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About AS Level

🗺 Cambridge AS Level Mathematics 9709 — Syllabus Map

Paper	ClipSAT Chapter	Key Topics
P1 Pure 1	Algebra — Indices, Surds & Quadratics	Laws of indices, surds, completing the square, discriminant
	Simultaneous Equations & Inequalities	Linear/nonlinear systems, modulus inequalities
	Graphs & Transformations	y=f(x) transformations: translations, stretches, reflections
	Coordinate Geometry	Line segments, circles, intersection
	Circular Measure	Radians, arc length, sector area, segment
	Trigonometry	Exact values, identities, solving equations in given range
	Series & Binomial Theorem	AP/GP, sum to infinity, $(a+b)^n$ expansion
	Differentiation & Integration	Gradient, tangent/normal, stationary points, definite integral, area
P1	Functions: Composite & Inverse	Domain, range, $fg$ , $f^{-1}$
M1 Mechanics	Kinematics	SUVAT, velocity-time graphs, calculus kinematics
	Forces & Equilibrium / Newton's Laws	Resultant, components, $F=ma$ , connected particles
	Energy, Work & Power	KE, PE, work-energy principle, power
S1 Statistics	Representation of Data	Histograms, cumulative freq., mean, variance, SD, box plots
	Permutations & Combinations	$n!$ , $^nP_r$ , $^nC_r$ , arrangement with restrictions
	Discrete Random Variables	Probability distribution, $E(X)$ , $\text{Var}(X)$
	Binomial Distribution	$X\sim B(n,p)$ , cumulative probabilities, mean & variance
	Normal Distribution	$X\sim N(\mu,\sigma^2)$ , standardisation, inverse normal

AS Level Mathematics is the first year of the A Level, offered by boards such as Cambridge (9709) and Edexcel. It blends Pure Mathematics with an introduction to Statistics and Mechanics.

What it covers

The Pure core runs through indices and surds, quadratics, coordinate geometry, graphs and transformations, trigonometry, and the first ideas of differentiation and integration. Applied units add probability, data handling and kinematics.

Pure

algebra · geometry · calculus

Statistics

data · probability

Mechanics

kinematics · forces

Calculator

allowed (most papers)

▲ How to use this course

Each chapter pairs a reference panel with worked examples; the explorers make slope, transformations and the gradient idea visual. Finish with the 50-problem set, then export it as a worksheet.

⚠ Cambridge 9709 Exam Rules & Accuracy Guidelines

Accuracy standard: Non-exact answers must be given to 3 significant figures, or 1 decimal place for angles in degrees, unless specified otherwise in the question. Do not round prematurely in intermediate steps.

Command words: 'Show that' demands complete algebraic working with no skipped steps. 'Hence' means you must use the preceding result — alternate methods receive no credit.

Angle modes: Circular measure and calculus strictly require radian mode. Working in degrees for arc length or calculus formulas voids the entire solution.

Cambridge AS Level Mathematics 9709 Paper Marks: Pure Mathematics 1 (75 marks, 60% of AS) paired with Mechanics (50 marks, 40%) or Probability & Statistics (50 marks, 40%).

A student computes the volume of a cylinder using

r = 3.2\text{ cm}

and

h = 7.85\text{ cm}

. The raw calculator display shows

252.518...

. How should the final answer be reported according to Cambridge AS Level guidelines?

Unless otherwise specified, non-exact numerical answers must be given to 3 significant figures. Rounding

252.518...

to 3 s.f. gives

253\text{ cm}^3

. Giving 1 or 2 s.f., or excessive decimal places without justification, incurs an accuracy penalty (A1).

📝 Chapter Quiz

Questions Level

1

Indices, Surds & Quadratics

📌 Algebra — Indices, Surds & Quadratics P1 · Laws of indices · Surds · Completing the square · Discriminant & nature of roots

The laws of indices, simplifying and rationalising surds, and the full toolkit for quadratics — factorising, completing the square and the formula.

▲ Reference — Algebra core

$a^{m}a^{n}=a^{m+n}$ , $a^{-n}=\tfrac{1}{a^{n}}$ , $a^{1/n}=\sqrt[n]{a}$ . Quadratic formula $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ ; the discriminant $b^{2}-4ac$ fixes the number of real roots.

Indices

a^{m}a^{n}=a^{m+n}

Surds

\sqrt{ab}=\sqrt{a}\sqrt{b}

Completing square

(x+\tfrac{b}{2})^{2}-\tfrac{b^{2}}{4}

Discriminant

b^{2}-4ac

Parabola

y = 2x^2 - 5x - 3

intersecting the

x

-axis at

x = -\frac{1}{2}

and

x = 3

, with vertex at

(1.25, -6.125)

.

⚠ Common Traps & Exam Pitfalls

Completing the square when $a \neq 1$ : Factor

a

from the quadratic and linear terms before halving the coefficient:

ax^2+bx+c = a\left[x^2+\tfrac{b}{a}x\right]+c = a\left(x+\tfrac{b}{2a}\right)^2 - a\left(\tfrac{b}{2a}\right)^2 + c

. Forgetting to multiply the subtracted square by

a

is the most frequent algebraic slip.

Discriminant phrasing: 'Real roots' means

b^2-4ac \ge 0

(includes repeated roots). 'Two distinct real roots' strictly requires

b^2-4ac > 0

. 'No real roots' means

b^2-4ac < 0

. If the leading coefficient contains a parameter

k

, remember

k \neq 0

for the equation to be quadratic.

Surd conjugates: To rationalise

\frac{a}{b+\sqrt{c}}

, multiply numerator and denominator by

b-\sqrt{c}

. Expanding

(b+\sqrt{c})(b-\sqrt{c}) = b^2 - c

eliminates the radical completely.

Worked example 1.A

Simplify $\sqrt{12}$ .

\sqrt{4\cdot3}=2\sqrt{3}.

Worked example 1.B

Complete the square for $x^{2}+4x$ .

(x+2)^{2}-4.

Solve the quadratic equation

2x^2 - 5x - 3 = 0

by completing the square. State the vertex of the corresponding parabola

y = 2x^2 - 5x - 3

.

Factor 2 from the first two terms:

2\left(x^2 - \tfrac{5}{2}x\right) - 3 = 0

. Halve the coefficient of

x

:

\tfrac{1}{2}\left(-\tfrac{5}{2}\right) = -\tfrac{5}{4}

. Complete the square:

2\left(x - \tfrac{5}{4}\right)^2 - 2\left(\tfrac{25}{16}\right) - 3 = 0\Rightarrow 2\left(x - \tfrac{5}{4}\right)^2 - \tfrac{25}{8} - \tfrac{24}{8} = 0\Rightarrow 2\left(x - \tfrac{5}{4}\right)^2 = \tfrac{49}{8}

. Divide by 2:

\left(x - \tfrac{5}{4}\right)^2 = \tfrac{49}{16}

. Take square roots:

x - \tfrac{5}{4} = \pm\tfrac{7}{4}\Rightarrow x = \tfrac{5+7}{4} = 3

or

x = \tfrac{5-7}{4} = -\tfrac{1}{2}

. The vertex is

\left(\tfrac{5}{4}, -\tfrac{49}{8}\right) = (1.25, -6.125)

.

Express

\dfrac{3+\sqrt{5}}{2-\sqrt{5}}

in the form

p + q\sqrt{5}

, where

p

and

q

are rational numbers to be found.

Multiply numerator and denominator by the conjugate of the denominator,

2+\sqrt{5}

:

\dfrac{3+\sqrt{5}}{2-\sqrt{5}} \times \dfrac{2+\sqrt{5}}{2+\sqrt{5}} = \dfrac{(3+\sqrt{5})(2+\sqrt{5})}{2^2 - (\sqrt{5})^2}

Expanding the numerator:

3(2) + 3\sqrt{5} + 2\sqrt{5} + (\sqrt{5})^2 = 6 + 5\sqrt{5} + 5 = 11 + 5\sqrt{5}

. The denominator is

4 - 5 = -1

. Therefore:

\dfrac{11+5\sqrt{5}}{-1} = -11 - 5\sqrt{5}

Thus

p = -11

and

q = -5

.

Find the set of values of the constant

k

for which the equation

kx^2 + 4x + (k-3) = 0

has real roots.

For real roots, the discriminant

\Delta = b^2 - 4ac \ge 0

. Here

a = k

,

b = 4

,

c = k-3

:

\Delta = 4^2 - 4(k)(k-3) = 16 - 4k^2 + 12k \ge 0

Divide the entire inequality by

-4

, reversing the inequality sign:

k^2 - 3k - 4 \le 0

Factor the quadratic:

(k - 4)(k + 1) \le 0

. The critical values are

k = -1

and

k = 4

. Since the parabola opens upward, the product is

\le 0

between the roots:

-1 \le k \le 4

. Note that if

k = 0

, the equation becomes the linear equation

4x - 3 = 0

, which has one real root

x = \tfrac{3}{4}

, so

k = 0

is included. Thus, the solution set is

-1 \le k \le 4

.

📝 Chapter Quiz

Questions Level

2

Simultaneous Equations & Inequalities

📌 Simultaneous Equations & Inequalities P1 · Linear & non-linear systems · Quadratic inequalities · Modulus function

Solving linear and linear-with-quadratic systems, and linear and quadratic inequalities.

▲ Reference — Systems & inequalities

Solve a linear system by substitution or elimination; for a linear–quadratic pair, substitute the linear equation into the quadratic. A quadratic inequality is solved by finding the roots and testing the sign of each interval.

Linear system

substitute / eliminate

Linear–quadratic

substitute the line

Sign rule

flip

\le

when

\div(-)

Quadratic ineq.

roots then test

Parabola

y = x^2 - 3

and line

y = 2x + 1

intersecting at

x = 1 \pm \sqrt{5}

.

⚠ Common Traps & Exam Pitfalls

Quadratic inequalities format: Never write

x^2 > 9 \Rightarrow x > \pm 3

. Always find critical roots and sketch the parabola. For

ax^2+bx+c > 0

(with

a > 0

), the solution consists of two separate inequalities connected by 'or':

x < r_1

or

x > r_2

. Combining them into

r_1 > x > r_2

loses the final mark.

Pairing coordinates in systems: After finding two values of

x

, substitute each into the linear equation to find the corresponding

y

. Pair them explicitly as

(x_1, y_1)

and

(x_2, y_2)

. Substituting into the quadratic equation often creates false phantom pairings.

Modulus validity check: In equations like

|ax+b| = cx+d

, any candidate solution that makes

cx+d < 0

is extraneous and must be discarded because absolute values are non-negative.

Worked example 2.A

Solve $x+y=5$ , $x-y=1$ .

Add: $2x=6$ :

x=3,\ y=2.

Worked example 2.B

Solve $x^{2}-4<0$ .

Roots $\pm2$ , curve below axis between them:

-2<x<2.

Solve the simultaneous equations:

2x + y = 7

x^2 + xy = 12

From the linear equation, express

y

in terms of

x

:

y = 7 - 2x

. Substitute into the second equation:

x^2 + x(7 - 2x) = 12 \Rightarrow x^2 + 7x - 2x^2 = 12 \Rightarrow -x^2 + 7x - 12 = 0

Multiply by

-1

:

x^2 - 7x + 12 = 0

. Factor:

(x - 3)(x - 4) = 0

, giving

x = 3

or

x = 4

. For

x = 3

:

y = 7 - 2(3) = 1

. For

x = 4

:

y = 7 - 2(4) = -1

. The solutions are

(3, 1)

and

(4, -1)

.

Find the set of values of

x

satisfying the inequality

2x^2 - 7x + 3 \le 0

.

Find the critical values where

2x^2 - 7x + 3 = 0

. Factor the quadratic:

(2x - 1)(x - 3) = 0

, so the roots are

x = \tfrac{1}{2}

and

x = 3

. Since the coefficient of

x^2

is positive (

a = 2 > 0

), the parabola opens upward. The quadratic is less than or equal to zero between and including the roots. Therefore, the solution set is

\tfrac{1}{2} \le x \le 3

.

Solve the inequality

|3x - 5| < 2|x + 1|

.

