📚 Case Study Practice for Year 11 CAIE Further Maths | Year 11 CAIE 进阶数学：案例分析实战演练
In CAIE IGCSE Additional Mathematics (or Year 11 Further Maths), many exam questions are set in a real-world context. Solving these case study problems requires not only solid algebraic and calculus skills but also the ability to translate a written scenario into mathematical models. This article walks through ten typical case studies, showing step-by-step approaches and key techniques.
在CAIE IGCSE 附加数学（或十一年级进阶数学）中，许多试题都以真实情境为背景。解答这些案例分析题不仅需要扎实的代数与微积分技能，还要能把文字描述转化为数学模型。本文通过十个经典案例，逐步展示解题方法和关键技巧。
1. Maximising Area Using Calculus | 利用微积分求最大面积
A farmer has 100 m of fencing and wants to enclose a rectangular pen against a barn (so only three sides need fencing). Find the maximum area of the pen.
一个农夫有100米栅栏，想靠着一间谷仓围一个矩形围栏（因此只需围三面）。求围栏的最大面积。
Let the side perpendicular to the barn be x m. Then the side parallel to the barn is (100 − 2x) m. Area A = x(100 − 2x) = 100x − 2x². Differentiate: dA/dx = 100 − 4x. Set dA/dx = 0 ⇒ 100 − 4x = 0 ⇒ x = 25. Then A = 25 × (100 − 50) = 1250 m². Second derivative d²A/dx² = −4 < 0 confirms a maximum. Always define variables clearly and use the constraint to reduce unknowns before differentiating.
设垂直于谷仓的边长为 x 米，则平行于谷仓的边长为 (100 − 2x) 米。面积 A = x(100 − 2x) = 100x − 2x²。求导：dA/dx = 100 − 4x。令导数为零 ⇒ x = 25。面积 A = 25 × 50 = 1250 m²。二阶导数 d²A/dx² = −4 < 0 确认是最大值。切记先定义变量，利用约束条件消元后再求导。
2. Exponential Decay of Radioactive Substance | 放射性物质的指数衰变
The mass M(t) of a radioactive substance decays according to M(t) = M₀ e⁻ᵏᵗ, where M₀ is initial mass and t is in hours. After 5 hours, 60% of the substance remains. (a) Find the decay constant k. (b) Calculate the half‑life (time for M = ½ M₀).
放射性物质的质量 M(t) 按 M(t) = M₀ e⁻ᵏᵗ 衰变，M₀ 为初始质量，t 以小时为单位。5小时后还剩60%的物质。(a) 求衰变常数 k。(b) 计算半衰期（质量减半所需时间）。
(a) 0.6 M₀ = M₀ e⁻⁵ᵏ ⇒ 0.6 = e⁻⁵ᵏ ⇒ ln 0.6 = −5k ⇒ k = −(ln 0.6)/5 ≈ 0.1022 h⁻¹. (b) ½ = e⁻ᵏᵗ ⇒ ln(0.5) = −k t ⇒ t = ln 2 / k ≈ 6.78 h. When using exponential models, take natural logs as soon as possible to isolate the exponent.
(a) 0.6 M₀ = M₀ e⁻⁵ᵏ ⇒ 0.6 = e⁻⁵ᵏ ⇒ ln 0.6 = −5k ⇒ k = −(ln 0.6)/5 ≈ 0.1022 h⁻¹。(b) ½ = e⁻ᵏᵗ ⇒ ln(0.5) = −k t ⇒ t = ln 2 / k ≈ 6.78 小时。处理指数模型时，尽早取自然对数以分离指数。
3. Solving a Trigonometric Equation in Navigation | 航海中的三角函数方程求解
A buoy bobs up and down with height h(t) = 2.5 sin(πt/4) + 5 metres, where t is time in seconds. Find the first two times when the height is exactly 6 m.
一个浮标上下浮动，高度函数为 h(t) = 2.5 sin(πt/4) + 5 米，t 以秒为单位。求高度首次两次达到6米的时间。
Set 2.5 sin(πt/4) + 5 = 6 ⇒ 2.5 sin(πt/4) = 1 ⇒ sin(πt/4) = 0.4. For 0 ≤ πt/4 < 2π, πt/4 = sin⁻¹(0.4) ≈ 0.4115 rad, and π − 0.4115 = 2.7301 rad. Thus t ≈ (0.4115×4)/π = 0.524 s, and t ≈ (2.7301×4)/π = 3.476 s. Always find both principal and supplementary solutions for sine within 0 to 2π.
