📚 Typical Exam Question Walkthroughs for CIE A-Level Mathematics | CIE A-Level 数学典型例题详解
Mastering CIE A-Level Mathematics requires not only understanding concepts but also applying them accurately under exam conditions. In this article, we dissect ten representative questions from Pure Mathematics, Statistics, and Mechanics, providing step-by-step solutions and key insights. Each example highlights common exam pitfalls and efficient problem‑solving strategies to help you achieve top marks.
掌握 CIE A-Level 数学不仅需要理解概念，更要在考试条件下准确应用。本文深入剖析十道来自纯数学、统计和力学的典型题目，提供逐步解答与核心要点。每个例题都突出常见易错点和高效的解题策略，助你斩获高分。
1. Functions and Inverses | 函数与反函数
The function f is defined by f(x) = (2x + 3)/(x – 1) for x ∈ ℝ, x ≠ 1. Find f⁻¹(x) and state its domain.
函数 f 定义为 f(x) = (2x + 3)/(x – 1)，x ∈ ℝ 且 x ≠ 1。求 f⁻¹(x) 并写出其定义域。
Start by writing y = f(x). Let y = (2x + 3)/(x – 1). To find the inverse, swap x and y after making x the subject.
首先写出 y = f(x)。令 y = (2x + 3)/(x – 1)。为求反函数，先解出 x，再交换变量。
Multiply both sides by (x – 1): y(x – 1) = 2x + 3. Expand to yx – y = 2x + 3. Collect terms involving x: yx – 2x = y + 3. Factorise: x(y – 2) = y + 3. Therefore x = (y + 3)/(y – 2).
两边同乘 (x – 1)：y(x – 1) = 2x + 3。展开得 yx – y = 2x + 3。将含 x 的项集中：yx – 2x = y + 3。提取公因式：x(y – 2) = y + 3。得 x = (y + 3)/(y – 2)。
Replacing y with x gives f⁻¹(x) = (x + 3)/(x – 2). The domain of f⁻¹ is the range of f, but we can find it directly: the denominator cannot be zero, so x – 2 ≠ 0 ⇒ x ≠ 2. Thus the domain is {x ∈ ℝ : x ≠ 2}.
将 y 换回 x 得 f⁻¹(x) = (x + 3)/(x – 2)。f⁻¹ 的定义域是 f 的值域，也可直接看出：分母不能为零，即 x – 2 ≠ 0 ⇒ x ≠ 2。因此定义域为 {x ∈ ℝ : x ≠ 2}。
2. Quadratics and the Discriminant | 二次方程与判别式
Find the set of values of k for which the equation 2x² + 3x + k = 0 has no real roots.
求使得方程 2x² + 3x + k = 0 没有实数根的 k 的取值范围。
A quadratic ax² + bx + c = 0 has no real roots when its discriminant Δ = b² – 4ac is negative. Here a = 2, b = 3, c = k.
二次方程 ax² + bx + c = 0 无实根当且仅当判别式 Δ = b² – 4ac 小于 0。此处 a = 2, b = 3, c = k。
Compute Δ = 3² – 4 × 2 × k = 9 – 8k. For no real roots, we require 9 – 8k < 0. Solve: -8k < -9 ⇒ k > 9/8. (Remember to reverse the inequality when dividing by a negative number.)
计算 Δ = 3² – 4 × 2 × k = 9 – 8k。无实根要求 9 – 8k < 0。解不等式：-8k < -9 ⇒ k > 9/8。（注意除以负数要变号。）
Therefore the solution set is {k ∈ ℝ : k > 9/8}, or in interval notation (9/8, ∞).
因此解集为 {k ∈ ℝ : k > 9/8}，或写成区间 (9/8, ∞)。
3. Arithmetic and Geometric Sequences | 等差数列与等比数列
In a geometric progression, the second term is 6 and the sum to infinity is 27. Find the first term a and the common ratio r (given |r| < 1).
一个等比数列的第二项为 6，无穷项和为 27。求首项 a 和公比 r（已知 |r| < 1）。
Let the first term be a and common ratio r. The second term is ar = 6. The sum to infinity of a geometric series is S∞ = a / (1 – r) = 27, provided |r| < 1. We solve the simultaneous equations.
设首项为 a，公比为 r。第二项满足 ar = 6。等比数列的无穷和公式为 S∞ = a / (1 – r) = 27，要求 |r| < 1。联立方程组求解。
From ar = 6 we have a = 6/r. Substitute into a/(1 – r) = 27: (6/r) / (1 – r) = 27 → 6/r = 27(1 – r). Multiply both sides by r: 6 = 27r(1 – r). Expand: 6 = 27r – 27r². Rearrange to 27r² – 27r + 6 = 0. Divide by 3: 9r² – 9r + 2 = 0.
由 ar = 6 得 a = 6/r。代入 a/(1 – r) = 27：6/[r(1 – r)] = 27 → 6 = 27r(1 – r)。两边乘 r：6 = 27r – 27r²。整理得 27r² – 27r + 6 = 0。约去 3：9r² – 9r + 2 = 0。
Factorise: (3r – 1)(3r – 2) = 0. So r = 1/3 or r = 2/3. Both satisfy |r| < 1. For r = 1/3, a = 6/(1/3) = 18. For r = 2/3, a = 6/(2/3) = 9. Hence two possible sequences exist.
