Practice | FMPR

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

(a)

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

(b)

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

(c)

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

Total5

Complex numbers

Roots of a cubic

MediumYear 12 · 5 marks

\mathrm{f}(z)

= z^3 - 3z^2 + z + 5

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

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

(a)

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

(b)

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

(c)

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

Total4

Proof

Induction with a sum

MediumYear 12 · 5 marks

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

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

Total5

Proof

Induction with divisibility

HardYear 12 · 6 marks

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

Show the worked answer

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

(a)

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

(b)

$\alpha^2$ $+ \beta^2$ $+ \gamma^2$ $= (\sum\alpha)^2$ $- 2\sum\alpha\beta$ Squaring the sum of the roots gives the squares plus twice every pair.
M1
$= \tfrac{9}{4} - 4$ $= -\tfrac{7}{4}$ Substitute the values from part (a).
A1
The squares of real numbers cannot add to a negative, so not all the roots are real.Two of the roots must be a complex conjugate pair.
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

(a)

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

(b)

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

(a)

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

(b)

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

.

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

Show the worked answer

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

.

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

Show the worked answer

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

Total5

Hyperbolic functions

Solving a hyperbolic equation

EasyYear 13 · 4 marks

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

Show the worked answer

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

Total4

Differential equations

Integrating factor

MediumYear 13 · 5 marks

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

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

(a)

$m^2 + 4m + 4$ $= 0$ $\Rightarrow m$ $= -2$ (repeated)Write the auxiliary equation by replacing each derivative with a power of $m$ .
M1
$x$ $= (A + Bt)\mathrm{e}^{-2t}$ A repeated root gives this form of general solution.
A1
$A$ $= 0$ ; $\dot{x}$ $= B\mathrm{e}^{-2t}$ $- 2Bt\mathrm{e}^{-2t}$ , so $B$ $= 2$ Use $x = 0$ at $t = 0$ , then differentiate and use $\dot{x} = 2$ at $t = 0$ .
M1
$x$ $= 2t\mathrm{e}^{-2t}$ Substitute the constants.
A1

(b)

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

Total8

Proof

Proof by deduction

EasyYear 12 · 3 marks

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

Show the worked answer

$n^2 - n$ $= n(n - 1)$ Factorise to see the structure of the expression.
M1
$n$ and $n - 1$ are consecutive integers, so one of them is even.Of any two numbers next to each other, one is a multiple of 2.
A1
An even number times any integer is even, so $n^2 - n$ is even.Finish with a clear statement of what has been proved.
A1

Total3

Algebra and functions

Equal roots

EasyYear 12 · 3 marks

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

Show the worked answer

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

(a)

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

(b)

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

Total4

Trigonometry

A quadratic in cosine

MediumYear 13 · 5 marks

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

Show the worked answer

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

(a)

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

(b)

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

(c)

It assumes the room stays at exactly 20 °C.As $t$ grows, $T$ approaches 20, so 20 °C must be the room temperature.
B1

Total7

Differentiation

Gradient of a curve

EasyYear 12 · 3 marks

A curve has equation

y

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

.

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

Show the worked answer

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

Total3

Differentiation

Stationary points

MediumYear 12 · 6 marks

A curve has equation

y

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

.

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

Show the worked answer

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

Total6

Integration

A definite integral

EasyYear 12 · 4 marks

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

Show the worked answer

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

(a)

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

(b)

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

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

Show the worked answer

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

(a)

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

(b)

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

.

Find the acceleration of the particle.
(4)

Show the worked answer

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

.

Calculate the resistance of the wire.
(2)

Show the worked answer

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

(a)

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

(b)

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

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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.

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

Show the worked answer

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

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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

(a)

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

(b)

Kinetic energy before $= \tfrac{1}{2} \times 2.0 \times 3.0^2$ $= 9.0\ \mathrm{J}$ Use $E_k = \tfrac{1}{2}mv^2$ before and after.
M1
After $= \tfrac{1}{2} \times 3.0 \times 2.0^2$ $= 6.0\ \mathrm{J}$ , so kinetic energy is lost.A collision that loses kinetic energy is inelastic.
A1

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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.

Calculate the speed of sound in air.
(2)

Show the worked answer

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

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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.

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

Show the worked answer

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

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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

(a)

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

(b)

The formula uses $\sin\theta$ $\approx \theta$ , which is only true for small angles.For large angles the motion is no longer simple harmonic.
B1

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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}}

.

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

$\dfrac{GMm}{r^2}$ $= mr\left(\dfrac{2\pi}{T}\right)^2$ Gravity provides the centripetal force that keeps the satellite in its circle.
M1
$r^3$ $= \dfrac{GMT^2}{4\pi^2}$ , with $T$ $= 86\,400$ sCancel $m$ , rearrange for $r^3$ , and convert 24 hours to seconds.
M1
$r$ $= 4.22 \times 10^{7}\ \mathrm{m}$ Take the cube root. This is measured from the centre of the Earth.
A1
Height $= 4.22 \times 10^{7}$ $- 6.37 \times 10^{6}$ $= 3.59 \times 10^{7}\ \mathrm{m}$ Subtract the Earth's radius to get the height above the surface.
A1

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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

(a)

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

(b)

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

(c)

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

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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

(a)

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

(b)

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

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