📚 A-Level CAIE Further Mathematics: Unit Test Mock Paper Solutions | A-Level CAIE进阶数学：单元测试模拟卷解析
This article steps through a full mock unit test for CAIE Further Mathematics (9231), covering Complex Numbers, Matrices, Differential Equations, Polar Coordinates, Hyperbolic Functions and Proof by Induction. Every part is explained in detail, pairing English exposition with its Chinese translation to help you master both the techniques and the reasoning behind them.
本文完整解析一份CAIE进阶数学(9231)单元测试模拟卷，涵盖复数、矩阵、微分方程、极坐标、双曲函数与归纳法证明。每个部分均搭配中英双语详解，帮助你掌握解题技巧与背后的逻辑。
1. Complex Numbers – Using de Moivre’s Theorem | 复数 – 应用棣莫弗定理
Given z = cos θ + i sin θ, we first express zn + z−n in terms of cos nθ. By de Moivre’s theorem, zn = cos nθ + i sin nθ and z−n = cos(−nθ) + i sin(−nθ) = cos nθ − i sin nθ.
给定 z = cos θ + i sin θ，我们先将 zn + z−n 用 cos nθ 表示。根据棣莫弗定理，zn = cos nθ + i sin nθ，且 z−n = cos nθ − i sin nθ。
Adding the two expressions cancels the imaginary parts, giving zn + z−n = 2 cos nθ. In particular, for n = 4 we obtain z4 + z−4 = 2 cos 4θ.
两式相加消去虚部，得 zn + z−n = 2 cos nθ。特别地，当 n = 4 时有 z4 + z−4 = 2 cos 4θ。
2. Solving z4 + z−4 = −1 | 解方程 z4 + z−4 = −1
Use the result from Section 1: 2 cos 4θ = −1 ⇒ cos 4θ = −½. The general solution for 4θ is 4θ = 2π/3 + 2kπ or 4θ = 4π/3 + 2kπ, where k ∈ ℤ.
利用第1节的结果：2 cos 4θ = −1 ⇒ cos 4θ = −½。4θ 的通解为 4θ = 2π/3 + 2kπ 或 4θ = 4π/3 + 2kπ，k ∈ ℤ。
Dividing by 4 gives θ = π/6 + kπ/2 and θ = π/3 + kπ/2 i sin θ, we list all distinct values of z for θ in [0, 2π). The eight distinct roots are:
除以4得 θ = π/6 + kπ/2 和 θ = π/3 + kπ/2。由于 z = cos θ + i sin θ，我们列出 θ 在 [0, 2π) 内所有不同的 z 值。8个不同的根为：
θ = π/6, 2π/3, 7π/6, 5π/3 → z = ±√3/2 ± i/2 (four combinations)
θ = π/6, 2π/3, 7π/6, 5π/3 → z = ±√3/2 ± i/2（四种组合）
θ = π/3, 5π/6, 4π/3, 11π/6 → z = ±1/2 ± i√3/2 (four combinations)
θ = π/3, 5π/6, 4π/3, 11π/6 → z = ±1/2 ± i√3/2（四种组合）
Consequently, the solution set consists of the eight vertices of a regular octagon on the unit circle.
因此，解集由单位圆上正八边形的八个顶点组成。
3. Matrices – Eigenvalues and Eigenvectors | 矩阵 – 特征值与特征向量
Let A = [[2,1],[1,2]]. The characteristic equation is det(A − λI) = (2−λ)² − 1 = 0, which simplifies to λ² − 4λ + 3 = 0. The eigenvalues are λ₁ = 1 and λ₂ = 3.
设矩阵 A = [[2,1],[1,2]]。特征方程为 det(A − λI) = (2−λ)² − 1 = 0，化简得 λ² − 4λ + 3 = 0。特征值为 λ₁ = 1 和 λ₂ = 3。
For λ = 1, solve (A − I)v = 0: [[1,1],[1,1]]v = 0 gives v₁ = t(1, −1)T. For λ = 3, (A − 3I)v = [[−1,1],[1,−1]]v = 0 gives v₂ = s(1, 1)T.
对于 λ = 1，解 (A − I)v = 0：[[1,1],[1,1]]v = 0，得 v₁ = t(1, −1)T。对于 λ = 3，(A − 3I)v = [[−1,1],[1,−1]]v = 0，得 v₂ = s(1, 1)T。
Normalising is not required for diagonalisation; we simply take basis eigenvectors (1, −1)T and (1, 1)T.
对角化时无需单位化，直接取基特征向量 (1, −1)T 和 (1, 1)T 即可。
4. Diagonalisation and Powers of a Matrix | 矩阵对角化与幂计算
Form the modal matrix P = [[1, 1],[−1, 1]]. Its inverse is P−1 = ½ [[1, −1],[1, 1]]. Then P−1AP = D = diag(1, 3).
构造模态矩阵 P = [[1, 1],[−1, 1]]，其逆为 P−1 = ½ [[1, −1],[1, 1]]。于是 P−1AP = D = diag(1, 3)。
To compute A5, use A5 = P D5 P−1. Since D5 = diag(15, 35) = diag(1, 243) Published by TutorHao | A-Level 进阶数学 Revision Series | aleveler.com
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