Practice questions

Complex numbers

Complex arithmetic

EasyYear 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)

Show the worked answer

Part (a)2 marks

Expand the brackets exactly as you would with $x$ : every term times every term. $zw$ ${}= 3$ ${}- 6\mathrm{i}$ ${}+ 4\mathrm{i}$ ${}- 8\mathrm{i}^2$
M1
Use $\mathrm{i}^2 = -1$ , so $-8\mathrm{i}^2 = +8$ , then collect real and imaginary parts. $= 11 - 2\mathrm{i}$
A1

Part (b)1 mark

The modulus is the distance from the origin on an Argand diagram, so use Pythagoras. $|z| = \sqrt{3^2 + 4^2} = 5$
B1

Part (c)2 marks

Multiply top and bottom by the conjugate of the denominator, which makes the denominator real. $\dfrac{(3 + 4\mathrm{i})(1 + 2\mathrm{i})}{(1 - 2\mathrm{i})(1 + 2\mathrm{i})}$ ${}= \dfrac{3 + 10\mathrm{i} + 8\mathrm{i}^2}{5}$
M1
Replace $\mathrm{i}^2$ with $-1$ and divide each part by 5. $= -1 + 2\mathrm{i}$
A1

Total5

Complex numbers

Roots of a cubic

MediumYear 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)

Show the worked answer

Part (a)5 marks

A polynomial with real coefficients has complex roots in conjugate pairs. $2 + \mathrm{i}$ is also a root.
B1
Two roots give two factors; multiplying them gives a real quadratic factor. $\bigl(z - (2 - \mathrm{i})\bigr)\bigl(z - (2 + \mathrm{i})\bigr)$ ${}= (z - 2)^2 - \mathrm{i}^2$
M1
Expand $(z-2)^2$ and use $\mathrm{i}^2 = -1$ . $= z^2 - 4z + 5$
A1
Divide, or compare coefficients: the constant term 5 = 5 × 1 gives the last factor. $z^3 - 3z^2 + z + 5$ ${}= (z^2 - 4z + 5)(z + 1)$
M1
Set each factor to zero to list all three roots. $z$ ${}= -1,\ 2$ ${}+ \mathrm{i},\ 2$ ${}- \mathrm{i}$
A1

Total5

Matrices

Inverse of a 2 × 2 matrix

EasyYear 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)

Show the worked answer

Part (a)1 mark

For $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ the determinant is $ad - bc$ . $\det\mathbf{A}$ ${}= 2 \times 3 - 1 \times 5$ ${}= 1$
B1

Part (b)2 marks

Swap the numbers on the leading diagonal, change the signs of the other two, then divide by the determinant. $\mathbf{A}^{-1}$ ${}= \dfrac{1}{1}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$
M1
Dividing by 1 changes nothing. $\mathbf{A}^{-1}$ ${}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$
A1

Part (c)1 mark

The determinant is the area scale factor of the transformation.Area $= 4 \times |\det\mathbf{A}| = 4$ square units
B1

Total4

Proof

Induction with a sum

MediumYear 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)

Show the worked answer

Part (a)5 marks

Basis: check the statement is true for the first value. $n = 1$ : left side $= 1 \times 2 = 2$ , right side $= \tfrac{1}{3}(1)(2)(3) = 2$
B1
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 $r = k + 1$ . $\sum_{r=1}^{k+1} r(r+1)$ ${}= \tfrac{1}{3}k(k+1)(k+2)$ ${}+ (k+1)(k+2)$
M1
Take out the common factor $\tfrac{1}{3}(k+1)(k+2)$ instead of expanding. $= \tfrac{1}{3}(k+1)(k+2)(k + 3)$
A1
Show it is the original formula with $n$ replaced by $k + 1$ .This is $\tfrac{1}{3}n(n+1)(n+2)$ with $n = k + 1$ .
A1
Conclusion: true for $n = 1$ , and true for $k$ means true for $k + 1$ .So it is true for all positive integers $n$ .
A1

Total5

Proof

Induction with divisibility

HardYear 12 · 6 marks

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

Show the worked answer

Part (a)6 marks

Basis: 9 is divisible by 9. $\mathrm{f}(1) = 4 + 6 - 1 = 9$
B1
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. $\mathrm{f}(k+1)$ ${}- 4\mathrm{f}(k)$ ${}= 4^{k+1}$ ${}+ 6k$ ${}+ 5$ ${}- 4^{k+1}$ ${}- 24k$ ${}+ 4$
M1
What is left is a multiple of 9. $= 9 - 18k = 9(1 - 2k)$
A1
Rearrange to write $\mathrm{f}(k+1)$ in terms of $\mathrm{f}(k)$ . $\mathrm{f}(k+1)$ ${}= 4\mathrm{f}(k) + 9(1 - 2k)$
M1
$\mathrm{f}(k)$ is a multiple of 9 by the assumption.Both terms are multiples of 9, so $\mathrm{f}(k+1)$ is too.
A1
Conclusion: true for $n = 1$ , and true for $k$ means true for $k + 1$ .So it is true for all positive integers $n$ .
A1

Total6

Further algebra and functions

Roots of a cubic and their squares

MediumYear 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)

Show the worked answer

Part (a)2 marks

For $ax^3 + bx^2 + cx + d$ : $\sum\alpha = -\tfrac{b}{a}$ , $\sum\alpha\beta = \tfrac{c}{a}$ , $\alpha\beta\gamma = -\tfrac{d}{a}$ . $\sum\alpha$ ${}= \tfrac{3}{2},$ $\quad \sum\alpha\beta$ ${}= 2,$ $\quad \alpha\beta\gamma$ ${}= \tfrac{5}{2}$
B1 B1

Part (b)3 marks

Squaring the sum of the roots gives the squares plus twice every pair. $\alpha^2 + \beta^2 + \gamma^2$ ${}= (\sum\alpha)^2$ ${}- 2\sum\alpha\beta$
M1
Substitute the values from part (a). $= \tfrac{9}{4} - 4 = -\tfrac{7}{4}$
A1
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.
B1

Total5

Further algebra and functions

Roots of a quartic

HardYear 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)

Show the worked answer

Part (a)3 marks

For $ax^4 + bx^3 + cx^2 + dx + e$ : $\sum\alpha\beta\gamma$ ${}= -\tfrac{d}{a}$ and $\alpha\beta\gamma\delta$ ${}= \tfrac{e}{a}$ . $\sum\alpha\beta\gamma$ ${}= 1,$ $\quad \alpha\beta\gamma\delta$ ${}= 3$
B1
Put the four fractions over the common denominator $\alpha\beta\gamma\delta$ . $\sum\dfrac{1}{\alpha}$ ${}= \dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}$
M1
Substitute the values. $= \tfrac{1}{3}$
A1

Part (b)3 marks

$\sum\alpha = -\tfrac{b}{a}$ and $\sum\alpha\beta = \tfrac{c}{a}$ ; there is no $x^2$ term, so $c = 0$ . $\sum\alpha$ ${}= -2,$ $\quad \sum\alpha\beta$ ${}= 0$
B1
The same identity as for a cubic: it works for any number of roots. $\sum\alpha^2$ ${}= (\sum\alpha)^2$ ${}- 2\sum\alpha\beta$
M1
Substitute the values. $= 4 - 0 = 4$
A1

Total6

Further algebra and functions

Method of differences

MediumYear 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)

Show the worked answer

Part (a)1 mark

Combine the right-hand side over a common denominator. $\dfrac{1}{r} - \dfrac{1}{r+1}$ ${}= \dfrac{(r + 1) - r}{r(r + 1)}$ ${}= \dfrac{1}{r(r+1)}$
B1

Part (b)3 marks

Write out the first few and last terms to see which parts cancel. $\left(1 - \tfrac{1}{2}\right)$ ${}+ \left(\tfrac{1}{2} - \tfrac{1}{3}\right)$ ${}+ \dots$ ${}+ \left(\tfrac{1}{n} - \tfrac{1}{n+1}\right)$
M1
Every term cancels except the first and the last. $= 1 - \dfrac{1}{n + 1}$
A1
Write it as a single fraction, as asked. $= \dfrac{n}{n + 1}$
A1

Total4

Further vectors

Angle between two lines

EasyYear 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)

Show the worked answer

Part (a)3 marks

The angle between lines is the angle between their direction vectors, found with the scalar product. $\mathbf{a}\cdot\mathbf{b}$ ${}= 2 - 2 + 4$ ${}= 4$
M1
Both vectors have magnitude $\sqrt{1 + 4 + 4} = 3$ . $\cos\theta$ ${}= \dfrac{4}{3 \times 3}$ ${}= \dfrac{4}{9}$
M1
$\cos\theta$ is positive, so this angle is already acute. $\theta = 63.6^\circ$
A1

Total3

Polar coordinates

Area inside a polar curve

MediumYear 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)

