📚 Typical Example Solutions for IB & CIE Mathematics | IB与CIE数学典型例题详解
Mastering the IB and CIE Mathematics examinations demands not only conceptual understanding but also the ability to apply techniques to standard problem types. This article presents a carefully chosen set of worked examples, each solved step by step with clear reasoning. The explanations are designed to mirror the dual-language approach used by many international learners, strengthening both mathematical fluency and examination confidence.
掌握 IB 与 CIE 数学考试不仅需要理解概念，还需要将技巧应用于典型题目。本文精选一组经典例题，逐步详解，剖析每一步的逻辑。双语讲解旨在帮助国际课程学生巩固数学思维，提升应试信心，从容面对各类核心考点。
1. Completing the Square and Vertex Form | 配方法与顶点式
Example: Express the quadratic function f(x) = 2x2 – 8x + 7 in the form a(x – h)2 + k and state the coordinates of its vertex.
例题：将二次函数 f(x) = 2x2 – 8x + 7 化为 a(x – h)2 + k 的形式，并写出顶点坐标。
Step 1: Factor out the coefficient of x2 from the first two terms: f(x) = 2(x2 – 4x) + 7.
步骤1：将前两项中 x2 的系数提取出来：f(x) = 2(x2 – 4x) + 7。
Step 2: Complete the square inside the parentheses. Half of -4 is -2, and (-2)2 = 4. Add and subtract 4: f(x) = 2[(x2 – 4x + 4) – 4] + 7.
步骤2：在括号内完成配方。x 项系数 -4 的一半为 -2，其平方为 4。添加并减去 4：f(x) = 2[(x2 – 4x + 4) – 4] + 7。
Step 3: Rewrite the perfect square trinomial as (x – 2)2 and simplify: f(x) = 2[(x – 2)2 – 4] + 7 = 2(x – 2)2 – 8 + 7.
步骤3：将完全平方三项式写为 (x – 2)2，然后化简：f(x) = 2[(x – 2)2 – 4] + 7 = 2(x – 2)2 – 8 + 7。
Step 4: Combine the constant terms to obtain the vertex form: f(x) = 2(x – 2)2 – 1.
步骤4：合并常数项得到顶点式：f(x) = 2(x – 2)2 – 1。
The vertex of the parabola is at (h, k) = (2, -1).
抛物线的顶点坐标为 (h, k) = (2, -1)。
2. Solving Exponential Equations | 解指数方程
Example: Solve the equation 32x+1 = 5x, giving your answer in terms of natural logarithms.
例题：解方程 32x+1 = 5x，答案用自然对数表示。
Step 1: Take the natural logarithm of both sides: ln(32x+1) = ln(5x).
步骤1：两边同时取自然对数：ln(32x+1) = ln(5x)。
Step 2: Apply the power rule of logarithms: (2x+1) ln 3 = x ln 5.
步骤2：利用对数幂法则：(2x+1) ln 3 = x ln 5。
Step 3: Expand and collect terms containing x: 2x ln 3 + ln 3 = x ln 5 → 2x ln 3 – x ln 5 = -ln 3.
步骤3：展开并合并含 x 的项：2x ln 3 + ln 3 = x ln 5 → 2x ln 3 – x ln 5 = -ln 3。
Step 4: Factor out x: x(2 ln 3 – ln 5) = -ln 3.
步骤4：提取 x：x(2 ln 3 – ln 5) = -ln 3。
Step 5: Solve for x: x = -ln 3 / (2 ln 3 – ln 5).
步骤5：求出 x：x = -ln 3 / (2 ln 3 – ln 5)。
This can also be written as x = ln 3 / (ln 5 – 2 ln 3) by multiplying numerator and denominator by -1.
也可将分子分母同乘 -1，写成 x = ln 3 / (ln 5 – 2 ln 3)。
3. Applying Logarithmic Properties | 对数性质应用
Example: Solve the equation log2(x) + log2(x – 2) = 3.
例题：解方程 log2(x) + log2(x – 2) = 3。
Step 1: Combine the two logarithms using the product rule: log2[x(x – 2)] = 3.
步骤1：利用加法法则合并两个对数：log2[x(x – 2)] = 3。
Step 2: Rewrite in exponential form: x(x – 2) = 23 = 8.
步骤2：改写为指数形式：x(x – 2) = 23 = 8。
Step 3: Expand and form a quadratic equation: x2 – 2x – 8 = 0.
步骤3：展开并整理成二次方程：x2 – 2x – 8 = 0。
Step 4: Factorise the quadratic: (x – 4)(x + 2) = 0, giving potential solutions x = 4 or x = -2.
