📚 Case Study Practice for CIE Year 12 Mathematics | CIE 12年级数学案例分析实战演练
Case study practice is one of the most effective ways to prepare for CIE Year 12 Mathematics. By working through realistic, exam‑style problems that integrate pure mathematics, mechanics and statistics, you develop deeper understanding and problem‑solving agility. This article presents ten carefully selected case studies, each broken down into clear steps with bilingual explanations, helping you master key techniques and avoid common pitfalls.
案例分析实战是备战 CIE 12 年级数学最有效的方法之一。通过演练融合纯数、力学与统计知识的真实考题风格问题，你能加深理解并提升解题灵活性。本文精选十个案例，每个都拆解为清晰的步骤并配以中英双语讲解，帮助你掌握关键技巧、避开常见陷阱。
1. Motion of a Particle (Kinematics) | 质点运动（运动学）
A particle moves along a straight line. Its displacement s metres from a fixed point at time t seconds is s(t) = t³ − 6t² + 9t for t ≥ 0.
一质点沿直线运动，其相对于固定点的位移 s（米）与时间 t（秒）的关系为 s(t) = t³ − 6t² + 9t，t ≥ 0。
First find the velocity v = ds/dt = 3t² − 12t + 9 and acceleration a = dv/dt = 6t − 12.
先求速度 v = ds/dt = 3t² − 12t + 9，加速度 a = dv/dt = 6t − 12。
The particle is at rest when v = 0: 3t² − 12t + 9 = 0 ⇒ t² − 4t + 3 = 0 ⇒ (t−1)(t−3) = 0, so t = 1 s and t = 3 s.
质点静止时 v = 0：3t² − 12t + 9 = 0 ⇒ t² − 4t + 3 = 0 ⇒ (t−1)(t−3) = 0，得 t = 1 秒和 t = 3 秒。
Calculate displacements: s(0)=0, s(1)=1−6+9=4, s(3)=27−54+27=0, s(5)=125−150+45=20. Distance travelled in first 5 s = |4−0| + |0−4| + |20−0| = 4 + 4 + 20 = 28 m.
计算位移：s(0)=0，s(1)=1−6+9=4，s(3)=27−54+27=0，s(5)=125−150+45=20。前 5 秒内的路程 = |4−0| + |0−4| + |20−0| = 4 + 4 + 20 = 28 米。
The acceleration is zero when 6t − 12 = 0 ⇒ t = 2 s, which corresponds to a point of inflection in the displacement graph.
加速度为零时 6t − 12 = 0 ⇒ t = 2 秒，此时位移图出现拐点。
2. Maximising Profit with Differentiation | 利用微分最大化利润
A company finds that its profit P(x) in thousands of dollars is modelled by P(x) = −x² + 12x − 20, where x is the number of items produced in hundreds.
某公司发现其利润 P(x)（千美元）可用 P(x) = −x² + 12x − 20 建模，其中 x 为以百件计的产品数量。
To locate the maximum, differentiate: P'(x) = −2x + 12. Setting P'(x) = 0 gives −2x + 12 = 0 ⇒ x = 6 (hundred items).
为求最大值，求导：P'(x) = −2x + 12。令 P'(x) = 0 得 −2x + 12 = 0 ⇒ x = 6（百件）。
Second derivative P”(x) = −2 < 0, confirming a maximum. Maximum profit is P(6) = −36 + 72 − 20 = 16 thousand dollars, or $16,000.
二阶导 P”(x) = −2 < 0，确认为最大值。最大利润为 P(6) = −36 + 72 − 20 = 16 千美元，即 $16,000。
The feasible domain for x is non‑negative, but the model only makes sense while profit remains non‑negative. Solving P(x) = 0 gives x = 2 and x = 10; outside this range the company makes a loss.
x 的可行域为非负数，但只有当利润非负时模型才有意义。解 P(x) = 0 得 x = 2 和 x = 10；在此范围外公司出现亏损。
3. Integration: Area Between Curves | 积分：曲线间的面积
Find the area enclosed by the parabola y = x² and the line y = 2x + 3.
