\int \mathrm{e}^x\,\mathrm{d}x$
$= \bigl[x\mathrm{e}^x - \mathrm{e}^x\bigr]_0^1$
$= (\mathrm{e} - \mathrm{e})$ ${}- (0 - 1)$
$= 1$

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Integration

Hard

Parts with a logarithm · Year 13 · 5 marks

(a)
Find the exact value of $\displaystyle\int_1^{\mathrm{e}} x\ln x\,\mathrm{d}x$ .
(5)

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Part (a)5 marks

Choose $u = \ln x$ : we can differentiate $\ln x$ but not easily integrate it. $u = \ln x$ , $v' = x$ : $\tfrac{1}{2}x^2\ln x$ ${}- \displaystyle\int \tfrac{1}{2}x\,\mathrm{d}x$
$= \left[\tfrac{1}{2}x^2\ln x - \tfrac{1}{4}x^2\right]_1^{\mathrm{e}}$
$\ln \mathrm{e} = 1$ , $\ln 1 = 0$ . $= \left(\tfrac{1}{2}\mathrm{e}^2 - \tfrac{1}{4}\mathrm{e}^2\right)$ ${}- \left(0 - \tfrac{1}{4}\right)$
$= \tfrac{1}{4}(\mathrm{e}^2 + 1)$

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Integration

Hard

A growth model · Year 13 · 6 marks

A population

P

grows at a rate proportional to its size. When

t = 0

,

P = 500

; when

t = 2

,

P = 800

.

(a)
Show that $P = 500\mathrm{e}^{kt}$ and find $k$ to 3 decimal places.
(4)
(b)
Predict the population when $t = 5$ .
(2)

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Part (a)4 marks

Separate the variables. $\dfrac{\mathrm{d}P}{\mathrm{d}t}$ ${}= kP$ ${}\Rightarrow \displaystyle\int \frac{1}{P}\,\mathrm{d}P$ ${}= \int k\,\mathrm{d}t$
$\ln P$ ${}= kt + c$ ${}\Rightarrow P$ ${}= A\mathrm{e}^{kt}$ ; $t = 0$ gives $A = 500$
$800$ ${}= 500\mathrm{e}^{2k}$ ${}\Rightarrow k$ ${}= \tfrac{1}{2}\ln 1.6$
$k = 0.235$

Part (b)2 marks

Use the exact value of $k$ . $500\mathrm{e}^{5k}$ ${}= 500 \times 1.6^{2.5}$
$= 1619$

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Numerical methods

Easy

Locating a root · Year 13 · 2 marks

\mathrm{f}(x) = x^3 - 2x - 5

(a)
Show that $\mathrm{f}(x) = 0$ has a root between $x = 2$ and $x = 3$ .
(2)

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Part (a)2 marks

Work out both values. $\mathrm{f}(2) = -1$ , $\mathrm{f}(3) = 16$
Mention both the sign change and continuity.There is a change of sign and $\mathrm{f}$ is continuous, so there is a root between 2 and 3

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Numerical methods

Easy

Using an iteration · Year 13 · 3 marks

(a)
Use $x_{n+1} = \sqrt[3]{2x_n + 5}$ with $x_0 = 2$ to find $x_1$ , $x_2$ and $x_3$ to 4 decimal places.
(3)

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Part (a)3 marks

Put each answer back in to get the next. $x_1 = \sqrt[3]{9} = 2.0801$
$x_2 = 2.0924$
$x_3 = 2.0942$

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Numerical methods

Easy

Newton–Raphson · Year 13 · 3 marks

(a)
Use the Newton–Raphson method with $x_0 = 2$ to find $x_2$ for $\mathrm{f}(x) = x^2 - 5$ .
(3)

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Part (a)3 marks

$x_{n+1}$ ${}= x_n$ ${}- \dfrac{\mathrm{f}(x_n)}{\mathrm{f}'(x_n)}$ . $x_{n+1}$ ${}= x_n$ ${}- \dfrac{x_n^2 - 5}{2x_n}$
$x_1 = 2 - \dfrac{-1}{4} = 2.25$
Close to $\sqrt{5} = 2.23607$ . $x_2$ ${}= 2.25 - \dfrac{0.0625}{4.5}$ ${}= 2.2361$

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Numerical methods

Easy

The trapezium rule · Year 13 · 4 marks

(a)
Use the trapezium rule with 4 strips to estimate $\displaystyle\int_0^2 \sqrt{1 + x^2}\,\mathrm{d}x$ .
(4)

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Part (a)4 marks

Strip width and the five $y$ -values. $h = 0.5$ ; $y$ ${}= 1,\ 1.1180,\ 1.4142,\ 1.8028,\ 2.2361$
$\tfrac{h}{2}\bigl[\text{ends} + 2(\text{middles})\bigr]$ . $\tfrac{0.5}{2}\bigl[1 + 2.2361 + 2(1.1180 + 1.4142 + 1.8028)\bigr]$
$= 2.977$

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Numerical methods

Easy

A root between 0 and 1 · Year 13 · 2 marks

(a)
Show that $\mathrm{e}^x = 3 - x$ has a root between 0 and 1.
(2)

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Part (a)2 marks

Rearrange to $\mathrm{f}(x) = 0$ first.Let $\mathrm{f}(x)$ ${}= \mathrm{e}^x + x - 3$ : $\mathrm{f}(0) = -2$ , $\mathrm{f}(1) = 0.718$
Sign change and $\mathrm{f}$ is continuous, so there is a root

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Numerical methods

Medium

Rearranging for an iteration · Year 13 · 4 marks

(a)
Show that $x^3 - 2x - 5 = 0$ can be written as $x = \sqrt{2 + \dfrac{5}{x}}$ .
(2)
(b)
Use $x_{n+1}$ ${}= \sqrt{2 + \dfrac{5}{x_n}}$ with $x_0 = 2$ to find $x_1$ , $x_2$ and $x_3$ .
(2)

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Part (a)2 marks

Divide by $x$ . $x^3$ ${}= 2x + 5$ ${}\Rightarrow x^2$ ${}= 2 + \dfrac{5}{x}$
Take the positive square root (the root is positive)

Part (b)2 marks

$x_1 = 2.1213$ , $x_2 = 2.0873$
The values swing either side of the root as they close in. $x_3 = 2.0965$

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Numerical methods

Medium

Newton–Raphson on a cubic · Year 13 · 4 marks

(a)
Use the Newton–Raphson method once with $x_0 = 2.1$ on $\mathrm{f}(x) = x^3 - 2x - 5$ , giving $x_1$ to 4 decimal places.
(4)

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Part (a)4 marks

$\mathrm{f}'(x) = 3x^2 - 2$
$\mathrm{f}(2.1) = 0.061$ , $\mathrm{f}'(2.1) = 11.23$
$x_1 = 2.1 - \dfrac{0.061}{11.23}$
$= 2.0946$

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Numerical methods

Medium

How good is the trapezium rule? · Year 13 · 5 marks

(a)
Use the trapezium rule with 4 strips to estimate $\displaystyle\int_1^3 \frac{1}{x}\,\mathrm{d}x$ .
(3)
(b)
The exact value is $\ln 3 = 1.099$ . Explain why the trapezium rule gives an overestimate here.
(2)

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Part (a)3 marks

$h = 0.5$ ; $y$ ${}= 1,\ 0.6667,\ 0.5,\ 0.4,\ 0.3333$
$0.25\bigl[1 + 0.3333 + 2(0.6667 + 0.5 + 0.4)\bigr]$
$= 1.117$

Part (b)2 marks

The curve $y = \tfrac{1}{x}$ bends upwards (it is convex)
So each trapezium's top lies above the curve

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Numerical methods

Hard

When Newton–Raphson fails · Year 13 · 5 marks

\mathrm{f}(x) = x^3 - 3x + 1

(a)
Explain why the Newton–Raphson method fails with $x_0 = 1$ .
(2)
(b)
Use $x_0 = 1.5$ to find $x_1$ .
(3)

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Part (a)2 marks

$\mathrm{f}'(1) = 3 - 3 = 0$
The tangent is horizontal, so it never meets the $x$ -axis (we would divide by zero)

Part (b)3 marks

$\mathrm{f}(1.5) = -0.125$ , $\mathrm{f}'(1.5) = 3.75$
$x_1 = 1.5 - \dfrac{-0.125}{3.75}$
$= 1.5333$

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Numerical methods

Hard

Will an iteration converge? · Year 13 · 5 marks

The equation

x^3 - 3x + 1 = 0

has roots near 0.35 and 1.53. It is rearranged as

x = \mathrm{g}(x)

, where

\mathrm{g}(x) = \dfrac{x^3 + 1}{3}

.

