IGCSE Additional Mathematics 0606 Paper 1 Practice Paper | 10 Pure Mathematics Questions, 100 Marks

2026-11-01

IGCSE Additional Mathematics

IGCSE Additional Mathematics 0606 Paper 1 practice paper: surds and indices, quadratic functions and the discriminant, logarithmic equations, the binomial theorem, differentiation and tangents, definite integrals and areas, coordinate geometry and circles, trigonometric equations, simultaneous equations and the linear law, and vectors; 10 questions for 100 marks in 2 hours with full answers and marking notes.

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Paper details　Level: IGCSE (S4-S5) | Subject: Additional Mathematics | Time: 2 hours | Marks: 100 | Questions: 10

Section A: Surds, indices, quadratic functions and logarithms (Questions 1-3, 30 marks)

1. Show all working; give numerical answers to 3 significant figures (angles to 1 decimal place). 2. This paper has 10 questions, 100 marks in total, time allowed 2 hours. 3. Do not use any equation-solving function of a calculator where the examination rules forbid it.

#	Question	Answer and marking notes
1	(a) Simplify √72 + √50 − √98, giving the answer in the form a√2. (3 marks) (b) Solve 2^(3x-1) = 16^(x-2). (3 marks) (c) Rationalise 5 divided by (3 – √2), giving the answer in the form p + q√2. (4 marks)	(a) √72 = 6√2, √50 = 5√2, √98 = 7√2, so the sum is 4√2. (b) Since 16 = 2^4, 3x – 1 = 4(x – 2), so 3x – 1 = 4x – 8 and x = 7. (c) Multiply top and bottom by (3 + √2): denominator 9 – 2 = 7, numerator 15 + 5√2, so the value is 2.1 + 0.714√2.
2	(a) Express 2x^2 – 8x + 5 in the form a(x – h)^2 + k and state the minimum value and the value of x at which it occurs. (4 marks) (b) Find the discriminant of 2x^2 – 8x + 5 = 0 and describe the nature of its roots. (3 marks) (c) Given that kx^2 + 4x + 2 = 0 has equal roots, find k. (3 marks)	(a) 2x^2 – 8x + 5 = 2[(x – 2)^2 – 4] + 5 = 2(x – 2)^2 – 3; the minimum value is -3, taken at x = 2. (b) The discriminant is 64 – 40 = 24, which is positive, so there are two distinct real roots. (c) Equal roots require a zero discriminant: 16 – 8k = 0, so k = 2.
3	(a) Find log base 2 of 32. (2 marks) (b) Solve log base 2 of (x – 1) equals 3. (3 marks) (c) Solve log base 5 of x plus log base 5 of (x – 4) equals 1. (3 marks) (d) Solve 3^x = 5, giving the answer to 3 significant figures. (2 marks)	(a) 32 = 2^5, so the logarithm is 5. (b) x – 1 = 8, so x = 9. (c) x(x – 4) = 5, so x^2 – 4x – 5 = 0 and (x – 5)(x + 1) = 0; x = -1 is rejected because the logarithm is undefined there, so x = 5. (d) Taking logarithms, x = ln 5 divided by ln 3 = 1.465.

Section B: Binomial theorem, differentiation and integration (Questions 4-6, 30 marks)

