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Cambridge A Level Further Mathematics 9231 Paper 1 Further Pure Mathematics 1 practice paper
A Level Further Mathematics 9231
A Level Further Mathematics 9231 Paper 1 practice paper: 8 questions, 75 marks, 2 hours on roots, matrices, series, complex numbers, vectors and polar area.
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Paper details Level: A Level | Subject: Further Mathematics 9231 | Paper 1 Further Pure Mathematics 1 | Time: 2 hours | Marks: 75 | Questions: 8
Answer all 8 questions. Total: 75 marks. Time allowed: 2 hours.
Show all necessary working; answers without working may not receive full marks.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | The cubic equation x³ − 6x² + 11x − 6 = 0 has roots α, β and γ. (a) Find Σα² and Σα²β². (b) Find the cubic equation, with integer coefficients, whose roots are α + 1, β + 1 and γ + 1. (8 marks) |
(a) By Vieta: Σα = 6, Σαβ = 11, αβγ = 6. Σα² = (Σα)² − 2Σαβ = 36 − 22 = 14. Σα²β² = (Σαβ)² − 2αβγΣα = 121 − 2(6)(6) = 121 − 72 = 49. (b) Sum of new roots = Σα + 3 = 9; sum of products taken two at a time = Σαβ + 2Σα + 3 = 11 + 12 + 3 = 26; product = αβγ + Σαβ + Σα + 1 = 6 + 11 + 6 + 1 = 24. Hence x³ − 9x² + 26x − 24 = 0 (roots 2, 3 and 4). Marking notes: (a) 2 marks per expression; (b) 1 mark per symmetric sum and 1 mark for the equation. |
| 2 | The matrix A = [[4, 1], [2, 3]]. (a) Find the two eigenvalues of A and an eigenvector for each. (b) Find A⁻¹ and hence solve 4x + y = 9, 2x + 3y = 7. (9 marks) |
(a) det(A − λI) = (4 − λ)(3 − λ) − 2 = λ² − 7λ + 10 = (λ − 5)(λ − 2), so λ = 5 or λ = 2. For λ = 5: (4 − 5)x + y = 0, giving the eigenvector (1, 1) (check A(1, 1) = (5, 5) = 5(1, 1)). For λ = 2: (4 − 2)x + y = 0, giving the eigenvector (1, −2) (check A(1, −2) = (2, −4) = 2(1, −2)). (b) det A = 12 − 2 = 10, so A⁻¹ = (1/10)[[3, −1], [−2, 4]]. x = (3 × 9 − 1 × 7) / 10 = 2 and y = (−2 × 9 + 4 × 7) / 10 = 1. Hence x = 2, y = 1. Marking notes: 2 marks for the characteristic equation, 1 mark per eigenvalue, 1 mark per eigenvector; 2 marks for A⁻¹ and 1 mark for the solution. |
| 3 | (a) Using the standard summation formulae, express Σ(r = 1 to n) r(2r + 3) in terms of n. (b) Using partial fractions and the method of differences, prove that Σ(r = 1 to n) 1/((2r − 1)(2r + 1)) = n/(2n + 1). (8 marks) |
(a) Σr(2r + 3) = 2Σr² + 3Σr = 2 × n(n + 1)(2n + 1)/6 + 3 × n(n + 1)/2. Combining over a common denominator gives n(n + 1)[2(2n + 1) + 9]/6 = n(n + 1)(4n + 11)/6. Check n = 4: 4 × 5 × 27/6 = 90, and 5 + 14 + 27 + 44 = 90. (b) 1/((2r − 1)(2r + 1)) = (1/2)[1/(2r − 1) − 1/(2r + 1)]. Summing cancels all the intermediate terms: (1/2)[1 − 1/(2n + 1)] = n/(2n + 1). Marking notes: (a) 1 mark per standard formula and 2 marks for the algebra; (b) 2 marks for partial fractions, 1 for the cancelling and 1 for the conclusion. |
| 4 | Let z = 2 + 2√3 i. (a) Find |z| and arg z, giving the argument as a multiple of π. (b) Find z³ in the form a + bi. (c) Hence solve w³ = −64, giving the three roots in the form r(cos θ + i sin θ). (10 marks) |
(a) |z| = √(4 + 12) = 4. tan θ = 2√3 / 2 = √3 and z is in the first quadrant, so arg z = π/3, i.e. z = 4(cos π/3 + i sin π/3). (b) z³ = 4³(cos π + i sin π) = 64(−1 + 0i) = −64. (c) From (b), w = 4(cos π/3 + i sin π/3) is one root; the others have arguments increased by 2π/3 and 4π/3: 4(cos π/3 + i sin π/3) = 2 + 2√3 i; 4(cos π + i sin π) = −4; 4(cos 5π/3 + i sin 5π/3) = 2 − 2√3 i. Marking notes: (a) 2 marks for the modulus, 1 for the argument; (b) 3 marks; (c) 1 mark per root and 1 mark for using multiples of π. |
