Year 12 WJEC Mathematics: Unit Test Mock Paper Walkthrough | WJEC 12 年级数学单元测试模拟卷解析

📚 Year 12 WJEC Mathematics: Unit Test Mock Paper Walkthrough | WJEC 12 年级数学单元测试模拟卷解析

This article provides a full walkthrough of a mock paper designed for a Year 12 WJEC Mathematics unit test. The questions cover essential AS Pure Mathematics topics such as algebra, trigonometry, differentiation, integration, coordinate geometry and sequences. Each solution is presented with clear step-by-step reasoning in both English and Chinese, helping you to consolidate methods and avoid typical examination pitfalls.

本文提供一套针对 WJEC 12 年级数学单元测试的模拟卷完整解析。题目涵盖 AS 纯数学的核心领域，包括代数、三角学、微分、积分、坐标几何和数列。每道题的解答都配有清晰的中英双语步骤详解，帮助你巩固解题方法，规避常见考试误区。

1. Q1: Simplifying Algebraic Expressions and Solving a Quadratic Inequality | 代数化简与二次不等式求解

Question: (a) Simplify (3x²y⁻³)² × (2x⁴y), expressing your answer with positive indices. (b) Solve the inequality 2x² – 5x – 3 ≤ 0.

题目：(a) 化简 (3x²y⁻³)² × (2x⁴y)，结果用正整数指数表示。(b) 解不等式 2x² – 5x – 3 ≤ 0。

(a) Start by applying the power to the first bracket: (3x²y⁻³)² = 3² · (x²)² · (y⁻³)² = 9x⁴y⁻⁶. Then multiply by (2x⁴y): 9x⁴y⁻⁶ × 2x⁴y = 18x⁴⁺⁴ y⁻⁶⁺¹ = 18x⁸y⁻⁵. To write with positive indices, bring y⁻⁵ to the denominator: 18x⁸ / y⁵.

(a) 首先对第一个括号进行乘方：(3x²y⁻³)² = 9x⁴y⁻⁶。再乘以 (2x⁴y) 得到 18x⁸y⁻⁵。变为正指数形式：18x⁸ / y⁵。

(b) Factorise the quadratic: 2x² – 5x – 3 = (2x + 1)(x – 3). The critical values occur where each factor equals zero: 2x + 1 = 0 ⇒ x = –½, and x – 3 = 0 ⇒ x = 3. Sketch a sign diagram or test intervals: for x < –½, both factors are negative, product positive; between –½ and 3, (2x+1) positive and (x–3) negative, product negative; for x > 3, both factors positive, product positive. Since the inequality is ≤ 0, we take the interval where the product is negative or zero, giving –½ ≤ x ≤ 3.

(b) 将二次式因式分解为 (2x + 1)(x – 3)。令各因式为零得临界值 x = –½ 和 x = 3。根据区间符号分析，乘积在 x ＜ –½ 和 x ＞ 3 时为正，在 –½ 到 3 之间为负。由于要求 ≤ 0，解集为 –½ ≤ x ≤ 3。

2. Q2: Factor Theorem and Polynomial Division | 因式定理与多项式除法

Question: Given f(x) = 2x³ + ax² + bx – 6, and that (x + 2) and (x – 3) are factors, find the values of a and b. Hence factorise f(x) completely.

题目：已知 f(x) = 2x³ + ax² + bx – 6，且 (x + 2) 和 (x – 3) 为其因式，求 a 和 b 的值，并由此将 f(x) 完全分解。

By the Factor Theorem, if (x + 2) is a factor, then f(–2) = 0. Substitute x = –2: f(–2) = 2(–8) + a(4) + b(–2) – 6 = –16 + 4a – 2b – 6 = 4a – 2b – 22 = 0 ⇒ 2a – b = 11 … (1). Similarly, f(3) = 0: f(3) = 2(27) + a(9) + b(3) – 6 = 54 + 9a + 3b – 6 = 9a + 3b + 48 = 0 ⇒ 3a + b + 16 = 0 ⇒ 3a + b = –16 … (2). Solve (1) and (2) simultaneously: adding (1) and (2) gives 5a = –5 ⇒ a = –1. Substitute into (2): –3 + b = –16 ⇒ b = –13. Now f(x) = 2x³ – x² – 13x – 6. Since (x+2) and (x–3) are factors, their product (x+2)(x–3) = x² – x – 6 is a factor. Divide f(x) by x² – x – 6 to find the remaining linear factor. Performing the division: 2x³ – x² – 13x – 6 divided by x² – x – 6 gives 2x + 1. Hence f(x) = (x + 2)(x – 3)(2x + 1).

根据因式定理，f(–2) = 0 得方程 2a – b = 11；f(3) = 0 得 3a + b = –16。联立解得 a = –1，b = –13。代回得 f(x) = 2x³ – x² – 13x – 6。因为已知 (x+2) 和 (x–3) 为因式，其乘积 x² – x – 6 也是一个因式。用多项式除法求出剩余因式为 (2x+1)。因此完全分解为 (x + 2)(x – 3)(2x + 1)。

3. Q3: Solving Trigonometric Equations | 三角方程求解

Question: Solve sin 2θ = cos θ for 0° ≤ θ ≤ 360°.

题目：在区间 0° ≤ θ ≤ 360° 内解方程 sin 2θ = cos θ。

Use the double-angle identity sin 2θ ≡ 2 sin θ cos θ. The equation becomes 2 sin θ cos θ = cos θ. Bring all terms to one side: 2 sin θ cos θ – cos θ = 0, then factorise: cos θ (2 sin θ – 1) = 0. Hence either cos θ = 0 or sin θ = ½. For cos θ = 0 in the given interval: θ = 90°, 270°. For sin θ = ½: reference angle is 30°, so θ = 30° and 180° – 30° = 150°. Collecting all solutions: θ = 30°, 90°, 150°, 270°.