步骤4：因式分解：(x – 4)(x + 2) = 0，得到可能的解 x = 4 或 x = -2。
Step 5: Check the domain of the original logarithms. The arguments must be positive, so x > 0 and x – 2 > 0 → x > 2. The solution x = -2 is extraneous. Therefore, the only valid solution is x = 4.
步骤5：检验原对数的定义域。真数必须为正，因此 x > 0 且 x – 2 > 0 → x > 2。x = -2 为增根，舍去。唯一有效解为 x = 4。
4. Solving Trigonometric Equations | 解三角方程
Example: Solve sin(2θ) = cos θ for 0 ≤ θ ≤ 2π.
例题：在 0 ≤ θ ≤ 2π 内解方程 sin(2θ) = cos θ。
Step 1: Use the double-angle identity sin(2θ) = 2 sin θ cos θ.
步骤1：利用倍角公式 sin(2θ) = 2 sin θ cos θ。
Step 2: Substitute into the equation: 2 sin θ cos θ = cos θ.
步骤2：代入方程：2 sin θ cos θ = cos θ。
Step 3: Bring all terms to one side: 2 sin θ cos θ – cos θ = 0 → cos θ (2 sin θ – 1) = 0.
步骤3：移项并提取公因式：2 sin θ cos θ – cos θ = 0 → cos θ (2 sin θ – 1) = 0。
Step 4: Set each factor equal to zero: cos θ = 0 or sin θ = 1/2.
步骤4：令各因式等于零：cos θ = 0 或 sin θ = 1/2。
Step 5: Solve cos θ = 0 for 0 ≤ θ ≤ 2π. The solutions are θ = π/2 and θ = 3π/2.
步骤5：解 cos θ = 0 在区间内的值：θ = π/2 和 θ = 3π/2。
Step 6: Solve sin θ = 1/2. The reference angle is π/6, and sine is positive in the first and second quadrants, giving θ = π/6 and θ = 5π/6.
步骤6：解 sin θ = 1/2。参考角为 π/6，正弦在第一、二象限为正，故 θ = π/6 和 θ = 5π/6。
Step 7: Collect all solutions in the interval: θ = π/6, π/2, 5π/6, 3π/2.
步骤7：合并区间内所有解：θ = π/6, π/2, 5π/6, 3π/2。
5. Differentiation: Tangents and Normals | 求导应用：切线与法线
Example: Find the equation of the tangent to the curve y = x3 – 3x2 + 2 at the point where x = 1.
例题：求曲线 y = x3 – 3x2 + 2 在 x = 1 处的切线方程。
Step 1: Differentiate to find the gradient function: dy/dx = 3x2 – 6x.
步骤1：求导得斜率函数：dy/dx = 3x2 – 6x。
Step 2: Evaluate the derivative at x = 1: m = 3(1)2 – 6(1) = 3 – 6 = -3.
步骤2：计算 x = 1 处的导数值：m = 3(1)2 – 6(1) = 3 – 6 = -3。
Step 3: Find the y-coordinate of the point: y = (1)3 – 3(1)2 + 2 = 1 – 3 + 2 = 0. The point is (1, 0).
步骤3：求该点的纵坐标：y = (1)3 – 3(1)2 + 2 = 1 – 3 + 2 = 0。点为 (1, 0)。
Step 4: Use the point-slope form y – y1 = m(x – x1). The tangent equation is y – 0 = -3(x – 1).
步骤4：运用点斜式 y – y1 = m(x – x1)。切线方程为 y – 0 = -3(x – 1)。
Step 5: Simplify to y = -3x + 3.
步骤5：化简得 y = -3x + 3。
6. Integration: Area Between Curves | 积分：曲线间面积
Example: Find the area enclosed by the curves y = x2 and y = x + 2.
例题：求由曲线 y = x2 和 y = x + 2 围成的面积。
Step 1: Find the points of intersection by setting x2 = x + 2 → x2 – x – 2 = 0.
步骤1：联立方程求交点：x2 = x + 2 → x2 – x – 2 = 0。
Step 2: Factorise: (x – 2)(x + 1) = 0 → x = -1 and x = 2.
步骤2：因式分解：(x – 2)(x + 1) = 0 → x = -1 与 x = 2。
Step 3: Determine which curve is above the other on the interval [-1, 2]. Test x = 0: y = x+2 gives 2, y = x2 gives 0, so the line is above the parabola.
步骤3：判断在区间 [-1, 2] 上谁在上方。取 x = 0：直线 y = 2，抛物线 y = 0，故直线在上方。
Step 4: Set up the integral for area: ∫-12 [(x + 2) – x2] dx.
步骤4：建立面积积分：∫-12 [(x + 2) – x2] dx。
Step 5: Integrate term by term: ∫ (x + 2 – x2) dx = (1/2)x2 + 2x – (1/3)x3 evaluated from -1 to 2.
