📚 AQA AS Maths Unit 2 January 2020 Question Paper: Key Topics and Exam Technique | AQA AS 数学第二单元 2020 年 1 月真题：核心考点与应试技巧
The January 2020 AQA AS Maths Unit 2 paper is a valuable resource for candidates preparing for the AS pure core assessment. It tests the second half of the pure mathematics syllabus, including algebra, coordinate geometry, trigonometry, logarithms and calculus. This article breaks down the high-yield topics, the common question types and the marking points, so that you can revise efficiently and avoid losing easy marks.
2020 年 1 月 AQA AS 数学第二单元试卷对准备 AS 纯数学核心考试的考生来说是一份很有价值的复习资料。它考查 AS 纯数学大纲的后半部分，包括代数、坐标几何、三角、对数和微积分。本文拆解高频考点、常见题型和评分要点，帮助你高效复习，避免在简单题上丢分。
1. Paper Structure and Assessment Focus | 试卷结构与考查重点
The January 2020 Unit 2 paper is designed to assess pure maths skills not fully covered in Unit 1. The paper is usually structured so that routine algebra and calculus appear early, while later questions require you to link ideas across topics such as logarithms, circles and sequences. Understanding the command words and the mark allocation is just as important as knowing the content itself.
2020 年 1 月第二单元试卷旨在考查第一单元未完全覆盖的纯数学技能。试卷通常结构安排为：较基础的代数和微积分出现在前半部分，后半部分则要求跨知识点综合运用，如对数、圆和数列。理解指令词和分值分布与掌握知识内容同样重要。
In a typical AS pure paper you have roughly 90 minutes for about 80 marks, which works out at just over one minute per mark. That budget sounds generous, but multi-step questions at the end of a section can easily consume three to four minutes each. The practical rule is to spend about one minute per mark on the first pass, then return to anything you have left blank.
在典型的 AS 纯数学试卷中，你大约有 90 分钟完成约 80 分，折算下来每分略多于一分钟。这个时间听起来宽裕，但每节末尾的多步骤题往往每题就要花三到四分钟。实用的策略是：第一轮按“每分一分钟”推进，再回头处理留空的题目。
| Question style | Typical focus | What markers look for |
| Short structured questions | Single skill application, such as differentiating a polynomial or finding a circle centre | Correct method, accurate substitution, exact answer |
| Multi-step questions | Linking two topics, for example logs with a quadratic, or a tangent to a circle | Clear intermediate working that can earn method marks |
| "Show that" questions | Guided algebra leading to a printed result | Every algebraic step shown, no skipping to the answer |
| Modelling questions | Applying calculus or exponentials to a real context | Correct interpretation, suitable accuracy, units where relevant |
2. Algebra, Surds and Function Skills | 代数、根式与函数技能
Algebra is the backbone of Unit 2 and it rarely appears as a standalone question. Instead, it hides inside calculus, coordinate geometry and logarithms. If your expansion or factorisation is slow, every later question becomes slower too. The exam expects confident manipulation of indices, surds, quadratics and simultaneous equations.
代数是第二单元的骨架，却很少以独立题目的形式出现，而是藏在微积分、坐标几何和对数题里。如果展开或因式分解速度慢，后面每道题都会跟着变慢。考试要求你能熟练处理指数、根式、二次方程和联立方程。
High-value skills include rationalising a denominator, simplifying surds such as √50 = 5√2, solving a quadratic by factorising or by the formula, and completing the square. Completing the square is particularly useful because it links algebra directly to graph work and to circle equations.
