Case Study Practice for OCR Year 13 Further Mathematics | OCR Year 13 进阶数学案例分析实战演练

📚 Case Study Practice for OCR Year 13 Further Mathematics | OCR Year 13 进阶数学案例分析实战演练

This article presents a series of real examination-style case studies designed for OCR Year 13 Further Mathematics. Each case integrates multiple pure and applied topics – including complex numbers, matrices, hyperbolic functions, differential equations, polar coordinates, series, mechanics and proof by induction. Worked solutions are broken down into clear steps, with explanations translated side by side to strengthen both technical understanding and bilingual confidence. Use these problems as a benchmark for your revision and as a template for tackling extended multi-part questions under timed conditions.

本文提供一系列针对 OCR Year 13 进阶数学的真实考试风格案例分析。每个案例综合了多个纯数和应用主题，包括复数、矩阵、双曲函数、微分方程、极坐标、级数、力学以及归纳法证明。详细解答拆分成清晰的步骤，中英对照说明，帮助巩固技术理解并提升双语应试信心。你可以把这些习题作为复习的标杆，也可以当作限时完成多问大题的答题模板进行练习。

1. Complex Roots of Unity and Argand Geometry | 复数的单位根与 Argand 图几何

Problem: Solve the equation z⁴ + 16 = 0. Express the roots in exponential form reⁱᶿ and plot them on an Argand diagram. Hence find the area of the quadrilateral formed by joining the roots in order.

问题：解方程 z⁴ + 16 = 0，将根表示为指数形式 reⁱᶿ，并在 Argand 图上标出它们。由此求出按顺序连接这些根所形成的四边形的面积。

Write z⁴ = -16 = 16eⁱ⁽π⁺²ⁿπ⁾ so that z = 2eⁱ⁽π/4 ⁺ ⁿπ/²⁾ for n = 0,1,2,3. The four roots are 2eⁱπ/4, 2eⁱ³π/4, 2eⁱ⁵π/4, 2eⁱ⁷π/4. These are equally spaced by π/2 and lie on a circle of radius 2.

由 z⁴ = -16 = 16eⁱ⁽π⁺²ⁿπ⁾ 得 z = 2eⁱ⁽π/₄ ⁺ ⁿπ/²⁾，n = 0,1,2,3。四个根为 2eⁱπ/₄、2eⁱ³π/₄、2eⁱ⁵π/₄、2eⁱ⁷π/₄，它们等间隔 π/2 并位于半径为 2 的圆上。

On the Argand diagram, the points form a square with vertices at (√2, √2), (-√2, √2), (-√2, -√2), (√2, -√2). The side length s is the distance between two consecutive vertices, e.g. |2eⁱπ/4 – 2eⁱ³π/4| = 2√2. The area is s² = (2√2)² = 8.

在 Argand 图上，这些点构成一个正方形，顶点坐标为 (√2, √2)、(-√2, √2)、(-√2, -√2)、(√2, -√2)。边长 s 为相邻顶点间的距离，如 |2eⁱπ/₄ – 2eⁱ³π/₄| = 2√2，面积 s² = (2√2)² = 8。

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2. Matrix Transformations and Invariant Lines | 矩阵变换与不变直线

Problem: A linear transformation T is represented by the matrix M = [2 1; 0 3]. Find the eigenvalues and eigenvectors of M. Hence identify the invariant lines through the origin, stating their Cartesian equations.

问题：线性变换 T 由矩阵 M = [2 1; 0 3] 表示。求 M 的特征值和特征向量，由此找出通过原点的不变直线，并写出其笛卡尔坐标方程。

Solve det(M – λI) = (2 – λ)(3 – λ) – 0 = 0 → λ = 2, 3. For λ = 2: (M – 2I)v = 0 gives [0 1; 0 1]v = 0, so v = k(1, 0)ᵀ. For λ = 3: [–1 1; 0 0]v = 0 gives v = k(1, 1)ᵀ.

解 det(M – λI) = (2 – λ)(3 – λ) – 0 = 0 得 λ = 2, 3。对于 λ = 2：(M – 2I)v = 0 → [0 1; 0 1]v = 0，得 v = k(1, 0)ᵀ。对于 λ = 3：[–1 1; 0 0]v = 0 → v = k(1, 1)ᵀ。

Invariant lines correspond to eigenvectors: y = 0 (the x‑axis) and y = x. Any point on y = x is mapped to a multiple of (1,1)ᵀ, so the line is invariant as a whole.

