Year 12 OCR Further Mathematics: Case Study Practical Drills | 案例分析实战演练

📚 Year 12 OCR Further Mathematics: Case Study Practical Drills | 案例分析实战演练

Mastery of Year 12 OCR Further Mathematics requires more than passive reading – it demands active engagement with carefully chosen case studies. This article presents ten in‑depth worked examples drawn from the core AS syllabus, spanning pure mathematics, mechanics and numerical methods. Each case breaks down a high‑frequency exam‑style problem into logical steps, highlights common pitfalls and reinforces the underlying theory. By working through these drills, you will sharpen your analytical skills, build confidence and learn to apply techniques flexibly in unfamiliar contexts.

要精通 Year 12 OCR 进阶数学，仅仅被动阅读是不够的，需要主动钻研精心挑选的案例。本文精选十个深入剖析的例题，覆盖 AS 核心大纲中的纯数、力学与数值方法。每个案例把一道高频考试题型拆解为逻辑步骤，突出常见误区，巩固理论基础。通过演练这些题目，你将提升分析能力，建立信心，并学会在陌生情境中灵活运用解题技术。

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1. Complex Numbers: Roots of Unity and Their Sum | 复数：单位根及其求和

: Solve the equation z³ = 1, giving your answers in exact Cartesian form. Let ω be the non‑real root with positive imaginary part. Without using a calculator, evaluate (1 + ω)⁵.

问题：解方程 z³ = 1，答案以精确的笛卡儿形式给出。记 ω 为虚部为正的非实数根。不使用计算器，求 (1 + ω)⁵ 的值。

Solution:

解答：

z = cos(2kπ/3) + i sin(2kπ/3) for k = 0, 1, 2. Hence z = 1, ω = cos(2π/3) + i sin(2π/3) = –½ + i(√3/2), ω² = cos(4π/3) + i sin(4π/3) = –½ – i(√3/2).

步骤 1：将 1 写成极坐标形式 1 = cos 0 + i sin 0。立方根由 z = cos(2kπ/3) + i sin(2kπ/3)，k = 0, 1, 2 给出。因此 z = 1，ω = cos(2π/3) + i sin(2π/3) = –½ + i(√3/2)，ω² = cos(4π/3) + i sin(4π/3) = –½ – i(√3/2)。

Step 2: Observe the identity 1 + ω + ω² = 0, which follows from the sum of the roots of z³ – 1 = 0. Consequently, 1 + ω = –ω².

步骤 2：注意到恒等式 1 + ω + ω² = 0，这源自方程 z³ – 1 = 0 的根之和。因此 1 + ω = –ω²。

Step 3: Compute (1 + ω)⁵ = (–ω²)⁵ = –ω¹⁰. Since ω³ = 1, we simplify ω¹⁰ = ω^(9+1) = (ω³)³ ω = 1·ω = ω. Hence (1 + ω)⁵ = –ω = ½ – i(√3/2).

步骤 3：计算 (1 + ω)⁵ = (–ω²)⁵ = –ω¹⁰。因为 ω³ = 1，化简得 ω¹⁰ = ω^(9+1) = (ω³)³ ω = ω。于是 (1 + ω)⁵ = –ω = ½ – i(√3/2)。

Key insight: Exploiting the cyclic property of roots of unity and the sum‑of‑roots identity avoids tedious expansion.

关键点拨：利用单位根的循环性质和根之和恒等式，可避免繁琐的展开。

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2. Matrices: Invariant Lines under Linear Transformations | 矩阵：线性变换下的不变线

sents a linear transformation. Find the invariant lines through the origin for this transformation.

问题：矩阵 M = [2 1; 1 2] 表示一个线性变换。求该变换下所有过原点的不变线。

Solution:

解答：

iant line through the origin can be written as y = m x. A general point (x, mx) maps to (x', y') = M (x, mx)ᵀ = (2x + mx, x + 2mx) = ( (2+m)x, (1+2m)x ).

