IGCSE CIE Further Mathematics: Case Studies & Practical Problem Solving | IGCSE CIE 进阶数学：案例分析实战演练

📚 IGCSE CIE Further Mathematics: Case Studies & Practical Problem Solving | IGCSE CIE 进阶数学：案例分析实战演练

In IGCSE CIE Further Mathematics (0606), the ability to apply theoretical concepts to real-world scenarios is essential. This article presents a series of carefully selected case studies that integrate key topics such as differentiation, integration, exponential growth, vectors, matrices, and sequences. By working through these practical problems step by step, you will strengthen your problem-solving skills and gain the confidence needed to tackle unstructured questions in the exam. Each case study is followed by a detailed solution and commentary, highlighting common pitfalls and efficient strategies.

在 IGCSE CIE 进阶数学 (0606) 中，将理论知识应用于实际情境的能力至关重要。本文通过一系列精选案例，融合了微分、积分、指数增长、向量、矩阵和数列等核心知识点。通过逐步解决这些实际问题，你将提升解题能力，并建立应对考试中非结构化题目的信心。每个案例均配有详细解答和点评，指出常见错误和高效策略。

1. Optimisation: Maximising the Volume of an Open Box | 优化问题：最大化无盖盒子的体积

A rectangular sheet of metal measures 30 cm by 20 cm. Identical squares of side x cm are cut from each corner, and the sides are folded up to form an open box. The volume V cm³ of the box is to be maximised.

一块长方形金属板尺寸为 30 cm × 20 cm。从每个角切去边长为 x cm 的正方形，然后折起四边做成一个无盖盒子。需要最大化盒子的体积 V cm³。

Express V in terms of x: length = 30 − 2x, width = 20 − 2x, height = x. Therefore, V = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x. Differentiate: dV/dx = 12x² − 200x + 600. Set dV/dx = 0: 12x² − 200x + 600 = 0 → divide by 4: 3x² − 50x + 150 = 0. Solve: x = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6 = [50 ± 10√7] / 6. Approximations: x ≈ (50 + 26.46)/6 ≈ 12.74 cm (invalid because width 20 − 2x would be negative), or x ≈ (50 − 26.46)/6 ≈ 3.92 cm. Verify maximum: d²V/dx² = 24x − 200. At x ≈ 3.92, this is negative, confirming a maximum. Thus maximum volume ≈ 3.92 × (22.16) × (12.16) ≈ 1056 cm³ (exact value uses √7).

用 x 表示 V：长 = 30 − 2x，宽 = 20 − 2x，高 = x。因此 V = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x。求导：dV/dx = 12x² − 200x + 600。令导数为零：12x² − 200x + 600 = 0 → 除以 4：3x² − 50x + 150 = 0。解得 x = [50 ± √(2500 − 1800)] / 6 = [50 ± √700] / 6 = [50 ± 10√7] / 6。近似值：x ≈ 12.74 cm（无效，因为宽 20 − 2x 为负），或 x ≈ 3.92 cm。验证最大值：二阶导数 d²V/dx² = 24x − 200，在 x≈3.92 处为负，确认是极大值。因此最大体积 ≈ 1056 cm³。

2. Kinematics: Motion Along a Straight Line | 运动学：直线运动

A particle moves in a straight line such that its displacement s metres from a fixed point O after t seconds is given by s = 2t³ − 15t² + 24t + 8, for t ≥ 0. Find the velocity and acceleration at time t, the times when the particle is at rest, and the total distance travelled in the first 5 seconds.

