Pre-U OCR Mathematics: Case Study Practical Drills | Pre-U OCR 数学：案例分析实战演练

📚 Pre-U OCR Mathematics: Case Study Practical Drills | Pre-U OCR 数学：案例分析实战演练

Case study questions lie at the heart of the Pre-U OCR Mathematics assessment, challenging students to apply a blend of pure, mechanics and statistics skills in unfamiliar contexts. This article provides a structured set of practical drills, each breaking down a typical case scenario into manageable steps, complete with full solutions and commentary. By working through these examples, you will sharpen your problem-solving toolkit, learn how to interpret complex prompts, and develop the confidence to tackle any exam case study.

案例分析题是 Pre-U OCR 数学考试的核心，要求学生在陌生情境中综合运用纯数学、力学和统计学的技能。本文提供了一套结构化的实战训练，每个训练都将一个典型情景拆解为可操作的步骤，并附有完整的解答和评注。通过这些例子，你将强化解题工具箱，学会如何解读复杂的题意，并建立起应对任何考试案例分析的信心。

1. Understanding the Pre-U OCR Mathematics Syllabus | 了解 Pre-U OCR 数学大纲

The Pre-U Mathematics qualification is composed of three compulsory components: Pure Mathematics with Statistics, Pure Mathematics with Mechanics, and a Comprehension paper. Within each, case study questions often integrate multiple topics, such as calculus with Newtonian motion, or probability with algebraic manipulation. Familiarity with the syllabus content – from hyperbolic functions to hypothesis testing – is essential before diving into drills.

Pre-U 数学资格证书由三个必修部分组成：纯数学与统计、纯数学与力学，以及阅读理解卷。其中，案例分析题经常综合多个主题，例如将微积分与牛顿运动定律结合，或概率与代数运算结合。在深入训练之前，熟悉大纲内容——从双曲函数到假设检验——是必不可少的。

The OCR specification emphasises modelling cycles, proof, and precise communication. Examiners look for logical flow, correct notation, and the ability to verify results. In the drills that follow, we will mirror this expectation by showing not only answers but also the reasoning behind each step.

OCR 的考纲强调建模循环、证明以及精确的表述。评分官看重逻辑流畅、符号正确以及验证结果的能力。在后续的操练中，我们将通过展示每一个步骤背后的推理来体现这一要求。

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2. Case Study Structure and Marking Criteria | 案例分析结构与评分标准

A typical Pre-U case study begins with a real-world scenario, followed by a series of guided questions. These questions gradually build a mathematical model, then ask for interpretations, refinements, or critical evaluation. Marks are awarded for method, accuracy, and clarity. Even if a final answer is wrong, a well-explained method can score significantly.

典型的 Pre-U 案例分析题以一个真实世界的情景开始，随后给出一系列引导性问题。这些问题逐步构建数学模型，然后要求进行解释、改进或批判性评估。评分依据方法、准确性和清晰度。即使最终答案错误，解释清楚的方法也能获得可观的分数。

In your practice, always write down the modelling assumptions you are making, define variables explicitly, and comment on the appropriateness of the model at the end. This shows the examiner you understand the bigger picture. Our drills will follow this template: problem statement, step-by-step solution, and reflective commentary.

在练习中，一定要写下你所做的建模假设，明确定义变量，并在最后评论模型的适用性。这向考官表明你理解了全局。我们的训练将遵循这一模板：问题陈述、逐步求解和反思性评注。

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3. Drills on Pure Mathematics: Algebraic Manipulation | 纯数学训练：代数运算

Problem: exact coordinates of the turning points of the curve y = P(x) and determine their nature.

问题：多项式 P(x) = x³ - 6x² + 11x - 6 经过点 (1,0), (2,0) 和 (3,0)。利用因式分解求出曲线 y = P(x) 的驻点精确坐标，并判断其性质。

-1)(x-2) = x² - 3x + 2, then multiply by (x-3) giving x³ - 6x² + 11x - 6. Differentiate: P'(x) = 3x² - 12x + 11. Set P'(x) = 0 → 3x² - 12x + 11 = 0. The quadratic formula gives x = [12 ± √144 - 132]/6 = [12 ± √12]/6 = [12 ± 2√3]/6 = 2 ± ½√3.

