Case Study Practical Workout for IGCSE CAIE Additional Mathematics | IGCSE CAIE 进阶数学：案例分析实战演练

📚 Case Study Practical Workout for IGCSE CAIE Additional Mathematics | IGCSE CAIE 进阶数学：案例分析实战演练

Practical case studies bridge the gap between abstract theory and real‑world problem solving. In IGCSE CAIE Additional Mathematics (0606), applying concepts such as calculus, exponentials, trigonometry, and coordinate geometry to contextual scenarios is essential for mastering the syllabus and acing the exam. This article presents a series of hands‑on case studies with step‑by‑step solutions, designed to reinforce your understanding and exam technique.

实际案例分析能够弥合抽象理论与现实问题解决之间的鸿沟。在 IGCSE CAIE 进阶数学（0606）中，将微积分、指数、三角学和坐标几何等概念应用于情境问题是掌握教学大纲并在考试中取得佳绩的关键。本文展示了一系列实践案例及其分步解答，旨在巩固你的理解并提升应试技巧。

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1. Optimising an Enclosure with a Wall | 借助围墙优化围栏面积

A farmer has 40 m of fencing and wishes to make a rectangular enclosure against a long straight wall. Only three sides need fencing. Find the dimensions that maximise the area, and calculate the maximum area.

一位农民有 40 m 的围栏，想借助一堵长长的直墙围一个矩形场地。只有三边需要围栏。求使面积最大的尺寸，并计算最大面积。

l fencing gives 2x + y = 40, so y = 40 − 2x. The area is A = x y = x(40 − 2x) = 40x − 2x².

设平行于墙的边长为 x 米，另一边长为 y 米。围栏总长给出 2x + y = 40，因此 y = 40 − 2x。面积为 A = x y = x(40 − 2x) = 40x − 2x²。

0. Then y = 40 − 2(10) = 20. The second derivative d²A/dx² = −4 < 0, confirming a maximum.

使用微积分求导：dA/dx = 40 − 4x。令 dA/dx = 0 得 40 − 4x = 0 → x = 10。于是 y = 40 − 2(10) = 20。二阶导数 d²A/dx² = −4 < 0，确认是极大值。

rallel to the wall and 20 m perpendicular to it. The maximum area is A = 10 × 20 = 200 m².

因此最优尺寸为平行于墙 10 m，垂直方向 20 m。最大面积为 A = 10 × 20 = 200 m²。

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2. Modelling Radioactive Decay | 放射性衰变建模

A radioactive substance decays according to m = m₀ e^-kt, where m₀ is the initial mass and k is a positive constant. After 5 years, the mass is 60% of the original. Find the value of k and the time when only 10% of the original mass remains.

一种放射性物质按 m = m₀ e^-kt 衰变，其中 m₀ 为初始质量，k 为正常数。5 年后，质量变为原来的 60%。求 k 的值，及质量仅剩原来的 10% 时所需的时间。

When t = 5, m = 0.6 m₀. Substitute into the model: 0.6 m₀ = m₀ e^-5k. Cancel m₀ (non‑zero): 0.6 = e^-5k 0.6 = -5k → k = - (ln 0.6) / 5 ≈ -(−0.5108) / 5 ≈ 0.10216 year⁻¹ (positive, as required).

当 t = 5 时，m = 0.6 m₀。代入模型：0.6 m₀ = m₀ e^-5k。约去非零的 m₀：0.6 = e^-5k。取自然对数：ln 0.6 = -5k → k = - (ln 0.6) / 5 ≈ -(−0.5108) / 5 ≈ 0.10216 年⁻¹（为正值，符合要求）。

For 10% remaining, set 0.1 m₀ = m₀ e^-kt → 0.1 = e^-kt -kt → t = - (ln 0.1) / k. Using k ≈ 0.10216 gives t = -(−2.3026) / 0.10216 ≈ 22.54 years.