Show the worked answer

Part (a)5 marks

Area in polar coordinates is $\tfrac{1}{2}\int r^2\,\mathrm{d}\theta$ over one full turn. $A$ ${}= \tfrac{1}{2}\int_0^{2\pi} 4(1 + \cos\theta)^2\,\mathrm{d}\theta$
M1
Expand the bracket before integrating. $= 2\int_0^{2\pi} \left(1 + 2\cos\theta + \cos^2\theta\right)\mathrm{d}\theta$
M1
$\cos^2\theta$ cannot be integrated directly, so use the double angle formula. $\cos^2\theta$ ${}= \tfrac{1}{2}(1 + \cos 2\theta)$
M1
Integrate each term. $= 2\left[\tfrac{3}{2}\theta + 2\sin\theta + \tfrac{1}{4}\sin 2\theta\right]_0^{2\pi}$
A1
The sine terms are zero at both limits. $= 6\pi$
A1

Total5

Hyperbolic functions

Solving a hyperbolic equation

EasyYear 13 · 4 marks

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

Show the worked answer

Part (a)4 marks

Replace $\cosh x$ with its definition in exponentials. $\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}$ ${}= 2$
M1
Multiply through by $2\mathrm{e}^x$ to get a quadratic in $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}- 4\mathrm{e}^x$ ${}+ 1$ ${}= 0$
M1
Solve the quadratic with the formula. $\mathrm{e}^x = 2 \pm \sqrt{3}$
A1
Both values of $\mathrm{e}^x$ are positive, so both give solutions. $x = \ln(2 + \sqrt{3})$ or $x$ ${}= \ln(2 - \sqrt{3})$ ${}= -\ln(2 + \sqrt{3})$
A1

Total4

Differential equations

Integrating factor

MediumYear 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)

Show the worked answer

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 $= \mathrm{e}^{\int \frac{1}{x}\,\mathrm{d}x}$ ${}= x$
M1
Multiplying by $x$ makes the left side the derivative of a product. $\dfrac{\mathrm{d}}{\mathrm{d}x}(xy)$ ${}= x^3$
A1
Integrate both sides; remember the constant. $xy = \tfrac{1}{4}x^4 + c$
M1
Use the given point to find $c$ . $1$ ${}= \tfrac{1}{4} + c$ ${}\Rightarrow c$ ${}= \tfrac{3}{4}$
M1
Divide by $x$ to make $y$ the subject. $y = \dfrac{x^3}{4} + \dfrac{3}{4x}$
A1

Total5

Differential equations

Modelling damped motion

HardYear 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)

Show the worked answer

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)
M1
A repeated root gives this form of general solution. $x = (A + Bt)\mathrm{e}^{-2t}$
A1
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 $B = 2$
M1
Substitute the constants. $x = 2t\mathrm{e}^{-2t}$
A1

Part (b)4 marks

The greatest displacement is where the particle stops, so $\dot{x} = 0$ . $\dot{x}$ ${}= 2\mathrm{e}^{-2t}(1 - 2t)$ ${}= 0$ ${}\Rightarrow t$ ${}= \tfrac{1}{2}$
M1
Substitute this time into $x$ . $x$ ${}= 2 \times \tfrac{1}{2} \times \mathrm{e}^{-1}$ ${}= \mathrm{e}^{-1}$
M1
Give a sensible number of significant figures.Greatest displacement $= 0.368$ m
A1
The auxiliary equation has a repeated root, so the particle returns to equilibrium without oscillating.Critical damping
B1

Total8

Proof

Proof by deduction

EasyYear 12 · 3 marks

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

Show the worked answer

Part (a)3 marks

Factorise to see the structure of the expression. $n^2 - n = n(n - 1)$
M1
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.
A1
Finish with a clear statement of what has been proved.An even number times any integer is even, so $n^2 - n$ is even.
A1

Total3

Algebra and functions

Equal roots

EasyYear 12 · 3 marks

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

Show the worked answer

Part (a)3 marks

Equal (repeated) roots happen when the discriminant is zero. $b^2 - 4ac = 0$
M1
Here $a = 1$ , $b = k$ and $c = 9$ . $k^2 - 4 \times 1 \times 9$ ${}= 0$ ${}\Rightarrow k^2$ ${}= 36$
A1
Square-rooting gives two answers: positive and negative. $k = 6$ or $k = -6$
A1

Total3

Sequences and series

Geometric series

MediumYear 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)

Show the worked answer

Part (a)2 marks

Use $S_\infty = \dfrac{a}{1 - r}$ . $\dfrac{12}{1 - r} = 48$
M1
Rearrange for $r$ . It is between −1 and 1, as it must be. $1 - r$ ${}= \tfrac{1}{4}$ ${}\Rightarrow r$ ${}= \tfrac{3}{4}$
A1

Part (b)2 marks

Use $S_n = \dfrac{a(1 - r^n)}{1 - r}$ . $S_{10}$ ${}= \dfrac{12\left(1 - 0.75^{10}\right)}{1 - 0.75}$
M1
Round to 3 significant figures as asked. $= 45.3$
A1

Total4

Trigonometry

A quadratic in cosine

MediumYear 13 · 5 marks

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

Show the worked answer

Part (a)5 marks

Use $\sin^2 x = 1 - \cos^2 x$ so there is only one trig function. $2(1 - \cos^2 x) = 3\cos x$
M1
Rearrange into a quadratic in $\cos x$ . $2\cos^2 x + 3\cos x - 2 = 0$
A1
Factorise as you would $2c^2 + 3c - 2$ . $(2\cos x - 1)(\cos x + 2) = 0$
M1
Cosine is always between −1 and 1. $\cos x = \tfrac{1}{2}$ ( $\cos x = -2$ has no solutions)
A1
Cosine is positive in the first and fourth quadrants. $x = 60^\circ,\ 300^\circ$
A1

Total5

Exponentials and logarithms

Modelling cooling

HardYear 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)

Show the worked answer

Part (a)3 marks

Substitute $t = 10$ and $T = 50$ , then isolate the exponential. $50$ ${}= 20 + 70\mathrm{e}^{-10k}$ ${}\Rightarrow \mathrm{e}^{-10k}$ ${}= \tfrac{3}{7}$
M1
Take natural logs of both sides to bring down the power. $-10k$ ${}= \ln\tfrac{3}{7}$ ${}\Rightarrow k$ ${}= \tfrac{1}{10}\ln\tfrac{7}{3}$
M1
Evaluate and round. $k = 0.0847$
A1

Part (b)3 marks

Substitute $T = 30$ and isolate the exponential. $\mathrm{e}^{-kt} = \tfrac{1}{7}$
M1
Take logs; $\ln\tfrac{1}{7} = -\ln 7$ . $t = \dfrac{\ln 7}{k}$
M1
Use the unrounded value of $k$ to avoid rounding errors. $t = 23.0$ minutes
A1

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.
B1

Total7

Differentiation

Gradient of a curve

EasyYear 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)

Show the worked answer

Part (a)3 marks

Differentiate each term: multiply by the power, then reduce the power by 1. The constant disappears. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 6x^2 - 10x$
M1 A1
The gradient at a point is the derivative evaluated there.At $x = 2$ : $24 - 20 = 4$
A1

Total3

Differentiation

Stationary points

MediumYear 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)

Show the worked answer

Part (a)6 marks

Stationary points are where the gradient is zero, so differentiate first. $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 3x^2 - 12x + 9$
M1
Set the derivative to zero and solve. $3(x - 1)(x - 3)$ ${}= 0$ ${}\Rightarrow x$ ${}= 1,\ x$ ${}= 3$
A1
Substitute each $x$ into the original equation to get $y$ . $(1, 5)$ and $(3, 1)$
A1
The second derivative tells you the shape at each point. $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ ${}= 6x - 12$
M1
Negative second derivative means the curve is bending down.At $x = 1$ : $-6 < 0$ , so $(1, 5)$ is a maximum.
A1
Positive second derivative means the curve is bending up.At $x = 3$ : $6 > 0$ , so $(3, 1)$ is a minimum.
A1

Total6

Integration

A definite integral

EasyYear 12 · 4 marks

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

Show the worked answer

Part (a)4 marks

Write $\frac{2}{x^2}$ as a power of $x$ so you can integrate it. $\int \left(3x^2 + 2x^{-2}\right)\mathrm{d}x$
M1
Add 1 to each power and divide by the new power. $= \left[x^3 - 2x^{-1}\right]_1^2$
A1
Substitute the top limit, then subtract the value at the bottom limit. $= (8 - 1) - (1 - 2)$
M1
Take care with the double negative. $= 8$
A1

Total4

Statistical distributions

Binomial probabilities

MediumYear 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)

Show the worked answer

Part (a)1 mark

Use the cumulative binomial function on your calculator with $n = 20$ , $p = 0.3$ . $\mathrm{P}(X \le 4) = 0.2375$
B1

Part (b)2 marks

The calculator gives 'less than or equal to', so use the complement. $\mathrm{P}(X \ge 8)$ ${}= 1 - \mathrm{P}(X \le 7)$
M1
Subtract from 1. $= 1 - 0.7723 = 0.2277$
A1

Total3

Statistical hypothesis testing

Is the coin biased?

