📚 Year 10 WJEC Further Mathematics: Unit Test Practice Paper Walkthrough | 威尔士联合考试局十年级进阶数学：单元测试模拟卷解析
Welcome to this walkthrough of a WJEC Year 10 Further Mathematics unit test practice paper. This resource breaks down key topics including quadratics, polynomials, binomial expansion, exponentials & logarithms, coordinate geometry, sequences, differentiation, and integration. By studying the solutions step by step, students can strengthen their problem-solving skills and build confidence for assessments that reflect typical WJEC Additional Maths style. Each question is carefully designed to mirror the depth and demand of a real unit test.
欢迎阅读威尔士联合考试局十年级进阶数学单元测试模拟卷的逐题解析。本资源涵盖了二次函数、多项式、二项式展开、指数与对数、坐标几何、数列、求导与积分等核心主题。通过逐步学习解题过程，同学们可以提升解题能力和应试技巧。每道题目均贴近 WJEC 进阶数学的典型考查方式，帮助大家查漏补缺。
1. Quadratic & Discriminant | 二次函数与判别式
Question: Solve the quadratic equation 2x2 – 3x – 5 = 0, and hence describe the nature of its roots using the discriminant.
问题：解二次方程 2x2 – 3x – 5 = 0，并利用判别式说明根的性质。
Step 1: Identify the coefficients a = 2, b = -3, c = -5.
步骤1：确定系数 a = 2, b = -3, c = -5。
Step 2: Write the discriminant formula Δ = b2 – 4ac.
步骤2：写出判别式公式 Δ = b2 – 4ac。
Substitute the values: Δ = (-3)2 – 4(2)(-5) = 9 + 40 = 49.
代入数值：Δ = (-3)2 – 4(2)(-5) = 9 + 40 = 49。
Since Δ > 0, the quadratic has two distinct real roots.
因为 Δ > 0，所以该二次方程有两个不相等的实根。
Step 3: Use the quadratic formula x = [-b ± √Δ] / (2a) to find the roots.
步骤3：使用求根公式 x = [-b ± √Δ] / (2a) 求根。
x = [3 ± √49] / (2×2) = (3 ± 7) / 4.
x = [3 ± √49] / (2×2) = (3 ± 7) / 4。
Thus, the two solutions are x = (3+7)/4 = 10/4 = 2.5 and x = (3-7)/4 = -4/4 = -1.
因此，两个解为 x = (3+7)/4 = 10/4 = 2.5 和 x = (3-7)/4 = -4/4 = -1。
Final answer: x = 2.5 or x = -1; the roots are real and distinct.
最终答案：x = 2.5 或 x = -1；根为两不等实根。
2. Factor Theorem & Polynomial Division | 因式定理与多项式除法
Question: Show that (x+2) is a factor of f(x) = x3 + 5x2 + 2x – 8, and fully factorise f(x).
问题：证明 (x+2) 是 f(x) = x3 + 5x2 + 2x – 8 的因式，并将 f(x) 完全因式分解。
Step 1: Apply the Factor Theorem – evaluate f(-2). If f(-2)=0, then (x+2) is a factor.
步骤1：应用因式定理 —— 计算 f(-2)。若 f(-2)=0，则 (x+2) 为因式。
f(-2) = (-2)3 + 5(-2)2 + 2(-2) – 8 = -8 + 20 – 4 – 8 = 0.
f(-2) = (-2)3 + 5(-2)2 + 2(-2) – 8 = -8 + 20 – 4 – 8 = 0。
Since the remainder is 0, (x+2) is indeed a factor.
由于余数为 0，确实 (x+2) 是其中一个因式。
Step 2: Perform polynomial division of f(x) by (x+2), either by long division or comparing coefficients.
步骤2：用长除法或系数比较法做 f(x) 与 (x+2) 的多项式除法。
Dividing x3 + 5x2 + 2x – 8 by (x+2) gives a quadratic quotient: x2 + 3x – 4.
x3 + 5x2 + 2x – 8 除以 (x+2) 得到二次商式：x2 + 3x – 4。
Step 3: Factorise the quadratic x2 + 3x – 4.
步骤3：将二次式 x2 + 3x – 4 因式分解。
We need two numbers that multiply to -4 and add to 3: these are 4 and -1.
寻找两数乘积为 -4、和为 3：这两个数是 4 和 -1。
So x2 + 3x – 4 = (x+4)(x-1).
因此 x2 + 3x – 4 = (x+4)(x-1)。
Full factorisation: f(x) = (x+2)(x+4)(x-1).
完全因式分解：f(x) = (x+2)(x+4)(x-1)。
3. Binomial Expansion | 二项式展开
Question: Find the first four terms in the expansion of (1 + 2x)5, simplifying the coefficients.
问题：写出 (1 + 2x)5 展开式的前四项，并化简系数。
Step 1: Use the binomial theorem: (a+b)n = Σ nCr an-r br. Here a=1, b=2x, n=5.
步骤1：运用二项式定理：(a+b)n = Σ nCr an-r br。此处 a=1, b=2x, n=5。
Step 2: Write the general term Tr+1 = 5Cr (1)5-r (2x)r = 5Cr 2r xr
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