📚 Year 12 CIE Maths: Case Study Practical Exercises | 案例分析实战演练
In Year 12 CIE Mathematics, case study questions bridge the gap between theoretical concepts and real-world problem solving. These exercises require you to apply your knowledge of pure mathematics, mechanics, or statistics to a given scenario, often involving multiple linked parts. Developing a systematic approach to such questions is essential for success in both school assessments and the final examination. This article will guide you through a selection of practical case studies, breaking down each one into clear steps and highlighting common pitfalls. You will learn how to extract mathematical models from a narrative, choose the correct technique, and present your reasoning in a way that meets CIE mark scheme requirements.
在Year 12 CIE数学课程中，案例分析题是连接理论概念与现实问题解决的桥梁。这些练习要求你将纯数学、力学或统计学的知识应用到一个给定的情景中，通常包含多个相互关联的部分。培养解决这类问题的系统方法对于在校内评估和最终考试中取得成功至关重要。本文将引导你完成一系列精选的实战案例，将每个案例分解为清晰的步骤，并指出常见的易错点。你将学会如何从叙述中提取数学模型，选择正确的解题技巧，并以符合CIE评分标准的方式呈现你的推理过程。
1. Understanding the Case Study Approach | 理解案例分析的方法
Case study questions in CIE Maths are designed to test your ability to translate a real-life situation into mathematical language. You will often be presented with a paragraph of text, a table of data, or a graph, and then asked a series of related questions. The key is to read the entire scenario carefully before attempting any calculations. Identify what the variables are, what information is given, and what assumptions you are allowed to make. Much like a detective, you need to spot clues: a phrase like “rate of change” suggests differentiation; “area under the curve” points to integration; “normally distributed” indicates a normal distribution model; and “test the claim” usually means a hypothesis test is required.
CIE数学的案例分析题旨在测试你将现实情境转化为数学语言的能力。题目通常会提供一段文字、一个数据表格或一张图表，然后要求你回答一系列相关问题。关键是在尝试任何计算之前，仔细阅读整个情景。确定变量是什么，给出了哪些信息，以及允许你做出的假设。就像侦探一样，你需要发现线索：比如“变化率”这样的措辞暗示需要使用微分； “曲线下的面积”指向积分； “正态分布”表明要使用正态分布模型；而“检验说法”通常意味着需要进行假设检验。
Always annotate the information provided. For instance, in an optimisation problem about fencing a field, highlight the fixed perimeter and the quantity to be maximised. Write down any formulas that connect the variables. This preliminary step transforms a wordy problem into a set of equations and constraints you can work with. Avoid the temptation to jump straight into algebra; a careful setup prevents mistakes later.
始终对提供的资料进行标注。例如，在一个关于围栏封地的优化问题中，突出显示固定的周长和需要最大化的量。写下连接各变量的任何公式。这个预备步骤将一个文字冗长的问题转化为你可以处理的一组方程和约束条件。避免直接跳入代数运算的冲动；仔细的设定过程可以防止之后出现错误。
2. Breaking Down Problems into Manageable Steps | 将问题分解为可管理的步骤
Once you have a clear picture of the scenario, break the question down into its individual parts. CIE examiners often design case studies so that earlier parts feed into later ones. For example, you might first be asked to find an expression for a function, then to differentiate it, and finally to interpret the result in context. Tackle each part in order, making sure your answer is labelled and clearly linked to the original problem. Use the word “so” or “therefore” to connect your mathematical statements to the real-world meaning.
一旦你对情景有了清晰的了解，就将问题分解为各个独立的部分。CIE考官通常设计案例分析题，使前面的部分为后面的部分提供铺垫。例如，你可能先被要求找出一个函数的表达式，然后对其进行微分，最后在给定背景下解释结果。按顺序处理每个部分，确保答案有标注，并清晰地与原始问题联系起来。使用“因此”或“所以”将你的数学陈述与现实世界的意义联系起来。
A helpful technique is to write a short plan before you compute. For a problem involving a falling object with air resistance, your plan might be: (1) write down the differential equation for velocity, (2) solve it using an integrating factor, (3) find the terminal velocity by letting t → ∞, (4) comment on whether the model is realistic. This structure keeps your working organised and makes it easier for the examiner to award marks for method, even if a numerical slip occurs.
