📚 Pre-U CAIE Mathematics Case Study: Practical Exam Techniques | Pre-U CAIE 数学：案例分析实战演练
Case study questions in the CAIE Pre-U Mathematics examination require you to apply a wide range of mathematical skills to a single extended scenario. They test not only your computational ability but also your capacity to model, interpret, and communicate mathematical reasoning clearly. This article presents practical strategies for tackling case studies effectively, using real-world inspired examples that mirror the style of Paper 3 and Paper 6 tasks. Whether you are optimizing a design, modelling population growth, or analysing data, these techniques will help you structure your solutions and avoid common mistakes.
在 CAIE Pre-U 数学考试中，案例分析题要求你将多种数学技能综合运用于一个复杂的现实情境中。它不仅考查计算能力，更检验你建模、解释和清晰表达数学推理的能力。本文提供实战技巧，通过贴近真实考试风格的案例，帮助你高效应对 Paper 3 和 Paper 6 中的类似任务。无论是优化设计方案、建立人口增长模型，还是分析统计数据，这些方法都能帮助你理清思路，避免易犯的错误。
1. Deconstructing the Problem | 拆解问题
Begin by reading the entire case study, not just the part you are about to answer. Identify the key quantities given, the unknowns to be found, and any constraints stated in words. Underline numerical data and conditions like ‘the rate of change is proportional to…’ or ‘the container must hold exactly 500 cm³’. Drawing a sketch that labels all relevant lengths, angles, and forces is almost always beneficial.
先通读整个案例，而不仅是你马上要回答的部分。找出现有的关键量和待求的未知量，并标记出文字中表述的约束条件，例如’变化率与……成正比’或’容器必须恰好容纳 500 cm³’。画出示意图，标出所有相关的长度、角度和力，这几乎总是有益的做法。
Example A rectangular tank with an open top is to be constructed from sheet metal. The base must be a square of side x cm, and the height h cm is such that the volume is 4000 cm³. The cost of material for the base is twice that for the sides. You need to find the dimensions that minimise the total cost.
示例 一个开口的长方体水箱由金属板制成，底面是边长为 x cm 的正方形，高度为 h cm，容积要求为 4000 cm³。底面的材料成本是侧面的两倍。你需要求出使总成本最低的尺寸。
After sketching, list the knowns: V = x²h = 4000, cost per cm² for sides = k, for base = 2k. The total cost C = (area of base)×2k + (area of four sides)×k = 2k x² + k(4xh). The objective is to minimise C subject to V = 4000.
画图后列出已知信息：V = x²h = 4000，侧面的单位面积成本为 k，底面成本为 2k。总成本 C = 底面积 × 2k + 四个侧面积 × k = 2k x² + k(4xh)。目标是在 V = 4000 的约束下使 C 最小。
2. Defining Variables and Stating Assumptions | 定义变量与明确假设
Explicitly define every variable you introduce. Use a short sentence: ‘Let x be the side length of the square base (cm) and h be the height (cm).’ Distinguish between constants and variables. Also list any assumptions you make, such as ‘the metal has negligible thickness’, ‘welding costs are ignored’, or ‘the tank is a perfect cuboid’. In statistical case studies, assumptions might include ‘the data are normally distributed’ or ‘samples are independent’.
明确地定义你引入的每一个变量。像这样用简短语句表述：’设 x 为正方形底面的边长（cm），h 为高度（cm）。’ 区分常量与变量。同时列出你所做的任何假设，例如’金属板的厚度可以忽略不计’、’焊接成本忽略’或’水箱是一个完美的长方体’。在统计学案例中，假设可能包括’数据服从正态分布’或’样本相互独立’等。
For the tank problem, you would write: ‘Assume the tank is made of uniform sheet metal with cost per unit area for the sides being £k per cm², and for the base £2k per cm². No material is wasted. The thickness of the metal is negligible. The volume is exactly 4000 cm³.’
针对水箱问题，你可以这样写：’假设水箱由材质均匀的金属板制成，侧面单位面积成本为 £k/cm²，底面为 £2k/cm²。材料无浪费。金属的厚度可忽略。容积恰好为 4000 cm³。’
These assumptions are critical because later you will need to comment on limitations. Without them, the model is incomplete.
