📚 Year 13 CCEA Further Mathematics: Unit Test Mock Paper Walkthrough | Year 13 CCEA 进阶数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 13 CCEA Further Mathematics, covering key topics from AS Pure and Applied modules. Each question is presented with a clear solution and commentary to reinforce understanding and exam technique.
本文为 Year 13 CCEA 进阶数学单元测试模拟卷提供详细解析，涵盖 AS 纯数学与应用数学的重要主题。每道题均配有清晰的解答与评注，以巩固理解与应试技巧。
1. Complex Numbers – Solving Quadratic Equations | 复数 – 解二次方程
Question: Solve the quadratic equation z² – (3 + i)z + (4 + 3i) = 0.
问题：求解二次方程 z² – (3 + i)z + (4 + 3i) = 0。
Solution: Use the quadratic formula z = [(3 + i) ± √Δ] / 2, where Δ = (3 + i)² – 4(4 + 3i).
解答：使用二次公式 z = [(3 + i) ± √Δ] / 2，其中 Δ = (3 + i)² – 4(4 + 3i)。
Calculate the discriminant: (3 + i)² = 9 + 6i + i² = 8 + 6i. Then Δ = (8 + 6i) – (16 + 12i) = –8 – 6i.
计算判别式：(3 + i)² = 9 + 6i + i² = 8 + 6i。于是 Δ = (8 + 6i) – (16 + 12i) = –8 – 6i。
Find the square root of Δ. Let √–8 – 6i = x + yi, so x² – y² = –8 and 2xy = –6.
求 Δ 的平方根。设 √–8 – 6i = x + yi，则 x² – y² = –8，2xy = –6。
Solving gives x = 1, y = –3 or x = –1, y = 3. Hence √Δ = ±(1 – 3i).
解得 x = 1, y = –3 或 x = –1, y = 3。因此 √Δ = ±(1 – 3i)。
z = [3 + i ± (1 – 3i)] / 2
Taking the plus sign: z₁ = (4 – 2i)/2 = 2 – i. Taking the minus sign: z₂ = (2 + 4i)/2 = 1 + 2i.
取加号：z₁ = (4 – 2i)/2 = 2 – i。取减号：z₂ = (2 + 4i)/2 = 1 + 2i。
The solutions are z = 2 – i and z = 1 + 2i.
方程的解为 z = 2 – i 和 z = 1 + 2i。
2. Matrices – Inverse and Linear Systems | 矩阵 – 逆与线性方程组
trix A = [ [2, 1], [3, 4] ], find A⁻¹ and hence solve the system 2x + y = 5, 3x + 4y = 13.
问题：已知矩阵 A = [ [2, 1], [3, 4] ]，求 A⁻¹ 并由此解方程组 2x + y = 5, 3x + 4y = 13。
Solution: First, det(A) = 2×4 – 1×3 = 8 – 3 = 5.
解答：首先，det(A) = 2×4 – 1×3 = 8 – 3 = 5。
The inverse of a 2×2 matrix is (1/det) × [ [d, –b], [–c, a] ]. Therefore:
2×2 矩阵的逆为 (1/det) × [ [d, –b], [–c, a] ]。因此：
A⁻¹ = (1/5) [ [4, –1], [–3, 2] ]
system in matrix form: A [x; y] = [5; 13]. Multiplying by A⁻¹ gives [x; y] = A⁻¹ [5; 13].
将方程组写为矩阵形式：A [x; y] = [5; 13]。左乘 A⁻¹ 得 [x; y] = A⁻¹ [5; 13]。
Compute: x = (1/5)(4×5 + (–1)×13) = (1/5)(20 – 13) = 7/5, y = (1/5)((–3)×5 + 2×13) = (1/5)(–15 + 26) = 11/5.
计算：x = (1/5)(4×5 + (–1)×13) = (1/5)(20 – 13) = 7/5，y = (1/5)((–3)×5 + 2×13) = (1/5)(–15 + 26) = 11/5。
Hence the solution is x = 7/5, y = 11/5.
因此解为 x = 7/5, y = 11/5。
3. Vectors – Angle Between Two Lines | 向量 – 直线间的夹角
Question: Find the acute angle between the lines r₁ = (i + 2j – k) + λ(i – 2j + 3k) and r₂ = (3i + k) + μ(2i + j – k).
问题：求直线 r₁ = (i + 2j – k) + λ(i – 2j + 3k) 与 r₂ = (3i + k) + μ(2i + j – k) 之间的锐角。
Solution: The direction vectors are a = (1, –2, 3) and b = (2, 1, –1).
解答：方向向量分别为 a = (1, –2, 3) 和 b = (2, 1, –1)。
Dot product: a·b = 1×2 + (–2)×1 + 3×(–1) = 2 – 2 – 3 = –3.
点积：a·b = 1×2 + (–2)×1 + 3×(–1) = 2 – 2 – 3 = –3。
Magnitudes: |a| = √(1² + (–2)² + 3²)
Published by TutorHao | Year 13 进阶数学 Revision Series | aleveler.com
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