HardYear 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)

Show the worked answer

Part (a)5 marks

$p$ is the probability of heads. The claim is 'towards heads', so the test is one-tailed. $\mathrm{H_0}: p$ ${}= 0.5,$ $\quad \mathrm{H_1}: p > 0.5$
B1
Assume the null hypothesis is true to find how likely the result is.Under $\mathrm{H_0}$ , $X \sim \mathrm{B}(20, 0.5)$
M1
Find the probability of a result at least as extreme as the one seen. $\mathrm{P}(X \ge 15)$ ${}= 1 - \mathrm{P}(X \le 14)$ ${}= 0.0207$
A1
Compare the probability with the significance level. $0.0207 < 0.05$ , so reject $\mathrm{H_0}$
M1
Always give the conclusion in the context of the question, without certainty.There is evidence that the coin is biased towards heads.
A1

Total5

Kinematics

Constant acceleration

EasyYear 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)

Show the worked answer

Part (a)2 marks

Use $v = u + at$ with $u = 4$ , $v = 20$ , $t = 8$ . $20 = 4 + 8a$
M1
Solve for $a$ . $a = 2\ \mathrm{m\,s^{-2}}$
A1

Part (b)2 marks

Use $s = \tfrac{1}{2}(u + v)t$ , which needs no acceleration. $s = \tfrac{1}{2}(4 + 20) \times 8$
M1
Include units. $s = 96$ m
A1

Total4

Forces and Newton's laws

Particle on an inclined plane

MediumYear 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)

Show the worked answer

Part (a)4 marks

Resolve along the plane and use $F = ma$ . The weight's component down the slope is $mg\sin 30^\circ$ . $15 - 2g\sin 30^\circ = 2a$
M1 A1
$2 \times 9.8 \times 0.5 = 9.8$ . $15 - 9.8 = 2a$
M1
The plane is smooth, so there is no friction to include. $a = 2.6\ \mathrm{m\,s^{-2}}$ up the plane
A1

Total4

Electric Circuits

Resistance of a wire

EasyYear 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)

Show the worked answer

Part (a)2 marks

Resistivity links resistance to the length and area of a wire. $R$ ${}= \dfrac{\rho L}{A}$ ${}= \dfrac{4.9 \times 10^{-7} \times 0.80}{2.0 \times 10^{-7}}$
M1
Give the unit, ohms. $R = 2.0\ \Omega$
A1

Total2

Electric Circuits

Series and parallel resistors

EasyYear 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)

Show the worked answer

Part (a)3 marks

Resistors in parallel combine with the reciprocal rule. $\dfrac{1}{R}$ ${}= \dfrac{1}{6.0}$ ${}+ \dfrac{1}{3.0}$ ${}\Rightarrow R$ ${}= 2.0\ \Omega$
M1
Resistances in series add.Total $= 2.0 + 4.0 = 6.0\ \Omega$
M1
Use $I = V/R$ for the whole circuit. $I$ ${}= \dfrac{12}{6.0}$ ${}= 2.0\ \mathrm{A}$
A1

Part (b)1 mark

The full current passes through the 2.0 Ω parallel pair, and both parallel resistors share its p.d. $V$ ${}= 2.0 \times 2.0$ ${}= 4.0\ \mathrm{V}$
B1

Total4

Mechanics

Horizontal projection

MediumYear 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)

Show the worked answer

Part (a)3 marks

Treat vertical motion on its own: it starts with zero vertical velocity and accelerates at $g$ . $1.8$ ${}= \tfrac{1}{2} \times 9.81 \times t^2$
M1
Solve for the time of flight. $t = 0.606\ \mathrm{s}$
A1
Horizontal velocity is constant, so distance = speed × time.Distance $= 5.0 \times 0.606$ ${}= 3.0\ \mathrm{m}$
A1

Total3

Mechanics

A collision

MediumYear 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)

Show the worked answer

Part (a)2 marks

Momentum is conserved: total before = total after. $2.0 \times 3.0 = 3.0 \times v$
M1
The trolleys move together, so they share one velocity. $v = 2.0\ \mathrm{m\,s^{-1}}$
A1

Part (b)2 marks

Use $E_k = \tfrac{1}{2}mv^2$ before and after.Kinetic energy before $= \tfrac{1}{2} \times 2.0 \times 3.0^2$ ${}= 9.0\ \mathrm{J}$
M1
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.
A1

Total4

Waves and Particle Nature of Light

The wave equation

EasyYear 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)

Show the worked answer

Part (a)2 marks

Wave speed = frequency × wavelength. $v = f\lambda = 440 \times 0.78$
M1
Round to match the data. $v = 343\ \mathrm{m\,s^{-1}}$
A1

Total2

Waves and Particle Nature of Light

The photoelectric effect

MediumYear 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)

Show the worked answer

Part (a)4 marks

Find the energy of one photon. Convert nm to m first. $E$ ${}= \dfrac{hc}{\lambda}$ ${}= \dfrac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{4.00 \times 10^{-7}}$ ${}= 4.97 \times 10^{-19}\ \mathrm{J}$
M1
Convert the work function to joules so the units match. $\phi$ ${}= 2.3 \times 1.60 \times 10^{-19}$ ${}= 3.68 \times 10^{-19}\ \mathrm{J}$
M1
$hf = \phi + E_{k(\max)}$ : the photon's energy pays the work function and the rest becomes kinetic energy. $E_{k(\max)}$ ${}= 4.97 \times 10^{-19}$ ${}- 3.68 \times 10^{-19}$ ${}= 1.29 \times 10^{-19}\ \mathrm{J}$
M1
Divide by $1.60 \times 10^{-19}$ to convert back to electronvolts. $= 0.81\ \mathrm{eV}$
A1

Total4

Oscillations

A simple pendulum

MediumYear 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)

Show the worked answer

Part (a)2 marks

For small swings a pendulum moves with simple harmonic motion. $T$ ${}= 2\pi\sqrt{\dfrac{l}{g}}$ ${}= 2\pi\sqrt{\dfrac{0.90}{9.81}}$
M1
Give 3 significant figures. $T = 1.90\ \mathrm{s}$
A1

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.
B1

Total3

Gravitational Fields

Geostationary orbit

HardYear 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)

Show the worked answer

Part (a)4 marks

Gravity provides the centripetal force that keeps the satellite in its circle. $\dfrac{GMm}{r^2}$ ${}= mr\left(\dfrac{2\pi}{T}\right)^2$
M1
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
M1
Take the cube root. This is measured from the centre of the Earth. $r = 4.22 \times 10^{7}\ \mathrm{m}$
A1
Subtract the Earth's radius to get the height above the surface.Height $= 4.22 \times 10^{7}$ ${}- 6.37 \times 10^{6}$ ${}= 3.59 \times 10^{7}\ \mathrm{m}$
A1

Total4

Electric and Magnetic Fields

Capacitor discharge

HardYear 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)

Show the worked answer

Part (a)1 mark

Convert kΩ and μF to base units before multiplying. $RC$ ${}= 10 \times 10^{3} \times 470 \times 10^{-6}$ ${}= 4.7\ \mathrm{s}$
B1

Part (b)2 marks

During discharge, $V = V_0\mathrm{e}^{-t/RC}$ . $V = 12\mathrm{e}^{-5.0/4.7}$
M1
Evaluate and round sensibly. $V = 4.1\ \mathrm{V}$
A1

Part (c)2 marks

Divide by 12, take natural logs, and rearrange for $t$ . $3.0$ ${}= 12\mathrm{e}^{-t/4.7}$ ${}\Rightarrow t$ ${}= 4.7\ln 4$
M1
Falling to a quarter takes two half-lives' worth of time: $2 \times 0.69RC$ . $t = 6.5\ \mathrm{s}$
A1

Total5

Nuclear Radiation

Half-life

MediumYear 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)

Show the worked answer

Part (a)2 marks

The decay constant and half-life are linked by $\lambda t_{1/2} = \ln 2$ . Convert hours to seconds. $\lambda$ ${}= \dfrac{\ln 2}{t_{1/2}}$ ${}= \dfrac{0.693}{6.0 \times 3600}$
M1
Give the unit. $\lambda$ ${}= 3.2 \times 10^{-5}\ \mathrm{s^{-1}}$
A1

Part (b)2 marks

Count how many half-lives fit into the time.24 hours = 4 half-lives
M1
Each half-life halves the number left.Fraction $= \left(\tfrac{1}{2}\right)^4$ ${}= \tfrac{1}{16}$
A1