令 2.5 sin(πt/4) + 5 = 6 ⇒ sin(πt/4) = 0.4。在 0 ≤ πt/4 < 2π 内，πt/4 = sin⁻¹(0.4) ≈ 0.4115 rad，以及 π − 0.4115 = 2.7301 rad。因此 t ≈ 0.524 s 和 t ≈ 3.476 s。解正弦方程时务必在 0 到 2π 范围内找出主值及补角解。
4. Proving Collinearity Using Vectors | 利用向量证明共线性
Three points have position vectors OA = 2i + j, OB = 5i + 7j, OC = 9i + 13j relative to origin O. Show that A, B and C are collinear.
三点相对于原点O的位置向量分别为 OA = 2i + j, OB = 5i + 7j, OC = 9i + 13j。证明A、B、C三点共线。
Find AB = OB − OA = (5−2)i + (7−1)j = 3i + 6j. AC = OC − OA = (9−2)i + (13−1)j = 7i + 12j. Check if AB is parallel to AC: 3/7 = 6/12 = 1/2, so AC = (7/3) AB. Since AB and AC share point A and are parallel, the points are collinear. The key is to show one vector is a scalar multiple of another.
计算 AB = OB − OA = 3i + 6j，AC = OC − OA = 7i + 12j。检验 AB 与 AC 是否平行：3/7 = 6/12 = 1/2，因此 AC = (7/3) AB。因 AB 与 AC 共起点A且平行，故三点共线。关键是证明一个向量是另一个向量的标量倍数。
5. Applying Polynomial Division to Find Unknown Coefficients | 应用多项式除法求未知系数
A cubic curve f(x) = 2x³ + ax² + bx − 6 passes through (1, −2) and has a factor of (x − 2). Determine a, b and fully factorise f(x).
三次曲线 f(x) = 2x³ + ax² + bx − 6 过点 (1, −2) 且含有因式 (x − 2)。求 a 和 b，并将 f(x) 完全因式分解。
From f(1) = −2: 2(1)³ + a(1)² + b(1) − 6 = −2 ⇒ a + b = 2. (1) Since (x−2) is a factor, f(2)=0: 2(8) + a(4) + 2b − 6 = 0 ⇒ 4a + 2b = −10 ⇒ 2a + b = −5. (2) Solve (1) and (2): subtract (1) from (2): a = −7, then b = 9. So f(x) = 2x³ − 7x² + 9x − 6. Divide by (x−2) to get quadratic 2x² − 3x + 3. Discriminant Δ = 9 − 24 < 0, so no further real factors. The factorised form is (x − 2)(2x² − 3x + 3). Using given points or roots to set up simultaneous equations is a powerful technique.
由 f(1)=−2 得 a + b = 2 (1)；由因式 (x−2) 得 f(2)=0 ⇒ 4a + 2b = −10 ⇒ 2a + b = −5 (2)。联立解得 a = −7，b = 9。故 f(x)=2x³−7x²+9x−6。除以 (x−2) 得 2x²−3x+3，其判别式Δ<0，无实根。完全因式分解为 (x−2)(2x²−3x+3)。利用已知点或根建立方程组是常用策略。
6. Solving Modulus Inequalities Graphically | 图解绝对值不等式
A manufacturing error tolerance is described by |2x − 7| ≤ 3, where x is a measurement in mm. Find the acceptable range of x.
制造误差容限为 |2x − 7| ≤ 3，x 为测量值（毫米）。求 x 的合格范围。
Remove modulus: −3 ≤ 2x − 7 ≤ 3. Add 7: 4 ≤ 2x ≤ 10 ⇒ 2 ≤ x ≤ 5. Therefore, measurements between 2 mm and 5 mm are acceptable. For |ax + b| < c, always rewrite as a double inequality. If the inequality was >, split into two separate inequalities.
去掉绝对值：−3 ≤ 2x − 7 ≤ 3。加7得 4 ≤ 2x ≤ 10 ⇒ 2 ≤ x ≤ 5。故合格范围为 2 mm 至 5 mm。处理 |ax+b| < c 型不等式时直接转化为双不等式；若为 > 则拆成两个独立不等式。
7. Connected Rates: Filling a Conical Tank | 相关变化率：圆锥水箱注水
Water is poured into an inverted conical tank (tip down) at 0.2 m³/min. The cone has height 4 m and base radius 2 m. Find the rate at which the water level rises when the depth is 1 m.