一个有用的技巧是在计算前写下简短的计划。对于一个涉及空气阻力的落体问题，你的计划可能是：(1) 写出关于速度的微分方程，(2) 用积分因子求解，(3) 令 t → ∞ 求出终端速度，(4) 评论该模型是否真实。这种结构可以使你的解题过程有条理，并且即使发生数字计算失误，也能使考官更容易为你的过程打分。
3. Case 1: Optimisation in Business – Maximising Revenue | 案例一：商业中的最优化—收益最大化
A company sells x units of a product per week. The price per unit, p (in pounds), is given by the demand equation p = 500 − 0.1x. The weekly cost C, in pounds, is C = 1000 + 200x. The business wishes to maximise weekly profit. First, define the profit function: Profit, Π, is revenue minus cost. Revenue R = xp = x(500 − 0.1x) = 500x − 0.1x². Thus Π = (500x − 0.1x²) − (1000 + 200x) = 300x − 0.1x² − 1000. This is a quadratic opening downwards, so its maximum occurs at the vertex.
一家公司每周销售 x 件产品。每件产品单价 p（英镑）由需求方程 p = 500 − 0.1x 给出。每周成本 C（英镑）为 C = 1000 + 200x。企业希望最大化每周利润。首先，定义利润函数：利润 Π 是收入减去成本。收入 R = xp = x(500 − 0.1x) = 500x − 0.1x²。因此 Π = (500x − 0.1x²) − (1000 + 200x) = 300x − 0.1x² − 1000。这是一个开口向下的二次函数，其最大值出现在顶点。
Differentiate Π with respect to x: dΠ/dx = 300 − 0.2x. Set this equal to zero: 300 − 0.2x = 0 ⇒ x = 1500. To confirm it is a maximum, check the second derivative: d²Π/dx² = −0.2, which is negative. So the maximum profit occurs when 1500 units are produced per week. Substituting back, maximum profit Π = 300(1500) − 0.1(1500)² − 1000 = 224,000 (pounds). This case study illustrates how calculus converts a business question into a simple mathematical problem. In the CIE exam, you would be expected to comment on the limitations of the model, such as the assumption that the demand equation is linear.
对 Π 关于 x 进行微分：dΠ/dx = 300 − 0.2x。令导数为零：300 − 0.2x = 0 ⇒ x = 1500。为了确认这是最大值，检查二阶导数：d²Π/dx² = −0.2，为负数。因此当每周生产1500件时获得最大利润。代回原式，最大利润 Π = 300(1500) − 0.1(1500)² − 1000 = 224,000（英镑）。这个案例研究说明了微积分如何将一个商业问题转化为一个简单的数学问题。在CIE考试中，你还需要对模型的局限性进行评论，例如假定需求方程是线性的这一条件。
4. Case 2: Modelling Motion with Integration | 案例二：用积分建模运动
An object moves along a straight line with velocity v ms⁻¹ at time t seconds given by v = 6t² − 4t + 2. Initially, at t = 0, the object is 3 metres from a fixed point O. Find the displacement from O after 4 seconds, and determine the distance travelled in that interval. While displacement is found by a definite integral of velocity, distance travelled must account for any changes of direction. First, check whether the velocity changes sign between 0 and 4.
一个物体沿直线运动，在时间 t 秒时的速度 v（单位 m·s⁻¹）为 v = 6t² − 4t + 2。初始时刻 t = 0，物体距离固定点 O 为 3 米。求 4 秒后物体相对于 O 的位移，并确定该时间间隔内所通过的路程。位移可以通过速度的定积分求得，但路程必须考虑任何方向变化。首先，检查速度在 0 到 4 之间是否会改变符号。
Solve v = 0: 6t² − 4t + 2 = 0. Discriminant = (−4)² − 4(6)(2) = 16 − 48 = −32 < 0. Hence v(t) > 0 for all real t. There is no change of direction, so distance travelled equals the displacement. Displacement s(t) is given by s(t) = ∫v dt + C. s(t) = ∫(6t² − 4t + 2) dt = 2t³ − 2t² + 2t + C. Using s(0) = 3, we get C = 3, so s(t) = 2t³ − 2t² + 2t + 3. After 4 seconds, s(4) = 2(64) − 2(16) + 2(4) + 3 = 128 − 32 + 8 + 3 = 107 m. Thus the displacement is 107 m from O, and the distance travelled is also 107 m. In your answers, always include units and interpret the result in the context of the problem.
