IGCSE Additional Mathematics 0606 Paper 1 Second Practice Paper (10 questions, 80 marks, 2 hours)

2026-10-31

Practice paper · IGCSE Additional Mathematics 0606

Free original TutorZone IGCSE Additional Mathematics 0606 Paper 1 second practice paper: 10 questions, 80 marks, 2 hours, covering Venn diagrams, quadratic inequalities, the remainder theorem, partial fractions, logarithmic equations, circles and tangents, trigonometric equations, the R-formula, kinematics and permutations, with full worked answers and marking notes in three languages.

Download　Question paper PDF　|　Answer key PDF

Paper details　Level: IGCSE | Subject: Additional Mathematics | Time: 120 minutes | Marks: 80 | Questions: 10

Section A: Short questions (32 marks)

Answer all questions. Show your working.

#	Question	Answer and marking notes
1	In a survey of 40 students, 24 like Mathematics (M), 18 like Physics (P) and 8 like both. (a) Draw a Venn diagram and find the number who like at least one subject. (3 marks) (b) Find the number who like neither subject. (2 marks) (c) A student is chosen at random. Find the probability that the student likes exactly one subject. (3 marks)	(a) n(M or P) = 24 + 18 – 8 = 34. (3 marks) (b) 40 – 34 = 6. (2 marks) (c) 16 like only Mathematics, 10 like only Physics, so 26 in total; probability = 26/40 = 13/20 = 0.65. (3 marks)
2	(a) Solve the inequality x^2 – 5x + 6 <= 0. (3 marks) (b) Find the range of values of k for which the quadratic equation 2x^2 + kx + 8 = 0 has real roots. (5 marks)	(a) (x – 2)(x – 3) <= 0, so 2 <= x <= 3. (3 marks) (b) Discriminant = k^2 - 64 >= 0, so k <= -8 or k >= 8. (5 marks: 2 for the discriminant, 3 for solving)
3	Let P(x) = 2x^3 + 3x^2 – 3x – 2. (a) Show that (x – 1) is a factor of P(x). (2 marks) (b) Factorise P(x) completely into a product of linear factors. (3 marks) (c) Solve P(x) = 0. (3 marks)	(a) P(1) = 2 + 3 – 3 – 2 = 0, so (x – 1) is a factor. (2 marks) (b) Division gives quotient 2x^2 + 5x + 2 = (2x + 1)(x + 2), so P(x) = (x – 1)(2x + 1)(x + 2). (3 marks) (c) x = 1, x = -1/2, x = -2. (3 marks)
4	(a) Express (5x + 7)/((x + 1)(x + 2)) in partial fractions. (4 marks) (b) Hence find the exact value of the integral of (5x + 7)/((x + 1)(x + 2)) from 0 to 1. (4 marks)	(a) Write it as A/(x + 1) + B/(x + 2), so A(x + 2) + B(x + 1) = 5x + 7. Comparing coefficients, A + B = 5 and 2A + B = 7, giving A = 2 and B = 3. (4 marks) (b) The integral is [2 ln(x + 1) + 3 ln(x + 2)] from 0 to 1 = (2 ln 2 + 3 ln 3) – 3 ln 2 = 3 ln 3 – ln 2 = ln(27/2), about 2.603. (4 marks)

Section B: Logarithms, circles and trigonometry (26 marks)

Answer all questions. Give answers to 3 significant figures unless stated otherwise.

#	Question	Answer and marking notes
5	Solve log_2 x + log_2 (x – 3) = 2. (a) Show that this equation can be written as x^2 – 3x – 4 = 0. (4 marks) (b) Hence solve the equation and explain why one root must be rejected. (5 marks)	(a) Combining the logarithms, log_2[x(x – 3)] = 2, so x(x – 3) = 4, i.e. x^2 – 3x – 4 = 0. (4 marks) (b) (x – 4)(x + 1) = 0 gives x = 4 or x = -1. Since logarithms need x > 0 and x – 3 > 0, x = -1 is rejected, so x = 4. Check: log_2 4 + log_2 1 = 2. (5 marks)
6	The circle C has equation x^2 + y^2 – 8x + 6y – 24 = 0. (a) Find the centre and radius of C. (4 marks) (b) The point P(10, 5) lies outside C. A tangent from P touches C at T. Find the length PT, exact and correct to 3 decimal places. (5 marks)	(a) (x – 4)^2 + (y + 3)^2 = 49, so the centre is (4, -3) and the radius is 7. (4 marks) (b) PC^2 = 36 + 64 = 100, so PT^2 = 100 – 49 = 51 and PT = sqrt(51), about 7.141. (5 marks)
7	(a) Solve 2cos^2(theta) – cos(theta) – 1 = 0 for 0 <= theta <= 360 degrees. (5 marks) (b) Hence solve 2cos^2(2x) - cos(2x) - 1 = 0 for 0 <= x <= 180 degrees. (3 marks)	(a) (2cos(theta) + 1)(cos(theta) – 1) = 0, so cos(theta) = 1 or -1/2, giving theta = 0, 120, 240, 360 degrees. (5 marks) (b) From (a), 2x = 0, 120, 240, 360 degrees within range, so x = 0, 60, 120, 180 degrees. (3 marks)

Section C: Applications (22 marks)

Answer all questions. Show full working and include units.

#	Question	Answer and marking notes
8	Let f(theta) = 3sin(theta) + 4cos(theta). (a) Express f(theta) in the form R sin(theta + alpha), where R > 0 and alpha lies between 0 and 90 degrees, giving alpha correct to 2 decimal places. (4 marks) (b) Hence find the maximum value of f(theta) and the corresponding theta for 0 <= theta <= 360 degrees. (3 marks)	(a) R = 5 and tan(alpha) = 4/3 gives alpha = 53.13 degrees, so f(theta) = 5sin(theta + 53.13 degrees). (4 marks) (b) The maximum is 5, when theta + 53.13 = 90, i.e. theta = 36.87 degrees. (3 marks)
9	A particle moves in a straight line. Its displacement s metres after t seconds is s = t^3 – 6t^2 + 9t, t >= 0. (a) Find expressions for the velocity v and the acceleration a. (2 marks) (b) Find the times at which the particle is instantaneously at rest. (2 marks) (c) Find the total distance travelled in the first 2 seconds. (4 marks)	(a) v = 3t^2 – 12t + 9 and a = 6t – 12. (2 marks) (b) 3(t – 1)(t – 3) = 0, so t = 1 s and t = 3 s. (2 marks) (c) s(1) = 4 and s(2) = 2, with a change of direction at t = 1; distance = 4 + 2 = 6 m. (4 marks)
10	Five boys and four girls stand in a row. (a) In how many ways can they be arranged if the four girls must stand together? (3 marks) (b) If all arrangements are equally likely, find the probability that the four girls stand together. (4 marks)	(a) Treat the four girls as one block, giving 6 units with the 5 boys: 6! x 4! = 720 x 24 = 17,280. (3 marks) (b) There are 9! = 362,880 arrangements, so the probability is 17,280/362,880 = 1/21, about 0.0476. (4 marks)

Find a tutor — free

A TutorZone practice paper; original questions and passages, not publisher material.