📚 A-Level WJEC Further Maths: Unit Test Mock Exam Breakdown | A-Level WJEC 进阶数学：单元测试模拟卷解析
This article breaks down a complete Unit Test mock paper for WJEC A-Level Further Mathematics, focusing on the core pure topics in Unit 3 (Further Pure A). Each section presents a representative exam question, a step-by-step solution, and targeted commentary to strengthen conceptual understanding and exam technique.
本文详细解析一套针对 WJEC A-Level 进阶数学单元测试的完整模拟卷，重点覆盖第三单元（Further Pure A）的核心纯数主题。每一节均提供一道典型考题、逐步求解过程以及有针对性的点评，以强化概念理解与应试技巧。
1. Complex Numbers: Cartesian Form Operations | 复数：笛卡尔形式运算
A unit test almost always begins with basic arithmetic in Cartesian form, such as addition, subtraction, multiplication, division and conjugation of complex numbers.
单元测试通常以笛卡尔形式的基本运算开场，例如复数的加法、减法、乘法、除法和取共轭。
Mock Question: Given z = 2 + 3i and w = 4 – 5i, calculate: (a) z + w, (b) z – w, (c) zw, (d) z/w, (e) the complex conjugate of z.
模拟题：已知 z = 2 + 3i，w = 4 – 5i，计算：(a) z + w，(b) z – w，(c) zw，(d) z / w，(e) z 的共轭复数。
Part (a): Add real and imaginary parts separately: (2 + 4) + (3 – 5)i = 6 – 2i.
(a) 解：实部与虚部分别相加：(2 + 4) + (3 – 5)i = 6 – 2i。
Part (b): Subtract real and imaginary parts: (2 – 4) + (3 – (–5))i = –2 + 8i.
(b) 解：实部与虚部分别相减：(2 – 4) + (3 – (–5))i = –2 + 8i。
Part (c): Multiply using i2 = –1: (2)(4) + (2)(–5i) + (3i)(4) + (3i)(–5i) = 8 – 10i + 12i – 15i2 = 8 + 2i + 15 = 23 + 2i.
(c) 解：利用 i2 = –1 展开相乘：… = 23 + 2i。
Part (d): Multiply numerator and denominator by the conjugate of the denominator (4 + 5i): (2+3i)(4+5i) / (42+52) = (8+10i+12i+15i2)/41 = (–7 + 22i)/41 = –7/41 + (22/41)i.
(d) 解：分子分母同乘分母的共轭复数 (4+5i)，得 –7/41 + (22/41)i。
Part (e): The conjugate z* = 2 – 3i.
(e) 解：共轭复数 z* = 2 – 3i。
Tip: Always express final answers in the form a + bi and double-check the sign of the imaginary part when dividing.
提示：最终答案务必写成 a + bi 的形式，并在做除法时仔细核对虚部的符号。
2. Matrix Algebra: Solving Linear Systems | 矩阵代数：解线性方程组
Many tests include a system of two linear equations in two unknowns that must be solved using the inverse matrix method.
许多测试包含需要利用逆矩阵方法求解的二元一次方程组。
Mock Question: Use matrices to solve the system: 2x + y = 5, 3x – 2y = 4.
模拟题：用矩阵求解方程组：2x + y = 5, 3x – 2y = 4。
Write the system as AX = B, where A =
| 2 | 1 |
| 3 | -2 |
, X =
| x |
| y |
, B =
| 5 |
| 4 |
.
将方程组写成矩阵形式 AX = B，其中 A、X、B 如上所示。
Find the determinant of A: det(A) = 2×(-2) – 1×3 = –4 – 3 = –7.
计算 A 的行列式：det(A) = 2×(-2) – 1×3 = –7。
Since det(A) ≠ 0, the inverse exists: A-1 = (1/–7)
| -2 | -1 |
| -3 | 2 |
.
因为 det(A) ≠ 0，逆矩阵存在：A-1 = (1/–7) 如上矩阵。
Then X = A-1B = (–1/7)
| -2 | -1 |
| -3 | 2 |
| 5 |
| 4 |
= (–1/7)
| -10 – 4 |
| -15 + 8 |
= (–1/7)
| -14 |
| -7 |
=
| 2 |
| 1 |
, so x = 2, y = 1.
于是 X = A-1B，计算得 x = 2，y = 1。
Tip: Always verify by substituting back into the original equations. Here 2(2)+1=5 and 3(2)-2(1)=4.
提示：务必代回原方程验证。这里 2(2)+1=5，3(2)-2(1)=4，成立。
3. Summation of Series: Method of Differences | 级数求和：差分法
Questions on series often test the method of differences to find an exact formula for the partial sum of a rational expression.
级数问题常考察利用差分法求有理分式的部分和精确公式。
Mock Question: Find an expression for Σr=1n 1/(r(r+1)) and use it to evaluate the sum from r=1 to 50.
