📚 Case Study Practice for WJEC Year 12 Further Maths | WJEC 12年级进阶数学：案例分析实战演练
Case study questions in WJEC Year 12 Further Maths challenge you to apply multiple mathematical techniques to realistic scenarios. This article walks you through carefully chosen worked examples from Pure, Mechanics, Statistics and Decision Mathematics. You will learn how to break down a complex problem, select the right mathematical models, and present your reasoning clearly — exactly what examiners look for in high-scoring answers.
WJEC 12年级进阶数学的案例分析题要求你将多种数学技巧综合运用于贴近现实的场景。本文将带你一步步演练来自纯数、力学、统计和决策数学的精选例题。你将学会如何拆解复杂问题、选择恰当的数学模型，并清晰地呈现推理过程——这正是阅卷老师在高分答案中所看重的。
1. Introduction to Case Study Methodology | 案例分析方法简介
Every case study begins by identifying what you are given, what you need to find, and which area of Further Maths applies. Start by underlining the key quantities, units and constraints. Draw a diagram wherever possible, as visual representation often reveals the best mathematical structure. Then, translate the words into equations, inequalities, or matrices before solving step by step.
每一个案例分析都始于识别已知条件、求解目标和所适用的进阶数学领域。先圈出关键量、单位和约束条件。尽可能画出示意图，因为图形往往能揭示最佳的数学结构。然后将文字翻译为方程、不等式或矩阵，再逐步求解。
A successful case study answer does more than compute — it interprets results in context. Always check whether your numerical answer makes sense (e.g. a probability must be between 0 and 1, a length cannot be negative). Finally, write a short concluding sentence that answers the original question directly.
一份成功的案例分析答案不只是计算，还要结合情境解读结果。务必检查数值答案是否合理（例如概率必须在0和1之间，长度不能为负）。最后，写一句简短的结论直接回答原问题。
2. Case 1: Designing a Logo Using Matrix Transformations | 案例一：使用矩阵变换设计标志
A graphic designer starts with a right-angled triangle having vertices at A(2, 1), B(5, 1) and C(2, 4). She first applies an anticlockwise rotation of 90° about the origin, then enlarges the result by a scale factor of 2. Find the coordinates of the final image and determine whether the combined transformation can be represented by a single matrix.
一名平面设计师从一个直角三角形入手，顶点坐标为 A(2, 1)、B(5, 1) 和 C(2, 4)。她先将图形绕原点逆时针旋转90°，再将结果放大2倍。求最终图像的坐标，并判断该复合变换能否用一个矩阵表示。
The rotation matrix R is [0 -1; 1 0] and the enlargement matrix E is [2 0; 0 2]. The combined matrix M is E × R = [0 -2; 2 0]. Applying M to each column vector yields A'(−2, 4), B'(−2, 10) and C'(−8, 4). The designer can now plot the enlarged, rotated logo. Because matrix multiplication is associative, the combined transformation is indeed represented by a single matrix.
旋转矩阵 R 为 [0 -1; 1 0]，放大矩阵 E 为 [2 0; 0 2]。复合矩阵 M = E × R = [0 -2; 2 0]。将 M 作用于各列向量得到 A'(−2, 4)、B'(−2, 10) 和 C'(−8, 4)。设计师现在可以绘制放大且旋转后的标志。由于矩阵乘法满足结合律，该复合变换确实可用单个矩阵表示。
- Always multiply transformation matrices in the order they are applied, reading from left to right.
- 变换矩阵按作用顺序从左到右相乘。
- Check determinant: det(M) = 4, non-zero, so the transformation is invertible — useful for recovering the original shape.
- 检查行列式：det(M) = 4 ≠ 0，变换可逆——便于恢复原始图形。
3. Case 2: Balancing Forces on a Signpost (Vectors and Statics) | 案例二：路标杆的受力平衡（向量与静力学）
A horizontal signpost of weight 150 N is supported by two light cables attached to a vertical wall. Cable 1 exerts a force with components (2i + 3j) × 10 N, and Cable 2 exerts an unknown force F = xi + yj N. The system is in equilibrium. Find x and y and determine the tension in each cable.
