📚 Case Study Practical Exercises for Edexcel Further Mathematics Year 13 | Edexcel 进阶数学 Year 13：案例分析实战演练
This article presents a series of case study exercises designed for Year 13 Edexcel Further Mathematics, covering topics such as complex numbers, matrices, differential equations, and series expansions. Each case study integrates multiple concepts to mirror the style of extended exam questions, encouraging deep understanding and problem-solving skills. Work through each example carefully, noting the step-by-step reasoning and the connection between algebraic manipulation and geometric interpretation.
本文为 Year 13 Edexcel 进阶数学课程设计了一系列案例分析实战演练，涵盖复数、矩阵、微分方程和级数展开等主题。每个案例融合多个知识点，模拟考试中综合大题的形式，旨在培养深层理解与解题能力。请仔细演练每个示例，关注逐步推理过程以及代数运算与几何意义之间的联系。
1. Complex Transformation of a Locus | 复数轨迹的变换
A transformation T: ℂ → ℂ is defined by w = (z – 2i)/(iz + 1). Find the image of the line Im(z) = 1 under this transformation. Begin by setting z = x + i, since Im(z) = 1. Substitute into the transformation and simplify the expression for w. Then separate real and imaginary parts. You will find that w satisfies the equation of a circle in the complex plane. Show that the image is the circle |w – (1/2) – (3/2)i| = √5/2. This illustrates how a Möbius transformation maps lines not passing through the pole to circles.
定义变换 T: ℂ → ℂ 为 w = (z – 2i)/(iz + 1)。求直线 Im(z) = 1 在此变换下的像。首先设 z = x + i，因为 Im(z) = 1。代入变换并化简 w 的表达式，然后分离实部和虚部。你会发现 w 满足复平面上的一个圆的方程。证明其像为圆 |w – (1/2) – (3/2)i| = √5/2。这体现了莫比乌斯变换将不通过极点的直线映射为圆的性质。
2. Matrix Representation of a Linear Transformation | 线性变换的矩阵表示
A linear transformation maps the vectors i, j, k to columns of matrix M. Given that M = [[2, -1, 3], [0, 4, 1], [-2, 5, 0]], find the image of the plane r·(1, 2, -1) = 3 under this transformation. First, express the plane in parametric form: r = a + λu + μv. Apply the transformation to each point on the plane by multiplying M by the position vector. The image is a plane spanned by Mu and Mv, passing through Ma. Determine its Cartesian equation. This exercise reinforces how matrices act on geometric objects, a key concept in Further Pure 1.
某线性变换将基向量 i, j, k 映射为矩阵 M 的列。已知 M = [[2, -1, 3], [0, 4, 1], [-2, 5, 0]]，求平面 r·(1, 2, -1) = 3 在该变换下的像。先将平面写成参数形式：r = a + λu + μv。用 M 乘各点位置向量，对平面上每一点施加变换。像平面由 Mu 和 Mv 张成，且经过 Ma。确定其笛卡尔方程。此练习巩固矩阵在几何对象上的作用，是 Further Pure 1 的核心概念。
3. First-Order Differential Equation with Integrating Factor | 一阶微分方程与积分因子
Consider the differential equation dy/dx + y tan x = sec x, with y(0) = 1. Identify the integrating factor e^(∫tan x dx) = sec x. Multiply through and rewrite the left side as d/dx (y sec x) = sec² x. Integrate both sides to get y sec x = tan x + C. Use the initial condition to find C. The particular solution is y = sin x + cos x. Verify your answer by substitution. This case highlights the use of trigonometric identities and the method of integrating factors in a non-separable first-order ODE.
考虑微分方程 dy/dx + y tan x = sec x，满足 y(0) = 1。找出积分因子 e^(∫tan x dx) = sec x。两边同乘后，将左侧改写为 d/dx (y sec x) = sec² x。两边积分得 y sec x = tan x + C。利用初始条件求 C，特解为 y = sin x + cos x。通过代入验证答案。本案例重点展示了在不可分离的一阶常微分方程中，积分因子法和三角恒等式的运用。
4. Second-Order Homogeneous ODE with Complex Roots | 具有复根的二阶齐次常微分方程
Solve the equation d²x/dt² + 2 dx/dt + 5x = 0, with x(0) = 2 and dx/dt = -2 at t = 0. The auxiliary equation m² + 2m + 5 = 0 gives complex roots m = -1 ± 2i. Hence the general solution is x = e⁻ᵗ (A cos 2t + B sin 2t). Apply initial conditions to determine A and B. You obtain A = 2 and B = 0, so the particular solution is x = 2e⁻ᵗ cos 2t. Interpret the motion as damped oscillations. This model appears in mechanics and electrical circuits, linking second-order ODEs with physical case studies.