由于已知零点，我们可以写成 P(x) = (x-1)(x-2)(x-3)。展开验证系数：(x-1)(x-2) = x² - 3x + 2，再乘以 (x-3) 得到 x³ - 6x² + 11x - 6。求导：P'(x) = 3x² - 12x + 11。令 P'(x) = 0 得到 3x² - 12x + 11 = 0。二次公式解得 x = [12 ± √144 - 132]/6 = [12 ± √12]/6 = [12 ± 2√3]/6 = 2 ± ½√3。

Substitute back to find y-coordinates: For x = 2 + ½√3, y = (1 + ½√3)(½√3)(-1 + ½√3) = ( (½√3)² - 1² ) × (½√3) = (¾ - 1)×½√3 = (-¼)×½√3 = -⅛√3. For x = 2 - ½√3, similarly y = +⅛√3. So turning points are (2 - √3/2, √3/8) and (2 + √3/2, -√3/8). Nature: check second derivative P''(x) = 6x - 12. At x = 2 - √3/2, P''(x) is negative, so a maximum; at the other, positive, so a minimum.

代入回求 y 坐标：对于 x = 2 + ½√3，y = (1 + ½√3)(½√3)(-1 + ½√3) = ( (½√3)² - 1² ) × (½√3) = (¾ - 1)×½√3 = (-¼)×½√3 = -⅛√3。对于 x = 2 - ½√3，类似得到 y = +⅛√3。因此驻点为 (2 - √3/2, √3/8) 和 (2 + √3/2, -√3/8)。性质：检查二阶导数 P''(x) = 6x - 12。在 x = 2 - √3/2 时，P''(x) 为负，故为极大值；另一点为正，故为极小值。

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4. Drills on Trigonometry and Identities | 三角学与恒等式训练

Case: ible values for angle B and discuss the ambiguous case. Then use an identity to find the exact value of sin 75°, and hence determine the exact length of side c in each case.

案例：在三角形 ABC 中，角 A = 75°，边 a = 12 cm，边 b = 10 cm。求角 B 的两个可能值并讨论不定情况。然后利用恒等式求出 sin 75° 的精确值，并据此确定每种情况下边 c 的精确长度。

= (10 sin 75°) / 12. First find sin 75° via sin(45°+30°) = sin45°cos30° + cos45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6+√2)/4.

使用正弦定理：sin B / b = sin A / a → sin B = (10 sin 75°) / 12。首先通过 sin(45°+30°) = sin45°cos30° + cos45°sin30° = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6+√2)/4 求出 sin 75°。

Then sin B = (10/12) × (√6+√2)/4 = (5/6)×(√6+√2)/4 = 5(√6+√2 + 106.6° = 181.6° > 180°, so B₂ is impossible. Thus only one valid triangle exists. However, the ambiguous case often yields two solutions when the given angle is acute and the opposite side is longer than the adjacent side.

则 sin B = (10/12) × (√6+√2)/4 = (5/6)×(√6+√2)/4 = 5(√6+√2)/24 ≈ 0.958。由于 sin B < 1，存在两个可能的角：B₁ ≈ 73.4°（锐角）和 B₂ ≈ 180° - 73.4° = 106.6°（钝角）。检查角度和：对于 B₂，A + B₂ = 75° + 106.6° = 181.6° > 180°，因此 B₂ 不成立。因此仅有一个有效三角形。不过，当已知角为锐角且对边比邻边长时，不定情况通常会给出两个解。

ssions, we can keep everything in terms of roots by using exact value of sin B: sin B = 5(√6+√2 ce. In an exam, approximate to 3 s.f. may be accepted: c ≈ (12 × 0.524) / 0.966 ≈ 6.51 cm.

当 B ≈ 73.4° 时，C = 180° - 75° - 73.4° = 31.6°。再次使用正弦定理：c / sin C = a / sin A → c = (12 × sin 31.6°) / sin 75°。为获得精确表达式，我们可以用根式保留 sin B 的精确值：sin B = 5(√6+√2)/24；但 B 不是标准角。改用余弦定理：c² = a² + b² - 2ab cos C，但 C 也不简洁。在考试中，保留三位有效数字的近似值可能被接受：c ≈ (12 × 0.524) / 0.966 ≈ 6.51 cm。

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5. Calculus in Context: Rates of Change | 微积分应用：变化率

Scenario: 2 m at a rate of 0.5 m³/min. How fast is the water level rising when the water is 3 m deep? The volume of a cone is V = ⅓πr²h.