(a)
Use $x_{n+1} = \mathrm{g}(x_n)$ with $x_0 = 0$ to find $x_1$ , $x_2$ and $x_3$ .
(2)
(b)
Explain why this iteration cannot find the root near 1.53.
(3)

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Part (a)2 marks

$x_1 = 0.3333$ , $x_2 = 0.3457$
$x_3 = 0.3471$

Part (b)3 marks

An iteration converges only where $|\mathrm{g}'(x)| < 1$ near the root. $\mathrm{g}'(x) = x^2$
$\mathrm{g}'(1.53) = 2.34 > 1$
So the values move away from that root: the iteration diverges there

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Vectors

Easy

Magnitude and unit vector · Year 12 · 2 marks

\mathbf{a}

{}= 3\mathbf{i} + 4\mathbf{j}

(a)
Find $|\mathbf{a}|$ and a unit vector in the direction of $\mathbf{a}$ .
(2)

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Part (a)2 marks

Pythagoras. $|\mathbf{a}|$ ${}= \sqrt{3^2 + 4^2}$ ${}= 5$
Divide by the magnitude to get length 1. $\tfrac{1}{5}(3\mathbf{i} + 4\mathbf{j})$

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Vectors

Easy

A vector between two points · Year 12 · 2 marks

A(1, 2)

and

B(4, -2)

(a)
Find $\overrightarrow{AB}$ and its magnitude.
(2)

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Part (a)2 marks

End minus start. $\overrightarrow{AB}$ ${}= \mathbf{b} - \mathbf{a}$ ${}= 3\mathbf{i} - 4\mathbf{j}$
$|\overrightarrow{AB}| = 5$

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Vectors

Easy

Direction of a vector · Year 12 · 2 marks

(a)
Find the angle between $5\mathbf{i} - 12\mathbf{j}$ and the positive $x$ -direction.
(2)

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Part (a)2 marks

Draw it: 5 across, 12 down. $\tan\theta = \dfrac{12}{5}$
$\theta = 67.4^\circ$ below the positive $x$ -direction

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Vectors

Easy

Distance in three dimensions · Year 13 · 2 marks

(a)
Find the distance between $(1, 2, 3)$ and $(3, 5, 9)$ .
(2)

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Part (a)2 marks

Pythagoras in three dimensions. $\sqrt{2^2 + 3^2 + 6^2}$
$= 7$

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Vectors

Easy

A midpoint · Year 12 · 2 marks

\mathbf{a}

{}= 2\mathbf{i} + \mathbf{j}

and

\mathbf{b}

{}= 6\mathbf{i} + 5\mathbf{j}

are the position vectors of

A

and

B

.

(a)
Find the position vector of the midpoint of $AB$ .
(2)

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Part (a)2 marks

$\tfrac{1}{2}(\mathbf{a} + \mathbf{b})$
$= 4\mathbf{i} + 3\mathbf{j}$

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Vectors

Medium

Parallel vectors · Year 12 · 2 marks

(a)
Find $k$ such that $2\mathbf{i} + k\mathbf{j}$ is parallel to $3\mathbf{i} - 6\mathbf{j}$ .
(2)

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Part (a)2 marks

Parallel vectors are multiples of each other. $3\mathbf{i} - 6\mathbf{j}$ ${}= 3(\mathbf{i} - 2\mathbf{j})$ , so we need $2(\mathbf{i} - 2\mathbf{j})$
$k = -4$

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Vectors

Medium

A resultant force · Year 12 · 4 marks

Forces

\mathbf{F}_1

{}= (3\mathbf{i} + 2\mathbf{j})

N and

\mathbf{F}_2

{}= (\mathbf{i} - 5\mathbf{j})

N act on a particle.

(a)
Find the magnitude and direction of the resultant force.
(4)

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Part (a)4 marks

Add the vectors. $\mathbf{R}$ ${}= 4\mathbf{i} - 3\mathbf{j}$
$|\mathbf{R}| = 5$ N
$\tan\theta = \tfrac{3}{4}$
$36.9^\circ$ below $\mathbf{i}$

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Vectors

Medium

A right-angled triangle in 3D · Year 13 · 4 marks

A(1, 1, 0)

,

B(3, 2, 2)

,

C(2, -1, 0)

(a)
Show that triangle $ABC$ has a right angle at $A$ .
(4)

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Part (a)4 marks

Find each side squared. $AB^2 = 2^2 + 1^2 + 2^2 = 9$
$AC^2 = 1^2 + 2^2 + 0^2 = 5$ , $BC^2 = 1^2 + 3^2 + 2^2 = 14$
Converse of Pythagoras. $9 + 5 = 14$ , so $AB^2 + AC^2 = BC^2$
The right angle is opposite the longest side $BC$ : at $A$

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Vectors

Hard

Vectors in a parallelogram · Year 12 · 6 marks

OABC

is a parallelogram with

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OC} = \mathbf{c}

.

M

is the midpoint of

AB

, and

OM

meets

AC

at

X

.

(a)
Find $\overrightarrow{OX}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ .
(6)

Worked answerHide answer

Part (a)6 marks

$\overrightarrow{AB}$ ${}= \overrightarrow{OC}$ ${}= \mathbf{c}$ . $\overrightarrow{OM}$ ${}= \mathbf{a}$ ${}+ \tfrac{1}{2}\mathbf{c}$
$X$ is on $OM$ . $\overrightarrow{OX}$ ${}= \lambda\left(\mathbf{a} + \tfrac{1}{2}\mathbf{c}\right)$
$X$ is also on $AC$ . $\overrightarrow{OX}$ ${}= \mathbf{a}$ ${}+ \mu(\mathbf{c} - \mathbf{a})$
Compare coefficients of $\mathbf{a}$ and $\mathbf{c}$ . $\lambda = 1 - \mu$ and $\tfrac{1}{2}\lambda = \mu$
$\lambda = \tfrac{2}{3}$ , so $\overrightarrow{OX}$ ${}= \tfrac{2}{3}\mathbf{a}$ ${}+ \tfrac{1}{3}\mathbf{c}$

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Vectors

Hard

Will the ships meet? · Year 12 · 6 marks

At noon, ship

A

is at

(2\mathbf{i} + 5\mathbf{j})

km moving with velocity

(3\mathbf{i} - \mathbf{j})

km h⁻¹. Ship

B

is at

(10\mathbf{i} + \mathbf{j})

km moving with velocity

(-\mathbf{i} + \mathbf{j})

km h⁻¹.

