Case Study in Action: A Comprehensive Edexcel Year 13 Mathematics Workout | 案例分析实战演练：Edexcel十三年级数学综合训练

📚 Case Study in Action: A Comprehensive Edexcel Year 13 Mathematics Workout | 案例分析实战演练：Edexcel十三年级数学综合训练

This revision workout gathers the core pure-mathematics topics of Year 13 Edexcel Mathematics into one continuous, case-study style sequence. Each section presents a typical exam-shaped challenge, from rational functions and partial fractions to binomial expansions, parametric equations, integration techniques, numerical methods and differential equations. Work through the paired English and Chinese explanations, attempt each stage before reading on, and build the fluency that the highest grades demand.

本次复习演练把 Edexcel 十三年级数学的核心纯数内容整合成一套连贯的案例分析序列。每个小节呈现一道典型的考试风格题目，从有理函数、部分分式，到二项展开、参数方程、积分技巧、数值方法与微分方程。请按照中英对照的解析逐步推进，先自己动手再往下阅读，从而打造高分所需的能力与熟练度。

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1. How to Use This Case-Study Workout | 如何使用这套案例分析训练

The workbook is designed as a single sitting of roughly ninety minutes, but it is equally effective when split across four short revision sessions. Every section follows the same rhythm: identify the structure of the expression or model, apply the standard technique, then check the answer against the domain, the range or the physical context.

这份训练设计为约九十分钟的一次完整练习，但拆成四次短时复习同样有效。每个小节遵循相同的节奏：先识别表达式或模型的结构，再套用标准方法，最后结合定义域、值域或实际情境检验答案。

Before starting, keep three things beside you: the official Edexcel formula booklet, a graphical calculator, and a single sheet of paper for error notes. Mistakes logged in the moment are worth far more than mistakes discovered in May.

开始之前，请在身边放好三样东西：官方 Edexcel 公式册、图形计算器，以及一张用于记录错误的纸。当场记录下来的错误，价值远高于五月才发现的错误。

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2. Rational Functions and Their Domains | 有理函数及其定义域

ssion at all, determine the domain by setting the denominator not equal to zero: x − 1 ≠ 0 ⇒ ecause x − 1 is not a factor of the numerator: substituting x = 1 gives 2 + 5 − 3 = 4 ≠ 0.

考虑函数 f(x) = (2x² + 5x − 3)/(x − 1)。在动手操作表达式之前，先令分母不为零以确定定义域：x − 1 ≠ 0 ⇒ x ≠ 1。尽管分子可分解为 (2x − 1)(x + 3)，但无法约分，因为 x − 1 并不是分子的因式：代入 x = 1 得 2 + 5 − 3 = 4 ≠ 0。

4/(x − 1). As x becomes large, the term 4/(x − 1) tends to 0, so the graph approaches the oblique asymptote y = 2x + 7, while the denominator vanishing at x = 1 produces a vertical asymptote.

长除法把假分式化为"商 + 余项"的形式。用 x − 1 除 2x² + 5x − 3，得商 2x + 7、余 4，因此 f(x) = 2x + 7 + 4/(x − 1)。当 x 变得很大时，4/(x − 1) 趋于 0，图像趋近斜渐近线 y = 2x + 7；而分母在 x = 1 处为零，形成垂直渐近线。

omain restrictions clearly before differentiating or integrating, since a missing restriction costs marks that are difficult to recover.

关键坐标很快就能得到：y 轴截距为 f(0) = 3；x 轴截距来自 2x − 1 = 0 与 x + 3 = 0，即 x = ½ 与 x = −3。在求导或积分之前，务必清楚写出定义域限制，因为漏掉限制条件所丢的分很难补回。

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3. Partial Fractions and Repeated Factors | 部分分式与重因子

factors, write (5x + 11)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2), then multiply through to obtain A(x + 2) + B(x + 1) = 5x + 11.

部分分式是"合并分式"的逆过程，是积分有理式的必备工具。对于互异的一次因式，先设 (5x + 11)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2)，再两边同乘分母得 A(x + 2) + B(x + 1) = 5x + 11。

is faster and less error-prone; both methods should be used as a mutual check in the exam.

