📚 Typical Example Problems in IB AQA Mathematics | IB AQA 数学典型例题详解
This article offers step-by-step solutions to typical problems encountered in the IB AQA Mathematics course, spanning algebra, functions, trigonometry, calculus, probability, and vectors. Each worked example is chosen to highlight essential techniques, common errors, and efficient problem-solving strategies that are regularly assessed.
本文针对 IB AQA 数学课程中的典型例题提供逐步解析，涵盖代数、函数、三角、微积分、概率和向量等内容。每道例题都旨在突出常考的核心方法、常见错误以及高效的解题策略。
1. Solving Quadratic Equations | 解二次方程
Problem: Solve the quadratic equation 2x² – 5x – 3 = 0 by factorisation and by the quadratic formula, and verify the solutions graphically.
例题：用因式分解和求根公式解二次方程 2x² – 5x – 3 = 0，并用图像验证解。
Begin with factorisation. Look for two numbers that multiply to 2 × (-3) = -6 and add to -5. These are -6 and 1. Rewrite: 2x² – 6x + x – 3 = 0. Group: 2x(x – 3) + 1(x – 3) = 0 ⇒ (2x + 1)(x – 3) = 0. Therefore x = -½ or x = 3.
首先用因式分解。寻找两个数，乘积为 2 × (-3) = -6，和为 -5，这两个数是 -6 和 1。改写：2x² – 6x + x – 3 = 0。分组：2x(x – 3) + 1(x – 3) = 0 ⇒ (2x + 1)(x – 3) = 0。因此 x = -½ 或 x = 3。
Using the quadratic formula x = [-b ± √(b² – 4ac)] / (2a) with a = 2, b = -5, c = -3. The discriminant: Δ = (-5)² – 4·2·(-3) = 25 + 24 = 49. Then x = (5 ± 7) / 4, giving x = 12/4 = 3 and x = -2/4 = -½.
用求根公式 x = [-b ± √(b² – 4ac)] / (2a)，其中 a = 2，b = -5，c = -3。判别式 Δ = (-5)² – 4·2·(-3) = 25 + 24 = 49。于是 x = (5 ± 7) / 4，得 x = 3 和 x = -½。
Graphically, the parabola y = 2x² – 5x – 3 crosses the x‑axis at x = -0.5 and x = 3, confirming the roots. Always check factorised form expands correctly.
从图像看，抛物线 y = 2x² – 5x – 3 在 x = -0.5 和 x = 3 处与 x 轴相交，验证了上述根。务必检查因式展开是否正确。
2. Function Transformations and Inverse | 函数变换与反函数
Problem: Given f(x) = √(x – 2) + 1 for x ≥ 2, sketch the graph and find the inverse function f⁻¹(x), stating its domain.
例题：已知 f(x) = √(x – 2) + 1，x ≥ 2，画出草图并求反函数 f⁻¹(x)，指出其定义域。
The graph starts at (2, 1) and increases gradually to the right. It is a horizontal translation of √x by 2 units right and 1 unit up. The range is y ≥ 1.
图像起始于点 (2, 1)，向右逐渐上升。这是 √x 向右平移 2 个单位、向上平移 1 个单位的结果。值域为 y ≥ 1。
To find the inverse, write y = √(x – 2) + 1. Swap x and y: x = √(y – 2) + 1. Isolate the radical: x – 1 = √(y – 2). Square both sides: (x – 1)² = y – 2. Hence y = (x – 1)² + 2. So f⁻¹(x) = (x – 1)² + 2. The domain of f⁻¹ is the range of f, so x ≥ 1.
求反函数时，设 y = √(x – 2) + 1，交换 x 和 y：x = √(y – 2) + 1。解出根式：x – 1 = √(y – 2)。两边平方：(x – 1)² = y – 2，得 y = (x – 1)² + 2。因此 f⁻¹(x) = (x – 1)² + 2，定义域为 x ≥ 1。
Note that the squared term removes the square root, but the restricted domain ensures the inverse remains a function.
注意平方会消去根号，但通过对定义域的限制保证了反函数仍是函数。
3. Trigonometric Equations and Identities | 三角方程与恒等式
Problem: Solve 2 sin²θ – cos θ = 1 for 0 ≤ θ < 2π, giving answers in radians.
