A-Level Maths Unit 5 Question Paper Jan 20: Key Topics and Revision | A-Level 数学 Unit 5 2020年1月试卷知识点精讲

📚 A-Level Maths Unit 5 Question Paper Jan 20: Key Topics and Revision | A-Level 数学 Unit 5 2020年1月试卷知识点精讲

The A-Level Mathematics Unit 5 paper, typically covering Pure Mathematics 3 content, requires confident handling of advanced algebra, trigonometry, exponentials, logarithms, differentiation and numerical methods. The January 2020 sitting tested students on a wide range of these skills through problem-solving and multi-step reasoning. In this article we revisit the core topics featured in that paper, offering clear explanations, worked-style insights and revision pointers to help you prepare for similar assessments.

A-Level 数学 Unit 5 试卷通常对应纯数学 3 的内容，要求考生熟练地处理高级代数、三角学、指数与对数、微分以及数值方法。2020 年 1 月的考试通过多步骤推理和实际问题，全面考查了这些技能。本文我们将重温该试卷所涉及的核心知识点，提供清晰的解释、类似解题思路和复习指引，帮助你为同类考试做好准备。

1. Algebraic Fractions and Partial Fractions | 分式代数与部分分式

Simplifying rational expressions by factorising and cancelling is always a starting point, but the Jan 20 paper pushed candidates towards partial fraction decomposition. You must be able to split a complex fraction into a sum of simpler fractions with distinct linear factors, repeated factors or an improper fraction that requires algebraic long division first.

通过因式分解和约分化简有理式往往是起点，但 2020 年 1 月的试卷要求考生掌握部分分式的分解。你必须能够将一个复杂的分式拆分成几个简单的分式之和，处理不同的线性因子、重复因子，或先通过代数长除法将假分式化为真分式。

Express 7−x/(x−3)(x+2) in partial fractions. → Write as A/(x−3) + B/(x+2), solve for A and B.
将 7−x/(x−3)(x+2) 表示成部分分式 → 设成 A/(x−3) + B/(x+2)，求出 A 和 B。
For repeated factor (x−1)², use A/(x−1) + B/(x−1)².
对于重复因子 (x−1)²，需要写成 A/(x−1) + B/(x−1)²。
If degree of numerator ≥ degree of denominator, divide first.
如果分子的次数 ≥ 分母的次数，必须先做长除法。

2. Functions, Domain and Range | 函数、定义域与值域

Understanding the language of functions was tested through mapping diagrams, domain restrictions and range determination. The paper required you to state the maximal domain of a function involving a square root or a denominator, and then find its range, often by sketching or by considering transformations.

试卷通过映射图、定义域限制和值域的确定考查了对函数语言的理解。你需要写出含有平方根或分母的函数的极大定义域，然后通过画草图或考虑变换来求出值域。

For f(x) = √(x−2), domain is x ≥ 2; range is f(x) ≥ 0.
对于 f(x) = √(x−2)，定义域为 x ≥ 2，值域为 f(x) ≥ 0。
For f(x) = 1/(x+3), domain is x ≠ −3; range is f(x) ≠ 0.
对于 f(x) = 1/(x+3)，定义域为 x ≠ −3，值域为 f(x) ≠ 0。

3. Composite Functions, Inverse Functions and Graphs | 复合函数、反函数及其图像

The January 2020 paper checked whether you can form fg(x) correctly and interpret f⁻¹(x) both algebraically and graphically. The inverse function reflects the original function in the line y = x, and its domain is the range of the original function. Be prepared to find the inverse by rearranging y = f(x) to make x the subject and then swapping x and y.

2020 年 1 月的试卷检查了你能否正确构造 fg(x)，并从代数与图像两个角度理解 f⁻¹(x)。反函数是原函数关于直线 y = x 的对称图形，其定义域即为原函数的值域。你需要先通过将 y = f(x) 变形使 x 成为主项，再交换 x 与 y 来求出反函数。

Given f(x) = 2x+3, g(x) = x², then fg(x) = 2x²+3.
已知 f(x) = 2x+3，g(x) = x²，则 fg(x) = 2x²+3。
To find f⁻¹(x) : write y = 2x+3 → x = (y−3)/2, so f⁻¹(x) = (x−3)/2.
求 f⁻¹(x)：设 y = 2x+3 → x = (y−3)/2，于是 f⁻¹(x) = (x−3)/2。
Restrict domain so the inverse is a function (one-to-one required).
需限制定义域以保证反函数仍为函数（要求原函数是一一映射）。

4. Modulus Functions and Graph Transformations | 绝对值函数与图像变换

Questions on the modulus function |f(x)| and the effect of |f(x)| vs f(|x|) appeared. The paper expected you to sketch graphs of y = |ax+b| and solve equations like |2x−1| = 5. You should also know how to combine transformations and use modulus inside quadratics or combined with inequalities.

