📚 AS AQA Maths Unit 2 January 2019: Pure Core 2 Question Paper Guide | AS AQA 数学 Unit 2 2019年1月：纯数核心2试卷解析指南
The January 2019 Unit 2 paper belongs to the legacy AQA AS Mathematics specification, where Unit 2 is Pure Core 2 (MPC2). It is a 75-mark, 1 hour 30 minute written paper covering algebra, sequences and series, trigonometry, exponentials and logarithms, differentiation, integration and numerical methods. This guide summarises the paper structure, typical question styles, key formulas, worked examples and the exam habits that earn marks, so that you can revise efficiently and walk into the hall knowing exactly what to expect.
2019年1月 Unit 2 试卷属于旧版 AQA AS 数学大纲，其中 Unit 2 是纯数核心2（MPC2）。这是一份满分75分、时长1小时30分钟的笔试，内容覆盖代数与函数、数列与级数、三角学、指数与对数、微分、积分以及数值方法。本指南总结了试卷结构、典型题型、关键公式、示例题以及能拿到分的答题习惯，帮助你高效复习，并在进入考场前清楚地知道会面对什么。
1. Paper Structure and Timing | 试卷结构与时间分配
The January 2019 Unit 2 paper is worth 75 marks and lasts 1 hour 30 minutes. It contains approximately 8 to 10 compulsory questions arranged roughly in increasing difficulty, mixing straightforward knowledge recall with multi-step problem solving. A calculator is permitted, and a formulae booklet is provided, so you never need to memorise the trapezium rule, the log laws or the binomial coefficients.
2019年1月 Unit 2 试卷满分75分，考试时间1小时30分钟。试卷约含8至10道必答题，难度大致递增，既有直接考查知识的题目，也有多步综合题。考试允许使用计算器，并会提供公式手册，因此梯形法则、对数运算律、二项式系数等公式无需死记。
A useful time rule is roughly one minute per mark plus ten minutes of checking at the end. That means a 4-mark question should take about four to five minutes, and a long 9-mark "show that and hence" question can legitimately take twelve minutes.
一个实用的时间法则是：大约每一分对应一分钟，最后再留出十分钟检查。也就是说，4分题大约用四到五分钟，而一道9分的"证明并由此求出"长题用十二分钟是完全合理的。
The table below shows the approximate mark allocation for each topic area in a typical Pure Core 2 paper. The balance can shift slightly from session to session, but the overall pattern is very stable.
下表展示了典型纯数核心2试卷中各主题的大致分值分布。不同考次的占比可能会有小幅波动，但整体格局非常稳定。
| Topic Area | 知识模块 | Typical Marks | 典型分值 |
|---|---|
| Algebra and functions | 代数与函数 | 15–20 |
| Sequences and series | 数列与级数 | 12–16 |
| Trigonometry | 三角学 | 14–18 |
| Exponentials and logarithms | 指数与对数 | 10–14 |
| Differentiation | 微分 | 12–16 |
| Integration | 积分 | 14–18 |
| Numerical methods | 数值方法 | 6–8 |
2. Algebra and Functions | 代数与函数
In Pure Core 2, algebra questions often involve the factor theorem, polynomial division, the remainder theorem and curve sketching. You may be asked to show that (x − a) is a factor of a cubic, then find the remaining quadratic factor and identify all the roots.
纯数核心2中，代数题常涉及因式定理、多项式除法、余数定理和曲线作图。你可能需要证明 (x − a) 是三次多项式的因式，然后求出剩余的二次因式并确定所有根。
en f(a) = 0; the remainder theorem states that when f(x) is divided by (x − a) the remainder is f(a).
需要准确表述的两个定理是：因式定理——若 f(a) = 0，则 (x − a) 是 f(x) 的因式，反之若 (x − a) 是因式则 f(a) = 0；余数定理——f(x) 除以 (x − a) 所得的余数等于 f(a)。
³ − 3x² − 11x + 6. Show that (x + 2) is a factor and express f(x) as three linear factors.