步骤5：逐项积分：∫ (x + 2 – x2) dx = (1/2)x2 + 2x – (1/3)x3，从 -1 到 2。
Step 6: Substitute the limits: At x=2: (1/2)(4) + 4 – (1/3)(8) = 2 + 4 – 8/3 = 6 – 8/3 = 10/3. At x=-1: (1/2)(1) – 2 – (1/3)(-1) = 0.5 – 2 + 1/3 = -1.5 + 1/3 = -3/2 + 1/3 = -9/6 + 2/6 = -7/6. Difference = 10/3 – (-7/6) = 20/6 + 7/6 = 27/6 = 9/2.
步骤6：代入上下限。x=2: (1/2)(4) + 4 – (1/3)(8) = 2 + 4 – 8/3 = 6 – 8/3 = 10/3。x=-1: (1/2)(1) – 2 – (1/3)(-1) = 0.5 – 2 + 1/3 = -3/2 + 1/3 = -7/6。面积 = 10/3 – (-7/6) = 20/6 + 7/6 = 27/6 = 9/2。
The bounded area is 9/2 square units.
围成的面积为 9/2 平方单位。
7. Binomial Distribution: Probability Calculation | 二项分布：概率计算
Example: A fair die is rolled 10 times. Find the probability of obtaining exactly four sixes.
例题：掷一粒公正骰子 10 次，求恰好得到 4 次六点的概率。
Step 1: Identify the parameters for the binomial distribution. n = 10 (trials), p = 1/6 (success of getting a six), q = 5/6.
步骤1：确定二项分布参数。n = 10（试验次数），p = 1/6（出现六点的成功概率），q = 5/6。
Step 2: Use the binomial probability formula: P(X = k) = nCk pk qn-k. Here k = 4.
步骤2：使用二项概率公式：P(X = k) = nCk pk qn-k。此处 k = 4。
Step 3: Calculate the combination: 10C4 = 10!/(4!6!) = (10×9×8×7)/(4×3×2×1) = 210.
步骤3：计算组合数：10C4 = 210。
Step 4: Substitute values: P(X = 4) = 210 × (1/6)4 × (5/6)6.
步骤4：代入数值：P(X = 4) = 210 × (1/6)4 × (5/6)6。
Step 5: Simplify: 210 × (56 / 610) = 210 × 15625 / 60466176 = 3281250 / 60466176 ≈ 0.0543 (4 s.f.).
步骤5：化简：210 × 15625 / 60466176 = 3281250 / 60466176 ≈ 0.0543（保留四位有效数字）。
The probability of exactly four sixes is approximately 0.0543.
恰好得到 4 次六点的概率约为 0.0543。
8. Normal Distribution and Inverse Normal | 正态分布与逆运算
Example: The weights of a certain fruit are normally distributed with mean μ = 200 g and standard deviation σ = 15 g. Find the weight below which the lightest 10% of fruit fall.
例题：某种水果的重量服从正态分布，均值 μ = 200 g，标准差 σ = 15 g。求最轻的 10% 的水果重量上限。
Step 1: We need the value w such that P(W < w) = 0.10. This is an inverse normal problem.
步骤1：需求出 w，使得 P(W < w) = 0.10。此为逆正态问题。
Step 2: Find the z-score corresponding to a left-tail probability of 0.10. From standard normal tables or calculator, z ≈ -1.2816.
步骤2：查标准正态分布表中左尾概率 0.10 对应的 z 值：z ≈ -1.2816。
Step 3: Use the standardisation formula z = (w – μ)/σ to link z and w.
步骤3：利用标准化公式 z = (w – μ)/σ 建立联系。
Step 4: Rearrange: w = μ + zσ = 200 + (-1.2816)(15).
步骤4：变形求解：w = μ + zσ = 200 + (-1.2816)(15)。
Step 5: Calculate w: -1.2816 × 15 = -19.224; w = 200 – 19.224 = 180.776 g.
步骤5：计算：-1.2816 × 15 = -19.224；w = 200 – 19.224 = 180.776 g。
The cut-off weight for the lightest 10% is approximately 181 g (3 s.f.).
最轻 10% 的重量上限约为 181 g（三位有效数字）。
9. Vectors: Angle Between Two Vectors | 向量：两向量夹角
Example: Given vectors a = 3i + 2j – k and b = i – 3j + 2k, find the acute angle between them.
例题：已知向量 a = 3i + 2j – k 与 b = i – 3j + 2k，求它们之间的锐角。
Step 1: Compute the dot product a · b: (3)(1) + (2)(-3) + (-1)(2) = 3 – 6 – 2 = -5.
步骤1：计算点积 a · b：(3)(1
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