求抛物线 y = x² 与直线 y = 2x + 3 所围区域的面积。
Intersections: x² = 2x + 3 ⇒ x² − 2x − 3 = 0 ⇒ (x−3)(x+1)=0, giving x = −1 and x = 3.
交点：x² = 2x + 3 ⇒ x² − 2x − 3 = 0 ⇒ (x−3)(x+1)=0，得 x = −1 和 x = 3。
Between these points the line lies above the parabola. Area = ∫−13 [(2x+3) − x²] dx = ∫−13 (−x² + 2x + 3) dx.
在此区间内直线位于抛物线上方。面积 = ∫−13 [(2x+3) − x²] dx = ∫−13 (−x² + 2x + 3) dx。
Integrate: [−x³/3 + x² + 3x]−13 = (−27/3 + 9 + 9) − (−(−1)/3 + 1 + (−3)) = (−9+18) − (1/3 − 2) = 9 − (−5/3) = 32/3 square units.
积分：[−x³/3 + x² + 3x]−13 = (−27/3 + 9 + 9) − (−(−1)/3 + 1 + (−3)) = (−9+18) − (1/3 − 2) = 9 − (−5/3) = 32/3 平方单位。
4. Probability Distribution: Binomial | 概率分布：二项分布
A fair six‑sided die is rolled 6 times. Let X be the number of times a ‘6’ appears. Find P(X = 2).
一个均匀六面骰子掷 6 次。设 X 为出现 ‘6’ 的次数。求 P(X = 2)。
This is a binomial experiment: n = 6, p = 1/6. Then P(X = 2) = 6C2 (1/6)² (5/6)⁴.
这是一个二项试验：n = 6，p = 1/6。于是 P(X = 2) = 6C2 (1/6)² (5/6)⁴。
6C2 = 15, (1/6)² = 1/36, (5/6)⁴ = 625/1296. Multiplying gives 15 × (1/36) × (625/1296) = (15 × 625) / (36 × 1296).
6C2 = 15，(1/6)² = 1/36，(5/6)⁴ = 625/1296。相乘得 15 × (1/36) × (625/1296) = (15 × 625) / (36 × 1296)。
15 × 625 = 9375; 36 × 1296 = 46656. Simplify dividing by 3: 9375/46656 = 3125/15552 ≈ 0.2009. So the probability of exactly two sixes is about 0.201.
15 × 625 = 9375；36 × 1296 = 46656。约分化简：9375/46656 除以 3 得 3125/15552 ≈ 0.2009。所以恰好出现两次 6 的概率约为 0.201。
It is useful to check using E(X) = np = 1, so two sixes is slightly above expectation.
用期望 E(X) = np = 1 可做检查，两次 6 略高于期望值。
5. Forces and Equilibrium on an Inclined Plane | 斜面上的力与平衡
A particle of mass 2 kg rests on a smooth plane inclined at 30° to the horizontal. It is held in equilibrium by a light inextensible string parallel to the plane. Find the tension in the string and the normal reaction.
一个质量为 2 kg 的质点静止在与水平面成 30° 的光滑斜面上，由一根平行于斜面的轻质不可伸长绳拉住以保持平衡。求绳中张力与斜面法向反力。
Resolve perpendicular to plane: R = 2g cos 30°. Using g = 9.8 m/s², cos 30° = √3/2 ≈ 0.866, so R ≈ 2 × 9.8 × 0.866 ≈ 16.97 N.
垂直斜面分解：R = 2g cos 30°。取 g = 9.8 m/s²，cos 30° = √3/2 ≈ 0.866，得 R ≈ 2 × 9.8 × 0.866 ≈ 16.97 N。
Resolve parallel to plane: T = 2g sin 30° = 2 × 9.8 × 0.5 = 9.8 N. The tension equals the component of weight down the slope.
平行斜面分解：T = 2g sin 30° = 2 × 9.8 × 0.5 = 9.8 N。张力等于重力沿斜面向下的分量。
If the plane were rough, an additional friction force would be included, but here ‘smooth’ means no friction.
