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About A2 Level

🗺 Cambridge A Level Mathematics 9709 — A2 Syllabus Map

Paper	ClipSAT Chapter	Key Topics
P3 Pure 3	Functions: Modulus & Inverse	Modulus function, inverse, domain restriction
	Exponentials & Logarithms	$e^x$ , $\ln x$ , solving exponential equations, log graphs
	Advanced Trigonometry	Compound angle, double angle, R formula $a\sin\theta+b\cos\theta$
	Differentiation	Implicit, parametric, related rates, product/quotient/chain
	Integration	By parts, substitution, partial fractions, volumes of revolution
	Sequences & Series	Maclaurin series, binomial expansion for fractional/negative powers
	Vectors	3D vectors, dot product, angle between lines, equation of a plane
	Algebra: Partial Fractions & Division	Partial fractions (all cases), polynomial long division
	Complex Numbers	Cartesian form, Argand diagram, modulus-argument, De Moivre
	Numerical Methods	Change of sign, iterative sequences, Newton-Raphson
	Differential Equations	Separable DEs, integrating factor, forming DEs
S2 Statistics 2	Poisson Distribution	$X\sim\text{Po}(\lambda)$ , mean = variance = $\lambda$ , approximation to binomial
	Continuous Random Variables	pdf, cdf, $E(X)$ , $\text{Var}(X)$ , median, quartiles
	Sampling & Estimation	Unbiased estimators, CLT, confidence intervals for $\mu$
	Hypothesis Testing	One/two-tailed tests for $\mu$ , Poisson mean, Binomial $p$

A2 is the second year of the A Level, completing courses such as Cambridge 9709 and Edexcel. It deepens Pure Mathematics and extends the applied content.

What it covers

Pure 2 & 3 brings modulus, composite and inverse functions, exponentials and logarithms, advanced trigonometry, the full rules of differentiation and integration, sequences and series (including the binomial expansion), and vectors.

Functions

modulus · inverse

Calculus

chain · product · quotient

Series

binomial · AP · GP

Vectors

2D & 3D

▲ How to use this course

Reference panels and worked examples sit beside three explorers — exponential behaviour, the tangent gradient and vector addition. Finish with the 50-problem set, then export it as a worksheet.

Cambridge International A Level Mathematics 9709 paper breakdown and raw mark allocations.

⚠️ Cambridge 9709 Traps — Exam Pacing & Standard Accuracy

Pacing on Paper 3: With 75 marks in 1 hour 50 minutes (110 minutes), you have roughly 1.45 minutes per mark. High-mark synoptic questions (differential equations, complex numbers with geometry, 3D vectors) typically appear toward the back of the paper and demand 15–20 minutes of sustained focus.

Cambridge Standard Accuracy Rule: Unless a question specifies otherwise, non-exact numerical values must be given correct to 3 significant figures, and angles in degrees to 1 decimal place. Premature rounding during intermediate steps is a frequent cause of losing the final A1 mark. Always store exact surd/fraction values or at least 4 significant figures in calculator memory registers.

A student calculates

\int_1^3 \frac{4}{2x+1}\,dx

by rounding

\ln 7 \approx 1.95

and

\ln 3 \approx 1.10

at the start, obtaining

2(1.95 - 1.10) = 1.70

. State the exact value of the integral and calculate the absolute error introduced by premature rounding.

Integrate exactly:

\int_1^3 \frac{4}{2x+1}\,dx = [2\ln(2x+1)]_1^3 = 2\ln 7 - 2\ln 3 = 2\ln\left(\frac{7}{3}\right) \approx 1.6946\dots

The exact value is $2\ln\left(\frac{7}{3}\right)$ . The premature rounding gives $1.70$ , yielding an absolute error of:

|1.70 - 1.6946| \approx 0.0054

On Cambridge Paper 3, this premature rounding costs the final accuracy mark (A1) because the 3 s.f. answer is 1.69, not 1.70.

📝 Chapter Quiz

Questions Level

1

Functions: Modulus & Inverse

📌 Functions: Modulus & Inverse P3 · Modulus graph · Inverse (domain restriction) · Composite functions · Graph of $|f(x)|$

The modulus function, composite functions, inverses, and domain and range.

▲ Reference — Functions

$|x|=c$ gives $x=\pm c$ . A composite $fg(x)=f(g(x))$ is applied inside-out. The inverse $f^{-1}$ reverses $f$ ; swap $x$ and $y$ and solve. The range of $f$ is the domain of $f^{-1}$ .

Modulus

|x|=\pm c

Composite

fg(x)=f(g(x))

Inverse

swap

x,y

& solve

Range

output values

Worked example 1.A

Find the inverse of $f(x)=2x+1$ .

Let $y=2x+1$ , swap and solve:

f^{-1}(x)=\tfrac{x-1}{2}.

Worked example 1.B

Solve $|x-1|=3$ .

x=4,\ -2.

Given

f(x)=2x-3

for

x\in\mathbb{R}

, sketch

y=|f(x)|

and solve

|2x-3|=5

.

Sketch: V-shape with vertex at

(1.5,0)

. Solve:

2x-3=5\Rightarrow x=4

or

2x-3=-5\Rightarrow x=-1

.

Graph of the modulus function y = |2x - 3|, showing the vertex on the x-axis and intersection points with y = 5.

⚠️ Cambridge 9709 Traps — Modulus Inequalities & Extraneous Solutions

Squaring Both Sides: When solving an inequality with moduli on both sides, such as $|ax+b| < |cx+d|$ , squaring both sides is fully valid because both quantities are non-negative: $(ax+b)^2 < (cx+d)^2$ . This quickly reduces to a standard quadratic inequality.

Single Modulus Pitfall: Do NOT square both sides if one side can be negative (e.g. $|2x-1| > x - 3$ ). Squaring can introduce extraneous solutions or flip inequality signs unpredictably. Instead, sketch the linear graphs or split into two linear cases: $2x-1 > x-3$ and $2x-1 < -(x-3)$ .

Solve the inequality

|3x - 2| < |x + 6|

.

Since both sides are non-negative, square both sides:

(3x - 2)^2 < (x + 6)^2

9x^2 - 12x + 4 < x^2 + 12x + 36

8x^2 - 24x - 32 < 0

x^2 - 3x - 4 < 0

Factor the quadratic:

(x - 4)(x + 1) < 0

The solution set is the interval between the roots:

-1 < x < 4

📝 Chapter Quiz

Questions Level

2

Exponentials & Logarithms

📌 Exponentials & Logarithms P3 · $e^x$ and $\ln x$ · Log laws · Solving $e^{ax+b}=k$ · Log-linear graph to find unknowns

The exponential function $e^{x}$ , logarithms and their laws, and solving exponential equations.

▲ Reference — Exp & logs

$\log_a(xy)=\log_a x+\log_a y$ ; $\log_a(x^{n})=n\log_a x$ . The natural log inverts the exponential: $\ln(e^{x})=x$ . To solve $a^{x}=b$ , take logs of both sides.

The curve is $y=e^{kx}$ : a positive $k$ grows, a negative $k$ decays. Slide $k$ through zero to see it flip.

⊙ Explorer · Exponential growth & decayinteractive

Rate k = 0.5

behaviour—

value at x = 1—

All curves pass through $(0,1)$ ; the sign of $k$ sets growth versus decay.

Product law

\log_a x+\log_a y

Power law

n\log_a x

Natural log

\ln(e^{x})=x

Solve

take logs

Worked example 2.A

Evaluate $\log_2 8$ .

2^{3}=8\Rightarrow 3.

Worked example 2.B

Solve $3^{x}=81$ .

3^{4}=81\Rightarrow x=4.

Given

\ln y = 2\ln x + \ln 3

, express

y

in terms of

x

. Hence sketch the graph.

\ln y=\ln x^2+\ln 3=\ln(3x^2)\Rightarrow y=3x^2

. Parabola through origin, scaled by 3.

Straight-line graph obtained by taking natural logarithms of the power relationship y = a x^b, with vertical intercept ln(a) and gradient b.

⚠️ Cambridge 9709 Traps — Linear Reduction of Non-linear Laws

A classic Cambridge P3 question asks students to reduce experimental data to a straight line $Y = mX + c$ . Be certain of the form:

Exponential form $y = Ab^x$ : Take logs: $\ln y = x\ln b + \ln A$ . Plot $\ln y$ against $x$ . The gradient is $\ln b$ (so $b = e^m$ ) and vertical intercept is $\ln A$ (so $A = e^c$ ).
Power form $y = Ax^n$ : Take logs: $\ln y = n\ln x + \ln A$ . Plot $\ln y$ against $\ln x$ . The gradient is the exponent $n$ directly, and the vertical intercept is $\ln A$ .

A fatal mistake is forgetting that the constant $A$ is $e^{\text{intercept}}$ , not the intercept itself!

Variables

x

and

y

satisfy

y = a x^b

. A graph of

\ln y

against

\ln x

is a straight line passing through

(0, 1.20)

and

(4, 6.80)

. Find the values of the constants

a

and

b

to 3 significant figures.

