Year 12 OCR Further Mathematics: Case Study Practical Exercises | Year 12 OCR 进阶数学：案例分析实战演练

📚 Year 12 OCR Further Mathematics: Case Study Practical Exercises | Year 12 OCR 进阶数学：案例分析实战演练

Welcome to a deep-dive revision resource where abstract Further Mathematics concepts are brought to life through carefully designed case studies. We will explore complex numbers, matrix algebra, series, vectors, mechanics, and statistical inference in the context of OCR Year 12 specification-style problems. Each section presents a realistic scenario, breaks down the method, and links directly to assessment objectives, helping you master the reasoning as well as the computation.

欢迎来到深度复习资源，这里通过精心设计的案例分析，将抽象的进阶数学概念生动呈现。我们将以OCR Year 12考试要求为背景，探索复数、矩阵代数、级数、向量、力学以及统计推断。每一节都展示一个真实场景，分解解题方法，并直接链接到评估目标，帮助你同时掌握推理过程和计算技巧。

1. Introduction to Case Studies | 案例分析导论

A case study in Further Mathematics is more than an exam question — it is a scaffolded investigation that connects multiple topic areas. In the OCR specification, these often appear in the applied modules (Mechanics and Statistics) or as multi-step pure problems demanding proof, calculation, and interpretation.

在进阶数学中，案例分析不仅仅是一道考题，它是一项有支架的探究，连接多个知识领域。在OCR考纲中，案例分析常出现在应用模块（力学和统计学）中，也以纯数学多步问题的形式出现，要求证明、计算与解释。

The key skills tested are: modelling a real-world situation mathematically, selecting the correct technique, executing accurate algebraic manipulation, and critically evaluating the result. We will practice these through ten targeted case studies covering pure, mechanics, and statistics.

考查的关键技能包括：将现实情境数学建模、选择正确的技巧、执行准确的代数运算，并批判性地评估结果。我们将通过十个有针对性的案例分析（涵盖纯数、力学和统计）来练习这些技能。

Approach: Model → Solve → Interpret → Refine

方法：建模 → 求解 → 解释 → 优化

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2. Complex Numbers and Polynomial Roots | 复数与多项式根

Scenario: An engineer analyses a cubic equation that models the resonance frequency of a damped oscillator: x³ − 5x² + 9x − 5 = 0. Given that 2 + i is a root, find all roots and express the polynomial as a product of linear factors over the complex field.

情境：一位工程师分析一个模拟阻尼振荡器共振频率的三次方程：x³ − 5x² + 9x − 5 = 0。已知 2 + i 是一个根，找出所有根并用复数域上的线性因子乘积表示该多项式。

Because coefficients are real, the complex conjugate 2 − i is also a root. The quadratic factor is therefore (x − (2 + i))(x − (2 − i)) = x² − 4x + 5. Dividing the original cubic by this quadratic gives the remaining linear factor (x − 1). Hence the roots are 1, 2 + i, 2 − i. Factorisation: (x − 1)(x − (2 + i))(x − (2 − i)).

由于系数为实数，共轭复数 2 − i 也是根。因此二次因子为 (x − (2 + i))(x − (2 − i)) = x² − 4x + 5。将原三次方程除以该二次式得到剩余的线性因子 (x − 1)。因此根为 1, 2 + i, 2 − i。因式分解：(x − 1)(x − (2 + i))(x − (2 − i))。

This case reinforces the conjugate root theorem and polynomial division, which are essential for argand diagrams and finding complex solutions in further pure core.

本案例强化了共轭根定理和多项式除法，这些是复平面图和进一步纯数核心中求解复数解的基础。

Sum of roots = 5, Product = 5

根之和 = 5，根之积 = 5

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3. Matrix Transformations in Geometry | 几何中的矩阵变换

Scenario: A graphic designer wants to rotate a shape by 120° anticlockwise about the origin and then reflect it in the line y = x. Determine the single 2 × 2 transformation matrix that represents this combined transformation, and describe its effect geometrically.