解 v = 0：6t² − 4t + 2 = 0。判别式 = (−4)² − 4(6)(2) = 16 − 48 = −32 < 0。因此对于所有实数 t，v(t) > 0。没有方向变化，所以通过的路程等于位移。位移 s(t) 由 s(t) = ∫v dt + C 给出。s(t) = ∫(6t² − 4t + 2) dt = 2t³ − 2t² + 2t + C。利用 s(0) = 3，得到 C = 3，所以 s(t) = 2t³ − 2t² + 2t + 3。4 秒后，s(4) = 2(64) − 2(16) + 2(4) + 3 = 128 − 32 + 8 + 3 = 107 m。因此位移为距 O 点 107 m，路程也是 107 m。在你的答案中，务必包含单位，并在问题背景下解释结果。
5. Case 3: Probability Distributions in Quality Control | 案例三：质量控制中的概率分布
A factory produces light bulbs whose lifetimes are normally distributed with mean 800 hours and standard deviation 40 hours. The quality control department tests a random sample of 25 bulbs. Find the probability that the sample mean lifetime is less than 790 hours. Here we apply the Central Limit Theorem: for a sample of size n, the sample mean X̄ follows N(μ, σ²/n). Thus X̄ ∼ N(800, 40²/25) = N(800, 64). The standard deviation of the sample mean is √64 = 8 hours.
一家工厂生产的灯泡寿命服从正态分布，均值为 800 小时，标准差为 40 小时。质检部门随机测试了 25 个灯泡的样本。求样本平均寿命低于 790 小时的概率。这里应用中心极限定理：对于样本量 n，样本均值 X̄ 服从 N(μ, σ²/n)。因此 X̄ ∼ N(800, 40²/25) = N(800, 64)。样本均值的标准差为 √64 = 8 小时。
Calculate the z-score: z = (790 − 800) / 8 = −10/8 = −1.25. Using the standard normal table, Φ(−1.25) = 1 − Φ(1.25) ≈ 1 − 0.8944 = 0.1056. So the probability that the sample mean is below 790 hours is about 10.6%. This kind of case study appears frequently in Statistics 1. Remember to check the assumptions: the population is normally distributed, so the sample mean is normal regardless of sample size. Always sketch a bell curve and shade the region of interest to avoid confusion between lower and upper tail probabilities.
计算 z 分数：z = (790 − 800) / 8 = −10/8 = −1.25。利用标准正态分布表，Φ(−1.25) = 1 − Φ(1.25) ≈ 1 − 0.8944 = 0.1056。因此样本均值低于 790 小时的概率约为 10.6%。这类案例分析在统计学1中经常出现。记得检查假设：总体服从正态分布，因此无论样本量大小，样本均值都服从正态分布。务必草绘一个钟形曲线，并为关心的区域涂上阴影，以避免混淆下尾概率和上尾概率。
6. Case 4: Hypothesis Testing for a New Drug | 案例四：新药的假设检验
A pharmaceutical company claims that a new drug is effective in 80% of patients. A doctor believes the true success rate is lower and conducts a trial with 30 patients, finding that 20 respond positively. Test at the 5% significance level whether there is evidence to support the doctor’s suspicion. This is a one-tailed binomial test with H₀: p = 0.8 against H₁: p < 0.8. Under H₀, X ∼ B(30, 0.8), where X is the number of successes. The observed value is x = 20. We need P(X ≤ 20 | p = 0.8).
一家制药公司声称一款新药对 80% 的患者有效。一位医生认为真实成功率更低，并对 30 名患者进行了试验，发现有 20 人出现积极效果。在 5% 的显著性水平下检验是否有证据支持医生的怀疑。这是一个单侧二项检验，H₀: p = 0.8 对 H₁: p < 0.8。在 H₀ 下，X ∼ B(30, 0.8)，其中 X 为成功数量。观测值为 x = 20。我们需要计算 P(X ≤ 20 | p = 0.8)。
Use the binomial formula or cumulative tables. P(X ≤ 20) = 1 − P(X ≥ 21). From tables, P(X ≥ 21) = 0.0995 (approximately), so the p-value = 0.0995. This is greater than 0.05, so we do not reject H₀. There is insufficient evidence at the 5% level to support the doctor’s claim. In a CIE answer, you must state your conclusion clearly in context: ‘There is not enough evidence to suggest the drug’s effectiveness is below 80%.’ Also mention that the result is based on the assumption that the sample is random and independent.
使用二项分布公式或累积表。P(X ≤ 20) = 1 − P(X ≥ 21)。查表，P(X ≥ 21) = 0.0995（大约），因此 p 值 = 0.0995。这大于 0.05，所以我们不拒绝 H₀。在 5% 的显著性水平下，没有足够证据支持医生的说法。在CIE的回答中，你必须清晰地结合背景陈述结论：“没有足够证据表明该药的有效性低于 80%。”同时也要提到这个结果基于样本是随机且独立的假设。
7. Case 5: Functions and Graphical Transformations | 案例五：函数与图像变换
Given the function f(x) = x² − 4x + 5, a case study might ask you to express it in completed square form, sketch the graph, and then apply a sequence of transformations to obtain the graph of y = 2f(x + 1) − 3. Completing the square: f(x) = (x − 2)² + 1. The vertex is at (2, 1), and it is a U-shaped parabola. This basic understanding helps when analysing composite transformations.