不变直线对应特征向量：y = 0（x 轴）和 y = x。位于 y = x 上的任意点均被映射至 (1,1)ᵀ 的倍数，因此该直线作为整体保持不变。

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3. First-Order Differential Equation Modelling | 一阶微分方程建模

Problem: The rate of cooling of a body is proportional to the excess of its temperature over the ambient temperature 20°C. Initially the body is at 100°C, and after 5 minutes it is at 60°C. Form and solve the differential equation. How long does it take to reach 30°C?

问题：一物体的冷却速率与其相对于环境温度 20°C 的超出量成正比。初始温度为 100°C，5 分钟后降至 60°C。建立并求解微分方程。问达到 30°C 需要多长时间？

Let θ be temperature, t in minutes. dθ/dt = –k(θ – 20). Separate variables: ∫ dθ/(θ – 20) = –k∫ dt → ln(θ – 20) = –kt + C. Using initial condition (t=0, θ=100): C = ln(80). So θ – 20 = 80e⁻ᵏᵗ.

设 θ 为温度，t 以分钟计。dθ/dt = –k(θ – 20)。分离变量：∫ dθ/(θ – 20) = –k∫ dt → ln(θ – 20) = –kt + C。代入初始条件 (t=0, θ=100) 得 C = ln(80)，故 θ – 20 = 80e⁻ᵏᵗ。

At t=5, θ=60: 40 = 80e⁻⁵ᵏ → e⁻⁵ᵏ = 1/2 → –5k = ln(1/2) → k = (ln 2)/5. Then set θ=30: 10 = 80e⁻ᵏᵗ → e⁻ᵏᵗ = 1/8 → –kt = –3 ln 2 → t = (3 ln 2)/k = 3 · 5 = 15 minutes.

t=5, θ=60 时：40 = 80e⁻⁵ᵏ → e⁻⁵ᵏ = 1/2 → k = (ln 2)/5。令 θ=30：10 = 80e⁻ᵏᵗ → e⁻ᵏᵗ = 1/8 → –kt = –3 ln 2 → t = (3 ln 2)/k = 15 分钟。

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4. Hyperbolic Functions and Integration | 双曲函数与积分

Problem: Evaluate ∫ (1/√(x² + 4x + 13)) dx using a hyperbolic substitution. Give your final answer in logarithmic form.

问题：用双曲代换计算 ∫ (1/√(x² + 4x + 13)) dx，将最终答案用对数形式表示。

Complete the square: x² + 4x + 13 = (x+2)² + 9. Let u = x+2, then ∫ du/√(u² + 9). Substitute u = 3 sinh t, du = 3 cosh t dt, √(u²+9) = 3 cosh t. The integral becomes ∫ (3 cosh t)/(3 cosh t) dt = ∫ 1 dt = t + C = arsinh(u/3) + C = arsinh((x+2)/3) + C.

配方：x² + 4x + 13 = (x+2)² + 9。令 u = x+2，积分变为 ∫ du/√(u² + 9)。代换 u = 3 sinh t，du = 3 cosh t dt，√(u²+9) = 3 cosh t。积分化为 ∫ 1 dt = t + C = arsinh(u/3) + C = arsinh((x+2)/3) + C。

Since arsinh y = ln(y + √(y²+1)), the final answer is ln( (x+2)/3 + √( ((x+2)²/9) + 1 ) ) + C = ln( (x+2)/3 + √(x²+4x+13)/3 ) + C = ln( x+2 + √(x²+4x+13) ) – ln 3 + C. Constant can absorb –ln 3.

由 arsinh y = ln(y + √(y²+1))，得最终答案为 ln( (x+2)/3 + √( ((x+2)²/9) + 1 ) ) + C = ln( x+2 + √(x²+4x+13) ) + C’。

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5. Polar Curves: Cardioid Area | 极坐标曲线：心形线面积

Problem: The curve C has polar equation r = 1 + cos θ for 0 ≤ θ ≤ 2π. Find the total area enclosed by C. Sketch the curve and set up the integral analytically.

问题：曲线 C 的极坐标方程为 r = 1 + cos θ，0 ≤ θ ≤ 2π。求 C 所围成的总面积，画出曲线示意图并分析性地建立积分式。

The cardioid is symmetric about the initial line θ=0. Total area A = (1/2) ∫₀²π r² dθ = (1/2) ∫₀²π (1 + cos θ)² dθ. Expand: (1 + cos θ)² = 1 + 2 cos θ + cos²θ. Use identity cos²θ = (1 + cos 2θ)/2.