(a)
Write down the position vectors of $A$ and $B$ after $t$ hours.
(2)
(b)
Show that the ships would collide, and find when and where.
(4)

Worked answerHide answer

Part (a)2 marks

Position = start + time × velocity. $\mathbf{r}_A$ ${}= (2 + 3t)\mathbf{i}$ ${}+ (5 - t)\mathbf{j}$
$\mathbf{r}_B$ ${}= (10 - t)\mathbf{i}$ ${}+ (1 + t)\mathbf{j}$

Part (b)4 marks

They collide if both components match at the same time. $\mathbf{i}$ : $2 + 3t = 10 - t \Rightarrow t = 2$
$\mathbf{j}$ : $5 - t = 1 + t \Rightarrow t = 2$
Same $t$ , so they collide at 2 pm
at $(8\mathbf{i} + 3\mathbf{j})$ km

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Statistical sampling

Easy

Census or sample? · Year 12 · 2 marks

(a)
Explain the difference between a census and a sample, and give one advantage of using a sample.
(2)

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Part (a)2 marks

A census uses every member of the population; a sample uses only some of them
Any one sensible advantage.A sample is quicker and cheaper (and does not destroy every item if testing is destructive)

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Statistical sampling

Easy

Simple random sampling · Year 12 · 2 marks

(a)
Describe how to take a simple random sample of 30 students from a school of 500.
(2)

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Part (a)2 marks

Every student needs a number.Number the students 1 to 500 (the sampling frame)
Use a random number generator to pick 30 different numbers, ignoring repeats

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Statistical sampling

Easy

Stratified sampling · Year 12 · 2 marks

A sixth form has 360 Year 12 and 240 Year 13 students.

(a)
A stratified sample of 50 students is taken. How many should come from each year?
(2)

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Part (a)2 marks

Each group's share of the sample matches its share of the population.Year 12: $\dfrac{360}{600} \times 50 = 30$
Year 13: 20

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Statistical sampling

Easy

Systematic sampling · Year 12 · 2 marks

(a)
Describe how to take a systematic sample of 60 from a list of 1200 names.
(2)

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Part (a)2 marks

$1200 \div 60 = 20$ , so take every 20th name
Start at a random name between 1 and 20

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Statistical sampling

Easy

Opportunity sampling · Year 12 · 2 marks

(a)
A student asks the first 20 people they meet in the canteen. Give one advantage and one disadvantage of this method.
(2)

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Part (a)2 marks

Advantage: quick and easy
Disadvantage: not random, so it may not represent the whole population

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Statistical sampling

Medium

Quota and stratified · Year 12 · 3 marks

(a)
State one similarity and one difference between quota sampling and stratified sampling.
(2)
(b)
Which of the two is a random sampling method?
(1)

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Part (a)2 marks

Similarity: both split the population into groups and sample each group in proportion
Difference: stratified picks at random within each group; quota does not

Part (b)1 mark

Stratified sampling

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Statistical sampling

Medium

Three strata · Year 12 · 3 marks

A company has 150 office staff, 90 drivers and 60 engineers.

(a)
A stratified sample of 40 is taken. Find how many of each group are chosen.
(3)

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Part (a)3 marks

Find the sampling fraction. $\dfrac{40}{300} = \dfrac{2}{15}$ of each group
Office $20$ , drivers $12$
Check: $20 + 12 + 8 = 40$ .Engineers $8$

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Statistical sampling

Medium

Spotting bias · Year 12 · 3 marks

To find out how much exercise people in a town do, a researcher asks people arriving at a gym at 7 am.

(a)
Explain why this sample is likely to be biased, and suggest a better method.
(3)

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Part (a)3 marks

The sample is not typical of the town.Gym users exercise more than most people
Early-morning visitors are a narrow group too
Better: a random (or stratified) sample from a list of all residents

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Statistical sampling

Hard

Designing a survey · Year 12 · 5 marks

A firm has 1000 employees in four departments. It wants to survey 100 of them about working hours.

(a)
Describe in detail how to take a stratified sample, and explain why it is better than a simple random sample here.
(5)

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Part (a)5 marks

Find the number of employees in each department
Sample $\dfrac{\text{department size}}{1000} \times 100$ from each department
Number the staff in each department and choose them using random numbers
Every department is represented in proportion
Working hours probably differ between departments, so this gives a more representative sample

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Statistical sampling

Hard

An online poll · Year 12 · 4 marks

A website asks visitors to vote on whether the voting age should be lowered. 80% of 5000 votes say yes.

(a)
Explain why this result may not reflect the views of the whole country, even though the sample is large.
(2)
(b)
Suggest how the poll could be improved.
(2)

Worked answerHide answer

Part (a)2 marks

Voters chose themselves (self-selected), so people with strong views are more likely to take part
A large biased sample is still biased.Visitors to one website are not typical of the country

Part (b)2 marks

Use a random sample from a list of the whole population (such as the electoral roll)
Make sure people cannot vote more than once

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Data presentation and interpretation

Easy

Averages · Year 12 · 3 marks

3,\ 7,\ 7,\ 8,\ 10,\ 12

(a)
Find the mean, median and mode.
(3)

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Part (a)3 marks

Add them up and divide by how many there are.Mean $= \dfrac{47}{6} = 7.83$
The middle of six values is halfway between the 3rd and 4th.Median $= \dfrac{7 + 8}{2} = 7.5$
The most common value.Mode $= 7$

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Data presentation and interpretation

Easy

Mean and standard deviation · Year 12 · 3 marks

For 12 values,

\sum x = 480

and

\sum x^2 = 19650

.

(a)
Find the mean and standard deviation.
(3)

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Part (a)3 marks

$\bar{x} = \dfrac{480}{12} = 40$
Mean of the squares minus the square of the mean.Variance $= \dfrac{19650}{12} - 40^2 = 37.5$
Standard deviation $= \sqrt{37.5} = 6.12$

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Data presentation and interpretation

Easy

Outliers · Year 12 · 3 marks

A data set has lower quartile 20 and upper quartile 30. An outlier is any value more than

1.5 \times

the IQR beyond a quartile.

(a)
Is the value 48 an outlier?
(3)

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Part (a)3 marks

IQR $= 30 - 20 = 10$
Upper limit $= 30 + 1.5 \times 10 = 45$
$48 > 45$ , so 48 is an outlier

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Data presentation and interpretation

Easy

Interpreting correlation · Year 12 · 2 marks

A scatter diagram of ice cream sales against sunburn cases shows strong positive correlation.

(a)
Describe the correlation, and explain why it does not show that ice cream causes sunburn.
(2)

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Part (a)2 marks

As one goes up, the other tends to go up
Correlation does not mean causation.Both are probably caused by a third factor: hot, sunny weather

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Data presentation and interpretation

Easy

Frequency density · Year 12 · 2 marks

(a)
In a histogram, the class 10–20 has frequency 30 and the class 20–40 has frequency 50. Find the frequency density of each bar.
(2)

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Part (a)2 marks

Frequency density = frequency ÷ class width. $30 \div 10 = 3$
$50 \div 20 = 2.5$

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Data presentation and interpretation

Medium

Grouped data · Year 12 · 6 marks

Times,

t

minutes:

0 \le t < 10

: 5;

10 \le t < 20

: 12;

20 \le t < 30

: 8.

(a)
Estimate the mean.
(3)
(b)
Use linear interpolation to estimate the median.
(3)

Worked answerHide answer

Part (a)3 marks

With grouped data, use each class's midpoint.Midpoints 5, 15, 25
$\dfrac{5(5) + 12(15) + 8(25)}{25}$ ${}= \dfrac{405}{25}$
$= 16.2$ minutes

Part (b)3 marks

5 values come before this class.The median is the 12.5th value, in the class $10 \le t < 20$
It is 7.5 of the 12 values into the class. $10$ ${}+ \dfrac{12.5 - 5}{12} \times 10$
$= 16.25$ minutes

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Data presentation and interpretation

Medium

Coding · Year 12 · 3 marks

Data are coded using

y = \dfrac{x - 50}{10}

. The coded data have mean 1.2 and standard deviation 0.8.

