Free A Level Mathematics 9709 Study Guide — Edvia College
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A Level Mathematics 9709 — all six components, free.

A complete study guide for Cambridge International AS & A Level Mathematics 9709, covering all 36 topics across all six components for exams in 2028–2030.

Sitting exams in 2026 or 2027? You are on the 2026–2027 version. Cambridge states there are no significant changes which affect teaching, so this guide covers both — but confirm your exam year with your school.

How to use it: 9709 is modular — you only sit some of these papers. Check the routes table first, then work only the components you need. Every topic below gives the method and a worked example.

📄 12 plain-English chapter handouts →✎ Practice & self-test →

CAIE 9709 · exams 2028–203036 topics · 6 componentsAS: 2 papers · A Level: 4 papersPure · Mechanics · StatsFree & shareable
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The papers

ComponentContent
Paper 1Pure Mathematics 1
Paper 2Pure Mathematics 2
Paper 3Pure Mathematics 3
Paper 4Mechanics
Paper 5Probability & Statistics 1
Paper 6Probability & Statistics 2

You take two components for AS Level and four for A Level — see the routes table below for the valid combinations.

A calculator is allowed on every 9709 paper, but exact answers are often demanded — "give your answer in terms of π", "leave in surd form", "show that". Read the instruction before reaching for the decimal.
Read this first

Which papers do I take?

RouteAS LevelA Level
Pure onlyPapers 1 and 2No progression to A Level from this route
Pure + MechanicsPapers 1 and 4Papers 1, 3, 4 and 5
Pure + StatisticsPapers 1 and 5Papers 1, 3, 5 and 6
The Papers 1 and 2 AS route is a dead end — it does not progress to A Level. If there is any chance you will continue to the full A Level, you must take Paper 1 with Paper 4 or Paper 5, not Paper 2.

Note that Paper 3 replaces Paper 2 at A Level: P3 contains everything in P2 plus vectors, differential equations and complex numbers.

Paper 1 · 8 topics

Pure Mathematics 1

1.1Quadratics

completed square: a(x + p)² + q, vertex (−p, q) · discriminant b² − 4ac

Use the discriminant to determine the number of real roots, and for tangency conditions (discriminant = 0). Solve quadratic inequalities, and equations reducible to quadratic form by substitution.

Worked example

Find the values of k for which y = kx − 2 is tangent to y = x² − 3x + 2.

  1. kx − 2 = x² − 3x + 2 → x² − (3 + k)x + 4 = 0
  2. Tangent → discriminant = 0: (3 + k)² − 16 = 0
  3. 3 + k = ±4 → k = 1 or k = −7

1.2Functions

Domain and range, composite functions fg(x) = f(g(x)), inverse functions (only for one-one functions), and the reflection of y = f(x) in y = x to give the inverse.

Transformations: y = f(x) + a translates up; y = f(x + a) translates left; y = af(x) stretches vertically by factor a; y = f(ax) stretches horizontally by factor 1/a.

A many-one function has no inverse until its domain is restricted. For f(x) = x² you must restrict to x ⩾ 0 (or x ⩽ 0) before an inverse exists — and questions test exactly this.

1.3Coordinate geometry

gradient, midpoint, distance · y − y₁ = m(x − x₁) · parallel m₁ = m₂ · perpendicular m₁m₂ = −1 circle: (x − a)² + (y − b)² = r²

Find intersections of lines and curves, and use the circle properties: the tangent is perpendicular to the radius at the point of contact, and the perpendicular from the centre to a chord bisects it.

1.4Circular measure

s = rθ · sector area = ½r²θ · segment = ½r²(θ − sin θ)  (θ in radians)
Worked example

A sector of radius 10 cm has angle 1.2 radians. Find the arc length, sector area and segment area.

  1. Arc = 10 × 1.2 = 12 cm
  2. Sector = ½ × 100 × 1.2 = 60 cm²
  3. Segment = ½ × 100 × (1.2 − sin 1.2) = 50 × (1.2 − 0.93204) = 13.4 cm² (3 s.f.)

