O Level Additional Maths 4037 — all 14 topics, free.

A complete study guide for Cambridge O Level Additional Mathematics 4037, covering all 14 topics of the official syllabus for exams in 2028–2030.

Sitting exams in 2026 or 2027? You are on the 2025–2027 version of the syllabus. Cambridge states there are no significant changes affecting teaching between the two versions, so this guide covers both — but check your exam year with your school.

How to use it: Add Maths is the bridge to A Level. It rewards fluent algebra above all — most lost marks are algebraic slips, not misunderstood topics. Work every example with the solution covered.

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

CAIE 4037 · exams 2028–203014 topicsBridges to A LevelFormula sheet analysisFree & shareable

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The papers

Paper	Format	Time / marks	Weight
Paper 1 — Non-calculator	Structured and unstructured questions	2 hours · 80 marks	50%
Paper 2 — Calculator	Structured and unstructured questions; scientific calculator required	2 hours · 80 marks	50%

Both papers cover the whole syllabus. The difference is the calculator — which means on Paper 1 you must be fluent with exact values, surds and standard angles.

Show every line of working. Add Maths marks are heavily weighted towards method: a differentiation question with a correct derivative and one arithmetic slip usually keeps most of its marks, while a bare wrong answer keeps none.

Know the difference

Given vs must-memorise formulas

Printed on the exam paper

Topic	Formula
Circle	(x − a)² + (y − b)² = r², centre (a, b), radius r
Quadratic formula	x = (−b ± √(b² − 4ac)) ÷ 2a
Binomial theorem	(a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + … , with ⁿC_r = n! ÷ ((n − r)! r!)
Arithmetic series	u_n = a + (n − 1)d · S_n = ½n(a + l) = ½n[2a + (n − 1)d]
Geometric series	u_n = arⁿ⁻¹ · S_n = a(1 − rⁿ) ÷ (1 − r) · S_∞ = a ÷ (1 − r) for \|r\| < 1
Trig identities	sin²A + cos²A = 1 · sec²A = 1 + tan²A · cosec²A = 1 + cot²A
Triangle formulas	Sine rule, cosine rule, area = ½ab sin C
Mensuration	Curved surface of cone πrl; sphere 4πr²; volume of pyramid/cone ⅓Ah; sphere 4/3 πr³

NOT given — you must know these

Topic	Formula
Discriminant	b² − 4ac: > 0 two distinct real roots; = 0 equal roots; < 0 no real roots
Completed square	a(x + p)² + q, vertex at (−p, q)
Logarithm laws	log a + log b = log ab · log a − log b = log(a/b) · log aⁿ = n log a · change of base
Straight line	gradient, midpoint, distance, y − y₁ = m(x − x₁), m₁m₂ = −1
Circular measure	s = rθ · area of sector = ½r²θ (θ in radians)
Differentiation	d/dx(xⁿ) = nxⁿ⁻¹ · sin, cos, tan · e^x · ln x · product, quotient and chain rules
Integration	∫xⁿ dx = xⁿ⁺¹/(n+1) + c, and the standard results
Kinematics	v = ds/dt · a = dv/dt · s = ∫v dt · v = ∫a dt

Exact trig values — Paper 1 essentials

θ	0°	30°	45°	60°	90°
radians	0	π/6	π/4	π/3	π/2
sin θ	0	½	1/√2	√3/2	1
cos θ	1	√3/2	1/√2	½	0
tan θ	0	1/√3	1	√3	undefined

Topic 1

Functions

1Functions

Key termsThe domain is the set of inputs; the range is the set of outputs. A function is one-one if each output comes from exactly one input — only one-one functions have an inverse.

Methods

Composite function fg(x) = f(g(x)) — apply the function nearest the x first.
Inverse: write y = f(x), swap x and y, make y the subject.
The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.
Modulus: |x| is the magnitude of x. To sketch y = |f(x)|, reflect any part of the curve below the x-axis up above it.

Worked example

f(x) = 2x + 3 and g(x) = x². Find (a) fg(x), (b) gf(x), (c) f⁻¹(x).

(a) fg(x) = f(x²) = 2x² + 3
(b) gf(x) = g(2x + 3) = (2x + 3)²
(c) y = 2x + 3 → swap: x = 2y + 3 → f⁻¹(x) = (x − 3)/2

Skill check: Solve |2x − 1| = 7.