这些假设至关重要，因为后续你可能需要对模型的局限性进行评论。如果没有这些假设，模型是不完整的。
3. Building a Mathematical Model | 构建数学模型
Translate the physical situation into equations. With the tank, use the volume constraint to eliminate h: 4000 = x²h ⇒ h = 4000/x². Substitute this into the cost function: C(x) = 2k x² + 4k x · (4000/x²) = 2k x² + 16000k / x. Since k is a positive constant, minimising C is equivalent to minimising f(x) = 2x² + 16000/x (for x > 0). This is the model to be optimised.
将物理情境转化为数学方程。在水箱问题中，用体积约束消去 h：4000 = x²h ⇒ h = 4000/x²。代入成本函数：C(x) = 2k x² + 4k x·(4000/x²) = 2k x² + 16000k / x。因为 k 是一个正常数，最小化 C 等价于最小化 f(x) = 2x² + 16000/x（x > 0）。这就是待优化的数学模型。
In modelling, always check that the domain of the independent variable makes sense. Here x must be positive, and h = 4000/x² must be a reasonable height for a tank. This domain awareness prevents algebraic solutions that are mathematically valid but physically impossible.
在建模过程中，务必检查自变量的定义域是否有实际意义。这里 x 必须为正数，且 h = 4000/x² 作为一个水箱的高度必须合理。这种定义域意识可以避免那些在数学上正确但在物理上不可能的代数解。
4. Solving Equations and Inequalities Strategically | 策略性求解方程与不等式
Once the model is set up, solving often reduces to algebraic manipulation and calculus. For the tank, differentiate f(x): f ‘(x) = 4x − 16000/x². Set f ‘(x) = 0: 4x = 16000/x² ⇒ 4x³ = 16000 ⇒ x³ = 4000 ⇒ x = ³√4000 = ³√(4000). Compute: ³√4000 ≈ 15.87 cm. Then h = 4000/x² ≈ 4000/(15.87²) ≈ 15.87 cm as well. Check the second derivative to confirm a minimum: f ”(x) = 4 + 32000/x³ > 0 for x > 0, so it is a minimum.
模型建立后，求解往往归结为代数运算和微积分。对水箱问题，求导 f(x)：f ‘(x) = 4x − 16000/x²。令 f ‘(x) = 0：4x = 16000/x² ⇒ 4x³ = 16000 ⇒ x³ = 4000 ⇒ x = ³√4000 ≈ 15.87 cm。此时 h = 4000/x² ≈ 4000/(15.87²) ≈ 15.87 cm。通过二阶导数验证其为最小值：f ”(x) = 4 + 32000/x³ > 0 (x > 0)，故确为最小值。
In longer case studies, you might encounter systems of equations or transcendental equations. Use clear logical steps, and if a numerical method is required, choose an appropriate one such as Newton-Raphson. Always state the iteration formula and initial guess.
在较长的案例分析中，你可能会遇到方程组或超越方程。采用清晰的逻辑步骤，若需要数值方法，应选择合适的方法，例如牛顿-拉夫森法。务必写出迭代公式和初始猜测值。
xₙ₊₁ = xₙ − f(xₙ)/f ‘(xₙ)
Label your working so the examiner can follow your reasoning easily.
标记你的计算步骤，以便考官能够轻松理解你的推理过程。
5. Differentiation and Optimisation in Context | 微分与实际优化
Optimisation is one of the most common themes in Pre-U case studies. Beyond stationary points, you must consider boundary behaviour. Sometimes the maximum or minimum occurs at an endpoint of the feasible domain. Always test endpoints when the domain is closed. Also, interpret the optimal values back in the context of the problem: state dimensions to a sensible degree of accuracy, and if necessary, round up or down according to practical constraints.