心形线关于极轴 θ=0 对称。总面积 A = (1/2) ∫₀²π (1 + cos θ)² dθ。展开：(1 + cos θ)² = 1 + 2 cos θ + cos²θ，用恒等式 cos²θ = (1 + cos 2θ)/2。

A = (1/2) ∫₀²π [1 + 2 cos θ + ½ + ½ cos 2θ] dθ = (1/2) ∫₀²π [3/2 + 2 cos θ + ½ cos 2θ] dθ. Integrate termwise: ∫₀²π (3/2) dθ = 3π, ∫ cos θ = 0, ∫ cos 2θ = 0. Hence A = (1/2) · 3π = 3π/2.

A = (1/2) ∫₀²π [1 + 2 cos θ + ½ + ½ cos 2θ] dθ = (1/2) ∫₀²π [3/2 + 2 cos θ + ½ cos 2θ] dθ。逐项积分：∫₀²π (3/2) dθ = 3π，∫ cos θ = 0，∫ cos 2θ = 0，故 A = (1/2)·3π = 3π/2。

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6. Maclaurin Series and Limit Evaluation | 麦克劳林级数与极限计算

Problem: Use Maclaurin series expansions up to and including the x³ term to evaluate lim_(x→0) (sin x – x) / (eˣ – 1 – x – x²/2). Confirm the limit analytically.

问题：利用含 x³ 项的麦克劳林展开式求极限 lim_(x→0) (sin x – x) / (eˣ – 1 – x – x²/2)，并分析性地确认结果。

Standard expansions: sin x = x – x³/3! + O(x⁵), so sin x – x = –x³/6 + O(x⁵). For denominator: eˣ = 1 + x + x²/2! + x³/3! + O(x⁴), thus eˣ – 1 – x – x²/2 = x³/6 + O(x⁴).

标准展开式：sin x = x – x³/3! + O(x⁵)，故 sin x – x = –x³/6 + O(x⁵)。分母：eˣ = 1 + x + x²/2! + x³/3! + O(x⁴)，因此 eˣ – 1 – x – x²/2 = x³/6 + O(x⁴)。

The limit becomes lim_(x→0) (–x³/6 + O(x⁵)) / (x³/6 + O(x⁴)) = –1. Cancelling the dominant x³ terms immediately yields –1. No need for L’Hôpital’s rule.

极限化为 lim_(x→0) (–x³/6 + O(x⁵)) / (x³/6 + O(x⁴)) = –1。消去主导项 x³ 后直接得 –1，无需使用洛必达法则。

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7. Mechanics: Oblique Collision and Impulse | 力学：斜碰撞与冲量

Problem: A smooth sphere of mass 0.2 kg moving at 10 m s⁻¹ strikes a fixed vertical wall at an angle of 60° to the wall. The coefficient of restitution is 0.5. Find the impulse exerted by the wall on the sphere.

问题：一个质量 0.2 kg 的光滑小球以 10 m s⁻¹ 的速度与固定竖直墙成 60° 角碰撞，恢复系数为 0.5。求墙对球施加的冲量。

Resolve velocity: perpendicular to wall: u_⊥ = 10 sin60° = 5√3; parallel stays unchanged. After impact, v_⊥ = e u_⊥ = 0.5 · 5√3 = (5√3)/2. The change in perpendicular velocity is Δv_⊥ = v_⊥ – (–u_⊥) = (5√3)/2 + 5√3 = (15√3)/2.

分解速度：垂直于墙的分量 u_⊥ = 10 sin60° = 5√3；平行分量不变。碰撞后 v_⊥ = e u_⊥ = 0.5 × 5√3 = (5√3)/2。垂直方向速度变化量 Δv_⊥ = v_⊥ – (–u_⊥) = (15√3)/2。

Impulse J = m Δv_⊥ = 0.2 × (15√3)/2 = 0.2 × 7.5√3 = 1.5√3 N s. The direction is perpendicular to the wall, away from it.

冲量 J = m Δv_⊥ = 0.2 × (15√3)/2 = 1.5√3 N s，方向垂直于墙且背离墙。

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8. Proof by Induction with Series | 归纳法证明级数求和

Problem: Prove by induction that Σ_(r=1)ⁿ r·2ʳ⁻¹ = (n – 1)2ⁿ + 1 for all positive integers n.

问题：用归纳法证明对于所有正整数 n，Σ_(r=1)ⁿ r·2ʳ⁻¹ = (n – 1)2ⁿ + 1。

Base case n=1: LHS = 1·2⁰ = 1. RHS = (1–1)·2¹ + 1 = 1. True. Assume true for n=k: Σ_(r=1)ᵏ r·2ʳ⁻¹ = (k–1)2ᵏ + 1.