给定函数 f(x) = x² − 4x + 5，一个案例分析题可能要求你用配方法表示该函数，绘制草图，然后应用一系列变换以得到 y = 2f(x + 1) − 3 的图像。配方：f(x) = (x − 2)² + 1。顶点在 (2, 1)，是一个U形抛物线。这种基本理解有助于分析复合变换。
To transform from y = f(x) to y = 2f(x + 1) − 3, follow the order of operations: start with the inside bracket, then multiplication, then subtraction. The transformation x → x + 1 shifts the graph 1 unit to the left. The multiplication f → 2f stretches it vertically by a factor of 2. Finally, subtracting 3 shifts the whole graph down by 3 units. The vertex moves from (2, 1) to (2 − 1, 2×1 − 3) = (1, −1). Always list the transformations step by step and draw a quick sketch to confirm. In CIE questions, you may be asked to find the coordinates of the image of a specific point, making this a very practical case study.
要从 y = f(x) 变换到 y = 2f(x + 1) − 3，遵循运算顺序：从括号内开始，然后乘法，最后减法。变换 x → x + 1 将图像向左平移 1 个单位。乘法 f → 2f 使其在垂直方向拉伸为原来的 2 倍。最后，减去 3 将整个图像向下平移 3 个单位。顶点从 (2, 1) 移动到 (2 − 1, 2×1 − 3) = (1, −1)。务必逐步列出各个变换，并快速绘图确认。在CIE的题目中，你可能会被要求求某个特定点映射后的坐标，这使得这成为一个非常实用的案例研究。
8. Case 6: Connecting Calculus with Real-World Data | 案例六：将微积分与真实数据联系起来
Imagine a case where the marginal cost of producing a good is given by C'(x) = 0.02x + 5, and the fixed cost is £200. You need to find the total cost function and then the total cost of producing 100 items. Integrate the marginal cost: C(x) = ∫(0.02x + 5) dx = 0.01x² + 5x + k. Using the condition C(0) = 200 gives k = 200, so C(x) = 0.01x² + 5x + 200. For 100 items, C(100) = 0.01(10000) + 500 + 200 = £800. This is a typical integration case study.
设想一个案例，生产一种商品的边际成本由 C'(x) = 0.02x + 5 给出，固定成本为 200 英镑。你需要求出总成本函数，然后求出生产 100 件产品的总成本。对边际成本积分：C(x) = ∫(0.02x + 5) dx = 0.01x² + 5x + k。利用条件 C(0) = 200 得到 k = 200，因此 C(x) = 0.01x² + 5x + 200。对于 100 件产品，C(100) = 0.01(10000) + 500 + 200 = 800 英镑。这是一个典型的积分案例研究。
Then you might be asked to compute the consumer surplus given a demand function p = 50 − 0.1q and equilibrium price £30. Equilibrium quantity: 30 = 50 − 0.1q ⇒ q = 200. Consumer surplus = ∫₀²⁰⁰ (50 − 0.1q) dq − 30 × 200 = [50q − 0.05q²]₀²⁰⁰ − 6000 = (10000 − 2000) − 6000 = 2000. This real-life application of definite integrals demonstrates how mathematics supports economic decision-making. Present the integral limits clearly and evaluate step by step.
然后你可能会被要求计算给定需求函数 p = 50 − 0.1q 和均衡价格 30 英镑下的消费者剩余。均衡数量：30 = 50 − 0.1q ⇒ q = 200。消费者剩余 = ∫₀²⁰⁰ (50 − 0.1q) dq − 30 × 200 = [50q − 0.05q²]₀²⁰⁰ − 6000 = (10000 − 2000) − 6000 = 2000。这个定积分的现实应用展示了数学如何支持经济决策。清晰地写出积分限，并逐步求值。
9. Common Pitfalls and How to Avoid Them | 常见误区及避免方法
One of the most common mistakes in case study questions is failing to distinguish between displacement and distance. Always check for changes of sign in velocity before integrating. Another pitfall is forgetting to include the constant of integration when finding a function from its derivative; the initial condition is there for a reason. In statistics, mixing up the standard deviation of the population and the standard error of the mean leads to incorrect z-scores. Pay close attention to whether the question asks about a single observation or a sample mean.