情景：水以 0.5 m³/min 的速率注入一个倒置的圆锥形水箱（顶点朝下），水箱高 6 m，底面半径 2 m。当水深为 3 m 时，水位上升速率是多少？圆锥体积 V = ⅓πr²h。

h = ⅓π (h²/9) h = π/27 h³. Now differentiate with respect to time t: dV/dt = (π/27) × 3h² × dh/dt = (π/9) h² (dh/dt).

由相似三角形，水面半径 r 与深度 h 成比例：r/h = 2/6 = 1/3，因此 r = h/3。代入体积公式：V = ⅓π (h/3)² h = ⅓π (h²/9) h = π/27 h³。现在关于时间 t 求导：dV/dt = (π/27) × 3h² × dh/dt = (π/9) h² (dh/dt)。

≈ 0.159 m/min. The water level is rising at about 0.159 metres per minute at that instant.

已知 dV/dt = 0.5 m³/min，当 h = 3 m 时，求解：0.5 = (π/9) × 9 × dh/dt = π × dh/dt。因此 dh/dt = 0.5/π ≈ 0.159 m/min。该时刻水位上升约每分钟 0.159 米。

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6. Differential Equations Modelling | 微分方程建模

Case: e the equation and find the time taken for the population to reach 800. Discuss the long-term behaviour.

案例：兔子种群 P(t) 满足逻辑斯特微分方程 dP/dt = 0.5P(1 - P/1000)，其中 t 以月为单位。初始数量 P(0) = 100。求解方程并找出种群达到 800 所需的时间。讨论长期行为。

ed, 1/P + (1/1000)/(1 - P/1000) = 1/P + 1/(1000 - P). Integrate: ∫ (1/P + 1/(1000 - P)) dP = ∫ 0.5 dt → ln|P| - ln|1000 - P| = 0.5t + C. Combine logs: ln|P/(1000 - P)| = 0.5t + C. Exponentiate: P/(1000 - P) = A e^0.5t, where A = e^C.

分离变量：dP / [P(1 - P/1000)] = 0.5 dt。重写左侧：利用部分分式 1/[P(1 - P/1000)] = (1/P) + (1/1000)/(1 - P/1000)。确实，1/P + (1/1000)/(1 - P/1000) = 1/P + 1/(1000 - P)。积分：∫ (1/P + 1/(1000 - P)) dP = ∫ 0.5 dt → ln|P| - ln|1000 - P| = 0.5t + C。合并对数：ln|P/(1000 - P)| = 0.5t + C。指数化：P/(1000 - P) = A e^0.5t，其中 A = e^C。

At t=0, P=100: 100/(900) = A → A = 1/9. So P/(1000 - P) = (1/9) e^0.5t. Solve for P: P = 1000 × (1/9)e^0.5t / (1 + (1/9)e^0.5t) = 1000 e^0.5t / (9 + e^0.5t). Set P=800: 800 = 1000 e^0.5t / (9 + e^0.5t) → 800(9 + e^0.5t) = 1000 e^0.5t → 7200 + 800 e^0.5t = 1000 e^0.5t → 7200 = 200 e^0.5t → e^0.5t = 36 → 0.5t = ln 36 → t = 2 ln 36 ≈ 2 × 3.5835 = 7.167 months. Long-term: as t → ∞, e^0.5t dominates, P → 1000. The population stabilises at the carrying capacity of 1000.

当 t=0，P=100：100/(900) = A → A = 1/9。因此 P/(1000 - P) = (1/9) e^0.5t。解得 P：P = 1000 × (1/9)e^0.5t / (1 + (1/9)e^0.5t) = 1000 e^0.5t / (9 + e^0.5t)。令 P=800：800 = 1000 e^0.5t / (9 + e^0.5t) → 800(9 + e^0.5t) = 1000 e^0.5t → 7200 + 800 e^0.5t = 1000 e^0.5t → 7200 = 200 e^0.5t → e^0.5t = 36 → 0.5t = ln 36 → t = 2 ln 36 ≈ 2 × 3.5835 = 7.167 个月。长期行为：当 t → ∞ 时，e^0.5t 占主导，P → 1000。种群稳定在环境承载量 1000。

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7. Probability and Statistical Inference Cases | 概率与统计推断案例

Situation: A manufacturer claims that the mean lifetime of their light bulbs is 1200 hours with standard deviation 80 hours. A consumer group tests a random sample of 36 bulbs and finds a sample mean of 1180 hours. Test at the 5% significance level whether there is evidence that the mean lifetime is less than claimed. State hypotheses and p-value.

