📚 IGCSE CAIE Additional Mathematics: Case Study Practical Workout | IGCSE CAIE 进阶数学：案例分析实战演练
This workout takes you through ten practical case studies that mirror the style of CAIE IGCSE Additional Mathematics (0606) exam questions. Each case links several syllabus areas, such as functions, quadratics, logarithms, trigonometry and calculus, so you can practise exactly how examiners expect you to reason, calculate and communicate.
本实战演练通过十个案例，模拟 CAIE IGCSE 进阶数学（0606）考试题型，把函数、二次式、对数、三角和微积分等核心考点串联起来，帮助你训练考纲要求的推理、计算与表达方式。
1. Functions and Inverse Functions | 函数与反函数
Case: A function is defined by f(x) = (2x + 3)/(x – 1), x ≠ 1. Find the inverse function f⁻¹(x) and state its domain.
案例：函数定义为 f(x) = (2x + 3)/(x – 1)，x ≠ 1。求反函数 f⁻¹(x) 并写出它的定义域。
Write y = (2x + 3)/(x – 1). Multiply both sides by (x – 1), then rearrange to make x the subject: y(x – 1) = 2x + 3, so yx – y = 2x + 3, hence x(y – 2) = y + 3.
设 y = (2x + 3)/(x – 1)。两边乘 (x – 1) 后整理出 x：y(x – 1) = 2x + 3，即 yx – y = 2x + 3，因此 x(y – 2) = y + 3。
Swapping x and y gives the inverse function. The denominator cannot be zero, so the domain is x ≠ 2.
交换 x 和 y 得到反函数。分母不能为零，所以定义域为 x ≠ 2。
f⁻¹(x) = (x + 3)/(x – 2), x ≠ 2
Always check that the domain of f⁻¹ matches the range of the original function. Here f(x) can never output 2, so the inverse domain x ≠ 2 is correct.
一定要检查 f⁻¹ 的定义域是否与原函数的值域一致。这里 f(x) 不可能等于 2，所以反函数定义域为 x ≠ 2 是正确的。
2. Quadratic Equations and the Discriminant | 二次方程与判别式
Case: Find the value of k for which the line y = 2x + k is a tangent to the curve y = x² + 3x – 1.
案例：求 k 的值，使得直线 y = 2x + k 与曲线 y = x² + 3x – 1 相切。
Equate the two expressions: x² + 3x – 1 = 2x + k. Rearranging gives x² + x – (1 + k) = 0. For a tangent, the quadratic must have exactly one real root, so its discriminant is zero.
令两个表达式相等：x² + 3x – 1 = 2x + k。整理得 x² + x – (1 + k) = 0。因为直线与曲线相切，二次方程必须有两个相等的实根，所以判别式为零。
Using Δ = b² – 4ac with a = 1, b = 1 and c = -(1 + k):
使用 Δ = b² – 4ac，其中 a = 1，b = 1，c = -(1 + k)：
Δ = 1² – 4(1)(-(1 + k)) = 1 + 4 + 4k = 0
Therefore 4k + 5 = 0, giving k = -5/4. At this value the line just touches the parabola at one point.
因此 4k + 5 = 0，得到 k = -5/4。在这个值下，直线恰好与抛物线在一个点相切。
3. Polynomials and the Factor Theorem | 多项式与因式定理
Case: f(x) = 2x³ + ax² + bx – 6 has (x – 1) as a factor. When f(x) is divided by (x + 2), the remainder is -30. Find a and b, then factorise f(x) fully.
案例：f(x) = 2x³ + ax² + bx – 6 含有因式 (x – 1)。当 f(x) 除以 (x + 2) 时，余数为 -30。求 a 和 b，然后将 f(x) 完全因式分解。
Since (x – 1) is a factor, f(1) = 0. Substituting gives 2 + a + b – 6 = 0, so a + b = 4. Since the remainder on division by (x + 2) is -30, f(-2) = -30.
