📚 Year 12 WJEC Further Maths: Unit Test Mock Paper Walkthrough | Year 12 WJEC 进阶数学：单元测试模拟卷解析
This walkthrough breaks down a mock unit test designed for the Year 12 WJEC Further Mathematics course. Each section presents a typical exam-style problem found in the Further Pure Mathematics unit and guides you through the key steps required to reach a full-mark solution. The selected topics include complex numbers, matrices, roots of polynomials, proof by induction, summation of series, hyperbolic functions, numerical methods and parametric equations — all core content for the AS qualification. Use this revision resource to consolidate your understanding and practise the precise reasoning expected by the exam board.
本文解析一套专为 Year 12 WJEC 进阶数学课程设计的单元模拟卷。每一节都呈现一道进阶纯数单元常见的考试题型，并带你梳理获得满分所需的关键步骤。所选主题涵盖复数、矩阵、多项式根、归纳法证明、级数求和、双曲函数、数值方法和参数方程——全部为 AS 资格证的核心内容。请利用这份复习资料巩固理解并练习考试局所要求的严谨推理。
1. Complex Numbers: Solving a Quadratic Equation | 复数：解二次方程
Problem: Solve the equation z² + 2z + 5 = 0. Express the solutions in the form a + bi, and determine the modulus and argument of each root.
题目：解方程 z² + 2z + 5 = 0。将解表示为 a + bi 的形式，并求每个根的模和辐角。
Apply the quadratic formula: z = [−b ± √(b² − 4ac)] / (2a). With a = 1, b = 2, c = 5, the discriminant Δ = 4 − 20 = −16. Therefore, z = (−2 ± √(−16)) / 2 = (−2 ± 4i) / 2 = −1 ± 2i.
运用求根公式：z = [−b ± √(b² − 4ac)] / (2a)。a = 1, b = 2, c = 5，判别式 Δ = 4 − 20 = −16。因此 z = (−2 ± √(−16)) / 2 = (−2 ± 4i) / 2 = −1 ± 2i。
For z₁ = −1 + 2i, modulus |z₁| = √((−1)² + 2²) = √(1 + 4) = √5. The point lies in the second quadrant, so argument θ₁ = arctan(2 / (−1)) + π ≈ 2.034 rad. For z₂ = −1 − 2i, the modulus is also √5; the argument is θ₂ = arctan((−2) / (−1)) − π ≈ −2.034 rad (or equivalently 2π − 2.034 rad).
对于 z₁ = −1 + 2i，模 |z₁| = √((−1)² + 2²) = √(1 + 4) = √5。该点位于第二象限，因此辐角 θ₁ = arctan(2 / (−1)) + π ≈ 2.034 弧度。对于 z₂ = −1 − 2i，模同样为 √5；辐角 θ₂ = arctan((−2) / (−1)) − π ≈ −2.034 弧度（或等价于 2π − 2.034 弧度）。
2. Matrices: Finding an Inverse and Solving a Linear System | 矩阵：求逆矩阵与解线性方程组
Problem: Given the matrix M = [ [3, 1], [2, 4] ], find M⁻¹. Hence solve the simultaneous equations 3x + y = 5 and 2x + 4y = 6.
题目：已知矩阵 M = [ [3, 1], [2, 4] ]，求 M⁻¹。进而解方程组 3x + y = 5 和 2x + 4y = 6。
For a 2×2 matrix [ [a, b], [c, d] ], the inverse is (1/(ad − bc)) [ [d, −b], [−c, a] ]. Here det = 3×4 − 1×2 = 12 − 2 = 10, so M⁻¹ = (1/10) [ [4, −1], [−2, 3] ] = [ [0.4, −0.1], [−0.2, 0.3] ].
对于 2×2 矩阵 [ [a, b], [c, d] ]，其逆为 (1/(ad − bc)) [ [d, −b], [−c, a] ]。此处行列式 det = 3×4 − 1×2 = 12 − 2 = 10，故 M⁻¹ = (1/10) [ [4, −1], [−2, 3] ] = [ [0.4, −0.1], [−0.2, 0.3] ]。
Write the system as M × [x, y]ᵀ = [5, 6]ᵀ. Multiplying by M⁻¹ gives [x, y]ᵀ = M⁻¹ [5, 6]ᵀ = [ [0.4, −0.1], [−0.2, 0.3] ] [5, 6]ᵀ = [0.4×5 − 0.1×6, −0.2×5 + 0.3×6]ᵀ = [2 − 0.6, −1 + 1.8]ᵀ = [1.4, 0.8]ᵀ. So x = 1.4, y = 0.8.
