📚 Typical Example Questions in IB & OCR Mathematics | IB OCR 数学典型例题详解
In both IB and OCR A‑Level Mathematics courses, students encounter a wide variety of problem types designed to assess conceptual understanding, algebraic manipulation, and the ability to apply methods in unfamiliar contexts. This article works through ten typical questions spanning functions, calculus, probability, vectors, trigonometry, sequences, complex numbers, kinematics and matrices. Each solution is shown step‑by‑step, with clear reasoning, to help reinforce exam‑ready techniques.
在 IB 和 OCR A‑Level 数学课程中，学生会遇到各式各样的典型题目，用以考察概念理解、代数运算以及在陌生情境中应用方法的能力。本文精选十个典型问题，涵盖函数、微积分、概率、向量、三角学、数列、复数、运动学和矩阵，每个解答都分步呈现并配有清晰的推理，帮助巩固应试技巧。
1. Function Composition | 复合函数
Question: Given f(x) = x² + 1 and g(x) = 2x − 3, find (a) f(g(x)) and (b) g(f(x)).
题目：已知 f(x) = x² + 1，g(x) = 2x − 3，求 (a) f(g(x)) 与 (b) g(f(x)).
Step 1: Replace the inner function. For f(g(x)), substitute g(x) into f.
步骤1：代入内层函数。对于 f(g(x))，将 g(x) 代入 f。
f(g(x)) = (2x − 3)² + 1
f(g(x)) = (2x − 3)² + 1
Step 2: Expand and simplify. (2x − 3)² = 4x² − 12x + 9, so f(g(x)) = 4x² − 12x + 10.
步骤2：展开并化简。(2x − 3)² = 4x² − 12x + 9，因此 f(g(x)) = 4x² − 12x + 10.
f(g(x)) = 4x² − 12x + 10
f(g(x)) = 4x² − 12x + 10
For g(f(x)): g(f(x)) = 2(x² + 1) − 3 = 2x² + 2 − 3 = 2x² − 1.
对于 g(f(x))：g(f(x)) = 2(x² + 1) − 3 = 2x² + 2 − 3 = 2x² − 1.
g(f(x)) = 2x² − 1
g(f(x)) = 2x² − 1
Notice that the two compositions produce different results, demonstrating that function composition is not commutative in general.
注意到两个复合结果不同，这说明函数的复合一般不满足交换律。
2. Differentiation from First Principles | 第一原理求导
Question: Using the definition of the derivative, find f′(x) for f(x) = x².
题目：用导数定义求 f(x) = x² 的导数 f′(x)。
The definition is f′(x) = limₕ→₀ [f(x+h) − f(x)] / h.
定义为 f′(x) = limₕ→₀ [f(x+h) − f(x)] / h。
Step 1: Compute f(x+h) = (x+h)² = x² + 2xh + h².
步骤1：计算 f(x+h) = (x+h)² = x² + 2xh + h².
Step 2: Form the difference quotient.
步骤2：构造差商。
[f(x+h) − f(x)] / h = (x² + 2xh + h² − x²) / h = (2xh + h²) / h
[f(x+h) − f(x)] / h = (x² + 2xh + h² − x²) / h = (2xh + h²) / h
Step 3: Factor h and cancel: = h(2x + h) / h = 2x + h, provided h ≠ 0.
步骤3：提取公因子 h 并约分：= h(2x + h) / h = 2x + h，当 h ≠ 0 时成立。
Step 4: Take the limit as h → 0.
步骤4：令 h → 0 取极限。
f′(x) = limₕ→₀ (2x + h) = 2x
f′(x) = limₕ→₀ (2x + h) = 2x
This rigorous approach underpins all differentiation rules used in IB and OCR exams.
这一严谨方法奠定了 IB 和 OCR 考试中所有求导法则的基础。
3. Integration by Substitution | 换元积分法
Question: Evaluate ∫ x√x² + 1 dx.
题目：求 ∫ x√x² + 1 dx.
Let u = x² + 1. Then du/dx = 2x, so du = 2x dx, and x dx = ½ du.
