Mathematics (9709) Past Papers & Revision Resources

Fix

Show ${x^2 - 5x + 6 }={ (x - 2)(x - 3)}$ or ${x }={ \dfrac{5 \pm \sqrt{25 - 24}}{2}}$ so ${x = 2}$ or ${3}$

Show ${\int_1^3 2x\,dx }={ \left[x^2\right]_1^3 }={ 3^2 - 1^2 }={ 8}$

Check on the calculator.

Fix

Factorise, never cancel: ${2\sin\theta\cos\theta = \sin\theta}$ gives ${\sin\theta(2\cos\theta - 1) }={ 0}$ so ${\sin\theta = 0}$ or ${\cos\theta = \tfrac12}$

After a square root, use ${\pm}$ ${\tan^2\theta = 3}$ gives ${\tan\theta = \pm\sqrt3}$

Fix

Name it exactly.

${y = f(3x)}$ is a stretch parallel to the ${x}$ scale factor ${\tfrac13}$

${y = f(x) + 3}$ is a translation by the vector ${\begin{pmatrix} 0 \\ 3 \end{pmatrix}}$

Fix

Put the whole term in brackets, sign included, before raising it to a power.

For example, the ${x^2}$ term of ${(1 - 2x)^5}$ is ${10(-2x)^2 }={ 40x^2}$

Fix

Keep ${\theta}$ in radians: ${s = r\theta}$ ${A = \tfrac12 r^2\theta}$ and segment area ${\tfrac12 r^2(\theta - \sin\theta)}$

With ${r = 6}$ and ${\theta = \tfrac{\pi}{3}}$ ${s = 2\pi}$ exactly; round only at the end.

Fix

Use ${\ln a - \ln b = \ln\dfrac{a}{b}}$ and ${\ln a + \ln b = \ln ab}$

Bring a power down whole, with brackets: ${\ln(2^{3x+1}) }={ (3x + 1)\ln 2}$

Fix

Keep ${\pi}$ ${e}$ ${\ln}$ and surds to the end: write ${\tfrac12\ln 3}$ not 0.549.

Use identities: ${\tan\theta = 2}$ gives ${\sec^2\theta = 1 + 4 = 5}$ not a calculator angle.

Fix

For ${x_{n+1} }={ \sqrt[3]{2x_n + 3}}$ from ${x_0 = 2}$ 1.9129, 1.8969, 1.8940, 1.8934, each to 4 decimal places.

Stop when two in a row agree to 2 decimal places: 1.89.

Fix

Treat ${y}$ as a function of ${x}$ the derivative of ${y^2}$ is ${2y\dfrac{dy}{dx}}$ and of ${xy}$ is ${y + x\dfrac{dy}{dx}}$

A constant differentiates to zero.

Fix

Dividing ${9x^2 + 0x + 2}$ by ${3x + 1}$ gives ${3x - 1 + \dfrac{3}{3x + 1}}$

${\displaystyle\int \frac{3}{3x + 1}\,dx }={ \ln|3x + 1| + c}$ since ${\displaystyle\int \frac{k}{ax + b}\,dx }={ \dfrac{k}{a}\ln|ax + b| + c}$

Fix

Use equal scales.

Plot the point before finding its argument: ${-1 - \sqrt3\,i}$ has argument ${-\dfrac{2\pi}{3}}$ not ${\dfrac{\pi}{3}}$

A half-line ${\arg(z - a) = \theta}$ starts at ${a}$

Fix

Write each line as ${\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}}$

Use only directions: for ${\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}$ and ${2\mathbf{i} - \mathbf{j} - 2\mathbf{k}}$ ${\cos\theta }={ \dfrac{|2 - 2 - 4|}{3 \times 3} }={ \dfrac49}$ giving the acute angle ${63.6^\circ}$

Fix

Substitute ${z = x + iy}$ and ${z^* = x - iy}$ using ${i^2 = -1}$

For ${2z - iz^* = 3}$ real parts give ${2x - y = 3}$ imaginary parts ${2y - x = 0}$ so ${z = 2 + i}$

Fix

Separate: ${\dfrac{dy}{dx} = \dfrac{x}{y}}$ gives ${\int y\,dy }={ \int x\,dx}$ so ${\tfrac12 y^2 }={ \tfrac12 x^2 + c}$

At ${x = 0}$ ${y = 3}$ ${c = 4.5}$ so ${y = \sqrt{x^2 + 9}}$ as ${y > 0}$

Fix

Start from the harder side, one valid step per line: ${\dfrac{1 - \cos^2 x}{\cos^2 x} }={ \dfrac{\sin^2 x}{\cos^2 x} }={ \tan^2 x}$

Use the given answer only as a target.

Fix

Suvat needs constant acceleration.

Otherwise use calculus.

For ${v = 3t^2 - 4t}$ ${a }={ \dfrac{dv}{dt} }={ 6t - 4}$

Then ${s }={ \int v\,dt }={ t^3 - 2t^2 + c}$ with ${c}$ from the starting values.