Solution: The modulus gives two cases. 2x − 1 = 7 → x = 4; or 2x − 1 = −7 → x = −3. Always check both in the original equation.

Topic 2

Quadratic functions

2Quadratic functions

Completed square: a(x + p)² + q → vertex (−p, q)
Discriminant: b² − 4ac

Discriminant	Roots	Graph
b² − 4ac > 0	Two distinct real roots	Crosses the x-axis twice
b² − 4ac = 0	Two equal roots (repeated)	Touches the x-axis — a tangent
b² − 4ac < 0	No real roots	Does not cross the x-axis

Worked example — completing the square

Express 2x² − 12x + 23 in the form a(x + p)² + q and state the minimum point.

Factor out the 2 from the x terms: 2(x² − 6x) + 23
Complete the square inside: x² − 6x = (x − 3)² − 9
2[(x − 3)² − 9] + 23 = 2(x − 3)² − 18 + 23 = 2(x − 3)² + 5
Minimum point (3, 5) — check: at x = 3, y = 2(9) − 36 + 23 = 5 ✓

Worked example — tangency condition

Find the value of k for which y = kx + 1 is a tangent to y = x² + 3x + 4.

Equate: kx + 1 = x² + 3x + 4 → x² + (3 − k)x + 3 = 0
Tangent means equal roots, so b² − 4ac = 0: (3 − k)² − 12 = 0
(3 − k)² = 12 → 3 − k = ±2√3 → k = 3 ∓ 2√3

"Tangent" always means the discriminant is zero. "Meets in two points" means it is positive; "does not meet" means negative. Translating the wording into a discriminant condition is the whole skill in this question type.

Topic 3

Factors of polynomials

3Factors of polynomials

The two theoremsRemainder theorem: when P(x) is divided by (x − a), the remainder is P(a).
Factor theorem: (x − a) is a factor of P(x) if and only if P(a) = 0.
For a divisor (bx − a), evaluate P(a/b).

Worked example

Show that (x − 2) is a factor of P(x) = x³ − 3x² − 4x + 12, and factorise fully.

P(2) = 8 − 12 − 8 + 12 = 0, so (x − 2) is a factor ✓
Divide: x³ − 3x² − 4x + 12 = (x − 2)(x² − x − 6)
Factorise the quadratic: x² − x − 6 = (x − 3)(x + 2)
Fully factorised: (x − 2)(x − 3)(x + 2), so roots are x = 2, 3, −2

Skill check: Find the remainder when x³ + 2x² − 5x + 1 is divided by (x + 3).

Solution: Divisor (x + 3) means evaluate at x = −3: (−27) + 2(9) − 5(−3) + 1 = −27 + 18 + 15 + 1 = 7.

Topic 4

Equations, inequalities and graphs

4Equations, inequalities and graphs

Quadratic inequalities — the reliable method

Rearrange so one side is zero.
Factorise and find the critical values (where it equals zero).
Sketch the parabola and read off where it is above or below the axis.

For a positive x² coefficient: "< 0" gives the region between the roots; "> 0" gives outside the roots.

Worked example

Solve x² − 5x + 6 < 0.

Factorise: (x − 2)(x − 3) < 0 → critical values 2 and 3
The parabola opens upwards, so it is below the axis between the roots
2 < x < 3

Worked example — modulus inequality

Solve |x − 4| < 3.

|A| < b means −b < A < b
−3 < x − 4 < 3
Add 4 throughout: 1 < x < 7

Also required: sketching cubic and reciprocal graphs, solving equations graphically, and using a substitution to reduce an equation to a quadratic (e.g. let u = x² to solve x⁴ − 5x² + 4 = 0, giving u = 1 or 4, so x = ±1, ±2).

Topic 5

Simultaneous equations

5Simultaneous equations

One linear, one non-linearAlways substitute the linear equation into the non-linear one — never the other way round. Rearrange the linear equation to make one variable the subject, substitute, solve the resulting quadratic, then find the paired values.

Worked example

Solve y = x + 2 and x² + y² = 10.

Substitute: x² + (x + 2)² = 10
x² + x² + 4x + 4 = 10 → 2x² + 4x − 6 = 0 → x² + 2x − 3 = 0
(x + 3)(x − 1) = 0 → x = −3 or x = 1
Pair them up: x = −3 → y = −1; x = 1 → y = 3
Solutions: (−3, −1) and (1, 3)

Giving the x values and y values as separate lists. They must be paired correctly — each x goes with its own y. Mismatched pairs lose the accuracy marks.