令 u = x² + 1，则 du/dx = 2x，因此 du = 2x dx，可得 x dx = ½ du.
Substitute into the integral: ∫ x√x² + 1 dx = ∫ √u · (½ du) = ½ ∫ u¹⁄₂ du.
代入积分：∫ x√x² + 1 dx = ∫ √u · (½ du) = ½ ∫ u¹⁄₂ du.
egrate using the power rule: ½ · (u³⁄₂) / (3/2) + C = ½ · (2/3) u³⁄₂ + C = (1/3) u³⁄₂ + C.
使用幂法则积分：½ · (u³⁄₂) / (3/2) + C = ½ · (2/3) u³⁄₂ + C = (1/3) u³⁄₂ + C.
Replace u with x² + 1: (1/3)(x² + 1)³⁄₂ + C.
将 u 替换回 x² + 1：(1/3)(x² + 1)³⁄₂ + C.
∫ x√x² + 1 dx = (1/3)(x² + 1)³⁄₂ + C
∫ x√x² + 1 dx = (1/3)(x² + 1)³⁄₂ + C
Always remember to add the constant of integration C, and check your result by differentiation.
永远记得加上积分常数 C，并通过求导验证结果。
4. Normal Distribution – Probability Calculation | 正态分布 – 概率计算
ble X follows a normal distribution with mean μ = 50 and variance σ² = 25. Find P(X < 55).
题目：随机变量 X 服从均值 μ = 50、方差 σ² = 25 的正态分布，求 P(X < 55).
First identify σ = √25 = 5. Standardise using Z = (X − μ) / σ.
首先确定 σ = √25 = 5。标准化：Z = (X − μ) / σ。
For X = 55, Z = (55 − 50) / 5 = 1.
当 X = 55 时，Z = (55 − 50) / 5 = 1.
Therefore, P(X < 55) = P(Z < 1). Using the standard normal table, Φ(1) ≈ 0.8413.
因此，P(X < 55) = P(Z < 1)。查标准正态分布表，Φ(1) ≈ 0.8413。
P(X < 55) ≈ 0.841
P(X < 55) ≈ 0.841
If a calculator is allowed, the exact value can be obtained directly; the standardisation method is essential for showing working.
若允许使用计算器，可直接得到精确值；但标准化方法是展示计算过程的关键。
5. Vector Equation of a Line | 直线的向量方程
tor equation of the line passing through A(1, 2, 3) with direction vector d = 2i − j + 4k.
题目：求过点 A(1, 2, 3) 且方向向量为 d = 2i − j + 4k 的直线的向量方程。
a + λd, where a is the position vector of a point on the line and λ is a scalar parameter.
直线的向量方程为 r = a + λd，其中 a 是线上一点的位置向量，λ 是标量参数。
Here, a = i + 2j + 3k and d = 2i − j + 4k. Hence,
此处，a = i + 2j + 3k，d = 2i − j + 4k。因此，
r = (1, 2, 3) + λ(2, −1, 4)
r = (1, 2, 3) + λ(2, −1, 4)
In component form: x = 1 + 2λ, y = 2 − λ, z = 3 + 4λ.
写成参数形式：x = 1 + 2λ，y = 2 − λ，z = 3 + 4λ。
This compact representation is widely used in IB and OCR vector problems involving intersections, angles, and distances.
这种紧凑的表示法广泛用于 IB 和 OCR 中涉及交点、角度与距离的向量问题。
6. Trigonometric Equations | 三角方程
Question: Solve sin(2x) = 1/2 for 0 ≤ x ≤ 2π.
题目：在区间 0 ≤ x ≤ 2π 内解方程 sin(2x) = 1/2。
t y = 2x. Then sin y = 1/2. The basic solutions for y are y = π/6 and y = 5π/6 in [0, 2π].
令 y = 2x，则 sin y = 1/2。在 [0, 2π] 中，y 的基础解为 y = π/6 和 y = 5π/6。
e the period of sin y is 2π, the general solutions are y = π/6 + 2kπ and y = 5π/6 + 2kπ, k ∈ ℤ.
