S6 Mathematics Practice Paper: Differentiation, Integration And Probability Distributions (12 Questions, 100 Marks, 75 Minutes, Answers Included)

S6 Mathematics Practice Paper: Differentiation, Integration and Probability Distributions (12 Questions, 100 Marks, 75 Minutes, Answers Included)

2026-11-07

S6 Maths: Differentiation, Integration and Probability Distributions

S6 Maths paper: 12 questions, 100 marks, 75 minutes on differentiation, tangents, related rates, definite integrals, areas, volumes and probability distributions.

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Section A: Differentiation and Its Applications (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	Let y = 2x³ − 9x² + 12x. (a) Find dy/dx. (b) Find the coordinates of all stationary points. (c) Use the second derivative to determine whether each stationary point is a maximum or a minimum, and state the corresponding extreme value.	(a) dy/dx = 6x² − 18x + 12. (b) Setting 6x² − 18x + 12 = 0 gives 6(x − 1)(x − 2) = 0, so x = 1 or x = 2; then y = 5 and y = 4, so the stationary points are (1, 5) and (2, 4). (c) d²y/dx² = 12x − 18. At x = 1 it is −6 < 0, so (1, 5) is a maximum point with maximum value 5; at x = 2 it is 6 > 0, so (2, 4) is a minimum point with minimum value 4.
2	The curve C has equation y = x³ − 4x and P(2, 0) lies on C. (a) Find the slope of the tangent to C at P. (b) Find the equation of the tangent to C at P. (c) Find the equation of the normal to C at P, giving your answer in the form ax + by + c = 0.	(a) dy/dx = 3x² − 4, so the slope at x = 2 is 8. (b) Tangent: y − 0 = 8(x − 2), i.e. y = 8x − 16. (c) The normal has slope −1/8, so y − 0 = −(1/8)(x − 2), i.e. x + 8y − 2 = 0.
3	The radius of a sphere increases at 0.5 cm per second and its volume is V = (4/3)πr³. (a) Find dV/dr. (b) When r = 6, find dV/dt in terms of π and correct to 2 decimal places. (c) Hence estimate the increase in volume when the radius grows from 6 cm to 6.1 cm.	(a) dV/dr = 4πr². (b) dV/dt = (dV/dr)(dr/dt) = 4π(6)²(0.5) = 72π ≈ 226.19 cm³/s. (c) Increase ≈ (dV/dr)(0.1) = 144π × 0.1 = 14.4π ≈ 45.24 cm³.
4	A piece of wire 40 cm long is bent into a rectangle. Let the length be x cm. (a) Express the area A in terms of x. (b) Find the value of x that maximises A and prove that it gives a maximum. (c) Find the maximum area and describe the shape of the rectangle at that point.	(a) 2x + 2y = 40, so y = 20 − x and A = x(20 − x) = 20x − x². (b) dA/dx = 20 − 2x; setting it to zero gives x = 10, and d²A/dx² = −2 < 0, so this is a maximum. (c) A = 10 × 10 = 100 cm², since length = width = 10 cm the rectangle is a square.