1.5Trigonometry

Exact values for 30°, 45°, 60°; graphs of sin, cos and tan with amplitude and period; the identities sin²θ + cos²θ = 1 and tan θ = sin θ / cos θ; solving equations in a given interval.

Worked example

Solve 3 sin²x = 1 + cos x for 0° ⩽ x ⩽ 360°.

  1. Replace sin²x = 1 − cos²x: 3(1 − cos²x) = 1 + cos x
  2. 3 − 3cos²x = 1 + cos x → 3cos²x + cos x − 2 = 0
  3. (3cos x − 2)(cos x + 1) = 0 → cos x = ⅔ or cos x = −1
  4. cos x = ⅔ → x = 48.2° or 311.8°; cos x = −1 → x = 180°
  5. x = 48.2°, 180°, 311.8°

1.6Series

Binomial (n positive integer): (a + b)ⁿ = Σ nCr · a^(n−r) · b^r AP: u_n = a + (n−1)d · S_n = ½n[2a + (n−1)d] GP: u_n = ar^(n−1) · S_n = a(1 − rⁿ)/(1 − r) · S∞ = a/(1 − r) for |r| < 1
Worked example

Find the coefficient of x⁴ in the expansion of (3 + 2x)⁶.

  1. Term = 6C4 × 32 × (2x)4
  2. = 15 × 9 × 16x⁴
  3. Coefficient = 2160

1.7Differentiation

Differentiate xn, sums and the chain rule for (ax + b)n. Applications: gradients, tangents and normals, stationary points and their nature, increasing/decreasing functions, connected rates of change, and small increments.

Worked example — connected rates of change

A spherical balloon is inflated so its volume increases at 50 cm³/s. Find the rate of increase of the radius when r = 5 cm. (V = 4/3 πr³)

  1. dV/dr = 4πr²; at r = 5 this is 100π
  2. Chain rule: dr/dt = dV/dt ÷ dV/dr = 50 ÷ 100π
  3. = 0.159 cm/s (3 s.f.)

1.8Integration

∫xⁿ dx = x^(n+1)/(n+1) + c (n ≠ −1) · ∫(ax+b)ⁿ dx = (ax+b)^(n+1)/[a(n+1)] + c

Applications: area under a curve, area between a curve and a line, and volumes of revolution about the x- or y-axis.

V = π ∫ y² dx  (about the x-axis)
Worked example

Find the volume generated when y = x² between x = 0 and x = 2 is rotated about the x-axis.

  1. V = π ∫₀² (x²)² dx = π ∫₀² x⁴ dx
  2. = π [x⁵/5]₀² = π × 32/5
  3. = 32π/5 ≈ 20.1 cubic units
Paper 2 · 6 topics

Pure Mathematics 2

2.1Algebra

Modulus — solve |ax + b| = c and inequalities, and sketch y = |f(x)|. Polynomials — division, the factor theorem and remainder theorem.

Worked example

P(x) = 2x³ + ax² − 5x + 6 has factor (x − 2). Find a.

  1. Factor theorem: P(2) = 0
  2. 16 + 4a − 10 + 6 = 0 → 4a + 12 = 0
  3. a = −3

2.2Logarithmic and exponential functions

Laws of logarithms, solving equations with unknown exponents, and reducing relationships to linear form to find constants graphically.

Worked example

Solve 32x+1 = 20, giving your answer to 3 significant figures.

  1. Take logs: (2x + 1) ln 3 = ln 20
  2. 2x + 1 = ln 20 ÷ ln 3 = 2.99573 ÷ 1.09861 = 2.72683
  3. 2x = 1.72683 → x = 0.863

2.3Trigonometry (P2)

sec²θ = 1 + tan²θ · cosec²θ = 1 + cot²θ sin(A ± B), cos(A ± B), tan(A ± B) · double angles: sin2A = 2sinAcosA, cos2A = 2cos²A − 1, etc. a sin θ + b cos θ = R sin(θ + α), where R = √(a² + b²)
Worked example — R-formula

Express 3 sin θ + 4 cos θ as R sin(θ + α), and state the maximum value.