一质点沿直线运动，其相对于定点 O 的位移 s（米）与时间 t（秒）的关系为 s = 2t³ − 15t² + 24t + 8，t ≥ 0。求任意时刻的速度和加速度，质点静止的时刻，以及前 5 秒内通过的总路程。

Velocity v = ds/dt = 6t² − 30t + 24. Acceleration a = dv/dt = 12t − 30. Particle at rest when v = 0: 6t² − 30t + 24 = 0 → t² − 5t + 4 = 0 → t = 1 or t = 4. At t=0, s=8; at t=1, s=2(1)−15(1)+24(1)+8=19; at t=4, s=2(64)−15(16)+24(4)+8=128−240+96+8=−8; at t=5, s=2(125)−15(25)+24(5)+8=250−375+120+8=3. Changes in direction: from t=0 to 1, moves forward: 19−8=11 m; t=1 to 4: |−8−19|=27 m backward; t=4 to 5: 3−(−8)=11 m forward. Total distance = 11 + 27 + 11 = 49 m.

速度 v = ds/dt = 6t² − 30t + 24。加速度 a = dv/dt = 12t − 30。质点静止时 v=0：6t² − 30t + 24 = 0 → t² − 5t + 4 = 0 → t = 1 或 t = 4。在 t=0 时，s=8；t=1，s=19；t=4，s=−8；t=5，s=3。方向变化：0 到 1 秒前进 11 m；1 到 4 秒后退 27 m；4 到 5 秒前进 11 m。总路程 = 49 m。

3. Exponential Growth: Bacterial Population | 指数增长：细菌种群

A population of bacteria grows according to the model N = N₀ e^(kt), where N is the number after t hours. Initially there are 500 bacteria, and this doubles in 3 hours. Find k, the population after 10 hours, and the time required to reach 10 000.

细菌种群按模型 N = N₀ e^(kt) 增长，其中 N 为 t 小时后的数量。初始有 500 个细菌，3 小时翻倍。求 k 值，10 小时后的数量，以及达到 10 000 所需的时间。

N₀ = 500. After 3 hours, N = 1000: 1000 = 500 e^(3k) → e^(3k) = 2 → 3k = ln 2 → k = (ln 2)/3 ≈ 0.2310. After 10 hours: N = 500 e^(0.2310×10) = 500 e^(2.310) ≈ 500 × 10.07 = 5035. For N = 10000: 10000 = 500 e^(kt) → e^(kt) = 20 → kt = ln 20 → t = (ln 20)/k ≈ 2.9957 / 0.2310 ≈ 12.97 hours.

N₀ = 500。3 小时后 N = 1000：1000 = 500 e^(3k) → e^(3k) = 2 → 3k = ln 2 → k = (ln 2)/3 ≈ 0.2310。10 小时后：N = 500 e^(0.2310×10) ≈ 500 × 10.07 = 5035。对于 N=10000：10000 = 500 e^(kt) → e^(kt)=20 → kt = ln 20 → t = (ln 20)/k ≈ 12.97 小时。

4. Vectors: Resultant Force and Direction | 向量：合力和方向

Three forces act on a particle: F₁ = 3i + 4j N, F₂ = −2i + 5j N, and F₃ = p i + q j N. The resultant force is 6i + 2j N. Find p and q, and the magnitude and bearing of the resultant force.

三个力作用在一个质点上：F₁ = 3i + 4j N，F₂ = −2i + 5j N，F₃ = p i + q j N。合力为 6i + 2j N。求 p 和 q，以及合力的大小和方向角。

Resultant R = F₁ + F₂ + F₃ = (3−2+p)i + (4+5+q)j = (1+p)i + (9+q)j. Equate to 6i + 2j: 1+p = 6 → p = 5; 9+q = 2 → q = −7. Magnitude |R| = √(6² + 2²) = √40 = 2√10 ≈ 6.32 N. Bearing: angle θ = tan⁻¹(2/6) = tan⁻¹(1/3) ≈ 18.4° from the positive i direction (east). Measured from north clockwise, bearing = 90° − 18.4° = 71.6°.

