📚 IB WJEC Mathematics: Worked Examples Explained | IB WJEC 数学：典型例题详解
Mastering IB Mathematics requires not only understanding concepts but also applying them to exam-style questions. This article presents a series of worked examples covering key topics from the IB syllabus, with clear step-by-step reasoning. These examples mirror the type of reasoning expected in WJEC-style assessments, helping you build confidence for both internal and external examinations.
掌握IB数学不仅需要理解概念，还需要将其应用到考试题型中。本文提供一系列涵盖IB大纲核心主题的典型例题，并配有清晰的逐步解析。这些例题反映了WJEC风格评估中所要求的推理方式，帮助你为校内及外部考试建立信心。
1. Functions and Equations | 函数与方程
Example: Let f(x) = x² − 4x + 3. Find the coordinates of the vertex and the x-intercepts.
例题：设 f(x) = x² − 4x + 3。求顶点坐标和 x 轴截距。
Complete the square: f(x) = (x − 2)² − 4 + 3 = (x − 2)² − 1. So the vertex is at (2, −1).
配方：f(x) = (x − 2)² − 4 + 3 = (x − 2)² − 1。因此顶点为 (2, −1)。
Set f(x) = 0: (x − 2)² − 1 = 0 ⇒ (x − 2)² = 1 ⇒ x − 2 = ± 1 ⇒ x = 1 or x = 3. The x-intercepts are (1,0) and (3,0).
令 f(x) = 0：(x − 2)² − 1 = 0 ⇒ (x − 2)² = 1 ⇒ x − 2 = ± 1 ⇒ x = 1 或 x = 3。x 轴截距为 (1,0) 和 (3,0)。
2. Differentiation: Tangents and Normals | 微分：切线与法线
Example: Find the equation of the tangent to the curve y = x³ − 3x at the point where x = 1.
例题：求曲线 y = x³ − 3x 在 x = 1 处的切线方程。
Differentiate: dy/dx = 3x² − 3. At x = 1, gradient m = 3(1)² − 3 = 0.
求导：dy/dx = 3x² − 3。在 x = 1 处，斜率 m = 3(1)² − 3 = 0。
When x = 1, y = (1)³ − 3(1) = −2. Tangent line: y − (−2) = 0(x − 1) ⇒ y = −2.
当 x = 1 时，y = (1)³ − 3(1) = −2。切线方程：y − (−2) = 0(x − 1) ⇒ y = −2。
3. Integration: Area Under a Curve | 积分：曲线下方面积
Example: Find the area bounded by the curve y = 4 − x² and the x-axis.
例题：求曲线 y = 4 − x² 与 x 轴所围成的面积。
Find x-intercepts: 4 − x² = 0 ⇒ x = ±2. The area is A = ∫₋₂² (4 − x²) dx.
求 x 轴截距：4 − x² = 0 ⇒ x = ±2。面积为 A = ∫₋₂² (4 − x²) dx。
Integrate: A = [4x − (1/3)x³]₋₂² = (8 − 8/3) − (−8 + 8/3) = (16/3) − (−16/3) = 32/3.
积分：A = [4x − (1/3)x³]₋₂² = (8 − 8/3) − (−8 + 8/3) = (16/3) − (−16/3) = 32/3。
The area is 32/3 square units.
面积为 32/3 平方单位。
4. Probability: Normal Distribution | 概率：正态分布
Example: on σ = 10 cm. Find the probability that a randomly selected student is taller than 185 cm.
例题：学生身高服从正态分布，均值 μ = 170 cm，标准差 σ = 10 cm。求随机选取一名学生身高超过 185 cm 的概率。
/10 = 1.5. Using the standard normal table, P(Z > 1.5) = 1 − Φ(1.5) = 1 − 0.9332 = 0.0668.
标准化：z = (185 − 170)/10 = 1.5。查标准正态分布表，P(Z > 1.5) = 1 − Φ(1.5) = 1 − 0.9332 = 0.0668。
So, about 6.68% of students are taller than 185 cm.
因此，约有 6.68% 的学生身高超过 185 cm。
5. Vectors: Scalar Product and Angle | 向量：数量积与夹角
Example: Given vectors a = 3i − j + 2k and b = i + 4j − k, find the angle between them.
例题：已知向量 a = 3i − j + 2k 和 b = i + 4j − k，求它们之间的夹角。
Dot product: a · b = (3)(1) + (−1)(4) + (2)(−1) = 3 − 4 − 2 = −3.
点乘：a · b = (3)(1) + (−1)(4) + (2)(−1) = 3 − 4 − 2 = −3。
Magnitudes: |a| = √(3² + (−1)² + 2²) = √9 + 1 + 4 = √14; |b| = √(1² + 4² + (−1)²) = √1 + 16 + 1 = √18.