比较系数可得 A + B = 5、2A + B = 11，故 A = 6、B = −1。直接代入 x = −1 与 x = −2 更快、更不易出错；考试中可两种方法互检。

When a factor repeats, the decomposition must include every power up to the repeat. For example, (x + 5)/((x − 2)(x + 1)²) takes the form A/(x − 2) + B/(x + 1) + C/(x + 1)². If the numerator has degree greater than or equal to the denominator, divide first, then decompose the remainder term.

当因式重复出现时，分解式必须包含直到该重数的每一幂次。例如 (x + 5)/((x − 2)(x + 1)²) 应设为 A/(x − 2) + B/(x + 1) + C/(x + 1)²。若分子次数大于或等于分母次数，应先做除法，再对余项部分分解。

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4. Binomial Expansion for Negative and Fractional Indices | 负整数与分数指数的二项展开

The general binomial series states that (1 + x)ⁿ = 1 + nx + [n(n − 1)/2!]x² + [n(n − 1)(n − 2)/3!]x³ + …, and this form remains valid for negative and fractional n provided |x| < 1.

一般二项级数为 (1 + x)ⁿ = 1 + nx + [n(n − 1)/2!]x² + [n(n − 1)(n − 2)/3!]x³ + …；只要 |x| < 1，该形式对负整数与分数指数 n 同样成立。

Take the expansion of (1 + 3x)^−½ = −5/16. Replacing x by 3x throughout gives 1 − (3/2)x + (27/8)x² − (135/16)x³, valid when |3x| < 1, that is |x| < ⅓.

以 (1 + 3x)^−½ 展开到含 x³ 项为例。取 n = −½，各项系数依次为 −½、(−½)(−3/2)/2 = 3/8、(−½)(−3/2)(−5/2)/6 = −5/16。把 x 全部换成 3x，得 1 − (3/2)x + (27/8)x² − (135/16)x³，成立条件为 |3x| < 1，即 |x| < ⅓。

stead of the transformed variable. Extracting a factor first, for example writing (4 + 3x)^1/2 = 2(1 + 3x/4)^1/2, keeps the algebra inside the |x| < 1 framework.

这里有两个反复出现的失分点：忘记把括号内的常数同步缩放，以及把成立范围写成 x 而不是变换后的变量。先提出因子可保持 |x| < 1 的框架，例如 (4 + 3x)^1/2 = 2(1 + 3x/4)^1/2。

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5. Parametric Equations: Gradients, Tangents and Normals | 参数方程：斜率、切线与法线

= y/4, substituting gives x = 2(y/4)² = y²/8, so y² = 8x, a parabola opening to the right.

由 x = 2t²、y = 4t 给出的参数曲线可通过消去 t 化为直角坐标形式。由 t = y/4，代入得 x = 2(y/4)² = y²/8，即 y² = 8x，是一条开口向右的抛物线。

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o dy/dx = 1/t. At the point where t = 2, the coordinates are (8, 8) and the gradient is ½.

斜率由链式法则的比值形式给出：dy/dx = (dy/dt) ÷ (dx/dt)。此处 dy/dt = 4、dx/dt = 4t，故 dy/dx = 1/t。当 t = 2 时，点为 (8, 8)，斜率为 ½。

), or 2x + y = 24. Writing both lines in a single consistent form avoids careless slips in the final answer line.

因此切线为 y − 8 = ½(x − 8)，整理得 y = ½x + 4。法线斜率为负倒数 −2，故 y − 8 = −2(x − 8)，即 2x + y = 24。把两条直线写成统一形式，可避免最后作答时的低级失误。

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6. Differentiation: Product, Quotient, Chain and Implicit | 求导：乘积、商、链式与隐函数

entiation questions in Year 13 rarely test one rule in isolation. A function such as y = e^2x sin 3x requires the product rule combined with the chain rule, giving dy/dx = 2e^2x sin 3x + 3e^2x cos 3x.

十三年级的求导题很少只考单一法则。像 y = e^2x sin 3x 这样的函数需要乘积法则结合链式法则，得 dy/dx = 2e^2x sin 3x + 3e^2x cos 3x。

1) can be integrated and differentiated more comfortably once written as y = 1 − 1/(x + 1).

遇到商式时，商法则往往比改写为负指数乘积更慢。例如 y = x/(x + 1)，若先写成 y = 1 − 1/(x + 1)，无论求导还是积分都更顺手。

y(dy/dx), gives 2x + 2y(dy/dx) = 0, so dy/dx = −x/y. The gradient at (3, 4) is therefore −3/4, and the tangent there is 3x + 4y = 25.