| 5 | The plane Π has equation 2x − y + 2z = 9. The point A is (1, −1, 3) and the line l has equation r = (3, 1, 2) + t(1, −2, 2). (a) Show that A lies on Π and find the point where l meets Π. (b) Find the perpendicular distance from B(5, 1, 4) to Π. (c) Find the angle θ between l and Π, giving sin θ. (10 marks) |
(a) At A: 2(1) − (−1) + 2(3) = 2 + 1 + 6 = 9, so A lies on Π. Substituting l into Π: 2(3 + t) − (1 − 2t) + 2(2 + 2t) = 9 + 8t. Setting 9 + 8t = 9 gives t = 0, so the intersection is (3, 1, 2). (b) At B: 2(5) − 1 + 2(4) = 17. The normal has length √(4 + 1 + 4) = 3, so the distance is |17 − 9| / 3 = 8/3. (c) With d = (1, −2, 2) and n = (2, −1, 2): d · n = 2 + 2 + 4 = 8 and |d| = |n| = 3, so sin θ = 8/9 (θ ≈ 62.7°). Marking notes: (a) 2 marks for verifying A and 2 for the intersection; (b) 2 marks for the substitution and normal length, 1 for the answer; (c) 1 mark for the dot product, 1 for the moduli, 1 for the answer. |
| 6 | Prove by induction that, for all positive integers n, Σ(r = 1 to n) r 2^r = (n − 1)2^(n+1) + 2. (10 marks) |
For n = 1: LHS = 1 × 2 = 2 and RHS = 0 × 4 + 2 = 2, so the statement holds. Assume it holds for some positive integer k, i.e. Σ(r = 1 to k) r 2^r = (k − 1)2^(k+1) + 2. Then Σ(r = 1 to k+1) r 2^r = (k − 1)2^(k+1) + 2 + (k + 1)2^(k+1) = 2^(k+1)[(k − 1) + (k + 1)] + 2 = 2k · 2^(k+1) + 2 = k 2^(k+2) + 2 = ((k + 1) − 1)2^((k+1)+1) + 2, so the statement holds for n = k + 1. By mathematical induction the statement holds for all positive integers n. Marking notes: 2 marks for the base case, 2 for the inductive assumption, 4 for the algebra and 2 for the concluding statement. |
| 7 | Let f(x) = √(1 + 2x). (a) Find f′(x), f″(x) and f‴(x). (b) Hence find the Maclaurin expansion of f(x), up to and including the term in x³. (c) Use the expansion in (b) to estimate √1.2 and state the difference from the true value in scientific notation. (10 marks) |
(a) f′(x) = (1 + 2x)^(−1/2), f″(x) = −(1 + 2x)^(−3/2) and f‴(x) = 3(1 + 2x)^(−5/2). Hence f(0) = 1, f′(0) = 1, f″(0) = −1 and f‴(0) = 3. (b) The Maclaurin expansion is 1 + x + (−1)x²/2! + 3x³/3! = 1 + x − x²/2 + x³/2. (c) With x = 0.1: 1 + 0.1 − 0.005 + 0.0005 = 1.0955. The true value is √1.2 = 1.095445…, so the difference is 5.5 × 10⁻⁵ (about 5.49 × 10⁻⁵). Marking notes: (a) 1 mark per derivative and 1 for the four values; (b) 3 marks for the expansion; (c) 1 mark each for the substitution, the true value and the difference. |
| 8 | The polar curve C has equation r = a(1 + cos θ), 0 ≤ θ ≤ 2π, where a > 0. (a) Show that the area enclosed by C is (3/2)πa². (b) Taking a = 2, find this area (exact value and to 3 significant figures) and the value of r at θ = π/3. (10 marks) |
(a) Area = (1/2)∫(0 to 2π) r² dθ = (a²/2)∫(0 to 2π) (1 + cos θ)² dθ. Expanding gives 1 + 2cos θ + cos²θ, with cos²θ = (1 + cos 2θ)/2. ∫(0 to 2π) 1 dθ = 2π, ∫(0 to 2π) 2cos θ dθ = 0 and ∫(0 to 2π) cos²θ dθ = π. The total is 3π, so the area is (a²/2)(3π) = (3/2)πa². (b) With a = 2: area = (3/2)π(4) = 6π ≈ 18.8. At θ = π/3, cos θ = 1/2, so r = 2(1 + 1/2) = 3. Marking notes: (a) 2 marks for the area formula, 2 for the expansion and half-angle, 1 for integrating term by term and 1 for the conclusion; (b) 2 marks for the exact value, 1 for the decimal and 1 for r = 3. |
A TutorZone practice paper; original questions and passages, not publisher material.