若斜面粗糙还需考虑摩擦力，但这里“光滑”表示无摩擦。
6. Related Rates: Filling a Conical Tank | 相关变化率：圆锥水箱注水
Water is poured into an inverted conical tank of height 4 m and base radius 2 m at a constant rate of 2 m³/min. Find the rate at which the water level is rising when the depth of water is 1 m.
水以 2 m³/min 的恒定速率注入一个高 4 m、底半径 2 m 的倒圆锥水箱。求水深为 1 m 时水位的上升速率。
By similar triangles, radius r at depth h satisfies r/h = 2/4 = 1/2, so r = h/2. Volume of water V = (1/3)π r² h = (1/3)π (h/2)² h = (π/12) h³.
由相似三角形，水深 h 时的半径 r 满足 r/h = 2/4 = 1/2，故 r = h/2。水的体积 V = (1/3)π r² h = (1/3)π (h/2)² h = (π/12) h³。
Differentiate with respect to time t: dV/dt = (π/4) h² (dh/dt). Given dV/dt = 2, when h = 1 we have 2 = (π/4) × 1² × (dh/dt).
对时间 t 求导：dV/dt = (π/4) h² (dh/dt)。已知 dV/dt = 2，当 h = 1 时，2 = (π/4) × 1² × (dh/dt)。
Solve: dh/dt = 8/π ≈ 2.546 m/min. Thus the water level rises at about 2.55 m/min at that instant.
解出：dh/dt = 8/π ≈ 2.546 m/min。因此该时刻水位上升速率约为 2.55 m/min。
7. Statistical Hypothesis Testing | 统计假设检验
A sample of 100 light bulbs has a mean lifetime of 1550 hours with a standard deviation of 120 hours. Test at the 5% significance level whether the population mean lifetime is greater than 1500 hours.
某批次 100 只灯泡的样本平均寿命为 1550 小时，标准差为 120 小时。在 5% 显著性水平下检验总体平均寿命是否大于 1500 小时。
H₀: μ = 1500; H₁: μ > 1500 (one‑tailed). Under H₀, the test statistic z = (x̄ − μ₀) / (σ/√n) = (1550 − 1500) / (120/√100) = 50 / 12 = 4.167.
H₀：μ = 1500；H₁：μ > 1500（单侧检验）。在 H₀ 下，检验统计量 z = (x̄ − μ₀) / (σ/√n) = (1550 − 1500) / (120/√100) = 50 / 12 = 4.167。
The critical z‑value for a one‑tailed 5% test is 1.645. Since 4.167 > 1.645, we reject H₀.
单侧 5% 检验的临界 z 值为 1.645。因为 4.167 > 1.645，故拒绝 H₀。
There is sufficient evidence at the 5% level to conclude that the mean lifetime exceeds 1500 hours. Since n=100 is large, the normal approximation is valid.
在 5% 水平下有充分证据表明平均寿命超过 1500 小时。由于 n=100 足够大，正态近似有效。
8. Optimisation of a Can Design | 罐头设计的优化
A cylindrical can without a lid is to be made to hold 1 litre (1000 cm³) of liquid. Find the radius and height that minimise the surface area of metal used.
制作一个无盖圆柱形罐子，容积为 1 升（1000 cm³）。求使所用金属表面积最小的底面半径与高。
Volume: V = π r² h = 1000, so h = 1000/(π r²). Surface area (no lid): A = base area + lateral area = π r² + 2π r h.
体积：V = π r² h = 1000，故 h = 1000/(π r²)。表面积（无盖）：A = 底面积 + 侧面积 = π r² + 2π r h。
Substitute h: A(r) = π r² + 2π r × (1000/(π r²)) = π r² + 2000/r. Differentiate: A'(r) = 2π r − 2000/r².
代入 h：A(r) = π r² + 2π r × (1000/(π r²)) = π r² + 2000/r。求导：A'(r) = 2π r − 2000/r²。
Set derivative to zero: 2π r = 2000/r² ⇒ r³ = 1000/π ⇒ r = (1000/π)^{1/3} ≈ (318.31)^{1/3} ≈ 6.83 cm. Then h = 1000/(π × 6.83²) ≈ 6.83 cm.