#	Question	Answer and marking notes
4	(a) Find the coefficient of x^3 in the expansion of (2 + x)^5. (4 marks) (b) Find the coefficient of x^2 in the expansion of (1 – 3x)^6. (4 marks) (c) State the constant term and the coefficient of x in that expansion. (2 marks)	(a) The general term gives C(5,3)(2^2)(x^3), so the coefficient is 10 times 4 = 40. (b) The general term gives C(6,2)(-3x)^2, so the coefficient is 15 times 9 = 135. (c) The constant term is 1 and the coefficient of x is 6 times (-3) = -18.
5	(a) Given y = 3x^3 – 4x^2 + 2x – 7, find dy/dx and its value at x = 2. (4 marks) (b) Find the tangent to the curve at x = 2, giving the equation in general form. (3 marks) (c) Find the stationary points of y = x^3 – 6x^2 + 9x and determine their nature. (3 marks)	(a) dy/dx = 9x^2 – 8x + 2; at x = 2 this is 36 – 16 + 2 = 22. (b) At x = 2, y = 24 – 16 + 4 – 7 = 5; the tangent is y – 5 = 22(x – 2), that is 22x – y – 39 = 0. (c) dy/dx = 3(x – 1)(x – 3), so the stationary points are x = 1 with y = 4 and x = 3 with y = 0; the second derivative 6x – 12 equals -6 at x = 1 (maximum) and 6 at x = 3 (minimum).
6	(a) Find the indefinite integral of 3x^2 – 2x + 1 with respect to x. (3 marks) (b) Hence evaluate the definite integral from 0 to 2 of 3x^2 – 2x + 1. (4 marks) (c) Find the area enclosed between y = x^2 and y = x from x = 0 to x = 1. (3 marks)	(a) The integral is x^3 – x^2 + x + C. (b) Evaluating gives (8 – 4 + 2) – 0 = 6. (c) The area is 1/2 – 1/3 = 1/6, about 0.167 square units.

Section C: Coordinate geometry and trigonometric equations (Questions 7-8, 20 marks)

#	Question	Answer and marking notes
7	A(1, 2) and B(7, 10) are given. (a) Find the midpoint and the length of AB. (4 marks) (b) Find the gradient of AB and the equation of its perpendicular bisector in general form. (3 marks) (c) Find the equation of the circle with AB as diameter. (3 marks)	(a) The midpoint is (4, 6); AB has length the square root of 36 + 64 = 10. (b) The gradient of AB is 4/3, so the perpendicular gradient is -3/4; y – 6 = -3/4(x – 4), that is 3x + 4y – 36 = 0. (c) The centre is (4, 6) and the radius is 5, so (x – 4)^2 + (y – 6)^2 = 25.
8	(a) Solve 2sin^2 x – sin x – 1 = 0 for 0 to 360 degrees. (5 marks) (b) Solve tan x = -1 for 0 to 360 degrees. (2 marks) (c) Solve 3sin x = 2, giving answers to 1 decimal place. (3 marks)	(a) With s = sin x, (2s + 1)(s – 1) = 0, so s = -1/2 or s = 1; the solutions are 90, 210 and 330 degrees. (b) x = 135 degrees and 315 degrees. (c) sin x = 2/3, so x = 41.8 degrees or 138.2 degrees.

Section D: Simultaneous equations, linear law and vectors (Questions 9-10, 20 marks)

#	Question	Answer and marking notes
9	(a) Solve the simultaneous equations y = x + 1 and x^2 + y^2 = 25. (5 marks) (b) The graph of y = a(x^n) becomes a straight line when log y is plotted against log x; state the gradient and the intercept. (5 marks)	(a) Substituting gives 2x^2 + 2x + 1 = 25, so x^2 + x – 12 = 0 and (x – 3)(x + 4) = 0; the solutions are (3, 4) and (-4, -3). (b) Taking logarithms gives log y = n log x + log a, so the gradient is n and the intercept is log a; then a equals 10 raised to the intercept.
10	Given a = 3i – j and b = i + 2j. (a) Find the magnitudes of a and b. (4 marks) (b) Find the vector 2a – 3b and its magnitude. (4 marks) (c) Find the unit vector in the direction of a. (2 marks)	(a) The magnitude of a is the square root of 10, about 3.162; that of b is the square root of 5, about 2.236. (b) 2a – 3b = 3i – 8j, with magnitude the square root of 73, about 8.544. (c) The unit vector is (3i – j) divided by the square root of 10, about 0.949i – 0.316j.

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