📚 A-Level AQA Maths: Worked Example Solutions | A-Level AQA 数学：典型例题详解
This article provides a selection of carefully worked examples covering key topics from the AQA A-Level Mathematics specification. Each problem is solved step-by-step, with clear explanations in both English and Chinese to help you master essential techniques and avoid common pitfalls.
本文精选了AQA A-Level数学考试大纲中的关键题型，进行逐步详细解答。每道题都配有中英文双语解释，帮助您掌握核心解题技巧，规避常见错误。
1. Differentiation from First Principles | 第一原理求导
Problem: Using first principles, find the derivative of f(x) = x² + 3x.
题目：利用第一原理求 f(x) = x² + 3x 的导数。
Step 1: Write the definition of the derivative from first principles.
步骤1：写出第一原理求导的定义式。
f'(x) = limh→0 [f(x+h) − f(x)] / h
Step 2: Substitute f(x+h). Compute f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h.
步骤2：代入 f(x+h)。计算得 f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h。
Step 3: Form the difference quotient. f(x+h) − f(x) = (x² + 2xh + h² + 3x + 3h) − (x² + 3x) = 2xh + h² + 3h.
步骤3：构造差商。f(x+h) − f(x) = (x² + 2xh + h² + 3x + 3h) − (x² + 3x) = 2xh + h² + 3h。
Step 4: Simplify by factoring h: f'(x) = limh→0 (h(2x + h + 3))/h = limh→0 (2x + h + 3).
步骤4：提取公因子 h 化简：f'(x) = limh→0 (h(2x + h + 3))/h = limh→0 (2x + h + 3)。
Step 5: Take the limit as h → 0: f'(x) = 2x + 0 + 3 = 2x + 3.
步骤5：让 h→0 取极限：f'(x) = 2x + 0 + 3 = 2x + 3。
Common mistake: forgetting to expand (x+h)² correctly or mishandling the limit.
常见错误：忘记正确展开 (x+h)² 或极限处理不当。
2. Integration by Substitution | 换元积分法
Problem: Evaluate ∫ 2x√(x²+1) dx.
题目：计算 ∫ 2x√(x²+1) dx。
Step 1: Choose substitution u = x² + 1, then du/dx = 2x ⇒ du = 2x dx.
步骤1：设 u = x² + 1，则 du/dx = 2x，因此 du = 2x dx。
Step 2: Rewrite the integral in terms of u: ∫ √u du.
步骤2：将积分转换为关于 u 的形式：∫ √u du。
Step 3: Integrate √u = u1/2: ∫ u1/2 du = (2/3) u3/2 + C.
步骤3：积分 u1/2：∫ u1/2 du = (2/3) u3/2 + C。
Step 4: Substitute back x: (2/3)(x² + 1)3/2 + C.
步骤4：回代 x：(2/3)(x² + 1)3/2 + C。
Note: Always check that the derivative of the inside function is present. Here the factor 2x matches exactly.
注意：务必检查内部函数的导数是否出现。此处因式 2x 恰好完全匹配。
3. Solving Trigonometric Equations | 解三角方程
Problem: Solve 2 sin²θ − sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.
题目：在 0° ≤ θ ≤ 360° 范围内解方程 2 sin²θ − sin θ − 1 = 0。
Step 1: Treat as a quadratic in sin θ. Let y = sin θ: 2y² − y − 1 = 0.
步骤1：将其视为关于 sin θ 的二次方程。令 y = sin θ：2y² − y − 1 = 0。
Step 2: Factorise: (2y + 1)(y − 1) = 0 ⇒ y = −1/2 or y = 1.
步骤2：因式分解：(2y + 1)(y − 1) = 0 ⇒ y = −1/2 或 y = 1。
Step 3: Solve sin θ = −1/2. Principal value: −30°, so in [0°,360°]: θ = 180° + 30° = 210°, and 360° − 30° = 330°.
步骤3：解 sin θ = −1/2。主值 −30°，在 [0°,360°] 内：θ = 180° + 30° = 210°，以及 360° − 30° = 330°。
Step 4: Solve sin θ = 1 ⇒ θ = 90°.
步骤4：解 sin θ = 1 ⇒ θ = 90°。
Solution set: θ = 90°, 210°, 330°.
解集：θ = 90°, 210°, 330°。
4. Vectors: Scalar Product and Angle | 向量：数量积与夹角
Problem: Find the angle between vectors a = 2i + 3j − k and b = i − j + 2k.
题目：求向量 a = 2i + 3j − k 与 b = i − j + 2k 之间的夹角。
Step 1: Compute the dot product a·b = (2)(1) + (3)(−1) + (−1)(2) = 2 − 3 − 2 = −3.
