Case Study Practice for Year 11 Cambridge Additional Mathematics | 剑桥Year 11进阶数学案例分析实战演练

📚 Case Study Practice for Year 11 Cambridge Additional Mathematics | 剑桥Year 11进阶数学案例分析实战演练

In Cambridge IGCSE Additional Mathematics, case study problems are designed to test your ability to link concepts across the syllabus. This article presents three worked case studies: composite functions with domain analysis, kinematics using differentiation, and matrix transformations. By following the step-by-step walkthroughs, you will develop a systematic approach and boost your confidence for exam-style questions.

在剑桥 IGCSE 进阶数学中，案例分析题目旨在考察你综合运用考纲概念的能力。本文展示三个精讲案例：带定义域分析的复合函数、运用微分的运动学、以及矩阵变换。通过逐步讲解，你将掌握系统化解题方法，增强应对类似考题的信心。

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1. What Makes a Good Case Study Approach? | 良好案例分析策略的要素

A strong case study response begins with careful reading, identifying the mathematical topics involved, and planning the order of operations. Always annotate given functions, unknowns and constraints before diving into calculations.

优秀的案例分析回答从仔细审题开始，确定涉及的数学主题，并规划运算顺序。在开始计算之前，务必标注给定函数、未知量和限制条件。

Check domain restrictions, variable definitions and possible simplifications early. This prevents lost marks due to invalid results or overlooked conditions.

尽早检查定义域限制、变量定义和可能的化简。这能避免因无效结果或忽略条件而失分。

Sketch a quick diagram if helpful, and break complex expressions into manageable steps. Use sign tables and graphs to interpret results clearly.

如有可能，快速绘制简图，并将复杂表达式分解为易于管理的步骤。使用符号表和图形清晰地解读结果。

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2. Case 1: Composite Functions – Core Thinking | 案例一：复合函数 – 核心思维

Let f(x) = √x − 1 − 2). The task is to form fg(x) and gf(x) and state the domain of each composite function.

设 f(x) = √x − 1，g(x) = 1/(x − 2)。任务为构造 fg(x) 和 gf(x) 并给出每个复合函数的定义域。

To form fg(x), substitute g(x) into f(x) for every x. Bear in mind that the output of the inner function must be a legal input for the outer function, and any original domain restrictions still apply.

要构造 fg(x)，将 g(x) 代入 f(x) 中的每一个 x。切记内部函数的输出必须是外部函数的合法输入，且原有的定义域限制依然适用。

: for f, x ≥ 1; for g, x ≠ 2. These will be the starting points for the composite domains.

首先写出每个函数各自的定义域：对于 f，x ≥ 1；对于 g，x ≠ 2。这些是复合定义域的出发点。

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3. Case 1: Step-by-Step Solution | 案例一：逐步解答

Step 1 – Find fg(x): f(g(x)) = f(1/(x − 2)) = √(1/(x − 2) − 1).

步骤1 – 求 fg(x)：f(g(x)) = f(1/(x − 2)) = √(1/(x − 2) − 1)。

Step 2 – The expression inside the square root must be ≥ 0: 1/(x − 2) − 1 ≥ 0 ⇒ 1/(x − 2) ≥ 1.

步骤2 – 平方根内的表达式必须 ≥ 0：1/(x − 2) − 1 ≥ 0 ⇒ 1/(x − 2) ≥ 1。

Step 3 – (x − 2))/(x − 2) ≥ 0 → (3 − x)/(x − 2) ≥ 0. A sign diagram reveals the solution 2 < x ≤ 3.

步骤3 – 解不等式。改写为 (1 − (x − 2))/(x − 2) ≥ 0 → (3 − x)/(x − 2) ≥ 0。符号表显示解为 2 < x ≤ 3。

Step 4 – Combine with g's domain x ≠ 2; the domain of fg is {x: 2 < x ≤ 3}.

步骤4 – 结合 g 的定义域 x ≠ 2；fg 的定义域为 {x: 2 < x ≤ 3}。

Now gf(x): g(f(x)) = 1/(√x − 1 − 2). The domain requires x ≥ 1 (from f) and √x − 1 − 2 ≠ 0 ⇒ √x − 1 ≠ 2 ⇒ x − 1 ≠ 4 ⇒ x ≠ 5. Thus domain of gf is {x: x ≥ 1, x ≠ 5}.

现在求 gf(x)：g(f(x)) = 1/(√x − 1 − 2)。定义域要求 x ≥ 1（来自 f）且 √x − 1 − 2 ≠ 0 ⇒ √x − 1 ≠ 2 ⇒ x − 1 ≠ 4 ⇒ x ≠ 5。因此 gf 的定义域为 {x: x ≥ 1, x ≠ 5}。

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4. Case 1: Common Pitfall – Domain and Range | 案例一：常见陷阱 – 定义域与值域

Many students forget that for gf(x), the output of f(x) must not make the denominator of g zero. They only consider the domain of f.

许多学生忘记对于 gf(x)，f(x) 的输出不能使 g 的分母为零。他们只考虑 f 的定义域。

2) without checking its sign can reverse the inequality. Always bring terms to one side and use common denominators.

解类似 1/(x − 2) ≥ 1 的不等式时，不经符号检查就两边同乘 (x − 2) 可能使不等号方向反转。应始终将各项移到一边并使用公分母。

pression to confirm they are valid. For fg(x) at x=2 or x=3, check that the output exists.

务必在原表达式中检验边界点以确认其有效性。对于 fg(x)，在 x=2 或 x=3 处应检查输出是否存在。

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5. Case 2: Kinematics and Differentiation – Core Thinking | 案例二：运动学与微分 – 核心思维

= 2t³ − 15t² + 36t, t ≥ 0. We need velocity, acceleration and a description of the motion.

一质点沿直线运动，位移 s = 2t³ − 15t² + 36t，t ≥ 0。求速度、加速度并描述运动。

Velocity v is the first derivative of displacement with respect to time; acceleration a is the second derivative. Their signs reveal direction and whether the particle is speeding up or slowing down.

速度 v 是位移对时间的一阶导数；加速度 a 是二阶导数。它们的符号表示运动方向以及质点是在加速还是减速。

ges direction. Examine intervals using sign tables to build a complete picture of the motion.

令 v = 0 求出质点改变方向的时刻。运用符号表分析区间，构建完整的运动图像。

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6. Case 2: Step-by-Step Solution | 案例二：逐步解答

Differentiate to find velocity and acceleration:

求导得速度和加速度：

v = ds/dt = 6t² − 30t + 36

a = dv/dt = 12t − 30

t² − 30t + 36 = 0 → divide by 6: t² − 5t + 6 = 0 → (t − 2)(t − 3) = 0, so t = 2 and t = 3.

令 v = 0：6t² − 30t + 36 = 0 → 除以 6：t² − 5t + 6 = 0 → (t − 2)(t − 3) = 0，解得 t = 2 和 t = 3。

moving forward); for 2 < t < 3, v < 0 (moving backward); for t > 3, v > 0 (forward again).

检验区间：0 ≤ t < 2 时，v > 0（正向运动）；2 < t < 3 时，v < 0（反向运动）；t > 3 时，v > 0（再次正向）。

12t − 30. At t = 2, a = −6 (negative); at t = 3, a = 6 (positive). This information helps

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