因式分解：(3r – 1)(3r – 2) = 0。得 r = 1/3 或 r = 2/3，均满足 |r| < 1。当 r = 1/3 时 a = 18；当 r = 2/3 时 a = 9。故存在两个可能的数列。
4. Differentiation and Stationary Points | 微分与驻点
Determine the coordinates and nature of the stationary points on the curve y = x⁴ – 4x³ + 4x² + 2.
求曲线 y = x⁴ – 4x³ + 4x² + 2 上驻点的坐标并判断其类型。
Differentiate: dy/dx = 4x³ – 12x² + 8x. Set dy/dx = 0: 4x³ – 12x² + 8x = 0. Factorise out 4x: 4x(x² – 3x + 2) = 0. So 4x = 0 or x² – 3x + 2 = 0. Solve x² – 3x + 2 = (x – 1)(x – 2) = 0, giving x = 1 and x = 2. Therefore stationary points occur at x = 0, 1, 2.
求导：dy/dx = 4x³ – 12x² + 8x。令 dy/dx = 0：4x³ – 12x² + 8x = 0。提取公因式 4x：4x(x² – 3x + 2) = 0。于是 4x = 0 或 x² – 3x + 2 = 0。解二次式得 (x-1)(x-2)=0，即 x = 1 和 x = 2。因此驻点出现在 x = 0, 1, 2。
Find y-coordinates: for x=0, y=2; x=1, y=1-4+4+2=3; x=2, y=16-32+16+2=2. Points: (0,2), (1,3), (2,2). To classify, use second derivative d²y/dx² = 12x² – 24x + 8.
求纵坐标：x=0 时 y=2；x=1 时 y=3；x=2 时 y=2。驻点为 (0,2), (1,3), (2,2)。用二阶导数 d²y/dx² = 12x² – 24x + 8 判断性质。
At x=0: d²y/dx² = 8 > 0 ⇒ local minimum. At x=1: d²y/dx² = 12 – 24 + 8 = -4 < 0 ⇒ local maximum. At x=2: d²y/dx² = 48 - 48 + 8 = 8 > 0 ⇒ local minimum.
x=0 处：二阶导数值 8>0 ⇒ 局部极小值。x=1 处：12-24+8=-4<0 ⇒ 局部极大值。x=2 处：48-48+8=8>0 ⇒ 局部极小值。
Thus (0,2) minimum, (1,3) maximum, (2,2) minimum. The curve has two “hollows” with the same height, separated by a peak.
因此 (0,2) 极小点，(1,3) 极大点，(2,2) 极小点。曲线有两个等高的“凹陷”，中间有一个峰值。
5. Integration and Area under Curves | 积分与曲线下方面积
Find the area of the region enclosed by the curve y = 4 – x² and the x-axis.
求曲线 y = 4 – x² 与 x 轴所围成区域的面积。
The curve meets the x-axis when y = 0: 4 – x² = 0 ⇒ x² = 4 ⇒ x = -2 or x = 2. The region is a symmetrical arch above the x‑axis between x = -2 and x = 2.
曲线与 x 轴相交于 y=0 时：4 – x² = 0 ⇒ x = ±2。该区域是 x = -2 到 x = 2 之间、x 轴上方的对称拱形。
Area = ∫₋₂² (4 – x²) dx. Integrate term by term: ∫4 dx = 4x, ∫x² dx = x³/3. So the antiderivative is [4x – x³/3] evaluated from -2 to 2.
面积 = ∫₋₂² (4 – x²) dx。逐项积分得 4x – x³/3，在 -2 到 2 上计算定积分。
Substitute upper limit: 4(2) – (2)³/3 = 8 – 8/3 = 16/3. Lower limit: 4(-2) – (-8)/3 = -8 + 8/3 = -16/3. Subtract: (16/3) – (-16/3) = 32/3 square units.
代入上限：8 – 8/3 = 16/3。下限：-8 + 8/3 = -16/3。相减得 16/3 – (-16/3) = 32/3 平方单位。
Alternatively, by symmetry, double the area from 0 to 2: 2 × ∫₀² (4 – x²) dx = 2[4x – x³/3]₀² = 2(16/3) = 32/3.
利用对称性，面积等于 0 到 2 的两倍：2 × ∫₀² (4 – x²) dx = 2[4x – x³/3]₀² = 2×16/3 = 32/3。
6. Trigonometric Equations and Identities | 三角方程与恒等式
Solve the equation 2 cos²θ + 3 sin θ = 3 for 0° ≤ θ ≤ 360°.
解方程 2 cos²θ + 3 sin θ = 3，其中 0° ≤ θ ≤ 360°。
Use the identity cos²θ = 1 – sin²θ. Substitute: 2(1 – sin²θ) + 3 sin θ = 3 ⇒ 2 – 2 sin²θ + 3 sin θ = 3. Rearrange to -2 sin²θ + 3 sin θ – 1 = 0, or multiply by -1: 2 sin²θ – 3 sin θ + 1 = 0.