Topic 6

Logarithmic and exponential functions

6Logarithmic and exponential functions

a^x = b ⟺ log_a b = x

The laws of logarithms

log a + log b = log(ab) log a − log b = log(a/b) log a^n = n log a log_a a = 1 · log_a 1 = 0 change of base: log_a b = (log b) / (log a)

e ≈ 2.718 is the natural base; ln x means log_e x, and ln e^x = x.

Worked example

Solve log₂(x) + log₂(x − 2) = 3.

Combine: log₂[x(x − 2)] = 3
Convert: x(x − 2) = 2³ = 8
x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or −2
Reject x = −2, since log of a negative number is undefined. Answer: x = 4

Forgetting to check solutions in the original equation. Log equations regularly produce an invalid root, and a mark is awarded specifically for rejecting it with a reason.

Worked example — reducing to linear form

Given y = axⁿ, explain how a straight-line graph finds a and n.

Take logs of both sides: lg y = lg a + n lg x
Compare with Y = mX + c where Y = lg y and X = lg x
So plotting lg y against lg x gives a straight line with gradient n and intercept lg a (so a = 10^intercept).

Topic 7

Straight-line graphs

7Straight-line graphs

gradient m = (y₂ − y₁)/(x₂ − x₁) · midpoint = ((x₁+x₂)/2, (y₁+y₂)/2) length = √[(x₂−x₁)² + (y₂−y₁)²] · y − y₁ = m(x − x₁) · parallel m₁ = m₂ · perpendicular m₁m₂ = −1

Worked example

A(1, 2) and B(7, 10). Find the equation of the perpendicular bisector of AB.

Midpoint = ((1+7)/2, (2+10)/2) = (4, 6)
Gradient of AB = (10 − 2)/(7 − 1) = 8/6 = 4/3
Perpendicular gradient = −3/4
Through (4, 6): y − 6 = −¾(x − 4) → 4y + 3x = 36

Also required: the area of a rectilinear figure from the coordinates of its vertices, and using the "reduce to linear form" technique from Topic 6 with relationships such as y = Ab^x.

Topic 8

Coordinate geometry of the circle

8Coordinate geometry of the circle

(x − a)² + (y − b)² = r² — centre (a, b), radius r (given in the exam)

From expanded form to centre and radiusGiven x² + y² + 2gx + 2fy + c = 0, complete the square in x and in y separately to reach the standard form.

Worked example

Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.

x² − 6x = (x − 3)² − 9 · y² + 4y = (y + 2)² − 4
(x − 3)² − 9 + (y + 2)² − 4 − 12 = 0
(x − 3)² + (y + 2)² = 25
Centre (3, −2), radius 5

A key property: the tangent to a circle is perpendicular to the radius at the point of contact. So to find a tangent's equation, find the gradient of the radius to that point, take the negative reciprocal, then use y − y₁ = m(x − x₁).

Topic 9

Circular measure

9Circular measure

π radians = 180° · arc length s = rθ · sector area = ½r²θ (θ in radians)

Worked example

A sector has radius 8 cm and angle 0.6 radians. Find the arc length, the sector area, and the area of the corresponding segment.

Arc s = rθ = 8 × 0.6 = 4.8 cm
Sector area = ½ × 64 × 0.6 = 19.2 cm²
Triangle area = ½r² sin θ = ½ × 64 × sin(0.6) = 32 × 0.5646 = 18.07 cm²
Segment = sector − triangle = 19.2 − 18.07 = 1.13 cm² (3 s.f.)

Using the formulas with the angle in degrees. Both s = rθ and ½r²θ require radians, and your calculator must be in radian mode for sin θ in the segment calculation.

Topic 10

Trigonometry

10Trigonometry

Identities

sin²A + cos²A = 1 (given) sec²A = 1 + tan²A (given) cosec²A = 1 + cot²A (given) tan A = sin A / cos A sec A = 1/cos A · cosec A = 1/sin A · cot A = 1/tan A

Solving trig equations — do not lose the extra solutions

Get the equation into the form sin/cos/tan of something = a value.
Find the principal value from the inverse function.
Use the graph or the CAST diagram to find all solutions in the required interval.
If the argument is like 2x or (x + 30°), widen the interval first, solve, then convert back.