一根水平路标杆重150 N，由两根轻质缆绳固定在竖直墙上。缆绳1施加的分力为 (2i + 3j) × 10 N，缆绳2施加未知力 F = xi + yj N。系统处于平衡。求 x 和 y，并确定每根缆绳的拉力。
Resolve forces: the weight acts downwards, so its vector is 0i − 150j N. Equilibrium demands ΣF_horizontal = 0 and ΣF_vertical = 0. Hence 20 + x = 0 → x = −20 N, and 30 + y − 150 = 0 → y = 120 N. The tension in Cable 1 is √(20² + 30²) = 10√13 ≈ 36.1 N; in Cable 2 is √((−20)² + 120²) = √14800 ≈ 121.7 N. The signs indicate directions — negative i means the second cable pulls towards the wall.
分解各力：重力竖直向下，其向量为 0i − 150j N。平衡要求 ΣF_水平 = 0 和 ΣF_竖直 = 0。因此 20 + x = 0 → x = −20 N，且 30 + y − 150 = 0 → y = 120 N。缆绳1的拉力为 √(20² + 30²) = 10√13 ≈ 36.1 N；缆绳2的拉力为 √((−20)² + 120²) = √14800 ≈ 121.7 N。正负号表示方向——i 分量为负意味着第二根绳朝墙壁方向拉。
| Force | i-component | j-component |
| Cable 1 | 20 | 30 |
| Cable 2 (unknown) | x | y |
| Weight | 0 | −150 |
4. Case 3: Call Centre Staffing with Poisson Distribution | 案例三：用泊松分布规划呼叫中心人员配置
A customer service centre receives an average of 3 calls every 5 minutes. The manager wants enough agents on duty so that when a call arrives, the probability that all agents are busy is less than 0.05. Model the number of calls per 5-minute interval as a Poisson random variable X ~ Po(3). Determine the minimum number of agents required.
某客服中心平均每5分钟接到3通来电。经理想安排足够的接线员，使得来电时所有接线员忙线的概率低于0.05。将每5分钟的来电数建模为泊松随机变量 X ~ Po(3)。求所需的最少接线员人数。
Let k be the number of agents. The condition ‘all agents are busy’ corresponds to X > k, so we need P(X > k) < 0.05. Using the cumulative Poisson table for λ = 3, we calculate: P(X ≤ 5) = 0.9161, P(X ≤ 6) = 0.9665. Thus P(X > 5) = 1 − 0.9161 = 0.0839 (>0.05), while P(X > 6) = 1 − 0.9665 = 0.0335 (<0.05). Therefore at least 6 agents are required.
令 k 为接线员人数。所有接线员忙线对应 X > k，因此需 P(X > k) < 0.05。查 λ = 3 的泊松累积表可得：P(X ≤ 5) = 0.9161，P(X ≤ 6) = 0.9665。因而 P(X > 5) = 1 − 0.9161 = 0.0839（>0.05），而 P(X > 6) = 1 − 0.9665 = 0.0335（<0.05）。因此至少需要6名接线员。
P(X ≤ r) = Σ (3ˣ e⁻³ / x!) from x=0 to r
- Poisson modelling assumes calls occur independently at a constant average rate.
- 泊松建模假设来电独立发生，且平均速率恒定。
- Sensitivity test: if the rate were 3.5 calls/5 min, recalculate — real-world planning often includes a safety margin.
- 敏感性测试：若速率为3.5通/5分钟，重新计算——实际规划通常包含安全裕度。
5. Case 4: Planning a School Event with Critical Path Analysis | 案例四：用关键路径分析策划学校活动
A school is organising a charity fair. The activities, dependencies and durations (in hours) are shown below. Draw an activity network, find the critical path and state the minimum project completion time.
某学校正在筹办一场慈善展。活动、依赖关系及时长（小时）如下表所示。绘制活动网络图，找出关键路径，并给出最短项目完成时间。
| Activity | Duration | Predecessors |
| A – Book venue | 4 | – |
| B – Hire stalls | 3 | A |
| C – Order supplies | 5 | A |
| D – Set up games | 2 | B, C |
| E – Decorate | 6 | B |
| F – Final checks | 1 | D, E |
Using a precedence diagram, the earliest start times are calculated. The path A(4) → C(5) → D(2) → F(1) gives a total of 12 hours, while A → B → E → F gives 4 + 3 + 6 + 1 = 14 hours. The critical path is A – B – E – F with a minimum completion time of 14 hours. Any delay in these activities delays the whole project.