优化是 Pre-U 案例分析中最常见的主题之一。除了寻找驻点，你还必须考虑边界情况。有时最大值或最小值出现在可行域的端点。当定义域为闭区间时，务必验算端点值。同时，要将最优值放回原问题情境中解释：以合理的精确度给出尺寸，如有实际限制，需向上或向下取整。
For the tank, the optimum is x = h ≈ 15.9 cm (to 3 s.f.). If sheet metal comes in widths of 30 cm, you might comment that this is a practical size. If the cost per unit area was a more complex function of x, you would still follow the same differentiation process.
对于水箱，最优解为 x = h ≈ 15.9 cm (三位有效数字)。若金属板的标准宽度为 30 cm，你可以说明这是一个可行的尺寸。假如单位面积成本是 x 的某个更复杂函数，同样的求导流程依然适用。
6. Integration and Accumulated Quantities | 积分与累积量
Many case studies involve finding total change from a rate, a volume of revolution, or an area under a curve. For example, the rate at which water flows into a tank might be given as R(t) = 12/(t+1) litres per minute, and you need the total volume after 5 minutes. Set up the definite integral: V = ∫₀⁵ 12/(t+1) dt = 12 ln|t+1| from 0 to 5 = 12 ln6 − 12 ln1 = 12 ln6 litres.
许多案例分析涉及从速率求总变化，旋转体体积，或曲线下面积。例如，水流进水箱的速率给定为 R(t) = 12/(t+1) 升/分钟，你需要计算 5 分钟后的总水量。建立定积分：V = ∫₀⁵ 12/(t+1) dt = 12 ln|t+1| 从 0 到 5 = 12 ln6 − 12 ln1 = 12 ln6 升。
In such problems, always include the limits correctly and interpret the result. If the integral is used to find work done or distance travelled, check that units are consistent. Numerical integration rules like Simpson’s rule may be needed when an exact antiderivative is not available.
在这类问题中，务必正确写出积分上下限并解释结果。如果用积分求做功或位移，需检查单位是否一致。当无法找到精确反导数时，可能需要使用数值积分法，如辛普森法则。
∫ₐᵇ f(x) dx ≈ (h/3)[y₀ + 4y₁ + 2y₂ + … + 4yₙ₋₁ + yₙ], h = (b−a)/n
State the formula you are using and choose an even number of strips for Simpson’s rule.
写出你使用的公式，并选择偶数个条带应用辛普森法则。
7. Probability Distributions and Statistical Inference | 概率分布与统计推断
Case studies with data often require you to select and justify a probability model. Suppose a manufacturer claims that the breaking strength of cables is normally distributed with mean 2400 N and standard deviation 150 N. You are given a sample of 10 cables with mean 2310 N. Test at the 5% significance level whether the mean has reduced. State H₀: μ = 2400, H₁: μ < 2400. The test statistic is Z = (x̄ − μ₀)/(σ/√n) = (2310 − 2400)/(150/√10) ≈ −1.897. The critical value is −1.645. Since −1.897 < −1.645, reject H₀ and conclude the mean has decreased.
涉及数据的案例分析通常要求你选择并论证一个概率模型。假设制造商声称电缆的断裂强力服从正态分布，均值为 2400 N，标准差为 150 N。现有一组 10 根电缆的样本，均值为 2310 N。在 5% 显著水平下检验均值是否下降。设 H₀: μ = 2400，H₁: μ < 2400。检验统计量 Z = (x̄ − μ₀)/(σ/√n) = (2310 − 2400)/(150/√10) ≈ −1.897。临界值为 −1.645。由于 −1.897 < −1.645，拒绝 H₀，认为均值有所下降。
To earn full marks in such tasks, you must interpret the conclusion in plain language: ‘There is sufficient evidence at the 5% level to suggest that the mean breaking strength is now less than 2400 N.’ Next, you might be asked to calculate a confidence interval, combine distributions, or assess the effect of an outlier. Always comment on assumptions like normality and independence.
