Free notes · Secondary 3 and 4
O-Level A-Math notes
Free O-Level A-Math notes on one page, for the 2026 O-Level (4049) and the 2027 SEC G3 Additional Mathematics (K341). Full notes for 10 of the 10 syllabus chapters, each with the key idea, the facts and formulas to know, a worked exam question, the common mistakes and exam tips. The other chapters have a checklist of what the syllabus asks for, so the page maps the whole syllabus.
Free to read, print or save as a PDF, with no sign-up. The print layout keeps every chapter and drops the menus. After each chapter, practise on the O-Level A-Math prelim papers with answers.
- 10 chapters
- Free to print
How to use these notes
- Pick one chapter. Read the key idea and the must-know points.
- Close the page and write down what you remember, then check and mark what you missed.
- Try the worked exam question before reading its steps, and read the common mistakes.
- Do questions on that chapter from a school paper and mark them. Redo the wrong ones a few days later.
Syllabus edition
These notes follow SEAB's O-Level Additional Mathematics syllabus 4049, examined in 2026, and SEC G3 Additional Mathematics K341, examined from 2027. The two documents list the same topics and the same two papers. Sources: SEAB, O-Level Additional Mathematics syllabus 4049 (2026), SEAB, SEC G3 Additional Mathematics syllabus K341 (2027).
Exam format
| Paper | What it is | Time | Marks | Weighting |
|---|---|---|---|---|
| Paper 1 | 12 to 14 questions of different lengths, up to 10 marks each. Answer all. | 2 h 15 min | 90 | 50% |
| Paper 2 | 9 to 11 questions of different lengths, up to 12 marks each. Answer all. | 2 h 15 min | 90 | 50% |
- Relevant formulae are printed. An approved calculator may be used in both papers.
- Missing essential working loses marks. E-Math content is assumed and can be needed inside A-Math questions.
The whole syllabus, chapter by chapter
All 10 chapters of the syllabus in order. 10 have full notes on this page. The rest have a checklist of what the syllabus asks you to do.
| Chapter | What it covers | On this page |
|---|---|---|
| 1. Quadratic Functions | Completing the square and when a quadratic is always positive or negative. | Full notes |
| 2. Equations and Inequalities | Nature of roots, lines and curves, simultaneous equations and quadratic inequalities. | Full notes |
| 3. Surds | Operations on surds, rationalising and surd equations. | Full notes |
| 4. Polynomials and Partial Fractions | Remainder and factor theorems, cubic equations and partial fractions. | Full notes |
| 5. Binomial Expansions | The binomial theorem and the general term. | Full notes |
| 6. Exponential and Logarithmic Functions | Graphs, laws of logarithms, change of base and models. | Full notes |
| 7. Trigonometric Functions, Identities and Equations | Six functions, graphs, identities, R-formula and equations. | Full notes |
| 8. Coordinate Geometry in Two Dimensions | Parallel and perpendicular lines, circles, areas and linear law. | Full notes, plus checklist |
| 9. Proofs in Plane Geometry | Geometric proofs using angle, triangle and circle properties. | Checklist |
| 10. Differentiation and Integration | Derivatives, stationary points, rates of change, integration and areas. | Differentiation, Applications of Differentiation, Integration, plus checklist |
O-Level A-Math at a glance
| Chapter | The one thing to remember |
|---|---|
| Quadratic Functions and Inequalities | y = a(x - h)² + k has turning point (h, k), a minimum if a > 0 and a maximum if a < 0 |
| Surds | √(ab) = √a × √b and √(a/b) = √a ÷ √b, but √(a + b) is not √a + √b |
| Polynomials and Partial Fractions | The remainder when f(x) is divided by (ax - b) is f(b/a) |
| Binomial Theorem | (a + b)^(n) has n + 1 terms, and the powers of a and b in each term add up to n |
| Exponential and Logarithmic Functions | logₐ(MN) = logₐ M + logₐ N and logₐ(M/N) = logₐ M - logₐ N |
| Trigonometric Identities and Equations | sin²A + cos²A = 1, sec²A = 1 + tan²A and cosec²A = 1 + cot²A |
| Coordinate Geometry of Circles | Standard form: (x - a)² + (y - b)² = r², with centre (a, b) and radius r |
| Differentiation | Power rule: the derivative of x^(n) is nx^(n-1), for any rational n |
| Applications of Differentiation | Tangent at x = a: gradient m = dy/dx at a. Normal: gradient -1/m |
| Integration | ∫x^(n) dx = x^(n+1)/(n + 1) + c, for n ≠ -1 |
Notes 1 of 10
Quadratic Functions and Inequalities
Key idea. A quadratic function has the form y = ax² + bx + c. Completing the square writes it as a(x - h)² + k, which gives the turning point (h, k). The discriminant b² - 4ac decides whether the roots are distinct, equal or not real, and whether a line cuts, touches or misses a curve. Solve quadratic inequalities by factorising and sketching the curve.
Must know
- y = a(x - h)² + k has turning point (h, k), a minimum if a > 0 and a maximum if a < 0
- b² - 4ac > 0: two distinct real roots. b² - 4ac = 0: equal roots. b² - 4ac < 0: no real roots
- Always positive: a > 0 and b² - 4ac < 0. Always negative: a < 0 and b² - 4ac < 0
- (x - a)(x - b) < 0, with a < b, gives a < x < b, between the roots
What the syllabus asks you to do
- Find the maximum or minimum value of a quadratic function by completing the square
- Find the conditions for ax² + bx + c to be always positive or always negative
- Find the conditions for a quadratic equation to have two real roots, two equal roots or no real roots
- Find the conditions for a given line to intersect a given curve, to be a tangent to it, or not to intersect it
- Solve simultaneous equations in two variables by substitution, where one equation is linear
- Solve quadratic inequalities and represent the solution on the number line
What the discriminant tells you
| Condition | Roots of the quadratic | Line and curve |
|---|---|---|
| b² - 4ac > 0 | Two real, distinct roots | The line cuts the curve at two points |
| b² - 4ac = 0 | Two real, equal roots | The line is a tangent to the curve |
| b² - 4ac < 0 | No real roots | The line does not meet the curve |
| b² - 4ac ≥ 0 | Real roots (distinct or equal) | The line meets the curve |
A U-shaped curve for x from 0 to 6. It starts at (0, 5) on the y-axis, falls to its lowest point (3, -13), then rises symmetrically back to (6, 5). It crosses the x-axis at about x = 0.45 and x = 5.55. The curve is symmetrical about the line x = 3.