Fix

Resolve perpendicular to the surface: a 16 N pull at ${22^\circ}$ above the horizontal makes ${R }={ mg - 16\sin 22^\circ}$

Friction reaches ${\mu R}$ only when sliding or about to slide.

Fix

Draw a force diagram; weight is ${mg}$

On a slope at ${\theta}$ ${mg\sin\theta}$ along, ${mg\cos\theta}$ perpendicular.

For 4 kg at ${15^\circ}$ with ${g = 10}$ 10.4 N and 38.6 N.

Fix

Make the 2 kg particle's direction positive.

At 3 m/s it meets 1 kg at 4 m/s: momentum ${2(3) + 1(-4) }={ 2}$

Joined, ${3v = 2}$ so ${v = \tfrac23\,\mathrm{m/s}}$ that way.

Fix

Work by driving force = KE gain + PE gain + work against resistance.

Up 50 m of a slope at ${6^\circ}$ PE uses height ${50\sin 6^\circ}$ but resistance uses 50 m.

Fix

Divide ${n!}$ by the factorial of each repeated group's size.

BANANA has three As and two Ns, so it has ${\dfrac{6!}{3! \times 2!} }={ 60}$ arrangements.

Fix

Plot cumulative frequency at each upper class boundary, such as 19.5 for a whole-number class 10 to 19.

A histogram's bar height is frequency ÷ class width.

Fix

Adjust only when a normal replaces a discrete distribution.

For binomial ${X}$ ${\mathrm{P}(X \le 12)}$ becomes ${\mathrm{P}(Y < 12.5)}$ and ${\mathrm{P}(X > 12)}$ becomes ${\mathrm{P}(Y > 12.5)}$

Fix

List the values.

In a binomial, 'more than 7' is 8, 9, …, so ${1 - \mathrm{P}(X \le 7)}$

'At least 7' is ${1 - \mathrm{P}(X \le 6)}$

'Fewer than 7' is 0 to 6.

Fix

Leave exact probabilities as fractions.

Otherwise keep at least 4 significant figures and round only the final answer.

${0.2347^2 }={ 0.0551}$ but rounding first gives ${0.235^2 }={ 0.0552}$

Fix

Under ${\mathrm{H}_0}$ ${X \sim \mathrm{B}(12, 0.25)}$ the 5% upper-tail region is ${X \ge 7}$

${\mathrm{P}(\text{Type I}) }={ \mathrm{P}(X \ge 7) }={ 0.0143}$ not 5%.

If the true ${p}$ is 0.5, ${\mathrm{P}(\text{Type II}) }={ \mathrm{P}(X \le 6) }={ 0.613}$

Fix

A 95% interval leaves 2.5% in each tail, so ${\Phi(z) = 0.975}$ and ${z = 1.96}$

The width is ${2z}$ × standard error.

The level is ${2\Phi(z) - 1}$ so ${z = 2.17}$ gives 97%.

Fix

Total area is 1: use geometry for straight-line graphs.

If ${f(x) }={ k(x - 2)(6 - x)}$ for ${2 \le x \le 6}$ say it is symmetric about ${x = 4}$ mean and median are 4.

Fix

In a binomial test, ${\mathrm{H}_0\colon p = 0.3}$ and ${\mathrm{H}_1\colon p < 0.3}$ use ${\lambda}$ for Poisson, ${\mu}$ for a mean.

Conclude in context: 'there is evidence that the proportion has fallen', never 'proves'.

Fix

If ${X \sim \mathrm{Po}(1.5)}$ under ${\mathrm{H}_0}$ and ${x = 4}$ use the tail.

${\mathrm{P}(X \ge 4) }={ 0.0656 > 0.05}$ not significant at 5%, one-tailed, though ${\mathrm{P}(X = 4) }={ 0.0471}$

Two-tailed: compare with half the level.

Fix

Write: 'The population of lengths is not known to be normal'.

Add: '${n = 100}$ is large, so the sample mean is approximately normal'.

Its standard deviation is ${\dfrac{\sigma}{\sqrt{100}} }={ \dfrac{\sigma}{10}}$

Fix

If ${\mathrm{Var}(X) = 4}$ then ${\mathrm{Var}(3X) }={ 9 \times 4 }={ 36}$

But three independent observations ${X_1 + X_2 + X_3}$ have variance ${3 \times 4 = 12}$

For independent ${X}$ and ${Y}$ ${\mathrm{Var}(X - Y) }={ \mathrm{Var}(X) + \mathrm{Var}(Y)}$

Fix

For ${f(x) = \dfrac{x}{12}}$ ${1 \le x \le 5}$ ${\mathrm{P}(X > 3) }={ \displaystyle\int_3^5 \frac{x}{12}\,dx }={ \dfrac23}$

Median: ${\displaystyle\int_1^m \frac{x}{12}\,dx }={ 0.5}$ gives ${m = \sqrt{13}}$ not ${\sqrt{12}}$ from lower limit 0.