Year 12 WJEC Further Mathematics: Unit Test Mock Paper Analysis | 12年级 WJEC 进阶数学：单元测试模拟卷解析

📚 Year 12 WJEC Further Mathematics: Unit Test Mock Paper Analysis | 12年级 WJEC 进阶数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a typical Year 12 WJEC Further Mathematics Unit Test mock paper, covering key topics from the Further Pure specification. Each section breaks down a core question, highlighting essential concepts, common pitfalls, and step-by-step solutions to help you strengthen your exam technique.

本文详细解析一份典型的12年级 WJEC 进阶数学单元测试模拟卷，涵盖进阶纯数课程的核心主题。每个小节分解一道核心题目，突出基本概念、常见错误和逐步解题过程，帮助你提升考试技巧。

1. Complex Numbers: Quadratic with Complex Roots | 复数：二次方程与复根

Question: Solve the equation z² − 4z + 13 = 0, giving your answers in the form a ± bi.

题目：解方程 z² − 4z + 13 = 0，将答案写成 a ± bi 的形式。

Step 1: Apply the quadratic formula z = [−b ± √b² − 4ac] / (2a) with a = 1, b = −4, c = 13. The discriminant is (−4)² − 4(1)(13) = 16 − 52 = −36.

步骤1：代入二次公式 z = [−b ± √b² − 4ac] / (2a)，其中 a=1, b=−4, c=13。判别式为 (−4)² − 4×1×13 = 16 − 52 = −36。

Step 2: Since the discriminant is negative, we use i² = −1. √−36 = √36 × √−1 = 6i. Hence z = [4 ± 6i] / 2 = 2 ± 3i.

步骤2：由于判别式为负，利用 i² = −1，得 √−36 = √36 × √−1 = 6i。于是 z = (4 ± 6i) / 2 = 2 ± 3i。

lify your final answer into a ± bi form and separate the real and imaginary parts clearly.

步骤3：最终答案务必化简为 a ± bi 形式，并明确分开实部和虚部。

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2. Modulus-Argument Form and Loci | 复数的模-辐角形式与轨迹

dulus-argument form r(cos θ + i sin θ), with −π < θ ≤ π. Hence describe the locus given by |z − (2 + i)| = 3.

题目：已知 z = −3 + 4i，将 z 表示为模-辐角形式 r(cos θ + i sin θ)，其中 −π < θ ≤ π。由此描述 |z − (2 + i)| = 3 所表示的轨迹。

Step 1: Compute modulus r = |z| = √((−3)² + 4²) = √9 + 16 = √25 = 5.

步骤1：计算模 r = |z| = √((−3)² + 4²) = √9 + 16 = √25 = 5。

ce z lies in the second quadrant, θ = π − tan⁻¹(4/3). In radians, approximately 2.214 rad.

步骤2：辐角 θ 满足 tan θ = 4/(−3)。因 z 位于第二象限，θ = π − tan⁻¹(4/3)，弧度近似为 2.214 rad。

Step 3: Thus z = 5 (cos θ + i sin θ). The locus |z − (2 + i)| = 3 represents a circle centred at (2, 1) with radius 3 in the complex plane.

步骤3：因此 z = 5(cos θ + i sin θ)。轨迹 |z − (2 + i)| = 3 表示复平面上以 (2, 1) 为圆心、半径为3的圆。

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3. Matrices: Determinant and Inverse of a 3×3 Matrix | 矩阵：3阶行列式与逆矩阵

Question: For matrix A =

2	1	−1
0	3	2
1	−1	2

, find det(A) and hence A⁻¹.

题目：已知矩阵 A =

2	1	−1
0	3	2
1	−1	2

，求 det(A) 及 A⁻¹。

)) − 1(0×2 − 2×1) + (−1)(0×(−1) − 3×1) = 2(6 + 2) − 1(0 − 2) − 1(0 − 3) = 16 + 2 + 3 = 21.

步骤1：按第一行计算行列式： − 2×(−1)) − 1(0×2 − 2×1) + (−1)(0×(−1) − 3×1) = 2(6+2) − 1(0−2) − 1(0−3) = 16 + 2 + 3 = 21。

then transpose to get the adjugate. Finally, divide each entry by det(A)=21 to obtain A⁻¹.

步骤2：求出余子式矩阵并转置得到伴随矩阵，再将每个元素除以行列式21，得到逆矩阵 A⁻¹。

erified by checking AA⁻¹ = I. This is a key skill for solving systems in the next section.

步骤3：可通过验证 AA⁻¹ = I 来检查。这是下一节中求解方程组的关键技能。

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4. Solving Linear Systems Using Matrices | 利用矩阵解线性方程组

Question: Use matrices to solve the system: 2x + y − z = 5, 3y + 2z = 4, x − y + 2z = 3.

题目：利用矩阵解方程组：2x + y − z = 5，3y + 2z = 4，x − y + 2z = 3。

Write the system as AX = B, where A is the coefficient matrix from Question 3, X = [x y z]ᵀ, and B = [5 4 3]ᵀ.

步骤1：将方程组写成 AX = B，其中 A 是第3题中的系数矩阵，X = [x y z]ᵀ，B = [5 4 3]ᵀ。

⁻¹: X = A⁻¹B. Using the inverse found earlier, compute the product to obtain the solution.

