📚 A-Level WJEC Mathematics: Worked Examples Explained | A-Level WJEC 数学：典型例题详解
Mastering A-Level WJEC Mathematics requires not only understanding theoretical concepts but also becoming fluent in applying them to exam-style problems. This article presents a carefully selected set of worked examples covering pure mathematics, mechanics, and statistics. Each example is unpacked step by step, with clear reasoning and dual-language explanations to help you build confidence and precision.
要掌握 A-Level WJEC 数学，不仅需要理解理论概念，还需要熟练地将它们应用到考试题型中。本文精选了一组涵盖纯数、力学和统计的典型例题，逐步展开详细解答，并提供中英双语解释，帮助你建立信心和准确性。
1. Solving Quadratic Equations | 解二次方程
Solve the quadratic equation 2x² − 5x − 3 = 0 by factorisation.
用因式分解法解二次方程 2x² − 5x − 3 = 0。
Step 1: Identify two numbers that multiply to a × c = 2 × (−3) = −6 and add to b = −5. The numbers are −6 and 1.
步骤1：找出两个数，乘积为 a × c = 2 × (−3) = −6，和为 b = −5。这两个数是 −6 和 1。
Step 2: Split the middle term: 2x² − 6x + x − 3 = 0.
步骤2：拆分中项：2x² − 6x + x − 3 = 0。
Step 3: Factor by grouping: 2x(x − 3) + 1(x − 3) = 0 → (2x + 1)(x − 3) = 0.
步骤3：分组因式分解：2x(x − 3) + 1(x − 3) = 0 → (2x + 1)(x − 3) = 0。
Step 4: Set each factor to zero: 2x + 1 = 0 ⇒ x = −½ ; x − 3 = 0 ⇒ x = 3.
步骤4：令每个因式为零：2x + 1 = 0 ⇒ x = −½ ; x − 3 = 0 ⇒ x = 3。
The solutions are x = −½ and x = 3.
解为 x = −½ 和 x = 3。
2. Differentiation from First Principles | 从第一原理求导
Given f(x) = x², find f ′(x) from first principles.
已知 f(x) = x²，用第一原理求 f ′(x)。
Step 1: Start with the definition f ′(x) = limh→0 [f(x+h) − f(x)] / h.
步骤1：由导数定义出发 f ′(x) = limh→0 [f(x+h) − f(x)] / h。
Step 2: Compute f(x+h) = (x+h)² = x² + 2xh + h².
步骤2：计算 f(x+h) = (x+h)² = x² + 2xh + h²。
Step 3: Form the difference f(x+h) − f(x) = (x² + 2xh + h²) − x² = 2xh + h².
步骤3：求差 f(x+h) − f(x) = (x² + 2xh + h²) − x² = 2xh + h²。
Step 4: Divide by h: (2xh + h²)/h = 2x + h.
步骤4：除以 h：(2xh + h²)/h = 2x + h。
Step 5: Take the limit as h → 0: f ′(x) = limh→0 (2x + h) = 2x.
步骤5：取 h → 0 的极限：f ′(x) = limh→0 (2x + h) = 2x。
Hence, the derivative of x² is 2x.
因此，x² 的导数为 2x。
3. Integration by Substitution | 代入法积分
Evaluate ∫ 2x (x² + 1)³ dx using a suitable substitution.
用合适的代入法求 ∫ 2x (x² + 1)³ dx。
Step 1: Let u = x² + 1. Then du/dx = 2x, so du = 2x dx.
步骤1：设 u = x² + 1，则 du/dx = 2x，因此 du = 2x dx。
Step 2: The integral becomes ∫ (x² + 1)³ · 2x dx = ∫ u³ du.
步骤2：原积分变为 ∫ (x² + 1)³ · 2x dx = ∫ u³ du。
Step 3: Integrate with respect to u: ∫ u³ du = ¼ u⁴ + C.
步骤3：对 u 积分：∫ u³ du = ¼ u⁴ + C。
Step 4: Substitute back u = x² + 1: ¼ (x² + 1)⁴ + C.
步骤4：代回 u = x² + 1：¼ (x² + 1)⁴ + C。
The final answer is ¼ (x² + 1)⁴ + C.
最终答案为 ¼ (x² + 1)⁴ + C。
4. Trigonometric Equations | 三角方程
Solve 2 sin x = 1 for 0° ≤ x ≤ 360°.
求解方程 2 sin x = 1，x 在 0° 到 360° 之间。
Step 1: Rearrange to sin x = ½.