由于 sin y 的周期为 2π，通解为 y = π/6 + 2kπ 与 y = 5π/6 + 2kπ，k ∈ ℤ。
Back‑substitute 2x for y: 2x = π/6 + 2kπ ⇒ x = π/12 + kπ; 2x = 5π/6 + 2kπ ⇒ x = 5π/12 + kπ.
将 2x 代回：2x = π/6 + 2kπ ⇒ x = π/12 + kπ；2x = 5π/6 + 2kπ ⇒ x = 5π/12 + kπ。
]: for k = 0, x = π/12, 5π/12; for k = 1, x = 13π/12, 17π/12. k = 2 gives x = 25π/12 > 2π.
筛选在 [0, 2π] 内的 x：k = 0 时 x = π/12, 5π/12；k = 1 时 x = 13π/12, 17π/12。k = 2 时 x = 25π/12 > 2π。
x = π/12, 5π/12, 13π/12, 17π/12
x = π/12, 5π/12, 13π/12, 17π/12
Understanding the effect of the coefficient inside the argument is crucial for trigonometric equations in these courses.
理解函数内部系数的影响对于解决这些课程中的三角方程至关重要。
7. Arithmetic Progression | 等差数列
Question: In an arithmetic progression, the 5th term is 20 and the 12th term is 41. Find the first term a, the common difference d, and the sum of the first n terms Sₙ.
题目：等差数列的第5项为20，第12项为41。求首项 a、公差 d 以及前 n 项和 Sₙ。
Use uₙ = a + (n − 1)d. The 5th term: a + 4d = 20. The 12th term: a + 11d = 41.
利用通项公式 uₙ = a + (n − 1)d。第5项：a + 4d = 20；第12项：a + 11d = 41。
Subtract the first equation from the second: (a + 11d) − (a + 4d) = 41 − 20 ⇒ 7d = 21 ⇒ d = 3.
将第二式减去第一式：(a + 11d) − (a + 4d) = 41 − 20 ⇒ 7d = 21 ⇒ d = 3。
Substitute d = 3 into a + 4d = 20: a + 12 = 20 ⇒ a = 8.
将 d = 3 代入 a + 4d = 20：a + 12 = 20 ⇒ a = 8。
m of the first n terms is Sₙ = n/2 [2a + (n − 1)d] = n/2 [16 + (n − 1)·3] = n/2 (3n + 13).
前 n 项和 Sₙ = n/2 [2a + (n − 1)d] = n/2 [16 + (n − 1)·3] = n/2 (3n + 13)。
Sₙ = (3n² + 13n)/2
Sₙ = (3n² + 13n)/2
Sequence problems frequently test the ability to set up and solve simultaneous equations.
数列问题常考察建立并求解联立方程组的能力。
8. Complex Numbers – De Moivre’s Theorem | 复数 – 棣莫弗定理
Question: Express (1 + i)⁶ in Cartesian form a + bi.
题目：将 (1 + i)⁶ 表示为笛卡尔形式 a + bi。
First write 1 + i in polar form. Modulus r = √1² + 1² = √2. Argument θ = arctan(1/1) = π/4.
先将 1 + i 写成极坐标形式。模长 r = √1² + 1² = √2，辐角 θ = arctan(1/1) = π/4。
So 1 + i = √2 (cos(π/4) + i sin(π/4)).
因此 1 + i = √2 (cos(π/4) + i sin(π/4))。
By De Moivre: (1 + i)⁶ = (√2)⁶ [cos(6·π/4) + i sin(6·π/4)] = 8 [cos(3π/2) + i sin(3π/2)].
根据棣莫弗定理：(1 + i)⁶ = (√2)⁶ [cos(6·π/4) + i sin(6·π/4)] = 8 [cos(3π/2) + i sin(3π/2)]。
Evaluate: cos(3π/2) = 0, sin(3π/2) = −1. Therefore (1 + i)⁶ = 8 (0 + i(−1)) = −8i.
求值：cos(3π/2) = 0，sin(3π/2) = −1。因此 (1 + i)⁶ = 8 (0 + i(−1)) = −8i。
(1 + i)⁶ = −8i
(1 + i)⁶ =
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