当 y 无法显式解出时，必须使用隐函数求导。逐项对 x² + y² = 25 求导，并记住 y² 的导数是 2y(dy/dx)，得 2x + 2y(dy/dx) = 0，即 dy/dx = −x/y。于是在 (3, 4) 处斜率为 −3/4，切线为 3x + 4y = 25。

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7. Integration Techniques: Substitution, By Parts and Partial Fractions | 积分技巧：换元、分部与部分分式

ake u = x² + 1, so du = 2x dx, and the integral becomes ∫ u⁵ du = u⁶/6 + C = (x² + 1)⁶/6 + C.

积分的关键在于"识别"。若被积函数是复合式且其内函数的导数也在场，就用换元：对 ∫ 2x(x² + 1)⁵ dx，令 u = x² + 1，则 du = 2x dx，积分化为 ∫ u⁵ du = u⁶/6 + C = (x² + 1)⁶/6 + C。

⁵ dx, u runs from 1 to 2, so the value is ½[u⁶/6] from 1 to 2 = (64 − 1)/12 = 21/4 = 5.25.

定积分还必须同步更换上下限。对 ∫₀¹ x(x² + 1)⁵ dx，u 从 1 变到 2，故结果为 ½[u⁶/6] 由 1 到 2 = (64 − 1)/12 = 21/4 = 5.25。

ike functions suggests integration by parts, ∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx. For ∫ x e^2x dx, choose u = x and dv/dx = e^2x, giving (x/2)e^2x − (1/4)e^2x tions convert a rational integrand into logarithms: ∫ (5x + 11)/((x + 1)(x + 2)) dx = 6 ln|x + 1| − ln|x + 2| + C.

两个"不同类"函数相乘提示使用分部积分：∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx。对 ∫ x e^2x dx，取 u = x、dv/dx = e^2x，得 (x/2)e^2x − (1/4)e^2x + C。最后，部分分式可把有理被积式化为对数：∫ (5x + 11)/((x + 1)(x + 2)) dx = 6 ln|x + 1| − ln|x + 2| + C。

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8. Volumes of Revolution and Modelling | 旋转体体积与建模

given limits. The square is applied to y before integrating, which is the single most common source of lost marks.

当曲线 y = f(x) 绕 x 轴旋转 2π 弧度时，所得旋转体体积为 V = π∫ y² dx，上下限由题目给出。必须先对 y 平方再积分，这是最常见的失分点。

For the region under y = √x the π outside the integral until the final line makes the arithmetic clearer and easier to check.

对 y = √x 在 x = 0 到 x = 4 之间的区域，y² = x，故 V = π∫₀⁴ x dx = π[x²/2] 由 0 到 4 = 8π。把 π 留到最后一并乘上，计算更清楚，也便于检查。

Modelling questions extend this idea to real objects, where assumptions such as constant density or a perfectly circular cross-section must be stated explicitly. Units deserve equal care: if x is measured in centimetres, the volume is in cm³, and examiners do award marks for correct units in context.

建模题会把这一思想延伸到真实物体，此时"密度恒定""横截面为正圆"等假设必须明确写出。单位同样重要：若 x 以厘米计，体积即为 cm³，在情境题中写对单位是可以得分的。

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9. Numerical Methods: Iteration and Newton–Raphson | 数值方法：迭代法与牛顿–拉夫森法

1 and f(3) = 16; the sign change plus continuity guarantees a root in the interval [2, 3].

数值方法考查的不只是运算，更是推理。为求 f(x) = x³ − 2x − 5 的根，先算 f(2) = −1、f(3) = 16；由符号改变与函数的连续性，可断定区间 [2, 3] 内必有根。

The Newton–Raphson formula is xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ) with f' 10 = 2.1, then x₂ = 2.1 − 0.061/11.23 ≈ 2.0946. The root to four decimal places is 2.0946.

牛顿–拉夫森公式为 xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)，其中 f'(x) = 3x² − 2。取 x₀ = 2，得 x₁ = 2 − (−1)/10 = 2.1，再得 x₂ = 2.1 − 0.061/11.23 ≈ 2.0946。故根的四位小数为 2.

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