例题：解方程 2 sin²θ – cos θ = 1，其中 0 ≤ θ < 2π，答案用弧度表示。
Use the identity sin²θ + cos²θ = 1 to rewrite sin²θ as 1 – cos²θ. Substitute: 2(1 – cos²θ) – cos θ = 1 → 2 – 2cos²θ – cos θ = 1 → -2cos²θ – cos θ + 1 = 0. Multiply by -1: 2cos²θ + cos θ – 1 = 0.
利用 sin²θ + cos²θ = 1，将 sin²θ 写成 1 – cos²θ。代入：2(1 – cos²θ) – cos θ = 1 → 2 – 2cos²θ – cos θ = 1 → -2cos²θ – cos θ + 1 = 0。乘以 -1 得：2cos²θ + cos θ – 1 = 0。
Treat as quadratic in cos θ. Factor: (2cos θ – 1)(cos θ + 1) = 0. So cos θ = ½ or cos θ = -1. For cos θ = ½, θ = π/3, 5π/3. For cos θ = -1, θ = π. All within [0, 2π).
把它看作关于 cos θ 的二次方程。因式分解：(2cos θ – 1)(cos θ + 1) = 0。得 cos θ = ½ 或 cos θ = -1。cos θ = ½ 对应 θ = π/3, 5π/3；cos θ = -1 对应 θ = π。所有解均在 [0, 2π) 内。
Always check solutions by substituting back or by considering the unit circle. The set of solutions is {π/3, π, 5π/3}.
务必通过代入原式或参考单位圆来验证解。解集为 {π/3, π, 5π/3}。
4. Differentiation from First Principles | 从第一原理求导
Problem: Using first principles, find the derivative of f(x) = 3x² – 2x.
例题：利用第一原理求 f(x) = 3x² – 2x 的导数。
First principles state f'(x) = lim_{h→0} [f(x+h) – f(x)] / h. Compute f(x+h) = 3(x+h)² – 2(x+h) = 3(x² + 2xh + h²) – 2x – 2h = 3x² + 6xh + 3h² – 2x – 2h.
第一原理公式为 f'(x) = lim_{h→0} [f(x+h) – f(x)] / h。计算 f(x+h) = 3(x+h)² – 2(x+h) = 3(x² + 2xh + h²) – 2x – 2h = 3x² + 6xh + 3h² – 2x – 2h。
Then f(x+h) – f(x) = (3x² + 6xh + 3h² – 2x – 2h) – (3x² – 2x) = 6xh + 3h² – 2h. Divide by h: (6xh + 3h² – 2h)/h = 6x + 3h – 2. Take the limit as h→0, giving f'(x) = 6x – 2.
于是 f(x+h) – f(x) = (3x² + 6xh + 3h² – 2x – 2h) – (3x² – 2x) = 6xh + 3h² – 2h。除以 h：(6xh + 3h² – 2h)/h = 6x + 3h – 2。令 h→0 取极限，得到 f'(x) = 6x – 2。
This result matches the standard rule: derivative of 3x² is 6x, derivative of -2x is -2. Understanding first principles reinforces the concept of instantaneous rate of change.
该结果与常规求导法则一致：3x² 的导数是 6x，-2x 的导数是 -2。理解第一原理有助于强化瞬时变化率的概念。
5. Integration and Definite Integrals | 积分与定积分
Problem: Evaluate ∫₁² (4x³ – 3x² + 2) dx and interpret as area.
例题：计算 ∫₁² (4x³ – 3x² + 2) dx 并解释其面积含义。
Integrate term by term: ∫4x³ dx = x⁴, ∫-3x² dx = -x³, ∫2 dx = 2x. So the indefinite integral is x⁴ – x³ + 2x + C.
逐项积分：∫4x³ dx = x⁴，∫-3x² dx = -x³，∫2 dx = 2x。因此不定积分为 x⁴ – x³ + 2x + C。
Apply limits: F(2) – F(1) where F(x) = x⁴ – x³ + 2x. F(2) = 16 – 8 + 4 = 12. F(1) = 1 – 1 + 2 = 2. So the definite integral equals 10.