Take natural logarithms of $y = a x^b$ :

\ln y = \ln a + b\ln x

This is a linear equation $Y = mX + c$ with $Y = \ln y$ , $X = \ln x$ , gradient $m = b$ , and intercept $c = \ln a$ .

Gradient:

b = \frac{6.80 - 1.20}{4 - 0} = \frac{5.60}{4} = 1.40

Vertical intercept gives $\ln a = 1.20$ :

a = e^{1.20} \approx 3.32

Therefore, $a = 3.32$ and $b = 1.40$ .

📝 Chapter Quiz

Questions Level

3

Advanced Trigonometry

📌 Advanced Trigonometry P3 · Compound angle formulae · Double angle · $R\sin(\theta+\alpha)$ form · Solving equations in a range

Reciprocal ratios, the double-angle formulae, and solving trigonometric equations.

▲ Reference — Identities

$\sec\theta=\dfrac{1}{\cos\theta}$ . Double angle: $\sin2\theta=2\sin\theta\cos\theta$ and $\cos2\theta=2\cos^{2}\theta-1$ . The tangent curve has period $180^{\circ}$ .

Reciprocal

\sec\theta=\tfrac{1}{\cos\theta}

Sine double

2\sin\theta\cos\theta

Cosine double

2\cos^{2}\theta-1

Period of tan

180^{\circ}

Worked example 3.A

Write $\cos2\theta$ in terms of $\cos\theta$ .

\cos2\theta=2\cos^{2}\theta-1.

Worked example 3.B

Solve $\cos\theta=0$ for $0^{\circ}\le\theta\le180^{\circ}$ .

\theta=90^{\circ}.

Express

3\sin\theta+4\cos\theta

in the form

R\sin(\theta+\alpha)

where

R>0

and

0<\alpha<90°

.

R=\sqrt{9+16}=5

.

\tan\alpha=4/3\Rightarrow\alpha=53.13°

. So

5\sin(\theta+53.13°)

.

Harmonic form 5 sin(θ + 53.1°), showing how a linear combination of sine and cosine forms a single shifted sinusoidal wave with amplitude R = 5.

⚠️ Cambridge 9709 Traps — Angle Mode, Domain Shifts & R-Formula

Angle Mode Discipline: Cambridge Paper 3 calculus questions and trigonometry in terms of $\pi$ require strict RADIAN mode. Only switch to degrees when the question explicitly gives an angle range like $0^\circ \le \theta \le 360^\circ$ .

Domain Adjustments for Shifted Arguments: When solving an equation like $R\sin(\theta - 30^\circ) = 2.5$ for $0^\circ \le \theta \le 360^\circ$ , first find the range of the argument: $-30^\circ \le \theta - 30^\circ \le 330^\circ$ . Looking only in $[0^\circ, 360^\circ]$ will cause you to miss negative principal values like $-30^\circ$ which map back to valid positive values of $\theta$ .

Solve the equation

2\cos 2x = 1 + \sin x

for

0 \le x \le 2\pi

radians, giving answers to 3 significant figures.

Substitute the double angle identity $\cos 2x = 1 - 2\sin^2 x$ :

2(1 - 2\sin^2 x) = 1 + \sin x

2 - 4\sin^2 x = 1 + \sin x \implies 4\sin^2 x + \sin x - 1 = 0

Apply quadratic formula for $\sin x$ :

\sin x = \frac{-1 \pm \sqrt{1 - 4(4)(-1)}}{8} = \frac{-1 \pm \sqrt{17}}{8}

Case 1: $\sin x = \frac{-1 + \sqrt{17}}{8} \approx 0.3904$ :

x = 0.401\text{ rad, } \pi - 0.401 = 2.74\text{ rad}

Case 2: $\sin x = \frac{-1 - \sqrt{17}}{8} \approx -0.6404$ :

x = \pi - (-0.695) = 3.84\text{ rad, } 2\pi + (-0.695) = 5.59\text{ rad}

Solutions in $[0, 2\pi]$ : $x \approx 0.401, 2.74, 3.84, 5.59$ .

📝 Chapter Quiz

Questions Level

4

Differentiation

📌 Differentiation P3 · Implicit differentiation · Parametric differentiation · Product/quotient rule · Related rates

Differentiating standard functions and applying the chain, product and quotient rules.

▲ Reference — Rules

$\tfrac{d}{dx}e^{x}=e^{x}$ , $\tfrac{d}{dx}\ln x=\tfrac1x$ , $\tfrac{d}{dx}\sin x=\cos x$ . Chain: $\tfrac{dy}{dx}=\tfrac{dy}{du}\tfrac{du}{dx}$ . Product: $(uv)'=u'v+uv'$ . Quotient: $\left(\tfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}$ .

Slide the point along $y=x^{3}-3x$ . The amber line is the tangent; its gradient is $3x^{2}-3$ .

⊙ Explorer · The tangent gradientinteractive

Point x = 1

gradient dy/dx—

The gradient is zero at the turning points, where $3x^{2}-3=0$ , i.e. $x=\pm1$ .

Exponential

\tfrac{d}{dx}e^{x}=e^{x}

Chain

\tfrac{dy}{du}\tfrac{du}{dx}

Product

u'v+uv'

Quotient

\dfrac{u'v-uv'}{v^{2}}

Worked example 4.A

Differentiate $y=(2x+1)^{3}$ .

Chain rule:

3(2x+1)^{2}\cdot2=6(2x+1)^{2}.

Worked example 4.B

Differentiate $y=\dfrac{x}{x+1}$ .

\frac{(x+1)-x}{(x+1)^{2}}=\frac{1}{(x+1)^{2}}.

A curve is defined parametrically by

x=t^2+1

,

y=2t^3-3t

. Find

\dfrac{dy}{dx}

in terms of

t

and the value at

t=1

.

\dfrac{dx}{dt}=2t

,

\dfrac{dy}{dt}=6t^2-3

.

\dfrac{dy}{dx}=\dfrac{6t^2-3}{2t}

. At

t=1

:

\dfrac{3}{2}

.

Cubic curve y = x^3 - 3x illustrating local turning points where dy/dx = 3x^2 - 3 = 0.

⚠️ Cambridge 9709 Traps — Second Derivative of Parametric Curves & Implicit Chain Rule

Second Derivative of Parametric Functions: When finding $\frac{d^2y}{dx^2}$ from parametric equations $x(t), y(t)$ , remember:

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}

Never simply differentiate $\frac{dy}{dx}$ with respect to $t$ and stop; you must divide by $\frac{dx}{dt}$ .

Implicit Chain Rule: Differentiating terms like $y^3$ or $x^2 y$ with respect to $x$ requires product and chain rules: $\frac{d}{dx}(x^2 y) = 2xy + x^2\frac{dy}{dx}$ . Forgetting $\frac{dy}{dx}$ on $y$ -terms collapses the implicit relationship.

The curve

C

has equation

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

. Find the coordinates of the two points on

C

where the tangent is parallel to the y-axis (vertical).

Differentiate implicitly with respect to $x$ :

3x^2 - 2\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0

3x^2 - 2y + (2y - 2x)\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{2y - 3x^2}{2y - 2x}

A vertical tangent occurs when the denominator is zero (and numerator non-zero):

2y - 2x = 0 \implies y = x

Substitute $y = x$ into the original curve equation:

x^3 - 2x(x) + x^2 = 4 \implies x^3 - x^2 - 4 = 0

Test integer factors: $x = 2$ gives $8 - 4 - 4 = 0$ . Factoring gives $(x - 2)(x^2 + x + 2) = 0$ . Since $x^2 + x + 2 = 0$ has no real roots ( $1 - 8 < 0$ ), $x = 2$ is the only real root.

Since $y = x$ , the point with vertical tangent is $(2, 2)$ .

📝 Chapter Quiz

Questions Level

5

Integration

📌 Integration P3 · Integration by parts · Substitution · Partial fractions · Volumes of revolution

Integrating standard functions, definite integrals, and the area under a curve.

▲ Reference — Standard integrals

$\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C$ ( $n\neq-1$ ); $\displaystyle\int e^{x}\,dx=e^{x}+C$ ; $\displaystyle\int\tfrac1x\,dx=\ln|x|+C$ . A definite integral gives the signed area: $F(b)-F(a)$ .

Power

\tfrac{x^{n+1}}{n+1}+C

Exponential

e^{x}+C

Reciprocal

\ln|x|+C

Area

F(b)-F(a)

Worked example 5.A

Find $\displaystyle\int 6x^{2}\,dx$ .

2x^{3}+C.

Worked example 5.B

Find the area under $y=x^{2}$ from $0$ to $3$ .

\left[\tfrac{x^{3}}{3}\right]_{0}^{3}=9.

Find

\displaystyle\int x\sin x\,dx

.

Integration by parts:

u=x

,

dv=\sin x\,dx\Rightarrow v=-\cos x

.

\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+c

.

Parabola y = x^2, with the shaded area under the curve from x = 0 to x = 3 evaluating to 9 square units.