解方程 d²x/dt² + 2 dx/dt + 5x = 0，初始条件为 x(0) = 2，dx/dt = -2 当 t = 0。辅助方程 m² + 2m + 5 = 0 给出复根 m = -1 ± 2i。因此通解为 x = e⁻ᵗ (A cos 2t + B sin 2t)。代入初始条件确定 A 和 B，得 A = 2，B = 0，特解为 x = 2e⁻ᵗ cos 2t。该运动可解释为阻尼振荡。此模型常出现在力学和电路中，将二阶常微分方程与物理案例分析联系起来。
5. Maclaurin Series for a Rational Function | 有理函数的麦克劳林级数
Find the Maclaurin series expansion of f(x) = ln(1 + x)/(1 – x) up to the term in x⁴. First expand ln(1 + x) = x – x²/2 + x³/3 – x⁴/4 + … and (1 – x)⁻¹ = 1 + x + x² + x³ + x⁴ + … . Then multiply the series carefully, collecting terms up to x⁴. The result is x + x²/2 + 5x³/6 + 13x⁴/12 + … . Alternatively, use the fact that f(x) = ln(1 + x) + ln(1 – x)⁻¹ = ln((1 + x)/(1 – x)) and differentiate first, but direct series multiplication is efficient here. Validate the series by evaluating the derivative at zero.
求 f(x) = ln(1 + x)/(1 – x) 的麦克劳林展开式，至 x⁴ 项。首先展开 ln(1 + x) = x – x²/2 + x³/3 – x⁴/4 + … 以及 (1 – x)⁻¹ = 1 + x + x² + x³ + x⁴ + … 。然后仔细将两级数相乘，收集到 x⁴ 项，结果为 x + x²/2 + 5x³/6 + 13x⁴/12 + … 。另一种方法是利用 f(x) = ln(1 + x) + ln(1 – x)⁻¹ = ln((1 + x)/(1 – x)) 并先求导，但直接级数乘法在此更高效。通过计算零点处的导数值验证该级数。
6. Polar Coordinates: Area and Tangent | 极坐标：面积与切线
The curve C has polar equation r = 2 + sin 3θ. Find the area of one loop of the curve. First determine the limits of θ where r = 0, giving sin 3θ = -2 (no solution) so the curve has no self-intersection? Actually, this curve does not pass through the pole; it is a dimpled limaçon. The whole area is found by integrating ½ r² dθ from 0 to 2π. Calculate the area as ∫₀²π ½ (2 + sin 3θ)² dθ = ½ ∫₀²π (4 + 4 sin 3θ + sin² 3θ) dθ = 9π/2. Also find the tangent at θ = π/6. The slope dy/dx = (r’ sin θ + r cos θ)/(r’ cos θ – r sin θ). Evaluate at θ = π/6: r = 2 + 1 = 3, r’ = 3 cos 3θ = 0, so slope = (0 * ½ + 3 * √3/2)/(0 * √3/2 – 3 * ½) = (3√3/2)/(-3/2) = -√3. This demonstrates polar calculus in the Edexcel syllabus.
曲线 C 的极坐标方程为 r = 2 + sin 3θ。求该曲线一个环的面积。首先确定 r = 0 的 θ 范围，得到 sin 3θ = -2（无解），因此该曲线不经过极点；它是一个有凹陷的蜗牛线。整个面积可通过从 0 到 2π 积分 ½ r² dθ 求得。计算面积为 ∫₀²π ½ (2 + sin 3θ)² dθ = ½ ∫₀²π (4 + 4 sin 3θ + sin² 3θ) dθ = 9π/2。再求在 θ = π/6 处的切线。斜率 dy/dx = (r’ sin θ + r cos θ)/(r’ cos θ – r sin θ)。在 θ = π/6 计值：r = 2 + 1 = 3，r’ = 3 cos 3θ = 0，于是斜率 = (0 * ½ + 3 * √3/2)/(0 * √3/2 – 3 * ½) = (3√3/2)/(-3/2) = -√3。这展示了 Edexcel 大纲中的极坐标微积分。
7. Hyperbolic Functions and Integration | 双曲函数与积分
Evaluate the integral ∫ dx/√(x² + 4x + 13). Complete the square inside the root: x² + 4x + 13 = (x + 2)² + 9. Use the substitution x + 2 = 3 sinh u, dx = 3 cosh u du. Then √((x+2)² + 9) = 3 cosh u. The integral reduces to ∫ du = u + C. Back-substituting gives arsinh((x+2)/3) + C. Alternatively, using the logarithmic form, we get ln|x + 2 + √(x² + 4x + 13)| + C. This exercise connects hyperbolic functions with integration techniques required for Further Pure and resembles typical case-study exam problems.
