Year 11 WJEC Further Maths: Unit Test Mock Paper Walkthrough | Year 11 WJEC 进阶数学：单元测试模拟卷解析

📚 Year 11 WJEC Further Maths: Unit Test Mock Paper Walkthrough | Year 11 WJEC 进阶数学：单元测试模拟卷解析

This article walks you through a typical WJEC Year 11 Further Mathematics unit test mock paper, covering core topics such as matrices, complex numbers, polynomials, roots of equations, proof by induction, trigonometry, vectors, and differentiation. Each section presents a sample question followed by a step-by-step solution explained in both English and Chinese, helping you master the skills needed for top marks.

本文带你逐题解析一套典型的WJEC Year 11 进阶数学单元测试模拟卷，内容涵盖矩阵、复数、多项式、方程根、归纳法证明、三角、向量和微分等核心主题。每个部分先呈现一道样题，然后以中英双语分步讲解，帮你掌握获得高分的关键技巧。

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1. Mock Paper Overview and Key Topics | 模拟卷概览与关键主题

The mock paper is designed to reflect the WJEC Year 11 Further Mathematics specification, mixing short structured questions and longer problem-solving tasks. It tests both fluency with algebraic manipulation and the ability to apply concepts in unfamiliar settings.

本模拟卷贴合WJEC Year 11 进阶数学大纲，融合了简短结构化问题和较长的综合题，既考查代数运算的熟练度，也评估在新情境中应用概念的能力。

You will encounter questions on matrix operations, complex number arithmetic, factor theorem, using relationships between roots and coefficients, proving sums by induction, solving trigonometric equations, working with vector geometry, and basic differentiation. A solid performance requires careful reading of each question, showing clear working, and double-checking for common sign errors or domain restrictions.

你将遇到的题目涉及矩阵运算、复数四则运算、因式定理、利用根与系数关系、归纳法证明求和公式、解三角方程、向量几何以及基础微分。想取得好成绩，需要仔细审题、展示清晰的解题步骤，并反复检查符号错误或定义域限制。

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2. Matrix Multiplication and Applications | 矩阵乘法及其应用

Question: Hence, or otherwise, find the matrix X such that AX = B, stating any necessary condition.

题目：已知 A = [2, 1; -1, 3], B = [0, 4; 2, -1], 求乘积 AB。由此求矩阵 X 使得 AX = B，并说明所需条件。

Solution: + 1×(-1) = 7; (AB)₂₁ = -1×0 + 3×2 = 6; (AB)₂₂ = -1×4 + 3×(-1) = -7. So AB = [2, 7; 6, -7].

解析：逐行乘列计算: + 1×2 = 2; (AB)₁₂ = 2×4 + 1×(-1) = 7; (AB)₂₁ = -1×0 + 3×2 = 6; (AB)₂₂ = -1×4 + 3×(-1) = -7。因此 AB = [2, 7; 6, -7]。

tible. X = A⁻¹B = (1/7)[3, -1; 1, 2] B. Compute X = [0.2857..., 1.857...; 0.5714..., 0.2857...] exactly as [2/7, 13/7; 4/7, 2/7] after simplification. Always verify by substitution.

求解 AX = B，两边左乘 A⁻¹（前提是行列式非零）。A 的行列式为 (2)(3) - (1)(-1) = 7，故 A 可逆。X = A⁻¹B = (1/7)[3, -1; 1, 2] B。计算后化简得 X = [2/7, 13/7; 4/7, 2/7]。务必代入原方程验证。

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3. Complex Numbers: Arithmetic and Modulus | 复数：四则运算与模

Question: = 1 - 2i. Find (a) z + w, (b) z w, (c) z / w, giving answers in the form a + bi. Determine |z| and arg(z) in degrees to one decimal place.

题目：设 z = 3 + 4i，w = 1 - 2i。求 (a) z + w, (b) z w, (c) z / w，结果以 a + bi 形式表示。求 |z| 和 arg(z)，角度精确到一位小数。

Solution: and denominator by the conjugate of denominator: (3+4i)(1+2i) / (1-2i)(1+2i) = (3+6i+4i+8i²) / (1 - 4i²) = (3+10i-8) / (1+4) = (-5+10i)/5 = -1 + 2i. Modulus |z| = √3² + 4² = 5. Argument arg(z) = tan⁻¹(4/3) ≈ 53.1°.

