📚 Case Study in Action: CCEA IGCSE Further Mathematics | IGCSE CCEA 进阶数学案例分析实战演练
This article works through exam-style case studies for CCEA IGCSE Further Mathematics, focusing on how pure and applied techniques are combined in real problem contexts. Each section presents a scenario, the key method, and a model solution, so you can see exactly what a full-mark answer looks like.
本文通过 CCEA IGCSE 进阶数学的考试风格案例，讲解如何在真实问题情境中综合运用纯数学与应用数学技巧。每节给出一个情境、关键方法和参考解答，让你看清满分答案究竟长什么样。
CCEA papers rarely test an isolated formula. Instead, they wrap algebraic, calculus, matrix, mechanics and statistical ideas inside one continuous story, and the marks are awarded for structure as much as for the final number.
CCEA 的试卷很少单独考一个公式，而是把代数、微积分、矩阵、力学与统计的思路塞进同一个连续情境里，而且得分不仅看最终数值，更看整体结构。
1. Case Study Overview | 案例总览
CCEA Further Mathematics questions often embed a single real-world setting across several parts. The marker expects clear structure: define variables, state formulas, substitute, solve, and interpret.
CCEA 进阶数学题目常把多个小题嵌入同一个真实情境。阅卷者希望看到清晰结构：设变量、列公式、代入、求解并解释结果。
In a case study, you may move from algebra to calculus to statistics within one problem, so you must keep units and labels consistent throughout. If part (a) calls the height h metres and the time t seconds, then parts (b) to (e) must use exactly the same letters.
在一个案例中，你可能需要在同一题里从代数转到微积分再到统计，因此单位与符号必须全程一致。如果 (a) 问把高度记为 h 米、时间记为 t 秒，那么 (b) 到 (e) 就必须沿用同样的字母。
Scenario → Variable → Formula → Substitute → Solve → Interpret in context
The five-step routine above is worth practising until it becomes automatic, because roughly half of the lost marks in case studies come from skipping the interpretation step.
上面的五步流程值得练到自动化，因为案例题丢分中大约有一半来自跳过"解释结果"这一步。
| Step 步骤 | What to write 要写什么 |
| Define 设变量 | "Let x m be the width and y m the depth." |
| Formula 列公式 | State the general result before substituting numbers. |
| Substitute 代入 | Show the numerical line, keep full accuracy. |
| Solve 求解 | Working shown; final answer rounded only at the end. |
| Interpret 解释 | Sentence with units and the word from the question. |
2. Modelling a Roller Coaster with Quadratics | 用二次函数建模过山车
.05x² + 1.2x + 8, where h is the height in metres and x is the horizontal distance in metres from the boarding platform. You are asked for the maximum height, the horizontal distance at which it occurs, and where the track returns to ground level.
某设计师用高度函数 h(x) = -0.05x² + 1.2x + 8 来建模过山车的第一段坡道，其中 h 为高度（米），x 为距离站台的横向距离（米）。题目要求求出最大高度、取得最大高度处的横向距离，以及轨道回到地面的位置。
Complete the square to reach vertex form. Factor out -0.05, then adjust the bracket to get -0.05(x - 12)² + 15.2, so the vertex is (12, 15.2). The maximum height is 15.2 m at a horizontal distance of 12 m.
配方化为顶点式：提出 -0.05，再调整括号，得到 -0.05(x - 12)² + 15.2，因此顶点为 (12, 15.2)。最大高度为 15.2 米，出现在横向距离 12 米处。
h(x) = -0.05(x - 12)² + 15.2 ⟹ max height 15.2 m at x = 12 m
3.04 is positive, so there are two roots, approximately x = 28.35 and x = -4.35. Reject the negative root because it lies outside the physical domain, and state the answer as about 28.4 m.
求轨道与地面交点时，令 h(x) = 0，用求根公式代入 a = -0.05、b = 1.2、c = 8。判别式 b² - 4ac = 1.44 + 1.6 = 3.04 为正，故有两个根，约为 x = 28.35 与 x = -4.35。负根在物理情境中不合理，应舍去，答案写约 28.4 米。
3. Matrices in a Transformation Puzzle | 矩阵与图形变换
, [0, 3]]. A case-study question may ask for the image of the triangle with vertices A(1, 1), B(3, 1) and C(1, 2), and then for the area scale factor.
某标识画在坐标网格上，经过矩阵 M = [[2, 1], [0, 3]] 变换。案例题可能要求求出顶点为 A(1, 1)、B(3, 1)、C(1, 2) 的三角形的像，并求面积缩放比。
1×1, 0×3 + 3×1) = (7, 3). For C: (2×1 + 1×2, 0×1 + 3×2) = (4, 6). The image triangle is A'(3, 3), B'(7, 3), C'(4, 6).
把矩阵乘以每个写成列向量的位置向量。对 A：(2×1 + 1×1, 0×1 + 3×1) = (3, 3)；对 B：(2×3 + 1×1, 0×3 + 3×1) = (7, 3)；对 C：(2×1 + 1×2, 0×1 + 3×2) = (4, 6)。像三角形为 A'(3, 3)、B'(7, 3)、C'(4, 6)。
nit, so the image has area 6 square units. Always check the determinant sign: a negative determinant means the orientation has been reversed.
M 的行列式为 2×3 - 1×0 = 6，所以面积被放大 6 倍。原三角形面积为 ½ × 2 × 1 = 1 平方单位，故像的面积为 6 平方单位。注意行列式的符号：为负表示方向发生了翻转。
e the signs of the off-diagonal entries and divide by the determinant, giving M⁻¹ = [[½, -⅙], [0, ⅓]].
求逆矩阵时，交换主对角线元素、改变副对角线元素的符号，再除以行列式，得 M⁻¹ = [[½, -⅙], [0, ⅓]]。
4. Complex Numbers and Rotation | 复数与旋转
otating the arm by 90° anticlockwise is equivalent to multiplying z by i, because i² = -1.
某机械臂在阿尔冈图上用复数 z = 2 + 3i 表示，单位为米。把机械臂逆时针旋转 90° 等价于把 z 乘以 i，因为 i² = -1。
Compute iz = i(2 + 3i) = 2i + 3i² = -3 + 2i. The modulus is unchanged at √13 ≈ 3.61 m, and the argument has increased by 90°, which matches the geometry of a rotation.
计算 iz = i(2 + 3i) = 2i + 3i² = -3 + 2i。模不变，仍为 √13 ≈ 3.61 米，而辐角增加了 90°，与旋转的几何意义完全吻合。
|z| = √2² + 3² = √13 ≈ 3.61, arg z ≈ 56.3°, arg(iz) ≈ 146.3°
For a general rotation by θ, multiply by cos θ + i sin θ. Multiplying by a complex number always scales by its modulus and rotates by its argument, which is why engineers use the modulus-argument form for repeated rotations.
一般的旋转 θ 角，只需乘以 cos θ + i sin θ。复数相乘总是按模缩放、按辐角旋转，这正是工程师在多次旋转中使用模辐角形式的原因。
5. Vectors in a Drone Navigation Scenario | 无人机导航中的向量
wind blows with velocity w = -2i + 3j. The resultant ground velocity is v + w = 4i + 11j.
一架送货无人机相对空气的速度为 v = 6i + 8j 米/秒，风速为 w = -2i + 3j。合速度为 v + w = 4i + 11j。
The speed over the ground is the magnitude, √4² + 11² s found from tan θ = 4 ÷ 11, giving θ ≈ 20.0° east of north, so the bearing is about 020°.
对地速度是该向量的模：√4² + 11² = √137 ≈ 11.7 米/秒。方位角由 tan θ = 4 ÷ 11 求得，θ ≈ 20.0°，即北偏东约 20.0°，故方位角约为 020°。
resultant velocity by the time: (4i + 11j) × 45 = 180i + 495j metres, which has magnitude √180² + 495² ≈ 526.7 m.
求 45 秒后的位移，只需用合速度乘以时间：(4i + 11j) × 45 = 180i + 495j 米，其模为 √180² + 495² ≈ 526.7 米。
the dot product: for u = 3i + 4j and v = -1i + 2j, u · v = -3 + 8 = 5, so cos θ = 5 ÷ (5 × √5) ≈ 0.447 and θ ≈ 63.4°.
若某小问要求两向量夹角，用点积：对 u = 3i + 4j 与 v = -1i + 2j，u · v = -3 + 8 = 5，因此 cos θ = 5 ÷ (5 × √5) ≈ 0.447，θ ≈ 63.4°。
6. Differentiation: Optimising a Fuel Tank | 微分：优化油箱容积
A cylindrical fuel tank must hold exactly 500 litres, that is 0.5 m³. The case study asks for the radius that minimises the amount of sheet metal used, and the corresponding height.
一个圆柱形油箱必须恰好装下 500 升，即 0.5 立方米。案例题要求求出使铁皮用料最少的半径以及对应的高。
e area of a closed cylinder is A = 2πr² + 2πrh, so substituting gives A = 2πr² + 1000 ÷ r.
由约束 V = πr²h = 500 得 h = 500 ÷ (πr²)。封闭圆柱的表面积为 A = 2πr² + 2πrh，代入后 A = 2πr² + 1000 ÷ r。
dA/dr = 4πr - 1000 ÷ r² = 0 ⟹ r³ = 250 ÷ π ⟹ r ≈ 4.30 m
r² = 4π + 2000 ÷ r³ is positive at this value, confirming a minimum rather than a maximum.
令导数为零得 r³ = 250 ÷ π ≈ 79.58，故 r ≈ 4.30 米，h = 500 ÷ (π × 18.50) ≈ 8.60 米。二阶导数 d²A/dr² = 4π + 2000 ÷ r³ 在此处为正，说明是极小值而非极大值。
Note the elegant result: the optimum height equals the diameter. Examiners often award a final mark for recognising and stating this relationship in words.
注意这个漂亮的结果：最优高度等于直径。阅卷方常常为用文字指出这一关系而给出最后一分。
7. Integration: Area Under a Velocity Curve | 积分：速度曲线下的面积
A cyclist' Integration of velocity gives displacement, and this is the standard applied use of definite integrals in the case-study paper.
某骑行者的速度模型为 v(t) = 6t - t² 米/秒，0 ≤ t ≤ 6 秒。对速度积分得到位移，这正是案例卷中定积分最典型的应用方式。
t³ ÷ 3 evaluated from 0 to 6. Substituting gives (3 × 36 - 216 ÷ 3) - 0 = 108 - 72 = 36 m.
逐项积分：位移 s = ∫(6t - t²) dt = 3t² - t³ ÷ 3，代入 0 到 6 的上下限，得 (3 × 36 - 216 ÷ 3) - 0 = 108 - 72 = 36 米。
s = ∫₀⁶ (6t - t²) dt = [3t² - t³ ÷ 3]₀⁶ = 36 m
e on the whole interval, the distance travelled equals the displacement, so no separate calculation is needed. If the sign changed, you would split the integral at the root and add the absolute values.
由于 v(t) = t(6 - t) 在整个区间上非负，路程与位移相等，无需另行计算。若符号发生变化，就要在零点处拆分积分并把绝对值相加。
d of a formula, use the trapezium rule with h = (b - a) ÷ n and the formula ½h[y₀ + 2(y₁ + … + yₙ₋₁) + yₙ].
若速度以数据表而非公式给出，就用梯形法：h = (b - a) ÷ n，公式为 ½h[y₀ + 2(y₁ + … + yₙ₋₁) + yₙ]。
8. Mechanics: Moments on a Balanced Beam | 力学：平衡横梁上的力矩
mass 30 kg is placed 1.5 m from A. Find the reaction at each support, taking g = 9.8 m/s².
一根长 4 米、质量 20 千克的均匀横梁 AB 两端 A、B 各有一个支点。在距 A 端 1.5 米处放一个 30 千克的箱子。取 g = 9.8 米/秒²，求各支点的反作用力。
he midpoint, 2 m from A. Take moments about A so that the reaction at A has zero moment: RB × 4 = 20g × 2 + 30g × 1.5 = 40g + 45g = 85g. Hence RB = 21.25g ≈ 208 N.
由于横梁均匀，其重力作用在中点，即距 A 端 2 米处。对 A 点取力矩，使 A 处反作用力的力矩为零：RB × 4 = 20g × 2 + 30g × 1.5 = 40g + 45g = 85g，故 RB = 21.25g ≈ 208 牛顿。
Resolving vertically gives RA + RB = 50g, so RA - 21.25g = 28.75g ≈ 282 N. Always verify with a second moment equation about B as a check.
竖直方向受力平衡给出 RA + RB = 50g，因此 RA = 50g - 21.25g = 28.75g ≈ 282 牛顿。最好再对 B 点取一次力矩方程作校验。
Σ moments = 0, Σ vertical forces = 0 ⟹ RB ≈ 208 N, RA ≈ 282 N
If the question instead asks where a third mass should be placed to make the beam tip, set the reaction at one support to zero and solve the single remaining moment equation.
若题目改问第三块重物应放在何处才能使横梁倾覆，只需令某一支点反作用力为零，再解剩下那一个力矩方程。
9. Applied Statistics: Hypothesis Testing on Battery Life | 统计：电池寿命的假设检验
A manufacturer claims that a battery lasts on average 40 hours. A sample of 25 batteries gives a mean of 38.2 hours. The population standard deviation is known to be 5 hours. Test the claim at the 5% significance level.
某厂商声称电池平均可用 40 小时。抽取 25 节电池，样本均值为 38.2 小时。已知总体标准差为 5 小时。在 5% 显著性水平下检验该声称。
s as H₀: μ = 40 and H₁: μ < 40, since the test is one-tailed. The standard error is 5 ÷ √25 = 1, so the test statistic is z = (38.2 - 40) ÷ 1 = -1.8.
由于是单侧检验，设 H₀: μ = 40，H₁: μ < 40。标准误为 5 ÷ √25 = 1，故检验统计量 z = (38.2 - 40) ÷ 1 = -1.8。
z = (x̄ - μ) ÷ (σ ÷ √n) = -1.8, critical value = -1.645
The critical value at the 5% level is -1.645. Since -1.8 < -1.645, the result lies in the critical region and we reject H₀. There is sufficient evidence at the 5% level to doubt the manufacturer's claim.
5% 水平的临界值为 -1.645。因为 -1.8 < -1.645，结果落在拒绝域内，故拒绝 H₀。在 5% 水平上有充分证据质疑厂商的声称。
Note the wording: never say "the claim is proved false". Say instead that the sample provides sufficient evidence to reject the null hypothesis in favour of the alternative. The conclusion must always refer back to the context of battery life.
注意措辞：不要说"该声称被证明是错的"，而要说样本提供了充分证据拒绝原假设、接受备择假设。结论必须回到电池寿命这一情境中表述。
10. Probability Trees and Conditional Probability | 概率树与条件概率
A factory uses two machines. Machine A makes 60% of the components and machine B makes 40%. Of A's output, 3% are faulty; of B's output, 5% are faulty. A component is chosen at random and found to be faulty.
某工厂使用两台机器。机器 A 生产 60% 的零件，机器 B 生产 40%。A 的产品中 3% 有缺陷，B 的产品中 5% 有缺陷。随机抽取一个零件，发现它有缺陷。
t branch for the machine and the second for faulty or not. The joint probabilities are P(A ∩ F) = 0.6 × 0.03 = 0.018 and P(B ∩ F) = 0.4 × 0.05 = 0.020.
画出概率树：第一层分支为机器，第二层为是否有缺陷。联合概率为 P(A ∩ F) = 0.6 × 0.03 = 0.018，P(B ∩ F) = 0.4 × 0.05 = 0.020。
en component is faulty is the sum, 0.018 + 0.020 = 0.038. The conditional probability that
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