Section B: Integration, Areas and Volumes (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	Let f(x) = 6x² − 4x + 3. (a) Find ∫f(x) dx. (b) Evaluate ∫₀² f(x) dx. (c) Hence find the area bounded by y = f(x), the x-axis, x = 0 and x = 2, and explain why the area equals the answer in (b).	(a) ∫(6x² − 4x + 3) dx = 2x³ − 2x² + 3x + C. (b) [2x³ − 2x² + 3x]₀² = (16 − 8 + 6) − 0 = 14. (c) The discriminant is (−4)² − 4(6)(3) = −56 < 0 and the leading coefficient 6 > 0, so f(x) is always positive and the graph lies above the x-axis; hence the area equals the definite integral, 14 square units.
6	The curve y = x² and the line y = 2x meet at two points. (a) Find the two points of intersection. (b) Find the area of the region enclosed by the curve and the line. (c) Find the volume of the solid formed when that region is rotated about the x-axis, in terms of π.	(a) x² = 2x gives x(x − 2) = 0, so the points are (0, 0) and (2, 4). (b) On 0 < x < 2, 2x > x², so area = ∫₀²(2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3 square units. (c) V = π∫₀²[4x² − x⁴] dx = π[4x³/3 − x⁵/5]₀² = π(32/3 − 32/5) = 64π/15 ≈ 13.40 cubic units.
7	Consider the curve y = x² + 1 together with the lines x = −1 and x = 2. (a) Evaluate ∫₋₁² (x² + 1) dx. (b) Hence find the area bounded by the curve, the x-axis and those two lines. (c) Estimate the same area by the trapezoidal rule with h = 1 and state the difference from the exact value.	(a) [x³/3 + x]₋₁² = (8/3 + 2) − (−1/3 − 1) = 14/3 + 4/3 = 6. (b) Since x² + 1 > 0 the curve is above the x-axis, so the area is 6 square units. (c) With h = 1 the values of y are 2, 1, 2 and 5, giving 1 × [(2 + 5)/2 + 1 + 2] = 6.5; the difference is 6.5 − 6 = 0.5, an overestimate because the curve is concave up.
8	A particle moves along a straight line with velocity v = 3t² − 12t + 9 m/s for 0 ≤ t ≤ 4 seconds. (a) Find the times when the particle is at rest. (b) Find its displacement from t = 0 to t = 4. (c) Find the total distance travelled from t = 0 to t = 4.	(a) 3t² − 12t + 9 = 0 gives 3(t − 1)(t − 3) = 0, so the particle is at rest at t = 1 s and t = 3 s. (b) Displacement = ∫₀⁴ v dt = [t³ − 6t² + 9t]₀⁴ = (64 − 96 + 36) − 0 = 4 m. (c) v is negative on (1, 3), so take absolute values: \|s(1) − s(0)\| = 4, \|s(3) − s(1)\| = 4 and \|s(4) − s(3)\| = 4, giving a total distance of 4 + 4 + 4 = 12 m.

Section C: Probability Distributions (40 marks)

Each question carries 10 marks; give full working or explanations.

#	Question	Answer and marking notes
9	Let X ~ B(10, 0.3). (a) Find P(X = 3), correct to 4 decimal places. (b) Find E(X) and Var(X). (c) Find P(X ≤ 1), correct to 4 decimal places.	(a) P(X = 3) = C(10,3)(0.3)³(0.7)⁷ = 120 × 0.027 × 0.0823543 ≈ 0.2668. (b) E(X) = 10 × 0.3 = 3 and Var(X) = 10 × 0.3 × 0.7 = 2.1. (c) P(X ≤ 1) = 0.7¹⁰ + 10(0.3)(0.7)⁹ ≈ 0.0282475 + 0.1210608 ≈ 0.1493.
10	An archer hits the target with probability 0.25 on each attempt, independently. He shoots 8 times and X is the number of hits. (a) Find P(X = 0). (b) Find P(X ≥ 2). (c) Find P(X ≤ 3) and decide whether “at least 4 hits” is a rare event.	(a) P(X = 0) = 0.75⁸ ≈ 0.1001. (b) P(X ≥ 2) = 1 − P(0) − P(1) = 1 − 0.1001 − 0.2670 ≈ 0.6329. (c) P(X ≤ 3) ≈ 0.1001 + 0.2670 + 0.3115 + 0.2076 ≈ 0.8862, so P(X ≥ 4) ≈ 0.1138; about 11.4%, which is not rare by the usual 5% standard.
11	Let X ~ N(50, 25). (a) Find P(X > 58). (b) Find P(44 < X < 58). (c) Find k such that P(X < k) = 0.9, given that P(Z < 1.28) = 0.8997.	(a) Z = (58 − 50)/5 = 1.6, so P(X > 58) = P(Z > 1.6) = 1 − 0.9452 = 0.0548. (b) Z₁ = (44 − 50)/5 = −1.2 and Z₂ = 1.6, so P = 0.9452 − 0.1151 = 0.8301. (c) Since P(Z < 1.28) = 0.8997, k = 50 + 1.28 × 5 = 56.4.
12	A discrete random variable X has probability distribution P(X = x) = k(x + 1) for x = 0, 1, 2, 3. (a) Find k. (b) Find E(X). (c) Find Var(X) and the standard deviation.	(a) The probabilities sum to 1: k(1 + 2 + 3 + 4) = 10k = 1, so k = 0.1. (b) E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4) = 2.0. (c) E(X²) = 0 + 0.2 + 4(0.3) + 9(0.4) = 5.0, so Var(X) = 5.0 − 2.0² = 1.0 and the standard deviation is 1.0.

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