Worked example

Solve 2 sin x = 1 for 0° ⩽ x ⩽ 360°.

sin x = ½
Principal value: x = 30°
Sine is also positive in the second quadrant: x = 180° − 30° = 150°
x = 30° or 150°

Worked example — using an identity

Solve 2 cos²x + sin x = 2 for 0° ⩽ x ⩽ 360°.

Replace cos²x with 1 − sin²x: 2(1 − sin²x) + sin x = 2
2 − 2sin²x + sin x = 2 → sin x(1 − 2 sin x) = 0
sin x = 0 → x = 0°, 180°, 360°; or sin x = ½ → x = 30°, 150°
x = 0°, 30°, 150°, 180°, 360°

Also required: graphs of sin, cos and tan including amplitude and period — for y = a sin(bx) + c, the amplitude is |a|, the period is 360°/b (or 2π/b), and c shifts the curve vertically.

Topic 11

Permutations and combinations

11Permutations and combinations

The distinctionA permutation counts arrangements where order matters (a race result). A combination counts selections where order does not matter (a committee).

ⁿP_r = n! ÷ (n − r)! · ⁿC_r = n! ÷ [(n − r)! r!]

Worked example

From 10 students: (a) how many ways can 3 be chosen for a committee? (b) how many ways can they be arranged as chairperson, secretary and treasurer?

(a) Order does not matter → combination: ¹⁰C₃ = (10 × 9 × 8) ÷ (3 × 2 × 1) = 120
(b) Order matters (the roles differ) → permutation: ¹⁰P₃ = 10 × 9 × 8 = 720
Note 720 = 120 × 3! — each selection of 3 can be arranged in 6 ways.

Skill check: How many different 4-letter arrangements can be made from the letters of the word MATHS?

Solution: Order matters and no letter repeats, so it is ⁵P₄ = 5 × 4 × 3 × 2 = 120.

Topic 12

Series

12Series

All given on the formula sheet

Binomial: (a + b)^n = a^n + nC1·a^(n−1)·b + nC2·a^(n−2)·b² + … Arithmetic: u_n = a + (n − 1)d · S_n = ½n[2a + (n − 1)d] = ½n(a + l) Geometric: u_n = ar^(n−1) · S_n = a(1 − r^n)/(1 − r) · S∞ = a/(1 − r) for |r| < 1

Worked example — binomial

Find the coefficient of x³ in the expansion of (2 + x)⁵.

General term: ⁵C_r × 2^5−r × x^r
For x³, take r = 3: ⁵C₃ × 2² = 10 × 4
Coefficient = 40

Worked example — arithmetic

An arithmetic progression has first term 5 and common difference 3. Find the 20th term and the sum of the first 20 terms.

u₂₀ = 5 + 19 × 3 = 62
S₂₀ = ½ × 20 × (5 + 62) = 10 × 67 = 670

Worked example — geometric and sum to infinity

A geometric progression has first term 12 and common ratio ⅓. Find the 5th term and the sum to infinity.

u₅ = 12 × (⅓)⁴ = 12/81 = 4/27
|r| < 1, so S_∞ = 12 ÷ (1 − ⅓) = 12 ÷ ⅔ = 18

Using S_∞ when |r| ⩾ 1. The sum to infinity only exists for −1 < r < 1; otherwise the series diverges. State the condition when you use it.

Topic 13

Vectors in two dimensions

13Vectors in two dimensions

Vectors may be written as a column, as AB, or as ai + bj. Add and subtract componentwise; multiply by a scalar to change magnitude.

magnitude |xi + yj| = √(x² + y²) · unit vector = vector ÷ its magnitude

Worked example

a = 3i + 4j. Find |a| and the unit vector in the direction of a.

|a| = √(9 + 16) = 5
Unit vector = (3i + 4j)/5 = 0.6i + 0.8j (check: √(0.36 + 0.64) = 1 ✓)

Worked example — relative velocity

A boat travels with velocity (6i + 8j) km/h. Find its speed and the distance travelled in 3 hours.

Speed = |6i + 8j| = √(36 + 64) = 10 km/h
Distance in 3 hours = 10 × 3 = 30 km; displacement = 3(6i + 8j) = 18i + 24j

Also required: showing points are collinear (if AB = kAC and they share a point, they lie on a straight line), and problems involving velocity, displacement and interception.