Worked exam question
Find the range of values of k for which the line y = kx - 1 does not meet the curve y = x² + x + 3.
- Substitute the line into the curve: x² + x + 3 = kx - 1.
- Collect the terms on one side: x² + (1 - k)x + 4 = 0.
- The line does not meet the curve, so this equation has no real roots and b² - 4ac < 0.
- Here a = 1, b = 1 - k and c = 4, so (1 - k)² - 16 < 0.
- Expand: 1 - 2k + k² - 16 < 0, which gives k² - 2k - 15 < 0.
- Factorise: (k - 5)(k + 3) < 0, so the roots are k = -3 and k = 5.
- The U-shaped graph of (k - 5)(k + 3) is below the axis between its roots, so -3 < k < 5.
Answer: -3 < k < 5
Common mistakes
- Writing the condition for real roots of x² + 6x + k = 0 as 36 - 4k > 0.Use b² - 4ac ≥ 0 for real roots, and > 0 only when the question says real and distinct.
- Writing 2x² - 12x + 5 as 2(x - 6)² plus a constant.Take the coefficient of x² out of the x terms first, 2(x² - 6x) + 5, then complete the square inside.
- Using the curve's own discriminant to decide whether a line meets the curve.Substitute the line into the curve, collect one quadratic, then use the discriminant of that new quadratic.
- Writing the solution of (x - 3)(x + 2) > 0 as -2 > x > 3.Sketch the curve and write the two outside regions with 'or': x < -2 or x > 3.
- Saying a quadratic is always positive because b² - 4ac < 0, without checking a.State both conditions every time: a > 0 and b² - 4ac < 0.
- Giving the minimum of y = (x + 4)² - 1 at x = 4.Solve x + 4 = 0 to find the turning point: x = -4, so the minimum point is (-4, -1).
Exam tips
- Underline the key word before writing the condition: 'meets' or 'intersects' means b² - 4ac ≥ 0, 'cuts at two distinct points' means > 0, 'tangent' means = 0 and 'does not meet' means < 0.
- For any inequality in k, finish with a quick sketch of the quadratic in k. It shows at a glance whether the answer lies between the roots or outside them.
- After completing the square, expand your answer in your head to check that it gives back the original expression.
Notes 2 of 10
Surds
Key idea. A surd is a root such as √2 or √5 that cannot be written as a rational number. Simplify a surd by taking out the largest square factor, so √72 = 6√2. Rationalise a denominator by multiplying the top and bottom by the surd, or by the conjugate when the denominator is a + √b. Solve surd equations by squaring, then check every answer.
Must know
- √(ab) = √a × √b and √(a/b) = √a ÷ √b, but √(a + b) is not √a + √b
- √a × √a = a, so (a + √b)(a - √b) = a² - b, which has no surd
- To rationalise 1/(a + √b), multiply the top and bottom by the conjugate a - √b
What the syllabus asks you to do
- Add and subtract surds, simplifying them first
- Multiply and divide expressions containing surds
- Rationalise the denominator of a fraction containing surds
- Solve equations involving surds
How to rationalise each type of denominator
| Denominator | Multiply top and bottom by | Example |
|---|---|---|
| √a | √a | 6/√3 = 6√3/3 = 2√3 |
| k√a | √a | 5/(2√5) = 5√5/10 = √5/2 |
| a + √b | a - √b | 4/(3 + √5) = 4(3 - √5)/4 = 3 - √5 |
| a - √b | a + √b | 1/(2 - √3) = (2 + √3)/(4 - 3) = 2 + √3 |
| √a + √b | √a - √b | 1/(√5 + √3) = (√5 - √3)/2 |
Isolate the square root
Get the root alone on one side, here √(2x + 3) = x.
Square both sides
2x + 3 = x², so x² - 2x - 3 = 0.
Solve the quadratic
(x - 3)(x + 1) = 0 gives x = 3 or x = -1.
Check in the original
x = 3 works. x = -1 gives √1 = 1, not -1.
State the valid solution
Reject x = -1, so x = 3 only.
Worked exam question
Without using a calculator, express (5 + 2√3)/(2 - √3) in the form a + b√3, where a and b are integers.
- The conjugate of the denominator 2 - √3 is 2 + √3, so multiply the numerator and the denominator by 2 + √3.
- Denominator: (2 - √3)(2 + √3) = 4 - 3 = 1.
- Numerator: (5 + 2√3)(2 + √3) = 10 + 5√3 + 4√3 + 2√3 × √3.
- Since √3 × √3 = 3, the last term is 2 × 3 = 6, so the numerator is 16 + 9√3.
- Divide by the denominator, which is 1: the result is 16 + 9√3.
- Check: (16 + 9√3)(2 - √3) = 32 - 16√3 + 18√3 - 27 = 5 + 2√3, which matches the original numerator.
Answer: 16 + 9√3, so a = 16 and b = 9
Common mistakes
- Writing √(9 + 16) as √9 + √16 = 7.Test any surd rule with small numbers first: √(9 + 16) = √25 = 5.
- Leaving 2√3 × √3 unsimplified instead of writing 6.Whenever a surd meets itself, replace √a × √a by a straight away.
- Multiplying only the denominator of 4/(3 + √5) by 3 - √5.Write the conjugate over itself, (3 - √5)/(3 - √5), before multiplying.
- Accepting x = 1 as a root of √(3x + 1) = x - 3 after squaring.Substitute every answer into the original equation and reject any that fail.
- Simplifying √48 as 2√12 and stopping.Look for the largest square factor first: √48 = √(16 × 3) = 4√3.