代入上下限：F(2) – F(1)，其中 F(x) = x⁴ – x³ + 2x。F(2) = 16 – 8 + 4 = 12，F(1) = 1 – 1 + 2 = 2，故定积分值为 10。
This value represents the net signed area between the curve y = 4x³ – 3x² + 2 and the x‑axis from x = 1 to x = 2. Since the function is positive on [1,2], the area is simply 10 square units.
该值表示曲线 y = 4x³ – 3x² + 2 与 x 轴之间在 x = 1 到 x = 2 上的带号面积。因函数在 [1,2] 上为正，故面积即为 10 平方单位。
6. Probability and Binomial Distribution | 概率与二项分布
Problem: A fair die is rolled 5 times. Find the probability of obtaining exactly two sixes.
例题：一枚均匀骰子掷 5 次，求恰好出现两次点数为六的概率。
This is a binomial situation: n = 5 trials, success probability p = 1/6, and we want k = 2 successes. The formula: P(X = k) = C(n, k) pᵏ (1 – p)ⁿ⁻ᵏ.
这是一个二项分布问题：n = 5 次试验，每次成功概率 p = 1/6，求 k = 2 次成功。公式：P(X = k) = C(n, k) pᵏ (1 – p)ⁿ⁻ᵏ。
Compute: C(5,2) = 10. p² = (1/6)² = 1/36. (1 – p)³ = (5/6)³ = 125/216. Multiply: 10 × (1/36) × (125/216) = 1250 / 7776 ≈ 0.1608. Simplify fraction: divide by 2 → 625/3888.
计算：C(5,2) = 10。p² = (1/6)² = 1/36。(1 – p)³ = (5/6)³ = 125/216。相乘：10 × (1/36) × (125/216) = 1250 / 7776 ≈ 0.1608。化简分数得 625/3888。
Thus the probability of exactly two sixes in five rolls is 625/3888. Remember to check that the number of trials is fixed and outcomes are independent.
所以，掷五次骰子恰好出现两次六点的概率是 625/3888。记得检查试验次数是否固定以及结果是否相互独立。
7. Vectors: Dot Product and Angle | 向量：点积与夹角
Problem: Given vectors a = 2i + j – 3k and b = -i + 4j + k, find the angle between them.
例题：已知向量 a = 2i + j – 3k，b = -i + 4j + k，求它们之间的夹角。
Use the dot product: a·b = |a||b| cos θ, so cos θ = (a·b) / (|a||b|). Compute a·b = (2)(-1) + (1)(4) + (-3)(1) = -2 + 4 – 3 = -1.
运用点积公式：a·b = |a||b| cos θ，得 cos θ = (a·b) / (|a||b|)。计算 a·b = (2)(-1) + (1)(4) + (-3)(1) = -2 + 4 – 3 = -1。
Magnitudes: |a| = √(2² + 1² + (-3)²) = √(4+1+9) = √14. |b| = √((-1)² + 4² + 1²) = √(1+16+1) = √18 = 3√2. Thus cos θ = -1 / (√14 · 3√2) = -1 / (3√28) = -1 / (6√7) after simplifying √28 = 2√7.
模长：|a| = √(2² + 1² + (-3)²) = √14。|b| = √((-1)² + 4² + 1²) = √18 = 3√2。于是 cos θ = -1 / (√14 × 3√2) = -1 / (3√28) = -1 / (6√7)（化简后 √28 = 2√7）。
Since cos θ is negative, the angle is obtuse. θ = cos⁻¹(-1/(6√7)) ≈ 93.6°. The dot product is invaluable for testing orthogonality and projections.
由于 cos θ 为负，夹角为钝角。θ = cos⁻¹(-1/(6√7)) ≈ 93.6°。点积在判断垂直关系和计算投影时非常有用。
8. Complex Numbers and Polar Form | 复数与极坐标形式
Problem: Express the complex number z = -√3 + i in polar form and find z⁴ using De Moivre’s theorem.