(a)
Find the mean and standard deviation of $x$ .
(3)

Worked answerHide answer

Part (a)3 marks

Rearrange the coding. $x = 10y + 50$
$\bar{x} = 10(1.2) + 50 = 62$
Adding 50 does not change the spread; multiplying by 10 does. $\sigma_x = 10 \times 0.8 = 8$

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Data presentation and interpretation

Medium

Comparing distributions · Year 12 · 4 marks

Group A: median 42 s, IQR 10 s. Group B: median 38 s, IQR 18 s. The times are for a puzzle.

(a)
Compare the two groups' times.
(4)

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Part (a)4 marks

Compare a measure of location...Group B has the lower median
...and say what it means in context.so group B were faster on average
Then compare a measure of spread...Group A has the smaller IQR
so group A's times were more consistent

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Data presentation and interpretation

Hard

Combining two groups · Year 12 · 6 marks

Group A: 10 values, mean 12, standard deviation 2. Group B: 15 values, mean 14, standard deviation 3.

(a)
Find the mean and standard deviation of all 25 values together.
(6)

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Part (a)6 marks

Total of each group = number × mean.Mean $= \dfrac{10(12) + 15(14)}{25}$ ${}= 13.2$
From $\sigma^2$ ${}= \dfrac{\sum x^2}{n}$ ${}- \bar{x}^2$ . $\sum x_A^2 = 10(2^2 + 12^2) = 1480$
$\sum x_B^2 = 15(3^2 + 14^2) = 3075$
$\sigma^2$ ${}= \dfrac{4555}{25} - 13.2^2$ ${}= 7.96$
$\sigma = 2.82$

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Data presentation and interpretation

Hard

Cleaning data · Year 12 · 4 marks

20 values have mean 42. One value, 95, is found to be a recording error and is removed.

(a)
Find the new mean.
(2)
(b)
State, with a reason, how removing the value changes the standard deviation.
(2)

Worked answerHide answer

Part (a)2 marks

Find the old total, take away 95, divide by 19. $\dfrac{20 \times 42 - 95}{19}$ ${}= \dfrac{745}{19}$
$= 39.2$

Part (b)2 marks

It decreases
95 was far from the mean, so it was adding a lot to the spread

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Probability

Easy

Exclusive or independent? · Year 12 · 3 marks

\mathrm{P}(A) = 0.3

and

\mathrm{P}(B) = 0.5

(a)
Find $\mathrm{P}(A \cup B)$ if $A$ and $B$ are mutually exclusive.
(1)
(b)
Find $\mathrm{P}(A \cup B)$ if instead $A$ and $B$ are independent.
(2)

Worked answerHide answer

Part (a)1 mark

Mutually exclusive events cannot both happen, so just add. $0.3 + 0.5 = 0.8$

Part (b)2 marks

Independent: multiply. $\mathrm{P}(A \cap B)$ ${}= 0.3 \times 0.5$ ${}= 0.15$
Take away the overlap so it is not counted twice. $\mathrm{P}(A \cup B)$ ${}= 0.3 + 0.5 - 0.15$ ${}= 0.65$

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Probability

Easy

Two dice · Year 12 · 2 marks

(a)
Two fair dice are rolled. Find the probability that the total is 7.
(2)

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Part (a)2 marks

There are $6 \times 6 = 36$ equally likely outcomes. $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ : 6 outcomes
$\dfrac{6}{36} = \dfrac{1}{6}$

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Probability

Easy

A Venn diagram · Year 12 · 3 marks

In a class of 30, 18 like maths, 12 like physics and 5 like both.

(a)
How many like neither?
(2)
(b)
Find the probability that a student chosen at random likes maths but not physics.
(1)

Worked answerHide answer

Part (a)2 marks

Fill in the overlap first.Maths only 13, physics only 7, both 5: total 25
$30 - 25 = 5$

Part (b)1 mark

$\dfrac{13}{30}$

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Probability

Easy

Without replacement · Year 12 · 4 marks

A bag holds 4 red and 6 blue counters. Two are taken without replacement.

(a)
Find the probability that both are red.
(2)
(b)
Find the probability that they are different colours.
(2)

Worked answerHide answer

Part (a)2 marks

After one red is taken, 3 reds are left out of 9. $\dfrac{4}{10} \times \dfrac{3}{9}$
$= \dfrac{2}{15}$

Part (b)2 marks

Red then blue, or blue then red. $\dfrac{4}{10} \times \dfrac{6}{9}$ ${}+ \dfrac{6}{10} \times \dfrac{4}{9}$
$= \dfrac{8}{15}$

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Probability

Easy

Testing independence · Year 12 · 2 marks

\mathrm{P}(A) = 0.4

,

\mathrm{P}(B) = 0.25

and

\mathrm{P}(A \cap B) = 0.1

.

(a)
Are $A$ and $B$ independent? Give a reason.
(2)

Worked answerHide answer

Part (a)2 marks

$\mathrm{P}(A) \times \mathrm{P}(B)$ ${}= 0.1$
This equals $\mathrm{P}(A \cap B)$ , so they are independent

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Probability

Medium

Conditional probability · Year 13 · 3 marks

\mathrm{P}(A) = 0.4

,

\mathrm{P}(B) = 0.5

and

\mathrm{P}(A \cap B) = 0.15

.

(a)
Find $\mathrm{P}(A \mid B)$ and $\mathrm{P}(B \mid A)$ .
(3)

Worked answerHide answer

Part (a)3 marks

“Given $B$ ” means we only look inside $B$ . $\mathrm{P}(A \mid B)$ ${}= \dfrac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)}$
$= \dfrac{0.15}{0.5} = 0.3$
$\mathrm{P}(B \mid A)$ ${}= \dfrac{0.15}{0.4}$ ${}= 0.375$

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Probability

Medium

Conditional probability from data · Year 13 · 4 marks

Of 100 students, 60 take maths, 45 take physics and 30 take both.

(a)
Find the probability that a student takes physics, given that they take maths.
(2)
(b)
Are taking maths and taking physics independent? Explain.
(2)

Worked answerHide answer

Part (a)2 marks

Only look at the 60 maths students. $\dfrac{30}{60}$
$= 0.5$

Part (b)2 marks

$\mathrm{P}(\text{physics}) = 0.45$
Knowing someone takes maths changes the chance they take physics. $0.5 \ne 0.45$ , so they are not independent

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Probability

Medium

A medical test · Year 13 · 5 marks

2% of people have a condition. A test is positive for 95% of people with it, and for 10% of people without it.

(a)
Find the probability that a person tests positive.
(2)
(b)
Given that a person tests positive, find the probability that they have the condition.
(3)

Worked answerHide answer

Part (a)2 marks

Draw a tree: has it and positive, or does not and positive. $0.02 \times 0.95$ ${}+ 0.98 \times 0.10$
$= 0.019 + 0.098 = 0.117$

Part (b)3 marks

P(has it and positive) ÷ P(positive). $\dfrac{0.019}{0.117}$
Most positive results are false, because the condition is rare. $= 0.162$

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Probability

Hard

Reversing a condition · Year 13 · 5 marks

\mathrm{P}(A) = 0.5

,

\mathrm{P}(B \mid A) = 0.4

and

\mathrm{P}(B \mid A') = 0.2

.