情境：一位图形设计师希望将图形绕原点逆时针旋转120°，再对直线 y = x 进行反射。确定代表此组合变换的单一 2 × 2 变换矩阵，并从几何上描述其效果。

The rotation matrix R for 120° (2π/3) is: [cos120° −sin120°; sin120° cos120°] = [−½ −√3/2; √3/2 −½]. The reflection in y = x is M = [0 1; 1 0]. The combined matrix T = M × R (apply rotation then reflection, so pre-multiply by M). Compute T = [0 1; 1 0] × [−½ −√3/2; √3/2 −½] = [√3/2 −½; −½ −√3/2]. This matrix corresponds to a rotation of 150° anticlockwise, because cos150° = −√3/2? Wait, check: T = [√3/2, −½; −½, −√3/2] has det = −¾ − ¼ = −1, so it is a reflection or roto-reflection? Actually det = (√3/2)(−√3/2) − (−½)(−½) = −3/4 − 1/4 = −1, so it is a reflection, not a pure rotation. Indeed, reflection in y=x followed by rotation? No, we applied rotation then reflection. The determinant of reflection is −1. So T is a reflection about a line through origin at 75° to the x-axis. Geometrically, it’s a reflection.

旋转120° (2π/3) 的旋转矩阵 R 为: [cos120° −sin120°; sin120° cos120°] = [−½ −√3/2; √3/2 −½]。关于 y = x 的反射矩阵 M = [0 1; 1 0]。组合矩阵 T = M × R（先旋转再反射，故左乘 M）。计算得 T = [0 1; 1 0] × [−½ −√3/2; √3/2 −½] = [√3/2 −½; −½ −√3/2]。该矩阵对应于一条过原点且与x轴成75°角的直线反射，因为行列式为 −1，表示反射而非旋转。

OCR examiners expect you to multiply matrices in the correct order and interpret the determinant to distinguish rotations (-1 for reflection, +1 for rotation). This case blends linear algebra with geometry.

OCR考官要求考生按正确顺序进行矩阵乘法，并通过行列式区分旋转（行列式为+1）和反射（行列式为−1）。该案例融合了线性代数与几何。

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4. Summation of Series Using Standard Results | 使用标准结果求和级数

Scenario: A programmer needs the sum of the first n terms of the sequence formed by multiplying triangular numbers by the corresponding odd numbers: Tₖ × (2k − 1), where Tₖ = k(k+1)/2. Derive an expression in terms of n.

情境：一位程序员需要求数列前 n 项之和，该数列由三角形数乘以对应的奇数构成：Tₖ × (2k − 1)，其中 Tₖ = k(k+1)/2。推导关于 n 的表达式。

The term uₖ = [k(k+1)/2] × (2k − 1) = ½ (k² + k)(2k − 1) = ½ (2k³ − k² + 2k² − k) = ½ (2k³ + k² − k). So sum Sₙ = ½ Σ(2k³ + k² − k) from k=1 to n. Split: Sₙ = ½ [2 Σk³ + Σk² − Σk]. Use standard results: Σk = n(n+1)/2, Σk² = n(n+1)(2n+1)/6, Σk³ = [n(n+1)/2]². Substitute and simplify: Sₙ = ½ { 2 × [n²(n+1)²/4] + n(n+1)(2n+1)/6 − n(n+1)/2 }. Factor n(n+1)/2: = ½ { [n²(n+1)²/2] + n(n+1)(2n+1)/6 − n(n+1)/2 }. Get common denominator 6 etc. Finally yields Sₙ = n(n+1)(3n² + n − 1)/12. Check small n to verify.

通项 uₖ = [k(k+1)/2] × (2k − 1) = ½ (k² + k)(2k − 1) = ½ (2k³ + k² − k)。因此求和 Sₙ = ½ Σ(2k³ + k² − k) 从 k=1 到 n。拆分：Sₙ = ½ [2 Σk³ + Σk² − Σk]。使用标准结果：Σk = n(n+1)/2，Σk² = n(n+1)(2n+1)/6，Σk³ = [n(n+1)/2]²。代入化简后可得 Sₙ = n(n+1)(3n² + n − 1)/12。用较小的 n 验证。

Mastering series manipulation is vital for OCR Further Pure, especially when combined with telescoping sums or proof by induction.