将方程组写作 M × [x, y]ᵀ = [5, 6]ᵀ。两边左乘 M⁻¹ 得 [x, y]ᵀ = M⁻¹ [5, 6]ᵀ = [ [0.4, −0.1], [−0.2, 0.3] ] [5, 6]ᵀ = [0.4×5 − 0.1×6, −0.2×5 + 0.3×6]ᵀ = [2 − 0.6, −1 + 1.8]ᵀ = [1.4, 0.8]ᵀ。因此 x = 1.4，y = 0.8。
3. Roots of Polynomials: Working with Sums of Powers | 多项式根：幂和的运用
Problem: The cubic equation 2x³ + 3x² − 4x + 1 = 0 has roots α, β, γ. Without solving the equation, calculate α² + β² + γ².
题目：三次方程 2x³ + 3x² − 4x + 1 = 0 的三个根为 α, β, γ。不解方程，计算 α² + β² + γ²。
Using the relationships between roots and coefficients for ax³ + bx² + cx + d = 0: Σα = −b/a = −3/2, Σαβ = c/a = −4/2 = −2. The required sum of squares is Σα² = (Σα)² − 2 Σαβ. Substitute: Σα² = (−3/2)² − 2(−2) = 9/4 + 4 = (9 + 16)/4 = 25/4.
利用根与系数的关系：对 ax³ + bx² + cx + d = 0，Σα = −b/a = −3/2，Σαβ = c/a = −4/2 = −2。所求平方和为 Σα² = (Σα)² − 2 Σαβ。代入得 Σα² = (−3/2)² − 2(−2) = 9/4 + 4 = (9 + 16)/4 = 25/4。
This method avoids explicitly finding the roots and is extremely efficient for symmetric expressions. Always check the sign of the sum of products pairwise: it is c/a, but the formula for the sum of squares uses exactly this coefficient.
该方法避免了显式求根，对对称式极为高效。务必检查两两乘积之和的符号：它等于 c/a，而平方和公式恰好用到该系数。
4. Proof by Induction: Sum of Squares Formula | 归纳法证明：平方和公式
Problem: Prove by induction that for all integers n ≥ 1, Σ_{r=1}^{n} r² = n(n+1)(2n+1)/6.
题目：用归纳法证明对所有整数 n ≥ 1，Σ_{r=1}^{n} r² = n(n+1)(2n+1)/6。
Base case n=1: LHS = 1² = 1, RHS = 1×2×3/6 = 1. Statement holds true. Inductive step: Assume true for n = k, i.e. Σ_{r=1}^{k} r² = k(k+1)(2k+1)/6. Then for n = k+1, Σ_{r=1}^{k+1} r² = Σ_{r=1}^{k} r² + (k+1)² = k(k+1)(2k+1)/6 + (k+1)².
基础情形 n=1：左边 = 1² = 1，右边 = 1×2×3/6 = 1。命题成立。归纳步骤：假设 n = k 时成立，即 Σ_{r=1}^{k} r² = k(k+1)(2k+1)/6。则 n = k+1 时，Σ_{r=1}^{k+1} r² = Σ_{r=1}^{k} r² + (k+1)² = k(k+1)(2k+1)/6 + (k+1)²。
Factor (k+1)/6: = (k+1)/6 [ k(2k+1) + 6(k+1) ] = (k+1)/6 [ 2k² + k + 6k + 6 ] = (k+1)/6 [ 2k² + 7k + 6 ] = (k+1)/6 (k+2)(2k+3). Since 2k+3 = 2(k+1)+1, the expression matches the formula for n = k+1, completing the induction.
提取 (k+1)/6：= (k+1)/6 [ k(2k+1) + 6(k+1) ] = (k+1)/6 [ 2k² + k + 6k + 6 ] = (k+1)/6 [ 2k² + 7k + 6 ] = (k+1)/6 (k+2)(2k+3)。因为 2k+3 = 2(k+1)+1，该式与 n = k+1 的公式一致，完成归纳。
5. Summation of Series: Using Standard Results | 级数求和：运用标准结果
Problem: Find an expression in terms of n for Σ_{r=1}^{n} r(r+2). Simplify your answer fully.
题目：求 Σ_{r=1}^{n} r(r+2) 关于 n 的表达式，并将答案完全化简。
Expand the term: r(r+2) = r² + 2r. So Σ_{r=1}^{n} r(r+2) = Σ_{r=1}^{n} r² + 2 Σ_{r=1}^{n} r. Using standard formulae: Σ r = n(n+1)/2 and Σ r² = n(n+1)(2n+1)/6.