高价值技能包括：分母有理化，化简根式（如 √50 = 5√2），用因式分解或公式法解二次方程，以及配方。配方尤其重要，因为它把代数与图像、圆的方程直接连接起来。
x = (−b ± √b² − 4ac) / 2a
(x + p)² + q , vertex at (−p, q)
- Indices: aᵐ × aⁿ = aᵐ⁺ⁿ ; aᵐ ÷ aⁿ = aᵐ⁻ⁿ ; (aᵐ)ⁿ = aᵐⁿ ; a⁰ = 1 ; a⁻ⁿ = 1 / aⁿ | 指数运算律需牢记
- Surds: √ab = √a × √b ; a / √b = a√b / b | 根式的乘法与有理化
- Discriminant: b² − 4ac > 0 two roots ; = 0 one repeated root ; < 0 no real roots | 判别式判断根的个数
- Function notation: f(x + 2), f⁻¹(x), composite functions and their domains | 函数记号、反函数与定义域
3. Coordinate Geometry: Lines and Circles | 坐标几何：直线与圆
Coordinate geometry questions in this paper usually combine a straight line with a circle. You might be asked to find the centre and radius of a circle from its expanded equation, then show that a given line is a tangent. The key idea is that a tangent is perpendicular to the radius at the point of contact, so the product of the two gradients is −1.
本卷的坐标几何题通常把直线与圆结合起来。可能要求你从圆的展开式写出圆心和半径，再证明某条直线是切线。核心思想是：切线与接触点处的半径垂直，因此两条直线的斜率之积为 −1。
= 0 becomes (x + f)² + (y + g)² = f² + g² − c, so the centre is (−f, −g) and the radius is √f² + g² − c. Always check whether the constant on the right is positive; if it is negative, the "circle" does not exist.
求圆心和半径要对 x 和 y 分别配方。圆方程 x² + y² + 2fx + 2gy + c = 0 可化为 (x + f)² + (y + g)² = f² + g² − c，因此圆心为 (−f, −g)，半径为 √f² + g² − c。务必检查等式右边是否为正；若为负，这个“圆”并不存在。
(x − a)² + (y − b)² = r² centre (a, b), radius r
m₁ × m₂ = −1 for perpendicular lines
- Line through two points: m = (y₂ − y₁) / (x₂ − x₁) | 两点求斜率
- Point-slope form: y − y₁ = m(x − x₁) | 点斜式
- Perpendicular bisector of a chord passes through the centre | 弦的垂直平分线过圆心
- Distance formula: d = √((x₂ − x₁)² + (y₂ − y₁)²) | 两点间距离公式
4. Sequences, Series and the Binomial Expansion | 数列、级数与二项式展开
Sequence questions in this paper are usually straightforward marks if you know the formulas, but they are also a common place to lose accuracy marks through careless substitution. Arithmetic and geometric sequences both appear, and the binomial expansion is almost always present because it combines algebra with a formal notation that examiners can mark precisely.
只要记住公式，本卷的数列题通常是容易拿分的题，但也常常因为代入粗心而丢掉精确分。等差数列和等比数列都会出现，而二项式展开几乎必考，因为它把代数与规范的记号结合起来，便于考官准确评分。
Arithmetic: uₙ = a + (n − 1)d , Sₙ = n/2 [2a + (n − 1)d]
Geometric: uₙ = arⁿ⁻¹ , Sₙ = a(1 − rⁿ) / (1 − r) for r ≠ 1
The binomial expansion of (a + b)ⁿ uses coefficients from Pascal's triangle or from the nCr notation. In AS papers n is normally a positive integer, so the expansion terminates. Write out the general term first, then substitute; this makes it much easier to earn method marks even if a later arithmetic slip occurs.
二项式展开 (a + b)ⁿ 的系数来自杨辉三角或组合数 nCr。在 AS 试卷中 n 通常是正整数，因此展开式是有限项。先把通项写出来再代入，这样即使后面算错，也能拿到方法分。
- nCr = n! / (r!(n − r)!) | 组合数公式
- Term in xʳ: nCr × aⁿ⁻ʳ × bʳ | 指定次幂项的求法
- Sum to infinity exists only when |r| < 1 | 只有 |r| < 1 时无穷级数才收敛
- Check whether the question asks for a term, a coefficient, or a full expansion | 区分“求某项”“求系数”“求展开式”
5. Trigonometry: Identities and Equations | 三角：恒等式与方程
in²θ + cos²θ = 1 to turn an equation into a quadratic in sin θ or cos θ, then solve and list every root inside the stated range.