Total4

Proof

Sum of odd numbers

EasyYear 12 · 4 marks

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

Show the worked answer

Part (a)4 marks

Start by checking the first case works. $n = 1$ : left side $= 1$ , right side $= 1^2 = 1$
B1
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 $r = k + 1$ . $\sum_{r=1}^{k+1} (2r - 1)$ ${}= k^2 + 2(k + 1) - 1$ ${}= k^2 + 2k + 1$
M1
This is the formula with $n = k + 1$ , which is what we wanted. $= (k + 1)^2$
A1
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 $n$ .
A1

Total4

Proof

Sum of the first n integers

EasyYear 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)

Show the worked answer

Part (a)4 marks

Check the first case. $n = 1$ : left side $= 1$ , right side $= \tfrac{1}{2}(1)(2) = 1$
B1
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 $\tfrac{1}{2}(k + 1)$ . $\sum_{r=1}^{k+1} r$ ${}= \tfrac{1}{2}k(k + 1)$ ${}+ (k + 1)$ ${}= \tfrac{1}{2}(k + 1)(k + 2)$
M1 A1
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 $n$ .
A1

Total4

Proof

Divisibility by 2

EasyYear 12 · 4 marks

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

Show the worked answer

Part (a)4 marks

Check the first case. $n = 1$ : $3 - 1 = 2$ , which is divisible by 2
B1
Suppose it is true when $n = k$ .Assume $3^k - 1$ is divisible by 2.
Write the next case using the one we assumed: $3 \times 3^k - 3 + 2$ . $3^{k+1} - 1 = 3(3^k - 1) + 2$
M1
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.
A1
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 $n$ .
A1

Total4

Proof

Induction with a sequence

EasyYear 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)

Show the worked answer

Part (a)4 marks

Check the formula gives the first term. $n = 1$ : $2^1 - 1 = 1 = u_1$
B1
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. $u_{k+1}$ ${}= 2(2^k - 1) + 1$ ${}= 2^{k+1} - 1$
M1 A1
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 $n$ .
A1

Total4

Proof

Induction with matrices

EasyYear 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)

Show the worked answer

Part (a)4 marks

The formula with $n = 1$ gives $\mathbf{M}$ itself. $n = 1$ : $\mathbf{M}^1$ ${}= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$
B1
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. $\mathbf{M}^{k+1}$ ${}= \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ ${}= \begin{pmatrix} 1 & k + 1 \\ 0 & 1 \end{pmatrix}$
M1 A1
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 $n$ .
A1

Total4

Proof

Induction with powers of 2

MediumYear 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)

Show the worked answer

Part (a)5 marks

Check the first case. $n = 1$ : left side $= 2$ , right side $= 2 + 0 = 2$
B1
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 $r = k + 1$ . $\sum_{r=1}^{k+1} r \times 2^r$ ${}= 2$ ${}+ (k - 1)2^{k+1}$ ${}+ (k + 1)2^{k+1}$
M1
Collect the $2^{k+1}$ terms: $(k - 1) + (k + 1) = 2k$ . $= 2 + 2k \times 2^{k+1}$ ${}= 2 + k \times 2^{k+2}$
A1
Rewrite it to match the formula exactly. $= 2$ ${}+ \bigl((k + 1) - 1\bigr)2^{(k+1)+1}$ , the formula with $n = k + 1$
A1
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 $n$ .
A1

Total5

Proof

Divisibility by 4

MediumYear 12 · 5 marks

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

Show the worked answer

Part (a)5 marks

Check the first case. $n = 1$ : $7 - 3 = 4$
B1
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 $3^k$ . $7^{k+1} - 3^{k+1}$ ${}= 7 \times 7^k - 3 \times 3^k$
M1
Split $7 \times 3^k$ off so the assumed expression appears: $-3 \times 3^k$ ${}= -7 \times 3^k + 4 \times 3^k$ . $= 7(7^k - 3^k) + 4 \times 3^k$
A1
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.
A1
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 $n$ .
A1

Total5

Proof

Induction with derivatives

HardYear 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)

Show the worked answer

Part (a)5 marks

Use the product rule to check the first case. $n = 1$ : $\dfrac{\mathrm{d}}{\mathrm{d}x}(x\mathrm{e}^x)$ ${}= \mathrm{e}^x + x\mathrm{e}^x$ ${}= (x + 1)\mathrm{e}^x$
B1
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. $\dfrac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}}(x\mathrm{e}^x)$ ${}= \dfrac{\mathrm{d}}{\mathrm{d}x}\bigl((x + k)\mathrm{e}^x\bigr)$
M1
Product rule again, then factorise. This is the formula with $n = k + 1$ . $= \mathrm{e}^x$ ${}+ (x + k)\mathrm{e}^x$ ${}= \bigl(x + (k + 1)\bigr)\mathrm{e}^x$
M1 A1
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 $n$ .
A1

Total5

Complex numbers

Complex roots of a quadratic

EasyYear 12 · 3 marks

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

Show the worked answer

Part (a)3 marks

Use the quadratic formula; the discriminant is negative, so the roots are complex. $z$ ${}= \dfrac{-6 \pm \sqrt{36 - 52}}{2}$
M1
$\sqrt{-16} = 4\mathrm{i}$ . $= \dfrac{-6 \pm \sqrt{-16}}{2}$ ${}= \dfrac{-6 \pm 4\mathrm{i}}{2}$
M1
The roots are a conjugate pair, as expected for real coefficients. $z = -3 + 2\mathrm{i}$ or $z = -3 - 2\mathrm{i}$
A1

Total3

Complex numbers

Modulus and argument

EasyYear 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)

Show the worked answer

Part (a)1 mark

The modulus is the distance from the origin. $|z|$ ${}= \sqrt{1^2 + (\sqrt{3})^2}$ ${}= 2$
B1

Part (b)1 mark

$z$ is in the first quadrant, so no adjustment is needed. $\arg z$ ${}= \tan^{-1}\dfrac{\sqrt{3}}{1}$ ${}= \dfrac{\pi}{3}$
B1

Part (c)1 mark

Modulus–argument form is $r(\cos\theta + \mathrm{i}\sin\theta)$ . $z$ ${}= 2\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right)$
B1

Total3

Complex numbers

The complex conjugate

EasyYear 12 · 3 marks

z = 5 - 2\mathrm{i}

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

Show the worked answer

Part (a)1 mark

The conjugate changes the sign of the imaginary part. $z^* = 5 + 2\mathrm{i}$
B1

Part (b)2 marks

The imaginary parts cancel, leaving twice the real part. $z + z^* = 10$
B1
This is a difference of two squares, and equals $|z|^2$ . $zz^* = 25 - 4\mathrm{i}^2 = 29$
B1

Total3

Complex numbers

A circle in the Argand diagram

EasyYear 12 · 2 marks

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

Show the worked answer

Part (a)2 marks

$|z - a|$ is the distance from $z$ to the point $a$ . $|z - (2 + \mathrm{i})| = 3$
M1
Every point exactly 3 away from $2 + \mathrm{i}$ .A circle with centre $(2, 1)$ and radius 3
A1

Total2

Complex numbers

De Moivre's theorem

MediumYear 13 · 4 marks

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

Show the worked answer

Part (a)4 marks

Modulus $\sqrt{2}$ , argument $\tfrac{\pi}{4}$ . $1 + \mathrm{i}$ ${}= \sqrt{2}\,\mathrm{e}^{\mathrm{i}\pi/4}$
M1 A1
Raise the modulus to the power 8 and multiply the argument by 8. $(1 + \mathrm{i})^8$ ${}= (\sqrt{2})^8\mathrm{e}^{2\pi\mathrm{i}}$
M1
$\mathrm{e}^{2\pi\mathrm{i}} = 1$ . $= 16$
A1

Total4

Complex numbers

Cube roots

MediumYear 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)

Show the worked answer

Part (a)4 marks

Write the right side in exponential form, including all the equivalent arguments. $8\mathrm{i}$ ${}= 8\mathrm{e}^{\mathrm{i}(\pi/2 + 2k\pi)}$
M1
Take the cube root of the modulus and divide the argument by 3. $z$ ${}= 2\mathrm{e}^{\mathrm{i}(\pi/6 + 2k\pi/3)}$
M1
Use $k = 0, 1, -1$ to keep the arguments in range. The three roots are equally spaced. $z$ ${}= 2\mathrm{e}^{\mathrm{i}\pi/6},\ 2\mathrm{e}^{5\mathrm{i}\pi/6},\ 2\mathrm{e}^{-\mathrm{i}\pi/2}$
A1 A1

Total4

Complex numbers

Multiple angles

HardYear 13 · 5 marks

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

Show the worked answer

Part (a)5 marks

De Moivre's theorem with $n = 3$ . $\cos 3\theta$ ${}+ \mathrm{i}\sin 3\theta$ ${}= (\cos\theta + \mathrm{i}\sin\theta)^3$
M1
Expand with the binomial theorem, writing $c = \cos\theta$ and $s = \sin\theta$ . $= c^3$ ${}+ 3c^2(\mathrm{i}s)$ ${}+ 3c(\mathrm{i}s)^2$ ${}+ (\mathrm{i}s)^3$
M1
$\mathrm{i}^2 = -1$ , so the real terms are $c^3$ and $-3cs^2$ .Real parts: $\cos 3\theta = c^3 - 3cs^2$
A1
Use $s^2 = 1 - c^2$ to write everything in cosines. $= c^3 - 3c(1 - c^2)$
M1
Collect the terms. $= 4\cos^3\theta - 3\cos\theta$
A1