⚠️ Cambridge 9709 Traps — Parts Signs & Substitution Limits

Signs in Integration by Parts: In $\int u \frac{dv}{dx}dx = uv - \int v \frac{du}{dx}dx$ , double negative signs are the primary source of lost marks. When $v$ involves a negative sign (e.g. $\int \sin x\,dx = -\cos x$ ), the minus-minus becomes a plus: $u(-\cos x) - \int(-\cos x)u'dx = -u\cos x + \int u'\cos x\,dx$ .

Definite Substitution Limits: When using substitution $u = g(x)$ on a definite integral, immediately transform both limits to $u$ -values: $u_1 = g(a)$ and $u_2 = g(b)$ . Evaluating a $u$ -integral using original $x$ -limits is an automatic method loss on Cambridge Paper 3.

Use the substitution

u = 2x + 1

to find the exact value of

\displaystyle\int_0^4 \frac{x}{\sqrt{2x+1}}\,dx

.

Let $u = 2x + 1$ , so $du = 2\,dx \implies dx = \frac{1}{2}\,du$ , and $x = \frac{u - 1}{2}$ .

Change limits:

x = 0 \implies u = 1,\quad x = 4 \implies u = 9

Transform the integral:

\int_0^4 \frac{x}{\sqrt{2x+1}}\,dx = \int_1^9 \frac{\frac{u - 1}{2}}{\sqrt{u}}\cdot\frac{1}{2}\,du = \frac{1}{4}\int_1^9 \left(u^{1/2} - u^{-1/2}\right)\,du

= \frac{1}{4}\left[ \frac{2}{3}u^{3/2} - 2u^{1/2} \right]_1^9 = \frac{1}{4}\left( \left(\frac{2}{3}(27) - 2(3)\right) - \left(\frac{2}{3}(1) - 2(1)\right) \right)

= \frac{1}{4}\left( (18 - 6) - \left(-\frac{4}{3}\right) \right) = \frac{1}{4}\left(12 + \frac{4}{3}\right) = \frac{1}{4}\left(\frac{40}{3}\right) = \frac{10}{3}

📝 Chapter Quiz

Questions Level

6

Sequences & Series

📌 Sequences & Series P3 · Maclaurin series · Binomial expansion (fractional & negative) · Convergence condition $|x|<1$

Arithmetic and geometric progressions, sums to infinity, and the binomial expansion.

▲ Reference — Progressions

AP term $a+(n-1)d$ ; GP term $ar^{\,n-1}$ . A convergent GP $(|r|<1)$ sums to $\dfrac{a}{1-r}$ . The sum of the first $n$ integers is $\dfrac{n(n+1)}{2}$ .

AP term

a+(n-1)d

GP term

ar^{\,n-1}

GP to infinity

\dfrac{a}{1-r}

First n integers

\dfrac{n(n+1)}{2}

Worked example 6.A

Find the 10th term of $2,5,8,\dots$

$a=2,\ d=3$ :

2+9(3)=29.

Worked example 6.B

Find the sum to infinity of $1+\tfrac12+\tfrac14+\dots$

\frac{1}{1-\tfrac12}=2.

Find the first three non-zero terms of the Maclaurin series for

e^{2x}

. State the range of validity.

e^{2x}=1+2x+\dfrac{(2x)^2}{2!}+\cdots=1+2x+2x^2+\cdots

. Valid for all

x\in\mathbb{R}

.

Validity interval |x| < 1/2 for the infinite binomial expansion of (1 + 2x)^n, with open endpoints indicating strict convergence bounds.

⚠️ Cambridge 9709 Traps — Binomial Expansion with Negative/Fractional Indices

Factoring Out the Leading Constant: To expand $(a + bx)^n$ where $n$ is not a positive integer, you MUST factor out $a^n$ :

(a + bx)^n = a^n \left(1 + \frac{b}{a}x\right)^n

A common error is forgetting to raise the factored $a$ to the power $n$ , e.g., factoring $4$ from $(4 - x)^{-1/2}$ gives $4^{-1/2} = \frac{1}{2}$ , not $4$ or $1$ .

Condition for Convergence: The expansion converges if and only if $\left|\frac{b}{a}x\right| < 1$ , meaning $|x| < \left|\frac{a}{b}\right|$ . Stating $|x| < 1$ by default forfeits the validity mark.

Find the binomial expansion of

\frac{1}{\sqrt{4 - x}}

in ascending powers of

x

up to and including the term in

x^2

. State the set of values of

x

for which the expansion is valid.

Rewrite with negative fractional index and factor out 4:

(4 - x)^{-1/2} = 4^{-1/2}\left(1 - \frac{x}{4}\right)^{-1/2} = \frac{1}{2}\left(1 - \frac{x}{4}\right)^{-1/2}

Apply $(1 + y)^n = 1 + ny + \frac{n(n-1)}{2}y^2 + \cdots$ with $n = -\frac{1}{2}$ and $y = -\frac{x}{4}$ :

\left(1 - \frac{x}{4}\right)^{-1/2} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}\left(-\frac{x}{4}\right)^2 + \cdots

= 1 + \frac{x}{8} + \frac{3}{8}\left(\frac{x^2}{16}\right) = 1 + \frac{x}{8} + \frac{3x^2}{128}

Multiply by $\frac{1}{2}$ :

\frac{1}{\sqrt{4 - x}} = \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256}

Validity condition: $\left|-\frac{x}{4}\right| < 1 \implies |x| < 4$ .

📝 Chapter Quiz

Questions Level

7

Vectors

📌 Vectors P3 · 3D column vectors · Dot product · Angle between lines · Vector equation of a plane

Vector components, addition and scalar multiples, and the magnitude of a vector.

▲ Reference — Vectors

Add componentwise: $(a_1,a_2)+(b_1,b_2)=(a_1+b_1,\,a_2+b_2)$ . A scalar multiple stretches: $k(a_1,a_2)=(ka_1,ka_2)$ . Magnitude $|\mathbf{v}|=\sqrt{a_1^{2}+a_2^{2}}$ .

Vector $\mathbf{a}=(3,1)$ is fixed (indigo). Slide the components of $\mathbf{b}$ (amber); the dashed line is the resultant $\mathbf{a}+\mathbf{b}$ .

⊙ Explorer · Adding vectorsinteractive

b₁ (x) = 1

b₂ (y) = 3

resultant a + b—

magnitude—

The resultant is the diagonal of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$ .

Add

componentwise

Scalar multiple

(ka_1,ka_2)

Magnitude

\sqrt{a_1^{2}+a_2^{2}}

Unit vector

\tfrac{\mathbf{v}}{|\mathbf{v}|}

Worked example 7.A

Find $(1,2)+(3,1)$ .

(4,\,3).

Worked example 7.B

Find the magnitude of $(6,8)$ .

\sqrt{36+64}=10.

Lines

\ell_1

and

\ell_2

have direction vectors

\mathbf{d_1}=\begin{pmatrix}2\\1\\-1\end{pmatrix}

and

\mathbf{d_2}=\begin{pmatrix}1\\-1\\3\end{pmatrix}

. Find the acute angle between them.

\mathbf{d_1}\cdot\mathbf{d_2}=2-1-3=-2

.

|\mathbf{d_1}|=\sqrt6

,

|\mathbf{d_2}|=\sqrt{11}

.

\cos\theta=\dfrac{2}{\sqrt{66}}\Rightarrow\theta=\arccos\!\left(\dfrac{2}{\sqrt{66}}\right)\approx75.7°

.

Vector triangle showing position vectors OA = a, OB = b, and the displacement vector AB = b - a.

⚠️ Cambridge 9709 Traps — Direction Vectors vs Position Vectors & Skew Lines

Angle Between Lines: When finding the angle between two 3D vector lines $\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{d}_1$ and $\mathbf{r} = \mathbf{a}_2 + \mu \mathbf{d}_2$ , you MUST use direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$ , NEVER position vectors $\mathbf{a}_1$ or $\mathbf{a}_2$ . Moreover, Cambridge asks for the acute angle: if the dot product $\mathbf{d}_1 \cdot \mathbf{d}_2 < 0$ , use $|\mathbf{d}_1 \cdot \mathbf{d}_2|$ in $\cos\theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$ .

Proving Skew Lines: To prove two lines are skew, showing they do not intersect is NOT enough; you must also explicitly prove their direction vectors are not parallel (not scalar multiples of each other).

Determine whether the lines

\mathbf{r}_1 = \begin{pmatrix}1\\2\\-1\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\3\end{pmatrix}

and

\mathbf{r}_2 = \begin{pmatrix}0\\5\\4\end{pmatrix} + \mu\begin{pmatrix}-1\\2\\1\end{pmatrix}

intersect or are skew.

1. Check parallelism: Direction vectors are $\mathbf{d}_1 = \begin{pmatrix}2\\1\\3\end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix}-1\\2\\1\end{pmatrix}$ . Since $\frac{2}{-1} \neq \frac{1}{2}$ , they are not scalar multiples, so the lines are not parallel.