掌握级数处理对于OCR进阶纯数至关重要，尤其是结合裂项求和或归纳证明时。

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5. Vector Geometry: Intersections and Angles | 向量几何：交点与角度

Scenario: Two lines L₁: r = (2i + 3j − k) + λ(i − 2j + 2k) and L₂: r = (4i + j + k) + μ(2i + j − 2k) are given. Determine whether they intersect, and if so, find the acute angle between their direction vectors.

情境：给定两条直线 L₁: r = (2i + 3j − k) + λ(i − 2j + 2k) 和 L₂: r = (4i + j + k) + μ(2i + j − 2k)。判断它们是否相交，如果相交，求其方向向量之间的锐角。

Set equal: 2 + λ = 4 + 2μ, 3 − 2λ = 1 + μ, −1 + 2λ = 1 − 2μ. Solve first two: λ − 2μ = 2, and −2λ − μ = −2 ⇒ 2λ + μ = 2. Add: 3λ = 4 ⇒ λ = 4/3, then μ = (2 − 8/3) = −2/3. Check third: LHS = −1 + 2(4/3) = 5/3; RHS = 1 − 2(−2/3) = 1 + 4/3 = 7/3, not equal. So no intersection. The angle between directions d₁ = (1, −2, 2) and d₂ = (2, 1, −2) is found using dot product: d₁·d₂ = 2 − 2 − 4 = −4. Magnitudes: |d₁| = √(1+4+4)=3, |d₂| = √(4+1+4)=3. cos θ = |−4|/(9) = 4/9 (taking acute). θ = arccos(4/9) ≈ 63.6°.

令位置向量相等：2 + λ = 4 + 2μ, 3 − 2λ = 1 + μ, −1 + 2λ = 1 − 2μ。解前两个方程得 λ = 4/3, μ = −2/3，代入第三式不成立，故不相交。方向向量 d₁ = (1, −2, 2) 与 d₂ = (2, 1, −2) 的点积为 −4，模长均为3，故锐角余弦值为 4/9，θ ≈ 63.6°。

This case assesses solving simultaneous vector equations, consistency checking, and 3D coordinate geometry, all heavily examined.

本案例考查向量方程联立求解、一致性检验及三维坐标几何，均为高频考点。

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6. Roots of Equations and Vieta’s Formulas | 方程的根与韦达定理

Scenario: The cubic equation 2x³ − 3x² + kx − 6 = 0 has roots α, β, γ. It is known that αβ = 2. Find k and the roots.

情境：三次方程 2x³ − 3x² + kx − 6 = 0 有根 α, β, γ。已知 αβ = 2。求 k 和所有的根。

Using Vieta: sum α+β+γ = 3/2, sum of product pairs αβ+βγ+γα = k/2, product αβγ = 6/2 = 3. From αβ = 2, product gives γ = 3/2. Then α+β = (3/2) − γ = 0, so β = −α. Then αβ = α(−α) = −α² = 2 ⇒ α² = −2, so α = i√2, β = −i√2. Then αβ+βγ+γα = 2 + βγ + γα. Since α+β=0, γ(α+β)=0, so sum of pairwise products = αβ = 2. Thus k/2 = 2 ⇒ k = 4. The roots are i√2, −i√2, 3/2. The equation is 2x³ −3x² +4x −6 = 0.

由韦达定理：α+β+γ = 3/2，αβ+βγ+γα = k/2，αβγ = 6/2 = 3。由 αβ = 2 得 γ = 3/2。进而 α+β = 0 ⇒ β = −α。于是 αβ = −α² = 2 ⇒ α² = −2，α = i√2，β = −i√2。然后 αβ+βγ+γα = 2 + γ(α+β) = 2，故 k/2 = 2 ⇒ k = 4。根为 i√2, −i√2, 3/2。方程为 2x³ −3x² +4x −6 = 0。

Vieta’s formulas efficiently link coefficients and roots, especially when pairwise product is given — a common trick in OCR papers.