(a)
Find $\mathrm{P}(B)$ .
(3)
(b)
Find $\mathrm{P}(A \mid B)$ .
(2)

Worked answerHide answer

Part (a)3 marks

$\mathrm{P}(A \cap B)$ ${}= 0.5 \times 0.4$ ${}= 0.2$
$\mathrm{P}(A' \cap B)$ ${}= 0.5 \times 0.2$ ${}= 0.1$
$\mathrm{P}(B) = 0.3$

Part (b)2 marks

$\dfrac{0.2}{0.3}$
$= \dfrac{2}{3}$

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Probability

Hard

At least one six · Year 12 · 5 marks

(a)
A fair die is rolled 4 times. Find the probability of at least one six.
(2)
(b)
Find the least number of rolls needed for the probability of at least one six to be more than 0.9.
(3)

Worked answerHide answer

Part (a)2 marks

“At least one” = 1 − “none”. $1 - \left(\tfrac{5}{6}\right)^4$
$= 0.518$

Part (b)3 marks

$1$ ${}- \left(\tfrac{5}{6}\right)^n > 0.9$ ${}\Rightarrow \left(\tfrac{5}{6}\right)^n < 0.1$
Dividing by $\ln\tfrac{5}{6} < 0$ reverses the inequality. $n > \dfrac{\ln 0.1}{\ln \tfrac{5}{6}}$ ${}= 12.6$
$n = 13$

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Statistical distributions

Easy

A binomial probability · Year 12 · 2 marks

X \sim \mathrm{B}(10, 0.3)

(a)
Find $\mathrm{P}(X = 3)$ .
(2)

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Part (a)2 marks

Choose which 3 trials succeed, then multiply the probabilities. $\dbinom{10}{3}(0.3)^3(0.7)^7$
$= 0.267$

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Statistical distributions

Easy

A discrete distribution · Year 12 · 3 marks

\mathrm{P}(X = x) = kx

for

x = 1, 2, 3, 4

.

(a)
Find $k$ .
(2)
(b)
Find $\mathrm{P}(X \ge 3)$ .
(1)

Worked answerHide answer

Part (a)2 marks

Probabilities add up to 1. $k + 2k + 3k + 4k = 1$
$k = 0.1$

Part (b)1 mark

$0.3 + 0.4 = 0.7$

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Statistical distributions

Easy

When to use a binomial model · Year 12 · 2 marks

(a)
State two conditions needed for a binomial model to be suitable.
(2)

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Part (a)2 marks

A fixed number of independent trials
Each trial has two outcomes, with the same probability of success every time

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Statistical distributions

Easy

A normal probability · Year 13 · 2 marks

X \sim \mathrm{N}(50, 4^2)

(a)
Find $\mathrm{P}(X < 56)$ .
(2)

Worked answerHide answer

Part (a)2 marks

Standardise, or use your calculator directly. $z = \dfrac{56 - 50}{4} = 1.5$
$\mathrm{P}(X < 56) = 0.933$

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Statistical distributions

Easy

Within one standard deviation · Year 13 · 2 marks

X \sim \mathrm{N}(50, 4^2)

(a)
Find $\mathrm{P}(46 < X < 54)$ .
(2)

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Part (a)2 marks

46 and 54 are each one standard deviation from the mean. $\mathrm{P}(-1 < Z < 1)$
About 68% of a normal distribution is within one standard deviation. $= 0.683$

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Statistical distributions

Medium

Cumulative binomial · Year 12 · 3 marks

X \sim \mathrm{B}(20, 0.25)

(a)
Find $\mathrm{P}(X \le 4)$ .
(1)
(b)
Find $\mathrm{P}(X \ge 8)$ .
(2)

Worked answerHide answer

Part (a)1 mark

Use the cumulative binomial function on your calculator. $0.415$

Part (b)2 marks

“At least 8” is everything except 0 to 7. $1 - \mathrm{P}(X \le 7)$
$= 1 - 0.898 = 0.102$

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Statistical distributions

Medium

Inverse normal · Year 13 · 3 marks

Adult heights are modelled as

\mathrm{N}(170, 8^2)

cm.

(a)
Find the height that only 10% of adults are taller than.
(3)

Worked answerHide answer

Part (a)3 marks

Find the $z$ -value with 0.9 below it. $\mathrm{P}(Z > z)$ ${}= 0.1$ ${}\Rightarrow z$ ${}= 1.2816$
Undo the standardising. $h = 170 + 1.2816 \times 8$
$= 180.3$ cm

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Statistical distributions

Hard

Finding the mean and standard deviation · Year 13 · 6 marks

X \sim \mathrm{N}(\mu, \sigma^2)

,

\mathrm{P}(X < 40) = 0.2

and

\mathrm{P}(X > 60) = 0.1

.

(a)
Find $\mu$ and $\sigma$ .
(6)

Worked answerHide answer

Part (a)6 marks

The $z$ -value with 0.2 below it is negative. $\dfrac{40 - \mu}{\sigma} = -0.8416$
The $z$ -value with 0.9 below it. $\dfrac{60 - \mu}{\sigma} = 1.2816$
Two simultaneous equations.Subtract: $20 = 2.1232\sigma$
$\sigma = 9.42$ , $\mu = 47.9$

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Statistical distributions

Hard

Boxes in a crate · Year 12 · 6 marks

4% of bags filled by a machine are underweight. Bags are packed in boxes of 50.

(a)
Find the probability that a box contains more than 3 underweight bags.
(3)
(b)
A crate holds 10 boxes. Find the probability that at least 2 boxes in the crate have more than 3 underweight bags.
(3)

Worked answerHide answer

Part (a)3 marks

$X \sim \mathrm{B}(50, 0.04)$
$\mathrm{P}(X > 3)$ ${}= 1 - \mathrm{P}(X \le 3)$
$= 0.139$

Part (b)3 marks

Now each box is a trial. $Y \sim \mathrm{B}(10, 0.1391)$
$1$ ${}- \mathrm{P}(Y = 0)$ ${}- \mathrm{P}(Y = 1)$
$= 0.415$

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Statistical hypothesis testing

Easy

Writing hypotheses · Year 12 · 2 marks

A student thinks a coin is biased, without knowing which way.

(a)
Write down suitable null and alternative hypotheses.
(2)

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Part (a)2 marks

The null hypothesis is “nothing unusual”: the coin is fair. $\mathrm{H}_0: p = 0.5$
Two-tailed, because the bias could be either way. $\mathrm{H}_1: p \ne 0.5$

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Statistical hypothesis testing

Easy

Key words · Year 12 · 2 marks

(a)
Explain what is meant by a critical region.
(1)
(b)
Explain what is meant by a 5% significance level.
(1)

Worked answerHide answer

Part (a)1 mark

The values of the test statistic that would lead you to reject $\mathrm{H}_0$

Part (b)1 mark

The probability of rejecting $\mathrm{H}_0$ when it is actually true is at most 5%

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Statistical hypothesis testing

Easy

A one-tailed test · Year 12 · 4 marks

A shop says 30% of customers buy a drink. A manager thinks it is more. In a sample of 20 customers, 10 buy a drink.

(a)
Test the manager's claim at the 5% level.
(4)

Worked answerHide answer

Part (a)4 marks

$\mathrm{H}_0: p = 0.3$ , $\mathrm{H}_1: p > 0.3$ ; $X \sim \mathrm{B}(20, 0.3)$
Find the probability of a result this extreme or more. $\mathrm{P}(X \ge 10)$ ${}= 1 - \mathrm{P}(X \le 9)$ ${}= 0.0480$
Always finish in context. $0.0480 < 0.05$ : reject $\mathrm{H}_0$ . There is evidence that more than 30% buy a drink

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Statistical hypothesis testing

Easy

Finding a critical region · Year 12 · 3 marks

X \sim \mathrm{B}(10, 0.5)

under

\mathrm{H}_0: p = 0.5

, and

\mathrm{H}_1: p < 0.5

.

(a)
Find the critical region at the 5% level.
(3)

Worked answerHide answer

Part (a)3 marks

Look for the biggest region with probability at most 0.05. $\mathrm{P}(X \le 1) = 0.0107$
$\mathrm{P}(X \le 2) = 0.0547$ , which is too big
Critical region: $X \le 1$

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Statistical hypothesis testing

Easy

Testing correlation · Year 13 · 3 marks

For 12 pairs of data, the product moment correlation coefficient is 0.62. The critical value for a one-tailed test at 5% with

n = 12

is 0.4973.