本卷的三角内容主要围绕精确值、恒等式和区间内解方程。题目常要求用 sin²θ + cos²θ = 1 把方程化为关于 sin θ 或 cos θ 的二次方程，然后求解并写出指定区间内所有的根。
ormally gives two answers and a cosine equation gives two as well, unless the interval is narrower. Sketching the curve or using the unit circle takes a few seconds and protects those marks.
漏掉第二个解是本卷最常见的失分之一。如果题目要求 0° ≤ θ ≤ 360° 范围内的解，正弦方程通常有两个答案，余弦方程也是两个，除非区间更窄。画一下曲线或用单位圆，只需几秒，就能保住这些分数。
sin²θ + cos²θ = 1 , tan θ = sin θ / cos θ
Exact values: sin 30° = ½ , cos 60° = ½ , tan 45° = 1
- Radian measure: π radians = 180° | 弧度与角度换算
- Arc length s = rθ ; sector area A = ½r²θ (θ in radians) | 弧长与扇形面积
- Sine and cosine rule for non-right-angled triangles: a / sin A = b / sin B | 正弦定理与余弦定理
- Graphs of y = sin x, y = cos x, y = tan x and their symmetries | 三个基本函数的图像与对称性
6. Exponentials and Logarithms | 指数与对数
e ability to switch between log form and index form, cover most of the marks. A typical question asks you to solve an equation such as 3ˣ = 20, which gives x = log 20 / log 3.
对数几乎出现在每一份 AS 纯数学试卷中，因为它能把指数方程转化为线性方程。三条对数运算律，加上对数式与指数式的互化，基本就能覆盖大部分分值。典型题目如解 3ˣ = 20，得到 x = log 20 / log 3。
ntext carefully: the question may ask for the value of t when the quantity halves, or for the initial value when t = 0, and those are different pieces of working.
建模题越来越常见。题目可能给出人口模型或放射性衰变模型，要求你求某个时间或某个变化率。仔细读题：题目可能问数量减半时的 t 值，也可能问 t = 0 时的初始值，两者需要的运算并不相同。
logₐ (xy) = logₐ x + logₐ y
logₐ (x / y) = logₐ x − logₐ y , logₐ (xⁿ) = n logₐ x
- Definition: y = aˣ ⇌ logₐ y = x | 指数式与对数式互化
- Special values: logₐ a = 1 , logₐ 1 = 0 | 特殊值
- Natural log: ln x = logₑ x , and eˣ is the inverse of ln x | 自然对数与 e 的互逆关系
- Solving: aˣ = b gives x = log b / log a | 换底公式解指数方程
7. Differentiation: Technique and Applications | 微分：技巧与应用
Differentiation carries a large share of the marks and appears in two forms: pure technique and applied problems. Pure technique means differentiating powers of x, including negative and fractional indices, after rewriting roots and fractions in index form. Applied problems mean tangents, normals, stationary points, increasing and decreasing functions and rates of change.
微分占的分值很大，主要分两类：纯技巧题和应用题。纯技巧题要求对 x 的幂函数求导，包括负指数和分数指数，因此要先把根式和分式改写成指数形式。应用题则涉及切线、法线、驻点、增减函数和变化率。
nd solve. Then use the second derivative to classify it: if d²y/dx² > 0 the point is a minimum, and if d²y/dx² < 0 it is a maximum. Many students find the x-value and stop, forgetting that the question usually wants the y-coordinate too.
求驻点时令 dy/dx = 0 求解，再用二阶导数判断类型：若 d²y/dx² > 0 为极小值点，若 d²y/dx² < 0 为极大值点。很多学生求出 x 值就停笔，忘记题目通常还要写出 y 坐标。
d/dx (xⁿ) = n·xⁿ⁻¹
Tangent: y − y₁ = m(x − x₁) with m = dy/dx at that point
- Normal gradient = −1 / (dy/dx) | 法线斜率与切线斜率互为负倒数
- Increasing where dy/dx > 0 , decreasing where dy/dx < 0 | 用一阶导数判断增减
- Second derivative test for maxima and minima | 二阶导数判别极值
- Rate of change: dy/dt = (dy/dx) × (dx/dt) | 链式法则处理变化率
8. Integration: Reversing the Process | 积分：微分的逆运算
Integration in AS pure maths is mostly the reverse of differentiation plus the calculation of areas. Indefinite integrals need the constant of integration, and that single symbol is worth a mark in almost every question. Definite integrals are evaluated by substituting the limits, and the answer may be negative if the curve lies below the x-axis.