- Giving 2.732 when the exact value 1 + √3 is asked for.Leave answers in surd form whenever the question says exact or without a calculator.
Exam tips
- During practice, do a quick decimal check on a calculator to confirm that your surd answer has the same value as the original expression. In the exam, only do this when calculators are allowed.
- When asked for a + b√c, compare rational parts and surd parts separately. One equation with surds gives you two equations without them.
- Write √3 × √3 = 3 as a separate line when you expand brackets. It is an easy place to drop a mark.
Notes 3 of 10
Polynomials and Partial Fractions
Key idea. A polynomial is a sum of whole-number powers of x. The remainder theorem says that f(x) divided by (x - a) leaves remainder f(a), and the factor theorem says (x - a) is a factor when f(a) = 0. Use them to factorise and solve cubic equations. Partial fractions split a proper fraction into simpler fractions, one for each factor of the denominator.
Must know
- The remainder when f(x) is divided by (ax - b) is f(b/a)
- (x - a) is a factor of f(x) exactly when f(a) = 0
- a³ + b³ = (a + b)(a² - ab + b²) and a³ - b³ = (a - b)(a² + ab + b²)
- If the numerator's degree is at least the denominator's, divide before using partial fractions
What the syllabus asks you to do
- Multiply and divide polynomials
- Use the remainder theorem and the factor theorem, including factorising polynomials
- Solve cubic equations
- Use a³ + b³ = (a + b)(a² - ab + b²) and a³ - b³ = (a - b)(a² + ab + b²)
- Express a fraction in partial fractions when the denominator is no more complicated than (ax + b)(cx + d), (ax + b)(cx + d)² or (ax + b)(x² + c²)
Partial fraction forms for each type of denominator
| Factor in the denominator | Partial fractions to write |
|---|---|
| Distinct linear factors (ax + b)(cx + d) | A/(ax + b) + B/(cx + d) |
| Repeated linear factor (cx + d)², with (ax + b) | A/(ax + b) + B/(cx + d) + C/(cx + d)² |
| Quadratic factor x² + c², with (ax + b) | A/(ax + b) + (Bx + C)/(x² + c²) |
| Improper fraction | Divide first: a polynomial plus a proper fraction |
Try factors of the constant
For x³ - 2x² - 5x + 6, try ±1, ±2, ±3 and ±6.
Find a root
f(1) = 0, so (x - 1) is a factor.
Divide to get a quadratic
Long division or comparing coefficients gives x² - x - 6.
Factorise the quadratic
x² - x - 6 = (x - 3)(x + 2).
State all the roots
x = 1, x = 3 or x = -2.
Worked exam question
Express (3x² - 5x + 8)/((x - 2)(x² + 1)) in partial fractions.
- The denominator has a linear factor and the quadratic factor x² + 1, so write the fraction as A/(x - 2) + (Bx + C)/(x² + 1).
- Multiply both sides by (x - 2)(x² + 1): 3x² - 5x + 8 = A(x² + 1) + (Bx + C)(x - 2).
- Put x = 2: 12 - 10 + 8 = 5A, so 10 = 5A and A = 2.
- Compare the coefficients of x²: 3 = A + B, so B = 1.
- Compare the constants: 8 = A - 2C, so 8 = 2 - 2C and C = -3.
- Check with the x terms: the coefficient of x on the right is -2B + C = -2 - 3 = -5, which matches the left side.
Answer: 2/(x - 2) + (x - 3)/(x² + 1)
Common mistakes
- Using f(2) for the remainder when dividing by (x + 2).Set the divisor to zero first: x + 2 = 0 gives x = -2, so use f(-2).
- Substituting x = 2 for the divisor (2x - 1).Solve 2x - 1 = 0 to get x = 1/2, and substitute that.
- Writing B/(x² + 4) over a quadratic factor.Give a quadratic factor the numerator Bx + C, so the number of unknowns matches the degree of the denominator.
- Splitting (x² + 1)/((x - 1)(x + 2)) straight into partial fractions.Compare degrees first. If the numerator's degree is not lower, divide before splitting.
- Factorising a³ - b³ as (a - b)(a² - ab + b²).Expand to check: (a - b)(a² + ab + b²) = a³ - b³, so the middle sign is opposite.
- Stopping at x = 1 for x³ + 2x² + 3x - 6 = 0.Divide to get x² + 3x + 6, then solve it or show b² - 4ac = 9 - 24 < 0.
Exam tips
- Substituting the root of each linear factor finds most constants in one line each. Use comparing coefficients for the rest, then use one unused coefficient as a check.
- In factor theorem questions with two unknowns, label your equations (1) and (2), solve them simultaneously, then write out the polynomial in full before doing the next part.
- When the quadratic factor of a cubic has a negative discriminant, write 'b² - 4ac < 0, so there are no further real roots' rather than leaving the quadratic unsolved.
Notes 4 of 10
Binomial Theorem
Key idea. The binomial theorem expands (a + b)^(n), for a positive integer n, as the sum of nCr a^(n-r) b^(r) for r = 0 to n. Use the general term T(r+1) = nCr a^(n-r) b^(r) to find one term without writing the full expansion. For the term independent of x, combine the powers of x into one index and set it equal to zero.
Must know
- (a + b)^(n) has n + 1 terms, and the powers of a and b in each term add up to n
- General term T(r+1) = nCr a^(n-r) b^(r), so the third term has r = 2
- nC0 = nCn = 1, nC1 = n and nCr = nC(n - r)
What the syllabus asks you to do
- Use the binomial theorem to expand (a + b)^(n) for a positive integer n
- Use the notations n factorial and nCr
- Use the general term nCr a^(n-r) b^(r), for 0 ≤ r ≤ n, to find a particular term
- Use the general term to find a coefficient, such as the term independent of x
Pascal's triangle gives the values of nCr
| n | Coefficients nC0 to nCn |
|---|---|
| 1 | 1, 1 |
| 2 | 1, 2, 1 |
| 3 | 1, 3, 3, 1 |
| 4 | 1, 4, 6, 4, 1 |
| 5 | 1, 5, 10, 10, 5, 1 |
| 6 | 1, 6, 15, 20, 15, 6, 1 |
| 7 | 1, 7, 21, 35, 35, 21, 7, 1 |
Two curves for x from 0 to 0.3. Both start at 1 when x = 0 and fall. At x = 0.02, used to estimate 0.98 to the power 5, they are almost identical. By x = 0.1 the approximation gives 0.6 against the exact 0.59, and by x = 0.3 it gives 0.4 against 0.17, so the approximation only works for small x.