2. Equate components:

1 + 2\lambda = -\mu \implies 2\lambda + \mu = -1 \quad (1)

2 + \lambda = 5 + 2\mu \implies \lambda - 2\mu = 3 \quad (2)

-1 + 3\lambda = 4 + \mu \implies 3\lambda - \mu = 5 \quad (3)

Add (1) and (3): $5\lambda = 4 \implies \lambda = \frac{4}{5}$ . From (1), $\mu = -1 - 2(4/5) = -\frac{13}{5}$ .

Substitute into (2):

\lambda - 2\mu = \frac{4}{5} - 2\left(-\frac{13}{5}\right) = \frac{4 + 26}{5} = 6 \neq 3

Since the values contradict equation (2), the lines do not intersect. Because they are neither parallel nor intersecting, the lines are skew.

📝 Chapter Quiz

Questions Level

8

Algebra: Partial Fractions & Division

📌 Algebra: Partial Fractions & Division P3 · Partial fractions (proper, improper, repeated, quadratic denominator) · Polynomial long division

Polynomial division and partial fractions break complicated algebraic fractions into simpler pieces.

▲ Reference — Partial fractions

A proper fraction $\tfrac{px+q}{(x+a)(x+b)}$ splits as $\tfrac{A}{x+a}+\tfrac{B}{x+b}$ . If the fraction is improper, divide first.

Linear factors

\tfrac{A}{x+a}

Repeated

\tfrac{A}{x+a}+\tfrac{B}{(x+a)^{2}}

Improper

divide first

Cover-up

find

A,B

Worked example 8.A

Express $\dfrac{1}{x(x+1)}$ in partial fractions.

\frac{1}{x}-\frac{1}{x+1}.

Worked example — Improper partial fractions

Express $\dfrac{x^3+2x}{x^2-1}$ in partial fractions (after division).

Degree of numerator ≥ denominator, so divide first: $x^3+2x=(x^2-1)\cdot x+3x$ . So:

\frac{x^3+2x}{x^2-1}=x+\frac{3x}{(x-1)(x+1)}=x+\frac{A}{x-1}+\frac{B}{x+1}.

Cover-up: $A=\tfrac{3}{2}$ , $B=\tfrac{3}{2}$ . Result: $x+\dfrac{3/2}{x-1}+\dfrac{3/2}{x+1}$ .

Express

\dfrac{3x+1}{(x+1)(x-2)}

in partial fractions.

\dfrac{A}{x+1}+\dfrac{B}{x-2}

. Cover-up: let

x=2

:

B=\tfrac{7}{3}

. Let

x=-1

:

A=\tfrac{-2}{-3}=\tfrac{2}{3}

. Answer:

\dfrac{2/3}{x+1}+\dfrac{7/3}{x-2}

.

Reciprocal rational function y = 1/(x-1) with vertical asymptote at x = 1, illustrating partial fraction components.

⚠️ Cambridge 9709 Traps — Improper Fractions & Repeated Factor Forms

Checking the Degree First: Before setting up partial fractions, always inspect the degrees: $\deg(\text{numerator})$ vs $\deg(\text{denominator})$ . If the degree of the top is greater than or equal to the bottom, the fraction is improper. You MUST perform algebraic long division first. Setting up partial fractions directly without dividing will yield impossible equations or wrong coefficients.

Repeated Linear Factors: For a denominator with a repeated term like $(x-1)^2(x+2)$ , the correct template is:

\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}

Omitting the single power $\frac{A}{x-1}$ is an immediate error that invalidates the rest of the problem.

Express

\frac{x^2 + 7x - 6}{(x-1)^2(x+2)}

in partial fractions.

Set up the partial fraction decomposition with repeated linear factor:

\frac{x^2 + 7x - 6}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}

Multiply both sides by $(x-1)^2(x+2)$ :

x^2 + 7x - 6 = A(x-1)(x+2) + B(x+2) + C(x-1)^2

Substitute $x = 1$ :

1 + 7 - 6 = B(3) \implies 2 = 3B \implies B = \frac{2}{3}

Substitute $x = -2$ :

4 - 14 - 6 = C(-3)^2 \implies -16 = 9C \implies C = -\frac{16}{9}

Equate coefficients of $x^2$ :

1 = A + C \implies A = 1 - \left(-\frac{16}{9}\right) = \frac{25}{9}

Final partial fractions:

\frac{25/9}{x-1} + \frac{2/3}{(x-1)^2} - \frac{16/9}{x+2}

📝 Chapter Quiz

Questions Level

9

Complex Numbers

📌 Complex Numbers P3 · Cartesian $a+bi$ · Argand diagram · Modulus-argument form · Roots of polynomials

The imaginary unit $i$ extends the reals; complex numbers add, multiply, and have a modulus and conjugate.

▲ Reference — Complex numbers

$i^{2}=-1$ ; $z=a+bi$ has modulus $|z|=\sqrt{a^{2}+b^{2}}$ and conjugate $\bar{z}=a-bi$ , with $z\bar{z}=|z|^{2}$ .

Unit

i^{2}=-1

Form

a+bi

Modulus

\sqrt{a^{2}+b^{2}}

Conjugate

a-bi

Worked example 9.A

Find $|3+4i|$ .

\sqrt{9+16}=5.

Worked example — Modulus-argument form

Write $z=-1+\sqrt{3}\,i$ in modulus-argument form and find $z^4$ using De Moivre's theorem.

$|z|=\sqrt{1+3}=2$ . $\arg z=\pi-\arctan\sqrt{3}=\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}$ .

$z=2\bigl(\cos\tfrac{2\pi}{3}+i\sin\tfrac{2\pi}{3}\bigr)$ . By De Moivre: $z^4=2^4\bigl(\cos\tfrac{8\pi}{3}+i\sin\tfrac{8\pi}{3}\bigr)=16\bigl(\cos\tfrac{2\pi}{3}+i\sin\tfrac{2\pi}{3}\bigr)=16(-\tfrac{1}{2}+\tfrac{\sqrt3}{2}i)=-8+8\sqrt3\,i$ .

Given that

z_1=3+4i

and

z_2=1-2i

, find

\dfrac{z_1}{z_2}

in the form

a+bi

.

Multiply by conjugate:

\dfrac{(3+4i)(1+2i)}{(1-2i)(1+2i)}=\dfrac{3+6i+4i-8}{1+4}=\dfrac{-5+10i}{5}=-1+2i

.

Argand diagram showing complex numbers plotted as points (Re, Im) in the complex plane with angle measured from positive real axis.

⚠️ Cambridge 9709 Traps — Principal Argument Interval & Geometric Loci

Principal Argument Range: Cambridge 9709 requires the principal argument to satisfy $-\pi < \arg(z) \le \pi$ . When $z$ lies in Quadrant 2 ( $x < 0, y > 0$ ), $\arg(z) = \pi - \arctan(|y/x|)$ . In Quadrant 3 ( $x < 0, y < 0$ ), $\arg(z) = -(\pi - \arctan(|y/x|))$ . Blindly using $\arctan(y/x)$ on a calculator yields the wrong quadrant.

Standard Geometric Loci:

$|z - z_1| = r$ : Circle of radius $r$ centred at $z_1$ .
$|z - z_1| = |z - z_2|$ : Perpendicular bisector of the line segment joining $z_1$ and $z_2$ .
$\arg(z - z_1) = \alpha$ : Half-line starting from (but not including) $z_1$ at angle $\alpha$ to the positive real direction.

On an Argand diagram, the point

P

represents the complex number

z

such that

|z - (2 + 3i)| \le 2

. Find the greatest possible value of

|z|

for points in this region.

The locus $|z - (2 + 3i)| \le 2$ represents a circular disc of radius $R = 2$ with center at $C(2, 3)$ .

The distance from the origin $O(0, 0)$ to the center $C$ is:

|C| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.6055

The point on the disc farthest from the origin lies on the line passing through $O$ and $C$ , on the far boundary of the circle:

|z|_{\max} = |C| + R = \sqrt{13} + 2 \approx 3.606 + 2 = 5.61\text{ (3 s.f.)}

📝 Chapter Quiz

Questions Level

10

Numerical Methods

📌 Numerical Methods P3 · Interval bisection / change of sign · Iterative formula $x_{n+1}=g(x_n)$ · Newton-Raphson

When an equation cannot be solved exactly, sign changes locate roots and iteration refines them.

▲ Reference — Finding roots

A sign change of $f$ across $[a,b]$ traps a root. Iteration $x_{n+1}=g(x_n)$ converges when it settles; the Newton–Raphson step is $x_{n+1}=x_n-\tfrac{f(x_n)}{f'(x_n)}$ .

Sign change

root in

[a,b]

Iteration

x_{n+1}=g(x_n)

Newton

x_n-\tfrac{f}{f'}

Converge

value settles

Worked example 10.A

For $f(x)=x^{2}-2$ , between which integers does a root lie?

1\ \text{and}\ 2,\ \text{since}\ f(1)<0<f(2).