要在这种任务中获得满分，你必须用通俗语言解释结论：’在 5% 显著性水平下，有足够证据表明平均断裂强力已降低至 2400 N 以下。’ 接下来，题目可能会要求你计算置信区间、组合分布或评估异常值的影响。务必对正态性和独立性等假设加以说明。
8. Vector Methods and Geometric Modelling | 向量方法与几何建模
Vector case studies often describe moving objects or forces in equilibrium. For instance, two boats A and B have position vectors r_A = (2i + 3j) + t(i + 2j) and r_B = (5i + j) + s(−i + 3j). To find whether they collide, set r_A = r_B for some t, s. Equating components gives: 2 + t = 5 − s, 3 + 2t = 1 + 3s. Solve: t + s = 3, 2t − 3s = −2. Substituting s = 3 − t yields 2t − 9 + 3t = −2 ⇒ 5t = 7 ⇒ t = 1.4, s = 1.6. Since both t and s are positive, the boats could meet if these times are simultaneous; but unless t equals s, they do not collide because the time parameters are measured from different starting references. Clarify that collision requires the same position at the same time using a common time parameter if needed.
向量案例分析通常描述运动物体或平衡力系。例如，两艘船 A 和 B 的位置向量分别为 r_A = (2i + 3j) + t(i + 2j) 和 r_B = (5i + j) + s(−i + 3j)。要检验它们是否相撞，令 r_A = r_B 寻求 t, s。由分量相等得：2 + t = 5 − s，3 + 2t = 1 + 3s。求解：t + s = 3，2t − 3s = −2。代入 s = 3 − t 得 2t − 9 + 3t = −2 ⇒ 5t = 7 ⇒ t = 1.4, s = 1.6。由于 t 和 s 均为正，如果在同一时间度量下它们相等，则相撞；但因 t 和 s 是各自船出发后的时间参数，除非它们使用共同的时间起点且相等，否则不会真正相撞。需要根据题意澄清这一点。
Dot products can find angles between paths or forces. In a truss bridge problem, use equilibrium of forces to write vector equations, then solve for tensions. Keep vectors in component form and show clear algebraic steps.
点积可用于求路径或力之间的夹角。在桁架桥问题中，利用力的平衡列出向量方程，然后求解拉力。将向量保持为分量形式，并展示清晰的代数步骤。
9. Numerical Methods and Error Analysis | 数值方法与误差分析
Some case studies demand numerical root-finding or iterative solutions where an algebraic approach is impossible. The Newton-Raphson method is a favourite. To solve e⁻ˣ = x, rewrite as f(x) = e⁻ˣ − x. The iteration is xₙ₊₁ = xₙ − (e⁻ˣₙ − xₙ)/(−e⁻ˣₙ − 1) = xₙ + (e⁻ˣₙ − xₙ)/(e⁻ˣₙ + 1). Starting with x₀ = 0.5 gives x₁ ≈ 0.5663, x₂ ≈ 0.5671, converging rapidly. Always show your working and round to the required accuracy.
有些案例分析需要数值求根或迭代解法，因为代数方法不可行。牛顿-拉夫森法是最常见的方法之一。为求解 e⁻ˣ = x，改写为 f(x) = e⁻ˣ − x。迭代公式为 xₙ₊₁ = xₙ − (e⁻ˣₙ − xₙ)/(−e⁻ˣₙ − 1) = xₙ + (e⁻ˣₙ − xₙ)/(e⁻ˣₙ + 1)。以 x₀ = 0.5 开始得 x₁ ≈ 0.5663，x₂ ≈ 0.5671，快速收敛。务必写出计算过程并按要求的精度取舍。
Error bounds can be estimated using the fact that the error halves approximately with each step in Newton-Raphson when near the root, or you can use the change in successive iterates. For a trapezium rule integration, the error E ≤ (b−a)³/(12n²) max|f ”(x)|. State the bound you used and comment on its reliability.
误差界可通过牛顿-拉夫森法在根附近的近似性质来估计（误差大约每步减半），或者用相邻迭代值之差来估算。对于梯形法求积，误差 E ≤ (b−a)³/(12n²) max|f ”(x)|。写出你所用的误差界，并讨论其可靠性。
10. Interpretation and Verification of Results | 结果解释与验证
After obtaining a solution, go back to the original problem and check whether it makes sense. Does the answer have the correct units? Is it within the expected range? Could there be multiple solutions, and if so, which one fits the context? In the tank example, x = h ≈ 15.9 cm gives a cube-like shape, which is intuitive for minimising surface area if the base cost is higher. Also, verify by substitution: f(15.9) = 2(15.9)² + 16000/15.9 ≈ 505.6 + 1006.3 ≈ 1511.9, while f(16) = 2(256) + 1000 = 1512; the function is indeed slightly higher. This quick sanity check builds confidence.
