Pre-U WJEC Further Mathematics: Case Study Walkthrough | Pre-U WJEC 进阶数学：案例分析实战演练

📚 Pre-U WJEC Further Mathematics: Case Study Walkthrough | Pre-U WJEC 进阶数学：案例分析实战演练

In the WJEC Pre-U Further Mathematics examination, applying abstract concepts to unfamiliar, multi-step problems is a skill that separates good candidates from excellent ones. This article presents a series of case studies—real exam-style questions—walked through step by step. Each case targets a key area from the specification: complex numbers, differential equations, matrices, hyperbolic functions, polar coordinates, series and vectors. By working through these examples, you will learn to decode problem statements, select appropriate techniques, structure logical solutions and avoid the most frequent pitfalls.

在 WJEC Pre-U 进阶数学考试中，将抽象概念应用于陌生的、多步骤的问题，是区分优秀考生与卓越考生的关键能力。本文提供一系列案例研究——真实的考试风格题目，并逐步进行演练。每个案例都针对考试大纲中的一个核心领域：复数、微分方程、矩阵、双曲函数、极坐标、级数和向量。通过逐一练习，你将学会解读问题要求、选择合适的解题方法、组织合乎逻辑的解答，并避开最常见的错误。

1. Approaching an Unseen Problem | 面对陌生问题的方法

Before diving into a problem, scan the command words and underline the given quantities and conditions. Identify the topic by spotting key phrases: ‘modulus and argument’ signals complex numbers; ‘rate of change’ points to differential equations; ‘area enclosed’ suggests polar coordinates. Always sketch a diagram where geometry is involved, and write down a plan that breaks the task into smaller, manageable stages.

在动手解题之前，先扫读指令词，并在已知量和条件下面划线。通过关键短语识别主题：”模和辐角” 往往意味着复数；”变化率” 指向微分方程；”所围面积” 提示极坐标。凡是涉及几何，都要画出示意图，并写下解题计划，将任务分解为几个更小、更易处理的阶段。

2. Complex Numbers: Solving z⁴ = −8 + 8√3 i | 复数：解方程 z⁴ = −8 + 8√3 i

First, express the right-hand side in polar form. Compute the modulus: |−8 + 8√3 i| = √( (−8)² + (8√3)² ) = √(64 + 192) = √256 = 16. The argument θ lies in the second quadrant: tan⁻¹(|8√3 / 8|) = π/3, so the principal value is π − π/3 = 2π/3. Hence, −8 + 8√3 i = 16 [cos(2π/3) + i sin(2π/3)].

首先，将方程右边化为极坐标形式。计算模长：|−8 + 8√3 i| = √( (−8)² + (8√3)² ) = √(64 + 192) = √256 = 16。辐角 θ 位于第二象限：tan⁻¹(|8√3 / 8|) = π/3，因此主值为 π − π/3 = 2π/3。于是 −8 + 8√3 i = 16 [cos(2π/3) + i sin(2π/3)]。

Using de Moivre’s theorem, the fourth roots are given by z = 16^(1/4) { cos[(2π/3 + 2kπ)/4] + i sin[(2π/3 + 2kπ)/4] } = 2 { cos[(π/6) + (kπ/2)] + i sin[(π/6) + (kπ/2)] } for k = 0, 1, 2, 3. Evaluate each root explicitly: for k=0, z₀ = 2(cos π/6 + i sin π/6) = 2(√3/2 + i/2) = √3 + i. Similarly, k=1 gives z₁ = 2(cos 2π/3 + i sin 2π/3) = 2(−1/2 + i√3/2) = −1 + i√3. k=2 gives z₂ = 2(cos 7π/6 + i sin 7π/6) = −√3 − i, and k=3 yields z₃ = 2(cos 5π/3 + i sin 5π/3) = 1 − i√3.