  1. R = √(3² + 4²) = 5
  2. tan α = 4/3 → α = 53.13°
  3. So 3 sin θ + 4 cos θ = 5 sin(θ + 53.13°)
  4. Maximum value = 5, when sin(θ + 53.13°) = 1

2.4Differentiation (P2)

Differentiate ex, ln x, sin x, cos x, tan x; the product, quotient and chain rules; and implicit and parametric differentiation.

product: (uv)′ = u′v + uv′ · quotient: (u/v)′ = (u′v − uv′)/v² parametric: dy/dx = (dy/dt) ÷ (dx/dt)
Worked example — quotient rule

Differentiate y = (2x + 1)/(x² + 3).

  1. u = 2x + 1, u′ = 2; v = x² + 3, v′ = 2x
  2. dy/dx = [2(x² + 3) − (2x + 1)(2x)] ÷ (x² + 3)²
  3. = (2x² + 6 − 4x² − 2x) ÷ (x² + 3)² = (−2x² − 2x + 6) ÷ (x² + 3)²

2.5Integration (P2)

∫e^(ax+b) dx = (1/a)e^(ax+b) + c · ∫1/(ax+b) dx = (1/a)ln|ax+b| + c ∫sin(ax+b) dx = −(1/a)cos(ax+b) + c · ∫sec²(ax+b) dx = (1/a)tan(ax+b) + c ∫ f′(x)/f(x) dx = ln|f(x)| + c

Also the trapezium rule for approximate areas, and knowing whether it over- or under-estimates depending on the curvature.

2.6Numerical solution of equations

Two required techniques
  1. Sign change: if f(a) and f(b) have opposite signs and f is continuous, a root lies between a and b.
  2. Iteration: rearrange to x = F(x) and use xn+1 = F(xn), repeating until the values agree to the required accuracy.
"Show that a root lies between 1 and 2" means evaluate f(1) and f(2), show the signs are opposite, and state that f is continuous. All three parts are needed for full marks.
Paper 3 · 9 topics

Pure Mathematics 3

3.1–3.3Algebra, logs and trigonometry (P3)

Paper 3 contains all of Paper 2 (topics 2.1–2.6 above) and adds the four topics below. In addition, P3 algebra includes partial fractions and the binomial expansion for negative and fractional indices.

Partial fractions — the three forms
Distinct linear factors: px+q / [(ax+b)(cx+d)] → A/(ax+b) + B/(cx+d) Repeated factor: → A/(ax+b) + B/(ax+b)² + C/(cx+d) Irreducible quadratic: → (Ax+B)/(x²+c) + C/(dx+e)
Worked example

Express (5x − 1)/[(x − 1)(x + 2)] in partial fractions.

  1. Let it equal A/(x − 1) + B/(x + 2), so 5x − 1 = A(x + 2) + B(x − 1)
  2. Put x = 1: 4 = 3A → A = 4/3
  3. Put x = −2: −11 = −3B → B = 11/3
  4. Answer: 4/[3(x − 1)] + 11/[3(x + 2)]

Binomial for any index: (1 + x)n = 1 + nx + [n(n−1)/2!]x² + …, valid only for |x| < 1 — always state the validity condition.

3.4Differentiation (P3) 3.6Numerical solution

As Paper 2, plus fuller treatment of implicit and parametric differentiation.

Worked example — implicit differentiation

Find dy/dx for x² + xy + y² = 7.

  1. Differentiate term by term: 2x + (y + x dy/dx) + 2y dy/dx = 0
  2. Collect: dy/dx(x + 2y) = −2x − y
  3. dy/dx = −(2x + y) ÷ (x + 2y)

3.5Integration (P3)

Adds integration by parts, integration by substitution, and integration using partial fractions.

∫ u dv = uv − ∫ v du
Worked example — by parts

Find ∫ x ex dx.

  1. Let u = x (differentiates to something simpler) and dv = ex dx
  2. du = dx, v = ex
  3. ∫ xex dx = xex − ∫ ex dx = xex − ex + c

Choose u by LATE: Logs, Algebraic, Trig, Exponential — whichever comes first in that list becomes u.

3.7Vectors

|a| = √(x² + y² + z²) · unit vector = a/|a| scalar (dot) product: a·b = x₁x₂ + y₁y₂ + z₁z₂ = |a||b| cos θ line: r = a + t·d
Worked example

Find the angle between a = 2i + j − 2k and b = 3i − 4k.