计算积分 ∫ dx/√(x² + 4x + 13)。将根号内配方：x² + 4x + 13 = (x + 2)² + 9。使用代换 x + 2 = 3 sinh u，dx = 3 cosh u du。则 √((x+2)² + 9) = 3 cosh u。积分简化为 ∫ du = u + C。回代得到 arsinh((x+2)/3) + C。另一种方法使用对数形式，结果为 ln|x + 2 + √(x² + 4x + 13)| + C。此练习将双曲函数与进阶纯数所需的积分技巧联系起来，类似典型的案例分析考题。
8. Series Solution of a Differential Equation | 微分方程的级数解
Find a series solution about x = 0 for the differential equation y” – xy’ + 2y = 0, up to the term in x⁵. Assume y = ∑ aₙ xⁿ. Differentiate and plug into the equation, aligning powers of x. Recurrence relation: (n+2)(n+1)aₙ₊₂ – n aₙ + 2aₙ = 0, which simplifies to aₙ₊₂ = (n – 2)aₙ/[(n+2)(n+1)]. With a₀ and a₁ arbitrary, we compute a₂ = -a₀, a₃ = -a₁/6, a₄ = 0, a₅ = a₁/120. Thus y = a₀(1 – x²) + a₁(x – x³/6 + x⁵/120) + … . Recognize the a₁ series as the expansion of sin x. This method is essential when closed-form solutions are not readily available.
求微分方程 y” – xy’ + 2y = 0 在 x = 0 附近的级数解，至 x⁵ 项。设 y = ∑ aₙ xⁿ。求导并代入方程，对齐 x 的幂次。递推关系为：(n+2)(n+1)aₙ₊₂ – n aₙ + 2aₙ = 0，化简得 aₙ₊₂ = (n – 2)aₙ/[(n+2)(n+1)]。a₀ 和 a₁ 为任意常数，计算得 a₂ = -a₀，a₃ = -a₁/6，a₄ = 0，a₅ = a₁/120。于是 y = a₀(1 – x²) + a₁(x – x³/6 + x⁵/120) + … 。可识别 a₁ 对应的级数正是 sin x 的展开式。当闭式解不易获得时，此方法至关重要。
9. Eigenvalues and Cayley-Hamilton Theorem | 特征值与凯莱-哈密顿定理
For matrix M = [[3, 2], [1, 4]], find eigenvalues and verify the Cayley-Hamilton theorem. The characteristic equation is λ² – (trace)λ + det = 0 → λ² – 7λ + 10 = 0, giving λ = 5, 2. Then compute M² – 7M + 10I. M² = [[3,2],[1,4]]² = [[11,14],[7,18]]. Subtract 7M = [[21,14],[7,28]] and add 10I = [[10,0],[0,10]] to get [[0,0],[0,0]], confirming the theorem. Use this to express M⁻¹ as (1/det)(7I – M) = (1/10)[[4,-2],[-1,3]] = [[0.4,-0.2],[-0.1,0.3]]. This case shows how the theorem simplifies matrix calculations, a valuable technique in Further Mathematics examinations.