利用倍角公式 sin 2θ = 2 sin θ cos θ，原方程化为 2 sin θ cos θ – cos θ = 0，提取公因式得 cos θ (2 sin θ – 1) = 0。故 cos θ = 0 或 sin θ = ½。cos θ = 0 给出 θ = 90°, 270°；sin θ = ½ 给出 θ = 30° 和 150°。全部解为 30°, 90°, 150°, 270°。

4. Q4a: Differentiation from First Principles | 第一原理求导

Question: Use differentiation from first principles to find the derivative of f(x) = 3x² + 5.

题目：利用第一原理求 f(x) = 3x² + 5 的导数。

The definition of the derivative is f'(x) = limₕ→₀ [f(x + h) – f(x)] / h. Compute f(x+h) = 3(x+h)² + 5 = 3(x² + 2xh + h²) + 5 = 3x² + 6xh + 3h² + 5. Subtract f(x) = 3x² + 5, giving [3x² + 6xh + 3h² + 5] – [3x² + 5] = 6xh + 3h². So the difference quotient is (6xh + 3h²) / h = 6x + 3h (for h ≠ 0). Taking the limit as h → 0, we obtain f'(x) = 6x.

导数的定义为 f'(x) = limₕ→₀ [f(x+h) – f(x)] / h。计算 f(x+h) = 3(x+h)² + 5 = 3x² + 6xh + 3h² + 5，减去 f(x) 得 6xh + 3h²。差商为 (6xh + 3h²)/h = 6x + 3h。取 h → 0 的极限即得 f'(x) = 6x。

5. Q4b: Equation of a Tangent to a Curve | 曲线的切线方程

Question: Find the equation of the tangent to the curve y = x³ – 2x + 1 at the point where x = 2.

题目：求曲线 y = x³ – 2x + 1 在 x = 2 处的切线方程。

First find the y-coordinate: y = (2)³ – 2(2) + 1 = 8 – 4 + 1 = 5. So the point is (2, 5). Differentiate to find the gradient function: dy/dx = 3x² – 2. At x = 2, the gradient m = 3(4) – 2 = 10. Using the point-slope form of a line, y – y₁ = m(x – x₁), we have y – 5 = 10(x – 2). Simplify to y – 5 = 10x – 20, giving y = 10x – 15. The equation of the tangent is y = 10x – 15.

先求对应 y 值：x=2 时 y = 8 – 4 + 1 = 5，切点为 (2, 5)。对函数求导得 dy/dx = 3x² – 2，代入 x=2 得斜率 m = 10。利用点斜式方程 y – 5 = 10(x – 2)，化简后切线方程为 y = 10x – 15。

6. Q5: Stationary Points and Their Nature | 驻点及其性质

Question: For the curve y = x⁴ – 4x³ + 4x², find the coordinates of any stationary points and determine their nature.

题目：求曲线 y = x⁴ – 4x³ + 4x² 的驻点坐标，并判断其性质。

Differentiate: dy/dx = 4x³ – 12x² + 8x. Set dy/dx = 0 to find stationary points: 4x³ – 12x² + 8x = 0 ⇒ 4x(x² – 3x + 2) = 0 ⇒ 4x(x – 1)(x – 2) = 0. Hence x = 0, 1, 2. Find the second derivative: d²y/dx² = 12x² – 24x + 8. Evaluate at each stationary point: at x = 0, f”(0) = 8 > 0 ⇒ minimum; at x = 1, f”(1) = 12 – 24 + 8 = –4 < 0 ⇒ maximum; at x = 2, f''(2) = 48 – 48 + 8 = 8 > 0 ⇒ minimum. Now find y-coordinates: for x=0, y=0; for x=1, y=1 – 4 + 4 = 1; for x=2, y=16 – 32 + 16 = 0. Stationary points are (0,0) minimum, (1,1) maximum, (2,0) minimum.

先求一阶导数：dy/dx = 4x³ – 12x² + 8x。令其为零，因式分解得 4x(x-1)(x-2)=0，故 x=0,1,2。求二阶导数：d²y/dx² = 12x² – 24x + 8。分别代入：x=0 时二阶导数为 8 > 0，故为极小点；x=1 时二阶导数为 –4 < 0，故为极大点；x=2 时二阶导数为 8 > 0，故为极小点。对应 y 坐标：x=0 得 y=0；x=1 得 y=1；x=2 得 y=0。因此驻点为极小点 (0,0)，极大点 (1,1) 和极小点 (2,0)。

7. Q6: Definite Integration and Area Under a Curve | 定积分与曲线下方面积

Question: (a) Evaluate ∫₀² (3x² – 2x + 1) dx. (b) Find the area bounded by the curve y = x² + 1, the x-axis, and the lines x = 1 and x = 3.

题目：(a) 计算 ∫₀² (3x² – 2x + 1) dx。(b) 求由曲线 y = x² + 1、x 轴以及直线 x = 1 和 x = 3 所围区域的面积。

(a) Integrate term by term: ∫ 3x² dx = x³, ∫ –2x dx = –x², ∫ 1 dx = x. So the indefinite integral is x³ – x² + x. Evaluate from 0 to 2: [2³ – 2² + 2] – [0] = (8 – 4 + 2) = 6.

(a) 逐项积分：∫3x² dx = x³，∫–2x dx = –x²，∫1 dx = x。不定积分为 x³ – x² + x。代入上下限：[(2³ – 2² + 2) – (0)] = 6。

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Year 12 WJEC Mathematics: Unit Test Mock Paper Walkthrough | WJEC 12 年级数学单元测试模拟卷解析