模长：|a| = √(3² + (−1)² + 2²) = √9 + 1 + 4 = √14；|b| = √(1² + 4² + (−1)²) = √1 + 16 + 1 = √18。
cos θ = (a · b) / (|a||b|) = −3 / (√14 × √18) = −3 / √252 = −3 / (6√7) ≈ −0.18898. θ ≈ arccos(−0.18898) ≈ 100.9°.
cos θ = (a · b) / (|a||b|) = −3 / (√14 × √18) = −3 / √252 = −3 / (6√7) ≈ −0.18898。θ ≈ arccos(−0.18898) ≈ 100.9°。
6. Complex Numbers: Modulus and Argument | 复数：模与辐角
Example: Express the complex number z = −1 + √3 i in polar form.
例题：将复数 z = −1 + √3 i 表示为极坐标形式。
Modulus: |z| = √((−1)² + (√3)²) = √1 + 3 = 2.
模：|z| = √((−1)² + (√3)²) = √1 + 3 = 2。
Argument: since the real part is negative and the imaginary part positive, the angle lies in the second quadrant. The reference angle is arctan(|√3 / −1|) = arctan(√3) = π/3. Therefore, arg(z) = π − π/3 = 2π/3.
辐角：实部为负，虚部为正，角度在第二象限。参考角为 arctan(|√3 / −1|) = arctan(√3) = π/3。因此，arg(z) = π − π/3 = 2π/3。
Polar form: z = 2(cos(2π/3) + i sin(2π/3)).
极坐标形式：z = 2(cos(2π/3) + i sin(2π/3))。
7. Sequences and Series: Geometric Sum | 数列与级数：等比数列求和
Example: The first term of a geometric series is 5, and the common ratio is 0.8. Find the sum to infinity.
例题：一个等比级数的首项为 5，公比为 0.8。求无穷项之和。
Sum to infinity exists because |r| = 0.8 < 1. S∞ = a / (1 − r) = 5 / (1 − 0.8) = 5 / 0.2 = 25.
因为 |r| = 0.8 < 1，无穷项之和存在。S∞ = a / (1 − r) = 5 / (1 − 0.8) = 5 / 0.2 = 25。
8. Trigonometry: Solving Equations | 三角学：解方程
Example: Solve 2 sin² x − sin x − 1 = 0 for 0 ≤ x ≤ 2π.
例题：解方程 2 sin² x − sin x − 1 = 0，其中 0 ≤ x ≤ 2π。
Let u = sin x. Then 2u² − u − 1 = 0. Factorise: (2u + 1)(u − 1) = 0 ⇒ u = −½ or u = 1.
令 u = sin x。则 2u² − u − 1 = 0。因式分解：(2u + 1)(u − 1) = 0 ⇒ u = −½ 或 u = 1。
sin x = −½ ⇒ x = 7π/6 or 11π/6. sin x = 1 ⇒ x = π/2.
sin x = −½ ⇒ x = 7π/6 或 11π/6。sin x = 1 ⇒ x = π/2。
Solutions: x = π/2, 7π/6, 11π/6.
解为：x = π/2, 7π/6, 11π/6。
9. Logarithms: Change of Base | 对数：换底公式
Example: Evaluate log₄ 8 without a calculator.
例题：不借助计算器，求 log₄ 8 的值。
Let log₄ 8 = x. Then 4ˣ = 8. Express both sides with base 2: (2²)ˣ = 2³ ⇒ 2²ˣ = 2³ ⇒ 2x = 3 ⇒ x = 1.5.
设 log₄ 8 = x。则 4ˣ = 8。两边均以 2 为底：(2²)ˣ = 2³ ⇒ 2²ˣ = 2³ ⇒ 2x = 3 ⇒ x = 1.5。
Thus, log₄ 8 = 3/2.
因此，log₄ 8 = 3/2。
10. Kinematics: Displacement and Velocity | 运动学：位移与速度
Example: traight line with velocity v(t) = 3t² − 6t + 2. Find its displacement from t = 0 to t = 2.
例题：一个质点沿直线运动，速度 v(t) = 3t² − 6t + 2。求从 t = 0 到 t = 2 的位移。
Displacement = ∫₀² (3t² − 6t + 2) dt = [t³ − 3t² + 2t]₀² = (8 − 12 + 4) − (0) = 0.
位移 = ∫₀² (3t² − 6t + 2) dt = [t³ − 3t² + 2t]₀² = (8 − 12 + 4) − (0) = 0。
The net displacement is 0; the particle returns to its starting position after 2 seconds.
净位移为 0；质点在 2 秒后回到起点。
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