IGCSE OCR Further Maths: Case Study Practice | IGCSE OCR 进阶数学：案例分析实战演练

📚 IGCSE OCR Further Maths: Case Study Practice | IGCSE OCR 进阶数学：案例分析实战演练

Bringing mathematical concepts to life through real-world scenarios is one of the most effective ways to prepare for the IGCSE OCR Further Mathematics examination. This article presents a series of carefully chosen case studies that mirror the style of applied questions you may encounter. Each case study is broken down step by step, integrating techniques from algebra, calculus, vectors, matrices, trigonometry, and statistics. By working through these practical drills, you will sharpen your problem-solving skills and gain the confidence to tackle complex, multi-step problems under timed conditions.

通过真实情境将数学概念带入生活，是备考 IGCSE OCR 进阶数学考试最有效的方法之一。本文精选了一系列贴近实际应用的案例分析，模拟你可能遇到的题目风格。每个案例都逐步分解，融合了代数、微积分、向量、矩阵、三角和统计等技巧。通过演练这些实战练习，你能提升解题能力，获得信心，从容应对考试中复杂的多步骤问题。

1. Exponential Growth of Bacteria | 细菌指数增长

A lab culture contains 2000 bacteria and grows at a rate proportional to its size, modelled by N(t) = 2000e^0.3t, where t is time in hours. Find the growth rate after 2 hours and the time needed for the population to double.

实验室培养物含有 2000 个细菌，其增长速率与自身数量成正比，模型为 N(t) = 2000e^0.3t，其中 t 是以小时计的时间。求 2 小时后的增长率以及数量翻倍所需的时间。

The instantaneous growth rate is given by the derivative dN/dt. Differentiating N(t): dN/dt = 2000 × 0.3 × e^0.3t = 600 e^0.3t. At t = 2, dN/dt = 600 e^0.6 ≈ 600 × 1.8221 ≈ 1093 bacteria per hour. This represents how quickly the population is increasing at that exact moment.

瞬时增长率由导数 dN/dt 给出。对 N(t) 求导：dN/dt = 2000 × 0.3 × e^0.3t = 600 e^0.3t。当 t = 2 时，dN/dt = 600 e^0.6 ≈ 600 × 1.8221 ≈ 每小时 1093 个细菌。这代表了那个精确时刻数量增加的速度。

To find the doubling time, set N(t) = 4000. Then 2000e^0.3t = 4000 → e^0.3t = 2. Taking natural logarithms: 0.3t = ln 2 → t = (ln 2)/0.3 ≈ 0.6931/0.3 ≈ 2.31 hours. Thus, the culture doubles in roughly 2 hours and 19 minutes. This example reinforces the importance of logarithmic manipulation and differentiation in modelling growth processes.

为求翻倍时间，令 N(t) = 4000。则 2000e^0.3t = 4000 → e^0.3t = 2。取自然对数：0.3t = ln 2 → t = (ln 2)/0.3 ≈ 0.6931/0.3 ≈ 2.31 小时。因此培养物约在 2 小时 19 分钟时翻倍。此例强化了对数运算和微分在增长建模中的重要性。

2. Maximising Enclosed Area with a Fixed Perimeter | 固定周长下最大化面积

A farmer has 100 m of fencing and wants to enclose a rectangular area against a straight wall (so only three sides need fencing). Determine the dimensions that maximise the area, and find the maximum area.

一位农民有 100 米长的篱笆，想靠着一面直墙围出一个矩形区域（因此只需围三边）。确定使面积最大化的尺寸，并求出最大面积。

Let the two identical sides perpendicular to the wall have length x metres, and the side parallel to the wall have length (100 − 2x) metres. The area A is given by A = x(100 − 2x) = 100x − 2x². This is a quadratic function, opening downwards, so its maximum occurs at the vertex.

设垂直于墙的两条等长边的长度为 x 米，则平行于墙的边长为 (100 − 2x) 米。面积 A 为 A = x(100 − 2x) = 100x − 2x²。这是一个开口向下的二次函数，因此最大值出现在顶点处。

Using calculus, differentiate A: dA/dx = 100 − 4x. Set dA/dx = 0 → 100 − 4x = 0 → x = 25. The second derivative d²A/dx² = −4 < 0 confirms a maximum. The parallel side is 100 − 2×25 = 50 m. The maximum area is 25 × 50 = 1250 m². Alternatively, you could complete the square to find the vertex, but calculus provides a more general tool for optimisation.