例题：将复数 z = -√3 + i 表示为极坐标形式，并用棣莫弗定理求 z⁴。
Plot the number: real part = -√3, imaginary part = 1. The modulus r = √[(-√3)² + 1²] = √(3 + 1) = 2. The argument (angle) θ must satisfy tan θ = 1/(-√3) = -1/√3. Since the point is in the second quadrant, θ = π – π/6 = 5π/6.
画出该点：实部 = -√3，虚部 = 1。模 r = √[(-√3)² + 1²] = √(3 + 1) = 2。辐角 θ 满足 tan θ = 1/(-√3) = -1/√3。由于该点位于第二象限，θ = π – π/6 = 5π/6。
Polar form: z = r(cos θ + i sin θ) = 2(cos(5π/6) + i sin(5π/6)). Alternatively, z = 2 cis(5π/6). For z⁴, apply De Moivre: z⁴ = 2⁴ [cos(4·5π/6) + i sin(4·5π/6)] = 16[cos(10π/3) + i sin(10π/3)].
极坐标形式：z = r(cos θ + i sin θ) = 2(cos(5π/6) + i sin(5π/6))，或记为 z = 2 cis(5π/6)。求 z⁴ 时，用棣莫弗定理：z⁴ = 2⁴ [cos(4 × 5π/6) + i sin(4 × 5π/6)] = 16[cos(10π/3) + i sin(10π/3)]。
Reduce the angle: 10π/3 – 2π = 10π/3 – 6π/3 = 4π/3. So z⁴ = 16[cos(4π/3) + i sin(4π/3)] = 16(-½ – i√3/2) = -8 – 8i√3. This matches expanding ( -√3 + i )⁴ directly.
化简角度：10π/3 – 2π = 4π/3。因此 z⁴ = 16[cos(4π/3) + i sin(4π/3)] = 16(-½ – i√3/2) = -8 – 8i√3。这与直接展开 (-√3 + i)⁴ 的结果一致。
9. Arithmetic and Geometric Sequences | 等差数列与等比数列
Problem: The 3rd term of an arithmetic sequence is 7 and the 8th term is 22. Find the sum of the first 20 terms.
例题：等差数列的第 3 项是 7，第 8 项是 22。求前 20 项的和。
For arithmetic sequences, uₙ = a + (n-1)d. Given u₃ = a + 2d = 7 and u₈ = a + 7d = 22. Subtract: (a+7d) – (a+2d) = 5d = 15 → d = 3. Substitute back: a + 2·3 = 7 → a = 1.
等差数列通项 uₙ = a + (n-1)d。已知 u₃ = a + 2d = 7，u₈ = a + 7d = 22。两式相减得 5d = 15 → d = 3。代回：a + 6 = 7 → a = 1。
Sum of first 20 terms: Sₙ = n/2 [2a + (n-1)d]. So S₂₀ = 20/2 [2·1 + 19·3] = 10 [2 + 57] = 10 × 59 = 590.
前 20 项和公式：Sₙ = n/2 [2a + (n-1)d]。故 S₂₀ = 20/2 [2·1 + 19·3] = 10 [2 + 57] = 590。
Always verify with a few terms: 1, 4, 7, 10, 13, 16, 19, 22, … The 3rd is 7, 8th is 22. Summation confirms the pattern.
不妨写出前几项验证：1, 4, 7, 10, 13, 16, 19, 22, … 第 3 项是 7，第 8 项是 22，求和正确。
10. Applications of Differentiation: Optimisation | 微分应用：最优化
Problem: A rectangular box with a square base and no lid is to be made from 600 cm² of material. Find the maximum possible volume.
例题：用 600 cm² 的材料制作一个无盖正方形底的长方体盒子，求可能的最大容积。
Let base side = x cm and height = h cm. Surface area: base area = x², four sides = 4xh. Total material: x² + 4xh = 600. So h = (600 – x²) / (4x). Volume V = x²h = x²·(600 – x²)/(4x) = (600x – x³)/4.
设底边长为 x cm，高为 h cm。表面积：底面积 = x²，四个侧面 = 4xh。所用材料：x² + 4xh = 600，因此 h = (600 – x²) / (4x)。体积 V = x²h = x²·(600 – x²)/(4x) = (600x – x³)/4。
Optimise: V(x) = 150x – (1/4)x³. Differentiate: V'(x) = 150 – (3/4)x². Set V'(x) = 0 → (3/4)x² = 150 → x² = 200 → x = √200 = 10√2 cm (positive root). Check second derivative V”(x) = -(3/2)x, which is negative for x>0, so maximum.