解析：(a) 实部与虚部分别相加得 4+2i。(b) 展开时注意 i² = -1: 3 - 6i + 4i - 8(-1) = 11 - 2i。(c) 分子分母同乘分母的共轭： (3+4i)(1+2i)/(1+4) = (3+10i-8)/5 = -1+2i。模长 |z| = √9+16=5。辐角 arg(z) = tan⁻¹(4/3) ≈ 53.1°。

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4. Polynomial Factorisation Using Factor Theorem | 利用因式定理分解多项式

Question: Show that (x - 2) is a factor of f(x) = x³ - 4x² + x + 6. Hence factorise f(x) completely.

题目：证明 (x - 2) 是 f(x) = x³ - 4x² + x + 6 的因式，并由此将 f(x) 完全分解。

Solution: 2) using synthetic or long division: coefficients 1, -4, 1, 6. Bring down 1, multiply by 2 → 2, add to -4 → -2, multiply by 2 → -4, add to 1 → -3, multiply by 2 → -6, add to 6 → 0. The quotient is x² - 2x - 3. Factorise the quadratic: (x - 3)(x + 1). Therefore f(x) = (x - 2)(x - 3)(x + 1).

解析：因式定理：若 f(2)=0，则 (x-2) 为因式。f(2)=8-16+2+6=0，成立。用综合除法：系数 1, -4, 1, 6，除以 (x-2) 得商 x² - 2x - 3。再分解二次式得 (x-3)(x+1)。故 f(x) = (x-2)(x-3)(x+1)。

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5. Roots and Coefficients: Quadratic Equations | 根与系数：二次方程

Question: as roots α and β. Without solving the equation, find (a) α + β, (b) αβ, (c) the quadratic equation whose roots are 2α + 1 and 2β + 1.

题目：二次方程 2x² + 3x - 4 = 0 有根 α 和 β。不解方程，求 (a) α + β, (b) αβ, (c) 以 2α+1 和 2β+1 为根的二次方程。

Solution: (-3/2)+2 = -3+2 = -1. Product = (2α+1)(2β+1) = 4αβ + 2(α+β) + 1 = 4(-2) + 2(-3/2) + 1 = -8 -3 + 1 = -10. New equation: x² - (sum)x + product = 0, i.e., x² - (-1)x + (-10) = 0 ⇒ x² + x - 10 = 0.

解析：对于 ax²+bx+c=0，根之和 = -b/a，积 = c/a。故 α+β = -3/2，αβ = -2。新根之和 = 2(α+β)+2 = -1，积 = 4αβ + 2(α+β) + 1 = -10。新方程为 x² - (和)x + 积 = 0 ⇒ x² + x - 10 = 0。

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6. Proof by Induction: Sum of Squares | 归纳法证明：平方和公式

Question: Prove by induction that for all positive integers n, 1² + 2² + ... + n² = n(n+1)(2n+1)/6.

题目：用归纳法证明对所有正整数 n，有 ∑_k=1ⁿ k² = n(n+1)(2n+1)/6。

Solution: n=1: LHS = 1² = 1. RHS = 1×2×3/6 = 1. True. Inductive step: Assume true for n = m, i.e., ∑_k=1^m k² = m(m+1)(2m+1)/6. For n = m+1, LHS = ∑_k=1^m : (2m+3)(m+2). So LHS = (m+1)(m+2)(2m+3)/6 = (m+1)[(m+1)+1][2(m+1)+1]/6, which matches the RHS form for n = m+1. Therefore, by induction, the statement holds for all n.

解析：基始 n=1：左 = 1，右 = 1，成立。假设 n=m 时成立，即 ∑₁ᵐ k² = m(m+1)(2m+1)/6。n=m+1 时，左 = m(m+1)(2m+1)/6 + (m+1)²。提出 (m+1)/6，得 (m+1)/6 [2m²+7m+6] = (m+1)(m+2)(2m+3)/6，恰为 n=m+1 的形式。由归纳法，对所有 n 成立。

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7. Solving Trigonometric Equations | 解三角方程

Question: - sin x - 1 = 0 for 0° ≤ x ≤ 360°, giving answers to the nearest degree where appropriate.