示例：f(x) = 2x³ − 3x² − 11x + 6。证明 (x + 2) 是它的因式，并把 f(x) 分解为三个一次因式。
3(−2)² − 11(−2) + 6 = −16 − 12 + 22 + 6 = 0, so by the factor theorem (x + 2) is a factor.
第一步：代入 x = −2。f(−2) = 2(−2)³ − 3(−2)² − 11(−2) + 6 = −16 − 12 + 22 + 6 = 0，故由因式定理可知 (x + 2) 是因式。
Step 2: divide 2x³ − 3x² − 11x + 6 by (x + 2) to obtain the quadratic 2x² − 7x + 3.
第二步：用 (x + 2) 除 2x³ − 3x² − 11x + 6，得到二次式 2x² − 7x + 3。
Step 3: factorise the quadratic, 2x² − 7x + 3 = (2x − 1)(x − 3).
第三步：分解二次式，2x² − 7x + 3 = (2x − 1)(x − 3)。
f(x) = (x + 2)(2x − 1)(x − 3), so the curve crosses the x-axis at x = −2, x = ½ and x = 3.
第四步：写出完整分解式与根。f(x) = (x + 2)(2x − 1)(x − 3)，因此曲线在 x = −2、x = ½ 和 x = 3 处与 x 轴相交。
f(x) = (x + 2)(2x − 1)(x − 3)
al shape and the behaviour as x → ±∞. A positive cubic goes from bottom-left to top-right.
作图时务必标出两轴上的截距（此处 y 轴截距为 f(0) = 6）、曲线的大致形状以及 x → ±∞ 时的走势。首项系数为正的三次曲线从左下延伸到右上。
3. Sequences and Series | 数列与级数
Sequences and series questions reward clean formula use. For an arithmetic series with first term a and common difference d, the nth term is a + (n − 1)d and the sum of the first n terms is ½n[2a + (n − 1)d], which can also be written ½n(a + l) where l is the last term.
数列与级数的题目重在规范使用公式。对于首项为 a、公差为 d 的等差数列，第 n 项为 a + (n − 1)d，前 n 项和为 ½n[2a + (n − 1)d]，也可写作 ½n(a + l)，其中 l 为末项。
uₙ = a + (n − 1)d Sₙ = ½n[2a + (n − 1)d]
Example: the 5th term of an arithmetic series is 17 and the 12th term is 45. Find a and d, then S₂₀.
示例：等差数列第5项为17，第12项为45。求 a、d 以及 S₂₀。
28, so d = 4, and a = 17 − 4 × 4 = 1. Then S₂₀ = ½ × 20 × [2(1) + 19(4)] = 10 × 78 = 780.
两式相减得 7d = 28，故 d = 4，a = 17 − 4 × 4 = 1。于是 S₂₀ = ½ × 20 × [2(1) + 19(4)] = 10 × 78 = 780。
he sum of the first n terms is a(1 − rⁿ)/(1 − r), and the sum to infinity exists only when |r| < 1, in which case S∞ = a/(1 − r).
对于首项为 a、公比为 r 的等比数列，第 n 项为 arⁿ⁻¹，前 n 项和为 a(1 − rⁿ)/(1 − r)，而无穷级数和只在 |r| < 1 时存在，此时 S∞ = a/(1 − r)。
S∞ = a / (1 − r), |r| < 1
and 13.5. Find the sum to infinity. Here r = 18 ÷ 24 = 0.75, so S∞ = 24 ÷ (1 − 0.75) = 96.
示例：等比数列前三项为 24、18、13.5，求无穷级数和。此处 r = 18 ÷ 24 = 0.75，故 S∞ = 24 ÷ (1 − 0.75) = 96。
of (a + b)ⁿ for positive integer n uses the coefficients ⁿCᵣ and the term formula ⁿCᵣ aⁿ⁻ʳ bʳ first three terms are 16 + 96x + 216x², obtained from 2⁴, 4 × 2³ × 3x and 6 × 2² × (3x)².
正整数幂 (a + b)ⁿ 的二项展开使用组合系数 ⁿCᵣ 和通项公式 ⁿCᵣ aⁿ⁻ʳ bʳ。对于 (2 + 3x)⁴，前三项为 16 + 96x +
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