步骤 1：过原点的不变线可写为 y = m x。一般点 (x, mx) 映射为 (x', y') = M (x, mx)ᵀ = (2x + mx, x + 2mx) = ( (2+m)x, (1+2m)x )。

Step 2: For the image point to lie on the same line, we need y' = m x' +2m)x = m (2+m)x. For x ≠ 0, this gives 1 + 2m = 2m + m², so m² = 1, yielding m = 1 or m = –1.

步骤 2：为使像点仍在同一直线上，要求 y' = m x'。因此 (1+2m)x = m (2+m)x。当 x ≠ 0 时，得 1 + 2m = 2m + m²，故 m² = 1，解得 m = 1 或 m = –1。

Step 3: The invariant lines are y = x and y = –x. Points on y = x are scaled by factor 3; points on y = –x are scaled by factor 1 (they are fixed).

步骤 3：不变线为 y = x 和 y = –x。y = x 上的点被缩放为原来的 3 倍；y = –x 上的点被缩放为原来的 1 倍（即保持不变）。

Key insight: Setting up the collinearity condition y' = m x' directly yields a simple quadratic for the gradient m.

关键点拨：直接建立共线条件 y' = m x'，即得到关于斜率 m 的简单二次方程。

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3. Series: The Method of Differences | 级数：差分法

Problem: Find an expression for the sum Sₙ = ∑_r=1ⁿ 1/(r(r+1)). Hence deduce the value of the infinite sum.

问题：求 Sₙ = ∑_r=1ⁿ 1/(r(r+1)) 的表达式，并由此推出无穷级数的值。

Solution:

解答：

Step 1: Split the term using partial fractions: 1/(r(r+1)) = 1/r – 1/(r+1).

步骤 1：用部分分式拆分项：1/(r(r+1)) = 1/r – 1/(r+1)。

Step 2: Write the sum explicitly and observe the telescoping: Sₙ = (1/1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n+1)) = 1 – 1/(n+1).

步骤 2：将和式明确写出并观察套叠相消：Sₙ = (1/1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n+1)) = 1 – 1/(n+1)。

Step 3: As n → ∞, 1/(n+1) → 0, so the infinite sum converges to 1.

步骤 3：当 n → ∞ 时，1/(n+1) → 0，因此无穷级数收敛于 1。

Key insight: The method of differences is ideal when terms can be rewritten as differences of successive expressions, allowing massive cancellation.

关键点拨：当项可重写为某个序列相邻项的差时，差分法极为有效，能实现大量相消。

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4. Vectors: Shortest Distance from a Point to a Line | 向量：点到直线的最短距离

nd the shortest distance from the point P(5, 2, 7) to the line r = (1, 0, 1) + λ(2, 1, 2).

问题：求点 P(5, 2, 7) 到直线 r = (1, 0, 1) + λ(2, 1, 2) 的最短距离。

Solution:

解答：

p 1: Let A = (1, 0, 1) and direction vector d = (2, 1, 2). Form the vector A to P: AP = (5–1, 2–0, 7–1) = (4, 2, 6).

步骤 1：取 A = (1, 0, 1)，方向向量 d = (2, 1, 2)。构造从 A 到 P 的向量 AP = (5–1, 2–0, 7–1) = (4, 2, 6)。

Step 2: The shortest distance is |AP × d| / |d|. Compute AP × d = (2·2 – 6·1, 6·2 – 4·2, 4·1 – 2·2) = (4–6, 12–8, 4–4) = (–2, 4, 0). Its magnitude is √((–2)² + 4² + 0²) = √4+16 = √20 = 2√5.

步骤 2：最短距离为 |AP × d| / |d|。计算向量积 AP × d = (2·2 – 6·1, 6·2 – 4·2, 4·1 – 2·2) = (4–6, 12–8, 4–4) = (–2, 4, 0)。其模为 √((–2)² + 4² + 0²) = √20 = 2√5。

Step 3: |d| = √2² + 1² + 2² = √9 = 3. Therefore the distance = (2√5)/3.