Total5

Complex numbers

Two loci

HardYear 12 · 4 marks

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

Show the worked answer

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 $x = 2$
M1 A1
Points at angle $\tfrac{\pi}{4}$ from the positive real axis. $\arg z = \tfrac{\pi}{4}$ : the half-line $y = x$ , $x > 0$
M1
Where the two loci cross: $x = 2$ , $y = 2$ . $z = 2 + 2\mathrm{i}$
A1

Total4

Matrices

Multiplying matrices

EasyYear 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)

Show the worked answer

Part (a)2 marks

Each entry is a row of $\mathbf{A}$ times a column of $\mathbf{B}$ . $\mathbf{AB}$ ${}= \begin{pmatrix} 1(0) + 2(1) & 1(1) + 2(0) \\ 3(0) + 4(1) & 3(1) + 4(0) \end{pmatrix}$
M1
Multiplying by $\mathbf{B}$ on the right swaps the columns. $= \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$
A1

Total2

Matrices

Describing a transformation

EasyYear 12 · 2 marks

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

Show the worked answer

Part (a)2 marks

The columns show where the unit vectors go. $\begin{pmatrix} 1 \\ 0 \end{pmatrix} \to \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ , $\begin{pmatrix} 0 \\ 1 \end{pmatrix} \to \begin{pmatrix} -1 \\ 0 \end{pmatrix}$
M1
Both unit vectors turn a quarter turn anticlockwise.A rotation of $90^\circ$ anticlockwise about the origin
A1

Total2

Matrices

Enlargement

EasyYear 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)

Show the worked answer

Part (a)1 mark

Each unit vector is multiplied by 3. $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$
B1

Part (b)1 mark

The area scale factor is the determinant: $3 \times 3 - 0 = 9$ . $\det = 9$
B1

Total2

Matrices

A singular matrix

EasyYear 12 · 2 marks

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

Show the worked answer

Part (a)2 marks

A matrix is singular when its determinant is zero. $6k - 6 = 0$
M1
Solve for $k$ . $k = 1$
A1

Total2

Matrices

Solving equations with an inverse

MediumYear 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)

Show the worked answer

Part (a)3 marks

Write the equations as $\mathbf{A}\mathbf{x} = \mathbf{b}$ , so $\mathbf{x}$ ${}= \mathbf{A}^{-1}\mathbf{b}$ . $\begin{pmatrix} x \\ y \end{pmatrix}$ ${}= \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}\begin{pmatrix} 5 \\ 13 \end{pmatrix}$
M1
Multiply out. $= \begin{pmatrix} 15 - 13 \\ -25 + 26 \end{pmatrix}$
M1
Check: $4 + 1 = 5$ and $10 + 3 = 13$ . $x = 2,\ y = 1$
A1

Total3

Matrices

A 3 × 3 determinant

MediumYear 12 · 3 marks

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

Show the worked answer

Part (a)3 marks

Expand along the top row, alternating signs $+, -, +$ . $2\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}$ ${}- 1\begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix}$ ${}+ 0$
M1
Work out each 2 × 2 determinant. $= 2(5) - 1(2)$
M1
$= 8$
A1

Total3

Matrices

Invariant lines

MediumYear 12 · 4 marks

\mathbf{M}

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

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

Show the worked answer

Part (a)4 marks

Transform a general point on the line. $\begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} x \\ mx \end{pmatrix}$ ${}= \begin{pmatrix} x + 2mx \\ 3mx \end{pmatrix}$
M1
For the line to be invariant, the image must also satisfy $y = mx$ . $3mx = m(x + 2mx)$
M1
Divide by $x$ and solve. $2m^2 - 2m = 0 \Rightarrow m = 0$ or $m = 1$
A1
$y = 0$ and $y = x$
A1

Total4

Matrices

Three planes

HardYear 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)

Show the worked answer

Part (a)2 marks

If the matrix of coefficients is singular, there is no unique solution. $\det\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{pmatrix}$ ${}= 1(-1) - 1(-2) + 1(-1)$ ${}= 0$
M1
The matrix is singular, so there is no single point of intersection.
A1

Part (b)3 marks

Look for a combination of equations that gives the third.The first equation plus the second gives $2x + 3y + 4z = 20$ .
M1
The third plane contains every point the first two share.This is the third equation, so the equations are consistent.
A1
Consistent but singular means infinitely many solutions on a line.The planes form a sheaf: they all meet in one common line.
A1

Total5

Matrices

Combining transformations

HardYear 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)

Show the worked answer

Part (a)3 marks

Write each matrix from where the unit vectors go. $\mathbf{P}$ ${}= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\ \mathbf{Q}$ ${}= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
B1
$\mathbf{P}$ first means $\mathbf{P}$ goes on the right: $\mathbf{QP}$ . $\mathbf{QP}$ ${}= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
M1 A1

Part (b)1 mark

It sends $x$ to $-x$ and keeps $y$ .A reflection in the $y$ -axis
B1

Total4

Further algebra and functions

Roots of a quadratic

EasyYear 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)

Show the worked answer

Part (a)1 mark

Sum $= -\tfrac{b}{a}$ , product $= \tfrac{c}{a}$ . $\alpha + \beta$ ${}= 5,\ \alpha\beta$ ${}= 7$
B1

Part (b)2 marks

Squaring the sum gives the squares plus $2\alpha\beta$ . $\alpha^2 + \beta^2$ ${}= (\alpha + \beta)^2$ ${}- 2\alpha\beta$
M1
$= 25 - 14 = 11$
A1

Total3

Further algebra and functions

Roots of a quartic

EasyYear 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)

Show the worked answer

Part (a)4 marks

$-\tfrac{b}{a}$ , with $a = 2$ , $b = -3$ . $\sum\alpha = \tfrac{3}{2}$
B1
$\tfrac{c}{a}$ : there is no $x^2$ term, so $c = 0$ . $\sum\alpha\beta = 0$
B1
$-\tfrac{d}{a}$ , with $d = 1$ . $\sum\alpha\beta\gamma$ ${}= -\tfrac{1}{2}$
B1
$\tfrac{e}{a}$ , with $e = -6$ . $\alpha\beta\gamma\delta = -3$
B1

Total4

Further algebra and functions

Using standard sums

EasyYear 12 · 4 marks

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

Show the worked answer

Part (a)4 marks

Expand, then split into standard sums. $\sum r(r + 2) = \sum r^2 + 2\sum r$
M1
Use the formulae for $\sum r^2$ and $\sum r$ . $= \tfrac{1}{6}n(n+1)(2n+1)$ ${}+ n(n+1)$
A1
Take out the common factor $\tfrac{1}{6}n(n+1)$ . $= \tfrac{1}{6}n(n+1)\bigl((2n + 1) + 6\bigr)$
M1
$= \tfrac{1}{6}n(n+1)(2n + 7)$
A1

Total4

Further algebra and functions

Sum of squares

EasyYear 12 · 2 marks

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

Show the worked answer

Part (a)2 marks

Use $\sum r^2$ ${}= \tfrac{1}{6}n(n+1)(2n+1)$ with $n = 10$ . $\tfrac{1}{6}(10)(11)(21)$
M1
$= 385$
A1

Total2

Further algebra and functions

A Maclaurin series

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

Replace $x$ with $2x$ in the series for $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}= 1$ ${}+ 2x$ ${}+ \dfrac{(2x)^2}{2!}$ ${}+ \dfrac{(2x)^3}{3!}$
M1
Simplify each term. $= 1 + 2x + 2x^2 + \tfrac{4}{3}x^3$
A1

Total2

Further algebra and functions

Transformed roots

MediumYear 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)

Show the worked answer

Part (a)3 marks

Each new root is double an old root.Let $w = 2x$ , so $x = \tfrac{w}{2}$
M1
Substitute into the original equation. $\dfrac{w^3}{8} - w + 5 = 0$
M1
Multiply by 8 to clear the fraction. $w^3 - 8w + 40 = 0$
A1

Total3

Further algebra and functions

Series for ln(cos x)

HardYear 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)

Show the worked answer

Part (a)5 marks

Write $\cos x$ as 1 plus a small quantity $u$ . $\cos x = 1 + u$ , where $u$ ${}= -\dfrac{x^2}{2}$ ${}+ \dfrac{x^4}{24}$
M1
Use the standard series; higher powers of $u$ give terms beyond $x^4$ . $\ln(1 + u)$ ${}= u - \dfrac{u^2}{2} + \dots$
M1
Only the $x^4$ term of $u^2$ is needed. $u^2 = \dfrac{x^4}{4} + \dots$
A1
Substitute. $\ln(\cos x)$ ${}= -\dfrac{x^2}{2}$ ${}+ \dfrac{x^4}{24}$ ${}- \dfrac{x^4}{8}$
M1
$\tfrac{1}{24} - \tfrac{3}{24}$ ${}= -\tfrac{2}{24}$ . $= -\dfrac{x^2}{2}$ ${}- \dfrac{x^4}{12}$
A1

