📚 IB and CCEA Mathematics: Typical Worked Examples Explained | IB与CCEA数学：典型例题详解
In this article, we explore high-yield worked examples that are common to both the IB Diploma Programme mathematics courses (Analysis & Approaches and Applications & Interpretation) and the CCEA GCE A-level mathematics specification. Each section breaks down a classic problem type, highlights subtle differences in expected solution methods, and provides step-by-step bilingual reasoning to deepen understanding. Whether you are preparing for IB examinations, CCEA modules, or simply strengthening your mathematical toolkit, these detailed solutions will help you recognise key patterns and avoid common pitfalls.
本文精选了IB文凭课程数学（分析与方法、应用与解释）和CCEA GCE A-level数学中常见的高频典型例题，逐步拆解解题过程，对比两种课程体系在解答要求上的细微差别，并配以中英双语推理，帮助深化理解。无论你正在备战IB考试、CCEA单元测验，还是在夯实数学功底，这些详尽例题都能助你识别核心模式、避开常见陷阱。
1. Polynomial Division and the Factor Theorem | 多项式除法与因子定理
Typical problem: Given f(x) = 2x³ − 5x² − 4x + 3, show that (x − 3) is a factor and hence solve f(x) = 0. In IB AA SL/HL this is often followed by sketching the graph; in CCEA C2 the focus is on finding all roots and factorising fully.
典型题目：已知 f(x) = 2x³ − 5x² − 4x + 3，证明 (x − 3) 是一个因子，并由此解方程 f(x) = 0。在IB AA SL/HL中, 此类题常要求后续绘制函数图像；在CCEA C2中，重点在于求全部实根并彻底因式分解。
First apply the Factor Theorem: f(3) = 2(27) − 5(9) − 12 + 3 = 54 − 45 − 12 + 3 = 0, so (x − 3) is indeed a factor.
首先应用因子定理：f(3) = 2(27) − 5(9) − 12 + 3 = 54 − 45 − 12 + 3 = 0，因此 (x − 3) 确为一个因子。
Now perform polynomial division or equate coefficients. Dividing 2x³ − 5x² − 4x + 3 by (x − 3) gives a quadratic quotient 2x² + x − 1.
现在进行多项式除法或待定系数法。将 2x³ − 5x² − 4x + 3 除以 (x − 3)，得到商 2x² + x − 1。
Factorising the quadratic: 2x² + x − 1 = (2x − 1)(x + 1). Hence the fully factorised form is f(x) = (x − 3)(2x − 1)(x + 1).
对二次式因式分解：2x² + x − 1 = (2x − 1)(x + 1)。因此完全分解式为 f(x) = (x − 3)(2x − 1)(x + 1)。
Solving f(x) = 0 yields x = 3, x = 1/2, and x = −1. In IB, you would then sketch the cubic with these intercepts and note the end behaviour; CCEA typically asks for the root set without further graphing, but sketching is still good practice.
解 f(x) = 0 得到 x = 3、x = 1/2 和 x = −1。在IB中，你会据此绘制三次函数图像，标明截距并分析当 x→±∞ 时的走势；CCEA通常只要求写出根集合，无需完整图像，但画图仍是良好的习惯。
2. Differentiation and Tangent Lines | 导数与切线方程
Problem: Find the equation of the tangent to the curve y = x³ − 4x² + 7 at the point where x = 2. Both IB and CCEA set this staple question, though IB often embeds it in a context problem or requires finding a normal line as well.
题目：求曲线 y = x³ − 4x² + 7 在 x = 2 处的切线方程。IB和CCEA都常考此类基础题，但IB有时会将之放置于应用情境中，或同时要求法线方程。
First differentiate: dy/dx = 3x² − 8x. At x = 2, the gradient is 3(4) − 8(2) = 12 − 16 = −4.
先求导：dy/dx = 3x² − 8x。当 x = 2 时，梯度为 3(4) − 8(2) = 12 − 16 = −4。
The y-coordinate is y = 8 − 16 + 7 = −1. So the point is (2, −1) and the gradient is −4.
y坐标为 y = 8 − 16 + 7 = −1。因此切点为 (2, −1)，梯度为 −4。
Equation of tangent: y − (−1) = −4(x − 2), which simplifies to y = −4x + 7. In CCEA C1 this would be the final answer; in IB you should state the equation clearly and perhaps verify with a GDC (graphing calculator).