(a)
Test, at the 5% level, whether there is positive correlation.
(3)

Worked answerHide answer

Part (a)3 marks

$\rho$ is the correlation in the whole population. $\mathrm{H}_0: \rho = 0$ , $\mathrm{H}_1: \rho > 0$
$0.62 > 0.4973$
Reject $\mathrm{H}_0$ : there is evidence of positive correlation

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Statistical hypothesis testing

Medium

A two-tailed test · Year 12 · 5 marks

A spinner should land on red with probability 0.4. In 20 spins it lands on red 13 times.

(a)
Test at the 5% level whether the probability of red is different from 0.4.
(5)

Worked answerHide answer

Part (a)5 marks

$\mathrm{H}_0: p = 0.4$ , $\mathrm{H}_1: p \ne 0.4$
$X \sim \mathrm{B}(20, 0.4)$ : $\mathrm{P}(X \ge 13)$ ${}= 1 - 0.9790$ ${}= 0.0210$
Split 5% between the two tails.Two-tailed, so compare with 0.025: $0.0210 < 0.025$
Reject $\mathrm{H}_0$ : there is evidence the probability of red is not 0.4

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Statistical hypothesis testing

Medium

Testing a normal mean · Year 13 · 5 marks

X \sim \mathrm{N}(\mu, 4^2)

. A sample of 16 has mean 51.8.

(a)
Test $\mathrm{H}_0: \mu = 50$ against $\mathrm{H}_1: \mu > 50$ at the 5% level.
(5)

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Part (a)5 marks

The mean of a sample has variance $\tfrac{\sigma^2}{n}$ . $\bar{X} \sim \mathrm{N}\left(50, \dfrac{4^2}{16}\right)$ ${}= \mathrm{N}(50, 1)$
$z = \dfrac{51.8 - 50}{1} = 1.8$
The 5% critical value for a one-tailed test. $1.8 > 1.6449$
Reject $\mathrm{H}_0$ : there is evidence the mean is greater than 50

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Statistical hypothesis testing

Medium

Two critical regions · Year 12 · 4 marks

X \sim \mathrm{B}(20, 0.5)

under

\mathrm{H}_0: p = 0.5

, with

\mathrm{H}_1: p \ne 0.5

.

(a)
Find the critical region at the 5% level, with about 2.5% in each tail.
(3)
(b)
State the actual significance level.
(1)

Worked answerHide answer

Part (a)3 marks

$\mathrm{P}(X \le 5) = 0.0207$ , $\mathrm{P}(X \le 6) = 0.0577$
Lower tail $X \le 5$
$p = 0.5$ makes the distribution symmetrical.Upper tail $X \ge 15$ , by symmetry

Part (b)1 mark

$2 \times 0.0207 = 0.0414$

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Statistical hypothesis testing

Hard

Are the bottles underfilled? · Year 13 · 6 marks

Bottles should hold 500 ml. The volume is normally distributed with standard deviation 6 ml. A sample of 25 bottles has mean 497.2 ml.

(a)
Test at the 5% level whether the mean volume is different from 500 ml.
(6)

Worked answerHide answer

Part (a)6 marks

$\mathrm{H}_0: \mu = 500$ , $\mathrm{H}_1: \mu \ne 500$
$\bar{X} \sim \mathrm{N}\left(500, \dfrac{6^2}{25}\right)$ , standard deviation 1.2
$z$ ${}= \dfrac{497.2 - 500}{1.2}$ ${}= -2.33$
Two-tailed at 5%.Critical values $\pm 1.96$ ; $-2.33 < -1.96$
Reject $\mathrm{H}_0$ : there is evidence the mean volume is not 500 ml

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Quantities and units in mechanics

Easy

SI units · Year 12 · 3 marks

(a)
Give the SI units of force, velocity and acceleration.
(3)

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Part (a)3 marks

Force: newtons, N
Velocity: m s⁻¹
Acceleration: m s⁻²

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Quantities and units in mechanics

Easy

Converting speeds · Year 12 · 2 marks

(a)
Convert 72 km h⁻¹ to m s⁻¹.
(2)

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Part (a)2 marks

km to m: × 1000; hours to seconds: ÷ 3600. $\dfrac{72 \times 1000}{3600}$
$= 20$ m s⁻¹

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Quantities and units in mechanics

Easy

Mass and weight · Year 12 · 2 marks

(a)
Find the weight of a 5 kg mass, and the mass of an object that weighs 98 N.
(2)

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Part (a)2 marks

$W = mg = 5 \times 9.8 = 49$ N
$m = 98 \div 9.8 = 10$ kg

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Quantities and units in mechanics

Easy

Base units · Year 12 · 2 marks

(a)
Write the newton in terms of the base units kg, m and s.
(1)
(b)
Explain why weight is a vector but mass is a scalar.
(1)

Worked answerHide answer

Part (a)1 mark

From $F = ma$ : kg × m s⁻². $\mathrm{kg\,m\,s^{-2}}$

Part (b)1 mark

Weight has a direction (downwards); mass has only a size

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Quantities and units in mechanics

Easy

Modelling words · Year 12 · 4 marks

(a)
State what each word means in a mechanics model: particle, light, inextensible, smooth.
(4)

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Part (a)4 marks

Particle: its size can be ignored
Light: its mass can be ignored
Inextensible: it does not stretch
Smooth: there is no friction

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Quantities and units in mechanics

Medium

Converting density · Year 12 · 2 marks

(a)
Aluminium has density 2.7 g cm⁻³. Convert this to kg m⁻³.
(2)

Worked answerHide answer

Part (a)2 marks

g to kg: ÷ 1000. There are $100^3$ cm³ in 1 m³. $2.7 \times \dfrac{1}{1000} \times 100^3$
$= 2700$ kg m⁻³

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Quantities and units in mechanics

Medium

Checking units · Year 12 · 3 marks

(a)
Show that every term in $s = ut + \tfrac{1}{2}at^2$ has units of metres.
(3)

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Part (a)3 marks

$ut$ : m s⁻¹ × s = m
$at^2$ : m s⁻² × s² = m
Numbers such as $\tfrac{1}{2}$ have no units, so every term is in metres

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Quantities and units in mechanics

Medium

Units in Newton's second law · Year 12 · 3 marks

(a)
A car of mass 1.2 tonnes has a resultant force of 3 kN on it. Find its acceleration.
(3)

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Part (a)3 marks

Convert to SI units first. $m = 1200$ kg, $F = 3000$ N
$a = \dfrac{3000}{1200}$
$= 2.5$ m s⁻²

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Quantities and units in mechanics

Hard

Questioning a model · Year 12 · 4 marks

A cyclist is modelled as a particle with no air resistance to predict their stopping distance.

(a)
Explain how each simplification affects the prediction, and whether it makes the model more or less safe to rely on.
(4)

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Part (a)4 marks

Ignoring air resistance: in reality it helps slow the cyclist
An overestimate is the cautious direction.so the model overestimates the stopping distance
As a particle, the bike's size and the rider's reaction time are ignored
Reaction time adds distance, so in that respect the model underestimates

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Quantities and units in mechanics

Hard

A train speeding up · Year 12 · 5 marks

(a)
A train speeds up uniformly from 18 km h⁻¹ to 90 km h⁻¹ in 1 minute 20 seconds. Find its acceleration and the distance in km.
(5)

Worked answerHide answer

Part (a)5 marks

Convert everything to SI units. $u = 5$ m s⁻¹, $v = 25$ m s⁻¹, $t = 80$ s
$a = \dfrac{25 - 5}{80} = 0.25$ m s⁻²
$s = \tfrac{(u + v)}{2}t$ . $s$ ${}= \dfrac{5 + 25}{2} \times 80$ ${}= 1200$ m
$= 1.2$ km

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Kinematics

Easy

Speeding up from rest · Year 12 · 3 marks

(a)
A car starts from rest and accelerates at 2 m s⁻² for 5 s. Find its final speed and the distance travelled.
(3)

Worked answerHide answer

Part (a)3 marks

$v = u + at$ . $v = 0 + 2 \times 5 = 10$ m s⁻¹
$s = ut + \tfrac{1}{2}at^2$ with $u = 0$ . $s$ ${}= \tfrac{1}{2} \times 2 \times 5^2$
$= 25$ m

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Kinematics

Easy

A falling stone · Year 12 · 3 marks

(a)
A stone is dropped from 45 m. Find the time it takes to reach the ground and its speed just before it lands.
(3)

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Part (a)3 marks

Dropped means $u = 0$ . $45 = \tfrac{1}{2}(9.8)t^2$
$t = 3.03$ s
$v = 9.8 \times 3.03 = 29.7$ m s⁻¹

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Kinematics

Easy

Distance from a graph · Year 12 · 3 marks

A car speeds up from rest to 12 m s⁻¹ in 4 s, travels at 12 m s⁻¹ for 10 s, then slows to rest in 6 s.

