Worked Examples for A-Level Edexcel Mathematics | A-Level Edexcel 数学典型例题详解

📚 Worked Examples for A-Level Edexcel Mathematics | A-Level Edexcel 数学典型例题详解

This article presents a collection of worked examples from across the Edexcel A-Level Mathematics specification. Each problem is solved in detail, with clear step-by-step reasoning to reinforce essential techniques and support exam preparation.

本文集合了 Edexcel A-Level 数学课程中的典型例题，每个问题均给出详细的逐步解答与清晰推理，帮助巩固核心解题技巧，为考试做好准备。

1. Quadratic Inequalities | 二次不等式

Solve the inequality x² – 5x + 6 > 0.

解不等式 x² – 5x + 6 > 0。

Step 1: Factorise the quadratic expression. We look for two numbers whose product is 6 and sum is -5; these are -2 and -3. Hence x² – 5x + 6 = (x – 2)(x – 3). The inequality becomes (x – 2)(x – 3) > 0.

步骤1：对二次式进行因式分解。寻找两个数，使其乘积为 6 且和为 -5，这两个数是 -2 与 -3。因此 x² – 5x + 6 = (x – 2)(x – 3)。不等式化为 (x – 2)(x – 3) > 0。

Step 2: Determine the critical values where the expression equals zero: x – 2 = 0 ⇒ x = 2; x – 3 = 0 ⇒ x = 3. These split the number line into three intervals: x < 2, 2 < x < 3, x > 3.

步骤2：求临界值，即表达式等于零的点：x – 2 = 0 ⇒ x = 2；x – 3 = 0 ⇒ x = 3。这些点将数轴分为三个区间：x < 2，2 < x < 3，x > 3。

Step 3: Test the sign of (x – 2)(x – 3) in each interval. For x = 0, (0-2)(0-3) = 6 > 0. For x = 2.5, (0.5)(-0.5) = -0.25 < 0. For x = 4, (2)(1) = 2 > 0. The inequality requires the product to be greater than zero, so the solution is the intervals where the product is positive.

步骤3：检验各区间的符号。取 x = 0，得 (0-2)(0-3) = 6 > 0；取 x = 2.5，得 (0.5)(-0.5) = -0.25 < 0；取 x = 4，得 (2)(1) = 2 > 0。不等式要求乘积大于零，因此解为使乘积为正的区间。

Step 4: Hence the solution is x < 2 or x > 3. In set notation: {x : x < 2} ∪ {x : x > 3}.

步骤4：因此解为 x < 2 或 x > 3。用集合表示为 {x : x < 2} ∪ {x : x > 3}。

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2. Trigonometric Equations | 三角方程

Solve 2 sin θ = 1 for 0° ≤ θ ≤ 360°.

在 0° ≤ θ ≤ 360° 范围内解方程 2 sin θ = 1。

Step 1: Divide both sides by 2 to obtain sin θ = 1/2.

步骤1：两边除以 2，得 sin θ = 1/2。

Step 2: The principal value is θ = sin⁻¹(1/2) = 30°.

步骤2：主值为 θ = sin⁻¹(1/2) = 30°。

Step 3: The sine function is positive in the first and second quadrants. The second solution in the given range is 180° – 30° = 150°.

步骤3：正弦函数在第一和第二象限为正。在给定范围内，第二个解为 180° – 30° = 150°。

Step 4: Therefore the solutions are θ = 30° and θ = 150°.

步骤4：因此解为 θ = 30° 与 θ = 150°。

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3. Differentiation: Stationary Points | 微分：驻点

Find the stationary points of the function f(x) = x³ – 3x² – 9x + 5 and determine their nature.

求函数 f(x) = x³ – 3x² – 9x + 5 的驻点，并判断其性质。

Step 1: Differentiate to find f'(x). f'(x) = 3x² – 6x – 9.

步骤1：求导得 f'(x) = 3x² – 6x – 9。

Step 2: Set f'(x) = 0 and solve: 3x² – 6x – 9 = 0 ⇒ x² – 2x – 3 = 0 ⇒ (x – 3)(x + 1) = 0. Thus the stationary points occur at x = 3 and x = -1.

步骤2：令 f'(x) = 0 并求解：3x² – 6x – 9 = 0 ⇒ x² – 2x – 3 = 0 ⇒ (x – 3)(x + 1) = 0。故驻点位于 x = 3 与 x = -1。

Step 3: Find the second derivative: f”(x) = 6x – 6. Then evaluate at each stationary point: f”(3) = 12 > 0 ⇒ minimum; f”(-1) = -12 < 0 ⇒ maximum.

步骤3：求二阶导数：f”(x) = 6x – 6。代入各驻点：f”(3) = 12 > 0 ⇒ 极小值；f”(-1) = -12 < 0 ⇒ 极大值。

Step 4: Calculate the corresponding y-coordinates: f(3) = 27 – 27 – 27 + 5 = -22; f(-1) = -1 – 3 + 9 + 5 = 10. Hence the minimum point is (3, -22) and the maximum point is (-1, 10).

步骤4：计算对应的 y 坐标：f(3) = 27 – 27 – 27 + 5 = -22；f(-1) = -1 – 3 + 9 + 5 = 10。因此极小值点为 (3, -22)，极大值点为 (-1, 10)。

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4. Integration: Area Under a Curve | 积分：曲线下方面积

Find the area bounded by the curve y = x² + 1, the x-axis, and the lines x = 0 and x = 2.

求由曲线 y = x² + 1、x 轴以及直线 x = 0 和 x = 2 所围成的面积。

Step 1: The area is given by the definite integral ∫₀² (x² + 1) dx.

步骤1：面积由定积分 ∫₀² (x² + 1) dx 给出。

Step 2: Integrate term by term: the antiderivative is (1/3)x³ + x.

步骤2：逐项积分，原函数为 (1/3)x³ + x。

Step 3: Evaluate from 0 to 2: [(1/3)(2)³ + 2]

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