  1. a·b = (2)(3) + (1)(0) + (−2)(−4) = 6 + 0 + 8 = 14
  2. |a| = √(4 + 1 + 4) = 3; |b| = √(9 + 0 + 16) = 5
  3. cos θ = 14 ÷ 15 = 0.9333 → θ = 21.0° (3 s.f.)

Also: determining whether lines are parallel, intersecting or skew, and finding the point of intersection when it exists.

3.8Differential equations

Separation of variablesWrite the equation as g(y) dy = f(x) dx, integrate both sides, include a single constant, then use the given condition to find it.
Worked example

Solve dy/dx = xy given y = 2 when x = 0.

  1. Separate: (1/y) dy = x dx
  2. Integrate: ln|y| = x²/2 + c
  3. x = 0, y = 2: ln 2 = c
  4. ln y = x²/2 + ln 2 → y = 2ex²/2

3.9Complex numbers

z = x + iy · modulus |z| = √(x² + y²) · argument arg z = arctan(y/x) in the correct quadrant polar: z = r(cos θ + i sin θ) · conjugate z* = x − iy multiplication: moduli multiply, arguments add
Worked example

Find the modulus and argument of z = 1 + i√3, and express it in polar form.

  1. |z| = √(1 + 3) = 2
  2. arg z = arctan(√3 / 1) = π/3 (first quadrant, since both parts are positive)
  3. z = 2(cos π/3 + i sin π/3)
Getting the argument in the wrong quadrant. Always sketch the point on an Argand diagram first — the calculator's arctan only returns values between −π/2 and π/2, so points in the second and third quadrants need adjusting by ±π.

Also: loci in the Argand diagram — |z − a| = r is a circle centre a radius r; |z − a| = |z − b| is the perpendicular bisector of the segment joining a and b; arg(z − a) = θ is a half-line from a.

Paper 4 · 5 topics

Mechanics

4.1Forces and equilibrium

Resolve forces into components, use equilibrium conditions (the resultant force is zero in both directions), and apply the model of a particle on a smooth or rough surface.

Friction: F ⩽ μR, with F = μR at the point of sliding
Worked example

A 5 kg block rests on a rough horizontal plane with coefficient of friction 0.4. Find the maximum horizontal force before it slides. (g = 10 m/s²)

  1. Normal reaction R = mg = 5 × 10 = 50 N
  2. Maximum friction = μR = 0.4 × 50 = 20 N
  3. Any horizontal force above 20 N causes sliding.

4.2Kinematics of motion in a straight line

Constant acceleration: v = u + at · s = ut + ½at² · v² = u² + 2as · s = ½(u + v)t Variable acceleration: v = ds/dt · a = dv/dt · s = ∫v dt · v = ∫a dt
Worked example

A particle moves with v = 3t² − 12t + 9 m/s. Find when it is at rest and the distance travelled in the first 2 s.

  1. At rest: 3(t − 1)(t − 3) = 0 → t = 1 s and 3 s
  2. Displacement = ∫₀² v dt = [t³ − 6t² + 9t]₀² = 8 − 24 + 18 = 2 m
  3. Careful: v changes sign at t = 1, so for total distance split the integral: 0→1 gives 4 m, and 1→2 gives −2 m, so distance = 4 + 2 = 6 m while displacement is only 2 m.

4.3Momentum

momentum = mv · conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Worked example

A 3 kg ball moving at 4 m/s collides with a stationary 2 kg ball. They coalesce. Find their common velocity.

  1. Momentum before = 3 × 4 + 2 × 0 = 12 kg m/s
  2. After: combined mass 5 kg at v: 5v = 12
  3. v = 2.4 m/s in the original direction

4.4Newton's laws of motion

F = ma applied to particles, including connected particles (over a pulley, or a car towing a trailer) and motion on inclined planes.