试卷中出现了有关绝对值函数 |f(x)| 以及 |f(x)| 与 f(|x|) 的差异的问题。你需要画出 y = |ax+b| 的图像，并求解类似 |2x−1| = 5 的方程。你还需要懂得如何复合多个变换，并将绝对值与二次式结合或用于解不等式。

|2x−1| = 5 → 2x−1 = 5 or 2x−1 = −5 → x = 3 or x = −2.
|2x−1| = 5 → 2x−1 = 5 或 2x−1 = −5 → x = 3 或 x = −2。
Sketch y = |x²−4| : reflect negative part of parabola above x-axis.
画 y = |x²−4| 的图像：将抛物线在 x 轴下方的部分对称翻折到上方。
Inequality |x+1| < 3 → −3 < x+1 < 3 → −4 < x < 2.
不等式 |x+1| < 3 → −3 < x+1 < 3 → −4 < x < 2。

5. Trigonometry: Compound Angles, Double Angles and R-form | 三角学：和角、倍角与 R 形式

This unit relies heavily on trigonometric identities. You must know sin(A±B), cos(A±B) and tan(A±B) by heart. Double-angle forms (sin2θ, cos2θ) and their rearrangements are essential for solving equations. The harmonic form Rsin(θ±α) or Rcos(θ±α) was key for solving equations like 3sinθ + 4cosθ = 2.

这个单元大量依赖三角恒等式。你必须熟记 sin(A±B)、cos(A±B) 和 tan(A±B) 公式。倍角公式（sin2θ、cos2θ）及其变形是解方程的关键。谐振形式 Rsin(θ±α) 或 Rcos(θ±α) 对于求解类似 3sinθ + 4cosθ = 2 的方程尤为重要。

3sinθ + 4cosθ = Rsin(θ+α), R = √(3²+4²)=5, α = arctan(4/3)

3sinθ + 4cosθ = Rsin(θ+α)，R = √(3²+4²) = 5，α = arctan(4/3)

Given the identity and a specified interval, solve 5sin(θ+α)=2, then find all solutions.
利用该恒等式和给定区间，解 5sin(θ+α)=2，然后求出所有解。

6. Exponentials and Natural Logarithms | 指数函数与自然对数

The exponential function eˣ and the natural logarithm ln x are inverses. Typical questions involved solving equations like e²ˣ − 5eˣ + 6 = 0 by treating it as a quadratic in eˣ, or taking logs to solve a⋅eᵏˣ = b. Differentiation and integration of eᵏˣ and 1/x also appeared.

指数函数 eˣ 与自然对数 ln x 互为反函数。典型题型包括通过将 eˣ 视为变量，求解类似 e²ˣ − 5eˣ + 6 = 0 的二次型方程，或通过取对数求解 a⋅eᵏˣ = b。试卷中还出现了对 eᵏˣ 和 1/x 的微分与积分。

Solve e²ˣ − 5eˣ + 6 = 0: let y = eˣ → y² − 5y + 6 = 0 → y = 2,3 → x = ln2, ln3.
解 e²ˣ − 5eˣ + 6 = 0：设 y = eˣ → y² − 5y + 6 = 0 → y = 2, 3 → x = ln 2, ln 3。
d/dx (e³ˣ) = 3e³ˣ, ∫ e⁻²ˣ dx = −½e⁻²ˣ + C.
d/dx (e³ˣ) = 3e³ˣ，∫ e⁻²ˣ dx = −½e⁻²ˣ + C。

7. Differentiation: Product, Quotient and Chain Rules | 微分：乘积法则、商法则与链式法则

Unit 5 demands fluency in differentiating products and quotients, often embedded in a chain of functions. The Jan 20 paper included differentiating expressions like x²ln x, sin2x/(1+cos2x) and composite functions such as (x²+1)⁵. Recognising which rule to apply first is essential.

Unit 5 要求对乘积和商的微分极为熟练，这些常常嵌套在复合函数中。2020 年 1 月的试卷包含了如 x²ln x、sin2x/(1+cos2x) 以及复合函数 (x²+1)⁵ 的求导。先识别用哪条规则至关重要。

Product: d/dx (x² sinx) = 2x sinx + x² cosx.
乘积法则：d/dx (x² sinx) = 2x sinx + x² cosx。
Quotient: d/dx (x/(x+1)) = [(x+1)·1 − x·1]/(x+1)² = 1/(x+1)².
商法则：d/dx (x/(x+1)) = [(x+1)·1 − x·1]/(x+1)² = 1/(x+1)²。
Chain: d/dx ( (2x+3)⁴ ) = 4(2x+3)³ × 2 = 8(2x+3)³.
链式法则：d/dx ( (2x+3)⁴ ) = 4(2x+3)³ × 2 = 8(2x+3)³。

8. Implicit and Parametric Differentiation | 隐函数微分与参数式微分

The paper tested implicit differentiation where you differentiate both sides of an equation like x² + y² = 25 with respect to x, using dy/dx. Parametric equations required using dy/dx = (dy/dt) / (dx/dt). Both often lead to finding gradients and equations of tangents.