因为 (x – 1) 是因式，所以 f(1) = 0。代入得 2 + a + b – 6 = 0，即 a + b = 4。又因为 f(x) 除以 (x + 2) 的余数为 -30，所以 f(-2) = -30。
Substituting x = -2 gives -16 + 4a – 2b – 6 = -30, which simplifies to 2a – b = -4. Solving a + b = 4 and 2a – b = -4 gives a = 0 and b = 4.
代入 x = -2 得 -16 + 4a – 2b – 6 = -30，化简为 2a – b = -4。解 a + b = 4 与 2a – b = -4，得 a = 0，b = 4。
Thus f(x) = 2x³ + 4x – 6 = 2(x³ + 2x – 3). Since (x – 1) is a factor, division gives x³ + 2x – 3 = (x – 1)(x² + x + 3), so f(x) = 2(x – 1)(x² + x + 3). The quadratic has no real roots.
因此 f(x) = 2x³ + 4x – 6 = 2(x³ + 2x – 3)。由于 (x – 1) 是因式，作除法得 x³ + 2x – 3 = (x – 1)(x² + x + 3)，所以 f(x) = 2(x – 1)(x² + x + 3)。这个二次式没有实数根。
4. Logarithmic Equations | 对数方程
Case: Solve the equation log₂(x + 1) – log₂(x – 2) = 3.
案例：解方程 log₂(x + 1) – log₂(x – 2) = 3。
Use the subtraction law of logarithms to combine the left-hand side: log₂((x + 1)/(x – 2)) = 3. Then rewrite in exponential form.
利用对数的减法法则把左边合并：log₂((x + 1)/(x – 2)) = 3。然后改写成指数形式。
(x + 1)/(x – 2) = 2³ = 8
Multiply both sides by (x – 2): x + 1 = 8x – 16. Rearranging gives 7x = 17, so x = 17/7. Check the original domain: x + 1 > 0 and x – 2 > 0, so x > 2. The solution x = 17/7 satisfies this.
两边乘 (x – 2)：x + 1 = 8x – 16。整理得 7x = 17，所以 x = 17/7。检验原方程的定义域：x + 1 > 0 且 x – 2 > 0，即 x > 2。解 x = 17/7 满足条件。
5. Trigonometric Identities and Equations | 三角恒等式与三角方程
Case: Solve 2 sin² θ – cos θ – 1 = 0 for 0° ≤ θ ≤ 360°.
案例：在 0° ≤ θ ≤ 360° 范围内解方程 2 sin² θ – cos θ – 1 = 0。
Use the identity sin² θ = 1 – cos² θ to rewrite the equation entirely in terms of cos θ: 2(1 – cos² θ) – cos θ – 1 = 0.
利用恒等式 sin² θ = 1 – cos² θ，把方程全部写成 cos θ 的形式：2(1 – cos² θ) – cos θ – 1 = 0。
2 – 2 cos² θ – cos θ – 1 = 0
2 cos² θ + cos θ – 1 = 0
Factorise: (2 cos θ – 1)(cos θ + 1) = 0. Therefore cos θ = 1/2 or cos θ = -1.
因式分解：(2 cos θ – 1)(cos θ + 1) = 0。因此 cos θ = 1/2 或 cos θ = -1。
For cos θ = 1/2 in the given interval, θ = 60° or 300°. For cos θ = -1, θ = 180°. The full solution set is θ = 60°, 180°, 300°.
在给定区间内，cos θ = 1/2 对应 θ = 60° 或 300°；cos θ = -1 对应 θ = 180°。完整解集为 θ = 60°，180°，300°。
6. Differentiation and Tangent/Normal | 导数与切线/法线
Case: For the curve y = x³ – 3x² + 2, find the equations of the tangent and the normal at x = 1.
案例：对于曲线 y = x³ – 3x² + 2，求在 x = 1 处的切线方程和法线方程。
First find the y-coordinate: y = 1 – 3 + 2 = 0, so the point is (1, 0). Differentiate to get dy/dx = 3x² – 6x. At x = 1, the gradient is 3 – 6 = -3.