Total5

Further calculus

Volume of revolution

EasyYear 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)

Show the worked answer

Part (a)3 marks

Volume about the $x$ -axis is $\pi\int y^2\,\mathrm{d}x$ , and $y^2 = x^4$ . $V = \pi\int_0^2 x^4\,\mathrm{d}x$
M1
Integrate. $= \pi\left[\dfrac{x^5}{5}\right]_0^2$
A1
$= \dfrac{32\pi}{5}$
A1

Total3

Further calculus

Volume with a square root

EasyYear 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)

Show the worked answer

Part (a)3 marks

Squaring $\sqrt{x}$ gives $x$ . $V = \pi\int_0^4 x\,\mathrm{d}x$
M1
$= \pi\left[\dfrac{x^2}{2}\right]_0^4$
A1
$= 8\pi$
A1

Total3

Further calculus

An improper integral

EasyYear 13 · 3 marks

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

Show the worked answer

Part (a)3 marks

Replace the infinite limit with $t$ and integrate. $\displaystyle\int_1^t x^{-2}\,\mathrm{d}x$ ${}= \left[-\frac{1}{x}\right]_1^t$ ${}= 1 - \frac{1}{t}$
M1 A1
The integral converges because the limit exists.As $t \to \infty$ , $\tfrac{1}{t} \to 0$ , so the integral is 1
A1

Total3

Further calculus

Mean value

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

Mean value $= \dfrac{1}{b - a}\int_a^b f(x)\,\mathrm{d}x$ . $\dfrac{1}{3 - 0}\displaystyle\int_0^3 x^2\,\mathrm{d}x$
M1
$= \tfrac{1}{3} \times 9 = 3$
A1

Total2

Further calculus

Differentiating arctan

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

Use $\dfrac{\mathrm{d}}{\mathrm{d}x}\arctan u$ ${}= \dfrac{1}{1 + u^2}$ and the chain rule. $\dfrac{1}{1 + (2x)^2} \times 2$
M1
$= \dfrac{2}{1 + 4x^2}$
A1

Total2

Further calculus

An integral giving π

MediumYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

$\dfrac{1}{1 + x^2}$ integrates to $\arctan x$ . $\bigl[\arctan x\bigr]_0^1$
M1
$\arctan 1 = \tfrac{\pi}{4}$ and $\arctan 0 = 0$ . $= \dfrac{\pi}{4}$
A1

Total2

Further calculus

Volume about the y-axis

MediumYear 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)

Show the worked answer

Part (a)3 marks

About the $y$ -axis use $\pi\int x^2\,\mathrm{d}y$ ; here $x^2 = y$ . $V$ ${}= \pi\int_0^4 x^2\,\mathrm{d}y$ ${}= \pi\int_0^4 y\,\mathrm{d}y$
M1
$= \pi\left[\dfrac{y^2}{2}\right]_0^4$
A1
$= 8\pi$
A1

Total3

Further calculus

An exponential improper integral

MediumYear 13 · 3 marks

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

Show the worked answer

Part (a)3 marks

Integrate up to $t$ first. $\displaystyle\int_0^t \mathrm{e}^{-2x}\,\mathrm{d}x$ ${}= \left[-\tfrac{1}{2}\mathrm{e}^{-2x}\right]_0^t$ ${}= \tfrac{1}{2}$ ${}- \tfrac{1}{2}\mathrm{e}^{-2t}$
M1 A1
As $t \to \infty$ , $\mathrm{e}^{-2t} \to 0$ , so the integral converges to $\tfrac{1}{2}$
A1

Total3

Further calculus

An inverse sine integral

HardYear 13 · 3 marks

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

Show the worked answer

Part (a)3 marks

$\int \dfrac{1}{\sqrt{a^2 - x^2}}\,\mathrm{d}x$ ${}= \arcsin\dfrac{x}{a}$ , with $a = 3$ . $\left[\arcsin\dfrac{x}{3}\right]_0^{3/2}$
M1 A1
$= \arcsin\tfrac{1}{2}$ ${}= \dfrac{\pi}{6}$
A1

Total3

Further calculus

Partial fractions with a quadratic factor

HardYear 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)

Show the worked answer

Part (a)3 marks

A quadratic factor that does not factorise needs a linear numerator. $\dfrac{2x + 3}{x(x^2 + 1)}$ ${}= \dfrac{A}{x}$ ${}+ \dfrac{Bx + C}{x^2 + 1}$
M1
Multiply through by the denominator. $2x + 3 = A(x^2 + 1) + (Bx + C)x$
M1
$x = 0$ gives $A$ ; compare $x^2$ and $x$ coefficients for $B$ and $C$ . $A = 3,\ B = -3,\ C = 2$
A1

Part (b)3 marks

Split the second fraction into two simpler ones. $\displaystyle\int \left(\frac{3}{x} - \frac{3x}{x^2 + 1} + \frac{2}{x^2 + 1}\right)\mathrm{d}x$
M1
The middle term is $\tfrac{f'(x)}{f(x)}$ form; the last is arctan. $= 3\ln|x|$ ${}- \tfrac{3}{2}\ln(x^2 + 1)$ ${}+ 2\arctan x$ ${}+ c$
A1 A1

Total6

Further vectors

Equation of a line

EasyYear 12 · 2 marks

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

Show the worked answer

Part (a)2 marks

The direction is $\mathbf{b} - \mathbf{a}$ . $\overrightarrow{AB}$ ${}= \begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$
M1
A point on the line plus any multiple of the direction. $\mathbf{r}$ ${}= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ ${}+ \lambda\begin{pmatrix} 3 \\ -2 \\ 2 \end{pmatrix}$
A1

Total2

Further vectors

Equation of a plane

EasyYear 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)

Show the worked answer

Part (a)2 marks

The normal's components are the coefficients of $x$ , $y$ and $z$ . $x + 3y + 2z = d$
M1
Substitute the point to find $d$ . $d = 2 + 3 - 2 = 3$ , so $x + 3y + 2z = 3$
A1

Total2

Further vectors

Perpendicular vectors

EasyYear 12 · 2 marks

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

Show the worked answer

Part (a)2 marks

Find the scalar product. $2(1) + (-1)(5) + 3(1) = 0$
M1
$\mathbf{a}\cdot\mathbf{b}$ ${}= |\mathbf{a}||\mathbf{b}|\cos\theta$ ${}= 0$ means $\theta = 90^\circ$ .The scalar product is zero, so the vectors are perpendicular.
A1

Total2

Further vectors

A line meeting a plane

EasyYear 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)

Show the worked answer

Part (a)3 marks

Substitute the line's coordinates into the plane's equation. $(1 + t) + t + (2 + t) = 9$
M1
$3t + 3 = 9 \Rightarrow t = 2$
A1
Put $t = 2$ back into the line. $(3, 2, 4)$
A1

Total3

Further vectors

Angle between a line and a plane

MediumYear 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)

Show the worked answer

Part (a)4 marks

First find the angle with the plane's normal $(2, -1, 2)$ . $\cos\theta$ ${}= \dfrac{(1)(2) + (2)(-1) + (2)(2)}{3 \times 3}$ ${}= \dfrac{4}{9}$
M1 A1
$\theta = 63.6^\circ$
The normal is at right angles to the plane, so subtract from $90^\circ$ .Angle with the plane $= 90^\circ - 63.6^\circ$ ${}= 26.4^\circ$
M1 A1

Total4

Further vectors

Distance from a point to a plane

MediumYear 12 · 3 marks

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

Show the worked answer

Part (a)3 marks

Substitute the point into the plane's equation and divide by the normal's length. $\dfrac{|2(1) - 2 + 2(3) - 10|}{\sqrt{2^2 + 1^2 + 2^2}}$
M1
$= \dfrac{|-4|}{3}$
A1
$= \dfrac{4}{3}$
A1

Total3

Further vectors

Skew lines

MediumYear 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)

Show the worked answer

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.
B1
Set the coordinates equal and solve two of the equations. $x$ : $1 + s = 2 \Rightarrow s = 1$ ; $y$ : $1 = t \Rightarrow t = 1$
M1
Check the third equation. $z$ : $s = 1$ but $1 + t = 2$ , and $1 \ne 2$
M1
The equations are inconsistent, so the lines do not meet: they are skew.
A1