步骤1：变形为 sin x = ½。
Step 2: The principal solution is x = 30° (since sin 30° = ½).
步骤2：主解为 x = 30°（因为 sin 30° = ½）。
Step 3: sin x is positive in the first and second quadrants. The second solution is 180° − 30° = 150°.
步骤3：sin x 在第一和第二象限为正。第二个解为 180° − 30° = 150°。
Step 4: Both solutions lie within 0° ≤ x ≤ 360°, so x = 30° and x = 150°.
步骤4：两个解都在给定区间内，故 x = 30° 和 x = 150°。
The solution set is {30°, 150°}.
解集为 {30°, 150°}。
5. Vectors: Angle Between Two Vectors | 向量：求两向量夹角
Find the angle between vectors a = 3i + 4j and b = 4i − 3j.
求向量 a = 3i + 4j 与 b = 4i − 3j 之间的夹角。
Step 1: Use the dot product formula: cos θ = (a · b) / (|a| |b|).
步骤1：使用点积公式：cos θ = (a · b) / (|a| |b|)。
Step 2: Compute a · b = (3)(4) + (4)(−3) = 12 − 12 = 0.
步骤2：计算 a · b = (3)(4) + (4)(−3) = 12 − 12 = 0。
Step 3: Find magnitudes: |a| = √(3² + 4²) = 5, |b| = √(4² + (−3)²) = 5.
步骤3：求模长：|a| = √(3² + 4²) = 5，|b| = √(4² + (−3)²) = 5。
Step 4: cos θ = 0 / (5 × 5) = 0, so θ = 90°.
步骤4：cos θ = 0 / (5 × 5) = 0，因此 θ = 90°。
The vectors are perpendicular.
这两个向量互相垂直。
6. Binomial Distribution | 二项分布
A fair coin is tossed 10 times. Find the probability of obtaining exactly 6 heads.
抛掷一枚公平硬币 10 次，求恰好出现 6 次正面的概率。
Step 1: Identify parameters: n = 10, p = 0.5, q = 1 − p = 0.5. Let X be the number of heads, X ~ B(10, 0.5).
步骤1：确定参数：n = 10，p = 0.5，q = 1 − p = 0.5。设 X 为正面次数，X ~ B(10, 0.5)。
Step 2: Use the probability mass function: P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ.
步骤2：使用概率质量函数：P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ。
Step 3: Plug in r = 6: P(X = 6) = ¹⁰C₆ (0.5)⁶ (0.5)⁴ = ¹⁰C₆ (0.5)¹⁰.
步骤3：代入 r = 6：P(X = 6) = ¹⁰C₆ (0.5)⁶ (0.5)⁴ = ¹⁰C₆ (0.5)¹⁰。
Step 4: ¹⁰C₆ = 210. Then P(X = 6) = 210 × (0.5)¹⁰ = 210 / 1024 ≈ 0.2051.
步骤4：¹⁰C₆ = 210。于是 P(X = 6) = 210 × (0.5)¹⁰ = 210 / 1024 ≈ 0.2051。
The probability of exactly 6 heads is about 0.205 (3 s.f.).
恰好 6 次正面的概率约为 0.205（三位有效数字）。
7. Projectile Motion | 抛体运动
A particle is projected from ground level with speed 20 m s⁻¹ at an angle of 30° to the horizontal. Find the time of flight and the horizontal range. (Take g = 9.8 m s⁻²)
一个质点从地面以 20 m s⁻¹ 的速度与水平方向成 30° 角抛出。求飞行时间和水平射程。（取 g = 9.8 m s⁻²）
Step 1: Resolve initial velocity: ux = 20 cos 30° = 20 × (√3 / 2) ≈ 17.32 m s⁻¹; uy = 20 sin 30° = 10 m s⁻¹.
步骤1：分解初速度：ux = 20 cos 30° = 20 × (√3 / 2) ≈ 17.32 m s⁻¹；uy = 20 sin 30° = 10 m s⁻¹。
Step 2: Time of flight is twice the time to reach maximum height. Use vertical motion: vy = uy − gt ⇒ 0 = 10 − 9.8 t ⇒ t = 10 / 9.8 ≈ 1.0204 s (to top). Total time T = 2 × 1.0204 ≈ 2.041 s.
步骤2：飞行时间是到达最高点时间的两倍。利用竖直运动：vy = uy − gt ⇒ 0 = 10 − 9.8 t ⇒ t = 10 / 9.8 ≈ 1.0204 s（到达最高点）。总时间 T = 2 × 1.0204 ≈ 2.041 s。
Step 3: Horizontal range = ux × T = 17.32 × 2.041 ≈ 35.4 m (3 s.f.).