利用前导网络图计算最早开始时间。路径 A(4) → C(5) → D(2) → F(1) 总计12小时，而 A → B → E → F 为 4 + 3 + 6 + 1 = 14 小时。关键路径为 A – B – E – F，最短完成时间为14小时。这些活动中任何一项延迟都将延误整个项目。
- Always double-check that dependencies are correctly mapped; a missing predecessor invalidates the network.
- 务必复核依赖关系是否已正确映射；遗漏前任活动会使网络失效。
- Float times: activities C and D have total float of 2 hours, meaning they can be delayed without affecting the overall deadline.
- 浮动时间：活动 C 和 D 的总时差为2小时，意味着它们可以延迟而不影响整体截止时间。
6. Case 5: Analysing an AC Circuit with Complex Numbers | 案例五：用复数分析交流电路
An alternating current circuit consists of a resistor R = 10 Ω and an inductor with reactance XL = 15 Ω in series. The supply voltage is v = 230 sin(100πt) volts. Using complex notation, the impedance Z = R + iXL = 10 + 15i Ω. Find the magnitude of the current and the phase angle between voltage and current.
某交流电路由电阻 R = 10 Ω 与感抗 XL = 15 Ω 串联而成。电源电压为 v = 230 sin(100πt) 伏。采用复数表示法，阻抗 Z = R + iXL = 10 + 15i Ω。求电流幅值以及电压与电流之间的相位角。
The peak voltage is V0 = 230 V. In complex form, voltage is represented as V = 230 + 0i (taking reference phase zero). The complex current I = V / Z = 230 / (10 + 15i). Multiply numerator and denominator by the conjugate 10 − 15i to obtain I = 230 × (10 − 15i) / (10² + 15²) = (2300 − 3450i) / 325 = 7.08 − 10.62i (approx). The magnitude |I| = √(7.08² + 10.62²) ≈ 12.7 A, and the phase angle φ = arctan(−10.62 / 7.08) ≈ −56.3°. The negative sign shows the current lags the voltage, consistent with an inductive circuit.
峰值电压 V0 = 230 V。复数形式下电压可表示为 V = 230 + 0i（以零相位为参考）。复电流 I = V / Z = 230 / (10 + 15i)。分子分母同乘共轭复数 10 − 15i，得 I = 230 × (10 − 15i) / (10² + 15²) = (2300 − 3450i) / 325 ≈ 7.08 − 10.62i。幅值 |I| = √(7.08² + 10.62²) ≈ 12.7 A，相位角 φ = arctan(−10.62 / 7.08) ≈ −56.3°。负号表明电流滞后于电压，这与电感电路的特性一致。
I = V / Z, |I| = |V| / |Z|, φ = −arctan(XL / R)
- In Further Pure, the imaginary unit is denoted by i, although engineers often use j. Exam questions will stick with i.
- 在进阶纯数中虚数单位用 i 表示，尽管工程师常用 j。考试题将统一使用 i。
- For a series R-L circuit, |Z| = √(R² + XL²). Always use the conjugate to divide complex numbers cleanly.
- 对于 R-L 串联电路，|Z| = √(R² + XL²)。务必用共轭复数进行整洁的复数除法。
7. Case 6: Saving for University using Summation of Series | 案例六：用级数求大学储蓄计划
A family saves for university by depositing £2,000 at the end of each year into an account paying 4% annual compound interest. What will be the total amount immediately after the 10th deposit? Use the sum of a geometric progression to express the total.
一个家庭为大学储蓄，在每年年末存入2000英镑，账户年复利4%。第10次存款后立即计算总金额。用等比数列的求和公式表示总额。
The first deposit grows for 9 years, the second for 8 years, …, the tenth for 0 years. Total S = 2000 × (1.04⁹ + 1.04⁸ + … + 1.04⁰). This is a geometric series with first term a = 1, common ratio r = 1.04, and n = 10 terms. Sum = a(rⁿ − 1) / (r − 1) = (1.04¹⁰ − 1) / 0.04 ≈ (1.48024 − 1)/0.04 = 12.0061. Hence S ≈ 2000 × 12.0061 = £24,012.20.
第一笔存款增长9年，第二笔增长8年，…，第十笔增长0年。总额 S = 2000 × (1.04⁹ + 1.04⁸ + … + 1.04⁰)。这是一个等比级数，首项 a = 1，公比 r = 1.04，项数 n = 10。和 = a(r
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