Both sides are non-negative, so we can square both sides:

|3x - 5|^2 < 4|x + 1|^2 \Rightarrow 9x^2 - 30x + 25 < 4(x^2 + 2x + 1)

Expand and simplify:

9x^2 - 30x + 25 < 4x^2 + 8x + 4 \Rightarrow 5x^2 - 38x + 21 < 0

Factor:

(5x - 3)(x - 7) < 0

. The critical values are

x = \tfrac{3}{5}

and

x = 7

. Since the parabola opens upward, the expression is strictly negative between the roots:

\tfrac{3}{5} < x < 7

.

📝 Chapter Quiz

Questions Level

3

Graphs & Transformations

📌 Graphs & Transformations P1 · Sketching curves · $f(x+a)$ , $f(ax)$ , $-f(x)$ , $f(-x)$ · Identifying transformations

Sketching standard curves and applying translations, reflections and stretches.

▲ Reference — Transformations of

y=f(x)

$f(x)+a$ shifts up by $a$ ; $f(x-a)$ shifts right by $a$ ; $-f(x)$ reflects in the $x$ -axis; $af(x)$ stretches vertically by factor $a$ .

The grey curve is $y=\sin x$ ; the indigo curve is $y=\sin(x-h)+k$ . Slide $h$ and $k$ to translate it.

⊙ Explorer · Translating a graphinteractive

Horizontal h = 0

Vertical k = 0

translation (h, k)—

Inside the function, $x-h$ moves right; outside, $+k$ moves up.

Up by a

f(x)+a

Right by a

f(x-a)

Reflect (x-axis)

-f(x)

Stretch

af(x)

Transformation of

y = x^2

to

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

: translated by

\begin{pmatrix}2\\-3\end{pmatrix}

and stretched vertically by scale factor 2.

⚠ Common Traps & Exam Pitfalls

Horizontal vs vertical operations: Operations inside the argument

f(\dots)

affect

x

inversely:

f(x-a)

translates right by

+a

, and

f(kx)

stretches horizontally by scale factor

\tfrac{1}{k}

. Operations outside affect

y

directly:

af(x)

stretches vertically by

a

, and

f(x)+c

shifts up by

c

.

Order of horizontal transformations: For

f(ax+b)

, always factor the bracket:

f\left(a(x + \tfrac{b}{a})\right)

. This represents a horizontal stretch by scale factor

\tfrac{1}{a}

followed by a translation by vector

\begin{pmatrix}-\frac{b}{a}\\0\end{pmatrix}

, OR a translation of

-b

followed by stretch of

\tfrac{1}{a}

. Stating transformations in the wrong order loses all marks.

Reflection axes:

y = -f(x)

is a reflection in the

x

-axis (

y \to -y

).

y = f(-x)

is a reflection in the

y

-axis (

x \to -x

).

Worked example 3.A

Describe $y=(x-2)^{2}$ relative to $y=x^{2}$ .

It is a translation $2$ units to the right.

Worked example 3.B

Describe $y=-x^{2}$ relative to $y=x^{2}$ .

A reflection in the $x$ -axis.

The curve

y = f(x)

has a stationary point at

(4, -6)

. State the coordinates of the corresponding stationary point on the curve: (a)

y = f(x+3) + 2

, (b)

y = 3f(2x)

, (c)

y = -f(-x)

.

(a)

y = f(x+3)+2

translates 3 units left and 2 units up:

(4-3, -6+2) = (1, -4)

.
(b)

y = 3f(2x)

stretches horizontally by

\tfrac{1}{2}

and vertically by 3:

\left(4 \times \tfrac{1}{2}, -6 \times 3\right) = (2, -18)

.
(c)

y = -f(-x)

reflects in both axes (point reflection through the origin):

(-4, -(-6)) = (-4, 6)

.

The curve

y = x^2 - 6x + 5

can be mapped to

y = x^2

by a sequence of two geometric transformations. Express the equation in vertex form, state the vertex, and describe fully each transformation.

Complete the square:

y = (x-3)^2 - 9 + 5 = (x-3)^2 - 4

. The vertex of the parabola is

(3, -4)

. To map

y = (x-3)^2 - 4

back to

y = x^2

, we must translate the graph 3 units to the left and 4 units upwards. In vector notation, this is a single translation by

\begin{pmatrix}-3\\4\end{pmatrix}

, or two sequential transformations: (1) Translation by vector

\begin{pmatrix}-3\\0\end{pmatrix}

(3 units left), and (2) Translation by vector

\begin{pmatrix}0\\4\end{pmatrix}

(4 units up).

📝 Chapter Quiz

Questions Level

4

Coordinate Geometry

📌 Coordinate Geometry P1 · Line equations · Circle equation & properties · Tangent to a circle

Straight lines — gradient, midpoint, distance and equations — and the equation of a circle.

▲ Reference — Lines & circles

Gradient $m=\dfrac{y_2-y_1}{x_2-x_1}$ ; line $y-y_1=m(x-x_1)$ . A circle with centre $(a,b)$ and radius $r$ is $(x-a)^{2}+(y-b)^{2}=r^{2}$ .

Adjust gradient $m$ and intercept $b$ of $y=mx+b$ ; the amber dots are the intercepts.

⊙ Explorer · Gradient & interceptsinteractive

Gradient m = 1

Intercept b = 0

gradient—

y-intercept—

x-intercept—

Perpendicular gradients satisfy $m_1m_2=-1$ .

Gradient

\dfrac{y_2-y_1}{x_2-x_1}

Line

y-y_1=m(x-x_1)

Distance

\sqrt{\Delta x^{2}+\Delta y^{2}}

Circle

(x-a)^{2}+(y-b)^{2}=r^{2}

Circle with centre

(3, -1)

and radius

r = 5

, passing through

P(7, 2)

with equation

(x-3)^2 + (y+1)^2 = 25

.

⚠ Common Traps & Exam Pitfalls

Circle center signs: The equation is

(x-a)^2 + (y-b)^2 = r^2

. In

(x-3)^2 + (y+1)^2 = 25

, the center is

(+3, -1)

, NOT

(-3, 1)

. Also, the right-hand side is

r^2

, so the radius is

\sqrt{25} = 5

, not 25.

Perpendicular lines: Gradients satisfy

m_1 m_2 = -1

, giving

m_{\perp} = -\frac{1}{m}

. A frequent error is forgetting the negative sign (reciprocal without negation).

Circle geometry properties: (1) The tangent to a circle is perpendicular to the radius at the point of contact. (2) The perpendicular bisector of any chord passes through the center of the circle. (3) The angle in a semicircle is a right angle.

Worked example 4.A

Find the gradient through $(1,2)$ and $(4,11)$ .

\frac{11-2}{4-1}=3.

Worked example 4.B

State the centre of $(x-2)^{2}+(y+3)^{2}=25$ .

(2,\,-3).

Find the equation of the perpendicular bisector of the line segment joining

A(-2, 5)

and

B(4, -3)

. Give your answer in the form

ax + by + c = 0

, where

a, b, c

are integers.

1. Midpoint of

AB

:

M = \left(\tfrac{-2+4}{2}, \tfrac{5-3}{2}\right) = (1, 1)

.
2. Gradient of

AB

:

m_{AB} = \tfrac{-3-5}{4-(-2)} = \tfrac{-8}{6} = -\tfrac{4}{3}

.
3. Gradient of perpendicular bisector:

m_{\perp} = -\tfrac{1}{-4/3} = \tfrac{3}{4}

.
4. Equation of line passing through

M(1, 1)

:

y - 1 = \tfrac{3}{4}(x - 1) \Rightarrow 4(y - 1) = 3(x - 1) \Rightarrow 4y - 4 = 3x - 3

Rearrange to standard integer form:

3x - 4y + 1 = 0

.

The circle

C

has equation

x^2 + y^2 - 6x + 4y - 12 = 0

. (a) Find the coordinates of the centre and the exact radius of

C

. (b) Find the equation of the tangent to

C

at the point

P(6, 2)

.

(a) Complete the squares for

x

and

y

:

(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 \Rightarrow (x - 3)^2 + (y + 2)^2 = 25

The centre is

(3, -2)

and the radius is

\sqrt{25} = 5

.
(b) The gradient of the radius from centre

(3, -2)

to

P(6, 2)

is:

m_{\text{rad}} = \dfrac{2 - (-2)}{6 - 3} = \dfrac{4}{3}

The tangent is perpendicular to the radius, so its gradient is

m_{\text{tan}} = -\tfrac{3}{4}

. Using point-slope form with

P(6, 2)

:

y - 2 = -\tfrac{3}{4}(x - 6) \Rightarrow 4(y - 2) = -3(x - 6) \Rightarrow 4y - 8 = -3x + 18 \Rightarrow 3x + 4y - 26 = 0

In slope-intercept form:

y = -\tfrac{3}{4}x + \tfrac{13}{2}

.

📝 Chapter Quiz

Questions Level

5

Trigonometry

📌 Trigonometry P1 · Exact trig values · $\sin^2\theta+\cos^2\theta=1$ · Solving trig equations in a range

Exact values, the identity linking sine and cosine, and solving simple trigonometric equations.

▲ Reference — Trig essentials

$\sin^{2}\theta+\cos^{2}\theta=1$ and $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ . Exact values: $\sin30^{\circ}=\tfrac12$ , $\cos30^{\circ}=\tfrac{\sqrt3}{2}$ , $\sin60^{\circ}=\tfrac{\sqrt3}{2}$ . The sine and cosine curves have period $360^{\circ}$ .

Identity

\sin^{2}\theta+\cos^{2}\theta=1

Tangent

\tfrac{\sin\theta}{\cos\theta}

Exact

\sin60^{\circ}=\tfrac{\sqrt3}{2}

Period

360^{\circ}

Unit circle showing angle

210^\circ

in Quadrant III with reference angle

30^\circ

and coordinates

\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)

.

⚠ Common Traps & Exam Pitfalls

Never divide by trig terms: In equations such as

2\sin\theta\cos\theta = \sin\theta

, dividing by

\sin\theta

destroys the roots where

\sin\theta = 0

. Always rearrange to one side and factor:

\sin\theta(2\cos\theta - 1) = 0

.

Compound angle domain: When solving an equation like

\cos(2x - 30^\circ) = \tfrac{1}{2}

for

0^\circ \le x \le 180^\circ

, first transform the domain for

u = 2x - 30^\circ

:

-30^\circ \le u \le 330^\circ

. Find all valid values for

u

within this adjusted window, then solve for

x

.

Identity substitution: Always check if replacing

\sin^2\theta

with

1 - \cos^2\theta

(or vice-versa) creates a single-variable quadratic in

\cos\theta

or

\sin\theta

.

Worked example 5.A

Solve $\sin\theta=\tfrac12$ for $0^{\circ}\le\theta\le180^{\circ}$ .

\theta=30^{\circ},\ 150^{\circ}.

Worked example 5.B

State the exact value of $\sin60^{\circ}$ .

\tfrac{\sqrt3}{2}.

Solve the equation

2\cos^2\theta - \sin\theta - 1 = 0

for

0^\circ \le \theta \le 360^\circ

.

Use the Pythagorean identity

\cos^2\theta = 1 - \sin^2\theta

:

2(1 - \sin^2\theta) - \sin\theta - 1 = 0 \Rightarrow 2 - 2\sin^2\theta - \sin\theta - 1 = 0 \Rightarrow -2\sin^2\theta - \sin\theta + 1 = 0

Multiply by

-1

:

2\sin^2\theta + \sin\theta - 1 = 0

. Factor:

(2\sin\theta - 1)(\sin\theta + 1) = 0

.
1.

2\sin\theta - 1 = 0 \Rightarrow \sin\theta = \tfrac{1}{2}

. In

[0^\circ, 360^\circ]

,

\theta = 30^\circ

or

180^\circ - 30^\circ = 150^\circ

.
2.

\sin\theta + 1 = 0 \Rightarrow \sin\theta = -1

. In

[0^\circ, 360^\circ]

,

\theta = 270^\circ

.
The complete solution set is

\theta = 30^\circ, 150^\circ, 270^\circ

.