展开通项：r(r+2) = r² + 2r。因此 Σ_{r=1}^{n} r(r+2) = Σ_{r=1}^{n} r² + 2 Σ_{r=1}^{n} r。利用标准公式：Σ r = n(n+1)/2，Σ r² = n(n+1)(2n+1)/6。
Substitute: S = n(n+1)(2n+1)/6 + 2 × n(n+1)/2 = n(n+1)(2n+1)/6 + n(n+1). Write n(n+1) as 6n(n+1)/6: S = [ n(n+1)(2n+1) + 6n(n+1) ] / 6 = n(n+1)[ (2n+1) + 6 ] / 6 = n(n+1)(2n+7) / 6.
代入：S = n(n+1)(2n+1)/6 + 2 × n(n+1)/2 = n(n+1)(2n+1)/6 + n(n+1)。将 n(n+1) 写作 6n(n+1)/6：S = [ n(n+1)(2n+1) + 6n(n+1) ] / 6 = n(n+1)[ (2n+1) + 6 ] / 6 = n(n+1)(2n+7) / 6。
The method of splitting the sum into simple polynomial terms is a fundamental skill in Further Mathematics; always keep the standard sigma results at hand.
将求和拆分为简单的多项式项是进阶数学中的一项基本技能；请始终牢记标准 sigma 结果。
6. Hyperbolic Functions: Solving an Equation Logarithmically | 双曲函数：用对数解方程
Problem: Solve the equation sinh x = 2, giving your answer in the form x = ln(a + √b) where a and b are integers.
题目：解方程 sinh x = 2，将答案写为 x = ln(a + √b) 的形式，其中 a 和 b 为整数。
Use the exponential definition: sinh x = (eˣ − e⁻ˣ)/2. Set (eˣ − e⁻ˣ)/2 = 2 ⇒ eˣ − e⁻ˣ = 4. Multiply throughout by eˣ: e²ˣ − 1 = 4eˣ ⇒ e²ˣ − 4eˣ − 1 = 0.
使用指数定义：sinh x = (eˣ − e⁻ˣ)/2。令 (eˣ − e⁻ˣ)/2 = 2 ⇒ eˣ − e⁻ˣ = 4。两边同乘 eˣ：e²ˣ − 1 = 4eˣ ⇒ e²ˣ − 4eˣ − 1 = 0。
Let y = eˣ, then y² − 4y − 1 = 0. By the quadratic formula, y = [4 ± √(16 + 4)] / 2 = [4 ± √20] / 2 = [4 ± 2√5] / 2 = 2 ± √5. Since y = eˣ > 0, we discard the negative root 2 − √5. Hence eˣ = 2 + √5, giving x = ln(2 + √5). Thus a = 2, b = 5.
设 y = eˣ，则 y² − 4y − 1 = 0。用求根公式，y = [4 ± √(16 + 4)] / 2 = [4 ± √20] / 2 = [4 ± 2√5] / 2 = 2 ± √5。因为 y = eˣ > 0，舍去负根 2 − √5。因此 eˣ = 2 + √5，得 x = ln(2 + √5)。于是 a = 2，b = 5。
7. Numerical Methods: Newton-Raphson Iteration | 数值方法：牛顿-拉夫森迭代
Problem: The equation x³ − 2x − 5 = 0 has a root near 2. Use the Newton-Raphson formula, x_{n+1} = x_n − f(x_n)/f'(x_n), starting with x₀ = 2, to perform two iterations and obtain an improved approximation. Round answers to 4 decimal places.
题目：方程 x³ − 2x − 5 = 0 在 2 附近有一个根。使用牛顿-拉夫森公式 x_{n+1} = x_n − f(x_n)/f'(x_n)，从 x₀ = 2 开始，进行两次迭代，求出改进的近似值。答案保留 4 位小数。
Define f(x) = x³ − 2x − 5, then f'(x) = 3x² − 2. Iteration 1: x₀ = 2. f(2) = 8 − 4 − 5 = −1; f'(2) = 12 − 2 = 10. Thus x₁ = 2 − (−1)/10 = 2 + 0.1 = 2.1000.
设 f(x) = x³ − 2x − 5，则 f'(x) = 3x² − 2。第一次迭代：x₀ = 2。f(2) = 8 − 4 − 5 = −1；f'(2) = 12 − 2 = 10。因此 x₁ = 2 − (−1)/10 = 2 + 0.1 = 2.1000。
Iteration 2: x₁ = 2.1. f(2.1) = (2.1)³ − 2(2.1) − 5 = 9.261 − 4.2 − 5 = 0.061; f'(2.1) = 3(4.41) − 2 = 13.23 − 2 = 11.
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