对于矩阵 M = [[3, 2], [1, 4]]，求特征值并验证凯莱-哈密顿定理。特征方程为 λ² – (迹)λ + det = 0 → λ² – 7λ + 10 = 0，得 λ = 5, 2。然后计算 M² – 7M + 10I。M² = [[3,2],[1,4]]² = [[11,14],[7,18]]。减去 7M = [[21,14],[7,28]] 并加 10I = [[10,0],[0,10]]，得到 [[0,0],[0,0]]，验证了定理。利用此结果将 M⁻¹ 表示为 (1/det)(7I – M) = (1/10)[[4,-2],[-1,3]] = [[0.4,-0.2],[-0.1,0.3]]。本例展示了该定理如何简化矩阵计算，是进阶数学考试中的一种重要技巧。
10. Further Complex Numbers: De Moivre’s Theorem Application | 进一步复数：棣莫弗定理的应用
Express cos 4θ in terms of cos θ using De Moivre’s theorem. Start with (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ. Expand the left side using the binomial theorem: cos⁴θ + 4i cos³θ sin θ – 6 cos²θ sin²θ – 4i cos θ sin³θ + sin⁴θ. Equate real parts: cos 4θ = cos⁴θ – 6 cos²θ sin²θ + sin⁴θ. Replace sin²θ with 1 – cos²θ to obtain cos 4θ = 8 cos⁴θ – 8 cos²θ + 1. This identity is useful in integration and series problems, linking De Moivre to trigonometric expansions, a staple in the Edexcel Core Pure specification.
利用棣莫弗定理用 cos θ 表示 cos 4θ。由 (cos θ + i sin θ)⁴ = cos 4θ + i sin 4θ 开始。用二项式定理展开左边：cos⁴θ + 4i cos³θ sin θ – 6 cos²θ sin²θ – 4i cos θ sin³θ + sin⁴θ。取实部相等：cos 4θ = cos⁴θ – 6 cos²θ sin²θ + sin⁴θ。将 sin²θ 替换为 1 – cos²θ，得 cos 4θ = 8 cos⁴θ – 8 cos²θ + 1。此恒等式在积分和级数问题中很有用，将棣莫弗定理与三角函数展开联系起来，是 Edexcel 核心纯数大纲的常见内容。
11. Reduction Formulae for Integration | 积分的归宿公式
Establish a reduction formula for Iₙ = ∫₀¹ xⁿ eˣ dx. Using integration by parts with u = xⁿ, dv = eˣ dx, we get Iₙ = [xⁿ eˣ]₀¹ – n ∫₀¹ xⁿ⁻¹ eˣ dx = e – n Iₙ₋₁. Apply this to find I₃. I₀ = e – 1. Then I₁ = e – 1·I₀ = e – (e – 1) = 1. I₂ = e – 2 I₁ = e – 2. I₃ = e – 3 I₂ = e – 3(e – 2) = 6 – 2e. This recurrence technique is a typical case study problem that tests integration skills and logical reasoning.
为 Iₙ = ∫₀¹ xⁿ eˣ dx 建立一个归宿公式。使用分部积分，令 u = xⁿ, dv = eˣ dx，得到 Iₙ = [xⁿ eˣ]₀¹ – n ∫₀¹ xⁿ⁻¹ eˣ dx = e – n Iₙ₋₁。应用此公式求 I₃。I₀ = e – 1。则 I₁ = e – 1·I₀ = e – (e – 1) = 1。I₂ = e – 2 I₁ = e – 2。I₃ = e – 3 I₂ = e – 3(e – 2) = 6 – 2e。这种递推方法是典型的案例分析题型，考查积分技能与逻辑推理。
12. Vector Cross Product and Plane Intersection | 向量叉积与平面交线
Find the line of intersection of the planes Π₁: r·(2i – j + 3k) = 5 and Π₂: r·(i + 2j – k) = 4. The direction vector of the line is parallel to the cross product of the normals: n₁ × n₂ = (2i – j + 3k) × (i + 2j – k) = (-5i + 5j + 5k). Simplify direction to (-1,1,1). To find a point on the line, set z = 0 (if not parallel to that plane) and solve: 2x – y = 5, x + 2y = 4 → solving yields x = 2, y = -1. So the line is r = (2i – j) + λ(-i + j + k). Check by substituting into both planes. This case study combines vector algebra and linear systems, typical of Further Pure 2.
求平面 Π₁: r·(2i – j + 3k) = 5 与 Π₂: r·(i + 2j – k) = 4 的交线。直线的方向向量与两法向量的叉积平行：n₁ × n₂ = (2i – j + 3k) × (i + 2j – k) = (-5i + 5j + 5k)。将方向化简为 (-1,1,1)。为在直线上找一点，设 z = 0（若不与该平面平行）并求解：2x – y = 5，x + 2y = 4 → 解得 x = 2，y = -1。因此直线为 r = (2i – j) + λ(-i + j + k)。代入两个平面检验。本案例融合了向量代数与线性方程组，是 Further Pure 2 的典型内容。
Published by TutorHao | Further Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导