Year 13 Edexcel Further Maths: Case Study Drills | Year 13 Edexcel 进阶数学：案例分析实战演练

📚 Year 13 Edexcel Further Maths: Case Study Drills | Year 13 Edexcel 进阶数学：案例分析实战演练

Case study style questions in Edexcel Year 13 Further Mathematics demand a confident blend of theory and problem-solving skills. This revision article walks you through ten targeted examples, covering Core Pure topics as well as selected applications from Further Mechanics and Further Statistics, to sharpen your exam technique.

Edexcel Year 13 进阶数学的案例分析题要求学生能够将理论与解题技巧有机融合。本文梳理了十个针对性案例，涵盖核心纯数内容以及部分力学与统计选修模块的典型应用题，帮助大家在练习中完善考试策略。

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1. Complex Loci & Apollonius Circle | 复数轨迹与阿波罗尼斯圆

Many Edexcel Further Maths questions ask you to identify the locus defined by an equation such as |z – a| = k|z – b|, where k is a positive constant and k ≠ 1. This always gives a circle.

很多Edexcel进阶数学题会要求识别由|z – a| = k|z – b|（k ≠ 1）所定义的轨迹，它一定是一个圆。

Let z = x + iy and consider the condition |z – 1| = 2|z + i|. Substituting gives |x + iy – 1| = 2|x + iy + i|.

设z = x + iy，考虑条件|z – 1| = 2|z + i|。代入后得到|x + iy – 1| = 2|x + iy + i|。

Square both sides and use the definition of modulus: (x – 1)² + y² = 4[x² + (y + 1)²]. Expand and collect terms.

两边平方并利用模的定义：(x – 1)² + y² = 4[x² + (y + 1)²]。展开并整理各项。

(x – 1)² + y² = 4x² + 4y² + 8y + 4 → 3x² + 3y² + 2x + 8y + 3 = 0

Divide by 3 and complete the square for x and y to obtain the circle’s standard form.

除以3后分别对x和y配方，得到圆的标准方程。

(x + 1/3)² + (y + 4/3)² = 8/9

Hence the locus is a circle with centre (-1/3, -4/3) and radius (2√2)/3. Sketching this on an Argand diagram always earns method marks.

因此轨迹是一个圆心为(-1/3, -4/3)、半径为(2√2)/3的圆。在复平面上画出该图像总能获得方法分。

When k = 1 the locus becomes the perpendicular bisector of the segment joining a and b – a straight line, not a circle.

当k=1时轨迹变成连接a与b线段的垂直平分线，这是一条直线而非圆。

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2. Matrices: Eigenvalues & Invariant Lines | 矩阵：特征值与不变直线

Interpreting a 2×2 matrix as a linear transformation is central to Edexcel Core Pure. A typical case study provides a matrix and asks for eigenvalues and invariant lines passing through the origin.

将2×2矩阵理解为线性变换是Edexcel核心纯数的核心内容。典型的案例分析给出一个矩阵，要求求特征值以及过原点的不变直线。

Let the matrix be M = [3 1; 4 2]. Solve det(M – λI) = 0: (3 – λ)(2 – λ) – 4 = 0 → λ² – 5λ + 2 = 0.

设矩阵M = [3 1; 4 2]。解特征方程det(M – λI) = 0：(3 – λ)(2 – λ) – 4 = 0 → λ² – 5λ + 2 = 0。

λ = (5 ± √17)/2

For each λ, solve (M – λI)v = 0 to find the corresponding eigenvector. The direction vectors are v₁ proportional to (1, (√17 – 1)/4) and v₂ proportional to (1, (-√17 – 1)/4).

对每一个λ，解(M – λI)v = 0得到对应的特征向量。方向向量分别为v₁正比于(1, (√17 – 1)/4)，v₂正比于(1, (-√17 – 1)/4)。

Lines through the origin in these directions are invariant lines because points on them are simply scaled by the eigenvalue under the transformation. In exam cases, you often label them y = m₁x and y = m₂x.