得到解后，回到原问题检查其是否合理。答案的单位是否正确？是否在预期范围内？是否可能存在多解？如果有，哪一个符合实际情境？在水箱示例中，x = h ≈ 15.9 cm 得到的形状近似正方体，这在底面成本较高时使总成本最小，合乎直觉。同时，通过代入验证：f(15.9) ≈ 1511.9，而 f(16) = 1512，函数值确实略高。这种快速检验能增强信心。
In statistical problems, compare the p-value to the significance level, but also comment on whether the sample size is adequate and if the test is valid. If the conclusion is to ‘accept’ the null hypothesis, phrase it carefully: ‘There is insufficient evidence to reject H₀ at the 5% level.’ Avoid saying ‘prove’ in statistics.
在统计问题中，要比较 p 值与显著水平，同时评论样本量是否足够、检验是否有效。如果结论是’接受’原假设，措辞要谨慎：’在 5% 水平下，没有足够证据拒绝 H₀。’ 在统计中避免使用’证明’一词。
11. Report Writing and Communication | 报告撰写与表达
In Pre-U case studies, especially those awarded marks for the quality of written communication, present your solution like a short technical report. Use clear headings, labelled diagrams, and a logical flow. Begin with a brief introduction restating the problem. State your assumptions in a bullet-like list. Use consistent notation. Separate mathematical working from commentary. End with a concise conclusion that answers the original question precisely.
在 Pre-U 案例分析中，尤其是那些对书面表达质量有评分的题目，你应该把解答写成一份简短的技术报告。使用清晰的标题、有标注的图示和清晰的逻辑流。开篇简要重述问题。用类似项目符号的方式列出假设。使用一致的符号。将数学演算与评论分开。以精确回答原问题的简明结论收尾。
When presenting a table of values or results, use a neatly formatted table with borders if possible. Labels and units must be included in the header. For example:
| x / cm | h / cm | C / pence |
|---|---|---|
| 15.0 | 17.8 | 1525 |
| 15.9 | 15.9 | 1512 |
| 17.0 | 13.8 | 1520 |
This approach helps the examiner see your systematic exploration and earns marks for presentation.
这种方式有助于考官看到你系统性的探索过程，并为表达部分赢得分数。
12. Common Pitfalls and Time Management | 常见陷阱与时间管理
Many candidates lose marks by rushing into calculations without a clear plan. Always spend the first few minutes reading and planning. A common mistake is forgetting to differentiate between a maximum and a minimum, or not checking the second derivative. Another is mishandling units – if the volume is given in cm³ and cost in £, your final cost should be in £. In probability, using the wrong standard error (population σ vs sample s) or misapplying a one-tailed test is frequent. Make a checklist: define variables, state assumptions, check domain, verify optimality, state conclusion.
许多考生因匆忙计算而缺乏清晰计划，从而失分。开头几分钟一定要用来阅读和规划。常见错误包括：未区分最大值与最小值、未检验二阶导数、忽略单位换算 —— 如果体积用 cm³ 而成本用 £，最终成本应以 £ 为单位。在概率题中，用错标准误（总体 σ 还是样本 s）或误用单尾检验也很常见。制作一个检查清单：定义变量、陈述假设、检查定义域、验证最优性、陈述结论。
Manage your time by allocating approximately 15–20 minutes for reading and modelling, 30–40 minutes for the main calculations, and 5–10 minutes for checking and writing the final conclusion. If a part seems too time-consuming, leave it and return later. Partial work often still gains marks. Keep your work legible and well-spaced.
通过合理分配时间来掌控全局：阅读和建模约 15–20 分钟，主要计算 30–40 分钟，最后 5–10 分钟用于检查并撰写结论。如果某一部分过于耗时，暂时放下，稍后回来再做。即使未完成，部分过程仍可得分。保持卷面清晰，行距适中。
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