(a)
Find the total distance travelled.
(3)

Worked answerHide answer

Part (a)3 marks

Distance = area under the velocity–time graph
Two triangles and a rectangle. $\tfrac{1}{2}(4)(12)$ ${}+ 10(12)$ ${}+ \tfrac{1}{2}(6)(12)$ ${}= 24 + 120 + 36$
$= 180$ m

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Kinematics

Easy

When is it at rest? · Year 12 · 3 marks

A particle has displacement

s = t^3 - 6t^2 + 9t

metres at time

t

seconds.

(a)
Find when the particle is at rest.
(3)

Worked answerHide answer

Part (a)3 marks

Velocity is the rate of change of displacement. $v$ ${}= \dfrac{\mathrm{d}s}{\mathrm{d}t}$ ${}= 3t^2 - 12t + 9$
At rest means $v = 0$ . $3(t - 1)(t - 3) = 0$
$t = 1$ s and $t = 3$ s

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Kinematics

Medium

Thrown upwards · Year 12 · 4 marks

(a)
A ball is thrown up at 14.7 m s⁻¹. Find the greatest height and the time until it returns to the same level.
(4)

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Part (a)4 marks

At the top, $v = 0$ . $0 = 14.7^2 - 2(9.8)s$
$s = 11.0$ m
Back at the start, $s = 0$ . $0 = 14.7t - 4.9t^2$
$t = 3$ s

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Kinematics

Medium

Variable velocity · Year 12 · 4 marks

A particle moves in a straight line with velocity

v = 6t - t^2

m s⁻¹ for

0 \le t \le 6

.

(a)
Find the distance travelled in the first 6 seconds.
(4)

Worked answerHide answer

Part (a)4 marks

Check it does not turn back. $v \ge 0$ for $0 \le t \le 6$ , so distance = displacement
$\displaystyle\int_0^6 (6t - t^2)\,\mathrm{d}t$ ${}= \left[3t^2 - \tfrac{1}{3}t^3\right]_0^6$
$= 108 - 72 = 36$ m

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Kinematics

Medium

A projectile · Year 13 · 5 marks

A ball is kicked from the ground at 20 m s⁻¹ at

30^\circ

above the horizontal.

(a)
Find the time of flight and the horizontal range.
(5)

Worked answerHide answer

Part (a)5 marks

Split the velocity into parts.Vertical speed $20\sin 30^\circ = 10$ ; horizontal $20\cos 30^\circ = 17.32$
It lands when the vertical displacement is 0. $0 = 10t - 4.9t^2$
$t = 2.04$ s
Horizontal speed is constant.Range $= 17.32 \times 2.04$
$= 35.3$ m

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Kinematics

Hard

Catching up · Year 12 · 5 marks

Car

A

passes a junction at a constant 20 m s⁻¹. At that moment car

B

starts from rest at the junction and accelerates at 2.5 m s⁻² in the same direction.

(a)
Find when and where $B$ catches $A$ .
(5)

Worked answerHide answer

Part (a)5 marks

Write each car's distance from the junction. $s_A = 20t$ , $s_B = 1.25t^2$
Same distance means they are side by side ( $t = 0$ is the start). $1.25t^2 = 20t \Rightarrow t = 16$ s
$320$ m from the junction

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Kinematics

Hard

A ball from a cliff · Year 13 · 6 marks

A ball is thrown from the top of a 40 m cliff at 15 m s⁻¹ at

20^\circ

above the horizontal. It lands in the sea.

(a)
Find how far from the foot of the cliff it lands.
(6)

Worked answerHide answer

Part (a)6 marks

Take up as positive; it ends 40 m below the start.Vertical: $-40 = 15\sin 20^\circ\,t - 4.9t^2$
$4.9t^2 - 5.130t - 40 = 0$
Take the positive root. $t = 3.43$ s
$x$ ${}= 15\cos 20^\circ \times 3.43$ ${}= 48.3$ m

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Forces and Newton's laws

Easy

Newton's second law · Year 12 · 2 marks

(a)
A car of mass 1000 kg has a driving force of 2500 N and resistance of 700 N. Find its acceleration.
(2)

Worked answerHide answer

Part (a)2 marks

Resultant force = mass × acceleration. $2500 - 700 = 1000a$
$a = 1.8$ m s⁻²

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Forces and Newton's laws

Easy

In a lift · Year 12 · 3 marks

(a)
A 60 kg person stands in a lift accelerating upwards at 0.5 m s⁻². Find the reaction force from the floor.
(3)

Worked answerHide answer

Part (a)3 marks

Resultant upwards = mass × acceleration. $R - 60(9.8) = 60(0.5)$
$R = 618$ N

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Forces and Newton's laws

Easy

Equilibrium · Year 12 · 2 marks

(a)
A particle is in equilibrium under forces $(3\mathbf{i} + 4\mathbf{j})$ N, $(-5\mathbf{i} + \mathbf{j})$ N and $\mathbf{F}$ . Find $\mathbf{F}$ .
(2)

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Part (a)2 marks

In equilibrium the forces add to zero. $\mathbf{F}$ ${}+ (-2\mathbf{i} + 5\mathbf{j})$ ${}= \mathbf{0}$
$\mathbf{F}$ ${}= (2\mathbf{i} - 5\mathbf{j})$ N

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Forces and Newton's laws

Easy

Towing a trailer · Year 12 · 4 marks

A 1200 kg car tows a 400 kg trailer. The driving force is 2400 N. Ignore resistances.

(a)
Find the acceleration and the tension in the tow bar.
(4)

Worked answerHide answer

Part (a)4 marks

Treat car and trailer as one.Whole system: $2400 = 1600a$
$a = 1.5$ m s⁻²
The only horizontal force on the trailer is the tension.Trailer: $T = 400 \times 1.5$
$T = 600$ N

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Forces and Newton's laws

Easy

Friction on a level surface · Year 13 · 3 marks

(a)
A 5 kg block on a rough horizontal surface ( $\mu = 0.3$ ) is pulled by a horizontal 20 N force. Find its acceleration.
(3)

Worked answerHide answer

Part (a)3 marks

On a level surface $R = mg$ . $F$ ${}= \mu R$ ${}= 0.3 \times 5 \times 9.8$ ${}= 14.7$ N
$20 - 14.7 = 5a$
$a = 1.06$ m s⁻²

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Forces and Newton's laws

Medium

A pulley · Year 12 · 5 marks

Particles of mass 3 kg and 5 kg hang from a light inextensible string over a smooth pulley.