Worked exam question
(a) Write down the first three terms in the expansion of (1 - 2x)^(7) in ascending powers of x. (b) Hence find the coefficient of x² in the expansion of (3 + x)(1 - 2x)^(7).
- Use the binomial theorem with a = 1, b = -2x and n = 7.
- First term: 7C0 (1)^(7) = 1.
- Second term: 7C1 (1)^(6) (-2x) = 7 × (-2x) = -14x.
- Third term: 7C2 (1)^(5) (-2x)² = 21 × 4x² = 84x², so (1 - 2x)^(7) = 1 - 14x + 84x² + ...
- For (3 + x)(1 - 14x + 84x² + ...), the x² terms come from 3 × 84x² and x × (-14x).
- Add them: 252x² - 14x² = 238x².
Answer: (a) 1 - 14x + 84x². (b) The coefficient of x² is 238.
Common mistakes
- Writing the x² term of (1 - 2x)^(7) as 21x² instead of 84x².Put brackets round the whole of b: 7C2(-2x)² = 21 × 4x² = 84x².
- Using r = 3 for the third term.Write T(r+1) and solve r + 1 = 3, so r = 2.
- Writing (-3x)³ as 27x³.Raise the sign separately: (-1)³ = -1, so (-3x)³ = -27x³.
- Giving 720x² when asked for the coefficient of x².Underline 'term' or 'coefficient' in the question. The coefficient of x² in 720x² is 720.
- Setting 12 - 2r = 0 for (x² + 2/x)^(6).Combine every power of x into one index, x^(12-3r), before solving for r.
- Missing the x × (-14x) pair when finding the x² coefficient of (3 + x)(1 - 2x)^(7).List every pair whose powers add up to the target power before multiplying.
Exam tips
- Write the general term with the powers of x simplified into a single index, such as x^(12-3r), before you look for a particular power. This one line does most of the work.
- Check nCr values with the calculator, but show the general term in your working so the method is clear even if a number slips.
- In 'hence' questions, expand only as far as the power you need. Extra terms take time and add chances to make a sign error.
Notes 5 of 10
Exponential and Logarithmic Functions
Key idea. A logarithm is a power: logₐ y = x means a^(x) = y, for a > 0 and a ≠ 1. Use the product, quotient and power laws to combine logarithms, and the change of base rule when bases differ. Solve exponential equations by matching bases or taking logarithms of both sides, and reject any answer that gives the logarithm of zero or a negative number.
Must know
- logₐ(MN) = logₐ M + logₐ N and logₐ(M/N) = logₐ M - logₐ N
- logₐ(M^(k)) = k logₐ M, logₐ 1 = 0 and logₐ a = 1
- Change of base: logₐ b = (lg b)/(lg a), or use ln on the top and bottom
- a^(x) > 0 for every x, so an equation such as 2^(x) = -1 has no solution
What the syllabus asks you to do
- Use the functions a^(x), e^(x), logₐ x and ln x and their graphs
- Use the laws of logarithms
- Use the equivalence of y = a^(x) and x = logₐ y
- Change the base of a logarithm
- Simplify expressions and solve simple equations involving exponential and logarithmic functions
- Use exponential and logarithmic functions as models
Laws of logarithms with examples
| Law | Rule | Example |
|---|---|---|
| Product | logₐ(MN) = logₐ M + logₐ N | lg 2 + lg 5 = lg 10 = 1 |
| Quotient | logₐ(M/N) = logₐ M - logₐ N | log₂ 48 - log₂ 3 = log₂ 16 = 4 |
| Power | logₐ(M^(k)) = k logₐ M | log₃ 81 = 4 log₃ 3 = 4 |
| Change of base | logₐ b = (lg b)/(lg a) | log₄ 8 = (3 lg 2)/(2 lg 2) = 3/2 |
| Special values | logₐ 1 = 0 and logₐ a = 1 | ln 1 = 0 and ln e = 1 |
A curve rising ever more steeply for t from 0 to 10. It starts at N = 500 when t = 0, reaches about 1359 at t = 5 and about 3695 at t = 10. A dashed horizontal line at N = 2000 meets the curve at t = 5 ln 4, which is about 6.93.
Worked exam question
Solve the equation 2^(2x+1) - 5(2^(x)) + 2 = 0.
- Use the index laws: 2^(2x+1) = 2 × 2^(2x) = 2(2^(x))².
- Let u = 2^(x). The equation becomes 2u² - 5u + 2 = 0.
- Factorise: (2u - 1)(u - 2) = 0, so u = 1/2 or u = 2.
- 2^(x) = 1/2 = 2^(-1) gives x = -1.
- 2^(x) = 2 = 2¹ gives x = 1.
- Check x = 1: 2³ - 5(2) + 2 = 8 - 10 + 2 = 0. Check x = -1: 2⁻¹ - 5(1/2) + 2 = 1/2 - 5/2 + 2 = 0.
Answer: x = -1 or x = 1
Common mistakes
- Writing lg(x + 3) as lg x + lg 3.Use the product law only for lg(MN). A sum inside a logarithm stays as it is.
- Writing (lg 20)/(lg 3) as lg(20/3).Only lg 20 - lg 3 equals lg(20/3). Evaluate (lg 20)/(lg 3) directly as 2.73.
- Keeping x = -9 as a solution of log₃(x + 6) + log₃ x = 3.Substitute each answer into every log term, and reject any that give the log of zero or a negative number.
- Rewriting 2^(2x+1) as (2^(x))² + 1.Use a^(m+n) = a^(m) × a^(n), so 2^(2x+1) = 2 × (2^(x))².