Prove the identity:

\dfrac{\cos\theta}{1 - \sin\theta} - \dfrac{\cos\theta}{1 + \sin\theta} \equiv 2\tan\theta

Start with the left-hand side and write over a common denominator:

\text{LHS} = \dfrac{\cos\theta(1 + \sin\theta) - \cos\theta(1 - \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)}

Expand the numerator:

\cos\theta + \cos\theta\sin\theta - \cos\theta + \cos\theta\sin\theta = 2\cos\theta\sin\theta

.
Expand the denominator:

1 - \sin^2\theta = \cos^2\theta

.
Substitute both back:

\text{LHS} = \dfrac{2\cos\theta\sin\theta}{\cos^2\theta} = \dfrac{2\sin\theta}{\cos\theta} = 2\tan\theta = \text{RHS}

Q.E.D.

Solve the equation

\tan(2x - 20^\circ) = \sqrt{3}

for

0^\circ \le x \le 180^\circ

.

Let

u = 2x - 20^\circ

. As

0^\circ \le x \le 180^\circ

, the domain for

u

is

-20^\circ \le u \le 340^\circ

.
Since

\tan 60^\circ = \sqrt{3}

, the principal solution is

u = 60^\circ

. Tangent has a period of

180^\circ

, so the solutions for

u

within

[-20^\circ, 340^\circ]

are:

u = 60^\circ, \quad u = 60^\circ + 180^\circ = 240^\circ

Now solve for

x = \tfrac{u + 20^\circ}{2}

:
1. For

u = 60^\circ

:

x = \tfrac{60^\circ + 20^\circ}{2} = 40^\circ

.
2. For

u = 240^\circ

:

x = \tfrac{240^\circ + 20^\circ}{2} = 130^\circ

.
Both solutions lie within

[0^\circ, 180^\circ]

. Thus

x = 40^\circ

or

x = 130^\circ

.

📝 Chapter Quiz

Questions Level

6

Differentiation & Integration

📌 Differentiation & Integration P1 · Gradient & stationary points · Tangent & normal · Definite integral & area under curve

The gradient of a curve from first principles, the power rule for differentiation, and reversing it to integrate.

▲ Reference — The power rule

$\dfrac{d}{dx}x^{n}=nx^{n-1}$ ; the gradient at a point is the value of $\tfrac{dy}{dx}$ there. Integration reverses this: $\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C$ for $n\neq-1$ .

The amber line is a secant from $x=1$ to $x=1+h$ on $y=x^{2}$ . Shrink $h$ and watch its slope approach the tangent gradient, $2$ .

⊙ Explorer · Secant → tangentinteractive

Gap h = 2

secant gradient—

The secant gradient is $2+h$ ; as $h\to0$ it tends to the tangent gradient $2$ .

Differentiate

nx^{n-1}

Gradient

value of

\tfrac{dy}{dx}

Integrate

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

Definite

F(b)-F(a)

Cubic curve

y = x^3 - 3x

showing local maximum at

(-1, 2)

and local minimum at

(1, -2)

.

⚠ Common Traps & Exam Pitfalls

Algebraic preparation: Express all algebraic fractions and surds in index form before differentiating or integrating:

\frac{3}{x^2} = 3x^{-2}

,

4\sqrt{x} = 4x^{1/2}

,

\frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-1/2}

. Differentiating denominators directly without rewriting produces invalid derivatives.

Tangent vs Normal: The tangent line has gradient

m_{\text{tan}} = \frac{dy}{dx}

. The normal line is perpendicular, having gradient

m_{\text{norm}} = -\frac{1}{m_{\text{tan}}}

. Both lines pass through the exact same point

(x_1, y_1)

on the curve.

Area between curves: Area is

\int_a^b (y_{\text{top}} - y_{\text{bottom}})\,dx

. If a curve crosses the

x

-axis within the limits of integration, integrating across the roots will result in positive and negative areas cancelling out. You must integrate each section separately and sum the positive areas.

Worked example 6.A

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

\frac{dy}{dx}=2x+3.

Worked example 6.B

Evaluate $\displaystyle\int_{0}^{1} 2x\,dx$ .

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

A curve has equation

y = 2x^3 - 9x^2 + 12x + 1

. (a) Find the coordinates of both stationary points. (b) Determine the nature of each stationary point using the second derivative

\frac{d^2y}{dx^2}

.

(a) Find

\frac{dy}{dx}

:

\dfrac{dy}{dx} = 6x^2 - 18x + 12

Set

\frac{dy}{dx} = 0

:

6(x^2 - 3x + 2) = 0 \Rightarrow 6(x - 1)(x - 2) = 0

, giving

x = 1

and

x = 2

.
When

x = 1

:

y = 2(1) - 9(1) + 12(1) + 1 = 6

. Point is

(1, 6)

.
When

x = 2

:

y = 2(8) - 9(4) + 12(2) + 1 = 16 - 36 + 24 + 1 = 5

. Point is

(2, 5)

.
(b) Second derivative:

\dfrac{d^2y}{dx^2} = 12x - 18

At

x = 1

:

\frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0

, so

(1, 6)

is a local maximum.
At

x = 2

:

\frac{d^2y}{dx^2} = 12(2) - 18 = +6 > 0

, so

(2, 5)

is a local minimum.

Find the exact area of the region enclosed between the curve

y = 4x - x^2

and the line

y = x

.

Find points of intersection:

4x - x^2 = x \Rightarrow 3x - x^2 = 0 \Rightarrow x(3 - x) = 0

The curves intersect at

x = 0

and

x = 3

. Between

0

and

3

, the curve

y = 4x - x^2

lies above the line

y = x

.
The enclosed area is:

A = \int_0^3 \left[(4x - x^2) - x\right]\,dx = \int_0^3 (3x - x^2)\,dx

Integrate:

A = \left[\dfrac{3x^2}{2} - \dfrac{x^3}{3}\right]_0^3 = \left(\dfrac{3(9)}{2} - \dfrac{27}{3}\right) - 0 = \dfrac{27}{2} - 9 = \dfrac{27 - 18}{2} = \dfrac{9}{2} = 4.5

Find the equation of the normal to the curve

y = \dfrac{4}{\sqrt{x}} + 2x

at the point where

x = 4

.

Rewrite in index form:

y = 4x^{-1/2} + 2x

.
At

x = 4

:

y = 4(4^{-1/2}) + 2(4) = 4\left(\tfrac{1}{2}\right) + 8 = 2 + 8 = 10

. Point of contact is

(4, 10)

.
Differentiate:

\dfrac{dy}{dx} = 4\left(-\tfrac{1}{2}x^{-3/2}\right) + 2 = -2x^{-3/2} + 2 = -\dfrac{2}{x^{3/2}} + 2

At

x = 4

:

x^{3/2} = (\sqrt{4})^3 = 8

. Thus:

m_{\text{tan}} = -\dfrac{2}{8} + 2 = -\dfrac{1}{4} + 2 = \dfrac{7}{4}

The normal has gradient

m_{\text{norm}} = -\dfrac{1}{7/4} = -\dfrac{4}{7}

.
Using point

(4, 10)

:

y - 10 = -\dfrac{4}{7}(x - 4) \Rightarrow 7(y - 10) = -4(x - 4) \Rightarrow 7y - 70 = -4x + 16 \Rightarrow 4x + 7y - 86 = 0

📝 Chapter Quiz

Questions Level

7

Statistics & Probability

📌 Statistics & Probability S1 · Basic probability · Conditional probability · Tree diagrams · Venn diagrams

Summarising data with averages and spread, and the probability of simple events.

▲ Reference — Data & chance

Mean is total over count; the median is the middle of the ordered data; range is largest minus smallest. Probability $=\dfrac{\text{favourable}}{\text{total}}$ .

Mean

\dfrac{\sum x}{n}

Median

middle value

Range

max − min

Probability

\dfrac{\text{fav}}{\text{total}}

Venn diagram showing two independent events

A

and

B

with

P(A) = 0.4

,

P(B) = 0.5

, and

P(A \cap B) = 0.2

.

⚠ Common Traps & Exam Pitfalls

Mutually exclusive vs Independent: These two concepts are mutually contradictory for non-trivial events. Mutually exclusive means they cannot happen together:

P(A \cap B) = 0

. Independent means the occurrence of one does not affect the probability of the other:

P(A \cap B) = P(A)P(B)

. If

P(A) > 0

and

P(B) > 0

, mutually exclusive events can never be independent.

Conditional probability denominator: In

P(A|B) = \frac{P(A \cap B)}{P(B)}

, the denominator is always the given condition

P(B)

, never

P(A)

.

Sampling without replacement: In successive draws without replacement, remember to reduce both the favorable count and the total count on second branches (e.g.

\frac{5}{9} \times \frac{4}{8}

).

Worked example 7.A

Find the mean of $2,\,4,\,6,\,8$ .

\frac{20}{4}=5.

Worked example 7.B

Find the median of $3,\,7,\,9,\,12,\,20$ .

9.

Events

A

and

B

are such that

P(A) = 0.6

,

P(B) = 0.5

, and

P(A \cup B) = 0.8

. Find

P(A \cap B)

, calculate

P(A|B)

, and determine whether

A

and

B

are independent.

1. Use the addition rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B) \Rightarrow 0.8 = 0.6 + 0.5 - P(A \cap B) \Rightarrow P(A \cap B) = 1.1 - 0.8 = 0.3

.
2. Conditional probability:

P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.3}{0.5} = 0.6

3. Check for independence:

P(A \cap B) = 0.3

and

P(A) \times P(B) = 0.6 \times 0.5 = 0.30

. Since

P(A \cap B) = P(A)P(B)

(or equivalently

P(A|B) = P(A) = 0.6

), events

A

and

B

are independent.

A bag contains 5 red and 4 green marbles. Three marbles are drawn at random without replacement. Find the probability that: (a) all three are red, (b) at least two are red.

Total marbles = 9. Number of ways to choose 3 marbles =

\binom{9}{3} = \tfrac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

.
(a) All 3 are red:

\binom{5}{3} = 10

. Probability:

P(\text{all red}) = \dfrac{10}{84} = \dfrac{5}{42} \approx 0.119

(b) 'At least two red' means exactly 2 red and 1 green, or exactly 3 red:
Ways for 2 red, 1 green:

\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40

.
Ways for 3 red:

\binom{5}{3} = 10

.
Total favorable outcomes:

40 + 10 = 50

.
Probability:

P(\ge 2 \text{ red}) = \dfrac{50}{84} = \dfrac{25}{42} \approx 0.595

📝 Chapter Quiz

Questions Level

8

Functions: Composite & Inverse

📌 Functions: Composite & Inverse P1 · Domain & range · Composite $fg(x)$ · Inverse $f^{-1}(x)$ · Modulus function

Beyond graphs, AS treats functions formally — domain, range, composition and inverses.

▲ Reference — Functions

$fg(x)=f(g(x))$ . The inverse $f^{-1}$ reverses $f$ and reflects its graph in the line $y=x$ ; the range of $f$ is the domain of $f^{-1}$ .

Composite

fg(x)

Inverse

swap & solve

Reflection

in

y=x

Range

domain of

f^{-1}

Linear function

f(x) = 2x - 3

and its inverse

f^{-1}(x) = \frac{x+3}{2}

, intersecting at

(3, 3)

along the reflection line

y = x

.

⚠ Common Traps & Exam Pitfalls

One-to-one condition for inverses: An inverse function

f^{-1}

exists if and only if

f

is one-to-one over its given domain. For a quadratic function

f(x) = a(x-h)^2 + k

, the domain must be restricted to one side of the vertex (either

x \ge h

or

x \le h

). If the domain is all real numbers, the inverse does not exist.

Domain of $f^{-1}$ : The domain of

f^{-1}

is always the range of

f

. Stating

x \in \mathbb{R}

when the original function had a bounded range (e.g.

f(x) \ge 4

) loses the final mark.

Order of composition:

fg(x)

denotes

f(g(x))

— evaluate

g

first, then insert into

f

. In general,

fg(x) \neq gf(x)

. Furthermore, for

fg

to be defined, the range of

g

must lie entirely within the domain of

f

.

Worked example 8.A

If $f(x)=3x-2$ , find $f^{-1}(x)$ .

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

Worked example — Composite functions

Given $f(x)=\sqrt{x+1}$ ( $x\geq-1$ ) and $g(x)=2x-3$ , find (a) $fg(x)$ , (b) the range of $fg$ .