题目：解方程 2 sin²x - sin x - 1 = 0，x 在 0° 到 360° 之间，答案精确到度。

Solution: Let y = sin x. Then 2y² - y - 1 = 0. Factorise: (2y + 1)(y - 1) = 0 ⇒ 180°+30°=210° and 360°-30°=330°. sin x = 1 gives x = 90°. Solutions: x = 90°, 210°, 330°.

解析：设 y = sin x，得 2y² - y - 1 = 0，分解为 (2y+1)(y-1)=0 ⇒ y = -½ 或 1。sin x = 1 ⇒ x = 90°。sin x = -½，参考角 30°，在区间内得 x = 210° 和 330°。解集 {90°, 210°, 330°}。

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8. Functions, Inverses and Graphs | 函数、反函数与图像

Question: 3 for all real x. (a) Find the inverse function f⁻¹(x). (b) Sketch the graphs of f and f⁻¹ on the same axes, indicating the line of symmetry.

题目：函数 f(x)=2x+3 对所有实数定义。求反函数 f⁻¹(x)，并在同一坐标系中画出 f 和 f⁻¹ 的图像，标出对称轴。

Solution: ine y = x. f has y-intercept (0,3) and slope 2; f⁻¹ has y-intercept (0, -1.5) and slope ½. Sketching clearly shows the reflection property.

解析：由 y=2x+3，交换 x,y 得 x=2y+3 ⇒ y=(x-3)/2，故 f⁻¹(x)=(x-3)/2。f 与 f⁻¹ 为直线，关于 y=x 对称。f 截距 (0,3) 斜率 2；f⁻¹ 截距 (0,-1.5) 斜率 ½。作图时标出对称轴 y=x。

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9. Vectors: Dot Product and Angle | 向量：点积与夹角

Question: nd (a) 2a - b, (b) the exact magnitude of a, (c) the angle between a and b to the nearest degree.

题目：已知向量 a = 2i + j, b = -i + 3j。求 (a) 2a - b, (b) a 的模的精确值, (c) a 与 b 的夹角（精确到度）。

Solution: (a) 2a - b = 2(2i + j) - (-i + 3j) = 4i + 2j + i - 3j = 5i - j. (b) |a| = √2² + 1² = √5. (c) Dot product a·b = (2)(-1) + (1)(3) = 1. |b| = √((-1)² + 3²) = √10. cos θ = (a·b)/(|a||b|) = 1/(√5√10) = 1/√50 = 1/(5√2). Thus θ = cos⁻¹(1/(5√2)) ≈ 81.9° → 82° to nearest degree.

解析：(a) 2a - b = 5i - j。(b) 模 |a| = √4+1=√5。(c) 点积 a·b = 1, |b|=√10, cos θ = 1/(√5×√10)=1/√50=1/(5√2)，θ ≈ cos⁻¹(0.1414) ≈ 82°。

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10. Differentiation: Tangent and Rate of Change | 微分：切线与变化率

Question: ere x = 1. (c) Find the x-coordinates of the stationary points and determine their nature.

题目：曲线 C: y = x³ - 3x + 2。(a) 求 dy/dx。(b) 求 x=1 处切线的方程。(c) 求驻点的 x 坐标并判断其性质。

Solution: =1, y = 1 - 3 + 2 = 0; gradient m = 3(1)² - 3 = 0. Tangent is horizontal: y - 0 = 0(x - 1) ⇒ minimum. At x=-1, d²y/dx² = -6 < 0 → maximum. So (1,0) is a minimum, (-1,4) is a maximum.

解析：(a) 导数为 3x² - 3。(b) x=1 处 y=0, 梯度 m=0，切线为 y=0。(c) 驻点 3x²-3=0 ⇒ x=±1。二阶导数 6x，x=1 处为正，极小值；x=-1 处为负，极大值。因此 (1,0) 为极小点，(-1,4) 为极大点。

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Published by TutorHao | Further Mathematics Revision Series | aleveler.com

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