应用棣莫弗定理，四个四次方根为 z = 16^(1/4) { cos[(2π/3 + 2kπ)/4] + i sin[(2π/3 + 2kπ)/4] } = 2 { cos[(π/6) + (kπ/2)] + i sin[(π/6) + (kπ/2)] }，其中 k = 0, 1, 2, 3。逐一求出各根：当 k=0 时，z₀ = 2(cos π/6 + i sin π/6) = 2(√3/2 + i/2) = √3 + i。类似地，k=1 得 z₁ = 2(cos 2π/3 + i sin 2π/3) = −1 + i√3；k=2 得 z₂ = 2(cos 7π/6 + i sin 7π/6) = −√3 − i；k=3 得 z₃ = 2(cos 5π/3 + i sin 5π/3) = 1 − i√3。

Common mistake: forgetting to add 2kπ before dividing the argument. Always plot the roots on an Argand diagram to verify that they are symmetrically spaced and have equal modulus. Here, all roots lie on a circle of radius 2, separated by π/2.

常见错误：在除以辐角之前忘记加上 2kπ。务必将这些根画在阿尔冈图上，检验它们是否呈对称分布且模长相等。此处，所有根都位于半径为 2 的圆周上，辐角间隔为 π/2。

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

A cup of coffee at 90 °C is placed in a room at 20 °C. Its temperature T(t) satisfies dT/dt = −k (T − 20), where t is in minutes. After 5 minutes the coffee has cooled to 60 °C. Find the temperature after 10 minutes, and the time taken to reach 30 °C.

一杯 90 °C 的咖啡放在 20 °C 的房间里。其温度 T(t) 满足 dT/dt = −k (T − 20)，t 以分钟计。5 分钟后咖啡降温至 60 °C。求 10 分钟后的温度，以及降至 30 °C 所需的时间。

This is a separable first-order ODE. Let y = T − 20, then dy/dt = −k y ⇒ ∫ (1/y) dy = −k ∫ dt, giving ln|y| = −kt + C. Exponentiating yields T = 20 + A e^(−kt), where A = e^C. Use the initial condition T(0)=90 to find A = 70. So T = 20 + 70 e^(−kt). Substitute T(5)=60: 60 = 20 + 70 e^(−5k) ⇒ e^(−5k) = 40/70 = 4/7 ⇒ k = −(1/5) ln(4/7) = (1/5) ln(7/4).

这是一个可分离的一阶常微分方程。令 y = T − 20，则 dy/dt = −k y，分离变量得 ∫ (1/y) dy = −k ∫ dt，故 ln|y| = −kt + C。指数化后得到 T = 20 + A e^(−kt)，其中 A = e^C。利用初始条件 T(0)=90 求出 A = 70，故 T = 20 + 70 e^(−kt)。代入 T(5)=60：60 = 20 + 70 e^(−5k) ⇒ e^(−5k) = 40/70 = 4/7 ⇒ k = −(1/5) ln(4/7) = (1/5) ln(7/4)。

Now T(10) = 20 + 70 e^(−10k) = 20 + 70 (e^(−5k))² = 20 + 70 (4/7)² = 20 + 70 × 16/49 = 20 + 1120/49 ≈ 20 + 22.857 = 42.86 °C. For T=30: 30 = 20 + 70 e^(−kt) ⇒ e^(−kt) = 1/7 ⇒ −kt = ln(1/7) ⇒ t = ln(7)/k = ln(7) / [ (1/5) ln(7/4) ] = 5 ln(7) / ln(7/4) ≈ 5 × 1.9459 / 0.5596 ≈ 17.4 minutes.

现在计算 T(10) = 20 + 70 e^(−10k) = 20 + 70 (e^(−5k))² = 20 + 70 (4/7)² = 20 + 70 × 16/49 = 20 + 1120/49 ≈ 20 + 22.857 = 42.86 °C。对于 T=30：30 = 20 + 70 e^(−kt) ⇒ e^(−kt) = 1/7 ⇒ −kt = ln(1/7) ⇒ t = ln(7)/k = ln(7) / [ (1/5) ln(7/4) ] = 5 ln(7) / ln(7/4) ≈ 5 × 1.9459 / 0.5596 ≈ 17.4 分钟。

4. Matrices: Solving Linear Systems and Geometric Interpretation | 矩阵：求解线性方程组及其几何意义

Consider the system x + 2y + 3z = 5, 2x + 5y + 3z = 3, x + 8z = 17. Form the augmented matrix and use row operations to find the values of x, y, z. Then comment on the geometrical configuration of the three planes.