情境：某制造商声称其灯泡的平均使用寿命为 1200 小时，标准差为 80 小时。一个消费者团体随机抽取了 36 只灯泡进行测试，得到样本平均寿命为 1180 小时。以 5% 的显著性水平检验是否有证据表明平均寿命低于声称值。说明假设和 p 值。

up: H₀: μ = 1200, H₁: μ < 1200 (one-tailed). Known σ = 80, n = 36. Under H₀, sample mean X̄ ~ N(1200, 80²/36) = N(1200, 13.333...²). Test statistic Z = (1180 - 1200) / (80/√36) = -20 / (80/6) = -20 / 13.333... = -1.5. Critical value for one-tailed 5% .0668. Since p-value > 0.05, we fail to reject H₀. There is insufficient evidence at the 5% level to conclude that the mean lifetime is less than 1200 hours.

设定：H₀: μ = 1200，H₁: μ < 1200（单尾）。已知 σ = 80，n = 36。在 H₀ 下，样本均值 X̄ ~ N(1200, 80²/36) = N(1200, 13.333…²)。检验统计量 Z = (1180 - 1200) / (80/√36) = -20 / (80/6) = -20 / 13.333… = -1.5。单尾 5% 的临界值为 -1.645（或使用 p 值）。p 值 = P(Z < -1.5) = 0.0668。因为 p 值 > 0.05，我们不能拒绝 H₀。在 5% 显著性水平下，没有足够证据表明平均寿命低于 1200 小时。

own, and the sample standard deviation is 85. Then we use t-test: t = (1180 - 1200) / (85/√36) = -20 / (85/6) ≈ -1.4118, degrees of freedom = 35. Critical t (one-tailed 5%) ≈ -1.69. p-value > 0.05, same conclusion.

现在假设总体标准差未知，样本标准差为 85。则我们使用 t 检验：t = (1180 - 1200) / (85/√36) = -20 / (85/6) ≈ -1.4118，自由度 = 35。单尾 5% 的 t 临界值 ≈ -1.69。p 值 > 0.05，结论相同。

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8. Mechanics: Projectile Motion Case | 力学：抛体运动案例

Problem: horizontal range, and the maximum height. Assume g = 9.8 m/s², and ignore air resistance.

问题：一个球从地面以 20 m/s 的速度、与水平面成 30° 角踢出。求飞行时间、水平射程和最大高度。设 g = 9.8 m/s²，忽略空气阻力。

Resolve velocity: u_x = 20 cos30° = 20 × √3/2 ≈ 17.32 m/s; u_y = 20 sin30° = 10 m/s. Time of flight: use vertical motion s = u_y 4.9 t² → t(10 - 4.9 t)=0 → t=0 (start) or t = 10/4.9 ≈ 2.0408 s. So flight time ≈ 2.04 s.

分解速度：u_x = 20 cos30° = 20 × √3/2 ≈ 17.32 m/s；u_y = 20 sin30° = 10 m/s。飞行时间：由竖直运动 s = u_y t - ½ g t²。落地时 s=0：0 = 10 t - 4.9 t² → t(10 - 4.9 t)=0 → t=0（起始）或 t = 10/4.9 ≈ 2.0408 s。因此飞行时间 ≈ 2.04 s。

Horizontal range = u_x × total time = 17.32 × 2.0408 ≈ 35.35 m. Maximum height occurs when v_y = 0: v_y = u_y ime: s = 10(1.0204) - 4.9(1.0204)² ≈ 10.204 - 5.102 = 5.102 m. So maximum height ≈ 5.10 m.

水平射程 = u_x × 总时间 = 17.32 × 2.0408 ≈ 35.35 m。最大高度发生在 v_y = 0 时：v_y = u_y - g t → 0 = 10 - 9.8 t → t = 10/9.8 ≈ 1.0204 s（总时间的一半）。该时刻高度：s = 10(1.0204) - 4.9(1.0204)² ≈ 10.204 - 5.102 = 5.102 m。因此最大高度 ≈ 5.10 m。

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9. Vectors in Geometry and Mechanics | 几何与力学中的向量

Case: - 3k N act on a particle. Find the resultant force and its magnitude. Then determine the work done by the resultant when the particle moves from point A (1,2,3) to point B (4,5,6) along a straight line, assuming distances in metres.