这两个方向所在的过原点直线就是不变直线，因为其上各点经变换后仅被特征值缩放。考试中通常标为y = m₁x与y = m₂x。

The case study may then ask you to describe the geometric effect of the transformation – a stretch by different factors along these two invariant directions.

案例还可能要求描述该变换的几何效果——即沿这两个不变方向进行不同比例的拉伸。

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3. Polar Coordinates & Area | 极坐标与面积

Polar curves such as cardioids appear regularly. Knowing how to set up the area integral correctly is critical.

心形线等极坐标曲线经常出现，正确建立面积积分至关重要。

Consider the cardioid r = 2(1 + cos θ). The full area is swept as θ runs from 0 to 2π.

考虑心形线 r = 2(1 + cos θ)。当θ从0变化到2π时扫出全部区域。

Area = ½ ∫₀²π r² dθ = ½ ∫₀²π 4(1 + cos θ)² dθ = 2 ∫₀²π (1 + 2cos θ + cos²θ) dθ

Use cos²θ = (1 + cos 2θ)/2 and integrate term by term: 2 ∫₀²π [3/2 + 2cos θ + ½ cos 2θ] dθ.

利用cos²θ = (1 + cos 2θ)/2，逐项积分：2 ∫₀²π [3/2 + 2cos θ + ½ cos 2θ] dθ。

The cosine terms integrate to zero over a full period, leaving 2 × [ (3/2)θ ]₀²π = 2 × 3π = 6π.

余弦项在完整周期内的积分为零，最终得到2 × [ (3/2)θ ]₀²π = 2 × 3π = 6π。

Always check symmetry: a cardioid is symmetric about the initial line, but the full-area integral from 0 to 2π accounts for this automatically.

务必留意对称性：心形线关于极轴对称，但从0到2π的全积分已自动包含这一点。

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4. Maclaurin Series & Limits | 麦克劳林级数与极限计算

Expanding a function like f(x) = ln(1 + sin x) as far as x⁴ tests differentiation and series manipulation.

将函数f(x) = ln(1 + sin x)展开到x⁴项，既考查微分又考查级数处理。

Start with f(0) = ln 1 = 0. f'(x) = cos x / (1 + sin x), so f'(0) = 1.

首先f(0)=ln1=0。f'(x) = cos x / (1 + sin x)，故f'(0)=1。

Compute f”(x) = [-sin x (1+sin x) – cos²x] / (1+sin x)². Evaluate at 0: f”(0) = -1.

计算f”(x) = [-sin x (1+sin x) – cos²x] / (1+sin x)²。代入x=0得f”(0) = -1。

Continuing, f”'(0) = 1 and f⁽⁴⁾(0) = -3. Thus the series to x⁴ is:

继续求导得f”'(0)=1，f⁽⁴⁾(0) = -3。因此级数到x⁴为：

ln(1 + sin x) ≈ x – ½ x² + (1/6)x³ – (1/8)x⁴ + …

A typical limit problem: limₓ→₀ [ln(1+sin x) – x + ½x²]/x³. Substitute the series: [x – ½x² + ⅙x³ – x + ½x²]/x³ = 1/6.

典型极限题：limₓ→₀ [ln(1+sin x) – x + ½x²]/x³。代入级数得[x – ½x² + ⅙x³ – x + ½x²]/x³ = 1/6。

This approach avoids multiple applications of L’Hôpital’s rule and demonstrates clear understanding.

该方法避免了多次使用洛必达法则，并展示出清晰的理解过程。

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5. Second-Order ODEs with Forcing | 受迫二阶常微分方程

Edexcel questions often feature a second-order linear ODE with constant coefficients and a trigonometric forcing term.

Edexcel考题常出现常系数二阶线性微分方程，并含有三角函数的强迫项。

Solve y” + 4y = 2 cos x. The auxiliary equation is m² + 4 = 0 → m = ±2i, so the complementary function is y_c = A cos 2x + B sin 2x.