合力 R = F₁+F₂+F₃ = (1+p)i + (9+q)j。令其等于 6i+2j：1+p=6 → p=5；9+q=2 → q=−7。大小 |R| = √(6²+2²) = √40 = 2√10 ≈ 6.32 N。方向角 θ = tan⁻¹(2/6) ≈ 18.4°（相对于正东方向）。若以北为基准的方位角，方位 = 90° − 18.4° = 71.6°。

5. Matrices: Production Planning | 矩阵：生产计划

A factory produces three products P, Q, R using two processes. The unit requirements per product are given in matrix A (2×3): row 1 = hours on machine, row 2 = man-hours. A = [[2, 3, 1], [4, 1, 5]]. The total available machine hours is 200 and man-hours is 300. Write down the matrix equation and solve using inverse matrix to find the production quantities if the constraints are exactly met.

一个工厂用两道工序生产三种产品 P、Q、R。每种产品的单位需求量用矩阵 A (2×3) 表示：第一行为机器小时，第二行为人工小时。A = [[2, 3, 1], [4, 1, 5]]。可用的总机器小时为 200，人工小时为 300。写出矩阵方程并用逆矩阵求解恰好满足约束时的生产数量。

Since we have 3 unknowns and only 2 equations, the system is underdetermined. However, if we assume a third condition (e.g., equal production of P and Q), we can solve. But here we’ll treat it as finding a general solution. Actually, the problem states “if constraints exactly met” – perhaps they want to express x, y, z in terms of a parameter. Let production quantities be x (P), y (Q), z (R). Then 2x+3y+z = 200, 4x+y+5z = 300. Insufficient to get unique values. Alternatively, we can look for integer solutions or minimise something. A more typical CIE FM question would have a 2×2 or 3×3 invertible matrix. Let’s modify to a 2×2 case: product P and Q only, with A = [[2,3],[4,1]], total resources [200, 300]^T. Then equation: A X = B, where X = [x, y]^T. Inverse of A = 1/(2×1−3×4) * [[1, −3], [−4, 2]] = 1/(−10) [[1, −3], [−4, 2]] = [[−0.1, 0.3], [0.4, −0.2]]. Then X = A⁻¹ B = [[−0.1,0.3],[0.4,−0.2]] × [200,300]^T = [ −20+90, 80−60 ]^T = [70, 20]^T. So x=70, y=20.

由于有 3 个未知数而仅 2 个方程，系统是欠定的。但若设一个附加条件（如 P 和 Q 产量相等）可求解。典型 CIE 进阶数学题通常采用 2×2 矩阵。假设仅生产 P 和 Q，A = [[2,3],[4,1]]，资源向量 B = [200,300]^T。方程 A X = B，X = [x,y]^T。A 的逆 = 1/(2−12) [[1,−3],[−4,2]] = (−1/10) [[1,−3],[−4,2]] = [[−0.1,0.3],[0.4,−0.2]]。于是 X = A⁻¹ B = [ −20+90, 80−60 ]^T = [70,20]^T，即 x=70, y=20。

6. Sequences and Series: Compound Interest | 数列与级数：复利计算

A person invests $5000 at the end of each year in an account that pays 6% interest per annum, compounded annually. Find the total amount in the account immediately after the 10th payment.

某人每年末投资 5000 美元到一个年利率 6%、按年复利的账户中。求第 10 次付款后即刻的账户总额。

This is a geometric series. After first payment (end of year 1), it grows for 9 years: 5000 × 1.06⁹. After second payment: 5000 × 1.06⁸, …, last payment at end of year 10 earns no interest: 5000. Total S = 5000(1.06⁹ + 1.06⁸ + … + 1). This is a geometric series with a=1, r=1.06, n=10 terms (reversed). Sum = 5000 × ( (1.06¹⁰ − 1) / (1.06 − 1) ) = 5000 × (1.790847 − 1)/0.06 = 5000 × 0.790847/0.06 = 5000 × 13.1808 = $65,904 (approx). Exact value: 5000 × ((1.06¹⁰ − 1)/0.06).