步骤 3：|d| = √2² + 1² + 2² = √9 = 3。因此距离为 (2√5)/3。

Key insight: The vector product formula gives the perpendicular distance efficiently without needing the foot of the perpendicular.

关键点拨：向量积公式能高效求出垂直距离，无需计算垂足。

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5. Differential Equations: Linear Integrating Factor | 微分方程：线性积分因子

Problem: Solve the differential equation dy/dx + y tan x = sec x, given that y(0) = 2.

问题：解微分方程 dy/dx + y tan x = sec x，已知 y(0) = 2。

Solution:

解答：

Step 1: The equation is first‑order linear. The integrating factor is I(x) = e^{∫ tan x dx} = e^{ln|sec x|} = sec x (taking x in (–π/2, π/2) where sec x > 0).

步骤 1：方程是一阶线性的。积分因子为 I(x) = e^{∫ tan x dx} = e^{ln|sec x|} = sec x（在区间 (–π/2, π/2) 上 sec x > 0）。

Step 2: Multiply through by sec x: sec x dy/dx + y sec x tan x = sec² x. The left‑hand side is d/dx (y sec x). Therefore d/dx (y sec x) = sec² x.

步骤 2：两边同乘 sec x：sec x dy/dx + y sec x tan x = sec² x。左边恰好是 d/dx (y sec x)。故 d/dx (y sec x) = sec² x。

Step 3: Integrate both sides: y sec x = ∫ sec² x dx = tan x + C. Thus y = sin x + C cos x.

步骤 3：两边积分：y sec x = ∫ sec² x dx = tan x + C。因此通解为 y = sin x + C cos x。

Step 4: Apply the condition y(0) = 2: 2 = sin 0 + C cos 0 = 0 + C·1 ⇒ C = 2. Hence y = sin x + 2 cos x.

步骤 4：代入初始条件 y(0) = 2：2 = sin 0 + C cos 0 = C，得 C = 2。于是特解为 y = sin x + 2 cos x。

Key insight: Recognising the product rule is crucial after multiplying by the integrating factor; always simplify the left‑hand side to a perfect derivative.

关键点拨：乘以积分因子后必须识别出乘法求导法则，将左边化为全导数形式。

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6. Mechanics: Resolving Forces and Equilibrium | 力学：力的分解与平衡

DB = 0.4 m. A particle of weight 30 N is hung from A. Find the reaction forces at C and D.

问题：一根匀质杆 AB 长 2 m，重 50 N，水平静置于两光滑支座 C 和 D 上，已知 AC = 0.5 m，DB = 0.4 m。在 A 点悬挂一个重 30 N 的质点。求支座 C 和 D 的反力。

Solution:

解答：

Step 1: Draw the rod with forces: weight 50 N acting at the midpoint (1 m from A), 30 N downward at A, and upward reactions R_C and R_D at C (0.5 m from A) and D (2 – 0.4 = 1.6 m from A).

步骤 1：画出杆的受力图：重力 50 N 作用于中点（距 A 1 m），A 点有 30 N 向下，C 点（距 A 0.5 m）和 D 点（距 A 1.6 m）有向上的支座反力 R_C 和 R_D。

Step 2: Resolve vertically: R_C + R_D = 50 + 30 = 80 N.

步骤 2：竖直方向受力平衡：R_C + R_D = 50 + 30 = 80 N。

Step 3: Take moments about A: (R_C × 0.5) + (R_D × 1.6) – (50 × 1) – (30 × 0) = 0 ⇒ 0.5 R_C + 1.6 R_D = 50.

步骤 3：对 A 点取矩：(R_C × 0.5) + (R_D × 1.6) – (50 × 1) – (30 × 0

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