Worked example — connected particles

Masses 5 kg and 3 kg hang either side of a smooth pulley. Find the acceleration and the tension. (g = 10 m/s²)

  1. For the whole system: resultant force = (5 − 3)g = 20 N; total mass = 8 kg
  2. a = 20 ÷ 8 = 2.5 m/s²
  3. For the 3 kg mass (moving up): T − 30 = 3 × 2.5 → T = 37.5 N
  4. Check with the 5 kg mass: 50 − 37.5 = 12.5 = 5 × 2.5 ✓

4.5Energy, work and power

work done = Fs cos θ · KE = ½mv² · PE = mgh power = Fv = work done ÷ time work–energy principle: work done by the resultant force = change in kinetic energy
Worked example

A car of mass 1200 kg travels at a constant 25 m/s against a resistance of 600 N. Find the power output of the engine.

  1. Constant velocity means the driving force equals the resistance: F = 600 N
  2. Power = Fv = 600 × 25 = 15 000 W = 15 kW
Paper 5 · 5 topics

Probability & Statistics 1

5.1Representation of data

Stem-and-leaf, box-and-whisker plots, histograms (frequency density on the vertical axis) and cumulative frequency graphs. Measures of central tendency and spread, including variance and standard deviation.

mean = Σx/n · variance = Σx²/n − (mean)² · standard deviation = √variance grouped: mean = Σfx/Σf · variance = Σfx²/Σf − (mean)²
Worked example

For the data 4, 7, 9, 10, 15: find the mean and standard deviation.

  1. Mean = 45 ÷ 5 = 9
  2. Σx² = 16 + 49 + 81 + 100 + 225 = 471
  3. Variance = 471/5 − 81 = 94.2 − 81 = 13.2
  4. Standard deviation = √13.2 = 3.63 (3 s.f.)

5.2Permutations and combinations

nPr = n!/(n−r)!  (order matters) · nCr = n!/[(n−r)!r!]  (order does not)

Includes arrangements with repeated items (divide by the factorials of the repeats), arrangements with restrictions, and arrangements in a circle.

Worked example

How many distinct arrangements are there of the letters of BANANA?

  1. 6 letters with A repeated 3 times and N repeated twice
  2. 6! ÷ (3! × 2!) = 720 ÷ 12 = 60

5.3Probability

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) independent: P(A ∩ B) = P(A)P(B) conditional: P(A | B) = P(A ∩ B) / P(B)
Worked example — conditional probability

P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Find P(A | B) and state whether A and B are independent.

  1. P(A | B) = 0.3 ÷ 0.5 = 0.6
  2. Independence test: P(A)P(B) = 0.6 × 0.5 = 0.3 = P(A ∩ B) ✓
  3. So A and B are independent — equivalently, P(A | B) = P(A).

5.4Discrete random variables

E(X) = Σ x·P(X = x) · Var(X) = Σ x²·P(X = x) − [E(X)]² Binomial B(n, p): P(X = r) = nCr · p^r · (1−p)^(n−r) · E(X) = np · Var(X) = np(1−p) Geometric Geo(p): P(X = r) = (1−p)^(r−1) · p · E(X) = 1/p
Worked example — binomial

A biased coin has P(head) = 0.3. It is tossed 5 times. Find P(exactly 2 heads), and the mean and variance.

  1. P(X = 2) = 5C2 × 0.3² × 0.7³ = 10 × 0.09 × 0.343
  2. = 0.3087
  3. Mean = np = 5 × 0.3 = 1.5; variance = np(1 − p) = 5 × 0.3 × 0.7 = 1.05

5.5The normal distribution

standardise: Z = (X − μ) ÷ σ, then use the standard normal tables
Worked example

X ~ N(50, 8²). Find P(X > 62).

  1. Z = (62 − 50) ÷ 8 = 1.5
  2. P(Z > 1.5) = 1 − Φ(1.5) = 1 − 0.9332
  3. = 0.0668

Normal approximation to the binomial is valid when np > 5 and n(1 − p) > 5; use μ = np, σ² = np(1−p) and apply a continuity correction of ±0.5.

Forgetting the continuity correction when approximating a discrete distribution with a continuous one. P(X ⩾ 20) becomes P(X > 19.5), not P(X > 20).
Paper 6 · 5 topics

Probability & Statistics 2

6.1The Poisson distribution

P(X = r) = e−λ λr ÷ r!  ·  mean = variance = λ

Use for events occurring randomly, independently and at a constant average rate. Poisson approximates the binomial when n is large and p small (roughly n > 50, np < 5), with λ = np. The normal approximates Poisson when λ > 15, with a continuity correction.

