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IGCSE Additional Mathematics 0606 Paper 1 (10 Questions, 100 Marks, 60 Minutes)
IGCSE Add Maths 0606 Paper 1
IGCSE Additional Mathematics 0606 Paper 1 practice paper (10 questions, 100 marks, 60 minutes): . All questions are original, with full answers and marking notes.
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Paper details Level: IGCSE | Subject: Additional Mathematics | Paper: 1 | Time: 60 minutes | Marks: 100 | Questions: 10
Section A Functions, Quadratics, Indices and Binomial (44 marks)
Questions 1-2
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | 1. f(x) = 2x + 3 and g(x) = x^2 minus 1. (a) Find gf(2). (2 marks) (b) Find f inverse of x and solve f inverse of x = 5. (3 marks) (c) Solve fg(x) = 15. (3 marks) |
(a) f(2) = 7, so g(7) = 48. (2 marks) (b) f^-1(x) = (x – 3)/2; f^-1(x) = 5 gives x = 13. (3 marks) (c) fg(x) = 2(x^2 – 1) + 3 = 2x^2 + 1 = 15, so x^2 = 7 and x = plus or minus sqrt 7. (3 marks) Marking notes: reversing the composition order scores 0; (b) answer only scores at most 2. |
| 2 | 2. The curve y = 3x^2 – 12x + 7. (a) Express it in the form a(x – h)^2 + k. (3 marks) (b) State the coordinates of the vertex and the range of y. (2 marks) (c) Find the set of values of k for which 3x^2 – 12x + 7 = k has two distinct real roots. (3 marks) |
(a) 3(x – 2)^2 – 5. (3 marks) (b) Vertex (2, -5); y >= -5. (2 marks) (c) k > -5. (3 marks) Marking notes: losing the factor 3 in completing the square scores at most 1; (c) needs a discriminant reason. |
Questions 3-4
| # | Question | Answer and marking notes |
|---|---|---|
| 3 | 3. (a) Simplify (27^(2/3) x 16^(3/4)) / 8^(1/3). (3 marks) (b) Rationalise 6/(sqrt 5 – sqrt 2). (3 marks) (c) Solve 2^(2x + 1) = 8^(x – 1). (3 marks) |
(a) 27^(2/3) = 9, 16^(3/4) = 8, 8^(1/3) = 2, so 9 x 8 / 2 = 35. (3 marks) (b) Multiply by (sqrt 5 + sqrt 2)/(sqrt 5 + sqrt 2): 6(sqrt 5 + sqrt 2)/3 = 2(sqrt 5 + sqrt 2). (3 marks) (c) The right side is 2^(3x – 3), so 2x + 1 = 3x – 3 and x = 4. (3 marks) Marking notes: (b) a denominator still containing surds scores 0; (c) no common base scores at most 1. |
| 4 | 4. f(x) = x^3 + 2x^2 – 5x – 6. (a) Show that (x – 2) is a factor of f(x). (3 marks) (b) Factorise f(x) completely into linear factors. (4 marks) (c) Solve f(x) = 0. (2 marks) |
(a) f(2) = 8 + 8 – 10 – 6 = 0, so (x – 2) is a factor. (3 marks) (b) Division gives x^2 + 4x + 3 = (x + 1)(x + 3), so f(x) = (x – 2)(x + 1)(x + 3). (4 marks) (c) x = 2, -1 or -3. (2 marks) Marking notes: (b) partial factorisation scores at most 2; (c) must agree with (b). |
Question 5
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | 5. (a) Write down the first three terms of the expansion of (1 + 2x)^6. (3 marks) (b) Find the coefficient of x^3. (3 marks) (c) Hence find the coefficient of x^3 in (1 + 2x)^6(1 – x). (4 marks) |
(a) 1 + 12x + 60x^2. (3 marks) (b) C(6,3) x 2^3 = 20 x 8 = 160. (3 marks) (c) 160 – 60 = 100. (4 marks) Marking notes: (b) omitting 2^3 and answering 20 scores 1; (c) full marks for a correct fresh expansion. |
Section B Logarithms, Circular Measure, Trigonometry and Calculus (56 marks)
Questions 6-7
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | 6. (a) Solve log base 3 of x + log base 3 of (x – 2) = 1. (4 marks) (b) Solve 5^x = 2^(x + 1), giving your answer to 3 significant figures. (3 marks) (c) Write log base a of the square root of (a^3 b) in terms of log base a of b. (3 marks) |