解y” + 4y = 2 cos x。辅助方程 m² + 4 = 0 → m = ±2i，因此补函数为y_c = A cos 2x + B sin 2x。

The forcing frequency 1 does not resonate with the natural frequency 2, so try y_p = C cos x + D sin x. Substitute into the ODE and equate coefficients.

强迫频率1与自然频率2不产生共振，可尝试特解y_p = C cos x + D sin x。代入方程并比较系数。

y_p” = -C cos x – D sin x ⇒ (-C + 4C)cos x + (-D + 4D)sin x = 2 cos x

This gives 3C = 2, 3D = 0, so C = 2/3, D = 0. The general solution is y = A cos 2x + B sin 2x + (2/3) cos x.

得到3C=2, 3D=0，因此C=2/3, D=0。通解为y = A cos 2x + B sin 2x + (2/3) cos x。

Given initial conditions, e.g. y(0)=1, y'(0)=0, determine A and B to complete the particular solution.

若给定初始条件（例如y(0)=1, y'(0)=0），求解A和B即可得出特解。

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6. Hyperbolic Functions & Integrals | 双曲函数与积分

Standard integrals involving √(a² + x²) are routinely handled via hyperbolic substitutions, which appears in Core Pure.

含有√(a² + x²)的标准积分通常通过双曲代换处理，这是核心纯数中的常规考点。

Evaluate ∫ dx/√(1 + 9x²). Let 3x = sinh u, then 3 dx = cosh u du, so dx = (1/3) cosh u du. The denominator √(1 + sinh²u) = cosh u.

求∫ dx/√(1 + 9x²)。令3x = sinh u，则3 dx = cosh u du，dx = (1/3) cosh u du。分母√(1 + sinh²u) = cosh u。

The integral reduces to ∫ (1/3) du = (1/3) u + C. Re-substitute: u = arsinh(3x). Hence the answer is (1/3) arsinh(3x) + C.

积分简化为∫ (1/3) du = (1/3) u + C。代回原变量：u = arsinh(3x)。因此答案为(1/3) arsinh(3x) + C。

In logarithmic form, arsinh(3x) = ln(3x + √(1+9x²)), matching the alternative expression often seen in mark schemes.

用对数形式表达，arsinh(3x) = ln(3x + √(1+9x²))，这与评分方案中常见的形式一致。

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7. Vectors: Shortest Distance Between Skew Lines | 向量：异面直线最短距离

Finding the shortest distance between two skew lines is a classic Further Pure problem that requires fluent vector product work.

求两异面直线的最短距离是经典的进阶纯数问题，需要熟练运用向量积。

Let L₁: r = i + 2j + λ(i – j + k) and L₂: r = j + 2k + μ(2i + j). Direction vectors are d₁ = i – j + k and d₂ = 2i + j.

设直线L₁: r = i + 2j + λ(i – j + k)，L₂: r = j + 2k + μ(2i + j)。方向向量为d₁ = i – j + k和d₂ = 2i + j。

Compute the cross product d₁ × d₂ = -i + 2j + 3k. Its magnitude is √(1+4+9) = √14.

计算叉积d₁ × d₂ = -i + 2j + 3k，其模长为√(1+4+9) = √14。

Take a point on each line: A₁=(1,2,0), A₂=(0,1,2). The vector A₁A₂ = -i – j + 2k. The shortest distance d = |(A₁A₂)·(d₁×d₂)| / |d₁×d₂|.

取每条直线上一点：A₁(1,2,0)，A₂(0,1,2)。向量A₁A₂ = -i – j + 2k。最短距离d = |(A₁A₂)·(d₁×d₂)| / |d₁×d₂|。

d = |(-1)(-1) + (-1)(2) + (2)(3)| / √14 = |1 – 2 + 6|/√14 = 5/√14

Writing the final answer in exact form and rationalising the denominator secures full marks.

将最终答案写成精确形式并有理化分母，可获得全部分数。

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8. Poisson Hypothesis Testing | 泊松假设检验

In Further Statistics, hypothesis tests for a Poisson mean arise frequently with industrial or biological data.

在进阶统计中，对泊松分布均值的假设检验在

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