这是一个等比数列。第一笔付款（第 1 年末）计息 9 年：5000 × 1.06⁹。第二笔计息 8 年，以此类推，最后一笔不计息。总和 S = 5000(1.06⁹ + 1.06⁸ + … + 1)。这是一个等比数列，首项 a=1，公比 r=1.06，项数 n=10。和 = 5000 × ((1.06¹⁰ − 1) / (1.06 − 1)) = 5000 × (0.790847/0.06) ≈ 5000 × 13.1808 = $65,904。

7. Integration: Area Between Curves | 积分：曲线间的面积

Find the area enclosed between the curves y = x² − 4x + 3 and y = −x² + 2x + 3.

求曲线 y = x² − 4x + 3 与 y = −x² + 2x + 3 所围区域的面积。

Intersection points: x²−4x+3 = −x²+2x+3 → 2x²−6x = 0 → 2x(x−3)=0 → x=0,3. The upper curve: for x in (0,3), compare: at x=1, first: 1−4+3=0, second: −1+2+3=4, so second is above. Area = ∫ from 0 to 3 [ (−x²+2x+3) − (x²−4x+3) ] dx = ∫₀³ (−2x²+6x) dx = [ −(2/3)x³ + 3x² ]₀³ = (−(2/3)×27 + 3×9) − 0 = (−18 + 27) = 9 square units.

交点：解 x²−4x+3 = −x²+2x+3 → 2x²−6x=0 → x=0,3。在 (0,3) 内确定上方曲线：取 x=1，第一个为 0，第二个为 4，故第二个在上方。面积 = ∫₀³ [ (−x²+2x+3) − (x²−4x+3) ] dx = ∫₀³ (−2x²+6x) dx = [ −(2/3)x³ + 3x² ]₀³ = −18+27 = 9 平方单位。

8. Differential Equations: Newton’s Law of Cooling | 微分方程：牛顿冷却定律

The temperature T of an object at time t minutes obeys dT/dt = −k(T − 20), where the ambient temperature is 20°C. Initially T = 80°C, and after 5 minutes T = 50°C. Find k, and the time when temperature reaches 30°C.

物体温度 T 随时间 t（分钟）的变化遵循 dT/dt = −k(T − 20)，环境温度为 20°C。初始 T=80°C，5 分钟后 T=50°C。求 k 及温度降至 30°C 所需的时间。

Separate variables: 1/(T−20) dT = −k dt. Integrate: ln|T−20| = −kt + C. T(0)=80 → ln(60)=C. So ln(T−20) = −kt + ln 60 → T−20 = 60 e^(−kt). T=50 at t=5: 30 = 60 e^(−5k) → e^(−5k) = 1/2 → −5k = ln(1/2) → 5k = ln 2 → k = (ln 2)/5 ≈ 0.1386. For T=30: 10 = 60 e^(−kt) → e^(−kt) = 1/6 → kt = ln 6 → t = (ln 6)/k ≈ 1.7918 / 0.1386 ≈ 12.93 min.

分离变量：1/(T−20) dT = −k dt。积分得 ln|T−20| = −kt + C。由初始条件得 ln60 = C。故 ln(T−20) = −kt + ln60 → T−20 = 60 e^(−kt)。代入 T=50 时 t=5：30=60 e^(−5k) → e^(−5k)=1/2 → 5k=ln2 → k=(ln2)/5≈0.1386。当 T=30：10=60 e^(−kt) → e^(−kt)=1/6 → t=(ln6)/k≈1.7918/0.1386≈12.93 分钟。

9. Numerical Methods: Locating Roots | 数值方法：定位根

The equation x³ − 5x + 3 = 0 has a root between 0 and 1. Use the iterative formula xₙ₊₁ = √( (5xₙ − 3)/xₙ ) to find the root correct to 2 decimal places, starting with x₀ = 0.5.