利用恒等式 cos²θ = 1 – sin²θ。代入得 2(1 – sin²θ) + 3 sin θ = 3 ⇒ 2 – 2 sin²θ + 3 sin θ = 3。移项得 -2 sin²θ + 3 sin θ – 1 = 0，乘以 -1：2 sin²θ – 3 sin θ + 1 = 0。
This is a quadratic in sin θ. Factorise: (2 sin θ – 1)(sin θ – 1) = 0. So sin θ = 1/2 or sin θ = 1.
这是关于 sin θ 的二次方程。因式分解：(2 sin θ – 1)(sin θ – 1) = 0。因此 sin θ = 1/2 或 sin θ = 1。
For sin θ = 1, θ = 90°. For sin θ = 1/2, the primary solutions in 0° to 360° are θ = 30° and θ = 150° (since sine is positive in 1st and 2nd quadrants). Combine all solutions: θ = 30°, 90°, 150°.
由 sin θ = 1 得 θ = 90°。由 sin θ = 1/2，在 0° 到 360° 内基本解为 θ = 30° 和 θ = 150°（正弦在第一、二象限为正）。合并得 θ = 30°、90°、150°。
7. Vector Equations of Lines and Dot Product | 直线的向量方程与点积
Two lines are given by L₁: r = (2, -1, 1) + λ(1, 2, -1) and L₂: r = (0, 3, 2) + μ(2, -1, 1). Find the acute angle between the lines and determine whether they intersect.
两直线方程为 L₁: r = (2, -1, 1) + λ(1, 2, -1) 和 L₂: r = (0, 3, 2) + μ(2, -1, 1)。求直线的锐角夹角并判断它们是否相交。
The direction vectors are d₁ = (1, 2, -1) and d₂ = (2, -1, 1). The angle θ between lines is given by cos θ = |d₁ · d₂| / (|d₁| |d₂|) for the acute angle.
方向向量为 d₁ = (1, 2, -1)，d₂ = (2, -1, 1)。锐角 θ 满足 cos θ = |d₁ · d₂| / (|d₁| |d₂|)。
Compute dot product: d₁ · d₂ = 1×2 + 2×(-1) + (-1)×1 = 2 – 2 – 1 = -1. Magnitudes: |d₁| = √(1²+2²+(-1)²) = √6; |d₂| = √(2²+(-1)²+1²) = √6.
计算点积：d₁ · d₂ = 1×2 + 2×(-1) + (-1)×1 = 2 – 2 – 1 = -1。模长：|d₁| = √6，|d₂| = √6。
So cos θ = |-1| / (√6 × √6) = 1/6. Hence θ = arccos(1/6) ≈ 80.4°.
于是 cos θ = |‑1| / (√6 × √6) = 1/6。因此 θ = arccos(1/6) ≈ 80.4°。
To check intersection, set (2+λ, -1+2λ, 1-λ) = (0+2μ, 3-μ, 2+μ). Equate components: 2+λ = 2μ …(1); -1+2λ = 3-μ …(2); 1-λ = 2+μ …(3). From (1): λ = 2μ – 2. Substitute into (2): -1+2(2μ-2) = 3-μ → -1+4μ-4 = 3-μ → 4μ-5 = 3-μ → 5μ = 8 → μ = 8/5, λ = 16/5 – 2 = 6/5. Check (3): LHS = 1 – 6/5 = -1/5; RHS = 2 + 8/5 = 18/5. Not equal, so lines do not intersect (they are skew).
检验是否相交：令坐标相等，由 (1) 得 λ = 2μ-2，代入 (2) 解得 μ=8/5, λ=6/5。代入 (3) 左边得 -1/5，右边 18/5，不相等。因此两直线为异面直线，永不相交。
8. Solving Differential Equations | 解微分方程
Solve the differential equation dy/dx = 3x² / (2y) given that y = 4 when x = 1. Express y in terms of x.
解微分方程 dy/dx = 3x² / (2y)，已知当 x=1 时 y=4。用 x 表示 y。
Separate the variables: 2y dy = 3x² dx. Integrate both sides: ∫ 2y dy = ∫ 3x² dx ⇒ y² = x³ + C, where C is the constant of integration.
分离变量得 2y dy = 3x² dx。两边积分：y² = x³ + C，其中 C 为积分常数。
Use the initial condition y(1) = 4: 4² = 1³ + C ⇒ 16 = 1 + C ⇒ C = 15. Therefore the particular solution is y² = x³ + 15. Since y = 4 > 0 at x=1, we take the positive square root: y = √(x³ + 15).
利用初始条件 y(1)=4：4² = 1³ + C ⇒ 16 = 1 + C ⇒ C = 15。特解为 y² = x³ + 15。由于 x=1 时 y>0，取正平方根：y = √(x³ + 15)。
The domain must satisfy x³ + 15 ≥ 0, i
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