切线方程：y − (−1) = −4(x − 2)，化简得 y = −4x + 7。在CCEA C1中这已是最终答案；在IB中你应清晰给出方程，并可用图形计算器(GDC)加以验证。
3. Definite Integration and Area Under a Curve | 定积分与曲线下方面积
Example: Calculate the finite area bounded by the curve y = x² − 5x + 6 and the x-axis from x = 1 to x = 4. This tests careful handling of regions where the curve lies below the axis, a classic trap in both IB and CCEA.
例题：计算曲线 y = x² − 5x + 6 与 x 轴在 x = 1 到 x = 4 之间所围成的有限面积。此题考验对函数位于轴下方区域的正确处理，这是IB和CCEA中的经典易错点。
Find where the curve crosses the x-axis: solve x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0, so roots at x = 2 and x = 3.
先求曲线与 x 轴的交点：解 x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0，根为 x = 2 和 x = 3。
Between x=2 and x=3 the graph is below the axis, so we split the integral into three parts and change the sign of the middle section.
在 x=2 到 x=3 之间，图像位于 x 轴下方，因此将积分分成三段，并对中间部分取相反数。
Area = ∫₁² (x²−5x+6) dx + |∫₂³ (x²−5x+6) dx| + ∫₃⁴ (x²−5x+6) dx.
面积 = ∫₁² (x²−5x+6) dx + |∫₂³ (x²−5x+6) dx| + ∫₃⁴ (x²−5x+6) dx。
Antiderivative: F(x) = x³/3 − (5/2)x² + 6x. Evaluate carefully: F(2)−F(1) gives a positive value; F(3)−F(2) is negative, we take its absolute value; F(4)−F(3) is positive again. Summing yields total area = 11/6 + 1/6 + 11/6 = 23/6 square units.
原函数：F(x) = x³/3 − (5/2)x² + 6x。仔细计算：F(2)−F(1) 为正；F(3)−F(2) 为负，取绝对值；F(4)−F(3) 再次为正。总和得面积为 11/6 + 1/6 + 11/6 = 23/6 平方单位。
In CCEA C2, leaving the answer as an exact fraction is expected; IB likewise values exact answers, though a decimal approximation from a GDC may be accepted for verification.
在CCEA C2中，答案需保留精确分数形式；IB同样强调精确值，但可辅以GDC得出近似值作为检验。
4. Trigonometric Equations | 三角函数方程求解
Solve: 2 sin²θ − sinθ − 1 = 0 for 0° ≤ θ ≤ 360°. Both specifications emphasise multiple solutions and the use of CAST or unit circle diagrams.
求在 0° ≤ θ ≤ 360° 范围内解方程 2 sin²θ − sinθ − 1 = 0。IB和CCEA均强调多解处理以及运用CAST图或单位圆。
Treat as a quadratic in sinθ: let u = sinθ, then 2u² − u − 1 = 0, factorising to (2u + 1)(u − 1) = 0, so u = 1 or u = −1/2.
把方程看作关于 sinθ 的二次式：设 u = sinθ，则 2u² − u − 1 = 0，分解为 (2u + 1)(u − 1) = 0，得 u = 1 或 u = −1/2。
sinθ = 1 gives θ = 90°. sinθ = −1/2 gives reference angle 30°; sine is negative in the third and fourth quadrants, so θ = 180°+30° = 210° and θ = 360°−30° = 330°.
sinθ = 1 给出 θ = 90°。sinθ = −1/2 的参考角为 30°；正弦在第三、四象限为负，故 θ = 180°+30° = 210° 以及 θ = 360°−30° = 330°。
Solution set: {90°, 210°, 330°}. In IB, radian measure (π/2, 7π/6, 11π/6) is often required; CCEA sets questions in both degrees and radians depending on the module.
解集：{90°, 210°, 330°}。在IB中，常要求用弧度表示 (π/2, 7π/6, 11π/6)；CCEA根据模块不同会同时考查角度制与弧度制。
5. Binomial Distribution | 二项分布
Typical CCEA S1 / IB AI SL question: The probability that a seed germinates is 0.7. If 10 seeds are planted, find the probability that exactly 8 germinate. Determine also the expected number and variance.
典型的CCEA S1 / IB AI SL 题目：一粒种子发芽的概率为 0.7。若种下10粒种子，求恰好8粒发芽的概率，并计算期望值与方差。
Let X ~ B(10, 0.7). P(X = 8) = C(10,8) × (0.7)⁸ × (0.3)² = 45 × 0.057648 × 0.09 ≈ 0.2335 (to 4 d.p.).
