Mark Scheme

1. From the first equation, y = 7 − 2x (M1). Substitute into the second: x² − x(7 − 2x) = −4 → x² − 7x + 2x² = −4 → 3x² − 7x + 4 = 0 (M1). Factorise: (3x − 4)(x − 1) = 0 (M1), so x = 4/3 or x = 1 (A1). Corresponding y-values: when x = 1, y = 5; when x = 4/3, y = 7 − 8/3 = 13/3 (A1). [5]

2. (a) 3x² − 12x + 7 = 3(x² − 4x) + 7 = 3[(x − 2)² − 4] + 7 (M1) = 3(x − 2)² − 12 + 7 (M1) = 3(x − 2)² − 5 (A1). [3] (b) Least value is −5 (B1), occurring at x = 2 (B1). [2]

3. No real roots requires discriminant < 0 (M1): (k − 2)² − 4(1)(4) < 0 (M1) → (k − 2)² < 16 → −4 < k − 2 < 4 (M1) → −2 < k < 6 (A1). [4]

4. (a) log₂ 45 = log₂(9 × 5) = log₂ 3² + log₂ 5 (M1) = 2 log₂ 3 + log₂ 5 (M1) = 2p + q (A1). [3] (b) Take logs: (2x + 1) log 3 = log 20 (M1). 2x + 1 = log 20 / log 3 = 2.7268 (M1). 2x = 1.7268, x = 0.863 to 3 s.f. (A1). [3]

5. (2 + kx)⁶: term in x⁰ is 2⁶ = 64 (B1). Term in x¹: C(6,1)·2⁵·(kx) = 6·32·k x = 192k x, so 192k = 576 → k = 3 (M1 A1). Term in x²: C(6,2)·2⁴·(kx)² = 15·16·k² x² = 240k² x². With k = 3, A = 240 × 9 = 2160 (M1 A1). [5]

6. (a) fg(x) = 2·g(x) + 3 = 2(x − 1)/(x + 2) + 3 (M1) = [2(x − 1) + 3(x + 2)]/(x + 2) = (5x + 4)/(x + 2) (A1). [2] (b) Let y = (x − 1)/(x + 2). Then y(x + 2) = x − 1 → yx + 2y = x − 1 (M1) → x(y − 1) = −1 − 2y (M1) → x = (−1 − 2y)/(y − 1). So g⁻¹(x) = (2x + 1)/(1 − x) (A1), with domain x ≠ 1 (B1). [4]

7. (a) Common denominator cos θ(1 + sin θ): numerator = (1 + sin θ)² + cos²θ (M1) = 1 + 2 sin θ + sin²θ + cos²θ = 1 + 2 sin θ + 1 = 2 + 2 sin θ = 2(1 + sin θ) (M1). So the expression = 2(1 + sin θ)/[cos θ(1 + sin θ)] (M1) = 2/cos θ = 2 sec θ (A1). [4] (b) 2 sec θ = 4 → sec θ = 2 → cos θ = 1/2 (M1). θ = 60° (A1) and θ = 300° (A1). [3]

8. (a) x² + 2 = 2x + 5 (M1) → x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 (M1) → x = 3 or x = −1 (A1). Points: A(−1, 3) and B(3, 11) (A1). [4] (b) Midpoint = ((−1 + 3)/2, (3 + 11)/2) (M1) = (1, 7) (A1). [2]

9. (a) At rest: 3t² − 12t + 9 = 0 → t² − 4t + 3 = 0 → (t − 1)(t − 3) = 0 (M1), so t = 1 and t = 3 (A1). [2] (b) a = dv/dt = 6t − 12 (M1). At t = 3: a = 18 − 12 = 6 m s⁻² (A1). [2] (c) Displacement function s = ∫v dt = t³ − 6t² + 9t (M1). The particle changes direction at t = 1, so evaluate separately. s(1) = 1 − 6 + 9 = 4; s(3) = 27 − 54 + 27 = 0 (M1). Distance 0→1 = |4 − 0| = 4 m; distance 1→3 = |0 − 4| = 4 m (M1 for splitting at t = 1). Total distance = 4 + 4 = 8 m (A1 A1). [5]

10. (a) Product rule: d/dx(x² ln x) = 2x·ln x + x²·(1/x) (M1 M1) = 2x ln x + x (A1). [3] (b) Set dy/dx = 0: 2x ln x + x = 0 → x(2 ln x + 1) = 0 (M1). Since x > 0, 2 ln x + 1 = 0 → ln x = −1/2 → x = e^(−1/2) (M1). y = (e^(−1/2))² · ln(e^(−1/2)) = e⁻¹ · (−1/2) = −1/(2e) (A1). Stationary point (e^(−1/2), −1/(2e)). Second derivative d²y/dx² = 2 ln x + 3; at x = e^(−1/2), = 2(−1/2) + 3 = 2 > 0 (M1), so it is a minimum (A1). [5]