剩余 10% 时，令 0.1 m₀ = m₀ e^-kt → 0.1 = e^-kt。取对数：ln 0.1 = -kt → t = - (ln 0.1) / k。代入 k ≈ 0.10216 得 t = -(−2.3026) / 0.10216 ≈ 22.54 年。

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3. Kinematics of a Particle | 质点运动学

pressions for velocity and acceleration, the times when the particle is at rest, and the distance travelled in the first 3 seconds.

一个质点沿直线运动。t 秒后相对于某固定点的位移 s（米）为 s(t) = t³ − 6t² + 9t + 2，定义域 0 ≤ t ≤ 5。求速度和加速度的表达式、质点处于静止的时刻，以及前 3 秒走过的距离。

Velocity v = ds/dt = 3t² − 12t + 9. Acceleration a = dv/dt = 6t − 12.

速度 v = ds/dt = 3t² − 12t + 9。加速度 a = dv/dt = 6t − 12。

: t² − 4t + 3 = 0 → (t − 1)(t − 3) = 0 → t = 1 s, t = 3 s. Both are in the given interval.

静止时 v = 0：3t² − 12t + 9 = 0 → 除以 3 得 t² − 4t + 3 = 0 → (t − 1)(t − 3) = 0 → t = 1 s, t = 3 s。两者均在给定区间内。

= 3. Calculate s(0) = 2, s(1) = 1 − 6 + 9 + 2 = 6, s(3) = 27 − 54 + 27 + 2 = 2. Distance = |s(1) − s(0)| + |s(3) − s(1)| = |6 − 2| + |2 − 6| = 4 + 4 = 8 m.

为求前 3 秒走过的距离，注意质点在 t = 1 和 t = 3 改变运动方向。计算 s(0) = 2，s(1) = 1 − 6 + 9 + 2 = 6，s(3) = 27 − 54 + 27 + 2 = 2。距离 = |s(1) − s(0)| + |s(3) − s(1)| = |6 − 2| + |2 − 6| = 4 + 4 = 8 m。

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4. Intersection of a Line and a Circle for a Bridge Alignment | 桥梁线形与圆的交点

about of radius 5 units is centred at (2, 3). Find the two points where the road meets the roundabout and the chord length between them.

一条桥梁引道沿直线 y = 2x + 1 布置。一个半径为 5 单位的圆形转盘中心在 (2, 3)。求道路与转盘的两个交点，以及它们之间的弦长。

)² = 25. Substitute y = 2x + 1: (x − 2)² + (2x + 1 − 3)² = 25 → (x − 2)² + (2x − 2)² = 25.

圆方程：(x − 2)² + (y − 3)² = 25。代入 y = 2x + 1：(x − 2)² + (2x + 1 − 3)² = 25 → (x − 2)² + (2x − 2)² = 25。

Expand: (x² − 4x + 4) + (4x² − 8x + 4) = 25 → 5x² − 12x + 8 = 25 → 5x² − 12x − 17 = 0.

展开：(x² − 4x + 4) + (4x² − 8x + 4) = 25 → 5x² − 12x + 8 = 25 → 5x² − 12x − 17 = 0。

Solve using the quadratic formula: x = [12 ± √144 + 340] / 10 = [12 ± √484] / 10 = [12 ± 22] / 10. So x = 34/10 = 3.4, or x = −10/10 = −1.

用求根公式：x = [12 ± √144 + 340] / 10 = [12 ± √484] / 10 = [12 ± 22] / 10。得 x = 34/10 = 3.4，或 x = −10/10 = −1。

3.4) + 1 = 7.8; if x = −1, y = 2(−1) + 1 = −1. The two points are (3.4, 7.8) and (−1, −1).

相应 y 值：若 x = 3.4，y = 2(3.4) + 1 = 7.8；若 x = −1，y = 2(−1) + 1 = −1。两个交点为 (3.4, 7.8) 和 (−1, −1)。

Chord length = √[(3.4 − (−1))² + (7.8 − (−1))²] = √[(4.4)² + (8.8)²] = √19.36 + 77.44 = √96.8 ≈ 9.84 units.

