📚 Pre-U WJEC Further Mathematics: Mock Unit Test Paper Walkthrough | Pre-U WJEC 进阶数学：单元测试模拟卷解析
Welcome to this comprehensive walkthrough of a mock unit test for the Pre-U WJEC Further Mathematics syllabus. This paper covers essential topics including complex numbers, matrices, hyperbolic functions, differential equations, polar coordinates, vectors, series, proof by induction, Maclaurin series and integration techniques. Each question is broken down step by step to reinforce key methods and exam technique. Whether you are revising for an end-of-unit assessment or preparing for the final examination, these detailed solutions will help you master the reasoning behind every answer.
欢迎阅读这份针对 Pre-U WJEC 进阶数学教学大纲的单元测试模拟卷详尽解析。试卷涵盖复数、矩阵、双曲函数、微分方程、极坐标、向量、级数、归纳法证明、麦克劳林级数和积分技巧等核心主题。每道题都逐步拆解，以巩固关键方法和考试技巧。无论你是在准备单元测验还是最终大考，这些详细解答都将帮助你掌握每题背后的推理。
1. Complex Numbers – De Moivre's Theorem and Roots of Unity | 复数 – 棣莫弗定理与单位根
em in the form cosθ + i sinθ. Hence determine the exact value of cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5).
题目：求方程 z⁵ = 1 的所有复数解，并用 cosθ + i sinθ 形式表示。由此求出 cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) 的精确值。
Apply de Moivre' er k. The fifth roots are therefore zₖ = cos(2kπ/5) + i sin(2kπ/5) with k = 0, 1, 2, 3, 4.
应用棣莫弗定理，将 1 写成 cos(2kπ) + i sin(2kπ)，k 为整数。于是五次方根为 zₖ = cos(2kπ/5) + i sin(2kπ/5)，其中 k = 0, 1, 2, 3, 4。
z₂ = cos(4π/5) + i sin(4π/5), z₃ = cos(6π/5) + i sin(6π/5), z₄ = cos(8π/5) + i sin(8π/5).
具体写出：z₀ = 1，z₁ = cos(2π/5) + i sin(2π/5)，z₂ = cos(4π/5) + i sin(4π/5)，z₃ = cos(6π/5) + i sin(6π/5)，z₄ = cos(8π/5) + i sin(8π/5)。
quals −1. Using symmetry, cos(6π/5) = cos(4π/5) and cos(8π/5) = cos(2π/5), giving the relation 2cos(2π/5) + 2cos(4π/5) = −1.
这些是 z⁵ − 1 = 0 的根，故根之和为零。对比实部得 1 + cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = 0，因此所求余弦和为 −1。利用对称性 cos(6π/5) = cos(4π/5) 且 cos(8π/5) = cos(2π/5)，得到 2cos(2π/5) + 2cos(4π/5) = −1。
2. Matrices – Eigenvalues and Diagonalisation | 矩阵 – 特征值与对角化
sponding eigenvectors. Hence diagonalise A by giving matrices P and D such that A = PDP⁻¹.
题目：对于矩阵 A = [[3, 2], [4, 1]]，求特征值和对应的特征向量。由此用矩阵 P 和 D 将 A 对角化，使得 A = PDP⁻¹。
Solve the characteristic equation det(A − λI) = 0:
求解特征方程 det(A − λI) = 0：
det[[3-λ, 2], [4, 1-λ]] = (3-λ)(1-λ) − 8 = λ² − 4λ − 5 = 0
This gives eigenvalues λ₁ = 5, λ₂ = −1.
得特征值 λ₁ = 5，λ₂ = −1。
For λ = 5, solve (A − 5I)v = 0: [[-2, 2], [4, -4]] gives v₁ = [1, 1]ᵀ (or any non-zero multiple).
对 λ = 5，解 (A − 5I)v = 0，得 v₁ = [1, 1]ᵀ（或任意非零倍数）。
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