AS 纯数学中的积分主要是微分的逆运算加上面积计算。不定积分必须写积分常数，而这一个符号几乎每道题都值一分。定积分通过代入上下限求值，如果曲线位于 x 轴下方，结果可能为负。
When a region is bounded by a curve and a straight line, the safest method is to subtract: integrate the upper function minus the lower function between the two intersection points. Sketching the region first tells you which curve is on top and prevents a sign error that would lose the final accuracy mark.
当区域由曲线和直线围成时，最稳妥的方法是相减：在两个交点之间对“上方函数 − 下方函数”积分。先画出区域草图，可以判断哪条曲线在上方，从而避免符号错误而丢掉最后的精确分。
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C , n ≠ −1
∫ₐᵇ f(x) dx = F(b) − F(a)
- Always write + C for an indefinite integral | 不定积分必须写 + C
- Area below the x-axis gives a negative value; area is its modulus | 轴下方的积分为负，面积取绝对值
- Area between two curves = ∫ (upper − lower) dx | 两曲线间面积等于上方减下方
- Use exact fractions when the question says "exact" | 题目要求 exact 时保留分数或根式
9. Proof and Mathematical Reasoning | 证明与数学推理
AQA papers include proof marks that are easy to earn and easy to throw away. A proof by exhaustion, a counter-example, or a completed algebraic argument all require clear structure rather than a correct answer alone. If the question says "show that", the printed result is given to you, so the marks come from the working in between.
AQA 试卷中包含证明分，既容易拿也容易丢。穷举法证明、反例，或一个完整的代数论证，都要求结构清晰，而不只是结论正确。如果题目写的是 show that，结果已经印在卷子上，分数完全来自中间过程。
For a "show that" question, start from one side of the statement and manipulate it until you reach the printed form. Never start from both sides at once and meet in the middle, because that is not a valid proof and examiners will not award full marks. Write each line as a new equation and keep the equals signs aligned.
遇到 show that 题时，从式子的一边出发，逐步变形，直到得到印出的形式。不要从两边同时开始、在中间会合，因为那不算有效证明，考官不会给满分。每一行写成新的等式，让等号整齐排列。
- Counter-example: one case that fails disproves a general statement | 一个反例即可推翻全称命题
- Exhaustion: check every possible case within a stated set | 穷举法需覆盖给定集合内所有情况
- Algebraic proof: define variables clearly and state what you are proving | 代数证明要先明确变量与目标
- Conclude with a statement such as "hence the statement is true" | 结尾要明确写出结论
10. Command Words and Mark Distribution | 指令词与分值分布
Command words tell you exactly how much working is expected. "Write down" or "state" usually needs no working at all, "find" needs method, and "show that" or "prove" needs a complete argument. Misreading a command word is a common reason for losing marks even when the mathematics is correct.
指令词会直接告诉你需要多少过程。write down 或 state 通常不需要过程，find 需要方法，show that 或 prove 需要完整论证。即使数学本身没错，误读指令词也是常见的失分原因。
Look also at how many marks a part is worth. A two-mark part rarely needs a long explanation, while a five-mark part almost always contains a hidden step such as finding an intersection, substituting a value, or interpreting a result in context.
还要看每一小问的分值。两分的小问很少需要长篇解释，而五分的小问几乎都藏着一个步骤，如求交点、代入求值，或结合情境解释结果。
| Command word | What to write |
| Write down / State | Just the answer, no working required |
| Find / Calculate | Clear method followed by the answer |
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