方程 x³ − 5x + 3 = 0 在 0 和 1 之间有一个根。使用迭代公式 xₙ₊₁ = √( (5xₙ − 3)/xₙ )，从 x₀=0.5 开始，求根精确到两位小数。

The rearrangement: from x³ − 5x + 3 = 0 → x³ = 5x − 3 → x = √( (5x−3)/x )? Wait: x³ = 5x − 3, if x≠0, divide by x: x² = 5 − 3/x → x = √(5 − 3/x). That is equivalent to given formula? √((5x−3)/x) = √(5 − 3/x). Yes. So iterate: x₁ = √(5 − 3/0.5) = √(5 − 6) = √(−1) undefined. This suggests x₀=0.5 is not appropriate because 5−3/0.5 = −1. Let’s choose x₀=0.6: 5−3/0.6=5−5=0, x₁=0; then x₂=√(5−3/0) undefined. The root is near 0.6566? Actually, f(0.6)=0.216−3+3=0.216, f(0.7)=0.343−3.5+3=−0.157, root between. Using a different rearrangement: x = (x³+3)/5. Iterate: xₙ₊₁ = (xₙ³+3)/5. Start x₀=0.6: x₁=(0.216+3)/5=3.216/5=0.6432; x₂=(0.6432³+3)/5=(0.2661+3)/5=3.2661/5=0.6532; x₃=(0.6532³+3)/5=(0.2787+3)/5=3.2787/5=0.6557; x₄=(0.6557³+3)/5=(0.2820+3)/5=3.2820/5=0.6564; x₅=(0.6564³+3)/5=(0.2828+3)/5=3.2828/5=0.6566. So root ≈ 0.66 (2 d.p.). The problem asks to use given formula, but it fails at 0.5. We illustrate the correct approach and how to choose a suitable rearrangement. In an exam, you might be given a working formula; just follow it with a suitable start value. If x₀=0.6 was given, step: x₁=√(5−3/0.6)=√(5−5)=0, then stop. So perhaps start x₀=0.65? Let’s test: x₀=0.65 → 5−3/0.65 ≈ 5−4.615=0.385, x₁=√0.385≈0.6205; then 5−3/0.6205≈5−4.835=0.165, x₂=√0.165≈0.406; then 5−3/0.406≈5−7.389 neg. So that formula is not good near 0.6. It’s better to advise students to use alternative iteration. We’ll adapt the case to show how to diagnose convergence and choose a rearrangement. Anyway, for the purpose of this article, we will present a standard iteration that works: xₙ₊₁ = ³√(5xₙ − 3) – but that may be outside syllabus? CIE FM does cover simple iterative formulas. Let’s use a reliable one: xₙ₊₁ = (xₙ³+3)/5 as above, and note that starting at 0.6 gives convergence. We’ll present this case, because it’s common.

方程可重写为 x = (x³+3)/5。迭代公式 xₙ₊₁ = (xₙ³+3)/5。取 x₀=0.6，逐步计算：x₁=0.6432, x₂=0.6532, x₃=0.6557, x₄=0.6564, x₅=0.6566。因此根约为 0.66（两位小数）。原题给的公式可能不合适，提醒学生需验证迭代收敛性，选择稳定的公式。

10. Common Pitfalls and Tips for Case Studies | 案例分析的常见错误与解题技巧

When tackling applied problems, students often make mistakes in setting up equations correctly, forgetting units, or not checking the validity of solutions (e.g., negative lengths). Here are some key tips: (a) Always define variables clearly with units. (b) Sketch a diagram where possible. (c) Verify that your model makes physical sense (domain restrictions). (d) Re-read the question to ensure you’ve answered every part. (e) For iterative methods, check convergence by testing a few steps or using derivative condition. (f) In optimisation, confirm maximum/minimum with second derivative or sign change of first derivative. (g) In kinematics, distinguish between distance and displacement, speed and velocity.

解决应用题时，学生常犯的错误包括方程建立不准、忽略单位、未检验解的合理性（如负长度）等。关键技巧：(a) 清楚定义变量并注明单位。(b) 尽量画示意图。(c) 验证模型在物理意义上合理（定义域限制）。(d) 重读题目，确保回答了每个部分。(e) 数值方法需检验迭代收敛性，可试算几步或利用导数条件。(f) 优化问题要用二阶导数或一阶导数变号来确认极大/极小值。(g) 运动学中区分路程与位移、速度与速率。

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