步骤2：两边同乘 A⁻¹：X = A⁻¹B。利用之前求出的逆矩阵计算乘积，得到解。

Check by substituting back: x = 2, y = 1, z = −1 often emerges from such a typical design.

步骤3：最终值应满足每个方程。可通过回代检验，例如 x = 2, y = 1, z = −1 是常见设计解。

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5. Roots of Polynomials: Sums and Products of Roots | 多项式的根：根的和与积

³ − 5x² + 8x − 6 = 0 has roots α, β, γ. Find ∑α, ∑αβ, αβγ and hence evaluate α² + β² + γ².

题目：三次方程 x³ − 5x² + 8x − 6 = 0 的根为 α, β, γ。求 ∑α, ∑αβ, αβγ，并由此计算 α² + β² + γ²。

= 0, ∑α = −p, ∑αβ = q, αβγ = −r. Here p = −5, q = 8, r = −6, so ∑α = 5, ∑αβ = 8, αβγ = 6.

步骤1：对于 x³ + px² + qx + r = 0，∑α = −p, ∑αβ = q, αβγ = −r。这里 p = −5, q = 8, r = −6，所以 ∑α = 5, ∑αβ = 8, αβγ = 6。

Step 2: Use the identity α² + β² + γ² = (∑α)² − 2∑αβ = 5² − 2×8 = 25 − 16 = 9.

步骤2：利用恒等式 α² + β² + γ² = (∑α)² − 2∑αβ = 5² − 2×8 = 25 − 16 = 9。

Step 3: This type of manipulation is frequently examined; memorising the symmetric sums saves time.

步骤3：这类变形经常考查，熟记对称和能节约时间。

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6. Proof by Induction: Summation of Squares | 数学归纳法：平方和

Question: Prove by induction that for all positive integers n, ∑_r=1ⁿ r² = n(n+1)(2n+1)/6.

题目：用数学归纳法证明：对所有正整数 n，有 ∑_r=1ⁿ r² = n(n+1)(2n+1)/6。

Step 1: Base case n=1: LHS = 1² = 1, RHS = 1×2×3/6 = 1. True.

步骤1：基础情况 n=1：左边 = 1，右边 = 1×2×3/6 = 1，成立。

Step 2: Assume true for n = k, i.e. ∑_r=1^k r² = k(k+1)(2k+1)/6. For n = k+1, LHS = ∑_r=1^k r² + (k+1)². Substitute the assumption and combine into a single fraction.

步骤2：假设 n = k 时成立，即 ∑_r=1^k r² = k(k+1)(2k+1)/6。对于 n = k+1，左边 = 假设和 + (k+1)²，代入猜想并通分合并。

aic simplification yields (k+1)(k+2)(2k+3)/6, which is the RHS with n = k+1. Hence proved.

步骤3：代数化简后得到 (k+1)(k+2)(2k+3)/6，正是 n = k+1 时的右边。证毕。

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7. Proof by Induction: Divisibility | 数学归纳法：整除性

Question: Prove that 5ⁿ − 1 is divisible by 4 for all positive integers n.

题目：证明对所有正整数 n，5ⁿ − 1 能被 4 整除。

Step 1: Base case n=1: 5¹ − 1 = 4, which is divisible by 4.

步骤1：基础情况 n=1：5¹ − 1 = 4，可被4整除。

ider 5ᵏ⁺¹ − 1 = 5×5ᵏ − 1 = 5(5ᵏ − 1) + 4 = 5×4m + 4 = 4(5m + 1), which is a multiple of 4.

步骤2：假设 n = k 时成立：5ᵏ − 1 = 4m。对于 n = k+1，考虑 5ᵏ⁺¹ − 1 = 5×5ᵏ − 1 = 5(5ᵏ − 1) + 4 = 5×4m + 4 = 4(5m + 1)，是4的倍数。

Step 3: Therefore, if the statement holds for k, it holds for k+1. By induction, it is true for all n ∈ ℕ.

步骤3：因此，若命题对 k 成立，则对 k+1 也成立。由归纳法，对所有正整数 n 均成立。

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8. Summation of Series: Using Standard Results | 级数求和：利用标准结果

Question: Evaluate ∑_r=1ⁿ (r+1)(r+3) and simplify your answer fully.

题目：计算 ∑_r=1ⁿ (r+1)(r+3)，并化简最终结果。

+1)(r+3) = r² + 4r + 3. Therefore the sum splits into three separate sums: ∑r² + 4∑r + ∑3.

步骤1：展开通项：(r+1)(r+3) = r² + 4r + 3。因此和式拆分为三个独立求和：∑r² + 4∑r + ∑3。

Step 2: Apply standard formulae: ∑_r=1ⁿ r = n(n+1)/2, ∑_r=1ⁿ r² = n(n+1)(2n+1)/6, and ∑_r=1ⁿ 3 = 3n.

步骤2：代入标准公式：∑_r=1ⁿ r = n(n+1)/2，∑_r=1ⁿ r² = n(n+1)(2n+1)/6，∑_r=1ⁿ 3 = 3n。

+1) + 3n. Put over a common denominator 6 to get n(2n² + 15n + 19)/6 after simplification.

步骤3：合并：n(n+1)(2n+1)/6 + 2n(n+1) + 3n。通分分母6，化简后得 n(2n² + 15n + 19)/6。

Step 4: Always check your expansion and factorisation to avoid algebraic errors.

步骤4：始终检查展开与因式分解，避免代数错误。

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