(a) $fg(x)=f(2x-3)=\sqrt{2x-3+1}=\sqrt{2x-2}$ , domain $x\geq1$ .

(b) As $x\geq1$ , $2x-2\geq0$ , so range of $fg$ is $fg(x)\geq0$ .

The function

f

is defined by

f(x) = x^2 - 6x + 13

for

x \ge 3

. (a) State the range of

f

. (b) Find an expression for

f^{-1}(x)

and state the domain and range of

f^{-1}

.

(a) Complete the square:

f(x) = (x - 3)^2 - 9 + 13 = (x - 3)^2 + 4

. Since

x \ge 3

,

(x - 3)^2 \ge 0

, so the minimum value is

f(3) = 4

. The range of

f

is

f(x) \ge 4

.
(b) Set

y = (x - 3)^2 + 4

. Rearrange for

x

:

y - 4 = (x - 3)^2 \Rightarrow x - 3 = +\sqrt{y - 4} \quad (\text{since } x \ge 3)

x = 3 + \sqrt{y - 4}

Therefore,

f^{-1}(x) = 3 + \sqrt{x - 4}

.
Domain of

f^{-1}

is

x \ge 4

(the range of

f

), and range of

f^{-1}

is

f^{-1}(x) \ge 3

(the domain of

f

).

Functions

g

and

h

are defined by

g(x) = 2x - 1

for

x \in \mathbb{R}

, and

h(x) = \dfrac{6}{x - 2}

for

x \neq 2

. (a) Find an expression for

hg(x)

in simplest form. (b) Solve the equation

hg(x) = g(x)

.

(a) Substitute

g(x)

into

h

:

hg(x) = h(2x - 1) = \dfrac{6}{(2x - 1) - 2} = \dfrac{6}{2x - 3} \quad \left(x \neq \tfrac{3}{2}\right)

(b) Set

hg(x) = g(x)

:

\dfrac{6}{2x - 3} = 2x - 1 \Rightarrow 6 = (2x - 1)(2x - 3)

Expand the right-hand side:

6 = 4x^2 - 8x + 3 \Rightarrow 4x^2 - 8x - 3 = 0

Factor:

(2x - 3)(2x + 1) = 0

.
Since

x \neq \tfrac{3}{2}

for

hg(x)

to be defined, the value

x = \tfrac{3}{2}

is excluded.
Therefore, the unique solution is

2x + 1 = 0 \Rightarrow x = -\tfrac{1}{2}

.

📝 Chapter Quiz

Questions Level

9

Circular Measure

📌 Circular Measure P1 · Radians ↔ degrees · Arc length $s=r\theta$ · Sector area $A=\tfrac{1}{2}r^2\theta$ · Segment area

Angles measured in radians give clean formulas for arc length and sector area.

▲ Reference — Radians

$180^{\circ}=\pi$ radians. Arc length $s=r\theta$ ; sector area $A=\tfrac12 r^{2}\theta$ , with $\theta$ in radians.

Convert

180^{\circ}=\pi

Arc length

s=r\theta

Sector

\tfrac12 r^{2}\theta

Full turn

2\pi

Circle with centre

O

and radius

r = 8\text{ cm}

, showing sector

OAB

subtending central angle

\theta = 1.2\text{ rad}

and chord

AB

.

⚠ Common Traps & Exam Pitfalls

Calculator in Radian mode: Formulas

s = r\theta

,

A = \tfrac{1}{2}r^2\theta

, and

A_{\Delta} = \tfrac{1}{2}r^2\sin\theta

are strictly valid only when

\theta

is in radians. If your calculator is in degree mode when evaluating

\sin(1.2)

, you will evaluate the sine of

1.2^\circ

instead of

1.2\text{ rad}

, wiping out all accuracy marks.

Perimeter of a sector: The perimeter of a sector consists of the curved arc length PLUS the two straight radii:

P = r\theta + 2r

. Overlooking the two radii (

+2r

) is the single most common slip in Cambridge circular measure questions.

Major vs Minor: Always double check whether the question specifies the minor or major segment/sector. For the major sector, the central angle is

2\pi - \theta

.

Worked example 9.A

Find the arc length when $r=5$ and $\theta=\tfrac{\pi}{3}$ .

s=5\cdot\frac{\pi}{3}=\frac{5\pi}{3}.

Worked example — Segment area

A sector OAB has radius 8 cm and angle $\angle AOB=1.2$ rad. Find the area of the minor segment cut off by chord AB.

\text{Sector area}=\tfrac{1}{2}(8)^2(1.2)=38.4\text{ cm}^2.

\text{Triangle area}=\tfrac{1}{2}(8)^2\sin(1.2)=32\sin(1.2)\approx29.56\text{ cm}^2.

\text{Segment}=38.4-29.56\approx8.84\text{ cm}^2.

A circle has radius

8\text{ cm}

. A sector of this circle has a total perimeter of

28\text{ cm}

. Find the angle of the sector in radians, and calculate the exact area of the sector.

The perimeter of a sector is given by

P = s + 2r = r\theta + 2r

.
Substitute

P = 28

and

r = 8

:

28 = 8\theta + 2(8) \Rightarrow 28 = 8\theta + 16 \Rightarrow 8\theta = 12 \Rightarrow \theta = \dfrac{12}{8} = 1.5\text{ radians}

The area of the sector is:

A = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(8^2)(1.5) = \dfrac{1}{2}(64)(1.5) = 32 \times 1.5 = 48\text{ cm}^2

In a circle of radius

10\text{ cm}

, chord

AB

subtends an angle of

\theta = \tfrac{2\pi}{3}

radians at the centre

O

. Find: (a) the exact length of chord

AB

, (b) the exact area of the minor segment bounded by chord

AB

and minor arc

AB

, in terms of

\pi

and

\sqrt{3}

.

(a) Use the cosine rule in triangle

OAB

with

OA = OB = 10

and

\angle AOB = \tfrac{2\pi}{3} = 120^\circ

:

AB^2 = 10^2 + 10^2 - 2(10)(10)\cos\left(\tfrac{2\pi}{3}\right) = 100 + 100 - 200\left(-\tfrac{1}{2}\right) = 200 + 100 = 300

Thus

AB = \sqrt{300} = 10\sqrt{3}\text{ cm}

.
(b) Area of sector

OAB

:

A_{\text{sector}} = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(10^2)\left(\tfrac{2\pi}{3}\right) = \dfrac{100\pi}{3}\text{ cm}^2

Area of triangle

OAB

:

A_{\Delta} = \dfrac{1}{2}r^2\sin\theta = \dfrac{1}{2}(10^2)\sin\left(\tfrac{2\pi}{3}\right) = 50\left(\dfrac{\sqrt{3}}{2}\right) = 25\sqrt{3}\text{ cm}^2

Area of minor segment:

A_{\text{segment}} = A_{\text{sector}} - A_{\Delta} = \dfrac{100\pi}{3} - 25\sqrt{3} = \dfrac{100\pi - 75\sqrt{3}}{3}\text{ cm}^2

📝 Chapter Quiz

Questions Level

10

Series & the Binomial Theorem

📌 Series & Binomial Theorem P1 · AP & GP formulas · Sum to infinity · $(a+b)^n$ expansion · General term

Arithmetic and geometric progressions, and the binomial expansion of $(a+b)^{n}$ .

▲ Reference — Series

AP sum $S_n=\tfrac{n}{2}\big(2a+(n-1)d\big)$ ; GP sum $S_n=\tfrac{a(1-r^{n})}{1-r}$ ; binomial $(1+x)^{n}=1+nx+\tfrac{n(n-1)}{2}x^{2}+\cdots$ .

AP term

a+(n-1)d

AP sum

\tfrac{n}{2}(2a+(n-1)d)

GP sum

\tfrac{a(1-r^{n})}{1-r}

Binomial

{}^nC_r

Decaying terms of a convergent geometric sequence with

a = 12

and

r = \frac{2}{3}

, approaching zero with sum to infinity

S_\infty = 36

.

⚠ Common Traps & Exam Pitfalls

Convergence condition for $S_\infty$ : The sum to infinity

S_\infty = \frac{a}{1-r}

is valid if and only if

|r| < 1

(

-1 < r < 1

). If a question states that a series has a sum to infinity, you can establish an inequality for unknown parameters:

-1 < r < 1

.

Binomial powers on coefficients: In

(a+bx)^n

, the term involving

x^r

contains

(bx)^r = b^r x^r

. Writing

3x^2

instead of

(3x)^2 = 9x^2

or

-2x^3

instead of

(-2x)^3 = -8x^3

destroys the coefficient.

Term formulas index offset: In an AP, the

n

-th term has

n-1

common differences:

u_n = a + (n-1)d

. In a GP,

u_n = ar^{n-1}

. Do not confuse the

n

-th term with the sum

S_n

.

Worked example 10.A

Find the sum of the first $10$ terms of $2,5,8,\dots$

S_{10}=\frac{10}{2}\big(4+27\big)=155.

Worked example — Geometric series & sum to infinity

A geometric series has first term 12 and common ratio $\tfrac{2}{3}$ . Find (a) the 5th term, (b) $S_\infty$ .

(a) $u_5=12\cdot(\tfrac{2}{3})^4=12\cdot\tfrac{16}{81}=\tfrac{64}{27}$ .

(b) $S_\infty=\dfrac{12}{1-\tfrac{2}{3}}=\dfrac{12}{\tfrac{1}{3}}=36$ .

In an arithmetic progression, the 3rd term is

14

and the 7th term is

34

. (a) Find the first term

a

and the common difference

d

. (b) Find the sum of the first

25

terms.

(a) Use

u_n = a + (n - 1)d

:

u_3 = a + 2d = 14

u_7 = a + 6d = 34

Subtract the first equation from the second:

4d = 20 \Rightarrow d = 5

.
Substitute back:

a + 2(5) = 14 \Rightarrow a = 4

.
(b) Use the sum formula

S_n = \tfrac{n}{2}\left(2a + (n - 1)d\right)

with

n = 25

:

S_{25} = \dfrac{25}{2}\left(2(4) + 24(5)\right) = \dfrac{25}{2}(8 + 120) = \dfrac{25}{2}(128) = 25 \times 64 = 1600

Find the coefficient of

x^3

in the expansion of

(2 - 3x)(1 + 2x)^6

.

Expand

(1 + 2x)^6

up to the

x^3

term:

\binom{6}{0} = 1

\binom{6}{1}(2x) = 6(2x) = 12x

\binom{6}{2}(2x)^2 = 15(4x^2) = 60x^2

\binom{6}{3}(2x)^3 = 20(8x^3) = 160x^3

So

(1 + 2x)^6 = 1 + 12x + 60x^2 + 160x^3 + \cdots

Now multiply by

(2 - 3x)

. The

x^3

term comes from:

2(160x^3) + (-3x)(60x^2) = 320x^3 - 180x^3 = 140x^3

Thus, the coefficient of

x^3

is

140

.

📝 Chapter Quiz

Questions Level

11

Kinematics

📌 Kinematics M1 · SUVAT equations · Velocity–time graphs · Calculus kinematics

Mechanics begins with motion in a straight line under constant acceleration.

▲ Reference — suvat equations

For constant acceleration: $v=u+at$ , $s=ut+\tfrac12 at^{2}$ , $v^{2}=u^{2}+2as$ . Velocity is the derivative of displacement; acceleration the derivative of velocity.

Velocity

v=u+at

Displacement

ut+\tfrac12at^{2}

No time

v^{2}=u^{2}+2as

Acceleration

\tfrac{dv}{dt}

Velocity–time trapezoid graph: uniform acceleration from

0

to

20\text{ m/s}

in

5\text{ s}

, constant velocity for

10\text{ s}

, and uniform deceleration to rest in

5\text{ s}

. Total displacement is the trapezoid area:

\frac{10+20}{2} \times 20 = 300\text{ m}

.

⚠ Common Traps & Exam Pitfalls

SUVAT vs Calculus: SUVAT formulas apply ONLY when acceleration is constant. If velocity or acceleration varies with time (e.g.

v = 3t^2 - 12t

), SUVAT yields zero marks — you must differentiate (

a = \frac{dv}{dt}

) or integrate (

s = \int v\,dt

).