Worked example — Newton-Raphson

Use Newton-Raphson to find the root of $f(x)=x^3-5$ near $x_0=1.7$ . Do one iteration.

$f(x)=x^3-5$ , $f'(x)=3x^2$ .

x_1=x_0-\frac{f(x_0)}{f'(x_0)}=1.7-\frac{1.7^3-5}{3(1.7)^2}=1.7-\frac{4.913-5}{8.67}=1.7-\frac{-0.087}{8.67}\approx1.710.

Show that

x^3+2x-6=0

has a root in

[1,2]

. Taking

x_0=1.5

, perform one Newton-Raphson iteration.

f(1)=-3<0

,

f(2)=6>0

: sign change confirms root.

f(1.5)=1.5^3+3-6=-0.625

.

f'(x)=3x^2+2\Rightarrow f'(1.5)=8.75

.

x_1=1.5-\tfrac{-0.625}{8.75}\approx1.571

.

Newton-Raphson approximation to the root of f(x) = x^3 - 5 showing rapid convergence from initial estimate x₀ = 1.7.

⚠️ Cambridge 9709 Traps — Continuity in Sign Change & Iteration Divergence

Sign Change Continuity Requirement: A change of sign $f(a)f(b) < 0$ guarantees a root ONLY if $f(x)$ is continuous on $[a, b]$ . Functions with vertical asymptotes (such as $f(x) = \tan x$ or $f(x) = \frac{1}{x}$ ) can change sign across a discontinuity without having any root. Cambridge always requires students to explicitly state that the function is continuous.

Fixed-Point Convergence Condition: An iteration $x_{n+1} = g(x_n)$ converges to a root $\alpha$ if and only if $|g'(\alpha)| < 1$ . If $|g'(\alpha)| > 1$ , the iteration will diverge or oscillate away from the root. If asked to explain why an iterative formula fails, evaluate $|g'(x)|$ near the root.

The equation

x^3 - x - 1 = 0

has a root near

x = 1.3

. (a) Show that the iterative formula

x_{n+1} = (x_n + 1)^{1/3}

converges by considering

g'(x)

. (b) Starting with

x_1 = 1.3

, calculate

x_2

and

x_3

to 4 decimal places.

(a) Let $g(x) = (x + 1)^{1/3}$ . Differentiate:

g'(x) = \frac{1}{3}(x + 1)^{-2/3} = \frac{1}{3(x + 1)^{2/3}}

Near the root $x \approx 1.3$ :

g'(1.3) = \frac{1}{3(2.3)^{2/3}} \approx \frac{1}{3(1.742)} \approx 0.191

Since $|g'(x)| < 1$ , the iteration is guaranteed to converge rapidly.

(b) Carry out iterations:

x_1 = 1.3

x_2 = (1.3 + 1)^{1/3} = (2.3)^{1/3} \approx 1.3200

x_3 = (1.3200 + 1)^{1/3} = (2.3200)^{1/3} \approx 1.3238

📝 Chapter Quiz

Questions Level

11

Differential Equations

📌 Differential Equations P3 · Separable variables · Integrating factor · Forming a DE from context

A differential equation relates a function to its rate of change; separable equations integrate directly.

▲ Reference — Separable equations

Write $\tfrac{dy}{dx}=f(x)g(y)$ as $\tfrac{1}{g(y)}\,dy=f(x)\,dx$ and integrate both sides; an initial condition fixes the constant.

Separate

variables

Integrate

both sides

Constant

+C

Condition

fixes

C

Worked example 11.A

Solve $\dfrac{dy}{dx}=y$ .

y=Ae^{x}.

Worked example — Integrating factor

Solve $\dfrac{dy}{dx}+2y=e^x$ .

Integrating factor: $\mu=e^{\int2\,dx}=e^{2x}$ . Multiply through: $\dfrac{d}{dx}(ye^{2x})=e^{3x}$ .

ye^{2x}=\int e^{3x}\,dx=\tfrac{1}{3}e^{3x}+c\;\Rightarrow\;y=\tfrac{1}{3}e^x+ce^{-2x}.

Solve the separable DE

\dfrac{dy}{dx}=\dfrac{x+1}{y}

given

y=2

when

x=0

.

Separate:

y\,dy=(x+1)\,dx

. Integrate:

\tfrac{y^2}{2}=\tfrac{x^2}{2}+x+c

. At

(0,2)

:

2=c

. So

y^2=x^2+2x+4

.

Direction field for the differential equation dy/dx = (x+1)/y showing the particular solution curve passing through (0, 2).

⚠️ Cambridge 9709 Traps — Timing of Constant of Integration & Modulus in Logs

Immediate Constant of Integration: The arbitrary constant $+C$ must be added at the exact line of integration, never tacked on at the end of subsequent algebraic simplification! Integrating $\int \frac{1}{y}dy = \int k\,dt$ gives $\ln|y| = kt + C \implies y = e^{kt+C} = Ae^{kt}$ . Writing $\ln y = kt \implies y = e^{kt} + C$ is mathematically invalid and earns zero marks for subsequent work.

Rates of Decay in Word Problems: When formulating a differential equation, words like 'rate of decrease' or 'decay' require an explicit negative sign: $\frac{dV}{dt} = -k\sqrt{V}$ where $k > 0$ . Omitting the negative turns decay into exponential explosion.

The temperature

\theta

of a cooling metal rod satisfies

\frac{d\theta}{dt} = -k(\theta - 20)

, where

t

is time in minutes and

k

is a constant. Given

\theta = 80

when

t = 0

and

\theta = 50

when

t = 10

, find the value of

t

when

\theta = 30

.

Separate variables and integrate:

\int \frac{1}{\theta - 20}\,d\theta = -\int k\,dt \implies \ln(\theta - 20) = -kt + C

Apply initial condition $t = 0, \theta = 80$ :

\ln(60) = C \implies \ln\left(\frac{\theta - 20}{60}\right) = -kt \implies \theta - 20 = 60e^{-kt}

Apply condition $t = 10, \theta = 50$ :

30 = 60e^{-10k} \implies e^{-10k} = \frac{1}{2} \implies k = \frac{\ln 2}{10} \approx 0.06931

Set $\theta = 30$ :

10 = 60e^{-kt} \implies e^{-kt} = \frac{1}{6} \implies t = \frac{\ln 6}{k} = \frac{10\ln 6}{\ln 2} \approx 25.8\text{ minutes}

📝 Chapter Quiz

Questions Level

S2·1

Poisson Distribution

📌 Poisson Distribution S2 · $X\sim\text{Po}(\lambda)$ · $P(X=r)=e^{-\lambda}\lambda^r/r!$ · Mean = Var = $\lambda$ · Poisson approx. to Binomial

Conditions

Events occur independently, at a constant average rate $\lambda$ , and singly in time/space. $X\sim\text{Po}(\lambda)$ .

Probability Formula

$P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!}$ for $r=0,1,2,\ldots$

Mean and Variance

$E(X)=\text{Var}(X)=\lambda$ . The mean equals the variance — a key identifying feature.

Cumulative Tables

Use Poisson tables for $P(X\leq r)$ . Note $P(X\geq r)=1-P(X\leq r-1)$ . Additive: $X+Y\sim\text{Po}(\lambda_1+\lambda_2)$ .

Changing the Interval

If $X\sim\text{Po}(\lambda)$ per hour, then in $t$ hours: $X\sim\text{Po}(\lambda t)$ .

Normal Approximation

If $\lambda>15$ : $\text{Po}(\lambda)\approx N(\lambda,\lambda)$ with continuity correction.

▲ Poisson Distribution — Reference

If events occur at a constant average rate $\lambda$ independently, $X\sim\text{Po}(\lambda)$ with: $P(X=r)=\dfrac{e^{-\lambda}\lambda^r}{r!}$ for $r=0,1,2,\ldots$ · $E(X)=\lambda$ , $\text{Var}(X)=\lambda$ . Poisson approximation to Binomial: if $n$ is large, $p$ is small, $np=\lambda$ is moderate ( $n>50, p<0.1$ ), then $B(n,p)\approx\text{Po}(\lambda)$ .

P(X=r)

\dfrac{e^{-\lambda}\lambda^r}{r!}

Mean = Var

\lambda

Sum of Poissons

X+Y\sim\text{Po}(\lambda_1+\lambda_2)

Binom approx

n>50,\,p<0.1,\,\lambda=np

Worked example — Exact probability

Faults in a wire occur at a rate of 2 per metre. Find the probability of (a) exactly 3 faults in 1 m, (b) fewer than 2 faults in 0.5 m.

(a) $\lambda=2$ : $P(X=3)=\dfrac{e^{-2}\cdot8}{6}\approx0.1804$ .

(b) $\lambda=1$ : $P(X<2)=P(0)+P(1)=e^{-1}+e^{-1}=2e^{-1}\approx0.7358$ .

Worked example — Binomial approximation

$X\sim B(200,0.02)$ . Use a Poisson approximation to find $P(X=5)$ .