韦达定理能高效关联系数与根，尤其在给出两两乘积时，这是OCR试卷中的常见技巧。

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7. Work, Energy and Power in Mechanics | 力学中的功、能与功率

Scenario: A parcel of mass 5 kg slides down a rough slope inclined at 30° to the horizontal. It starts from rest and after moving 4 m its speed is 3 m s⁻¹. Find the work done against friction and the coefficient of friction.

情境：一个质量为5 kg 的包裹沿与水平面成30°的粗糙斜面下滑。从静止开始，移动4 m 后速度为 3 m s⁻¹。求克服摩擦力所做的功及摩擦系数。

Change in GPE = mgh = 5 × 9.8 × (4 sin30°) = 5 × 9.8 × 2 = 98 J. Increase in KE = ½ × 5 × 3² = 22.5 J. Work done against friction = loss in mechanical energy = 98 − 22.5 = 75.5 J. Friction force F = work done / distance = 75.5 / 4 = 18.875 N. Normal reaction R = 5g cos30° = 5 × 9.8 × √3/2 = 42.435 N. Coefficient of friction μ = F/R = 18.875 / 42.435 ≈ 0.445.

重力势能减少 = mgh = 5 × 9.8 × (4 sin30°) = 98 J。动能增加 = ½ × 5 × 3² = 22.5 J。克服摩擦力做功 = 机械能损失 = 98 − 22.5 = 75.5 J。摩擦力 F = 做功 / 距离 = 75.5 / 4 = 18.875 N。法向反作用力 R = 5g cos30° = 5 × 9.8 × √3/2 = 42.435 N。摩擦系数 μ = F/R ≈ 0.445。

The principle of conservation of energy is a foundation of OCR Mechanics Major. Always check if there is any other work done, such as by a pulling force.

能量守恒原理是OCR力学主修的基础。务必检查是否有其他力做功，例如拉力。

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8. Collisions and Coefficient of Restitution | 碰撞与恢复系数

Scenario: Two smooth spheres A (0.2 kg) and B (0.3 kg) move on a horizontal table with speeds 4 m s⁻¹ and 2 m s⁻¹ in opposite directions. After a direct collision, B reverses with speed 1 m s⁻¹. Find the speed of A after impact and the coefficient of restitution e.

情境：两个光滑球体A（0.2 kg）和B（0.3 kg）在水平桌面上以4 m s⁻¹ 和 2 m s⁻¹ 的速度相向运动。发生正碰后，B以1 m s⁻¹的速度反向运动。求碰撞后A的速度和恢复系数 e。

Take direction of A as positive. Initial momentum: (0.2 × 4) + (0.3 × (−2)) = 0.8 − 0.6 = 0.2 kg m s⁻¹. Let vₐ be velocity of A after. Momentum after: 0.2 vₐ + 0.3 × (1) (since B’s velocity is now +1 m s⁻¹). Conservation: 0.2 vₐ + 0.3 = 0.2 ⇒ vₐ = −0.1/0.2 = −0.5 m s⁻¹, i.e., A reverses at 0.5 m s⁻¹. Then e = (speed of separation) / (speed of approach) = (v_B − v_A) after / (u_A − u_B) before. Note separation velocity = 1 − (−0.5) = 1.5 m s⁻¹, approach velocity = 4 − (−2) = 6 m s⁻¹. So e = 1.5/6 = 0.25.

取A的方向为正。初始动量：0.8 − 0.6 = 0.2 kg m s⁻¹。设碰撞后A的速度为vₐ，B的速度为+1 m s⁻¹。动量守恒：0.2 vₐ + 0.3 = 0.2 ⇒ vₐ = −0.5 m s⁻¹，即A以0.5 m s⁻¹反向。分离速度 = 1 − (−0.5) = 1.5 m s⁻¹，接近速度 = 4 − (−2) = 6 m s⁻¹，故 e = 1.5/6 = 0.25。

Always define a positive direction and keep signs consistent. OCR expects clear working for impulse as well, which is change in momentum.

务必规定正方向并保持符号一致。OCR也要求清晰计算冲量，即动量变化量。

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9. Hypothesis Testing with Poisson Distribution | 泊松分布的假设检验

Scenario: A call centre receives an average of 8 complaints per day. After a new system, over 10 days the total complaints are 95. Test at the 5% significance level whether there is evidence that the rate has changed.