Velocity-Time graphs: Gradient equals acceleration (

a = \frac{\Delta v}{\Delta t}

). Total area under the graph equals displacement. For distance vs displacement: if the graph drops below the time axis (negative velocity), distance sums the absolute areas, whereas displacement nets them.

Vertical motion sign convention: Fix a reference direction at the start (usually upwards positive). Under this convention, acceleration is

a = -g = -9.8\text{ m/s}^2

(or

-9.81

). Any position below the launch point corresponds to negative displacement

s < 0

.

Worked example 11.A

With $u=0$ , $a=2$ , $t=3$ , find $v$ .

v=0+2(3)=6.

Worked example — Calculus kinematics

A particle moves so that its velocity at time $t$ s is $v=3t^2-12t+9$ m/s. Find (a) the acceleration at $t=2$ , (b) when the particle is momentarily at rest.

(a) $a=\dfrac{dv}{dt}=6t-12$ . At $t=2$ : $a=0$ m/s².

(b) $v=0\Rightarrow3(t-1)(t-3)=0\Rightarrow t=1$ s or $t=3$ s.

A train accelerates uniformly from rest to

30\text{ m/s}

in

20\text{ s}

, travels at constant speed

30\text{ m/s}

for

60\text{ s}

, and then decelerates uniformly to rest in

15\text{ s}

. Find: (a) the acceleration in the first phase, (b) the total distance travelled, (c) the average speed for the entire journey.

(a)

a_1 = \dfrac{v - u}{t} = \dfrac{30 - 0}{20} = 1.5\text{ m/s}^2

.
(b) Total distance is the area of the trapezoid on the

v

-

t

graph with parallel sides

60\text{ s}

and

20 + 60 + 15 = 95\text{ s}

, and height

30\text{ m/s}

:

s = \dfrac{60 + 95}{2} \times 30 = \dfrac{155}{2} \times 30 = 155 \times 15 = 2325\text{ m}

(c) Average speed:

\text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{2325\text{ m}}{95\text{ s}} \approx 24.47\text{ m/s} \approx 24.5\text{ m/s (3 s.f.)}

A stone is thrown vertically upwards with speed

15\text{ m/s}

from the top of a cliff

40\text{ m}

above sea level. Taking

g = 9.8\text{ m/s}^2

and upwards as positive: (a) find the maximum height reached above the cliff, (b) calculate the time taken for the stone to reach the sea.

(a) At maximum height,

v = 0

.

u = +15\text{ m/s}

,

a = -9.8\text{ m/s}^2

:

v^2 = u^2 + 2as \Rightarrow 0 = 15^2 + 2(-9.8)s \Rightarrow 19.6s = 225 \Rightarrow s = \dfrac{225}{19.6} \approx 11.48\text{ m} \approx 11.5\text{ m}

(b) At the sea, the displacement relative to the launch point is

s = -40\text{ m}

. Use

s = ut + \tfrac{1}{2}at^2

:

-40 = 15t + \tfrac{1}{2}(-9.8)t^2 \Rightarrow -40 = 15t - 4.9t^2 \Rightarrow 4.9t^2 - 15t - 40 = 0

Solve via quadratic formula:

t = \dfrac{15 \pm \sqrt{(-15)^2 - 4(4.9)(-40)}}{2(4.9)} = \dfrac{15 \pm \sqrt{225 + 784}}{9.8} = \dfrac{15 \pm \sqrt{1009}}{9.8}

Since

\sqrt{1009} \approx 31.765

, and

t > 0

:

t = \dfrac{15 + 31.765}{9.8} = \dfrac{46.765}{9.8} \approx 4.77\text{ s (3 s.f.)}

📝 Chapter Quiz

Questions Level

12

Forces & Equilibrium

📌 Forces & Equilibrium M1 · Resultant force · Resolving into components · Lami's theorem · Friction $F=\mu R$

Types of Forces

Weight $W=mg$ , Normal reaction $N$ , Tension $T$ , Friction $F\leq\mu N$ . Forces are vectors — resolve horizontally and vertically.

Equilibrium

A particle is in equilibrium when the resultant force is zero: $\sum F_x=0$ and $\sum F_y=0$ . Use triangle/polygon of forces.

Resolving Forces

Force $F$ at angle $\theta$ : horizontal component $F\cos\theta$ , vertical component $F\sin\theta$ .

Friction

Limiting friction: $F=\mu N$ . When moving: $F=\mu N$ . Static: $F\leq\mu N$ . $\mu$ is the coefficient of friction.

Lami's Theorem

For three concurrent forces in equilibrium: $\dfrac{F_1}{\sin\alpha}=\dfrac{F_2}{\sin\beta}=\dfrac{F_3}{\sin\gamma}$

Inclined Planes

On slope at angle $\alpha$ : component along slope $=mg\sin\alpha$ , perpendicular $=mg\cos\alpha$ .

▲ Forces & Equilibrium — Reference

A force is a vector (magnitude + direction). The resultant of forces is found by vector addition; for equilibrium the resultant is zero. Resolve forces into horizontal and vertical components: $F_x=F\cos\theta$ , $F_y=F\sin\theta$ . Friction force satisfies $F\leq\mu R$ ; limiting equilibrium: $F=\mu R$ .

Resolving

F_x=F\cos\theta,\quad F_y=F\sin\theta

Equilibrium

\sum F_x=0\quad\&\quad\sum F_y=0

Friction (limiting)

F=\mu R

Resultant

R=\sqrt{F_x^2+F_y^2}

Closed vector triangle of forces in equilibrium: Weight

W = 20\text{ N}

, Tension

T

, and Normal reaction

R

summing to zero resultant.

⚠ Common Traps & Exam Pitfalls

Normal reaction is perpendicular to the plane: On an inclined plane of angle

\alpha

, the normal reaction is

R = mg\cos\alpha

(if no other perpendicular forces act). Assuming

R = mg

on an inclined plane is an instant fatal error.

Gravity components on a slope: Gravity acts vertically downward. Its resolved components are

mg\sin\alpha

acting down the slope and

mg\cos\alpha

acting perpendicularly into the plane.

Friction is reactive: The friction force obeys

F \le \mu R

. It only equals

\mu R

when motion is impending (limiting equilibrium) or actively occurring. If a system is stationary and not on the verge of slipping,

F

balances whatever force opposes it, satisfying

F < \mu R

.

Worked example — Resolving forces in equilibrium

A particle is in equilibrium under three forces: 10 N acting due East, 8 N acting due North, and a third force $F$ . Find the magnitude and direction of $F$ .

For equilibrium: $F_x=-10$ N, $F_y=-8$ N.

|F|=\sqrt{100+64}=\sqrt{164}\approx12.8\text{ N},\quad\tan\theta=\frac{8}{10}\Rightarrow\theta=38.7°\text{ W of S}.

Worked example — Friction on an inclined plane

A block of mass 5 kg lies on a rough plane inclined at 30° to the horizontal ( $\mu=0.4$ , $g=10$ m/s²). Find the friction force acting when the block is on the point of sliding down.

Normal reaction: $R=5g\cos30°=50\times\tfrac{\sqrt3}{2}\approx43.3$ N.

Friction (limiting, up the slope): $F=0.4\times43.3\approx17.3$ N.

Check: Gravity component down slope $=50\sin30°=25$ N $>F$ , so block would slide — friction acts up slope.

A block of mass

8\text{ kg}

rests in limiting equilibrium on a rough plane inclined at

30^\circ

to the horizontal. Taking

g = 9.8\text{ m/s}^2

: (a) calculate the normal reaction

R

, (b) find the coefficient of friction

\mu

between the block and the plane.

(a) Resolve perpendicular to the inclined plane:

R = mg\cos 30^\circ = 8(9.8)\cos 30^\circ = 78.4 \times \dfrac{\sqrt{3}}{2} = 39.2\sqrt{3} \approx 67.9\text{ N (3 s.f.)}

(b) Resolve parallel to the plane. Since the block is in limiting equilibrium on the verge of sliding down, friction

F = \mu R

acts up the plane:

F = mg\sin 30^\circ = 8(9.8)\sin 30^\circ = 78.4 \times 0.5 = 39.2\text{ N}

Setting

F = \mu R

:

\mu = \dfrac{F}{R} = \dfrac{mg\sin 30^\circ}{mg\cos 30^\circ} = \tan 30^\circ = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \approx 0.577\text{ (3 s.f.)}

A small particle of weight

24\text{ N}

is held in equilibrium by two light inextensible strings. String

A

makes an angle of

30^\circ

with the upward vertical, and string

B

makes an angle of

60^\circ

with the upward vertical. Find the tensions

T_A

and

T_B

in the two strings.

Notice that the angle between string

A

and string

B

is

30^\circ + 60^\circ = 90^\circ

, so the strings are perpendicular.
Resolve horizontally:

T_A\sin 30^\circ = T_B\sin 60^\circ \Rightarrow T_A(0.5) = T_B\left(\tfrac{\sqrt{3}}{2}\right) \Rightarrow T_A = \sqrt{3}T_B

Resolve vertically:

T_A\cos 30^\circ + T_B\cos 60^\circ = 24 \Rightarrow T_A\left(\tfrac{\sqrt{3}}{2}\right) + T_B(0.5) = 24

Substitute

T_A = \sqrt{3}T_B

:

(\sqrt{3}T_B)\left(\tfrac{\sqrt{3}}{2}\right) + 0.5T_B = 24 \Rightarrow \tfrac{3}{2}T_B + \tfrac{1}{2}T_B = 24 \Rightarrow 2T_B = 24 \Rightarrow T_B = 12\text{ N}

Then:

T_A = 12\sqrt{3} \approx 20.8\text{ N (3 s.f.)}

🔗 See also: Newton's Laws · Energy, Work & Power

Questions Level

13

Newton's Laws of Motion

📌 Newton's Laws of Motion M1 · $F=ma$ · Connected particles & pulleys · Tension & reaction forces

Newton's First Law

A body remains at rest or in uniform motion unless acted upon by a resultant force. Inertia.

Newton's Second Law

$F=ma$ where $F$ is resultant force (N), $m$ is mass (kg), $a$ is acceleration (m/s²).

Newton's Third Law

For every action there is an equal and opposite reaction. Forces occur in equal pairs on different bodies.

Connected Particles

Particles connected by strings over pulleys: apply $F=ma$ to each particle separately, then solve simultaneously.

Acceleration on Slopes

Net force along slope $=mg\sin\alpha - F_{friction}$ . Apply $F=ma$ to find acceleration.

Momentum

Momentum $p=mv$ . Newton's 2nd law: $F=\dfrac{d(mv)}{dt}$ . Conservation: $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ .

▲ Newton's Laws — Reference

N1: A body remains at rest or in uniform motion unless acted on by a net force. N2: $F=ma$ (net force = mass × acceleration). N3: For every action there is an equal and opposite reaction. For connected particles (e.g. Atwood's machine), consider each particle separately; the tension $T$ is the same throughout an inextensible string over a smooth pulley.

Newton's 2nd Law

F_\text{net}=ma

Atwood's machine

a=\dfrac{(m_1-m_2)g}{m_1+m_2}

Tension (same string)

Same

T

throughout

Weight

W=mg

Velocity–time graph of connected particles

m_1 = 5\text{ kg}

and

m_2 = 3\text{ kg}

accelerating at

a = \frac{5-3}{5+3}(9.8) = 2.45\text{ m/s}^2

.

⚠ Common Traps & Exam Pitfalls

Separate equations for connected bodies: Never attempt to guess internal forces in one step. Always write

\Sigma F = ma

separately for each particle in its direction of motion. Adding the equations cancels the internal tension

T

and yields

a

immediately.

Assumptions and vocabulary: Cambridge papers frequently ask for the physical significance of modeling assumptions: 'Light' means string/pulley mass is negligible; 'Inextensible' means accelerations of both particles have the same magnitude; 'Smooth' means tension

T

is uniform across both sides.

Force exerted on the pulley: For two vertical hanging strings, the total downward force on the pulley is

2T

. If one string is horizontal along a table and the other hangs vertically, the resultant force on the pulley is

\sqrt{T^2 + T^2} = T\sqrt{2}

at

45^\circ

.