弦长 = √[(3.4 − (−1))² + (7.8 − (−1))²] = √[(4.4)² + (8.8)²] = √19.36 + 77.44 = √96.8 ≈ 9.84 单位。

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5. Height of a Ferris Wheel | 摩天轮的高度

time t minutes can be modelled by h(t) = 25 − 20 cos(2πt), assuming the capsule starts at the lowest point. Find the height at t = 0.5 min and determine the first time (t > 0) when the capsule reaches a height of 30 m.

一个半径为 20 m 的摩天轮，中心离地 25 m。它每分钟转一圈。乘客吊舱在 t 分钟时的高度 h（米）可建模为 h(t) = 25 − 20 cos(2πt)，假设吊舱从最低点出发。求 t = 0.5 分钟时的高度，并确定吊舱首次达到 30 m 高度的时间（t > 0）。

At t = 0.5: cos(2π × 0.5) = cos π = −1, so h = 25 − 20(−1) = 45 m. (The highest point.)

t = 0.5 时：cos(2π × 0.5) = cos π = −1，因此 h = 25 − 20(−1) = 45 m（最高点）。

Set h = 30: 30 = 25 − 20 cos(2πt) → 5 = −20 cos(2πt) → cos(2πt) = −0.25.

令 h = 30：30 = 25 − 20 cos(2πt) → 5 = −20 cos(2πt) → cos(2πt) = −0.25。

on is 2π − 1.8235 = 4.4597 rad. The smallest positive t comes from 2πt = 1.8235 → t = 1.8235 / (2π) ≈ 0.290 min.

解 2πt = cos⁻¹(−0.25)。主值为 arccos(−0.25) ≈ 1.8235 弧度。余弦在第三象限也为负，另一解为 2π − 1.8235 = 4.4597 弧度。最小的正 t 来自 2πt = 1.8235 → t = 1.8235 / (2π) ≈ 0.290 分钟。

Thus the capsule first reaches 30 m at about 0.290 min (≈ 17.4 seconds).

因此吊舱在大约 0.290 分钟（≈ 17.4 秒）时首次达到 30 m。

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6. Solving a Mixture Problem with Simultaneous Equations | 混合物问题的联立方程

Two alloys are available: Alloy A contains 60% gold and 40% silver; Alloy B contains 30% gold and 70% silver. A jeweller wants to produce 100 g of an alloy containing exactly 45% gold. How many grams of each alloy should be used?

有两种合金：合金 A 含 60% 金和 40% 银；合金 B 含 30% 金和 70% 银。一位珠宝匠想要制作 100 g 含金量为 45% 的合金。需要各用多少克？

d. The total mass gives x + y = 100. The gold content gives 0.6x + 0.3y = 0.45 × 100 = 45.

设用 x g 合金 A，y g 合金 B。总质量有 x + y = 100。金的含量有 0.6x + 0.3y = 0.45 × 100 = 45。

itute: 0.6x + 0.3(100 − x) = 45 → 0.6x + 30 − 0.3x = 45 → 0.3x = 15 → x = 50. Then y = 50.

联立求解。由第一式得 y = 100 − x。代入：0.6x + 0.3(100 − x) = 45 → 0.6x + 30 − 0.3x = 45 → 0.3x = 15 → x = 50。于是 y = 50。

So the jeweller should use 50 g of Alloy A and 50 g of Alloy B.

因此珠宝匠应使用 50 g 合金 A 和 50 g 合金 B。

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7. Polynomial Remainder Theorem in Volume Design | 体积设计中的多项式余式定理

− 6x² + 11x − 6, where x is a dimension in decimetres. A known linear factor is (x − 1). Fully factorise V(x) and determine all integer dimensions. If the box must have a volume of 6 dm³, find the possible values of x.

一个盒子的体积由 V(x) = x³ − 6x² +

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