11. (a) ∫(6x² − 4x⁻²) dx = 6·x³/3 − 4·x⁻¹/(−1) + c (M1 M1) = 2x³ + 4/x + c (A1). [3] (b) f(x) = ∫(6x − 4) dx = 3x² − 4x + c (M1 M1). Through (2, 5): 5 = 12 − 8 + c → c = 1 (M1). f(x) = 3x² − 4x + 1 (A1). [4] (c) ∫₁³(6x − 4) dx = [3x² − 4x]₁³ = (27 − 12) − (3 − 4) = 15 − (−1) = 16 (M1 A1). This value is f(3) − f(1) = the change in f(x) between x = 1 and x = 3 (i.e. the net area under f′(x)) (B1). [3]

12. (a) AB→ = b − a = (8 − 2)i + (−1 − 3)j = 6i − 4j (M1). |AB→| = √(6² + (−4)²) = √52 (M1) = 7.21 to 3 s.f. (A1). [3] (b) C divides AB in ratio 1 : 2, so c = a + (1/3)(AB→) (M1) = (2i + 3j) + (1/3)(6i − 4j) = (2 + 2)i + (3 − 4/3)j (M1) = 4i + (5/3)j (A1). [3]

13. (a) Total ways to choose 4 from 12 = C(12, 4) (M1) = 495 (A1). [2] (b) Ways = C(7, 2) × C(5, 2) (M1) = 21 × 10 (M1) = 210 (A1). [3] (c) "At least 3 women" = (3 women, 1 man) + (4 women, 0 men) (M1). = C(7,3)×C(5,1) + C(7,4)×C(5,0) = 35 × 5 + 35 × 1 (M1) = 175 + 35 = 210 (A1). [3]

14. (a) At t = 0, m = 80 e⁰ = 80 g (B1). [1] (b) m = 80 e^(−0.03 × 20) = 80 e^(−0.6) (M1) = 80 × 0.5488 = 43.9 g (to 3 s.f.) (A1). [2] (c) Half of 80 = 40, so 40 = 80 e^(−0.03t) (M1) → 0.5 = e^(−0.03t) → ln 0.5 = −0.03t (M1) → t = ln 0.5 / (−0.03) = 23.10 (M1) → t = 23 days (A1). [4]

15. (a) Midpoint of PQ = ((1 + 5)/2, (4 + 2)/2) = (3, 3) (M1). Gradient of PQ = (2 − 4)/(5 − 1) = −2/4 = −1/2 (M1). Gradient of perpendicular bisector = 2 (negative reciprocal) (M1). Line: y − 3 = 2(x − 3) → y = 2x − 3 (M1) → 2x − y = 3 (A1). [5] (b) At the x-axis, y = 0: 2x − 0 = 3 → x = 1.5 (M1). R = (1.5, 0) (A1). [2]

Sample Answers with Examiner Commentary

Question 3 — Sample Answers

Grade A response. "For no real roots the discriminant must be negative: b² − 4ac < 0. Here a = 1, b = (k − 2), c = 4, so (k − 2)² − 16 < 0. This gives (k − 2)² < 16, so −4 < k − 2 < 4, therefore −2 < k < 6."

Mark: 4/4. Examiner commentary: A model response. The candidate states the discriminant condition explicitly before substituting, which secures the method mark even if an arithmetic slip followed. Crucially they take the square root of the inequality correctly, writing a two-sided inequality −4 < k − 2 < 4 rather than the common error k − 2 < 4 alone, and present the final answer as a clean interval. Every mark is earned and the working is fully self-explanatory.

Grade C response. "Discriminant < 0. (k − 2)² − 16 < 0, so (k − 2)² < 16, so k − 2 < 4, so k < 6."

Mark: 3/4. Examiner commentary: The method is correct and three marks are earned — the discriminant condition, the substitution, and the simplification to (k − 2)² < 16. The final mark is lost because the candidate takes only the positive square root, giving k < 6 and missing the lower bound. Taking the square root of a squared expression in an inequality always produces two bounds; a quick sketch of y = (k − 2)² against the line 16 would have revealed the interval −2 < k < 6.

Question 9(c) — Sample Answers

Grade A response. "s = ∫v dt = t³ − 6t² + 9t (taking s = 0 at t = 0). The particle is at rest at t = 1 and t = 3, so it may change direction at t = 1 within the interval. s(1) = 1 − 6 + 9 = 4 m and s(3) = 27 − 54 + 27 = 0 m. Between t = 0 and t = 1 it moves +4 m; between t = 1 and t = 3 it moves from +4 to 0, i.e. −4 m, a distance of 4 m. Total distance = 4 + 4 = 8 m."

Mark: 5/5. Examiner commentary: Full marks, and a textbook demonstration of the key discriminator on this question. The candidate recognises that displacement and distance differ when the particle reverses, identifies t = 1 as the turning point inside the interval, and evaluates the two legs separately before adding their magnitudes. Simply computing s(3) − s(0) = 0 would have scored only the first method mark, since it gives net displacement, not total distance.

Grade C response. "s = t³ − 6t² + 9t. At t = 3, s = 27 − 54 + 27 = 0. So the distance travelled is 0 m."

Mark: 2/5. Examiner commentary: The candidate integrates correctly and evaluates s(3) accurately, earning the first two marks, but concludes that the distance is 0 m. This is the classic error: 0 m is the net displacement — the particle returns to O — not the distance travelled. The candidate has not checked whether the velocity changes sign in the interval. Solving v = 0 to find t = 1 and t = 3, then splitting the motion at t = 1, is exactly what separates this from a full-mark answer.