Worked example — Single particle on a surface

A force of 30 N acts on a 6 kg box on a rough horizontal surface ( $\mu=0.3$ , $g=10$ m/s²). Find the acceleration.

R=mg=60\text{ N},\quad F_\text{friction}=0.3\times60=18\text{ N}.

F_\text{net}=30-18=12\text{ N},\quad a=\frac{12}{6}=2\text{ m/s}^2.

Worked example — Connected particles over a pulley

Two particles of mass 4 kg and 6 kg are connected by a light inextensible string over a smooth pulley. Find the acceleration and the tension in the string ( $g=10$ m/s²).

Taking down as positive for the 6 kg particle:

6g-T=6a,\quad T-4g=4a.

\text{Adding: }2g=10a\Rightarrow a=2\text{ m/s}^2,\quad T=4(10+2)=48\text{ N}.

Two particles

P

of mass

3\text{ kg}

and

Q

of mass

5\text{ kg}

are connected by a light inextensible string passing over a smooth fixed pulley. The system is held at rest and released. Taking

g = 9.8\text{ m/s}^2

: (a) find the acceleration

a

of the system, (b) calculate the tension

T

in the string, (c) find the force exerted by the string on the pulley.

(a) Write

F = ma

for each particle in its direction of motion:
For

Q

(moving downwards):

5g - T = 5a

For

P

(moving upwards):

T - 3g = 3a

Add the equations:

2g = 8a \Rightarrow a = \dfrac{2(9.8)}{8} = 2.45\text{ m/s}^2

(b) Substitute

a

into the equation for

P

:

T = 3g + 3a = 3(9.8 + 2.45) = 3(12.25) = 36.75\text{ N} \approx 36.8\text{ N}

(c) The pulley supports two vertical segments of string with tension

T

, so the total downward force on the pulley is:

F_{\text{pulley}} = 2T = 2(36.75) = 73.5\text{ N}

A block of mass

4\text{ kg}

is pulled along a rough horizontal plane (

\mu = 0.25

) by a pulling force of magnitude

25\text{ N}

acting at

30^\circ

above the horizontal. Taking

g = 9.8\text{ m/s}^2

: (a) show that the normal reaction

R = 26.7\text{ N}

, (b) find the acceleration of the block.

(a) Resolve vertically (equilibrium in vertical direction):

R + 25\sin 30^\circ = mg \Rightarrow R + 25(0.5) = 4(9.8) \Rightarrow R + 12.5 = 39.2 \Rightarrow R = 26.7\text{ N}

(b) Limiting friction force opposes motion:

F = \mu R = 0.25 \times 26.7 = 6.675\text{ N}

Resolve horizontally using

F_{\text{net}} = ma

:

25\cos 30^\circ - F = ma \Rightarrow 25\left(\dfrac{\sqrt{3}}{2}\right) - 6.675 = 4a

21.651 - 6.675 = 4a \Rightarrow 14.976 = 4a \Rightarrow a = \dfrac{14.976}{4} \approx 3.74\text{ m/s}^2\text{ (3 s.f.)}

🔗 See also: Forces & Equilibrium · Energy, Work & Power

Questions Level

14

Energy, Work & Power

📌 Energy, Work & Power M1 · Kinetic energy · Gravitational PE · Work-energy principle · Power $P=Fv$

Work Done

$W=Fd\cos\theta$ where $F$ is force, $d$ distance, $\theta$ angle between force and displacement. Units: Joules (J).

Kinetic Energy

$KE=\tfrac{1}{2}mv^2$ . Work-energy theorem: $W_{net}=\Delta KE=\tfrac{1}{2}mv^2-\tfrac{1}{2}mu^2$ .

Potential Energy

Gravitational PE: $GPE=mgh$ . Change in GPE: $\Delta GPE=mg\Delta h$ .

Conservation of Energy

$KE+PE=\text{constant}$ (no friction). With friction: $\Delta KE+\Delta PE+W_{friction}=0$ .

Power

$P=\dfrac{W}{t}=Fv$ . Units: Watts (W). Maximum speed when driving force = resistance: $P=F_{resistance}\cdot v_{max}$ .

Efficiency

$\text{efficiency}=\dfrac{\text{useful power output}}{\text{power input}}\times100\%$

▲ Energy, Work & Power — Reference

Kinetic energy: $\text{KE}=\tfrac{1}{2}mv^2$ . Gravitational PE: $\text{GPE}=mgh$ . Work-energy principle: net work done = change in KE. Work done by a force: $W=Fd\cos\theta$ . Power: $P=\dfrac{W}{t}=Fv$ . Conservation of energy: $\text{KE}_1+\text{GPE}_1=\text{KE}_2+\text{GPE}_2$ (no friction).

Kinetic energy

\tfrac{1}{2}mv^2

Gravitational PE

mgh

Work done

Fd\cos\theta

Power

P=Fv

Quadratic relationship between kinetic energy and speed

KE = \frac{1}{2}mv^2

for a

2\text{ kg}

mass. Doubling speed quadruples the kinetic energy.

⚠ Common Traps & Exam Pitfalls

Double-counting gravity: In energy equations, you either account for gravity through Gravitational Potential Energy (

\Delta KE + \Delta PE = W_{\text{external}}

) OR through work done by gravity (

\Delta KE = W_g + W_{\text{friction}} + \dots

). Including both simultaneously counts gravitational work twice.

Driving force vs Net force in $P = Fv$ : The force in

P = Fv

is the tractive/driving force produced by the engine, NOT the resultant force. At maximum or steady velocity, acceleration is zero, so the driving force exactly equals the total opposing resistance (

F_{\text{drive}} = R + mg\sin\alpha

).

Work against friction: Work done against friction is

W = F \times d

(where

d

is the distance along the surface). This mechanical energy converts irreversibly into heat.

Worked example — Work-energy principle

A 2 kg ball rolls down a smooth slope of height 5 m from rest. Find its speed at the bottom ( $g=10$ m/s²).

\text{Loss in GPE}=\text{Gain in KE}:\quad2\times10\times5=\tfrac{1}{2}\times2\times v^2\Rightarrow v^2=100\Rightarrow v=10\text{ m/s}.

Worked example — Power

A car of mass 900 kg moves at constant speed 30 m/s against a resistance of 500 N. Find the power output of the engine.

At constant speed, driving force = resistance = 500 N.

P=Fv=500\times30=15\,000\text{ W}=15\text{ kW}.

A car of mass

1200\text{ kg}

ascends a hill of inclination

\alpha

, where

\sin\alpha = \tfrac{1}{20}

, against a constant resistance to motion of

500\text{ N}

. Taking

g = 9.8\text{ m/s}^2

: (a) find the driving force required to travel at a steady speed of

25\text{ m/s}

, (b) calculate the power developed by the engine in kilowatts.

(a) Component of weight down the slope:

W_{\parallel} = mg\sin\alpha = 1200(9.8)\left(\tfrac{1}{20}\right) = 60(9.8) = 588\text{ N}

Since the car travels at steady speed (acceleration

a = 0

), the engine driving force balances all resistance:

F_{\text{drive}} = \text{Resistance} + mg\sin\alpha = 500 + 588 = 1088\text{ N}

(b) Power output:

P = F_{\text{drive}} \times v = 1088 \times 25 = 27\,200\text{ W} = 27.2\text{ kW}

A block of mass

5\text{ kg}

is released from rest at the top of a rough plane inclined at

30^\circ

to the horizontal. The length of the incline is

8\text{ m}

and the coefficient of friction is

\mu = 0.2

. Using the work-energy principle and taking

g = 9.8\text{ m/s}^2

, calculate the speed of the block at the bottom of the incline.

1. Vertical height descended:

h = 8\sin 30^\circ = 8(0.5) = 4\text{ m}

.
Initial gravitational potential energy:

\text{PE} = mgh = 5(9.8)(4) = 196\text{ J}

2. Normal reaction:

R = mg\cos 30^\circ = 5(9.8)\left(\tfrac{\sqrt{3}}{2}\right) = 24.5\sqrt{3}\text{ N}

.
Friction force:

F = \mu R = 0.2(24.5\sqrt{3}) = 4.9\sqrt{3}\text{ N}

.
Work done against friction over distance

d = 8\text{ m}

:

W_{\text{fric}} = F \times d = (4.9\sqrt{3}) \times 8 = 39.2\sqrt{3} \approx 67.896\text{ J}

3. By work-energy principle:

\text{Gain in KE} = \text{Loss in PE} - W_{\text{fric}} \Rightarrow \dfrac{1}{2}mv^2 = 196 - 67.896 = 128.104\text{ J}

Substitute

m = 5\text{ kg}

:

2.5v^2 = 128.104 \Rightarrow v^2 = \dfrac{128.104}{2.5} = 51.2416 \Rightarrow v = \sqrt{51.2416} \approx 7.16\text{ m/s (3 s.f.)}

🔗 See also: Forces · Newton's Laws

Questions Level

15

Representation of Data

📌 Representation of Data S1 · Histograms & frequency density · Cumulative frequency · Mean, variance & SD · Box-and-whisker plots

Stem-and-Leaf Diagrams

Display raw data in order. Back-to-back stem plots compare two datasets. Median and quartiles easy to read.

Box-and-Whisker Plots

Shows minimum, Q1, median, Q3, maximum. IQR = Q3 − Q1. Outliers: values outside $Q_1-1.5\times IQR$ or $Q_3+1.5\times IQR$ .

Histograms

Frequency density = frequency ÷ class width. Area of bar represents frequency. Used for continuous grouped data.

Cumulative Frequency

Ogive (cumulative frequency curve). Read median at $n/2$ , Q1 at $n/4$ , Q3 at $3n/4$ .

Measures of Location

Mean $\bar{x}=\dfrac{\sum fx}{\sum f}$ , Median (middle value), Mode (most frequent). For grouped data use midpoint.

Measures of Spread

Range, IQR, Variance $\sigma^2=\dfrac{\sum fx^2}{n}-\bar{x}^2$ , Standard deviation $\sigma$ .

▲ Representation of Data — Reference

For grouped data with $n$ values: mean $\bar{x}=\dfrac{\sum fx}{\sum f}$ ; variance $\sigma^2=\dfrac{\sum fx^2}{\sum f}-\bar{x}^2$ ; standard deviation $\sigma=\sqrt{\text{Var}}$ . In a histogram (unequal class widths), plot frequency density $=\text{freq}/\text{class width}$ . In a cumulative frequency graph, the median is the $\tfrac{n}{2}$ th value, $Q_1$ is the $\tfrac{n}{4}$ th, $Q_3$ is the $\tfrac{3n}{4}$ th.

Mean (grouped)

\bar{x}=\dfrac{\sum fx}{\sum f}

Variance

\dfrac{\sum fx^2}{\sum f}-\bar{x}^2

Freq. density

freq ÷ class width

IQR

Q_3-Q_1

Box-and-whisker plot displaying Five-Number Summary: Min = 12, Q1 = 28, Median = 44, Q3 = 62, Max = 85 with IQR = 34.

⚠ Common Traps & Exam Pitfalls

Histograms with unequal class widths: The vertical axis must ALWAYS represent frequency density (

\text{frequency} \div \text{class width}

), never raw frequency. The area of each rectangle represents frequency.

Linear coding of data: If data values

x

are transformed via

y = ax + b

:
• Mean scales and shifts:

\bar{y} = a\bar{x} + b

• Variance scales by

a^2

only:

\text{Var}(y) = a^2\text{Var}(x)

(the additive constant

b

does not affect spread)
• Standard deviation scales by

|a|

:

\sigma_y = |a|\sigma_x

.

Variance formula accuracy: When calculating

\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2

, never round

\bar{x}

prematurely before squaring; keep exact fractions or high precision to avoid severe cancellation errors.

Worked example — Mean and variance from table

Scores: 1 (freq 2), 2 (freq 5), 3 (freq 8), 4 (freq 3), 5 (freq 2). Find the mean and standard deviation.

$\sum f=20$ , $\sum fx=2+10+24+12+10=58$ . Mean $=58/20=2.9$ .

$\sum fx^2=2+20+72+48+50=192$ . Var $=192/20-2.9^2=9.6-8.41=1.19$ . SD $\approx1.09$ .