Total4

Further vectors

Foot of a perpendicular

HardYear 12 · 5 marks

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

Show the worked answer

Part (a)5 marks

The perpendicular from $P$ goes in the direction of the plane's normal. $\mathbf{r}$ ${}= \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}$ ${}+ t\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$
M1
Substitute into the plane to find where it lands. $(2 + t)$ ${}+ 2(3 + 2t)$ ${}+ 2(4 + 2t)$ ${}= 3$
M1
$16 + 9t$ ${}= 3$ ${}\Rightarrow t$ ${}= -\tfrac{13}{9}$
A1
Put $t$ back into the line.Foot $= \left(\tfrac{5}{9}, \tfrac{1}{9}, \tfrac{10}{9}\right)$
M1 A1

Total5

Further vectors

Line of intersection of two planes

HardYear 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)

Show the worked answer

Part (a)5 marks

Use a parameter for one variable, then solve for the other two.Let $z = \lambda$ : $x + y = 6 - \lambda$ , $x - y = 5 - 2\lambda$
M1
Add and subtract the two equations. $x$ ${}= \dfrac{11 - 3\lambda}{2},\ y$ ${}= \dfrac{1 + \lambda}{2}$
M1 A1
Separate the constants from the $\lambda$ terms. Any multiple of the direction is fine. $\mathbf{r}$ ${}= \begin{pmatrix} 5.5 \\ 0.5 \\ 0 \end{pmatrix}$ ${}+ \lambda\begin{pmatrix} -1.5 \\ 0.5 \\ 1 \end{pmatrix}$
M1 A1

Total5

Polar coordinates

Polar to Cartesian

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

Use $x = r\cos\theta$ and $y = r\sin\theta$ . $x$ ${}= 4\cos\dfrac{\pi}{3}$ ${}= 2,\ y$ ${}= 4\sin\dfrac{\pi}{3}$ ${}= 2\sqrt{3}$
M1 A1

Total2

Polar coordinates

Cartesian to polar

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

$r$ is the distance from the origin. $r = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$
B1
The point is in the second quadrant, so $\theta = \pi - \tfrac{\pi}{4}$ . $\theta = \dfrac{3\pi}{4}$
B1

Total2

Polar coordinates

A circle in polar form

EasyYear 13 · 3 marks

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

Show the worked answer

Part (a)3 marks

Multiply by $r$ so you can use $r^2 = x^2 + y^2$ and $r\cos\theta = x$ . $r^2 = 4r\cos\theta$
M1
$x^2 + y^2 = 4x$
A1
Complete the square. $(x - 2)^2 + y^2 = 4$ : a circle, centre $(2, 0)$ , radius 2
A1

Total3

Polar coordinates

A line in polar form

EasyYear 13 · 1 marks

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

Show the worked answer

Part (a)1 mark

Every point on it is at angle $\tfrac{\pi}{4}$ from the initial line, whatever $r$ is. $\theta = \dfrac{\pi}{4}$
B1

Total1

Polar coordinates

Area of a polar sector

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

Area $= \tfrac{1}{2}\int r^2\,\mathrm{d}\theta$ . $\tfrac{1}{2}\displaystyle\int_0^{\pi/3} 9\,\mathrm{d}\theta$
M1
Matches the sector formula $\tfrac{1}{2}r^2\theta$ . $= \dfrac{3\pi}{2}$
A1

Total2

Polar coordinates

One loop of a rose curve

MediumYear 13 · 4 marks

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

Show the worked answer

Part (a)4 marks

Use $\tfrac{1}{2}\int r^2\,\mathrm{d}\theta$ . $A$ ${}= \tfrac{1}{2}\displaystyle\int_0^{\pi/2} \sin^2 2\theta\,\mathrm{d}\theta$
M1
$\sin^2 A$ ${}= \tfrac{1}{2}(1 - \cos 2A)$ . $= \tfrac{1}{4}\displaystyle\int_0^{\pi/2} (1 - \cos 4\theta)\,\mathrm{d}\theta$
M1
$= \tfrac{1}{4}\left[\theta - \tfrac{1}{4}\sin 4\theta\right]_0^{\pi/2}$
A1
$= \dfrac{\pi}{8}$
A1

Total4

Polar coordinates

Where two curves meet

MediumYear 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)

Show the worked answer

Part (a)3 marks

Set the two expressions for $r$ equal. $1 + \cos\theta$ ${}= 3\cos\theta$ ${}\Rightarrow \cos\theta$ ${}= \tfrac{1}{2}$
M1 A1
Find $r$ from either equation. $\left(\tfrac{3}{2}, \tfrac{\pi}{3}\right)$ and $\left(\tfrac{3}{2}, -\tfrac{\pi}{3}\right)$
A1

Total3

Polar coordinates

Tangents parallel to the initial line

HardYear 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)

Show the worked answer

Part (a)5 marks

A tangent parallel to the initial line is where $y$ is stationary. $y$ ${}= r\sin\theta$ ${}= \sin\theta$ ${}+ \sin\theta\cos\theta$
M1
$\sin\theta\cos\theta$ ${}= \tfrac{1}{2}\sin 2\theta$ . $\dfrac{\mathrm{d}y}{\mathrm{d}\theta}$ ${}= \cos\theta + \cos 2\theta$
A1
Set to zero and use $\cos 2\theta = 2\cos^2\theta - 1$ . $2\cos^2\theta + \cos\theta - 1$ ${}= 0$ ${}\Rightarrow (2\cos\theta - 1)(\cos\theta + 1)$ ${}= 0$
M1
$\cos\theta = -1$ gives $\theta = \pi$ , outside the range. $\cos\theta$ ${}= \tfrac{1}{2}$ ${}\Rightarrow \theta$ ${}= \tfrac{\pi}{3}$
A1
The point $\left(\tfrac{3}{2}, \tfrac{\pi}{3}\right)$
A1

Total5

Polar coordinates

Area of a limaçon

HardYear 13 · 5 marks

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

Show the worked answer

Part (a)5 marks

Square $r$ first. $A$ ${}= \tfrac{1}{2}\displaystyle\int_0^{2\pi} (4 + 4\sin\theta + \sin^2\theta)\,\mathrm{d}\theta$
M1
Needed to integrate $\sin^2\theta$ . $\sin^2\theta$ ${}= \tfrac{1}{2}(1 - \cos 2\theta)$
M1
A full turn of sine integrates to zero. $\displaystyle\int_0^{2\pi} 4\sin\theta\,\mathrm{d}\theta$ ${}= 0$
A1
$4$ gives $8\pi$ , and $\sin^2\theta$ gives $\pi$ . $A = \tfrac{1}{2}(8\pi + \pi)$
M1
$= \dfrac{9\pi}{2}$
A1

Total5

Hyperbolic functions

Evaluating sinh

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

Use $\sinh x$ ${}= \tfrac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})$ ; $\mathrm{e}^{-\ln 3} = \tfrac{1}{3}$ . $\dfrac{\mathrm{e}^{\ln 3} - \mathrm{e}^{-\ln 3}}{2}$ ${}= \dfrac{3 - \tfrac{1}{3}}{2}$
M1
$= \dfrac{4}{3}$
A1

Total2

Hyperbolic functions

The hyperbolic identity

EasyYear 13 · 3 marks

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

Show the worked answer

Part (a)3 marks

Square the definition; $\mathrm{e}^x\mathrm{e}^{-x} = 1$ . $\cosh^2 x$ ${}= \tfrac{1}{4}(\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x})$
M1
$\sinh^2 x$ ${}= \tfrac{1}{4}(\mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x})$
A1
Everything cancels except the constants.Subtracting leaves $\tfrac{1}{4}(4) = 1$
A1

Total3

Hyperbolic functions

Differentiating hyperbolic functions

EasyYear 13 · 3 marks

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

Show the worked answer

Part (a)1 mark

$\cosh$ differentiates to $\sinh$ (no minus sign), times 3 by the chain rule. $3\sinh 3x$
B1

Part (b)2 marks

Product rule: differentiate $x$ , then $\sinh x$ . $\sinh x + x\cosh x$
M1 A1

Total3

Hyperbolic functions

Inverse tanh as a logarithm

EasyYear 13 · 2 marks

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

Show the worked answer

Part (a)2 marks

Use $\operatorname{artanh} x$ ${}= \tfrac{1}{2}\ln\dfrac{1 + x}{1 - x}$ . $\tfrac{1}{2}\ln\dfrac{1 + \tfrac{1}{2}}{1 - \tfrac{1}{2}}$
M1
$= \tfrac{1}{2}\ln 3$
A1

Total2

Hyperbolic functions

A hyperbolic equation

MediumYear 13 · 5 marks

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

Show the worked answer

Part (a)5 marks

Write both in exponentials. $\tfrac{3}{2}(\mathrm{e}^x + \mathrm{e}^{-x})$ ${}- \tfrac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})$ ${}= 3$
M1
Collect like terms and halve. $\mathrm{e}^x$ ${}+ 2\mathrm{e}^{-x}$ ${}= 3$
A1
Multiply by $\mathrm{e}^x$ to get a quadratic in $\mathrm{e}^x$ . $\mathrm{e}^{2x}$ ${}- 3\mathrm{e}^x$ ${}+ 2$ ${}= 0$
M1
$(\mathrm{e}^x - 1)(\mathrm{e}^x - 2)$ ${}= 0$
M1
$x = 0$ or $x = \ln 2$
A1