水以0.2 m³/min 注入顶角在下的倒圆锥水箱。锥高4 m，底面半径2 m。求水深1 m 时水面上升的速率。
Volume V = ⅓πr²h. By similar triangles, r/h = 2/4 = ½ ⇒ r = h/2. Substitute: V = ⅓π (h/2)² h = (π/12) h³. dV/dt = (π/4) h² (dh/dt). Given dV/dt = 0.2, when h = 1: 0.2 = (π/4)(1)² (dh/dt) ⇒ dh/dt = 0.8/π ≈ 0.255 m/min. Always express V in terms of a single variable before differentiating with respect to time.
体积 V = ⅓πr²h。由相似三角形 r/h = 2/4 = ½ ⇒ r = h/2。代入得 V = ⅓π (h/2)² h = (π/12) h³。求导得 dV/dt = (π/4) h² (dh/dt)。代入 dV/dt=0.2 及 h=1：0.2 = (π/4) × 1 × (dh/dt) ⇒ dh/dt = 0.8/π ≈ 0.255 m/min。切记在关于时间求导前，将体积表示为单一变量的函数。
8. Geometric Series in Compound Interest | 复利中的等比数列
£5000 is invested at 4% annual compound interest. (a) Find the value after 10 years. (b) How many whole years are needed for the investment to exceed £8000?
投入5000英镑，年利率4%复利。(a) 求10年后的价值。(b) 至少需要多少整年才能使投资超过8000英镑？
(a) Value = 5000 × (1.04)¹⁰ ≈ 5000 × 1.4802 = £7401. (b) Solve 5000 × (1.04)ⁿ > 8000 ⇒ (1.04)ⁿ > 1.6. Take logs: n ln(1.04) > ln(1.6) ⇒ n > ln(1.6)/ln(1.04) ≈ 11.9. So 12 years are needed. Always round up to the next whole number for ‘exceeds’ or ‘at least’ questions.
(a) 终值 = 5000 × (1.04)¹⁰ ≈ £7401。(b) 解 5000×(1.04)ⁿ > 8000 ⇒ (1.04)ⁿ > 1.6。取对数：n > ln(1.6)/ln(1.04) ≈ 11.9，因此需要12整年。遇到“超过”或“至少”问题时，务必向上取整。
9. Inverse Functions and Their Domains | 反函数及其定义域
Given f(x) = (2x + 3)/(x − 1) for x ≠ 1. Find f⁻¹(x) and state its domain and range.
已知 f(x) = (2x + 3)/(x − 1)，x ≠ 1。求 f⁻¹(x) 并写出其定义域和值域。
Write y = (2x+3)/(x−1). Swap x and y: x = (2y+3)/(y−1). Multiply: x(y−1) = 2y+3 ⇒ xy − x = 2y + 3 ⇒ xy − 2y = x + 3 ⇒ y(x−2) = x+3 ⇒ y = (x+3)/(x−2). So f⁻¹(x) = (x+3)/(x−2). Domain of f⁻¹ is all real numbers except x=2 (the range of original f). Range of f⁻¹ is y ≠ 1 (the domain of f). The procedure is algebraic, but always check the domain–range swap.
令 y = (2x+3)/(x−1)。交换 x 和 y 得 x = (2y+3)/(y−1)。交叉相乘：x(y−1)=2y+3 ⇒ xy−x=2y+3 ⇒ xy−2y=x+3 ⇒ y(x−2)=x+3 ⇒ y=(x+3)/(x−2)。故 f⁻¹(x)=(x+3)/(x−2)。f⁻¹ 的定义域为 x ≠ 2（原函数的值域），值域为 y ≠ 1（原函数的定义域）。过程纯代数化，但务必验证定义域与值域的互换。
10. Strategy Summary for Case Study Questions | 案例分析题的策略总结
Across these examples, a common thread emerges: translate the scenario into mathematical language by defining variables, writing equations or functions, applying the correct topic technique (differentiation, logs, vectors, etc.), and interpreting the solution back in context. Practice drawing diagrams, checking constraints, and verifying second‑derivative or domain restrictions.
纵观以上案例，一条主线清晰可见：将情境转化为数学语言——定义变量、写出方程或函数、选择正确的方法（微分、对数、向量等），并将结果带回原情境解释。平时多练习画图、核查约束条件、检验二阶导数或定义域限制。
Case study questions test both your mathematical fluency and your ability to think critically about real‑world conditions. Break the problem into small steps, and you will find them manageable.
案例分析题既考查数学的熟练程度，也考验对现实条件的批判性思考。将问题分解为小步骤，就会发现它们完全可以驾驭。
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