求最值：V(x) = 150x – (1/4)x³。求导：V'(x) = 150 – (3/4)x²。令导数为零：(3/4)x² = 150 → x² = 200 → x = 10√2 cm（仅取正根）。检查二阶导数 V”(x) = -(3/2)x，当 x>0 时为负，故为极大值。
Then h = (600 – 200) / (4·10√2) = 400 / (40√2) = 10/√2 = 5√2 cm. Maximum volume = x²h = (200) × (5√2) = 1000√2 cm³ ≈ 1414.2 cm³. Always confirm the dimensions satisfy the original constraint.
此时 h = (600 – 200) / (4 × 10√2) = 400 / (40√2) = 10/√2 = 5√2 cm。最大容积 = x²h = 200 × 5√2 = 1000√2 cm³ ≈ 1414.2 cm³。务必验证所得尺寸满足原始约束条件。
11. Integration by Substitution | 换元积分法
Problem: Find ∫ 2x √(x² + 1) dx using substitution.
例题：利用换元法求 ∫ 2x √(x² + 1) dx。
Let u = x² + 1. Then du/dx = 2x, so du = 2x dx. The integral becomes ∫ √u du = ∫ u^(1/2) du.
令 u = x² + 1，则 du/dx = 2x，故 du = 2x dx。原积分化为 ∫ √u du = ∫ u^(1/2) du。
Integrate: (2/3) u^(3/2) + C. Substitute back: (2/3) (x² + 1)^(3/2) + C. Check by differentiating: derivative of (2/3)(x²+1)^(3/2) is (2/3)·(3/2)(x²+1)^(1/2)·2x = 2x√(x²+1), which matches the integrand.
积分得 (2/3) u^(3/2) + C。再代回：(2/3)(x² + 1)^(3/2) + C。验证：对 (2/3)(x²+1)^(3/2) 求导得 (2/3)·(3/2)(x²+1)^(1/2)·2x = 2x√(x²+1)，与被积函数一致。
Substitution is powerful when a function and its derivative appear together. Always remember to express the final answer in terms of the original variable.
当被积函数中出现一个函数及其导数时，换元法非常有效。切记最终答案要换回原变量。
12. Probability Tree Diagrams and Conditional Probability | 概率树状图与条件概率
Problem: Bag A contains 3 red and 2 blue pens. Bag B contains 4 red and 1 blue pen. A bag is chosen at random and a pen is drawn. If the pen is red, what is the probability it came from Bag A?
例题：A 袋有 3 红 2 蓝笔，B 袋有 4 红 1 蓝笔。随机选一个袋子再从中抽一支笔。若抽到红色，求它来自 A 袋的概率。
Use Bayes’ theorem or tree diagram. P(A) = P(B) = ½. P(Red | A) = 3/5, P(Red | B) = 4/5. Total probability of red: P(Red) = P(A)P(Red|A) + P(B)P(Red|B) = (½)(3/5) + (½)(4/5) = (3/10) + (4/10) = 7/10.
使用贝叶斯定理或树状图。P(A) = P(B) = ½。P(红|A) = 3/5，P(红|B) = 4/5。抽到红色的全概率：P(红) = (½)(3/5) + (½)(4/5) = 3/10 + 4/10 = 7/10。
Desired conditional probability: P(A | Red) = [P(A)P(Red|A)] / P(Red) = (½ × 3/5) / (7/10) = (3/10) / (7/10) = 3/7. So there is a 3/7 chance the red pen came from Bag A.
所求条件概率：P(A|红) = [P(A)P(红|A)] / P(红) = (½ × 3/5) / (7/10) = (3/10) / (7/10) = 3/7。因此该红色笔来自 A 袋的概率为 3/7。
This illustrates how prior probabilities update with evidence. Drawing a tree can help visualise the branches and computations.
此题展示了先验概率如何随证据更新。画树状图有助于直观展示各个分支和计算过程。
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