- Trying to solve 2^(x) = -1 for x.Reject u = a^(x) ≤ 0 as soon as the quadratic in u is solved.
- Writing log₃(x²) as (log₃ x)².Bring the index down as a multiplier: log₃(x²) = 2 log₃ x.
Exam tips
- Look for a hidden quadratic whenever you see a^(2x) or a^(2x+k) together with a^(x). Substitute u = a^(x) and remember that u must be positive.
- When logarithms have different bases, convert everything to one base with the change of base rule before combining. Choosing the smaller base, such as 2 instead of 4, usually keeps the numbers whole.
- Give final answers from logarithms to 3 significant figures unless told otherwise, but keep full calculator values in the intermediate steps.
Notes 6 of 10
Trigonometric Identities and Equations
Key idea. Trigonometric identities are equations that are true for every angle, such as sin²A + cos²A = 1. Use them, with the addition and double angle formulae, to rewrite an expression in a single trigonometric function. To solve an equation, find the basic angle, use the quadrant signs to list every angle in the interval, and adjust the range first for angles such as 2x.
Must know
- sin²A + cos²A = 1, sec²A = 1 + tan²A and cosec²A = 1 + cot²A
- sin 2A = 2 sin A cos A and cos 2A = 2cos²A - 1 = 1 - 2sin²A
- a cos θ + b sin θ = R cos(θ - α), where R = √(a² + b²) and tan α = b/a
- y = a sin(bx) + c has amplitude |a| and period 360°/b
What the syllabus asks you to do
- Use the six trigonometric functions for angles of any size, in degrees or radians
- Use the principal values of sin⁻¹x, cos⁻¹x and tan⁻¹x
- Use exact values for 30°, 45° and 60°, or π/6, π/4 and π/3
- Use amplitude, period and symmetry to draw y = a sin(bx) + c, y = a cos(bx) + c and y = a tan(bx)
- Use sin²A + cos²A = 1, sec²A = 1 + tan²A, cosec²A = 1 + cot²A, the addition formulae and the double angle formulae
- Express a cos θ + b sin θ in the form R cos(θ ± α) or R sin(θ ± α)
- Simplify trigonometric expressions and prove simple trigonometric identities
- Solve simple trigonometric equations in a given interval
Exact values of the trigonometric ratios
| Angle | sin and cos | tan |
|---|---|---|
| 0° | sin = 0, cos = 1 | 0 |
| 30° (π/6) | sin = 1/2, cos = √3/2 | 1/√3 = √3/3 |
| 45° (π/4) | sin = cos = √2/2 | 1 |
| 60° (π/3) | sin = √3/2, cos = 1/2 | √3 |
| 90° (π/2) | sin = 1, cos = 0 | Undefined |
A cosine-shaped wave for θ from 0° to 360°. It starts at 3, rises to a maximum of 5 at θ = 53.1°, falls to a minimum of -5 at θ = 233.1° and returns to 3 at 360°. The dashed line y = 2 cuts the wave twice, at θ = 119.6° and θ = 346.7°, the two solutions.
Worked exam question
Express 3 cos θ + 4 sin θ in the form R cos(θ - α), where R > 0 and α is acute. Hence solve 3 cos θ + 4 sin θ = 2 for 0° ≤ θ ≤ 360°.
- Write 3 cos θ + 4 sin θ = R cos θ cos α + R sin θ sin α, so R cos α = 3 and R sin α = 4.
- R = √(3² + 4²) = 5 and tan α = 4/3, so α = 53.13°.
- The equation becomes 5 cos(θ - 53.13°) = 2, so cos(θ - 53.13°) = 0.4.
- Since 0° ≤ θ ≤ 360°, the angle θ - 53.13° lies between -53.13° and 306.87°.
- The basic angle is cos⁻¹ 0.4 = 66.42°. Cosine is positive in the first and fourth quadrants, so θ - 53.13° = 66.42° or 360° - 66.42° = 293.58°. The value -66.42° is outside the range.
- Add 53.13°: θ = 119.55° or θ = 346.71°.
Answer: 3 cos θ + 4 sin θ = 5 cos(θ - 53.1°), and θ = 119.6° or θ = 346.7° to 1 decimal place
Common mistakes
- Dividing sin x cos x = cos x by cos x to get sin x = 1.Bring all terms to one side and factorise: cos x(sin x - 1) = 0, then solve each factor.
- Solving sin 2x = 0.5 for 0° ≤ x ≤ 360° and giving only 15° and 75°.Write the new range, 0° ≤ 2x ≤ 720°, and list every angle in it before halving.
- Writing cos(A + B) = cos A cos B + sin A sin B.Learn the sign change, and test with A = B = 45°: cos 90° must come out as 0.
- Giving x = π/6 and x = 150° in the same answer.Read the units of the interval first and set the calculator mode to match before solving.
- Proving an identity by cross-multiplying both sides.Start from one side, usually the more complicated, and transform it step by step into the other.
- Writing tan α = 3/4 for 3 cos θ + 4 sin θ = R cos(θ - α).Expand R cos(θ - α) and compare: R sin α = 4 and R cos α = 3, so tan α = 4/3.
Exam tips
- Write down the range of the transformed angle, such as 0° ≤ 2x ≤ 720°, before solving. It tells you how many solutions to expect.
- When an equation mixes sin x with cos²x, or tan x with sec²x, use a Pythagorean identity to change it into a quadratic in one function.
- In R-formula questions, keep α to at least 2 decimal places in your working, and round only the final angles.
Notes 7 of 10
Coordinate Geometry of Circles
Key idea. A circle with centre (a, b) and radius r has equation (x - a)² + (y - b)² = r². The general form x² + y² + 2gx + 2fy + c = 0 has centre (-g, -f) and radius √(g² + f² - c). A tangent is perpendicular to the radius at the point of contact, and the perpendicular bisector of any chord passes through the centre.