先求 y 坐标：y = 1 – 3 + 2 = 0，所以点是 (1, 0)。求导得 dy/dx = 3x² – 6x。在 x = 1 处，斜率为 3 – 6 = -3。
The tangent has gradient -3 through (1, 0), so its equation is y = -3(x – 1), or y = -3x + 3. The normal is perpendicular, so its gradient is the negative reciprocal: 1/3.
切线经过 (1, 0) 且斜率为 -3，所以切线方程为 y = -3(x – 1)，即 y = -3x + 3。法线与切线垂直，因此法线斜率为负倒数：1/3。
The normal equation is y = (1/3)(x – 1), or 3y = x – 1.
法线方程为 y = (1/3)(x – 1)，即 3y = x – 1。
7. Integration and Area Under a Curve | 积分与曲线下方面积
Case: Find the area enclosed by the curve y = 4x – x² and the x-axis between x = 0 and x = 4.
案例：求曲线 y = 4x – x² 与 x 轴在 x = 0 到 x = 4 之间围成的面积。
The curve meets the x-axis when 4x – x² = 0, that is x(x – 4) = 0, giving x = 0 and x = 4. The required area is the definite integral of y from 0 to 4.
曲线与 x 轴的交点满足 4x – x² = 0，即 x(x – 4) = 0，得 x = 0 和 x = 4。所求面积是 y 从 0 到 4 的定积分。
Area = ∫₀⁴ (4x – x²) dx
Integrate: ∫(4x – x²) dx = 2x² – x³/3. Substitute the limits: 2(16) – 64/3 = 32 – 64/3 = 32/3 square units.
积分得 ∫(4x – x²) dx = 2x² – x³/3。代入上下限：2(16) – 64/3 = 32 – 64/3 = 32/3 平方单位。
8. Kinematics with Calculus | 运动学中的微积分
Case: A particle moves in a straight line so that its displacement s metres from a fixed point at time t seconds is s = t³ – 6t² + 9t. Find the velocity and acceleration, and determine when the particle is at rest.
案例：一个质点沿直线运动，它在 t 秒时相对于固定点的位移 s 米为 s = t³ – 6t² + 9t。求速度和加速度，并判断质点何时静止。
Velocity is the first derivative of displacement: v = ds/dt = 3t² – 12t + 9. Acceleration is the derivative of velocity: a = dv/dt = 6t – 12.
速度是位移的一阶导数：v = ds/dt = 3t² – 12t + 9。加速度是速度的导数：a = dv/dt = 6t – 12。
The particle is at rest when v = 0, so 3t² – 12t + 9 = 0. Dividing by 3 gives t² – 4t + 3 = 0, hence (t – 1)(t – 3) = 0. The particle is at rest at t = 1 s and t = 3 s.
质点静止时 v = 0，所以 3t² – 12t + 9 = 0。两边除以 3 得 t² – 4t + 3 = 0，即 (t – 1)(t – 3) = 0。质点在 t = 1 秒和 t = 3 秒时静止。
9. Simultaneous Equations with a Line and a Curve | 直线与曲线的联立方程
Case: Solve the simultaneous equations y = x² – 2x – 3 and y = 2x – 3. Interpret the solutions geometrically.
案例：解联立方程 y = x² – 2x – 3 与 y = 2x – 3，并从几何上解释这些解。
Equate the two right-hand sides: x² – 2x – 3 = 2x – 3. Simplify to x² – 4x = 0, so x(x – 4) = 0. Thus x = 0 or x = 4.
令两个等式的右边相等：x² – 2x – 3 = 2x – 3。化简得 x² – 4x = 0，即 x(x – 4) = 0。因此 x = 0 或 x = 4。
Substitute into y = 2x – 3: when x = 0, y = -3; when x = 4, y = 5. The solutions are (0, -3) and (
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