令导数为零：2π r = 2000/r² ⇒ r³ = 1000/π ⇒ r = (1000/π)^{1/3} ≈ (318.31)^{1/3} ≈ 6.83 cm。则 h = 1000/(π × 6.83²) ≈ 6.83 cm。
Second derivative A”(r) = 2π + 4000/r³ > 0, confirming a minimum. The optimal can has equal radius and height.
二阶导 A”(r) = 2π + 4000/r³ > 0，确认为最小值。最优罐子的半径与高相等。
9. Curve Sketching and Transformations | 曲线绘制与变换
Sketch the graph of f(x) = (x−1)²(x+2), showing all intercepts and stationary points.
绘制 f(x) = (x−1)²(x+2) 的图像，标出所有截距和驻点。
Find intercepts: y‑intercept when x=0: f(0) = (1)²×2 = 2. x‑intercepts: f(x)=0 ⇒ x=1 (double root) and x=−2.
求截距：y 轴截距 x=0 时 f(0)=2；x 轴截距：f(x)=0 ⇒ x=1（二重根）和 x=−2。
Expand to differentiate: f(x) = (x² −2x +1)(x+2) = x³ −2x² +x +2x² −4x +2 = x³ −3x +2. Then f'(x) = 3x² −3 = 3(x²−1) = 3(x−1)(x+1).
展开后求导：f(x) = (x² −2x +1)(x+2) = x³ −3x +2。则 f'(x) = 3x² −3 = 3(x²−1) = 3(x−1)(x+1)。
Stationary points at x=1 and x=−1. f(1)=0, f(−1)= (−2)²×1 = 4. Test: x=1 is a double root of f and root of f’, hence a repeated stationary point (point of inflection). x=−1 gives local maximum 4.
驻点在 x=1 和 x=−1。f(1)=0，f(−1)=(−2)²×1=4。检验：x=1 是 f 的二重根且为 f’ 的根，故为重复驻点（拐点）。x=−1 给出局部极大值 4。
As x→∞, f(x)→∞; as x→−∞, f(x)→−∞. The graph crosses at −2, touches at 1 with inflection, and peaks at (−1,4).
当 x→∞ 时 f(x)→∞；x→−∞ 时 f(x)→−∞。图像穿过 −2，在 1 处接触横轴且带拐点，在 (−1,4) 处取得峰值。
10. Connected Particles and Newton’s Laws | 连接体与牛顿定律
Two boxes of masses 3 kg and 5 kg are connected by a light inextensible string passing over a smooth pulley. The 5 kg mass rests on a smooth horizontal table, while the 3 kg mass hangs freely. Determine the acceleration of the system and the tension in the string. (Take g=9.8 m/s²)
两个质量分别为 3 kg 和 5 kg 的箱子通过轻质不可伸长绳跨过光滑滑轮连接。5 kg 箱子置于光滑水平桌面，3 kg 箱子自由悬挂。求系统的加速度和绳中张力。（取 g=9.8 m/s²）
For the 3 kg hanging mass: weight − tension = 3a ⇒ 3g − T = 3a.
对 3 kg 悬挂物体：重力 − 张力 = 3a ⇒ 3g − T = 3a。
For the 5 kg mass on the table, the only horizontal force is tension: T = 5a.
对 5 kg 桌面物体，水平方向只有张力：T = 5a。
Substitute T into the first equation: 3g − 5a = 3a ⇒ 3g = 8a ⇒ a = (3 × 9.8)/8 = 29.4/8 = 3.675 m/s².
将 T 代入第一式：3g − 5a = 3a ⇒ 3g = 8a ⇒ a = (3 × 9.8)/8 = 29.4/8 = 3.675 m/s²。
Then T = 5 × 3.675 = 18.375 N. The system accelerates to the right (table) and downward (hanging).
则 T = 5 × 3.675 = 18.375 N。系统加速度方向为桌面向右、悬挂物向下。
Notice that the pulley is smooth, so tension is uniform throughout the string. Frictionless surfaces simplify the model.
注意滑轮光滑，因此绳上各处张力相等。无摩擦表面简化了模型。
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