Worked example

Faults occur at an average of 2.5 per metre. Find the probability of exactly 3 faults in one metre.

  1. P(X = 3) = e−2.5 × 2.5³ ÷ 3!
  2. = 0.082085 × 15.625 ÷ 6
  3. = 0.214 (3 s.f.)

6.2Linear combinations of random variables

E(aX + b) = aE(X) + b · Var(aX + b) = a²Var(X) E(aX + bY) = aE(X) + bE(Y) Var(aX + bY) = a²Var(X) + b²Var(Y) — for INDEPENDENT X and Y
Writing Var(X − Y) = Var(X) − Var(Y). Variances always add for independent variables: Var(X − Y) = Var(X) + Var(Y).

6.3Continuous random variables

total probability: ∫ f(x) dx = 1 over the whole range E(X) = ∫ x·f(x) dx · Var(X) = ∫ x²·f(x) dx − [E(X)]² median m: ∫ from lower limit to m of f(x) dx = 0.5
Worked example

f(x) = kx for 0 ⩽ x ⩽ 2, and zero elsewhere. Find k and E(X).

  1. ∫₀² kx dx = k[x²/2]₀² = 2k = 1 → k = ½
  2. E(X) = ∫₀² x × ½x dx = ½[x³/3]₀² = ½ × 8/3 = 4/3

6.4Sampling and estimation

sample mean distribution: X̄ ~ N(μ, σ²/n) unbiased estimate of variance: s² = [Σx² − (Σx)²/n] / (n − 1) confidence interval for μ: x̄ ± z × (σ/√n)

Know why a random sample matters, and the Central Limit Theorem: for large n, the sample mean is approximately normally distributed whatever the population distribution.

Worked example

A sample of 100 has mean 52 from a population with σ = 10. Find a 95% confidence interval for μ.

  1. Standard error = 10 ÷ √100 = 1
  2. For 95%, z = 1.96
  3. Interval = 52 ± 1.96 × 1 = (50.04, 53.96)

6.5Hypothesis tests

The five-step method — follow it every time
  1. State H₀ and H₁ clearly, with the parameter defined.
  2. State the significance level and whether the test is one- or two-tailed.
  3. Calculate the test statistic (or the probability of the observed result or more extreme).
  4. Compare with the critical value and state whether to reject H₀.
  5. Write a conclusion in context — never just "reject H₀".
Type I and Type II errorsA Type I error is rejecting H₀ when it is true — its probability equals the significance level. A Type II error is failing to reject H₀ when it is false. Reducing the significance level reduces Type I risk but increases Type II risk.
The final conclusion must be in context and suitably tentative: "There is sufficient evidence at the 5% level to conclude that the mean weight has increased" — not "the mean weight has increased". Examiners deduct for over-claiming.
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Reference

Free past papers & how to revise

Official (free)

  • Cambridge International — 9709 subject page: syllabus, specimen papers, past papers, mark schemes and examiner reports.
  • Examiner reports name the exact questions candidates got wrong each series — read them for every paper you attempt.

Free archives

How to revise this subject

  1. Know which papers you are entered for and ignore the rest — there is no credit for studying components you will not sit.
  2. Fix algebraic fluency first. At this level, most lost marks are sign errors, dropped brackets and mishandled fractions, not conceptual gaps.
  3. Learn the formula list that comes with the papers, and — more importantly — the results that are not on it.
  4. "Show that" questions are free marks: work towards the printed answer line by line. Never write the given result and stop.
  5. Always state the units and the interval. Mechanics answers need units; trigonometric solutions need every value in the stated range.

Edvia Free Resources — AS & A Level Mathematics 9709. Original notes and worked examples written for the Cambridge AS & A Level Mathematics 9709 syllabus for examination in 2028–2030. An independent free study resource, not affiliated with or endorsed by Cambridge University Press & Assessment. Syllabus reference codes are used for navigation. Share it freely — it will always be free.

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