$\lambda=200\times0.02=4$ . $P(X=5)=\dfrac{e^{-4}\cdot4^5}{120}=\dfrac{1024e^{-4}}{120}\approx0.1563$ .

Cars arrive at a toll booth at an average rate of 3 per minute. Find the probability that (a) exactly 5 arrive in one minute, (b) at least 2 arrive in 30 seconds.

(a)

P(X=5)=\dfrac{e^{-3}\cdot243}{120}\approx0.1008

. (b)

\lambda=1.5

:

P(X\geq2)=1-P(0)-P(1)=1-e^{-1.5}-1.5e^{-1.5}=1-2.5e^{-1.5}\approx0.4422

.

Discrete probability distribution for X ~ Po(3), highlighting equal peak probabilities at r = 2 and r = 3.

⚠️ Cambridge 9709 Traps — Linear Combinations vs Scaled Multiples in Poisson

Sum of Independent Poisson Variables: If $X_1 \sim \text{Po}(\lambda_1)$ and $X_2 \sim \text{Po}(\lambda_2)$ are independent, then their sum is Poisson: $X_1 + X_2 \sim \text{Po}(\lambda_1 + \lambda_2)$ .

Scalar Multiples Are NOT Poisson: In contrast, $2X_1$ is NOT a Poisson random variable because it can only take even integer values (0, 2, 4, ...). Its variance is $\text{Var}(2X_1) = 4\text{Var}(X_1) = 4\lambda_1 \neq E(2X_1) = 2\lambda_1$ . Always distinguish between observing two separate intervals (use $X_1 + X_2$ ) versus doubling the count of one interval (use $2X_1$ ).

The number of accidents on a highway follows a Poisson distribution with mean 1.6 per week. (a) Find the probability of at least 3 accidents in a 2-week period. (b) Given that at least one accident occurs in a week, find the probability that there are exactly 2 accidents in that week.

(a) In a 2-week period, the parameter scales to $\lambda = 1.6 \times 2 = 3.2$ :

P(X \ge 3) = 1 - P(X \le 2) = 1 - e^{-3.2}\left(1 + 3.2 + \frac{3.2^2}{2}\right) = 1 - 9.32e^{-3.2} \approx 1 - 0.3799 = 0.620\text{ (3 s.f.)}

(b) For 1 week, $\lambda = 1.6$ . Use conditional probability:

P(X = 2 \mid X \ge 1) = \frac{P(X = 2)}{P(X \ge 1)} = \frac{\frac{e^{-1.6}(1.6)^2}{2}}{1 - e^{-1.6}} = \frac{1.28e^{-1.6}}{1 - e^{-1.6}} \approx \frac{0.2584}{0.7981} \approx 0.324\text{ (3 s.f.)}

🔗 See also: Binomial Distribution · Hypothesis Testing

Questions Level

S2·2

Continuous Random Variables

📌 Continuous Random Variables S2 · Probability density function · CDF $F(x)$ · $E(X)$ & $\text{Var}(X)$ · Median & quartiles

Probability Density Function

$f(x)\geq0$ for all $x$ and $\displaystyle\int_{-\infty}^{\infty}f(x)\,dx=1$ . $P(a\leq X\leq b)=\displaystyle\int_a^b f(x)\,dx$ .

Cumulative Distribution Function

$F(x)=P(X\leq x)=\displaystyle\int_{-\infty}^{x}f(t)\,dt$ . Differentiating: $f(x)=F'(x)$ .

Mean (Expectation)

$E(X)=\displaystyle\int_{-\infty}^{\infty}xf(x)\,dx$ . $E(g(X))=\displaystyle\int g(x)f(x)\,dx$ .

Variance

$\text{Var}(X)=E(X^2)-[E(X)]^2$ where $E(X^2)=\displaystyle\int x^2f(x)\,dx$ .

Median and Percentiles

Median $m$ : $F(m)=0.5$ . $p$ -th percentile $x_p$ : $F(x_p)=p/100$ .

Uniform Distribution

$X\sim U(a,b)$ : $f(x)=\dfrac{1}{b-a}$ , $E(X)=\dfrac{a+b}{2}$ , $\text{Var}(X)=\dfrac{(b-a)^2}{12}$ .

▲ Continuous Random Variables — Reference

A CRV $X$ has a probability density function $f(x)\geq0$ with $\int_{-\infty}^{\infty}f(x)\,dx=1$ . CDF: $F(x)=P(X\leq x)=\int_{-\infty}^{x}f(t)\,dt$ . Mean: $E(X)=\int x\,f(x)\,dx$ . Variance: $\text{Var}(X)=\int x^2f(x)\,dx-[E(X)]^2$ . Median $m$ : solve $F(m)=0.5$ . Mode: maximum of $f(x)$ .

pdf condition

\int f(x)\,dx=1

CDF

F(x)=\int_{-\infty}^{x}f(t)\,dt

E(X)

\int xf(x)\,dx

Var(X)

E(X^2)-[E(X)]^2

Worked example — Finding the constant

$f(x)=kx(2-x)$ for $0\leq x\leq2$ , zero otherwise. Find $k$ , then $E(X)$ .

\int_0^2 kx(2-x)\,dx=k\int_0^2(2x-x^2)\,dx=k\left[x^2-\tfrac{x^3}{3}\right]_0^2=k\cdot\tfrac{4}{3}=1\Rightarrow k=\tfrac{3}{4}.

E(X)=\int_0^2 x\cdot\tfrac{3}{4}x(2-x)\,dx=\tfrac{3}{4}\int_0^2(2x^2-x^3)\,dx=\tfrac{3}{4}\left[\tfrac{2x^3}{3}-\tfrac{x^4}{4}\right]_0^2=\tfrac{3}{4}\cdot\tfrac{4}{3}=1.

Worked example — CDF and median

With the same $f(x)=\tfrac{3}{4}x(2-x)$ on $[0,2]$ , find the CDF $F(x)$ and the median.

F(x)=\int_0^x \tfrac{3}{4}t(2-t)\,dt=\tfrac{3}{4}\left[t^2-\tfrac{t^3}{3}\right]_0^x=\tfrac{3x^2}{4}-\tfrac{x^3}{4}.

Median: $F(m)=\tfrac{1}{2}\Rightarrow\tfrac{3m^2}{4}-\tfrac{m^3}{4}=\tfrac{1}{2}\Rightarrow m^3-3m^2+2=0$ . Try $m=1$ : $1-3+2=0$ ✓. So median $=1$ .

f(x)=cx^2

for

0\leq x\leq3

, zero otherwise. Find

c

,

E(X)

, and

P(1\leq X\leq2)

.

\int_0^3 cx^2\,dx=9c=1\Rightarrow c=\tfrac{1}{9}

.

E(X)=\tfrac{1}{9}\int_0^3 x^3\,dx=\tfrac{1}{9}\cdot\tfrac{81}{4}=\tfrac{9}{4}

.

P(1\leq X\leq2)=\tfrac{1}{9}\left[\tfrac{x^3}{3}\right]_1^2=\tfrac{1}{27}(8-1)=\tfrac{7}{27}

.

Probability density function f(x) = (3/4)x(2 - x) for 0 ≤ x ≤ 2, with total area equal to 1 and symmetric peak mode at x = 1.

⚠️ Cambridge 9709 Traps — Defining the Complete CDF & Point Probabilities

Specifying the CDF Across the Entire Real Line: A cumulative distribution function $F(x)$ is defined for ALL real numbers $x$ . Cambridge will penalize solutions that write only the piecewise middle formula. You must include:

F(x) = \begin{cases} 0 & x < a \\ \int_a^x f(t)\,dt & a \le x \le b \\ 1 & x > b \end{cases}

Continuous Point Probabilities: For any continuous random variable, $P(X = c) = 0$ for any individual point $c$ . Hence $P(a < X < b) = P(a \le X \le b) = F(b) - F(a)$ . Do not apply discrete continuity corrections unless using a normal approximation to a discrete distribution.

A continuous random variable

X

has PDF given by

f(x) = \frac{k}{x^2}

for

1 \le x \le 4

, and 0 elsewhere. (a) Show that

k = \frac{4}{3}

. (b) Find the median of

X

.

(a) Integrate over the support and equate to 1:

\int_1^4 \frac{k}{x^2}\,dx = k\left[-\frac{1}{x}\right]_1^4 = k\left(-\frac{1}{4} - (-1)\right) = \frac{3}{4}k = 1 \implies k = \frac{4}{3}

(b) Find the CDF for $1 \le x \le 4$ :

F(x) = \int_1^x \frac{4}{3t^2}\,dt = \frac{4}{3}\left(1 - \frac{1}{x}\right)

Set $F(m) = 0.5$ to find the median $m$ :

\frac{4}{3}\left(1 - \frac{1}{m}\right) = \frac{1}{2} \implies 1 - \frac{1}{m} = \frac{3}{8} \implies \frac{1}{m} = \frac{5}{8} \implies m = \frac{8}{5} = 1.6

🔗 See also: Discrete RVs · Integration

Questions Level

S2·3

Sampling & Estimation

📌 Sampling & Estimation S2 · Random sampling · Unbiased estimators $\bar{x}$ and $s^2$ · Central Limit Theorem · Confidence interval for $\mu$

Populations and Samples

Population: entire group of interest. Sample: subset used to make inferences. Random sampling ensures unbiased estimates.