奠基 n=1：左式 = 1·2⁰ = 1，右式 = (1–1)·2¹ + 1 = 1，成立。假设 n=k 时成立：Σ_(r=1)ᵏ r·2ʳ⁻¹ = (k–1)2ᵏ + 1。

For n=k+1: LHS = Σ_(r=1)ᵏ r·2ʳ⁻¹ + (k+1)2ᵏ = (k–1)2ᵏ + 1 + (k+1)2ᵏ = (2k)2ᵏ + 1 = k·2ᵏ⁺¹ + 1. RHS = (k+1–1)2ᵏ⁺¹ + 1 = k·2ᵏ⁺¹ + 1. Thus true for k+1. Hence by induction, the statement holds for all n ∈ ℕ.

当 n=k+1：左式 = Σ_(r=1)ᵏ r·2ʳ⁻¹ + (k+1)2ᵏ = (k–1)2ᵏ + 1 + (k+1)2ᵏ = (2k)2ᵏ + 1 = k·2ᵏ⁺¹ + 1。右式 = (k+1–1)2ᵏ⁺¹ + 1 = k·2ᵏ⁺¹ + 1。因此 n=k+1 时成立。由归纳法，命题对所有正整数 n 成立。

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9. Second-Order Differential Equations and Damping | 二阶微分方程与阻尼

Problem: A particle moves under a damping force proportional to velocity with equation x″ + 4x′ + 13x = 0. Initially x(0)=2, x′(0)=0. Solve for x(t) and describe the motion.

问题：一质点受到与速度成正比的阻尼力，运动方程为 x″ + 4x′ + 13x = 0，初始条件 x(0)=2, x′(0)=0。求解 x(t) 并描述运动。

Auxiliary equation: m² + 4m + 13 = 0 → m = –2 ± 3i. Thus the general solution is x(t) = e⁻²ᵗ (A cos 3t + B sin 3t). Apply initial conditions: x(0)=A=2. x′(t) = e⁻²ᵗ[(–2A+3B) cos 3t + (–2B–3A) sin 3t]; x′(0) = –2A+3B = 0 → –4+3B=0 → B=4/3.

辅助方程：m² + 4m + 13 = 0 → m = –2 ± 3i。通解为 x(t) = e⁻²ᵗ (A cos 3t + B sin 3t)。代入初始条件：x(0)=A=2。x′(0) = –2A+3B = 0 → B=4/3。

Hence x(t) = e⁻²ᵗ (2 cos 3t + (4/3) sin 3t). This represents damped oscillatory motion with exponentially decaying amplitude. The system is underdamped, with angular frequency 3 rad/s and decaying envelope 2e⁻²ᵗ.

因此 x(t) = e⁻²ᵗ (2 cos 3t + (4/3) sin 3t)，表示幅值指数衰减的振荡运动。系统为欠阻尼，角频率为 3 rad/s，衰减包络为 2e⁻²ᵗ。

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10. Numerical Methods: Newton-Raphson for Systems | 数值方法：牛顿-拉夫森法求解方程组

Problem: Use the Newton-Raphson method for simultaneous equations to find the positive intersection of y = x² and y = ln(x+2), starting from x₀ = 1. Perform two iterations.

问题：用解方程组的牛顿-拉夫森方法求 y = x² 与 y = ln(x+2) 的正交点，初始值 x₀ = 1，进行两次迭代。

Set F(x) = x² – ln(x+2) = 0. F'(x) = 2x – 1/(x+2). Iteration: xₙ₊₁ = xₙ – F(xₙ)/F'(xₙ). For x₀=1: F(1) = 1 – ln 3 ≈ 1 – 1.0986 = –0.0986. F'(1) = 2 – 1/3 ≈ 1.6667. Then x₁ = 1 – (–0.0986)/1.6667 ≈ 1.0592.

设 F(x) = x² – ln(x+2) = 0，F'(x) = 2x – 1/(x+2)。迭代式：xₙ₊₁ = xₙ – F(xₙ)/F'(xₙ)。x₀=1：F(1) ≈ –0.0986，F'(1) ≈ 1.6667，得 x₁ ≈ 1.0592。

Second iteration: F(x₁) = 1.0592² – ln(3.0592) ≈ 1.1218 – 1.1180 = 0.0038. F'(x₁) = 2·1.0592 – 1/3.0592 ≈ 2.1184 – 0.3268 = 1.7916. x₂ = 1.0592 – 0.0038/1.7916 ≈ 1.0571. The root is near 1.057.

第二次迭代：F(x₁) ≈ 0.0038，F'(x₁) ≈ 1.7916，x₂ ≈ 1.0571。根约在 1.057 附近。

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