Must know
- Standard form: (x - a)² + (y - b)² = r², with centre (a, b) and radius r
- General form: x² + y² + 2gx + 2fy + c = 0, centre (-g, -f), radius √(g² + f² - c)
- Tangent gradient = -1 ÷ (gradient of the radius) at the point of contact
What the syllabus asks you to do
- Use the equation of a circle in the form (x - a)² + (y - b)² = r²
- Use the equation of a circle in the form x² + y² + 2gx + 2fy + c = 0
- Use the conditions for two lines to be parallel or perpendicular
- Find the midpoint of a line segment
- Solve problems involving one circle and straight lines (problems involving two circles are not included)
Reading the centre and radius from a circle equation
| Equation | Centre | Radius |
|---|---|---|
| x² + y² = 25 | (0, 0) | 5 |
| (x - 3)² + (y + 2)² = 25 | (3, -2) | 5 |
| (x + 4)² + (y - 1)² = 9 | (-4, 1) | 3 |
| x² + y² - 6x + 4y - 12 = 0 | (3, -2) | √(9 + 4 + 12) = 5 |
| x² + y² + 8x - 2y - 8 = 0 | (-4, 1) | √(16 + 1 + 8) = 5 |
| x² + y² - 4x - 2y - 5 = 0 | (2, 1) | √(4 + 1 + 5) = √10 |
A quarter circle of radius 5 centred at the origin runs from (0, 5) to (5, 0). A dashed radius joins the origin to P(3, 4), with gradient 4/3. A straight tangent touches the circle only at P and crosses the axes at (0, 6.25) and about (8.33, 0). Its gradient is -3/4, so it is perpendicular to the radius.
Worked exam question
A circle passes through the points A(1, 4) and B(5, 2), and its centre lies on the line y = x - 1. Find the equation of the circle.
- The centre lies on the perpendicular bisector of the chord AB. The midpoint of AB is ((1 + 5)/2, (4 + 2)/2) = (3, 3).
- The gradient of AB is (2 - 4)/(5 - 1) = -1/2, so the perpendicular gradient is 2.
- The perpendicular bisector is y - 3 = 2(x - 3), which is y = 2x - 3.
- Solve with y = x - 1: 2x - 3 = x - 1, so x = 2 and y = 1. The centre is (2, 1).
- Radius squared = (1 - 2)² + (4 - 1)² = 1 + 9 = 10. Check with B: (5 - 2)² + (2 - 1)² = 9 + 1 = 10.
- The equation is (x - 2)² + (y - 1)² = 10, which expands to x² + y² - 4x - 2y - 5 = 0.
Answer: (x - 2)² + (y - 1)² = 10, or x² + y² - 4x - 2y - 5 = 0
Common mistakes
- Reading the centre of (x + 4)² + (y - 1)² = 9 as (4, -1).Set each bracket to zero: x + 4 = 0 and y - 1 = 0 give the centre (-4, 1).
- Giving the radius of (x + 4)² + (y - 1)² = 9 as 9.Always take the square root of the right side: r = √9 = 3.
- Finding the radius of x² + y² - 6x + 4y - 12 = 0 as √(9 + 4 - 12) = 1.Substitute c with its sign into √(g² + f² - c): √(9 + 4 + 12) = 5.
- Using 4/3 as the tangent gradient at (3, 4) on x² + y² = 25.The tangent is perpendicular to the radius, so take the negative reciprocal: -3/4.
- Reading 2g = -8 from 2x² + 2y² - 8x + 4y = 0.Divide every term by 2 first: x² + y² - 4x + 2y = 0, so g = -2 and f = 1.
- Expanding (x + 1)² as x² + 1 after substituting y = x + 1 into a circle.Write (x + 1)² = x² + 2x + 1 in full every time.
Exam tips
- Draw a quick sketch with the centre, the radius and any given points marked. It shows which fact to use, the perpendicular bisector or the tangent.
- When a circle touches the x-axis, the radius equals the size of the y-coordinate of the centre, and when it touches the y-axis, the radius equals the size of the x-coordinate. Questions often give the radius this way.
- After finding an equation, substitute one of the given points back in. It catches sign errors in the centre straight away.
Notes 8 of 10
Differentiation
Key idea. Differentiation finds the gradient function dy/dx, the rate at which y changes with x. For y = x^(n), dy/dx = nx^(n-1). Use the chain rule for a function inside another function, the product rule for uv and the quotient rule for u/v. With x in radians, sin x differentiates to cos x, e^(x) to e^(x) and ln x to 1/x.
Must know
- Power rule: the derivative of x^(n) is nx^(n-1), for any rational n
- Chain rule: dy/dx = dy/du × du/dx
- Product rule: (uv)' = uv' + vu'. Quotient rule: (u/v)' = (vu' - uv')/v²
- sin x gives cos x, cos x gives -sin x, tan x gives sec²x, e^(x) gives e^(x), ln x gives 1/x
What the syllabus asks you to do
- Understand the derivative as the gradient of the tangent at a point and as a rate of change
- Use the notations f'(x), f''(x), dy/dx and d²y/dx²
- Differentiate x^(n) for any rational n, sin x, cos x, tan x, e^(x) and ln x, with constant multiples, sums and differences
- Differentiate products and quotients of functions
- Use the chain rule
Standard derivatives, with x in radians
| y | dy/dx | With (ax + b) inside |
|---|---|---|
| x^(n) | nx^(n-1) | an(ax + b)^(n-1) |
| sin x | cos x | a cos(ax + b) |
| cos x | -sin x | -a sin(ax + b) |
| tan x | sec²x | a sec²(ax + b) |
| e^(x) | e^(x) | ae^(ax + b) |
| ln x | 1/x | a/(ax + b) |
A parabola y = x² rises from the origin to (4, 16). A straight tangent line touches it at the point (2, 4) and runs from (1, 0) to (4, 12), with gradient 4. Since dy/dx = 2x, the derivative at x = 2 gives the same gradient, 4.
Worked exam question
Given that y = x/√(2x + 1), show that dy/dx = (x + 1)/(2x + 1)^(3/2). Hence find the gradient of the curve at the point where x = 4.