步骤3：水平射程 = ux × T = 17.32 × 2.041 ≈ 35.4 m（三位有效数字）。
The time of flight is 2.04 s and the range is 35.4 m.
飞行时间为 2.04 秒，射程为 35.4 米。
8. Arithmetic Series | 等差数列
The first term of an arithmetic series is 5 and the common difference is 3. Find the sum of the first 20 terms.
一个等差数列的首项为 5，公差为 3。求前 20 项的和。
Step 1: Use the sum formula Sn = n/2 [2a + (n − 1)d]. Here a = 5, d = 3, n = 20.
步骤1：使用求和公式 Sn = n/2 [2a + (n − 1)d]。这里 a = 5, d = 3, n = 20。
Step 2: Compute inside the brackets: 2a = 10, (n−1)d = 19 × 3 = 57. Sum = 10 + 57 = 67.
步骤2：计算括号内：2a = 10，(n−1)d = 19 × 3 = 57。和为 10 + 57 = 67。
Step 3: S20 = (20/2) × 67 = 10 × 67 = 670.
步骤3：S20 = (20/2) × 67 = 10 × 67 = 670。
The sum of the first 20 terms is 670.
前 20 项的和为 670。
9. Exponential Growth and Decay | 指数增长与衰减
A bacterial population is modelled by N = 200 e0.05t, where N is the number of bacteria and t is time in hours. Find the population after 10 hours and the rate of growth at that instant.
某细菌数量模型为 N = 200 e0.05t，其中 N 表示细菌数量，t 为时间（小时）。求 10 小时后的数量及该时刻的增长率。
Step 1: Substitute t = 10: N = 200 e0.5. Using e0.5 ≈ 1.6487, N ≈ 200 × 1.6487 ≈ 329.7. So approximately 330 bacteria.
步骤1：代入 t = 10：N = 200 e0.5。e0.5 ≈ 1.6487，N ≈ 200 × 1.6487 ≈ 329.7，约为 330 个细菌。
Step 2: The rate of growth is dN/dt = 200 × 0.05 e0.05t = 10 e0.05t.
步骤2：增长率 dN/dt = 200 × 0.05 e0.05t = 10 e0.05t。
Step 3: At t = 10, dN/dt = 10 e0.5 ≈ 10 × 1.6487 ≈ 16.5 bacteria per hour.
步骤3：当 t = 10 时，dN/dt = 10 e0.5 ≈ 10 × 1.6487 ≈ 16.5 个/小时。
After 10 hours, the population is about 330 and it is increasing by approximately 16.5 bacteria per hour.
10 小时后数量约为 330，并以每小时约 16.5 个的速度增长。
10. Hypothesis Testing for a Mean | 均值的假设检验
A manufacturer claims that the mean lifetime of a light bulb is 800 hours. A sample of 36 bulbs has a mean of 780 hours. The population standard deviation is known to be 50 hours. Test at the 5% significance level whether the claim should be rejected.
某制造商声称灯泡的平均寿命为 800 小时。随机抽取 36 个灯泡，样本均值为 780 小时，总体标准差已知为 50 小时。在 5% 的显著性水平下检验该声称是否应被拒绝。
Step 1: State hypotheses: H₀: μ = 800, H₁: μ ≠ 800 (two-tailed test).
步骤1：提出假设：H₀: μ = 800，H₁: μ ≠ 800（双尾检验）。
Step 2: Significance level α = 0.05, critical z-values are ±1.96.
步骤2：显著性水平 α = 0.05，临界 z 值为 ±1.96。
Step 3: Calculate the test statistic: z = (x̄ − μ) / (σ / √n) = (780 − 800) / (50 / √36) = −20 / (50/6) = −20 / 8.333… = −2.4.
步骤3：计算检验统计量：z = (x̄ − μ) / (σ / √n) = (780 − 800) / (50 / √36) = −20 / (50/6) = −20 / 8.333… = −2.4。
Step 4: Compare with critical region: Since −2.4 < −1.96, the test statistic falls in the rejection region.
步骤4：与临界域比较：因为 −2.4 < −1.96，检验统计量落入拒绝域。
Step 5: Conclusion: Reject H₀. There is sufficient evidence at the 5% level to conclude that the mean lifetime is not 800 hours.
步骤5：结论：拒绝 H₀。在 5% 的显著性水平下，有充分证据表明平均寿命不是 800 小时。
The manufacturer’s claim is rejected.
制造商的声称被拒绝。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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