OCR Year 13 Further Maths Mock Test Solution Walkthrough | OCR 进阶数学单元测试模拟卷解析

📚 OCR Year 13 Further Maths Mock Test Solution Walkthrough | OCR 进阶数学单元测试模拟卷解析

This article provides a detailed step-by-step walkthrough of a practice unit test designed for Year 13 students studying OCR Further Mathematics. The mock paper covers essential topics from the Further Pure Core syllabus, including complex numbers, matrices, proof by induction, hyperbolic functions, polar coordinates, differential equations, Maclaurin series, and vectors. Each solution is presented with paired explanations in English and Chinese to support bilingual learners and reinforce exam technique.

本文为学习 OCR 进阶数学的 Year 13 学生提供一套单元测试模拟卷的详细逐步解析。模拟试卷涵盖了 Further Pure Core 大纲中的核心主题，包括复数、矩阵、归纳法证明、双曲函数、极坐标、微分方程、麦克劳林级数以及向量。每道题的解答都配有中英双语解释，以辅助双语学习者并强化考试技巧。

1. Complex Numbers: Roots of a Cubic Equation | 复数：三次方程的根

Question: Solve the equation z³ = -8i, giving your answers in the form re^iθ where r > 0 and -π < θ ≤ π. Hence sketch the roots on an Argand diagram.

题目：解方程 z³ = -8i，答案以 re^iθ 的形式给出，其中 r > 0 且 -π < θ ≤ π。据此在阿尔冈图上画出这些根。

Step 1: Express the right-hand side in modulus-argument form. The complex number -8i lies on the negative imaginary axis, so its modulus is 8 and its principal argument is -π/2. Therefore, -8i = 8e^{i(-π/2 + 2kπ)} for any integer k.

步骤1：将右边表示为模长-辐角形式。复数 -8i 位于负虚轴上，因此其模长为 8，辐角主值为 -π/2。故对于任意整数 k，有 -8i = 8e^{i(-π/2 + 2kπ)}。

Step 2: Take the cube root of both sides. Using de Moivre’s theorem, z = [8e^{i(-π/2 + 2kπ)}]^1/3 = 2e^{i(-π/6 + 2kπ/3)}. Substitute k = 0, 1, 2 to obtain the three distinct cube roots.

步骤2：等式两边开立方。根据棣莫弗定理，z = [8e^{i(-π/2 + 2kπ)}]^1/3 = 2e^{i(-π/6 + 2kπ/3)}。代入 k = 0, 1, 2 得到三个不同的立方根。

Step 3: Calculate each root and ensure the argument lies in the required interval.

• k = 0: z₀ = 2e^-iπ/6 (θ = -π/6, within (-π, π])
• k = 1: z₁ = 2e^i(π/2) (θ = π/2, valid)
• k = 2: z₂ = 2e^i(7π/6) → argument 7π/6 > π, so we subtract 2π to get 2e^-i5π/6 (θ = -5π/6, valid).

Thus the three roots are 2e^-iπ/6, 2e^iπ/2, and 2e^-i5π/6.

步骤3：计算每一个根，并确保辐角在要求的区间内。

• k = 0：z₀ = 2e^-iπ/6 (θ = -π/6，在 (-π, π] 内)
• k = 1：z₁ = 2e^i(π/2) (θ = π/2，有效)
• k = 2：z₂ = 2e^i(7π/6) → 辐角 7π/6 > π，故减去 2π 得 2e^-i5π/6 (θ = -5π/6，有效)。

因此三个根为 2e^-iπ/6、2e^iπ/2 和 2e^-i5π/6。

Sketch: The roots lie on a circle of radius 2 centred at the origin. They are equally spaced at angular intervals of 2π/3, with arguments -5π/6, -π/6, and π/2. Mark these points symmetrically on the Argand diagram.

草图：这些根位于以原点为圆心、半径为 2 的圆上。它们以 2π/3 的等角距分布，辐角分别为 -5π/6、-π/6 和 π/2。在阿尔冈图上对称地标出这些点。

2. Matrices: Determinant, Inverse and Linear Systems | 矩阵：行列式、逆与线性方程组

Question: Given matrix A =
[2, -1, 3; 1, 0, -2; 4, -1, 1].
(a) Evaluate det(A).
(b) Find the inverse matrix A^-1.
(c) Hence solve the system of equations:
2x – y + 3z = 7
x – 2z = -3
4x – y + z = 5.

题目：已知矩阵 A =
[2, -1, 3; 1, 0, -2; 4, -1, 1]。
(a) 计算 det(A)。
(b) 求逆矩阵 A^-1。
(c) 由此解方程组：
2x – y + 3z = 7
x – 2z = -3
4x – y + z = 5。

Part (a): Expand along the second row (containing a zero) to simplify.
det(A) = 1 × C₂₁ + 0 × C₂₂ + (-2) × C₂₃. First find the cofactors. C₂₁ = -det([-1,3; -1,1]) = -((-1)*1 – 3*(-1)) = -( -1 + 3) = -2. C₂₃ = -det([2,-1; 4,-1]) = -(2*(-1) – (-1)*4) = -(-2 + 4) =

Published by TutorHao | Year 13 进阶数学 Revision Series | aleveler.com

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