Worked example — Cumulative frequency

Data range 0–50. Cumulative frequencies at 10,20,30,40,50 are 5,18,34,44,50. Estimate the median and IQR.

Median at $n/2=25$ th value: read from cf graph at cf=25 → $\approx27$ . $Q_1$ at cf=12.5 → $\approx18$ . $Q_3$ at cf=37.5 → $\approx33$ . IQR $\approx15$ .

A data set has mean 12 and standard deviation 4. A new value of 20 is added to

n=9

existing values. Without finding the new variance, find the new mean.

Old sum

=9\times12=108

. New sum

=108+20=128

. New mean

=128/10=12.8

.

A continuous dataset is grouped into class intervals. The interval

20 \le x < 35

has a frequency of 45. Find the frequency density for this class.

Class width

w = 35 - 20 = 15

. Frequency density

= \frac{\text{frequency}}{\text{class width}} = \frac{45}{15} = 3.0

. In a histogram, the height of the bar drawn on the interval

[20, 35)

is 3.0.

Questions Level

16

Permutations & Combinations

📌 Permutations & Combinations S1 · Factorial · $^nP_r$ · $^nC_r$ · Arrangements with identical items or restrictions

Factorial

$n! = n\times(n-1)\times\cdots\times2\times1$ . Convention: $0!=1$ .

Permutations

Ordered arrangements: $^nP_r=\dfrac{n!}{(n-r)!}$ . Order matters. $n$ objects in a line: $n!$ ways.

Combinations

Unordered selections: $^nC_r=\dbinom{n}{r}=\dfrac{n!}{r!(n-r)!}$ . Order does not matter.

Restrictions

Always together: treat as one unit, then $(n-1)!\times r!$ . Never together: total − (always together).

Circular Arrangements

$(n-1)!$ ways to arrange $n$ objects in a circle (one position is fixed).

Identical Objects

Arrangements of $n$ objects with repeats: $\dfrac{n!}{a!\,b!\,c!\cdots}$

▲ Permutations & Combinations — Reference

Permutation (order matters): $^nP_r=\dfrac{n!}{(n-r)!}$ . Combination (order does not matter): $^nC_r=\dbinom{n}{r}=\dfrac{n!}{r!\,(n-r)!}$ . Arrangements with identical items: divide by the factorial of each repeated item. With restrictions: fix the restricted items first, then arrange the rest.

Permutation

^nP_r=\dfrac{n!}{(n-r)!}

Combination

^nC_r=\dfrac{n!}{r!(n-r)!}

Identical items

\dfrac{n!}{a!\,b!\cdots}

Fix & arrange

Fix restricted → count rest

Combinations

\binom{6}{r}

for

r = 0, 1, \ldots, 6

, displaying the symmetric binomial coefficient distribution peaking at

r = 3

.

⚠ Common Traps & Exam Pitfalls

Permutation vs Combination: If order or role matters (e.g. President and Vice President, line order, distinct positions), use permutations

^nP_r

. If order does not matter (e.g. selecting a committee of 3 students), use combinations

^nC_r

.

Identical items vs Distinct slots: Arranging

n

objects where some are identical requires dividing by each repeat count:

\frac{n!}{p!\,q!\cdots}

.

'Together' vs 'Separated': For items that must stay together, glue them as a single super-element and arrange

(n - k + 1)!

, then multiply by

k!

for internal order. For items that must NEVER be adjacent, arrange the unrestricted items first, creating slots

\underline{\quad}\star\underline{\quad}\star\dots

, and choose slot positions.

Worked example — Combinations with restrictions

A committee of 4 is chosen from 5 men and 6 women. Find the number of ways if the committee must contain at least 2 women.

Exactly 2W: $\binom{6}{2}\binom{5}{2}=15\times10=150$ . Exactly 3W: $\binom{6}{3}\binom{5}{1}=20\times5=100$ . Exactly 4W: $\binom{6}{4}\binom{5}{0}=15$ . Total: $265$ .

Worked example — Arrangements of letters

How many different arrangements are there of the letters of the word BANANA?

6 letters; N appears 2 times, A appears 3 times: $\dfrac{6!}{3!\,2!}=\dfrac{720}{12}=60$ .

In how many ways can 5 people be seated in a row if two particular people (A and B) must not sit next to each other?

Total

=5!=120

. Cases where A & B are adjacent: treat AB as a block →

4!\times2=48

. Answer:

120-48=72

.

Find the number of ways to arrange the 7 letters in the word SUCCESS.

The word SUCCESS has 7 letters with 3 S's, 2 C's, 1 U, and 1 E. The number of distinct permutations is

\frac{7!}{3!\,2!} = \frac{5040}{6 \times 2} = \frac{5040}{12} = 420

.

🔗 See also: Binomial Distribution

Questions Level

17

Discrete Random Variables

📌 Discrete Random Variables S1 · Probability distribution table · $E(X)$ · $\text{Var}(X)$ · Linear transforms $E(aX+b)$

Probability Distribution

Table of $x$ values and corresponding $P(X=x)$ . Must satisfy: $\sum P(X=x)=1$ and $P(X=x)\geq0$ .

Expectation

$E(X)=\sum x\cdot P(X=x)=\mu$ . Also $E(aX+b)=aE(X)+b$ .

Variance

$\text{Var}(X)=E(X^2)-[E(X)]^2$ where $E(X^2)=\sum x^2P(X=x)$ . Also $\text{Var}(aX+b)=a^2\text{Var}(X)$ .

Standard Deviation

$\sigma=\sqrt{\text{Var}(X)}$ . Measures spread around the mean.

Uniform Distribution

Each value equally likely. If $X$ takes values $1,2,\ldots,n$ : $E(X)=\dfrac{n+1}{2}$ , $\text{Var}(X)=\dfrac{n^2-1}{12}$ .

Combined Variables

$E(X+Y)=E(X)+E(Y)$ . If independent: $\text{Var}(X+Y)=\text{Var}(X)+\text{Var}(Y)$ .

▲ Discrete Random Variables — Reference

A discrete random variable $X$ has a probability distribution with $\sum P(X=x)=1$ . Expectation: $E(X)=\sum x\,P(X=x)$ . Variance: $\text{Var}(X)=E(X^2)-[E(X)]^2$ where $E(X^2)=\sum x^2 P(X=x)$ . For linear transforms: $E(aX+b)=aE(X)+b$ ; $\text{Var}(aX+b)=a^2\text{Var}(X)$ .

E(X)

\sum x\,P(X=x)

Var(X)

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

E(aX+b)

aE(X)+b

Var(aX+b)

a^2\text{Var}(X)

Probability distribution of a discrete random variable

X

taking values

\{1, 2, 3, 4\}

with

\sum P(X=x) = 1.0

.

⚠ Common Traps & Exam Pitfalls

Variance formula trap: Remember

\text{Var}(X) = E(X^2) - [E(X)]^2

. Do not forget to square

E(X)

, and do not compute

E(X^2) - E(X)

.

Linear coding of variance:

\text{Var}(aX + b) = a^2 \text{Var}(X)

. Notice that the constant

b

is eliminated completely, and a negative multiplier

a

becomes positive due to squaring (e.g.

\text{Var}(5 - 2X) = (-2)^2 \text{Var}(X) = 4\text{Var}(X)

).

Sum of probabilities: The total probability MUST sum to 1:

\sum P(X = x) = 1

. Always verify this condition first when solving for unknown parameters

p

or

k

.

Worked example — Finding E(X) and Var(X)

A biased die shows: $P(1)=0.1,P(2)=0.2,P(3)=0.3,P(4)=0.3,P(5)=0.1$ . Find $E(X)$ and $\text{Var}(X)$ .

E(X)=0.1(1)+0.2(2)+0.3(3)+0.3(4)+0.1(5)=0.1+0.4+0.9+1.2+0.5=3.1.

E(X^2)=0.1(1)+0.2(4)+0.3(9)+0.3(16)+0.1(25)=0.1+0.8+2.7+4.8+2.5=10.9.

\text{Var}(X)=10.9-3.1^2=10.9-9.61=1.29.

Worked example — Unknown probability

$X$ takes values 0, 1, 2 with $P(X=0)=k$ , $P(X=1)=2k$ , $P(X=2)=3k$ . Find $k$ and $E(X)$ .

k+2k+3k=1\Rightarrow k=\tfrac{1}{6}.\quad E(X)=0\cdot\tfrac{1}{6}+1\cdot\tfrac{2}{6}+2\cdot\tfrac{3}{6}=\tfrac{8}{6}=\tfrac{4}{3}.

E(X)=3

and

\text{Var}(X)=2

. Find

E(3X-1)

and

\text{Var}(3X-1)

.

E(3X-1)=3(3)-1=8

.

\text{Var}(3X-1)=9\times2=18

.

A discrete random variable

Y

has

E(Y) = 5

and

\text{Var}(Y) = 4

. Calculate

E(Y^2)

and

\text{Var}(10 - 3Y)

.

Using

\text{Var}(Y) = E(Y^2) - [E(Y)]^2

, we get

E(Y^2) = \text{Var}(Y) + [E(Y)]^2 = 4 + 5^2 = 4 + 25 = 29

. For the linear transformation:

\text{Var}(10 - 3Y) = (-3)^2 \text{Var}(Y) = 9 \times 4 = 36

.

🔗 See also: Binomial Distribution · A2 Continuous RVs

Questions Level

18

Binomial Distribution

📌 Binomial Distribution S1 · $X\sim B(n,p)$ · $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ · Mean $np$ · Variance $np(1-p)$

Conditions

Fixed number of trials $n$ ; each trial has two outcomes (success/failure); constant probability $p$ ; trials independent. Then $X\sim B(n,p)$ .

Probability Formula

$P(X=r)=\dbinom{n}{r}p^r(1-p)^{n-r}$ for $r=0,1,2,\ldots,n$ .

Mean and Variance

$E(X)=np$ , $\text{Var}(X)=np(1-p)$ , $\sigma=\sqrt{np(1-p)}$ .

Cumulative Probabilities

Use tables or calculator: $P(X\leq r)$ . Note: $P(X\geq r)=1-P(X\leq r-1)$ .

Normal Approximation

If $n$ large and $p$ close to 0.5: $X\approx N(np,\,np(1-p))$ with continuity correction.

Finding Unknown n or p

Set up equations using given probabilities and solve. May require trial and error or logarithms for $n$ .

▲ Binomial Distribution — Reference

If $X\sim B(n,p)$ : $n$ independent trials, each with probability $p$ of success. $P(X=r)=\dbinom{n}{r}p^r(1-p)^{n-r}$ . Mean: $E(X)=np$ . Variance: $\text{Var}(X)=np(1-p)$ . Use the binomial distribution when: fixed $n$ , trials independent, constant $p$ , only 2 outcomes (success/failure).

P(X=r)

\binom{n}{r}p^r(1-p)^{n-r}

Mean

np

Variance

np(1-p)

Conditions

Fixed

n

, independent, constant

p

Binomial distribution

X \sim B(5, 0.4)

probability distribution across

r = 0, 1, 2, 3, 4, 5

.

⚠ Common Traps & Exam Pitfalls

Inequality bounds for discrete distributions: Unlike continuous variables, strict inequalities make a critical difference:
•

P(X \ge r) = 1 - P(X \le r - 1)

(NOT

1 - P(X \le r)

)
•

P(X > r) = 1 - P(X \le r)

•

P(X < r) = P(X \le r - 1)

.

'At least one' complement shortcut:

P(X \ge 1) = 1 - P(X = 0) = 1 - (1-p)^n

. Always look for this time-saving identity in exam questions.

Four binomial conditions: Cambridge exam questions frequently ask you to state assumptions justifying the binomial model: (1) Fixed number of trials

n

, (2) Constant probability of success

p

, (3) Independent trials, (4) Exactly two mutually exclusive outcomes (success/failure).

Worked example — Exact probability

$X\sim B(8,0.3)$ . Find (a) $P(X=3)$ , (b) $P(X\leq1)$ .

(a) $P(X=3)=\binom{8}{3}(0.3)^3(0.7)^5=56\times0.027\times0.16807\approx0.2541$ .

(b) $P(X=0)=(0.7)^8\approx0.0576$ ; $P(X=1)=8(0.3)(0.7)^7\approx0.1977$ . $P(X\leq1)\approx0.2553$ .