使用微积分，对 A 求导：dA/dx = 100 − 4x。令 dA/dx = 0 → 100 − 4x = 0 → x = 25。二阶导数 d²A/dx² = −4 < 0 确认为极大值。平行边长为 100 − 2×25 = 50 m。最大面积为 25 × 50 = 1250 m²。当然你也可以用配方法找顶点，但微积分提供了更通用的优化工具。

3. Matrix Encryption of a Message | 矩阵加密信息

A secret message “HELP” is encoded by converting each letter to its position in the alphabet (A=1, B=2, …) and grouping into 2×1 vectors. Use the encoding matrix M = [3 2; 5 3] to encrypt the message, and demonstrate how the inverse matrix can decrypt it.

机密信息 “HELP” 的加密过程是：将每个字母转换为字母表中的位置（A=1, B=2, …），并组成 2×1 向量。使用加密矩阵 M = [3 2; 5 3] 对信息加密，并演示如何用逆矩阵解密。

First, convert ‘H’=8, ‘E’=5, ‘L’=12, ‘P’=16. Form two column vectors: p₁ = [8; 5] and p₂ = [12; 16]. Multiply each by M: c₁ = M p₁ = [3×8+2×5; 5×8+3×5] = [34; 55]. c₂ = M p₂ = [3×12+2×16; 5×12+3×16] = [68; 108]. The encrypted message is the sequence 34,55,68,108. To decipher, we need the inverse of M. det(M) = 3×3 − 2×5 = 9 − 10 = −1. Then M⁻¹ = (1/det) × [3, −2; −5, 3] = [−3, 2; 5, −3].

首先，将 ‘H’=8, ‘E’=5, ‘L’=12, ‘P’=16。组成两个列向量：p₁ = [8; 5] 和 p₂ = [12; 16]。分别与 M 相乘：c₁ = M p₁ = [3×8+2×5; 5×8+3×5] = [34; 55]。c₂ = M p₂ = [3×12+2×16; 5×12+3×16] = [68; 108]。加密后的信息是序列 34,55,68,108。要解密，我们需要 M 的逆矩阵。det(M) = 3×3 − 2×5 = 9 − 10 = −1。于是 M⁻¹ = (1/det) × [3, −2; −5, 3] = [−3, 2; 5, −3]。

Applying M⁻¹ to c₁: [−3×34+2×55; 5×34+−3×55] = [−102+110; 170−165] = [8; 5], recovering ‘H’ and ‘E’. Similarly for c₂, we get [12; 16], i.e., ‘L’ and ‘P’. This case illustrates matrix operations and the concept of inverse matrices in a practical context, a key topic in the OCR Further Maths syllabus.

将 M⁻¹ 应用于 c₁：[−3×34+2×55; 5×34+−3×55] = [−102+110; 170−165] = [8; 5]，恢复出 ‘H’ 和 ‘E’。同理 c₂ 得到 [12; 16]，即 ‘L’ 和 ‘P’。这个案例在实际背景下展示了矩阵运算及逆矩阵的概念，这是 OCR 进阶数学课程中的重要主题。

4. River Crossing with Vectors | 向量渡河问题

A boat can travel at 3 m/s in still water and aims to cross a river flowing at 1.5 m/s due east. If the boat points directly north, find its resultant velocity vector, actual speed, and direction (bearing).

一艘船在静水中的速度为 3 m/s，要渡一条水流以 1.5 m/s 正东向流动的河。如果船头指向正北，求它的合速度向量、实际速率以及方向（方位角）。

Represent the boat’s velocity relative to water as v_bw = [0, 3] (north) and the water’s velocity as v_w = [1.5, 0] (east). The resultant velocity v_r = v_bw + v_w = [1.5, 3]. The actual speed is |v_r| = √(1.5² + 3²) = √(2.25+9) = √11.25 ≈ 3.354 m/s.