模拟题：求 Σr=1n 1/(r(r+1)) 的表达式，并用它计算 r=1 到 50 的和。
First, decompose into partial fractions: 1/(r(r+1)) ≡ 1/r – 1/(r+1).
首先，分解为部分分式：1/(r(r+1)) ≡ 1/r – 1/(r+1)。
Write the sum: Σr=1n (1/r – 1/(r+1)). Expand terms: (1/1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + … + (1/n – 1/(n+1)).
写出求和式，展开成 (1/1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n+1))。
The terms telescope: everything cancels except the first and last terms, leaving 1 – 1/(n+1) = n/(n+1).
中间项全部消去，仅剩首尾两项，求和结果为 1 – 1/(n+1) = n/(n+1)。
For n = 50, the sum equals 50/51.
当 n = 50 时，和为 50/51。
Tip: Always state the final telescoped form clearly and check with a small n value, e.g., n=2 gives 1/2 + 1/6 = 2/3, and formula gives 2/3.
提示：明确写出化简后的通项，并用小的 n 值验证，如 n=2 时和为 2/3。
4. Proof by Induction: Divisibility | 归纳法证明：整除性
Induction is a staple topic; a divisibility proof appears frequently in WJEC Unit 3 papers.
归纳法是必考主题；WJEC 第三单元试卷中频繁出现整除性证明。
Mock Question: Prove by induction that 32n – 1 is divisible by 8 for all positive integers n.
模拟题：用数学归纳法证明对所有正整数 n，32n – 1 能被 8 整除。
Basis (n = 1): 32 – 1 = 9 – 1 = 8, which is divisible by 8.
奠基 (n = 1)：32 – 1 = 8，能被 8 整除。
Inductive hypothesis: Assume true for n = k, so 32k – 1 = 8m for some integer m.
归纳假设：假设 n = k 时命题成立，即存在整数 m 使得 32k – 1 = 8m。
Inductive step (n = k+1): Consider 32(k+1) – 1 = 32k+2 – 1 = 9·32k – 1. Rewrite as 9(32k – 1) + 9 – 1 = 9(8m) + 8 = 8(9m + 1), clearly divisible by 8.
归纳递推 (n = k+1)：考虑 32(k+1) – 1 = 9·32k – 1，变形为 9(32k – 1) + 8 = 9·8m + 8 = 8(9m + 1)，显然能被 8 整除。
Hence, by mathematical induction, the statement is true for all positive integers n.
因此，由数学归纳法可知对所有正整数 n 命题成立。
Tip: The key step is to extract the term 32k – 1 and add a compensating term; always clearly state the integer multiplier.
提示：关键步骤是提取 32k – 1 并添加补偿项；务必明确指出整除后的整数乘积式。
5. Roots of Polynomials: Transformation of Roots | 多项式根：根的变换
Cubic and quartic root relations often involve transforming the roots to obtain a new polynomial equation without solving the original.
三次或四次方程的根关系常涉及对根进行变换以得到新多项式方程，而无需求解原方程。
Mock Question: The cubic equation x3 – 3x2 + 4 = 0 has roots α, β, γ. Find the cubic equation whose roots are α + 2, β + 2, γ + 2.
模拟题：已知三次方程 x3 – 3x2 + 4 = 0 的根为 α, β, γ，求以 α+2, β+2, γ+2 为根的三次方程。
Let the new variable be y = x + 2, so x = y – 2. Substitute into the original equation: (y – 2)3 – 3(y – 2)2 + 4 = 0.
设新变量 y = x + 2，则 x = y – 2。代入原方程：(y – 2)3 – 3(y – 2)2 + 4 = 0。
Expand: (y3 – 6y2 + 12y – 8) – 3(y2 – 4y + 4) + 4 = 0.
展开：y3 – 6y2 + 12y – 8 – 3y2 + 12y – 12 + 4 = 0。
Simplify: y3 – 9y2 + 24y – 16 = 0.
整理得所需方程：y3 – 9y2 + 24y – 16 = 0。
Thus, the required equation is x3 – 9x2 + 24x – 16 = 0.
因此，所求方程为 x3 – 9x2 + 24x – 16 = 0。
Tip: Substitution method avoids symmetric sum calculations; always expand carefully and collect like terms.
提示：代换法避免了对称和计算；一定要仔细展开并合并同类项。
6. Argand Diagram: Loci and Half-Lines | 阿尔冈图：轨迹与射线
Locus questions test the ability to interpret modulus and argument equations geometrically and sketch them on an Argand diagram.
轨迹问题考查对模和辐角方程进行几何解释并在阿尔冈图上作图的能力。
Mock Question: Sketch the locus of points z satisfying |z – 3| = |z + i|, and describe it geometrically.
模拟题：画出满足 |z – 3| = |z + i| 的点 z 的轨迹，并从几何角度
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