Sample Mean Distribution

If $X\sim N(\mu,\sigma^2)$ then $\bar{X}\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)$ . Standard error $=\dfrac{\sigma}{\sqrt{n}}$ .

Unbiased Estimators

$\bar{X}$ is unbiased for $\mu$ . Unbiased variance: $s^2=\dfrac{1}{n-1}\sum(x_i-\bar{x})^2$ .

Confidence Intervals

95% CI for $\mu$ : $\bar{x}\pm z_{0.025}\cdot\dfrac{\sigma}{\sqrt{n}}$ where $z_{0.025}=1.96$ .

Interpretation

A 95% CI means: 95% of all such intervals constructed would contain the true $\mu$ . Not a probability statement about $\mu$ .

Sample Size

Width of CI $\propto 1/\sqrt{n}$ . To halve width, multiply $n$ by 4. Required $n\geq\left(\dfrac{z\sigma}{E}\right)^2$ .

▲ Sampling & Estimation — Reference

An unbiased estimator of $\mu$ is $\bar{X}$ . An unbiased estimator of $\sigma^2$ is $s^2=\dfrac{\sum(x_i-\bar{x})^2}{n-1}=\dfrac{\sum x_i^2-n\bar{x}^2}{n-1}$ . CLT: For large $n$ , $\bar{X}\sim N\!\left(\mu,\dfrac{\sigma^2}{n}\right)$ approximately. A 95% confidence interval for $\mu$ (known $\sigma$ ): $\bar{x}\pm1.96\dfrac{\sigma}{\sqrt{n}}$ . Use $z=2.576$ for 99% CI.

Unbiased

\hat{\mu}

\bar{x}=\dfrac{\sum x}{n}

Unbiased

\hat{\sigma}^2

s^2=\dfrac{\sum x^2-n\bar{x}^2}{n-1}

Dist. of

\bar{X}

N\!\left(\mu,\dfrac{\sigma^2}{n}\right)

95% CI

\bar{x}\pm1.96\dfrac{\sigma}{\sqrt{n}}

Worked example — Confidence interval

A sample of 36 light bulbs has mean lifetime $\bar{x}=1200$ hours. The population SD is known to be $\sigma=60$ hours. Find a 95% CI for the population mean.

\bar{x}\pm1.96\cdot\frac{\sigma}{\sqrt{n}}=1200\pm1.96\cdot\frac{60}{\sqrt{36}}=1200\pm19.6.

95% CI: $(1180.4,\;1219.6)$ hours.

Worked example — Unbiased estimators

A sample of 5 values: 12, 15, 11, 14, 13. Find unbiased estimates of $\mu$ and $\sigma^2$ .

$\bar{x}=65/5=13$ . $\sum x^2=144+225+121+196+169=855$ . $s^2=\dfrac{855-5(169)}{4}=\dfrac{855-845}{4}=\dfrac{10}{4}=2.5$ .

A sample of size 100 from a population (known

\sigma=20

) gives

\bar{x}=142

. Find a 99% CI for

\mu

and state what "99% confidence" means.

142\pm2.576\cdot\dfrac{20}{\sqrt{100}}=142\pm5.15=(136.85,\,147.15)

. "99% confidence" means: if this process were repeated many times, 99% of the intervals produced would contain the true

\mu

.

Sampling distribution of the mean, with shaded central 95% area corresponding to margin of error ± 1.96 standard errors.

⚠️ Cambridge 9709 Traps — Unbiased Sample Variance (n-1) & CLT Applicability

Divisor $n-1$ for Unbiased Variance: When estimating population variance from sample data, you MUST divide by $n-1$ , not $n$ :

s^2 = \frac{1}{n-1}\left(\sum x^2 - n\bar{x}^2\right)

Using $n$ in the denominator gives the biased sample variance and forfeits marks.

Central Limit Theorem Meaning: The CLT states that the distribution of the sample mean $\bar{X}$ tends toward normality as sample size $n$ grows large ( $n \ge 30$ ), regardless of the parent distribution. It does NOT make the individual data points normally distributed! If the parent population is already normal, $\bar{X}$ is exactly normal for *any* sample size, and CLT is not required.

The masses,

x

grams, of 50 packets of tea gave

\sum x = 5090

and

\sum x^2 = 518380

. (a) Calculate unbiased estimates of the population mean and variance. (b) Calculate a 95% confidence interval for the population mean.

(a) Sample mean:

\bar{x} = \frac{5090}{50} = 101.8\text{ g}

Unbiased sample variance:

s^2 = \frac{1}{50 - 1}\left(518380 - 50(101.8)^2\right) = \frac{1}{49}(518380 - 518162) = \frac{218}{49} \approx 4.449\text{ g}^2

(b) Standard error: $\text{SE} = \sqrt{\frac{s^2}{n}} = \sqrt{\frac{4.449}{50}} \approx 0.2983$ .

\bar{x} \pm 1.96(0.2983) = 101.8 \pm 0.585 = (101.2,\; 102.4)\text{ g (1 d.p.)}

🔗 See also: Hypothesis Testing

Questions Level

S2·4

Hypothesis Testing

📌 Hypothesis Testing S2 · Null & alternative hypothesis · Critical region · One- & two-tailed tests · Type I & Type II errors

Hypotheses

$H_0$ (null hypothesis): assumed true. $H_1$ (alternative): what we test for. One-tailed or two-tailed test.

Test Statistic

$Z=\dfrac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}\sim N(0,1)$ under $H_0$ . Compare with critical value $z_\alpha$ .

Critical Region

Range of test statistic that leads to rejection of $H_0$ . For two-tailed 5%: $|Z|>1.96$ . One-tailed 5%: $Z>1.645$ or $Z<-1.645$ .

p-value

Probability of observing a result as extreme as the sample, assuming $H_0$ . Reject $H_0$ if $p\text{-value}<\alpha$ .

Type I and II Errors

Type I: reject $H_0$ when true (probability = significance level $\alpha$ ). Type II: fail to reject $H_0$ when false (probability $\beta$ ).

Testing Poisson/Binomial

Find $P(X\geq x_0)$ (or $P(X\leq x_0)$ ) under $H_0$ . If $<\alpha$ , reject $H_0$ . State conclusion in context.

▲ Hypothesis Testing — Reference

Steps: (1) State $H_0$ and $H_1$ . (2) Specify significance level $\alpha$ . (3) Identify test statistic and its distribution under $H_0$ . (4) Find the critical region or $p$ -value. (5) Compare and conclude in context.
One-tailed: $H_1:\mu>\mu_0$ or $\mu<\mu_0$ . Two-tailed: $H_1:\mu\neq\mu_0$ (split $\alpha$ equally). Type I error: reject $H_0$ when it is true (probability = $\alpha$ ). Type II error: fail to reject $H_0$ when it is false.

Test statistic (

\sigma

known)

Z=\dfrac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}

Critical value (5%, 1-tail)

z=1.645

Critical value (5%, 2-tail)

z=\pm1.96

Type I error prob.

P(\text{reject }H_0\mid H_0\text{ true})=\alpha

Worked example — One-tailed test for

\mu

Bags of flour are labelled 500 g. A sample of 25 bags gives $\bar{x}=497$ g. Population SD is 8 g. Test at 5% whether there is evidence that bags are underweight.

$H_0:\mu=500$ , $H_1:\mu<500$ (one-tailed). $Z=\dfrac{497-500}{8/5}=\dfrac{-3}{1.6}=-1.875$ . Critical value: $-1.645$ . Since $-1.875<-1.645$ , reject $H_0$ . There is significant evidence at 5% that bags are underweight.

Worked example — Poisson hypothesis test

Road accidents used to occur at a rate of 4 per week. After safety measures, only 1 accident is observed in one week. Test at 5% whether the rate has decreased.

$H_0:\lambda=4$ , $H_1:\lambda<4$ . $P(X\leq1)=e^{-4}(1+4)=5e^{-4}\approx0.0916$ . Since $0.0916>0.05$ , do not reject $H_0$ . Insufficient evidence that the rate has decreased.

A manufacturer claims that the mean breaking strength of cables is at least 80 kN. A sample of 16 cables gives

\bar{x}=77.5

kN with population SD 5 kN. Test this claim at 1% significance.

H_0:\mu=80

,

H_1:\mu<80

.

Z=\dfrac{77.5-80}{5/4}=-\dfrac{2.5}{1.25}=-2

. Critical value at 1%:

-2.326

. Since

-2>-2.326

, do not reject

H_0

. Insufficient evidence at 1% that mean breaking strength is below 80 kN.

Standard normal rejection region under H₀ for a two-tailed test at the 5% level, showing the lower 2.5% rejection tail below z = -1.96.

