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IB Maths AA SL Paper 2 Mock Paper (10 questions, 80 marks, calculator)
IB Mathematics AA
IB Mathematics Analysis and Approaches SL Paper 2 mock paper: 10 questions, 80 marks, 90 minutes, calculator allowed, covering sequences, exponentials and logarithms, quadratics, trigonometry, probability, calculus, the normal distribution, vectors, discrete distributions and exponential models, with full worked answers and marking notes.
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Paper details Level: IBDP | Subject: Mathematics AA SL | Time: 90 minutes | Marks: 80 | Questions: 10 | Calculator allowed
Section A (Questions 1 to 5, 40 marks)
Answer ALL questions. A calculator (including a graphical calculator) is allowed. Unless told otherwise, give numerical answers to 3 significant figures and angles to 1 decimal place.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | The arithmetic sequence {un} has u1 = 7 and common difference d = 5; the geometric sequence {vn} has v1 = 200 and common ratio r = 0.8. (a) Find u20.(b) Find the sum of the first 20 terms, S20. (c) Find v5. (d) Find the sum to infinity Sinf of the geometric sequence. | (a) u20 = 7 + 19 x 5 = 102.(b) S20 = 20 / 2 x (2 x 7 + 19 x 5) = 10 x 109 = 1090. (c) v5 = 200 x 0.84 = 200 x 0.4096 = 81.92. (d) Since |r| = 0.8 is less than 1, Sinf = 200 / (1 – 0.8) = 1000. |
| 2 | (a) Solve 32x-1 = 27x-2. (b) Solve log2x + log2(x – 2) = 3. (c) Find the value of log5(1/125). (d) Simplify (ln 8 + ln 2) / ln 4. (e) Solve 2x = 50, giving your answer to 3 significant figures. | (a) Since 27 = 33, 2x – 1 = 3x – 6, so x = 5. (b) The domain needs x > 2; x(x – 2) = 8 gives x2 – 2x – 8 = 0, (x – 4)(x + 2) = 0, so x = 4 (x = -2 rejected), hence x = 4. (c) 1/125 = 5-3, so the value is -3. (d) ln 8 + ln 2 = ln 16 and ln 16 / ln 4 = log416 = 2. (e) x = log250 = ln 50 / ln 2 = 5.64 (3 s.f.). |
| 3 | Let f(x) = 2x2 – 8x + 5. (a) Find the coordinates of the vertex. (b) Find the discriminant and state the number of real roots of f(x) = 0. (c) Solve f(x) = 0, giving answers to 3 significant figures. (d) Write down the equation of the axis of symmetry and the minimum value. | (a) x = 8 / 4 = 2 and f(2) = 8 – 16 + 5 = -3, so the vertex is (2, -3). (b) Discriminant = 64 – 40 = 24, which is greater than 0, so there are two distinct real roots. (c) x = (8 +/- sqrt 24) / 4 = 2 +/- 1.2247, so x = 3.22 or x = 0.775 (3 s.f.). (d) Axis of symmetry x = 2; minimum value -3. |
| 4 | In triangle ABC, a = 8 cm, b = 11 cm and angle C = 35 degrees. (a) Find the length of c (3 s.f.). (b) Find the area of the triangle (3 s.f.). (c) Find angle A and angle B (1 d.p.). | (a) By the cosine rule, c2 = 64 + 121 – 176 cos 35 = 40.829, so c = 6.39 cm (3 s.f.). (b) Area = 0.5 x 8 x 11 x sin 35 = 25.2 cm2 (3 s.f.). (c) By the sine rule, sin A = 8 sin 35 / 6.390 = 0.7181, so A = 45.9 degrees and B = 180 – 35 – 45.87 = 99.1 degrees. |
| 5 | A bag contains 6 red balls and 4 blue balls. (a) Two balls are drawn without replacement; find the probability that both are red. (b) Hence find the probability that at least one ball is blue. (c) If draws are made with replacement, find the probability that exactly 2 of 3 draws are red. (d) For 200 draws with replacement, find the expected number of red balls. | (a) 6/10 x 5/9 = 1/3 = 0.333. (b) 1 – 1/3 = 2/3 = 0.667. (c) 3 x 0.62 x 0.4 = 0.432. (d) 200 x 0.6 = 120. |
Section B (Questions 6 to 10, 40 marks)
Answer ALL questions. A calculator (including a graphical calculator) is allowed. Unless told otherwise, give numerical answers to 3 significant figures and angles to 1 decimal place.
| # | Question | Answer and marking notes |
|---|---|---|
| 6 | The curve has equation y = x3 – 6x2 + 9x. (a) Find dy/dx and the x-coordinates of the stationary points. (b) Use the second derivative to classify each stationary point and find its y value. (c) Evaluate the integral from 0 to 2 of (3x2 – 4x + 1) dx. | (a) dy/dx = 3x2 – 12x + 9 = 3(x – 1)(x – 3), so stationary points at x = 1 and x = 3. (b) y” = 6x – 12; at x = 1, y” = -6 (maximum, y = 4); at x = 3, y” = 6 (minimum, y = 0). (c) An antiderivative is x3 – 2x2 + x, giving 2 – 0 = 2. |
| 7 | The random variable X is normally distributed with mean 50 and standard deviation 8. (a) Find P(X > 62). (b) Find P(42 < X < 58). (c) Find the 90th percentile of X (2 d.p.). | (a) Z = (62 – 50) / 8 = 1.5, so P(Z > 1.5) = 0.0668. (b) Z lies between -1 and 1, so the probability is 0.6827. (c) z = 1.2816, so X = 50 + 8 x 1.2816 = 60.25. |
| 8 | The vectors are a = (3, -2) and b = (-1, 4). (a) Find a + b and its magnitude. (b) Find the scalar product a . b. (c) Find the angle between a and b (1 d.p.). (d) Find the unit vector in the direction of a (3 s.f.). | (a) a + b = (2, 2) with magnitude sqrt 8 = 2.83. (b) a . b = 3 x (-1) + (-2) x 4 = -11. (c) |a| = 3.606 and |b| = 4.123, so cos(theta) = -11 / 14.866 = -0.7399 and theta = 137.7 degrees. (d) The unit vector is (3, -2) / 3.606 = (0.832, -0.555). |
| 9 | The discrete random variable X has P(X = 0) = 0.1, P(X = 1) = 0.3, P(X = 2) = 0.4 and P(X = 3) = 0.2. (a) Find E(X). (b) Find E(X2). (c) Find Var(X). (d) Find the standard deviation of X. | (a) E(X) = 0 x 0.1 + 1 x 0.3 + 2 x 0.4 + 3 x 0.2 = 1.7. (b) E(X2) = 0 + 0.3 + 1.6 + 1.8 = 3.7. (c) Var(X) = 3.7 – 2.89 = 0.81. (d) Standard deviation = sqrt 0.81 = 0.9. |
| 10 | The population of a town grows at 4% per year and is now 1 200. (a) Find the population after 10 years (nearest integer). (b) Find the number of years needed for the population first to exceed 2 000. (c) 5 000 dollars is invested at 3.6% per year compounded quarterly; find the amount after 4 years (2 d.p.). | (a) 1200 x 1.0410 = 1776 (nearest integer). (b) 1200 x 1.04t > 2000 gives t > ln(5/3) / ln 1.04 = 13.02, so 14 years are needed. (c) 5000 x (1 + 0.036/4)16 = 5770.70 dollars. |
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