情境：某呼叫中心平均每天接到8起投诉。新系统实施后，10天内共接到95起投诉。在5%的显著性水平下检验是否有证据表明投诉率发生了变化。

Null H₀: λ = 8 per day. Over 10 days, λ = 80. Test statistic X ~ Po(80). Approximate with normal: X ~ N(80, 80). Two-tail test. Observed x = 95. z = (95 − 80) / √80 ≈ 15 / 8.944 = 1.677. Critical value for 5% two-tail is ±1.96. Since 1.677 < 1.96, we do not reject H₀. Insufficient evidence that rate has changed. (Alternatively use exact Poisson tails, but normal approximation is expected in Year 12 OCR).

原假设 H₀: λ = 8 每天。10天 λ = 80。检验统计量 X ~ Po(80)。用正态近似：X ~ N(80, 80)。双侧检验。观测值 x = 95。z = (95 − 80) / √80 ≈ 1.677。双侧5%临界值为 ±1.96。因 1.677 < 1.96，不拒绝 H₀。证据不足以说明投诉率改变。（亦可用精确泊松尾部，但OCR Year 12要求正态近似。）

Remember to state conclusion in context. OCR Statistics Minor heavily tests Poisson modelling and distributional approximations.

记住要在语境中陈述结论。OCR统计辅修重点考查泊松建模和分布近似。

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10. Chi-Squared Tests for Independence | 卡方独立性检验

Scenario: A survey of 200 people investigates whether preference for three brands (A, B, C) is independent of age group (Under 30, 30-50, Over 50). The observed frequencies are given in a table. Perform a chi-squared test at the 1% level, stating hypotheses and conclusion.

情境：一项200人的调查探究三个品牌（A, B, C）的偏好是否与年龄组（30岁以下、30-50岁、50岁以上）独立。观测频数如表所示。在1%显著性水平下进行卡方检验，陈述假设和结论。

Observed table (rows: age groups, columns: brands):

Age \ Brand	A	B	C	Total
<30	30	45	25	100
30-50	20	30	10	60
>50	10	5	25	40
Total	60	80	60	200

H₀: Brand preference and age group are independent. H₁: They are associated. Expected frequencies = (row total × column total) / grand total. Compute χ² = Σ (O−E)² / E. Degrees of freedom = (3−1)×(3−1)=4. Critical value at 1% (4 df) is 13.277. Observed χ² value will be calculated; if larger, reject H₀. This example teaches layout and interpretation.

H₀：品牌偏好与年龄组独立。H₁：两者相关。期望频数 = (行合计 × 列合计) / 总计。χ² = Σ (O−E)² / E。自由度 = (3−1)×(3−1)=4。1%水平下临界值为13.277。若计算出的χ²值更大，则拒绝H₀。该案例教格式和解读。

OCR expects clear tables, correct degrees of freedom, and a conclusion in non-technical language.

OCR要求清晰的表格、正确的自由度，以及用非技术语言陈述结论。

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11. Final Synthesis and Exam Tips | 综合归纳与考试技巧

These case studies mirror the integrated nature of OCR Year 12 Further Mathematics. Success relies on recognising the underlying mathematical structure, not just plugging numbers. Always state assumptions, show diagrams, and validate your answer against the context.

这些案例分析反映了OCR Year 12进阶数学的综合性。成功依赖于识别背后的数学结构，而不仅仅是代入数字。务必陈述假设、画图，并根据语境验证答案。

• In pure topics, algebraic manipulation and exact values are rewarded over decimals.

  在纯数课题中，代数运算和精确值比小数更受青睐。

• In mechanics, draw a force diagram and define positive direction before writing equations.

  力学中，在列方程前先画受力图并规定正方向。

• In statistics, interpret p-values or test statistics in the context of the problem.

  统计学中，结合问题背景解释p值或检验统计量。

Practise these case studies repeatedly, varying parameters to build confidence. Your ability to apply knowledge flexibly is what will earn top marks in OCR Further Mathematics.

反复练习这些案例，改变参数以建立信心。灵活应用知识的能力将帮助你在OCR进阶数学中获得高分。

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