Topic 14 — the biggest topic

Calculus

14Calculus

Differentiation — none of these are given

y	dy/dx
xⁿ	nxⁿ⁻¹
sin x	cos x
cos x	−sin x
tan x	sec²x
e^x	e^x
ln x	1/x

Product rule: (uv)′ = u′v + uv′ · Quotient rule: (u/v)′ = (u′v − uv′)/v² · Chain rule: dy/dx = dy/du × du/dx

Worked example — stationary points

Find and classify the stationary points of y = x³ − 3x² + 4.

dy/dx = 3x² − 6x = 3x(x − 2) = 0 → x = 0 or x = 2
y-values: at x = 0, y = 4; at x = 2, y = 8 − 12 + 4 = 0
Second derivative: d²y/dx² = 6x − 6
At x = 0: 6(0) − 6 = −6 < 0 → maximum at (0, 4)
At x = 2: 6(2) − 6 = 6 > 0 → minimum at (2, 0)

Integration — the reverse process

∫x^n dx = x^(n+1)/(n+1) + c (n ≠ −1) ∫sin x dx = −cos x + c · ∫cos x dx = sin x + c ∫sec²x dx = tan x + c · ∫e^x dx = e^x + c · ∫(1/x) dx = ln|x| + c ∫(ax + b)^n dx = (ax + b)^(n+1) / [a(n+1)] + c

Never forget "+ c" for indefinite integrals — it is a mark in itself.

Worked example — definite integral and area

Find the area under y = x² between x = 1 and x = 3.

Area = ∫₁³ x² dx = [x³/3]₁³
= 27/3 − 1/3 = 9 − ⅓
= 26/3 ≈ 8.67 square units

Worked example — kinematics

A particle moves with velocity v = 3t² − 12t + 9 m/s. Find (a) when it is instantaneously at rest, (b) its acceleration at t = 1, (c) the displacement in the first 2 seconds.

(a) At rest when v = 0: 3(t² − 4t + 3) = 0 → (t − 1)(t − 3) = 0 → t = 1 s and t = 3 s
(b) a = dv/dt = 6t − 12; at t = 1: 6 − 12 = −6 m/s² (decelerating)
(c) s = ∫₀² (3t² − 12t + 9) dt = [t³ − 6t² + 9t]₀²
= (8 − 24 + 18) − 0 = 2 m

Rates of changeUse the chain rule to connect rates: dV/dt = dV/dr × dr/dt. Identify which rate you are given and which you want, then write the chain that links them.

Confusing displacement with distance travelled. If the velocity changes sign in the interval, the definite integral gives net displacement — for total distance you must split the integral at the times when v = 0 and add the magnitudes.

Skill check: Differentiate y = x² sin x.

Solution: Product rule with u = x², v = sin x: u′ = 2x, v′ = cos x. So dy/dx = u′v + uv′ = 2x sin x + x² cos x.

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Reference

Free past papers & how to revise

Official (free)

Cambridge International — 4037 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

GCE Guide · PastPapers.co — full CAIE past-paper archives.
Physics & Maths Tutor — topic-sorted questions.

How to revise this subject

Fix your algebra first. Almost every lost mark at this level is a sign error, a bracket error or a lost fraction — not a misunderstood topic.
Learn the exact trig values for 0°, 30°, 45°, 60°, 90° and their surd forms. Paper 1 assumes them.
Know which formulas are given (see the reference section) and drill the ones that are not — especially the differentiation and integration rules.
Practise "show that" questions. They are common and generous: work towards the given result line by line, and never just restate it.
Always check whether an angle question wants degrees or radians, and give all solutions in the stated interval.

O Level Additional Maths 4037 — all 14 topics, free.

The papers

Given vs must-memorise formulas

Printed on the exam paper

NOT given — you must know these

Functions

1Functions

Quadratic functions

2Quadratic functions

Factors of polynomials

3Factors of polynomials

Equations, inequalities and graphs

4Equations, inequalities and graphs

Simultaneous equations

5Simultaneous equations

Logarithmic and exponential functions

6Logarithmic and exponential functions

Straight-line graphs

7Straight-line graphs

Coordinate geometry of the circle

8Coordinate geometry of the circle

Circular measure

9Circular measure

Trigonometry

10Trigonometry

Permutations and combinations

11Permutations and combinations

Series

12Series

Vectors in two dimensions

13Vectors in two dimensions

Calculus

14Calculus

Study planner & progress

Free past papers & how to revise

Official (free)

Free archives

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