将船对水的速度表示为 v_bw = [0, 3]（北），水流速度 v_w = [1.5, 0]（东）。合速度 v_r = v_bw + v_w = [1.5, 3]。实际速率为 |v_r| = √(1.5² + 3²) = √(2.25+9) = √11.25 ≈ 3.354 m/s。

To find the bearing, calculate the angle θ measured from north (the y-axis). tan φ = (east component)/(north component) = 1.5/3 = 0.5, where φ is the angle east of north. Thus φ = tan⁻¹(0.5) ≈ 26.57°. The bearing is 026.6° (to 1 decimal place). The boat travels faster than its still-water speed due to the current, but it is pushed off course. Advanced problems may ask for the heading needed to move directly north, which involves solving a vector triangle.

为求方位角，计算从北（y 轴）起测量的角度 θ。tan φ = （东分量）/（北分量）= 1.5/3 = 0.5，其中 φ 是东偏北的角度。于是 φ = tan⁻¹(0.5) ≈ 26.57°。方位角为 026.6°。由于水流，船的实际速度比静水速度更快，但偏离了原定方向。更高级的问题可能要求船头方向需如何调整才能沿正北移动，这需要解向量三角形。

5. Polynomial Curve Sketching Using Calculus | 用微积分绘制多项式曲线

A manufacturer’s cost function is modelled by C(x) = 2x³ − 9x² + 12x + 1, where x is output in hundreds of units. Use differentiation to find the stationary points and their nature, and sketch the curve for x ≥ 0.

某制造商的成本函数模型为 C(x) = 2x³ − 9x² + 12x + 1，其中 x 是以百为单位计算的产量。用微分求驻点及其性质，并画出 x ≥ 0 的曲线草图。

First derivative: C'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x−1)(x−2). Set C'(x) = 0 ⇒ x = 1 or x = 2. The second derivative is C”(x) = 12x − 18. Evaluate at x=1: C”(1) = 12−18 = −6 < 0 ⇒ local maximum. At x=2: C”(2) = 24−18 = 6 > 0 ⇒ local minimum.

一阶导数：C'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x−1)(x−2)。令 C'(x) = 0 ⇒ x = 1 或 x = 2。二阶导数 C”(x) = 12x − 18。计算 x=1 处：C”(1) = 12−18 = −6 < 0 ⇒ 极大值。x=2 处：C”(2) = 24−18 = 6 > 0 ⇒ 极小值。

Find the corresponding C-values: C(1) = 2−9+12+1 = 6; C(2) = 16−36+24+1=5. Also note C(0)=1. The curve rises from (0,1), reaches a local maximum at (1,6), then falls to a local minimum at (2,5), afterwards rising. No inflection is needed here, but knowing the shape of a cubic helps. This case study connects differentiation to economic modelling, illustrating how calculus identifies cost tipping points.

求相应的 C 值：C(1) = 2−9+12+1 = 6；C(2) = 16−36+24+1=5。另注意 C(0)=1。曲线从 (0,1) 上升，在 (1,6) 达到局部极大，然后下降到 (2,5) 的局部极小，之后上升。这里无需拐点，但了解三次函数形状有帮助。本案例将微分与经济建模相联系，展示了微积分如何识别成本的转折点。

6. Trigonometric Tidal Height Modelling | 三角函数潮汐高度模型

The water depth at a harbour entrance is modelled by h(t) = 5 + 3 sin(πt/6), where h is in metres and t is the time in hours after midnight. Determine the maximum and minimum depths, the times of the first high and low tides, and when the depth is exactly 4 metres during the morning.

某海港入口的水深可用模型 h(t) = 5 + 3 sin(πt/6) 表示，其中 h 单位为米，t 为午夜后的小时数。求最大和最小水深、首次高潮和低潮的时间，以及上午何时水深恰好为 4 米。

The sine function oscillates between −1 and 1, so h_max = 5+3=8 m, h_min = 5−3=2 m. High tide occurs when sin(πt/6)=1, i.e., πt/6 = π/2 + 2kπ. The first positive solution is t = 3 (3 a.m.). Low tide occurs when sin(πt/6)=−1, i.e., πt/6 = 3π/2 + 2kπ, giving t = 9 (9 a.m.). This annual model is common in OCR questions.