Total5

Hyperbolic functions

Logarithmic form of arsinh

MediumYear 13 · 4 marks

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

Show the worked answer

Part (a)4 marks

Rewrite the inverse as the original function.Let $y = \operatorname{arsinh} x$ , so $x$ ${}= \sinh y$ ${}= \tfrac{1}{2}(\mathrm{e}^y - \mathrm{e}^{-y})$
M1
Multiply by $2\mathrm{e}^y$ . $\mathrm{e}^{2y}$ ${}- 2x\mathrm{e}^y$ ${}- 1$ ${}= 0$
A1
Quadratic formula in $\mathrm{e}^y$ . $\mathrm{e}^y$ ${}= x \pm \sqrt{x^2 + 1}$
M1
$x - \sqrt{x^2 + 1}$ is always negative, so reject it. $\mathrm{e}^y > 0$ , so $y$ ${}= \ln\bigl(x + \sqrt{x^2 + 1}\bigr)$
A1

Total4

Hyperbolic functions

An arsinh integral

MediumYear 13 · 3 marks

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

Show the worked answer

Part (a)3 marks

$\int \dfrac{1}{\sqrt{x^2 + a^2}}\,\mathrm{d}x$ ${}= \operatorname{arsinh}\dfrac{x}{a}$ . $\left[\operatorname{arsinh}\dfrac{x}{2}\right]_0^2$
M1 A1
Use the logarithmic form. $= \operatorname{arsinh} 1$ ${}= \ln(1 + \sqrt{2})$
A1

Total3

Hyperbolic functions

An equation with cosh 2x

HardYear 13 · 5 marks

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

Show the worked answer

Part (a)5 marks

Use $\cosh 2x = 1 + 2\sinh^2 x$ to get one function. $1 + 2\sinh^2 x - 3\sinh x = 0$
M1
Factorise the quadratic in $\sinh x$ . $(2\sinh x - 1)(\sinh x - 1) = 0$
M1 A1
$x$ ${}= \operatorname{arsinh} 1$ ${}= \ln(1 + \sqrt{2})$
A1
$\tfrac{1}{2}$ ${}+ \sqrt{\tfrac{1}{4} + 1}$ ${}= \tfrac{1 + \sqrt{5}}{2}$ . $x$ ${}= \operatorname{arsinh}\tfrac{1}{2}$ ${}= \ln\dfrac{1 + \sqrt{5}}{2}$
A1

Total5

Hyperbolic functions

A hyperbolic substitution

HardYear 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)

Show the worked answer

Part (a)6 marks

$\sqrt{1 + \sinh^2 u} = \cosh u$ and $\mathrm{d}x = \cosh u\,\mathrm{d}u$ . $\displaystyle\int \cosh u \cdot \cosh u\,\mathrm{d}u$ ${}= \int \cosh^2 u\,\mathrm{d}u$
M1
Use $\cosh 2u = 2\cosh^2 u - 1$ . $= \tfrac{1}{2}\displaystyle\int (1 + \cosh 2u)\,\mathrm{d}u$
M1
$\tfrac{1}{2}\sinh 2u$ ${}= \sinh u\cosh u$ . $= \tfrac{1}{2}(u + \sinh u\cosh u)$
A1
Change the limits to $u$ .Limits: $u = 0$ to $u = \operatorname{arsinh} 1$ , where $\sinh u = 1$ , $\cosh u = \sqrt{2}$
M1
$= \tfrac{1}{2}\bigl(\ln(1 + \sqrt{2}) + \sqrt{2}\bigr)$
A1 A1

Total6

Differential equations

First-order linear equation

EasyYear 13 · 4 marks

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

Show the worked answer

Part (a)4 marks

The equation is $y' + P(x)y = Q(x)$ with $P = 2$ .Integrating factor $= \mathrm{e}^{\int 2\,\mathrm{d}x}$ ${}= \mathrm{e}^{2x}$
M1
Multiply through by the integrating factor. $\dfrac{\mathrm{d}}{\mathrm{d}x}(y\mathrm{e}^{2x})$ ${}= \mathrm{e}^{3x}$
A1
Integrate both sides. $y\mathrm{e}^{2x}$ ${}= \tfrac{1}{3}\mathrm{e}^{3x}$ ${}+ c$
M1
Divide by $\mathrm{e}^{2x}$ . $y$ ${}= \tfrac{1}{3}\mathrm{e}^x$ ${}+ c\mathrm{e}^{-2x}$
A1

Total4

Differential equations

Two real roots

EasyYear 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)

Show the worked answer

Part (a)3 marks

Auxiliary equation. $m^2 - 5m + 6 = 0$
M1
$m = 2$ or $m = 3$
A1
Two different real roots give two exponentials. $y$ ${}= A\mathrm{e}^{2x}$ ${}+ B\mathrm{e}^{3x}$
A1

Total3

Differential equations

Imaginary roots

EasyYear 13 · 3 marks

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

Show the worked answer

Part (a)3 marks

Auxiliary equation. $m^2 + 9$ ${}= 0$ ${}\Rightarrow m$ ${}= \pm 3\mathrm{i}$
M1 A1
Roots $\pm q\mathrm{i}$ give $A\cos qx + B\sin qx$ . $y = A\cos 3x + B\sin 3x$
A1

Total3

Differential equations

A repeated root

EasyYear 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)

Show the worked answer

Part (a)3 marks

Auxiliary equation. $m^2 - 4m + 4 = 0 \Rightarrow m = 2$ (repeated)
M1 A1
A repeated root needs the extra factor $x$ . $y = (A + Bx)\mathrm{e}^{2x}$
A1

Total3

Differential equations

A particular integral

EasyYear 13 · 4 marks

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

Show the worked answer

Part (a)4 marks

Solve $m^2 + 1 = 0$ first.Complementary function: $A\cos x + B\sin x$
M1 A1
The right side is linear, so try a linear particular integral.Try $y = ax + b$ : $0 + ax + b = 3x + 1$ , so $a = 3,\ b = 1$
M1
General solution = complementary function + particular integral. $y = A\cos x + B\sin x + 3x + 1$
A1

Total4

Differential equations

Complex roots with conditions

MediumYear 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)

Show the worked answer

Part (a)5 marks

Auxiliary equation, by the quadratic formula. $m^2 + 2m + 5$ ${}= 0$ ${}\Rightarrow m$ ${}= -1 \pm 2\mathrm{i}$
M1 A1
Roots $p \pm q\mathrm{i}$ give $\mathrm{e}^{px}(A\cos qx + B\sin qx)$ . $y$ ${}= \mathrm{e}^{-x}(A\cos 2x + B\sin 2x)$
A1
Use $y(0)$ , then differentiate and use $y'(0)$ . $A = 1$ ; $y'(0)$ ${}= -A + 2B$ ${}= 1$ ${}\Rightarrow B$ ${}= 1$
M1
$y$ ${}= \mathrm{e}^{-x}(\cos 2x + \sin 2x)$
A1

Total5

Differential equations

Simple harmonic motion

MediumYear 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)

Show the worked answer

Part (a)3 marks

Auxiliary equation $m^2 + 4 = 0$ gives $m = \pm 2\mathrm{i}$ . $x = A\cos 2t + B\sin 2t$
M1
Use the starting conditions. $A = 3$ , and $\dot{x}(0)$ ${}= 2B$ ${}= 0$ ${}\Rightarrow B$ ${}= 0$
M1
$x = 3\cos 2t$
A1

Part (b)1 mark

The period is $\tfrac{2\pi}{\omega}$ with $\omega = 2$ . $\dfrac{2\pi}{2} = \pi$ seconds
B1

Total4

Differential equations

Modelling speed

HardYear 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)

Show the worked answer

Part (a)5 marks

Standard first-order linear form.Integrating factor $= \mathrm{e}^{\int \frac{1}{t+1}\,\mathrm{d}t}$ ${}= t + 1$
M1 A1
Multiply through by the integrating factor. $\dfrac{\mathrm{d}}{\mathrm{d}t}\bigl(v(t + 1)\bigr)$ ${}= 2(t + 1)$
M1
Integrate, then use the starting condition. $v(t + 1) = (t + 1)^2 + c$ ; $t = 0$ , $v = 0$ gives $c = -1$
M1
$v = t + 1 - \dfrac{1}{t + 1}$
A1

Part (b)2 marks

Substitute $t = 3$ . $v$ ${}= 4 - \tfrac{1}{4}$ ${}= 3.75\ \mathrm{m\,s^{-1}}$
B1
A good model would level off at a top speed.The speed keeps increasing without limit, which is unrealistic for a cyclist.
B1

Total7