考虑方程组 x + 2y + 3z = 5, 2x + 5y + 3z = 3, x + 8z = 17。写出增广矩阵并通过行变换求出 x, y, z 的值，然后阐述这三个平面的几何构型。

Write the coefficient matrix A and vector b, then perform Gaussian elimination. The first pivot row is (1, 2, 3 | 5). Eliminate x from row 2: R₂ − 2R₁ → (0, 1, –3 | –7). Eliminate x from row 3: R₃ − R₁ → (0, –2, 5 | 12). Next, eliminate y from row 3 using the new pivot in row 2: R₃ + 2R₂ → (0, 0, –1 | –2). This gives –z = –2 ⇒ z = 2. Back-substitute: from row 2, y – 6 = –7 ⇒ y = –1; from row 1, x – 2 + 6 = 5 ⇒ x = 1. The unique solution is (1, –1, 2). The three planes intersect at a single point, so they form a sheaf but with a common point.

写出系数矩阵 A 与向量 b，然后进行高斯消元。第一个主元行为 (1, 2, 3 | 5)。消去第二行的 x：R₂ − 2R₁ → (0, 1, –3 | –7)。消去第三行的 x：R₃ − R₁ → (0, –2, 5 | 12)。接着，用第二行的新主元消去第三行的 y：R₃ + 2R₂ → (0, 0, –1 | –2)。得到 –z = –2 ⇒ z = 2。回代得：从第二行，y – 6 = –7 ⇒ y = –1；从第一行，x – 2 + 6 = 5 ⇒ x = 1。唯一解为 (1, –1, 2)，三个平面交于一点。

A common pitfall is misinterpreting an all-zero row. If the last row had been 0 = 1, the system would be inconsistent—the planes would not share a common point. Also, note that the determinant of A is –1 (non-zero), confirming a unique point of intersection.

常见的易错点是误读全零行。若最后一行变成 0 = 1，则方程组不相容，意味着平面没有公共交点。此外，注意到 A 的行列式为 –1（非零），这证实了存在唯一的交点。

5. Hyperbolic Functions: Integration and Osborn’s Rule | 双曲函数：积分与奥斯本法则

Evaluate ∫ sinh²x dx. Many students reach for the identity sinh²x = (cosh 2x − 1)/2, which mirrors the trigonometric identity sin²θ = (1 − cos 2θ)/2 but with a sign change. This is a direct consequence of Osborn’s rule: replace cos²θ by cosh²x and change the sign of any product or implied product of two sines.

计算 ∫ sinh²x dx。许多学生会直接想到恒等式 sinh²x = (cosh 2x − 1)/2，这与三角恒等式 sin²θ = (1 − cos 2θ)/2 相似，但符号不同。这是奥斯本法则的直接推论：将 cos²θ 换成 cosh²x，并将任意两个正弦的乘积或隐含乘积的符号反转。

Using the identity, ∫ sinh²x dx = ½ ∫ (cosh 2x − 1) dx = ½ ( (1/2) sinh 2x − x ) + C = ¼ sinh 2x − ½ x + C. An alternative approach is integration by parts: let u = sinh x, dv = sinh x dx, leading to the same result via the identity cosh²x − sinh²x = 1. Always check by differentiating your answer.

利用该恒等式，∫ sinh²x dx = ½ ∫ (cosh 2x − 1) dx = ½ ( (1/2) sinh 2x − x ) + C = ¼ sinh 2x − ½ x + C。另一种方法是分部积分：令 u = sinh x，dv = sinh x dx，并利用 cosh²x − sinh²x = 1 即可得到相同结果。务必通过求导来检验答案。

When dealing with hyperbolic integrals, keep a table of key identities handy: