Case Study Practical Drills for Year 11 Cambridge Further Mathematics | 剑桥 Year 11 进阶数学案例分析实战演练

📚 Case Study Practical Drills for Year 11 Cambridge Further Mathematics | 剑桥 Year 11 进阶数学案例分析实战演练

Case studies in Further Mathematics develop the ability to model real-world situations, apply advanced techniques, and interpret results critically. This article presents a series of practical drills covering calculus, mechanics, statistics, sequences, matrices, differential equations, vectors, and more. Each case is broken down into a problem statement, step-by-step reasoning, and a final answer. Working through these examples will strengthen your problem-solving fluency and prepare you for the demands of the Cambridge examination.

进阶数学中的案例分析能够培养建模能力、应用高级技巧以及批判性地解释结果的能力。本文提供一系列实战演练，涵盖微积分、力学、统计、数列、矩阵、微分方程、向量等多个主题。每个案例均包含问题陈述、分步推理和最终答案。通过这些练习，你将强化解题的熟练度，为应对剑桥考试做好充分准备。

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1. Maximising the Volume of an Open Box (Calculus) | 最大化无盖盒的容积（微积分）

Problem: A square sheet of cardboard with side length 30 cm has squares of side x cm cut from each corner. The sides are then folded up to form an open box. Find the value of x that maximises the volume of the box, and calculate the maximum volume.

问题：一张边长为 30 cm 的正方形纸板，从每个角剪去边长为 x cm 的小正方形，然后折起四边做成一个无盖盒子。求使盒子容积最大的 x 值，并计算最大容积。

Let the side of the removed square be x. After folding, the base is a square of side (30 – 2x) cm, and the height is x cm. The volume V (in cm³) is given by V = x(30 – 2x)². Expand: V = x(900 – 120x + 4x²) = 4x³ – 120x² + 900x.

设剪去的小正方形边长为 x。折起后，底面是边长为 (30 – 2x) cm 的正方形，高为 x cm。体积 V（单位 cm³）为 V = x(30 – 2x)²。展开得 V = 4x³ – 120x² + 900x。

Differentiate with respect to x: dV/dx = 12x² – 240x + 900. Set dV/dx = 0: 12x² – 240x + 900 = 0 → divide by 12: x² – 20x + 75 = 0. Factorise: (x – 5)(x – The domain is 0 < x < 15; x = 15 gives zero volume. Second derivative test: d²V/dx² = 24x – 240. At x = 5, d²V/dx² = 120 – 240 = –120 < 0 → maximum. The maximum volume is V(5) = 5(30 – 10)² = 5 × 400 = 2000 cm³.

对 x 求导：dV/dx = 12x² – 240x + 900。令导数为零：12x² – 240x + 900 = 0 → 除以 12 得 x² – 20x + 75 = 0。因式分解：(x – 5)(x – 15) = 0，故 x = 5 或 x = 15。定义域为 0 < x < 15；x = 15 时体积为零。用二阶导数检验：d²V/dx² = 24x – 240。当 x = 5 时，d²V/dx² = –120 < 0，所以是极大值点。最大体积 V(5) = 5(30 – 10)² = 2000 cm³。

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2. Projectile Motion: Maximum Range on a Horizontal Plane (Mechanics) | 抛体运动：水平面上的最大射程（力学）

Problem: ngle that gives the maximum horizontal range and calculate that range. Take g = 9.8 m s⁻².

问题：一个质点从地面以初速度 u = 25 m s⁻¹、与水平面成 θ 角抛出。忽略空气阻力，求使水平射程最大的角度并计算该射程。取 g = 9.8 m s⁻²。

Resolve initial velocity: horizontal component u_x = u cos θ, vertical u_y = u sin θ. Time of flight T: using s_y = u_y t – ½ gt², set s_y = 0 (returns to ground) → 0 = (u sin θ)T – ½ g T² → T = (2u sin θ)/g. Range R = u_x /g = (u² sin 2θ)/g. R is maximised when sin 2θ = 1, so 2θ = 90° → θ = 45°. Maximum range R_max = u²/g = (25²)/9.8 ≈ 63.8 m (to 3 s.f.).

分解初速度：水平分量 u_x = u cos θ，竖直分量 u_y = u sin θ。飞行时间 T：由 s_y = u_y t – ½ gt²，落地时 s_y = 0，得 0 = (u sin θ)T – ½ g T² → T = (2u sin θ)/g。射程 R = u_x × T = (u cos θ) × (2u sin θ)/g = (u² sin 2θ)/g。当 sin 2θ = 1 时 R 最大，即 2θ = 90° → θ = 45°。最大射程 R_max = u²/g = 625/9.8 ≈ 63.8 m（保留三位有效数字）。

This classic result shows that the optimal angle is 45° in the absence of air resistance, regardless of the initial speed. The derivation also illustrates how parametric equations and trigonometric identities combine in mechanics modelling.

这个经典结果表明，在没有空气阻力时最佳抛射角恒为 45°，与初速度无关。推导过程也展示如何在力学建模中结合参数方程与三角恒等式。

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3. Hypothesis Test for a Coin’s Fairness (Statistics) | 硬币公平性的假设检验（统计）

Problem: A coin is tossed 100 times and lands heads 60 times. Test at the 5% significance level whether the coin is biased in favour of heads. State your hypotheses, calculate the p-value (using a normal approximation), and give a conclusion in context.

问题：一枚硬币抛掷 100 次，出现 60 次正面。在 5% 的显著性水平下检验该硬币是否偏向正面。写出假设，用正态近似计算 p 值，并在实际语境中给出结论。

bability of heads. H₀: p = 0.5; H₁: p > 0.5 (one-tailed test). Under H₀, number of heads X ~ B(100, 0.5). Mean μ = np = 50, variance σ² = np(1– tion σ = 5. Using normal approximation with continuity correction: P(X ≥ 60) ≈ P(Z ≥ (59.5 – 50)/5) = P(Z ≥ 1.9). From normal tables, P(Z ≥ 1.9) = 1 – is p-value (0.0287) is less than 0.05; we reject H₀. There is sufficient evidence at the 5% level to conclude that the coin is biased in favour of heads.

设 p 为正面朝上的概率。H₀: p = 0.5；H₁: p > 0.5（单尾检验）。在 H₀ 下，正面次数 X ~ B(100, 0.5)。均值 μ = 50，方差 σ² = 25，标准差 σ = 5。使用带连续性校正的正态近似：P(X ≥ 60) ≈ P(Z ≥ (59.5 – 50)/5) = P(Z ≥ 1.9)。查表得 P(Z ≥ 1.9) = 1 – 0.9713 = 0.0287。p 值 0.0287 < 0.05，拒绝 H₀。在 5% 水平下有充分证据表明该硬币偏向正面。

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4. Loan Repayment with a Recurrence Relation (Sequences) | 贷款偿还与递推关系（数列）

Problem: Anna borrows £8000 at an annual interest rate of 6%, compounded yearly. She repays £1500 at the end of each year. Let u_n be the amount owed after n years (just after the n-th payment). Write a recurrence relation for u_n, find a closed-form expression, and determine the number of years required to clear the debt.

问题：安娜以年利率 6% 借款 £8000，每年复利一次。她在每年末偿还 £1500。令 u_n 为 n 年后（刚完成第 n 次还款后）的欠款额。写出 u_n 的递推关系，求其闭合表达式，并确定还清债务所需的年数。

Recurrence: u₀ = 8000; u_n+1 = 1.06 u_n – 1500. This is a first-order linear recurrence. Homogeneous solution: u_n^h = A(1.06)ⁿ. Particular solution: try constant C satisfying C = 1.06C – 1500 → 0.06C = 1500 → C = 25000. Thus u_n = A(1.06)ⁿ + 25000. Using u₀ = 8000: 8000 = A + 25000 → A = –17000. Hence u_n = 25000 – 17000(1.06)ⁿ. We want smallest n such that u_n ≤ 0. Solve 25000 – 17000(1.06)ⁿ ≤ 0 → (1.06)ⁿ 5827 ≈ 6.63. Therefore, n = 7 years are needed. After the 7th payment the debt is cleared.

递推关系：u₀ = 8000；u_n+1 = 1.06 u_n – 1500。这是一阶线性递推。齐次解为 u_n^h = A(1.06)ⁿ。特解：设常数 C 满足 C = 1.06C – 1500 → 0.06C = 1500 → C = 25000。故 u_n = A(1.06)ⁿ + 25000。利用初始值：8000 = A + 25000 → A = –17000，因此 u_n = 25000 – 17000(1.06)ⁿ。要求最小的 n 使 u_n ≤ 0。解不等式 25000 – 17000(1.06)ⁿ ≤ 0 → (1.06)ⁿ ≥ 25000/17000 ≈ 1.4706。取自然对数：n ln 1.06 ≥ ln 1.4706 → n ≥ 0.3862/0.05827 ≈ 6.63。因此需要 n = 7 年。第 7 次还款后债务还清。

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5. Matrix Transformations for a Logo Design (Matrices) | 标志设计中的矩阵变换（矩阵）

Problem: se about the origin. Find the single 2×2 matrix that represents the combined transformation. Hence determine the image of the point (3, –2) under this combined transformation.

问题：一位平面设计师对一个点依次施加两次变换：先关于直线 y = x 作反射，再绕原点逆时针旋转 90°。求表示这一复合变换的单一 2×2 矩阵。据此求出点 (3, –2) 在该复合变换下的像。

x has matrix R = [[0, 1], [1, 0]]. Rotation by 90° anticlockwise has matrix T = [[cos90°, –sin90°], [sin90°, cos90°]] = [[0, –1], [1, 0]]. The combined transformation is T ◦ R (first R, then T): multiply T × R = [[0, –1], [1, 0]] × [[0, 1], [1, 0]] = [[0·0 + (–1)·1, 0·1 + (–1)·0], [1·0 + 0·1, 1·1 + 0·0]] = [[–1, 0], [0, 1]]. This is a reflection in the y-axis. Apply this to (3, –2): [[–1, 0], [0, 1]] [3; –2] = [–3; –2]. So the image is (–3, –2).

关于 y = x 的反射矩阵为 R = [[0, 1], [1, 0]]。逆时针旋转 90° 的矩阵为 T = [[0, –1], [1, 0]]。复合变换为先 R 后 T：计算 T × R = [[0, –1], [1, 0]] × [[0, 1], [1, 0]] = [[–1, 0], [0, 1]]。这正是关于 y 轴的反射。作用到点 (3, –2)：得到 (–3, –2)。因此像点为 (–3, –2)。

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6. Logistic Population Growth (Differential Equations) | 逻辑斯蒂人口增长（微分方程）

Problem: A population P(t) in thousands grows according to dP/dt = 0.8P(1 – rential equation to find P(t) explicitly. Determine the limiting population size as t → ∞.

问题：某人口 P(t)（以千人为单位）的增长满足 dP/dt = 0.8P(1 – P/100)。初始条件 P(0) = 20。求解该微分方程，显式表达 P(t)。并确定当 t → ∞ 时的人口极限规模。

Separate variables: dP / [P(1 – P/100)] = 0.8 dt. Write 1/[P(1 – P/100)] = 1/P + 1/(100 – P). Integrate both sides: ∫ (1/P + 1/(100 – P)) dP = ∫ 0.8 dt → ln|P| – ln|100 – P| = 0.8t + C. Combine logs: ln|P/(100 – P)| = 0.8t + C. Exponentiate: P/(100 – P) = Ae^0.8t, where A = e^C. Use P(0)=20: 20/(80) = A → A = 0.25. So P/(100 – P) = 0.25 e^0.8t. Solve for P: P = (100 × 0.25 e^0.8t)/(1 + 0.25 e^0.8t) = 25 e^0.8t / (1 + 0.25 e^0.8t) = 100 / (4 e^–0.8t + 1). As t → ∞, e^–0.8t → 0, so P → 100. The limiting population is 100 thousand, which is the carrying capacity.

分离变量：dP / [P(1 – P/100)] = 0.8 dt。部分分式：1/[P(1 – P/100)] = 1/P + 1/(100 – P)。两边积分：∫ (1/P + 1/(100 – P)) dP = ∫ 0.8 dt → ln|P| – ln|100 – P| = 0.8t + C。合并对数：ln|P/(100 – P)| = 0.8t + C。取指数：P/(100 – P) = A e^0.8t，其中 A = e^C。代入 P(0)=20：20/80 = A → A = 0.25。故 P/(100 – P) = 0.25 e^0.8t。解出 P：P = 25 e^0.8t / (1 + 0.25 e^0.8t) = 100 / (4 e^–0.8t + 1)。当 t → ∞ 时 e^–0.8t → 0，所以 P → 100。极限人口为 10 万人，即环境承载量。

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7. Safe Cornering on a Friction-Dependent Bend (Mechanics) | 依赖摩擦的弯道安全过弯（力学）

Problem: A car of mass 1200 kg travels around a horizontal circular bend of radius 50 m. The coefficient of static friction between tyres and road is 0.72. Determine the maximum speed the car can have without skidding. What would happen if the track is banked at an angle of 15° instead? Compute the ideal speed for no sideways friction in the banked case.

问题：一辆质量为 1200 kg 的汽车在半径为 50 m 的水平圆形弯道上行驶。轮胎与路面的静摩擦系数为 0.72。求汽车不致侧滑的最大速度。如果将弯道改为 15° 的倾斜弯道，情况会如何？计算在倾斜路面无横向摩擦力时的理想速度。

lely by friction: f = μR = μmg. Set mv²/r = μmg → v² = μgr = 0.72 × 9.8 × 50 = 352.8. So v_max e without relying on friction. Resolving: N sin θ = mv²/r and N cos θ = mg. Divide: tan θ = v²/(gr). Hence ideal v = √gr tan θ = √9.8 × 50 × tan15°. tan15° ≈ 0.2679, so v = √9.8 × 50 × 0.2679 = √131.27 ≈ 11.5 m s⁻¹. This is the speed at which no lateral friction is needed.

对于水平弯道，向心力完全由摩擦力提供：f = μR = μmg。代入 mv²/r = μmg 得 v² = μgr = 0.72 × 9.8 × 50 = 352.8，故 v_max ≈ 18.8 m s⁻¹（约 67.7 km/h）。对于倾角 θ = 15° 的弯道，路面法向反力的水平分量提供向心力而不依赖摩擦。分解：N sin θ = mv²/r，N cos θ = mg。相除得 tan θ = v²/(gr)。因此理想速度 v = √gr tan θ = √9.8 × 50 × tan15° ≈ √131.27 ≈ 11.5 m s⁻¹。在此速度下无需横向摩擦力。

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8. Intersection of Two Lines in 3D and Distance from a Point to a Line (Vectors) | 三维空间中两条直线的交点及点到直线的距离（向量）

Problem: Line L₁ passes through A(1, 0, 2) with direction vector d₁ = (2, –1, 1). Line L₂ passes through B(3, 2, 0) with direction vector d₂ = (– ection. Hence calculate the shortest distance from the origin to the line of intersection.

问题：直线 L₁ 通过点 A(1, 0, 2)，方向向量为 d₁ = (2, –1, 1)。直线 L₂ 通过点 B(3, 2, 0)，方向向量为 d₂ = (–1, 2, 1)。证明 L₁ 与 L₂ 相交，并求其交点。进而计算原点到该交线的最短距离。

Parametric forms: L₁: r = (1, 0, 2) + λ(2, –1, 1); L₂: r = (3, 2, 0) + μ(–1, 2, 1). For intersection, equate coordinates: 1+2λ = 3 – μ, 0 – λ = 2 + 2μ, 2+λ = 0 + μ. From the second equation: –λ = 2 + 2μ → λ = –2 – 2μ. Substitute into third: 2 + (–2 – 2μ) = μ → 0 – 2μ = μ → 3μ = 0 → μ = 0. Then λ = –2 – 0 = –2. Check first equation: 1+2(–2) = –3, and 3 – 0 = 3. Wait, –3 ≠ 3; something is wrong. Let's re-check. Actually solving: second eq: –λ = 2 + 2μ → λ = –2 – 2μ. Third eq: 2+λ = μ. Sub λ: 2 + (–2 – 2μ) = μ → –2μ = μ → 3μ = 0 → μ = 0, λ = –2. Plug into first: LHS = 1 + 2(–2) = –3; RHS = 3 – 0 = 3. They do not match, so lines do not intersect. Let' adjust the problem to ensure they intersect. I need a consistent case. Choose d₂ = (1, 2, – r to construct a simple intersection: Let L₁: r = (1,0,2) + λ(1,1,1); L₂: r = (3,2,3) + μ(–1,–1,–1) they are parallel. Not good. I'll redefine: Lines should intersect. Make L₁: A(1,0,2), d₁=(2,1,1); L₂: B(3,1,3), d₂=(–1,2,–2). Solve: 1+2λ = 3 – μ; 0+λ = 1+2μ; 2+λ = 3 – 2μ. From second: λ = 1+2μ. Sub in third: 2+1+2μ = 3 – 2μ → 3+2μ = 3–2μ → 4μ=0 → μ=0, λ=1. Check first: 1+2=3, 3–0=3. Works. So point of intersection is (1+2, 0+1, 2+1) = (3,1,3). So I' ta. Modify problem: L₁ through A(1,0,2) direction (2,1,1); L₂ through B(3,1,3) direction (–1,2,–2). Show they intersect at (3,1,3). Then find distance from origin to this point. Actually "line of intersection" is ambiguous; perhaps distance from origin to the common point? Or distance from origin to the line L₁? I' ask: find shortest distance from origin to L₁. That is a vector problem. Modify question: "Hence calculate the shortest distance from the origin to L₁." That's fine.

重新设计问题：L₁ 通过 (1,0,2) 方向 (2,1,1)；L₂ 通过 (3,1,3) 方向 (–1,2,–2)。解方程：1+2λ = 3 – μ；0+λ = 1+2μ；2+λ = 3 – 2μ。由第二式 λ = 1+2μ。代入第三式：2+1+2μ = 3–2μ → 3+2μ = 3–2μ → μ=0，λ=1。代入第一式验证：左=3，右=3。交点即 (1+2, 0+1, 2+1) = (3,1,3)。然后求原点到 L₁ 的最短距离。计算距离 d = |(OA) × d₁| / |d₁|。OA = (1,0,2)，d₁=(2,1,1)。叉积：OA × d₁ = determinant |i j k; 1 0 2; 2 1 1| = i(0·1 – 2·1) – j(1·1 – 2·2) + k(1·1 – 0·2) = i(–2) – j(1 – 4) + k(1) = (–2, 3, 1)。其模 = √4+9+1=√14。|d₁|=√4+1+1=√6。所以 d = √14/√6 = √7/3 ≈ 1.53。答案合理。

现在写出完整的解答。我会将问题陈述为修改后的版本。注意保持一致。

Parametric equations: L₁: r = (1,0,2) + λ(2,1,1); L₂: r = (3,1,3) + μ(–1,2,–2). Equate: 1+2λ = 3 – μ, 0+λ = 1+2μ, 2+λ = 3 – 2μ. From second, λ = 1+2μ. Substitute into third: 2+1+2μ = 3 – 2μ →

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