- Use the quotient rule with u = x and v = (2x + 1)^(1/2).
- du/dx = 1. By the chain rule, dv/dx = 1/2 (2x + 1)^(-1/2) × 2 = (2x + 1)^(-1/2).
- dy/dx = [(2x + 1)^(1/2) × 1 - x(2x + 1)^(-1/2)]/(2x + 1), since v² = 2x + 1.
- Multiply the numerator and the denominator by (2x + 1)^(1/2) to clear the negative power: dy/dx = [(2x + 1) - x]/(2x + 1)^(3/2).
- Simplify the numerator: 2x + 1 - x = x + 1, so dy/dx = (x + 1)/(2x + 1)^(3/2), as required.
- At x = 4: dy/dx = 5/9^(3/2) = 5/27, since 9^(3/2) = (√9)³ = 27.
Answer: dy/dx = (x + 1)/(2x + 1)^(3/2), and the gradient at x = 4 is 5/27
Common mistakes
- Differentiating sin 4x as cos 4x.Ask 'what is inside?' every time and multiply by its derivative, giving 4 cos 4x.
- Differentiating x² sin x as 2x cos x.Name u and v first, then write u dv/dx + v du/dx = x² cos x + 2x sin x.
- Writing the quotient rule numerator as u dv/dx - v du/dx.Learn it as 'v du minus u dv', and test with y = x/(x + 1), whose derivative must be positive.
- Writing the derivative of cos x as sin x.Remember the cycle sin, cos, -sin, -cos, where each step is one derivative.
- Differentiating 1/x³ as 1/(3x²).Rewrite as x^(-3) first, giving -3x^(-4) = -3/x⁴.
- Differentiating ln(2x) as 1/(2x).Either use the chain rule, 2/(2x) = 1/x, or split ln(2x) = ln 2 + ln x first.
Exam tips
- Rewrite every root and fraction as a power of x before differentiating. Many power rule slips come from skipping this step.
- After using the product or quotient rule, take out common factors. 'Show that' answers are usually factorised, and it makes solving dy/dx = 0 later much easier.
- Use the laws of logarithms to split ln of a product, quotient or power before differentiating. It turns one long chain or quotient rule problem into simple terms.
Notes 9 of 10
Applications of Differentiation
Key idea. Applications of differentiation use dy/dx as a gradient and as a rate. The tangent at a point has gradient dy/dx there, and the normal has gradient -1/(dy/dx). Stationary points occur where dy/dx = 0: d²y/dx² < 0 gives a maximum and d²y/dx² > 0 a minimum. Connected rates of change use the chain rule, dy/dt = dy/dx × dx/dt.
Must know
- Tangent at x = a: gradient m = dy/dx at a. Normal: gradient -1/m
- Stationary point: dy/dx = 0. d²y/dx² < 0 for a maximum, d²y/dx² > 0 for a minimum
- Increasing where dy/dx > 0, decreasing where dy/dx < 0
- Connected rates of change: dy/dt = dy/dx × dx/dt
What the syllabus asks you to do
- Determine where a function is increasing or decreasing
- Find stationary points: maximum and minimum turning points and stationary points of inflexion
- Use the second derivative test to tell maxima from minima
- Apply differentiation to gradients, tangents and normals
- Apply differentiation to connected rates of change and to maximum and minimum problems
Classifying stationary points
| Test | Maximum | Minimum |
|---|---|---|
| Second derivative | d²y/dx² < 0 | d²y/dx² > 0 |
| Sign of dy/dx, left to right | Positive, zero, negative | Negative, zero, positive |
| Shape of the curve | Rises then falls | Falls then rises |
| If d²y/dx² = 0 | Use the first derivative test | Use the first derivative test |
A U-shaped curve. A is about 810 cm² when r = 2, falls to its lowest value of 150π, about 471 cm², at r = 5, then rises to about 785 cm² when r = 10. The lowest point at r = 5 is where dA/dr = 0, and the curve opening upwards matches d²A/dr² > 0.
Worked exam question
A closed cylinder has radius r cm, height h cm and volume 250π cm³. Show that its total surface area A cm² is given by A = 2πr² + 500π/r. Find the value of r for which A is stationary, show that this gives a minimum, and find the minimum surface area in terms of π.
- Volume: πr²h = 250π, so h = 250/r².
- Surface area: A = 2πr² + 2πrh = 2πr² + 2πr(250/r²) = 2πr² + 500π/r, as required.
- Differentiate: dA/dr = 4πr - 500π/r².
- For a stationary value, 4πr - 500π/r² = 0, so 4r³ = 500, r³ = 125 and r = 5.
- Second derivative: d²A/dr² = 4π + 1000π/r³. At r = 5 this is 4π + 8π = 12π > 0, so A is a minimum.
- Minimum A = 2π(25) + 500π/5 = 50π + 100π = 150π.
Answer: r = 5 cm, and the minimum surface area is 150π cm², about 471 cm² (the height is then 10 cm)
Common mistakes
- Giving the normal a gradient of -10 where the tangent gradient is 10.Use normal gradient = -1/(tangent gradient), here -1/10, and check that the two gradients multiply to -1.
- Stopping at x = 3 after solving dy/dx = 0.Substitute x into y for the coordinates, then find d²y/dx² to state the nature.
- Substituting r = 10 into A = πr² before differentiating.Differentiate the general formula first, then substitute the particular value.
- Trying to maximise A = xy with both x and y still unknown.Make one variable the subject of the constraint and substitute before differentiating.
- Calling a point an inflexion because d²y/dx² = 0 there.Check the sign of dy/dx on each side. For y = x⁴ at x = 0, this shows a minimum.
- Answering x = 20 when asked for the maximum area of the pen.Reread the question and substitute back: A = 20 × 40 = 800 m².
Exam tips
- In 'show that' maximum problems the expression is given, so even if you cannot show it, use the given expression to finish the rest of the question.
- Write the chain rule in full for rates of change, such as dr/dt = dV/dt ÷ dV/dr, and put units on the answer. It makes the method clear even if a number slips.