步骤1：计算点积 a·b = (2)(1) + (3)(−1) + (−1)(2) = 2 − 3 − 2 = −3。
Step 2: Find magnitudes |a| = √(2²+3²+(−1)²) = √(4+9+1) = √14; |b| = √(1²+(−1)²+2²) = √(1+1+4) = √6.
步骤2：求模长 |a| = √(2²+3²+(−1)²) = √(4+9+1) = √14；|b| = √(1²+(−1)²+2²) = √(1+1+4) = √6。
Step 3: Use formula cos θ = (a·b) / (|a||b|) = −3 / (√14·√6) = −3 / √84 = −3 / (2√21) = −√21 / 14 after rationalising.
步骤3：利用公式 cos θ = (a·b) / (|a||b|) = −3 / (√14·√6) = −3 / √84 = −3 / (2√21)，有理化后得 −√21 / 14。
Step 4: θ = arccos(−√21/14) ≈ 109.1°.
步骤4：θ = arccos(−√21/14) ≈ 109.1°。
5. Probability: Binomial Distribution | 概率：二项分布
Problem: A fair die is rolled 8 times. Find the probability of getting a six exactly 3 times.
题目：一枚均匀骰子投掷8次。求恰好出现3次6点的概率。
Step 1: Identify n = 8, p = 1/6, q = 5/6. Let X ~ B(8, 1/6).
步骤1：确定参数 n = 8，p = 1/6，q = 5/6。设 X ~ B(8, 1/6)。
Step 2: Use binomial formula P(X=r) = ⁿCᵣ pʳ qⁿ⁻ʳ.
步骤2：使用二项式公式 P(X=r) = ⁿCᵣ pʳ qⁿ⁻ʳ。
Step 3: r = 3, P(X=3) = ⁸C₃ (1/6)³ (5/6)⁵.
步骤3：r = 3，P(X=3) = ⁸C₃ (1/6)³ (5/6)⁵。
Step 4: ⁸C₃ = 56. Compute: 56 × (1/216) × (3125/7776) = 56 × 3125 / (216×7776). Simplify: 216×7776 = 1,679,616; 56×3125 = 175,000, so probability = 175000/1679616 ≈ 0.1042.
步骤4：⁸C₃ = 56。计算：56 × (1/216) × (3125/7776) = 56 × 3125 / (216×7776)。简化：216×7776=1,679,616；56×3125=175,000，故概率 = 175000/1679616 ≈ 0.1042。
Tip: Always check your calculator input; the probability should be between 0 and 1.
提示：务必检查计算器输入；概率值应在0到1之间。
6. Hypothesis Testing for the Mean | 均值的假设检验
Problem: A sample of 50 students has a mean score of 72 with a known population standard deviation of 8. Test at the 5% significance level whether the mean is different from 70.
题目：某样本50名学生的平均分为72，已知总体标准差为8。在5%显著性水平下检验总体均值是否不同于70。
Step 1: State hypotheses: H₀: μ = 70, H₁: μ ≠ 70 (two-tailed).
步骤1：提出假设：H₀：μ = 70，H₁：μ ≠ 70（双侧检验）。
Step 2: Calculate test statistic: z = (x̄ − μ) / (σ/√n) = (72 − 70) / (8/√50) = 2 / (8/7.071) ≈ 2 / 1.131 = 1.768.
步骤2：计算检验统计量：z = (x̄ − μ) / (σ/√n) = (72 − 70) / (8/√50) ≈ 2 / 1.131 ≈ 1.768。
Step 3: For α=0.05 two-tailed, critical z-value = ±1.96.
步骤3：当 α=0.05 双侧时，临界 z 值 = ±1.96。
Step 4: Since |1.768| < 1.96, we do not reject H₀. There is insufficient evidence to say the mean differs from 70.
步骤4：由于 |1.768| < 1.96，我们不拒绝 H₀。没有足够证据表明均值不同于70。
7. Kinematics: SUVAT Equations | 运动学：SUVAT 方程
Problem: A particle is projected vertically upwards with speed 20 m/s from ground level. Find the time taken to reach maximum height and the maximum height. Use g = 9.8 m/s².
题目：一质点从地面以20 m/s的速度竖直向上抛出。求到达最高点所需时间及最大高度。取 g = 9.8 m/s²。
Step 1: At maximum height, velocity v = 0. Use v = u + at ⇒ 0 = 20 − 9.8t ⇒ t = 20/9.8 ≈ 2.04 s.
步骤1：在最高点，速度 v = 0。使用 v = u + at，得 0 = 20 − 9.8t ⇒ t = 20/9.8 ≈ 2.04 s。
Step 2: Find height using s = ut + ½at²: s = 20×2.04 − ½×9.8×(2.04)² = 40.8 − 4.9×4.1616 = 40.8 − 20.4 ≈ 20.4 m. Alternatively use
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