案例：两个力 F₁ = 3i - 2j + k N 和 F₂ = -i + 4j - 3k N 作用在一个质点上。求合力及其大小。然后确定当质点沿直线从点 A (1,2,3) 移动到点 B (4,5,6) 时合力所做的功，假设距离单位为米。

Resultant force R = F₁ + F₂ = (3-1)i + (-2+4)j + (1-3)k = 2i + 2j - 2k N. Magnitude |R| = √(2² + 2² + (-2)²) = √4+4+4 = √12 = 2√3 ≈ 3.464 N.

合力 R = F₁ + F₂ = (3-1)i + (-2+4)j + (1-3)k = 2i + 2j - 2k N。大小 |R| = √(2² + 2² + (-2)²) = √4+4+4 = √12 = 2√3 ≈ 3.464 N。

e dot product captures the component of force in the direction of displacement. In a mechanics case study, you might be asked to verify that the force field is conservative, or to find potential energy function.

位移向量 d = B - A = (4-1)i + (5-2)j + (6-3)k = 3i + 3j + 3k m。做功 W = R · d = (2)(3) + (2)(3) + (-2)(3) = 6 + 6 - 6 = 6 J。注意功是标量；点乘捕获了力在位移方向上的分量。在力学案例分析中，你还可能被要求验证力场是否是保守的，或找出势能函数。

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10. Complex Number Scenarios | 复数情景分析

Problem: Given z = 1 + i√3, express z in polar form r(cosθ + i sinθ). Hence find z⁶ and all fourth roots of z. Represent the roots on an Argand diagram.

问题：已知 z = 1 + i√3，将 z 表示为极形式 r(cosθ + i sinθ)。由此求 z⁶ 和 z 的所有四次方根。在阿尔冈图上表示这些根。

Modulus r = √(1² + (√3)²) = √1+3 = 2. Argument θ = arctan(√3 de Moivre, z⁶ = 2⁶ (cos(6×π/3) + i sin(6×π/3)) = 64 (cos 2π + i sin 2π) = 64(1 + 0i) = 64.

模长 r = √(1² + (√3)²) = √1+3 = 2。辐角 θ = arctan(√3/1) = π/3（因为实部和虚部均为正）。因此 z = 2(cos(π/3) + i sin(π/3))。由德莫弗尔定理，z⁶ = 2⁶ (cos(6×π/3) + i sin(6×π/3)) = 64 (cos 2π + i sin 2π) = 64(1 + 0i) = 64。

Fourth roots of z: we need w such that w⁴ = z = 2(cos(π/3) + i sin(π/3)). The roots are w_k = 2^1/4 [ cos((π/3 + 2kπ)/4) + i sin((π/3 + 2kπ)/4) ] for k = 0,1,2,3. 2^1/4 = √√2 (π/3 + 18π/3)/4 = 19π/12. These four roots are equally spaced around a circle of radius 2^1/4.

z 的四次方根：我们需要 w 满足 w⁴ = z = 2(cos(π/3) + i sin(π/3))。根为 w_k = 2^1/4 [ cos((π/3 + 2kπ)/4) + i sin((π/3 + 2kπ)/4) ]，k = 0,1,2,3。2^1/4 = √√2 ≈ 1.189。计算角度：k=0：(π/3)/4 = π/12；k=1：(π/3 + 2π)/4 = (π/3 + 6π/3)/4 = 7π/12；k=2：(π/3 + 4π)/4 = (π/3 + 12π/3)/4 = 13π/12；k=3：(π/3 + 6π)/4 = (π/3 + 18π/3)/4 = 19π/12。这四个根在一个半径为 2^1/4 的圆上等距分布。

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11. Exam Tips and Time Management | 考试技巧与时间管理

Pre-U case study papers are lengthy; typical marks range from 20 to 30 per question, with a recommended time of about 30-40 minutes. Start by scanning the whole case to understand the storyline. Then read each part carefully, highlighting key numbers and conditions. If you get stuck on a part, move on and return later – later parts often rely on earlier answers, so state your previous result clearly to avoid losing method marks.

Pre-U 案例分析试卷篇幅较长；每题通常 20 到 30 分，建议用时 30 到 40 分钟。开始时快速浏览整个案例以了解故事情节。然后仔细阅读每一部分，标出关键数字和条件。如果在某一部分卡住，先跳过，稍后再回过来——后面的部分往往依赖前面的答案，因此明确陈述你此前的答案以免丢失方法分。

Always show your modelling assumptions, such as 'assume no air resistance' or 'assume the population is isolated'. When interpreting results, comment on realism and possible improvements. In statistics cases, mention the Central Limit Theorem if you use normal approximation. Finally, reserve two minutes at the end to check for arithmetic errors and

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