设 X ~ B(10, 0.7)。P(X = 8) = C(10,8) × (0.7)⁸ × (0.3)² = 45 × 0.057648 × 0.09 ≈ 0.2335 (保留四位小数)。
Expected value E(X) = np = 10 × 0.7 = 7; variance Var(X) = np(1−p) = 10 × 0.7 × 0.3 = 2.1. IB may additionally ask for a hypothesis test or to compare with a normal approximation; CCEA S1 focuses on exact binomial calculations from formula or tables.
期望值 E(X) = np = 10 × 0.7 = 7；方差 Var(X) = np(1−p) = 10 × 0.7 × 0.3 = 2.1。IB可能进一步要求假设检验或正态近似比较；CCEA S1侧重于使用公式或表格进行精确的二项计算。
6. Vectors and the Scalar Product | 向量与点积应用
Problem: Given vectors a = 2i − j + 3k and b = i + 4j − 2k, find the angle between a and b. This is standard across IB HL and CCEA FP3/M3.
题目：已知向量 a = 2i − j + 3k，b = i + 4j − 2k，求 a 与 b 之间的夹角。这是IB HL和CCEA FP3/M3中的标准题型。
Compute dot product: a · b = (2)(1) + (−1)(4) + (3)(−2) = 2 − 4 − 6 = −8.
计算点积：a · b = (2)(1) + (−1)(4) + (3)(−2) = 2 − 4 − 6 = −8。
Magnitudes: |a| = √(4+1+9) = √14, |b| = √(1+16+4) = √21.
模长：|a| = √(4+1+9) = √14，|b| = √(1+16+4) = √21。
cosθ = (a·b) / (|a||b|) = −8 / (√14 √21) = −8 / √294. Simplify √294 = 7√6, so cosθ = −8 / (7√6). Rationalising: cosθ = −8√6 / 42 = −4√6 / 21. Hence θ = arccos(−4√6/21) ≈ 118.1° or 2.06 rad.
cosθ = (a·b) / (|a||b|) = −8 / (√14 √21) = −8 / √294。化简 √294 = 7√6，故 cosθ = −8 / (7√6)。有理化：cosθ = −8√6 / 42 = −4√6 / 21。因此 θ = arccos(−4√6/21) ≈ 118.1° 或 2.06 弧度。
In IB, you might also be asked to determine if the vectors are perpendicular (clearly not, as dot product ≠ 0) or to find a vector perpendicular to both (cross product). CCEA often stays with dot product applications and work on angles.
在IB中，你可能还需判断向量是否垂直（显然不是，因点积≠0）或求同时垂直两者的向量（叉积）。CCEA通常止于点积的应用和夹角计算。
7. Complex Numbers – Operations and Quadratic Equations | 复数运算与二次方程
Example: Solve z² − 4z + 13 = 0 over the complex numbers, and express the roots in the form a ± bi. Then find the modulus and argument of one root. (IB AA HL) / CCEA FP1 may ask for roots and subsequent plotting on an Argand diagram.
例题：在复数范围内解方程 z² − 4z + 13 = 0，并将根表示为 a ± bi 形式，进而求其中一个根的模与辐角。(IB AA HL) / CCEA FP1可能要求求根并在阿干特图上标出。
Using the quadratic formula: z = [4 ± √(16 − 52)] / 2 = [4 ± √(−36)] / 2 = [4 ± 6i] / 2 = 2 ± 3i.
使用求根公式：z = [4 ± √(16 − 52)] / 2 = [4 ± √(−36)] / 2 = [4 ± 6i] / 2 = 2 ± 3i。
Pick z₁ = 2 + 3i. Modulus r = √(2² + 3²) = √13. Argument θ = arctan(3/2) ≈ 56.3° or 0.983 rad. These are fundamental skills for both syllabi.
取 z₁ = 2 + 3i。模 r = √(2² + 3²) = √13。辐角 θ = arctan(3/2) ≈ 56.3° 或 0.983 弧度。这是两个课程体系共通的基本功。
8. Arithmetic and Geometric Sequences | 等差与等比数列求和
Scenario: The sixth term of an arithmetic sequence is 23 and the tenth term is 39. Find the first term and common difference, and calculate the sum of the first 30 terms. Both IB SL and CCEA C1/C2 set such problems frequently.
情景：一个等差数列的第六项为23，第十项为39。求首项、公差以及前30项之和。IB SL和CCEA C1/C2都频繁考查此类问题。
Using uₙ = a + (n−1)d: for n=6, a + 5d = 23; for n=10, a + 9d = 39. Subtracting gives 4d = 16, so d = 4. Then a = 23 − 5(4) = 3.
根据公式 uₙ = a + (n−1)d：当 n=6 时，a + 5d = 23；当 n=10 时，a + 9d = 39。相减得 4d = 16，故 d = 4。进而 a = 23 − 5(4) = 3。
Sum S₃₀ = (30/2)[2a + (29)d] = 15[6 + 116] = 15 × 122 = 1830.