Worked example — Finding n from mean

The mean of $X\sim B(n,0.4)$ is 6. Find $n$ and the standard deviation.

np=6\Rightarrow n=15.\quad\sigma=\sqrt{15\times0.4\times0.6}=\sqrt{3.6}\approx1.90.

A fair coin is tossed 10 times. Find the probability of getting (a) exactly 6 heads, (b) fewer than 3 heads.

(a)

\binom{10}{6}(0.5)^{10}=210\times\tfrac{1}{1024}\approx0.2051

. (b)

P(X\leq2)=P(0)+P(1)+P(2)=\tfrac{1+10+45}{1024}=\tfrac{56}{1024}\approx0.0547

.

In a manufacturing batch, 5% of components are defective. A sample of 12 components is inspected. Find the probability that at least 2 components are defective.

Let

X \sim B(12, 0.05)

. We need

P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]

.

P(X = 0) = (0.95)^{12} \approx 0.5404

.

P(X = 1) = 12(0.05)(0.95)^{11} \approx 0.3413

. Thus,

P(X \ge 2) = 1 - (0.5404 + 0.3413) = 1 - 0.8817 = 0.118

(3 s.f.).

🔗 See also: Permutations & Combinations · Normal Distribution · A2 Poisson

Questions Level

19

Normal Distribution

📌 Normal Distribution S1 · $X\sim N(\mu,\sigma^2)$ · Standardising $Z=\tfrac{X-\mu}{\sigma}$ · Tables & symmetry · Finding $\mu$ or $\sigma$

The Normal Curve

Bell-shaped, symmetric about mean $\mu$ . $X\sim N(\mu,\sigma^2)$ . Approximately 68%/95%/99.7% within 1/2/3 standard deviations.

Standardising

$Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)$ . Use $Z$ -tables (standard normal) to find probabilities.

Finding Probabilities

$P(X<a)=\Phi\!\left(\dfrac{a-\mu}{\sigma}\right)$ . Use symmetry: $P(Z<-z)=1-P(Z<z)$ .

Inverse Normal

Given probability, find the value. $P(X<a)=p\Rightarrow a=\mu+z\sigma$ where $z=\Phi^{-1}(p)$ .

Finding μ or σ

Set up two equations from given probabilities, solve simultaneously for unknown parameters.

Normal Approximation

To Binomial $B(n,p)$ : if $n$ large, $np>5$ , $n(1-p)>5$ : use $N(np,np(1-p))$ with continuity correction $\pm0.5$ .

▲ Normal Distribution — Reference

If $X\sim N(\mu,\sigma^2)$ : standardise with $Z=\dfrac{X-\mu}{\sigma}$ where $Z\sim N(0,1)$ . Use tables (or calculator) for $\Phi(z)=P(Z\leq z)$ . Key symmetry: $P(Z\leq -z)=1-P(Z\leq z)$ . For inverse normal: given a probability, find the corresponding $x$ value using $x=\mu+z\sigma$ .

Standardise

Z=\dfrac{X-\mu}{\sigma}

Symmetry

P(Z\leq-z)=1-\Phi(z)

Inverse normal

x=\mu+z\sigma

Normal approx to B

Use when

n

large,

p\approx0.5

Standard normal distribution curve

Z \sim N(0, 1)

with shaded region representing

P(-1.5 \le Z \le 1.5) \approx 0.8664

.

⚠ Common Traps & Exam Pitfalls

Variance vs Standard Deviation: Cambridge notation specifies

X \sim N(\mu, \sigma^2)

. When standardising

Z = \frac{X - \mu}{\sigma}

, remember to divide by

\sigma = \sqrt{\sigma^2}

, NOT by

\sigma^2

.

Negative z-values and tables: Standard normal tables provide

\Phi(z)

only for

z > 0

. By symmetry:

\Phi(-z) = 1 - \Phi(z)

. For lower-tail probabilities

P(Z < -1.25) = 1 - \Phi(1.25)

.

Continuity correction (Normal approx to Binomial): When approximating discrete

X \sim B(n, p)

by continuous

Y \sim N(np, np(1-p))

:
•

P(X \ge a) \to P(Y > a - 0.5)

•

P(X > a) \to P(Y > a + 0.5)

•

P(X \le a) \to P(Y < a + 0.5)

•

P(X = a) \to P(a - 0.5 < Y < a + 0.5)

.

Worked example — Finding a probability

$X\sim N(50,16)$ so $\sigma=4$ . Find $P(44\leq X\leq58)$ .

P\!\left(\frac{44-50}{4}\leq Z\leq\frac{58-50}{4}\right)=P(-1.5\leq Z\leq 2)=\Phi(2)-\Phi(-1.5).

=0.9772-(1-0.9332)=0.9772-0.0668=0.9104.

Worked example — Finding

\mu

$X\sim N(\mu,25)$ . Given $P(X>43)=0.2119$ , find $\mu$ .

$P(Z>\tfrac{43-\mu}{5})=0.2119\Rightarrow\tfrac{43-\mu}{5}=0.8$ (from tables). $\mu=43-4=39$ .

X\sim N(30,\sigma^2)

and

P(X<25)=0.0668

. Find

\sigma

.

P(Z<\tfrac{25-30}{\sigma})=0.0668\Rightarrow\tfrac{-5}{\sigma}=-1.5\Rightarrow\sigma=\tfrac{5}{1.5}\approx3.33

.

The random variable

X \sim N(\mu, \sigma^2)

satisfies

P(X < 65) = 0.8413

and

P(X < 45) = 0.0228

. Find the values of

\mu

and

\sigma

.

Standardising:

\Phi\left(\frac{65 - \mu}{\sigma}\right) = 0.8413 \Rightarrow \frac{65 - \mu}{\sigma} = 1.0

, so

65 - \mu = \sigma

.
For the lower tail:

P(X < 45) = 0.0228 \Rightarrow \Phi\left(\frac{45 - \mu}{\sigma}\right) = 0.0228 \Rightarrow \frac{45 - \mu}{\sigma} = -2.0

, so

\mu - 45 = 2\sigma

.
Adding the two equations:

65 - 45 = 3\sigma \Rightarrow 20 = 3\sigma \Rightarrow \sigma = 6.67

. Then

\mu = 65 - 6.67 = 58.33

.

🔗 See also: Binomial Distribution · A2 Hypothesis Testing

Questions Level

§

Practice set

Fifty AS-style problems across the Pure core plus statistics, tagged by difficulty. Click any problem for a full worked solution.

1

Simplify $x^{3}\cdot x^{4}$ .

Basic

x^{7}.

2

Evaluate $2^{-3}$ .

Intermediate

\frac{1}{2^{3}}=\tfrac{1}{8}.

3

Simplify $\sqrt{12}$ .

Intermediate

\sqrt{4\cdot3}=2\sqrt{3}.

4

Rationalise $\dfrac{1}{\sqrt{2}}$ .

Advanced

Multiply by $\tfrac{\sqrt2}{\sqrt2}$ :

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

5

Simplify $(x^{2})^{5}$ .

Intermediate

x^{10}.

6

Evaluate $27^{1/3}$ .

Advanced

\sqrt[3]{27}=3.

7

Solve $x^{2}=16$ .

Basic

x=\pm4.

8

Factor $x^{2}+5x+6$ .

Intermediate

(x+2)(x+3).

9

Solve $x^{2}-6x+8=0$ .

Intermediate

$(x-2)(x-4)=0$ :

x=2,\ 4.

10

Complete the square for $x^{2}+4x$ .

Intermediate

(x+2)^{2}-4.

11

How many real roots does $x^{2}+x+1=0$ have?

Advanced

Discriminant $=1-4=-3<0$ :

\text{none.}

12

Find the sum of the roots of $x^{2}-5x+6=0$ .

Intermediate

2+3=5.

13

Solve $2x^{2}-5x-3=0$ .

Advanced

$(2x+1)(x-3)=0$ :

x=3,\ -\tfrac12.

14

Find the vertex of $y=x^{2}-4x+3$ .

Intermediate

$(x-2)^{2}-1$ :

(2,\,-1).

15

Solve $x+y=5$ , $x-y=1$ .

Intermediate

Add: $2x=6$ :

x=3,\ y=2.

16

Solve $2x+y=8$ , $y=x-1$ .

Intermediate

$2x+x-1=8 \Rightarrow x=3$ :

x=3,\ y=2.

17

Solve $y=x^{2}$ , $y=x+2$ .

Advanced

$x^{2}-x-2=0\Rightarrow(x-2)(x+1)=0$ :

(2,4),\ (-1,1).

18

Solve $2x-3>5$ .

Intermediate

$2x>8$ :

x>4.

19

Solve $3x+1\le10$ .

Intermediate

$3x\le9$ :

x\le3.

20

Solve $x^{2}-4<0$ .

Advanced

-2<x<2.

21

Find the gradient through $(1,2)$ and $(4,11)$ .

Intermediate

\frac{11-2}{4-1}=3.

22

Find the midpoint of $(2,3)$ and $(8,7)$ .

Intermediate

(5,\,5).

23

Find the equation of the line with gradient $2$ through $(1,3)$ .

Advanced

$y-3=2(x-1)$ :

y=2x+1.

24

Find the distance between $(1,1)$ and $(4,5)$ .

Intermediate

\sqrt{9+16}=5.

25

State the centre of $(x-2)^{2}+(y+3)^{2}=25$ .

Advanced

(2,\,-3).

26

State the radius of $x^{2}+y^{2}=49$ .

Intermediate

r=7.

27

Describe $y=x^{2}+3$ relative to $y=x^{2}$ .

Intermediate

A translation $3$ units up.

28

Describe $y=(x-2)^{2}$ relative to $y=x^{2}$ .

Intermediate

A translation $2$ units to the right.

29

Describe $y=-x^{2}$ relative to $y=x^{2}$ .

Advanced

A reflection in the $x$ -axis.

30

How does $y=f(x)+k$ transform the graph of $y=f(x)$ ?

Intermediate

A vertical translation by $k$ .

31

Evaluate $\sin30^{\circ}$ .

Intermediate

\tfrac12.

32

Evaluate $\cos0^{\circ}$ .

Intermediate

1.

33

Evaluate $\tan45^{\circ}$ .

Intermediate

1.

34

Solve $\sin\theta=\tfrac12$ for $0^{\circ}\le\theta\le180^{\circ}$ .

Advanced

\theta=30^{\circ},\ 150^{\circ}.

35

Simplify $\sin^{2}\theta+\cos^{2}\theta$ .

Intermediate

1.

36

State the exact value of $\sin60^{\circ}$ .

Advanced

\tfrac{\sqrt{3}}{2}.

37

Evaluate $\cos90^{\circ}$ .

Intermediate

0.

38

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

Advanced

360^{\circ}.

39

Differentiate $x^{2}$ .

Intermediate

2x.

40

Differentiate $x^{3}$ .

Intermediate

3x^{2}.

41

Differentiate $5x$ .

Intermediate

5.

42

Find $\dfrac{dy}{dx}$ for $y=x^{2}+3x$ .

Advanced

2x+3.

43

Find $\displaystyle\int 2x\,dx$ .

Intermediate

x^{2}+C.

44

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

Advanced

\tfrac{x^{3}}{3}+C.

45

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

Intermediate

$\tfrac{dy}{dx}=2x=6$ :

6.

46

Evaluate $\displaystyle\int_{0}^{1} 2x\,dx$ .

Advanced

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

47

Find the mean of $2,\,4,\,6,\,8$ .

Intermediate

\frac{20}{4}=5.

48

Find the median of $3,\,7,\,9,\,12,\,20$ .

Intermediate

9.

49

A fair die is rolled. Find $P(\text{even})$ .

Advanced

\tfrac{3}{6}=\tfrac12.

50

Find the range of $5,\,9,\,2,\,11$ .

Intermediate

11-2=9.

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AS Level Mathematics

About AS Level

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Simultaneous Equations & Inequalities

Graphs & Transformations

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Probability Distribution

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Uniform Distribution

Combined Variables

Binomial Distribution

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Probability Formula

Mean and Variance

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Normal Distribution

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Finding Probabilities

Inverse Normal

Finding μ or σ

Normal Approximation

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