(a) x(x – 2) = 3, so x^2 – 2x – 3 = 0 and x = 3 or -1; x = -1 is invalid, so x = 3. (4 marks) (b) x log 5 = (x + 1) log 2, so x = log 2/(log 5 – log 2) = 0.756. (3 marks) (c) 3/2 + (1/2) log base a of b. (3 marks) Marking notes: (a) failing to reject x = -1 loses 1; (c) leaving (3/2) log base a of a unsimplified scores at most 2. |
| 7 | 7. A sector has radius 10 cm and angle 1.2 radians. (a) Find the arc length and the perimeter. (3 marks) (b) Find the area of the sector. (3 marks) (c) Find the area of the corresponding segment, to 3 significant figures. (4 marks) |
(a) Arc = 10 x 1.2 = 12 cm; perimeter = 12 + 20 = 32 cm. (3 marks) (b) Area = (1/2) x 100 x 1.2 = 60 cm^2. (3 marks) (c) Triangle area = (1/2) x 100 x sin 1.2 = 46.6 cm^2, so segment = 60 – 46.6 = 13.4 cm^2. (4 marks) Marking notes: the calculator must be in radian mode; (c) degree mode scores 1. |
Questions 8-9
| # | Question | Answer and marking notes |
|---|---|---|
| 8 | 8. (a) Show that 2sin^2(theta) + 3cos(theta) – 3 = 0 can be written as 2cos^2(theta) – 3cos(theta) + 1 = 0. (3 marks) (b) Solve 2cos^2(theta) – 3cos(theta) + 1 = 0 for 0 <= theta <= 360 degrees. (5 marks) (c) Hence solve 2sin^2(2x) + 3cos(2x) - 3 = 0 for 0 <= x <= 180 degrees. (4 marks) |
(a) Using sin^2 = 1 – cos^2: 2 – 2cos^2 + 3cos – 3 = 0, that is 2cos^2 – 3cos + 1 = 0. (3 marks) (b) (2cos – 1)(cos – 1) = 0, so cos(theta) = 1/2 or 1, giving 60, 300, 0 and 360 degrees. (5 marks) (c) 2x = 0, 60, 300 or 360 degrees, so x = 0, 30, 150 or 180 degrees. (4 marks) Marking notes: (b) missing 0 and 360 scores at most 3; (c) ignoring the range for 2x scores at most 2. |
| 9 | 9. The curve C has equation y = x^3 – 6x^2 + 9x + 1. (a) Find dy/dx. (2 marks) (b) Find the coordinates of the stationary points of C. (4 marks) (c) Determine the nature of each stationary point. (3 marks) (d) Find the equation of the tangent to C at x = 2. (3 marks) |
(a) dy/dx = 3x^2 – 12x + 9. (2 marks) (b) 3x^2 – 12x + 9 = 0 gives (x – 1)(x – 3) = 0, so the points are (1, 5) and (3, 1). (4 marks) (c) d2y/dx2 = 6x – 12, which is -6 at x = 1 (maximum) and +6 at x = 3 (minimum). (3 marks) (d) At x = 2, y = 3 and the gradient is -3, so y = -3x + 9. (3 marks) Marking notes: (b) x-values without y-coordinates lose 2; (c) needs a second derivative or sign table. |
Question 10
| # | Question | Answer and marking notes |
|---|---|---|
| 10 | 10. A particle has velocity v = (3t^2 – 8t + 4)i + 2t j m/s and is at the origin when t = 0. (a) Find the acceleration vector at t = 2 and its magnitude to 3 significant figures. (4 marks) (b) Find the times when the i-component of the velocity is zero. (4 marks) (c) Find the displacement vector at t = 3 and the distance from the origin. (4 marks) |
(a) a = (6t – 8)i + 2j; at t = 2, a = 4i + 2j and the magnitude is sqrt 20 = 4.47 m/s^2. (4 marks) (b) 3t^2 – 8t + 4 = (3t – 2)(t – 2) = 0, so t = 2/3 or t = 2. (4 marks) (c) x = t^3 – 4t^2 + 4t gives x = 3 at t = 3; y = t^2 = 9, so the displacement is 3i + 9j and the distance is sqrt 90 = 9.49 m. (4 marks) Marking notes: (b) a single time scores at most 2; (c) substituting t directly without integrating scores 0. |
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