试卷考查了隐函数微分，例如对 x² + y² = 25 的等式两边关于 x 求导，并用到 dy/dx。参数方程则需要使用 dy/dx = (dy/dt) / (dx/dt)。两种方法通常都会引出求切线斜率及方程的问题。

Implicit: 2x + 2y (dy/dx) = 0 → dy/dx = −x/y.
隐函数：2x + 2y (dy/dx) = 0 → dy/dx = −x/y。
Parametric: x = t²+1, y = 2t → dx/dt = 2t, dy/dt = 2, so dy/dx = 2/(2t) = 1/t.
参数式：x = t²+1，y = 2t → dx/dt = 2t，dy/dt = 2，所以 dy/dx = 2/(2t) = 1/t。

9. Differentiation of Exponential, Logarithmic and Trigonometric Functions | 指数、对数与三角函数的微分

Beyond basic powers, you must confidently differentiate eᵏˣ, aˣ, ln|x|, sin(kx), cos(kx), tan(kx), sec(kx), cosec(kx), cot(kx) and their combinations. The Jan 20 paper required quick recall of these standard derivatives, especially when they were embedded in product or chain rules.

除了基本的幂函数外，你还必须能熟练地对 eᵏˣ、aˣ、ln|x|、sin(kx)、cos(kx)、tan(kx)、sec(kx)、cosec(kx)、cot(kx) 及其组合进行微分。2020 年 1 月的试卷要求快速回忆这些标准导数，尤其当它们嵌套在乘积或链式法则中时。

d/dx (aˣ) = aˣ ln a, d/dx (ln|2x+1|) = 2/(2x+1).
d/dx (aˣ) = aˣ ln a，d/dx (ln|2x+1|) = 2/(2x+1)。
d/dx (tan3x) = 3sec²3x, d/dx (cot x) = −cosec² x.
d/dx (tan3x) = 3sec²3x，d/dx (cot x) = −cosec² x。

10. Numerical Methods: Iteration and Newton-Raphson | 数值方法：迭代与牛顿-拉夫森法

The January 2020 paper included an iterative formula question where you rearranged an equation into the form x = g(x) and used it to find a root to a specified accuracy. The Newton-Raphson method xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ) was also a possible feature, requiring you to differentiate correctly and apply the formula repeatedly.

2020 年 1 月的试卷中有迭代公式的题目，要求你将方程改写为 x = g(x) 的形式，并利用它求出达到指定精确度的根。牛顿-拉夫森法 xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ) 也可能出现，这需要你正确求导并反复使用该公式。

x₁ = x₀ − f(x₀)/f'(x₀)

Show that x³ − 3x − 1 = 0 can be written as x = ∛(3x+1). Use x₀ = 2 to find x₁, x₂.
证明 x³ − 3x − 1 = 0 可改写为 x = ∛(3x+1)。取 x₀ = 2 求 x₁ 和 x₂。

11. Connected Rates of Change and Differential Equations | 相关变化率与微分方程

Questions on rates of change involved using the chain rule to relate different rates, typically dy/dt = (dy/dx)·(dx/dt). A simple separable differential equation, like dy/dx = f(x)g(y), required separation of variables and integration. The Jan 20 paper might have given a context such as a growth/decay model or a geometric volume change.

变化率的问题需要利用链式法则关联不同的变化率，典型形式为 dy/dt = (dy/dx)·(dx/dt)。简单的可分离变量型微分方程，例如 dy/dx = f(x)g(y)，要求分离变量并积分。2020 年 1 月的试卷很可能以增长/衰变模型或几何体积变化为背景出题。

Given dy/dx = 2x/y, separate: y dy = 2x dx → ½y² = x² + C.
已知 dy/dx = 2x/y，分离变量：y dy = 2x dx → ½y² = x² + C。
If a cylindrical volume V=πr²h and dr/dt is known, find dV/dt.
若圆柱体积 V=πr²h 且已知 dr/dt，求 dV/dt。

12. Exam Technique and Common Pitfalls | 考试技巧与常见失分点

Many students lost marks on the Jan 20 Unit 5 paper not through lack of knowledge but by missing interval boundaries, forgetting ± signs when using modulus, or failing to check domain restrictions for inverses. Always verify that your solution lies in the required range, and present working clearly so that method marks can be awarded even if the final answer is wrong.

在 2020 年 1 月的 Unit 5 试卷中，许多学生丢分并非因为知识欠缺，而是忽略了区间的边界、使用绝对值时漏写了 ± 号，或者忘记检查反函数的定义域限制。务必验证你的解落在要求的范围内，并清晰地展示解题步骤，这样即便最终答案有误也能获得方法分。

When solving trig equations, always use CAST diagrams or graphs to find all solutions.
解三角方程时，务必用 CAST 图或图像求出所有解。
For domain of f⁻¹, state range of f explicitly.
写 f⁻¹ 的定义域时，要明确写出 f 的值域。
Check approximation accuracy after iteration: two successive x-values agree to, say, 3 decimal places.
迭代后检查近似精度：相邻两次 x 值需在小数点后若干位一致。

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A-Level Maths Unit 5 Question Paper Jan 20: Key Topics and Revision | A-Level 数学 Unit 5 2020年1月试卷知识点精讲