⚠️ Cambridge 9709 Traps — One-Tailed vs Two-Tailed Tests & Critical Values

Deciding Tail Direction: The alternative hypothesis $H_1$ must be determined strictly from the context before looking at sample numbers. Neutral words like 'has changed' or 'is different' require a two-tailed test ( $H_1: \mu \neq \mu_0$ ), splitting $\alpha$ into $\alpha/2$ in each tail (e.g. $z = \pm 1.96$ at 5%). Directional words like 'has decreased' require a one-tailed test ( $z = -1.645$ at 5%).

Contextual Conclusion: Cambridge requires two parts in any hypothesis test conclusion: (1) A statistical decision: 'Reject $H_0$ ' or 'Do not reject $H_0$ '. (2) An in-context non-assertive summary: 'There is sufficient evidence at the 5% level to conclude that the mean weight has changed.' Never write 'proved that $H_0$ is true'.

A random variable

X

has distribution

\text{Po}(\lambda)

. A single observation of

X

is taken to test

H_0: \lambda = 6

against

H_1: \lambda < 6

at the 5% level. (a) Find the critical region. (b) State the actual significance level. (c) Find the probability of a Type II error if the true value of

\lambda

is 3.

(a) Under $H_0$ , $X \sim \text{Po}(6)$ . Calculate cumulative probabilities:

P(X = 0) = e^{-6} \approx 0.00248

P(X \le 1) = e^{-6}(1 + 6) = 7e^{-6} \approx 0.01735

P(X \le 2) = e^{-6}(1 + 6 + 18) = 25e^{-6} \approx 0.06197

Since $P(X \le 1) < 0.05$ and $P(X \le 2) > 0.05$ , the critical region is $X \le 1$ (i.e. $X = 0$ or $1$ ).

(b) Actual significance level = $P(X \le 1 \mid \lambda = 6) = 7e^{-6} \approx 0.0174$ (or 1.74%).

(c) Type II error occurs when we fail to reject $H_0$ when $\lambda = 3$ . We fail to reject when $X \ge 2$ :

P(X \ge 2 \mid \lambda = 3) = 1 - P(X \le 1 \mid \lambda = 3) = 1 - e^{-3}(1 + 3) = 1 - 4e^{-3} \approx 1 - 0.1991 = 0.801\text{ (3 s.f.)}

🔗 See also: Sampling & Estimation · Poisson Distribution

Questions Level

§

Practice set

Fifty A2-style problems across functions, logarithms, trigonometry, calculus, series and vectors, tagged by difficulty. Click any problem for a full worked solution.

1

Find the inverse of $f(x)=2x+1$ .

Basic

f^{-1}(x)=\tfrac{x-1}{2}.

2

If $f(x)=x^{2}$ and $g(x)=x+1$ , find $fg(2)$ .

Intermediate

$g(2)=3$ , then $f(3)=9$ :

9.

3

Solve $|x|=4$ .

Intermediate

x=\pm4.

4

Solve $|x-1|=3$ .

Advanced

$x-1=\pm3$ :

x=4,\ -2.

5

If $f(x)=x^{2}$ and $g(x)=x+1$ , find $gf(x)$ .

Intermediate

$g(f(x))=f(x)+1$ :

x^{2}+1.

6

State the range of $f(x)=x^{2}+1$ .

Advanced

f(x)\ge 1.

7

Solve $2^{x}=8$ .

Basic

2^{3}=8\Rightarrow x=3.

8

Evaluate $\log_2 8$ .

Intermediate

3.

9

Evaluate $\log_{10}1000$ .

Intermediate

3.

10

Solve $e^{x}=1$ .

Advanced

x=0.

11

Evaluate $\ln e$ .

Intermediate

1.

12

Solve $\log_{10}x=2$ .

Advanced

x=100.

13

Write $\log a+\log b$ as a single logarithm.

Intermediate

\log(ab).

14

Solve $3^{x}=81$ .

Advanced

3^{4}=81\Rightarrow x=4.

15

State the formula for $\sin2\theta$ .

Intermediate

2\sin\theta\cos\theta.

16

Evaluate $\cos0^{\circ}$ .

Intermediate

1.

17

Solve $\cos\theta=0$ for $0^{\circ}\le\theta\le180^{\circ}$ .

Advanced

\theta=90^{\circ}.

18

Evaluate $\tan45^{\circ}$ .

Intermediate

1.

19

State the exact value of $\sin45^{\circ}$ .

Advanced

\tfrac{\sqrt{2}}{2}.

20

Express $\sec\theta$ in terms of $\cos\theta$ .

Intermediate

\sec\theta=\tfrac{1}{\cos\theta}.

21

Write $\cos2\theta$ in terms of $\cos\theta$ .

Advanced

2\cos^{2}\theta-1.

22

State the period of $y=\tan x$ .

Intermediate

180^{\circ}.

23

Differentiate $x^{4}$ .

Intermediate

4x^{3}.

24

Differentiate $\sin x$ .

Intermediate

\cos x.

25

Differentiate $e^{x}$ .

Intermediate

e^{x}.

26

Differentiate $\ln x$ .

Intermediate

\tfrac{1}{x}.

27

Differentiate $x^{2}e^{x}$ .

Advanced

Product rule:

2xe^{x}+x^{2}e^{x}=(x^{2}+2x)e^{x}.

28

Differentiate $(2x+1)^{3}$ .

Intermediate

Chain rule:

6(2x+1)^{2}.

29

Differentiate $\cos x$ .

Advanced

-\sin x.

30

Find the gradient of $y=x^{3}$ at $x=2$ .

Intermediate

$\tfrac{dy}{dx}=3x^{2}=12$ :

12.

31

Differentiate $\dfrac{x}{x+1}$ .

Advanced

Quotient rule:

\frac{1}{(x+1)^{2}}.

32

Find the second derivative of $x^{3}$ .

Intermediate

$\tfrac{dy}{dx}=3x^{2}$ , again:

6x.

33

Find $\displaystyle\int x^{3}\,dx$ .

Intermediate

\tfrac{x^{4}}{4}+C.

34

Find $\displaystyle\int e^{x}\,dx$ .

Intermediate

e^{x}+C.

35

Find $\displaystyle\int \cos x\,dx$ .

Intermediate

\sin x+C.

36

Find $\displaystyle\int \tfrac{1}{x}\,dx$ .

Advanced

\ln|x|+C.

37

Evaluate $\displaystyle\int_{0}^{1} x\,dx$ .

Intermediate

\left[\tfrac{x^{2}}{2}\right]_{0}^{1}=\tfrac12.

38

Evaluate $\displaystyle\int_{0}^{\pi} \sin x\,dx$ .

Advanced

\left[-\cos x\right]_{0}^{\pi}=2.

39

Find $\displaystyle\int 6x^{2}\,dx$ .

Intermediate

2x^{3}+C.

40

Find the area under $y=x^{2}$ from $0$ to $3$ .

Advanced

\left[\tfrac{x^{3}}{3}\right]_{0}^{3}=9.

41

State the sum of the first $n$ integers.

Intermediate

\tfrac{n(n+1)}{2}.

42

Find the 10th term of the AP $2,5,8,\dots$

Intermediate

$a=2,\ d=3$ :

2+9(3)=29.

43

Find the 5th term of the GP $3,6,12,\dots$

Advanced

$a=3,\ r=2$ :

3\cdot2^{4}=48.

44

Expand $(1+x)^{2}$ .

Intermediate

1+2x+x^{2}.

45

Find the sum $1+2+\dots+100$ .

Advanced

\tfrac{100\cdot101}{2}=5050.

46

Find the sum to infinity of a GP with $a=1$ , $r=\tfrac12$ .

Intermediate

\frac{1}{1-\tfrac12}=2.

47

Find the magnitude of $(3,4)$ .

Intermediate

\sqrt{9+16}=5.

48

Find $(1,2)+(3,1)$ .

Intermediate

(4,\,3).

49

Find $2(1,3)$ .

Advanced

(2,\,6).

50

Find the magnitude of $(6,8)$ .

Intermediate

\sqrt{36+64}=10.

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Based on Cambridge International A Level Mathematics 9709 Paper 3: Pure Mathematics 3. Advanced topics: complex numbers, series, differential equations, vectors, numerical methods.

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A2 Level Mathematics

About A2 Level

Functions: Modulus & Inverse

Exponentials & Logarithms

Advanced Trigonometry

Differentiation

Integration

Sequences & Series

Vectors

Algebra: Partial Fractions & Division

Complex Numbers

Numerical Methods

Differential Equations

Poisson Distribution

Conditions

Probability Formula

Mean and Variance

Cumulative Tables

Changing the Interval

Normal Approximation

Continuous Random Variables

Probability Density Function

Cumulative Distribution Function

Mean (Expectation)

Variance

Median and Percentiles

Uniform Distribution

Sampling & Estimation

Populations and Samples

Sample Mean Distribution

Unbiased Estimators

Confidence Intervals

Interpretation

Sample Size

Hypothesis Testing

Hypotheses

Test Statistic

Critical Region

p-value

Type I and II Errors

Testing Poisson/Binomial

Practice set

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