在案例分析题中最常见的错误之一是没有区分位移和路程。务必在积分前检查速度符号的变化。另一个误区是在从导函数求原函数时忘记加上积分常数；初始条件的存在是有原因的。在统计学中，混淆总体标准差和均值的标准误会得出错误的 z 分数。要密切关注题目问的是关于单个观测值还是样本均值。
Also, when performing transformations of graphs, many students apply the horizontal shift in the wrong direction or reverse the order of operations. Remember: f(x + a) moves the graph left by a units if a > 0, and vertical stretch comes before vertical translation. Finally, in hypothesis testing, always write your conclusion in the context of the problem and include the significance level. Never just say ‘reject H₀’; explain what that means for the real-world claim being tested. Keeping these points in mind will dramatically improve your case study marks.
此外，在进行图像变换时，许多学生将水平平移方向搞反，或者颠倒了运算顺序。请记住：如果 a > 0，f(x + a) 会将图像向左平移 a 个单位，而垂直伸缩在垂直平移之前进行。最后，在假设检验中，务必在问题背景下写出你的结论，并包含显著性水平。绝不要只说“拒绝 H₀”；要解释这对被检验的现实说法意味着什么。牢记这几点将显著提高你的案例分析得分。
10. Exam Tips for CIE Case Study Questions | CIE案例分析题的应试技巧
When you open your CIE Maths paper, case study questions often appear towards the end of the structured section. Allow sufficient time to read and plan: at least 3-4 minutes of a 15-minute problem should be spent on understanding the stem. Use a highlighter to underline key numerical values and verbs like ‘show’, ‘find’, ‘prove’, or ‘determine’. In multi-part questions, even if you cannot answer an earlier part, you can still attempt later parts by using the given result; CIE often awards method marks for correct application of a wrong value.
当你打开CIE数学试卷时，案例分析题通常出现在结构化题目的靠后部分。要留出足够的时间阅读和计划：对于一道 15 分钟的题目，至少应花 3-4 分钟来理解题干。用荧光笔标出关键的数值和诸如“证明”、“求”、“证明”或“确定”等动词。在多个小问的题目中，即使你无法回答前面的部分，仍然可以尝试用给定的结果回答后续部分；CIE常常对错误数值的正确应用方法给予步骤分。
Present your working logically, with clear labelling of each step. If the question asks ‘Comment on the validity of the model’, prepare to mention assumptions such as no air resistance, constant demand, or random sampling. Use technical vocabulary correctly: ‘rate of change’, ‘stationary point’, ‘significance level’, etc. Finally, check your answer: does it make sense in the context? Could a probability be greater than 1? Could a length be negative? A quick sense-check can catch many errors.
有条理地呈现你的解题过程，并清楚地标注每一步。如果题目要求“评论模型的有效性”，准备好提及诸如无空气阻力、需求不变或随机抽样等假设。正确使用技术词汇：“变化率”、“驻点”、“显著性水平”等。最后，检查你的答案：在相关背景下它是否合理？概率可能大于 1 吗？长度可能为负吗？快速进行一次常识性检查就能发现许多错误。
11. Practice Makes Perfect: Building Your Own Cases | 熟能生巧：构建你自己的案例
An excellent way to master case studies is to create your own. Take a real-world scenario—like the cooling of a cup of coffee, the spread of a social media post, or the trajectory of a basketball shot—and try to model it mathematically. For example, use Newton’s law of cooling: dT/dt = −k(T − Tₐ), and solve the differential equation. Anticipate what questions an examiner might ask: find the time to reach a certain temperature, or explain why the model breaks down after a long time. This active practice deepens your understanding far beyond textbook exercises.
掌握案例分析的一个绝佳方法是创建你自己的案例。选取一个真实世界的情景——比如一杯咖啡的冷却、一条社交媒体帖子的传播或篮球投篮的轨迹——并尝试用数学进行建模。例如，使用牛顿冷却定律：dT/dt = −k(T − Tₐ)，并求解微分方程。预想考官可能会问哪些问题：求达到某特定温度所需的时间，或者解释为什么该模型在长时间后失效。这种主动练习远胜过课本练习，能加深你的理解。
Swap your self-made case studies with classmates and try to solve each other’s problems. This will expose you to different mathematical topics and improve your ability to interpret a variety of descriptions. Over time, you will notice that the underlying structure of CIE case study questions is quite predictable: set up the model, perform a calculation, interpret the result, and critique the model. Once you have internalised this framework, you can tackle any problem with confidence.
与同学交换你自己创编的案例分析题，并尝试互相解答。这将使你接触到不同的数学主题，并提高你解释各种描述的能力。随着时间的推移，你会注意到CIE案例分析题的基本结构是相当可预测的：建立模型、执行计算、解释结果、评论模型。一旦你内化了这个框架，你就能自信地处理任何问题。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导