- To show that a function is increasing for all x, show that dy/dx is always positive, often by completing the square. For example, 3x² - 6x + 5 = 3(x - 1)² + 2, which is always at least 2.
Notes 10 of 10
Integration
Key idea. Integration reverses differentiation. For powers, ∫x^(n) dx = x^(n+1)/(n + 1) + c when n ≠ -1, and for a linear inside function you also divide by a, the coefficient of x. A definite integral F(b) - F(a) gives the area between a curve and the x-axis, with regions below the axis counting as negative, so split the integral where the curve crosses the axis.
Must know
- ∫x^(n) dx = x^(n+1)/(n + 1) + c, for n ≠ -1
- ∫(ax + b)^(n) dx = (ax + b)^(n+1)/(a(n + 1)) + c, for n ≠ -1
- Area between y = f(x) above and y = g(x) below is the integral of f(x) - g(x) from a to b
- An area below the x-axis gives a negative integral, so take its absolute value
What the syllabus asks you to do
- Understand integration as the reverse of differentiation
- Integrate x^(n) for any rational n (n ≠ -1), sin x, cos x, sec²x and e^(x), with constant multiples, sums and differences
- Integrate (ax + b)^(n) for any rational n (n ≠ -1), sin(ax + b), cos(ax + b) and e^(ax + b)
- Evaluate definite integrals and use them for the area under a curve
- Find the area of a region bounded by a curve and one or more lines, including regions below the x-axis
Standard integrals (add + c to each)
| f(x) | ∫f(x) dx | Example |
|---|---|---|
| x^(n), n ≠ -1 | x^(n+1)/(n + 1) | ∫x³ dx = x⁴/4 + c |
| (ax + b)^(n), n ≠ -1 | (ax + b)^(n+1)/(a(n + 1)) | ∫(2x - 5)⁴ dx = (2x - 5)⁵/10 + c |
| sin(ax + b) | -(1/a) cos(ax + b) | ∫sin 3x dx = -(1/3) cos 3x + c |
| cos(ax + b) | (1/a) sin(ax + b) | ∫cos 2x dx = ½ sin 2x + c |
| sec²x | tan x | ∫3 sec²x dx = 3 tan x + c |
| e^(ax + b) | (1/a) e^(ax + b) | ∫e^(3x) dx = e^(3x)/3 + c |
An upside-down U-shaped curve rises from (0, -5) to its highest point (3, 4) and falls to (5, 0). A straight line rises from (0, -1) to (5, 4). They cross at (1, 0) and (4, 3). Between these points the curve lies above the line, and the region between them has area 9/2.
Worked exam question
The line y = x - 1 meets the curve y = 6x - x² - 5 at the points A and B. Find the x-coordinates of A and B, and find the area of the region enclosed by the line and the curve.
- At the points of intersection, 6x - x² - 5 = x - 1, so x² - 5x + 4 = 0.
- Factorise: (x - 1)(x - 4) = 0, so A and B have x = 1 and x = 4.
- Between x = 1 and x = 4 the curve is above the line. For example, at x = 2 the curve gives 3 and the line gives 1.
- Area = integral from 1 to 4 of [(6x - x² - 5) - (x - 1)] dx = integral from 1 to 4 of (5x - x² - 4) dx.
- Integrate: [5x²/2 - x³/3 - 4x] from 1 to 4.
- At x = 4: 40 - 64/3 - 16 = 8/3. At x = 1: 5/2 - 1/3 - 4 = -11/6.
- Area = 8/3 - (-11/6) = 16/6 + 11/6 = 27/6 = 9/2.
Answer: x = 1 and x = 4, and the enclosed area is 9/2 = 4.5 units²
Common mistakes
- Writing ∫2x dx = x² with no + c.Add + c to every indefinite integral as you write the final line.
- Writing ∫cos 2x dx = sin 2x + c.Differentiate your answer to check: sin 2x gives 2 cos 2x, so the answer must be ½ sin 2x + c.
- Writing ∫sin x dx = cos x + c.Check by differentiating: -cos x differentiates to sin x, so ∫sin x dx = -cos x + c.
- Finding the area between y = x² - 4x and the x-axis from x = 0 to x = 5 in one integral.Sketch first, split at each crossing point, and add the sizes of the separate areas.
- Treating ∫x(x + 1) dx as ∫x dx × ∫(x + 1) dx.Expand to x² + x first, then integrate term by term.
- Applying the power rule to ∫x^(-1) dx and dividing by zero.Check the power before integrating. The power rule for integration does not cover 1/x.
Exam tips
- Differentiate your answer as a check. It takes a few seconds and catches a missing 1/a or a wrong sign.
- For area questions, sketch the curve and mark where it crosses the x-axis before you integrate. The sketch tells you whether to split the integral.
- Keep exact values such as 9/2 or π/4 until the end, and round only if the question asks for a decimal.
Syllabus checklist for the rest
What the syllabus asks you to do in the chapters, and parts of chapters, that do not have full notes yet. Each point is the syllabus learning outcome in plain words. Tick each one you can do without looking.
Chapter 8. Coordinate Geometry in Two Dimensions: also in the syllabus
2 points to check
- Find the area of a rectilinear figure from the coordinates of its vertices.
- Change relationships such as y = axⁿ and y = kbˣ to a straight-line form and find the unknown constants from the graph.
Chapter 9. Proofs in Plane Geometry
1 point to check
- Write proofs using the properties of parallel lines, perpendicular and angle bisectors, triangles, special quadrilaterals and circles, congruent and similar triangles, the midpoint theorem and the tangent-chord (alternate segment) theorem.
Chapter 10. Differentiation and Integration: also in the syllabus
1 point to check
- Apply differentiation and integration to the displacement, velocity and acceleration of a particle moving in a straight line.
If your child is struggling
- Find the exact chapter and point. The checklists and common mistakes on this page show where an answer breaks down.
- Go back to that chapter's worked question and do it again without the steps.
- Ask the school teacher which chapters the class has covered and which are tested next.
- If several chapters are weak, work on one at a time, with short timed questions, rather than rereading everything.