(a)
Find the acceleration and the tension in the string.
(5)

Worked answerHide answer

Part (a)5 marks

Write an equation for each particle, in its own direction of motion.5 kg: $5g - T = 5a$
3 kg: $T - 3g = 3a$
Add: $2g = 8a \Rightarrow a = 2.45$ m s⁻²
$T = 3(9.8 + 2.45) = 36.8$ N

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Forces and Newton's laws

Medium

Sliding down a rough slope · Year 13 · 5 marks

(a)
A 4 kg block is released on a rough plane inclined at $30^\circ$ ( $\mu = 0.25$ ). Find its acceleration.
(5)

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Part (a)5 marks

Resolve perpendicular to the slope. $R = 4g\cos 30^\circ$
Friction is at its limit because the block slides. $F = 0.25 \times 4g\cos 30^\circ$
Resolve down the slope. $4g\sin 30^\circ - F = 4a$
$a$ ${}= g\sin 30^\circ$ ${}- 0.25g\cos 30^\circ$
$= 2.78$ m s⁻²

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Forces and Newton's laws

Hard

Pushing up a rough slope · Year 13 · 6 marks

(a)
A 10 kg box is on a rough plane inclined at $20^\circ$ ( $\mu = 0.4$ ). Find the force parallel to the plane needed to make the box just about to move up.
(6)

Worked answerHide answer

Part (a)6 marks

$R = 10g\cos 20^\circ = 92.1$ N
Friction opposes the movement that is about to happen.Friction acts down the slope: $F = 0.4R$
Forces up the slope balance forces down it. $P$ ${}= 10g\sin 20^\circ$ ${}+ 0.4 \times 10g\cos 20^\circ$
$P = 70.4$ N

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Forces and Newton's laws

Hard

A block pulled off a table · Year 13 · 8 marks

Block

A

(2 kg) is on a rough horizontal table (

\mu = 0.2

). A string over a smooth pulley at the edge joins it to block

B

(3 kg), which hangs 1.5 m above the floor. The system is released from rest.

(a)
Find the acceleration and the speed of $B$ as it hits the floor.
(5)
(b)
After $B$ hits the floor, $A$ slides on. Find how much further it travels, assuming it does not reach the pulley.
(3)

Worked answerHide answer

Part (a)5 marks

Friction on $A$ is $\mu R = 0.2 \times 2g$ . $3g - T = 3a$ and $T - 0.2(2g) = 2a$
$a = \dfrac{3g - 0.4g}{5} = 5.10$ m s⁻²
$v^2$ ${}= 2(5.096)(1.5)$ ${}\Rightarrow v$ ${}= 3.91$ m s⁻¹

Part (b)3 marks

The string goes slack, so only friction acts.Deceleration $= \mu g = 1.96$ m s⁻²
$0 = 15.29 - 2(1.96)s$
$s = 3.9$ m

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Moments

Easy

Balancing a see-saw · Year 13 · 2 marks

(a)
A 30 kg child sits 2 m from the pivot of a see-saw. Where must a 40 kg child sit to balance it?
(2)

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Part (a)2 marks

Clockwise moments = anticlockwise moments. $30g \times 2 = 40g \times d$
$d = 1.5$ m on the other side

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Moments

Easy

A beam on two supports · Year 13 · 4 marks

A uniform beam

AB

, 4 m long and of mass 10 kg, rests on supports at

A

and

B

. A 20 kg mass is placed 1 m from

A

.

(a)
Find the reaction at each support.
(4)

Worked answerHide answer

Part (a)4 marks

The beam's weight acts at its middle. Taking moments about $A$ removes $R_A$ .Moments about $A$ : $4R_B = 10g(2) + 20g(1)$
$R_B = 98$ N
Up = down. $R_A = 30g - 98 = 196$ N

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Moments

Easy

Balancing a rod · Year 13 · 3 marks

(a)
A uniform 6 m rod of mass 8 kg is pivoted 2 m from end $A$ . Find the mass hung at $A$ that keeps it level.
(3)

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Part (a)3 marks

Its centre is 3 m from $A$ .The rod's weight acts 1 m beyond the pivot
$m g \times 2 = 8g \times 1$
$m = 4$ kg

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Moments

Easy

Resultant moment · Year 13 · 2 marks

(a)
About a point $P$ , a 5 N force acts clockwise at 2 m, and an 8 N force acts anticlockwise at 1.5 m. Find the resultant moment.
(2)

Worked answerHide answer

Part (a)2 marks

Moment = force × perpendicular distance. $8 \times 1.5 - 5 \times 2 = 2$
2 N m anticlockwise

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Moments

Easy

About to tilt · Year 13 · 3 marks

A uniform 5 m plank of mass 20 kg rests on supports 1 m from each end.

(a)
A 50 kg man walks along the plank past one support. How far beyond it can he go before the plank tips?
(3)

Worked answerHide answer

Part (a)3 marks

So take moments about the support he passed.About to tip: the reaction at the other support is zero
The plank's centre is 1.5 m from that support. $50g \times x = 20g \times 1.5$
$x = 0.6$ m

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Moments

Medium

A non-uniform rod · Year 13 · 4 marks

A non-uniform rod

AB

, 3 m long and of mass 6 kg, rests on supports at

A

and

B

. The reaction at

A

is twice the reaction at

B

.

(a)
Find the distance of the centre of mass from $A$ .
(4)

Worked answerHide answer

Part (a)4 marks

$R_A + R_B = 6g$ and $R_A = 2R_B$
$R_B = 2g$
Moments about $A$ : $6g \times x = 2g \times 3$
$x = 1$ m

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Moments

Medium

Supports not at the ends · Year 13 · 4 marks

A uniform 6 m plank of mass 30 kg rests on supports at

C

, 1 m from

A

, and

D

, 2 m from

B

.

(a)
Find the reactions at $C$ and $D$ .
(4)

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Part (a)4 marks

The centre is 3 m from $A$ : 2 m from $C$ and 1 m from $D$
Moments about $C$ : $3R_D = 30g \times 2$
$R_D = 196$ N
$R_C = 294 - 196 = 98$ N

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Moments

Medium

A rod held by a string · Year 13 · 4 marks

A uniform 2 m rod of weight 40 N has one end on the ground. It is held at

30^\circ

to the horizontal by a string at the other end, at right angles to the rod.

(a)
Find the tension in the string.
(4)

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Part (a)4 marks

This avoids the unknown forces there.Moments about the end on the ground
The weight's perpendicular distance is $1\cos 30^\circ$ . $T \times 2$ ${}= 40 \times 1\cos 30^\circ$
$T = 17.3$ N

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Moments

Hard

A ladder · Year 13 · 7 marks

A uniform ladder, 4 m long and of mass 15 kg, rests against a smooth wall with its foot on rough ground. It makes

60^\circ

with the ground and is about to slip.

(a)
Find the coefficient of friction between the ladder and the ground.
(7)

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Part (a)7 marks

The smooth wall gives only a horizontal force $S$ .Vertically: $R = 15g = 147$ N
Horizontally: $F = S$
Use perpendicular distances.Moments about the foot: $S \times 4\sin 60^\circ$ ${}= 15g \times 2\cos 60^\circ$
$S = 42.4$ N
About to slip, so friction is at its limit. $\mu$ ${}= \dfrac{F}{R}$ ${}= \dfrac{42.4}{147}$
$\mu = 0.289$

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Moments

Hard

How far can he walk? · Year 13 · 5 marks

A uniform plank

AB

, 8 m long and of mass 40 kg, rests on a support at

A

and a support at

C

, 6 m from

A

. A man of mass 80 kg walks from

A

towards

B

.

(a)
Find how far from $A$ he can walk before the plank starts to tip.
(3)
(b)
Find the reaction at $C$ at that moment.
(2)

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Part (a)3 marks

About to tip about $C$ : the reaction at $A$ is zero
The plank's centre is 4 m from $A$ , 2 m before $C$ .Moments about $C$ : $80g \times d = 40g \times 2$
$d = 1$ m beyond $C$ : 7 m from $A$

Part (b)2 marks

It holds up both weights on its own. $R_C = 120g$
$= 1176$ N

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