正弦函数在 −1 到 1 之间振荡，因此 h_max = 5+3=8 m，h_min = 5−3=2 m。高潮出现在 sin(πt/6)=1 时，即 πt/6 = π/2 + 2kπ。第一个正解为 t = 3（凌晨 3 点）。低潮出现在 sin(πt/6)=−1，即 πt/6 = 3π/2 + 2kπ，得 t = 9（早上 9 点）。此周期模型在 OCR 考题中常见。

To find when depth = 4 m, solve 5 + 3 sin(πt/6) = 4 → 3 sin(πt/6) = −1 → sin(πt/6) = −1/3. Let α = sin⁻¹(1/3) ≈ 0.3398 rad. Solutions: πt/6 = π + α, 2π − α, etc. The first solution in the morning (t > 0) is t = (6/π)(π + α) ≈ 6(1 + 0.108) = 6.65 hours, i.e., about 6:39 a.m. The next one after 9 a.m. would be t = (6/π)(2π − α) ≈ 12 − 0.65 = 11.35 hours (11:21 a.m.). This illustrates solving trigonometric equations with a non-standard value in a practical setting.

求水深为 4 米的时刻：解 5 + 3 sin(πt/6) = 4 → 3 sin(πt/6) = −1 → sin(πt/6) = −1/3。设 α = sin⁻¹(1/3) ≈ 0.3398 rad。解为：πt/6 = π + α, 2π − α 等。上午（t > 0）的首个解为 t = (6/π)(π + α) ≈ 6(1 + 0.108) = 6.65 小时，即约 6 点 39 分。9 点之后的另一个解为 t = (6/π)(2π − α) ≈ 12 − 0.65 = 11.35 小时（11 点 21 分）。这展示了在实际情境下求解含非特殊值的三角方程。

7. Intersecting Lines and Perpendicularity | 相交直线与垂直判定

Two delivery drones follow straight-line paths: Drone A along 2x − y = 4, Drone B passes through (1,5) and is perpendicular to Drone A’s path. Find the coordinates where their paths cross.

两架送货无人机沿直线飞行：无人机 A 沿 2x − y = 4，无人机 B 经过点 (1,5) 且与 A 的路径垂直。求它们路径交点的坐标。

Rewrite line A in slope-intercept form: y = 2x − 4, so its gradient m₁ = 2. A perpendicular line has gradient m₂ = −1/m₁ = −1/2. Line B passes through (1,5), so its equation is y − 5 = −½(x − 1) → y = −½x + ½ + 5 → y = −½x + 5.5 or, in standard form, x + 2y = 11.

将 A 的方程改写为斜截式：y = 2x − 4，故斜率 m₁ = 2。垂直线的斜率 m₂ = −1/m₁ = −1/2。直线 B 过 (1,5)，其方程为 y − 5 = −½(x − 1) → y = −½x + ½ + 5 → y = −½x + 5.5，或化为标准式 x + 2y = 11。

To find the intersection, solve the system: y = 2x − 4 and y = −½x + 5.5. Equate: 2x − 4 = −½x + 5.5 → 2.5x = 9.5 → x = 9.5/2.5 = 3.8. Substitute back: y = 2(3.8)−4 = 7.6−4 = 3.6. The paths cross at (3.8, 3.6). This case reinforces coordinate geometry, perpendicular gradients, and simultaneous equations, all of which appear regularly in the IGCSE Further Maths exam.

求交点，解方程组：y = 2x − 4 与 y = −½x + 5.5。联立：2x − 4 = −½x + 5.5 → 2.5x = 9.5 → x = 9.5/2.5 = 3.8。代回：y = 2(3.8)−4 = 7.6−4 = 3.6。路径交点为 (3.8, 3.6)。本案例巩固了坐标几何、垂直斜率与联立方程等知识，这些在 IGCSE 进阶数学考试中频繁出现。

8. Circle Equation and Radar Coverage | 圆的方程与雷达覆盖

A radar station located at the origin has a range of 20 km. Its coverage area can be modelled by x² + y² = 400. A ship reporting its position at (12, 16) asks whether it is within radar coverage. Also, determine the actual distance from the station and the shortest distance from the ship to the edge of the coverage boundary.

位于原点的雷达站覆盖半径为 20 km，其覆盖区域可用方程 x² + y² = 400 表示。某艘船报告其位置为 (12, 16)，请问它是否处于雷达覆盖范围内？此外，求它距雷达站的实际距离，以及到覆盖边界的最短距离。

Calculate the ship’s

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IGCSE OCR Further Maths: Case Study Practice | IGCSE OCR 进阶数学：案例分析实战演练