求和 S₃₀ = (30/2)[2a + (29)d] = 15[6 + 116] = 15 × 122 = 1830。
For IB, the question might be extended to model a real-world context (e.g., seating rows in a theatre); CCEA keeps it purely algebraic, but both require confident manipulation of the formula booklet.
在IB中，题目可能延伸至实际情境建模（如剧院座位排列）；CCEA保持纯代数形式，但两者都要求学生熟练运用公式表中的公式。
9. Function Transformations and Composite Functions | 函数变换与复合函数
Given f(x) = ln(x) for x > 0. Describe the sequence of transformations that map f(x) to g(x) = 2 ln(3x − 1) + 4, and find the domain of g. This type of reasoning appears in IB AA/AI and in CCEA C3.
已知 f(x) = ln(x)，x > 0。描述将 f(x) 映射到 g(x) = 2 ln(3x − 1) + 4 的一系列变换，并求 g 的定义域。这类推理出现在IB AA/AI和CCEA C3中。
Start from inside: replace x with 3x − 1: horizontal transformation. First, 3x indicates a horizontal stretch by factor 1/3 (IB often specifies ‘horizontal shrink by factor 1/3’), then shift right by 1/3? Actually, rewrite: ln(3(x − 1/3)). So stretch in x-direction by factor 1/3, translate right by 1/3.
从内向外分析：将 x 替换为 3x − 1，即水平变换。首先，3x 表示水平方向拉伸系数为 1/3（IB常说“水平方向收缩至1/3倍”），然后向右平移 1/3。改写为 ln(3(x − 1/3))：水平方向乘1/3，右移1/3。
Then multiply by 2: vertical stretch by factor 2. Finally add 4: vertical translation upwards by 4.
接着乘以2：垂直方向拉伸为原来2倍。最后加4：垂直向上平移4个单位。
Domain: 3x − 1 > 0 → x > 1/3. Both IB and CCEA require clear domain statements; IB might further ask for the range and sketch.
定义域：3x − 1 > 0 → x > 1/3。IB和CCEA均要求清晰写出定义域；IB可能进一步要求值域并作图。
10. Mathematical Induction | 数学归纳法证明
Prove by induction that Σᵣ₌₁ⁿ r² = n(n+1)(2n+1)/6 for all positive integers n. A core proof topic in IB HL and an optional but highly useful skill for CCEA FP3.
用归纳法证明对于所有正整数 n，Σᵣ₌₁ⁿ r² = n(n+1)(2n+1)/6。这是IB HL的核心证明题，也是CCEA FP3中可选但极为有用的技能。
Base case (n=1): LHS = 1² = 1; RHS = 1×2×3/6 = 1. True.
奠基步 (n=1)：左边 = 1² = 1；右边 = 1×2×3/6 = 1。成立。
Inductive step: Assume true for n = k, so Σᵣ₌₁ᵏ r² = k(k+1)(2k+1)/6. For n = k+1, LHS = Σᵣ₌₁ᵏ r² + (k+1)².
递推步：假设 n = k 时成立，即 Σᵣ₌₁ᵏ r² = k(k+1)(2k+1)/6。对于 n = k+1，左边 = Σᵣ₌₁ᵏ r² + (k+1)²。
Substitute the assumption: = k(k+1)(2k+1)/6 + (k+1)². Factor (k+1): = (k+1)[k(2k+1)/6 + (k+1)] = (k+1)[(2k² + k + 6k + 6)/6] = (k+1)(2k² + 7k + 6)/6.
代入假设：= k(k+1)(2k+1)/6 + (k+1)²。提取因式 (k+1)：= (k+1)[k(2k+1)/6 + (k+1)] = (k+1)[(2k² + k + 6k + 6)/6] = (k+1)(2k² + 7k + 6)/6。
Factor the quadratic: 2k² + 7k + 6 = (k+2)(2k+3). So expression = (k+1)(k+2)(2k+3)/6, which is exactly the formula with n = k+1. Hence true for all n.
对二次式因式分解：2k² + 7k + 6 = (k+2)(2k+3)。于是表达式 = (k+1)(k+2)(2k+3)/6，恰为 n = k+1 时的公式。因此对全体正整数成立。
IB expects a formal concluding statement; CCEA likewise rewards clarity and logical flow. Notice the precise algebraic manipulation is essential in both.
IB期望规范的结论语句；CCEA同样看重清晰的表述和逻辑连贯。注意精确的代数操作在两个体系中都至关重要。
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