📚 Mastering OxfordAQA International AS Mathematics 9660 Pure Mathematics PSM1 | 攻克牛津AQA国际AS数学9660纯数学PSM1
This revision guide is designed for students sitting the OxfordAQA International AS Mathematics 9660 Pure Mathematics PSM1 paper. It brings together the essential pure mathematics content, worked-style reminders and exam-focused strategies you need to build confidence and accuracy in algebra, coordinate geometry, trigonometry, sequences and introductory calculus.
本复习指南专为参加牛津AQA国际AS数学9660纯数学PSM1考试的学生编写。它汇集了代数、坐标几何、三角学、数列和入门微积分的核心内容、解题方法提醒与应试策略，帮助你建立信心并提高准确度。
1. PSM1 Exam Blueprint | PSM1 考试蓝图
The PSM1 unit forms the core of the International AS Mathematics 9660 qualification. It assesses algebraic manipulation, functions, coordinate geometry, sequences, trigonometry and introductory calculus. Questions mix straightforward skills with multi-step problem solving, so you need confident recall of methods and the ability to choose the most efficient route.
PSM1 单元是国际AS数学9660资格的核心组成部分。它考查代数运算、函数、坐标几何、数列、三角学以及入门微积分。题目既有直接的技能题，也有多步骤问题求解，因此你需要牢固掌握方法，并能选择最高效的解题路径。
Most questions carry method marks, so even if a numerical slip occurs, clear logical working can still earn substantial credit. Practise showing every substitution, rearrangement and conclusion clearly, especially under timed conditions.
大多数题目都有方法分，因此即使出现数值失误，清晰有逻辑的解题步骤仍能获得可观的分数。练习在限时条件下清楚地写出每一步代入、变形和结论。
2. Surds and Indices | 根式与指数
Surds are irrational roots such as √2 or √3. You must be able to simplify expressions like √50 = 5√2 and rationalise denominators, for example 1/(2+√3) is rewritten using the conjugate 2−√3. Rationalising is usually tested in the first part of a question before further manipulation.
根式是无理根，如√2或√3。你必须能够化简如√50=5√2的表达式，并将分母有理化，例如利用共轭式2−√3重写1/(2+√3)。分母有理化通常出现在题目的第一部分，之后还会继续进行运算。
laws are essential: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, a⁻ⁿ = 1/aⁿ, and a1/n = ⁿ√a. Use them to solve equations such as 2x+1 = 8x by expressing both sides with the same base.
指数法则是必备知识：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，(aᵐ)ⁿ = aᵐⁿ，a⁰=1，a⁻ⁿ=1/aⁿ，a1/n=ⁿ√a。通过将两边写成同底数，利用它们解方程，如2x+1=8x。
√50 = √25 × √2 = 5√2
1 ÷ (2+√3) = (2−√3) ÷ [(2+√3)(2−√3)] = 2−√3
3. Quadratic Functions | 二次函数
d Δ < 0 gives no real roots. This is often tested in questions about intersections between curves and lines.
对于 ax²+bx+c=0，判别式 Δ=b²−4ac 说明根的性质：Δ>0 有两个不同实根，Δ=0 有一个重根，Δ<0 无实根。这一知识点常用于考查曲线与直线交点的题目。
−3, −8). This is the fastest way to find the turning point and range of a quadratic, and it is also useful for solving quadratic equations when factorising is not obvious.
配方法可将 y=x²+6x+1 变形为 y=(x+3)²−8，因此顶点为 (−3,−8)。这是求二次函数转折点和值域最快的方法，在不易因式分解时也可用于解二次方程。
| Δ > 0 | two distinct real roots |
| Δ = 0 | one repeated real root |
| Δ < 0 | no real roots |
The table above summarises how the discriminant controls the number of real solutions. Always relate the sign of Δ back to the context of the question, for example a line missing a curve entirely.
上表总结了判别式如何决定实数解的个数。务必把 Δ 的符号与题目情境联系起来，例如一条直线与曲线完全没有交点。
4. Simultaneous Equations and Inequalities | 联立方程与不等式
resulting quadratic. Check both solutions in the original equations to avoid sign errors.
对于一次与二次联立方程，将一次表达式代入二次方程。例如 y=x+1 与 x²+y²=13 可化为 x²+(x+1)²=13，然后解所得二次方程。将两个解代回原方程检验，以避免符号错误。
Quadratic inequalities are tackled by finding critical values where the expression equals zero, then testing intervals. For x² − 5x + 6 > 0, factor to (x−2)(x−3) > 0, giving x < 2 or x > 3. Remember to reverse the inequality when multiplying or dividing by a negative number.
二次不等式通过找出表达式等于零的临界值，再检验区间来求解。对于 x²−5x+6>0，因式分解为 (x−2)(x−3)>0，得到 x<2 或 x>3。记住当乘以或除以负数时，不等号要反向。
x² − 5x + 6 > 0 → critical values x = 2 and x = 3 → solution x < 2 or x > 3
5. Polynomials and Algebraic Division | 多项式与代数除法
0. The remainder theorem gives the remainder when f(x) is divided by (x − a) as f(a). These two theorems let you check factors quickly before performing long division.
因式定理指出，当且仅当 f(a)=0 时，(x−a) 是 f(x) 的因式。余式定理给出 f(x) 除以 (x−a) 的余数为 f(a)。这两个定理可让你在进行长除法之前快速检验因式。
Use algebraic long division to divide a cubic by a linear factor. For example, dividing x³ − 4x² + x + 6 by (x−2) gives x² − 2x − 3, which then factorises to (x−3)(x+1). State the full factorisation clearly in your final answer.
使用代数长除法将三次多项式除以一次因式。例如，x³−4x²+x+6 除以 (x−2) 得 x²−2x−3，再因式分解为 (x−3)(x+1)。在最终答案中要清楚地写出完整的因式分解形式。
x³ − 4x² + x + 6 = (x−2)(x−3)(x+1)
6. Coordinate Geometry and Circles | 坐标几何与圆
(y₁+y₂)/2). Lines are parallel if their gradients are equal, and perpendicular if the product of their gradients is −1.
关键公式：斜率 m=(y₂−y₁)/(x₂−x₁)，距离 d=√[(x₂−x₁)²+(y₂−y₁)²]，中点 M=((x₁+x₂)/2,(y₁+y₂)/2)。斜率相等则两线平行，斜率乘积为−1则两线垂直。
d the centre and radius. Tangent questions often require using the perpendicular gradient to the radius.
圆方程 (x−a)²+(y−b)²=r² 的圆心为 (a,b)，半径为 r。你可能需要对 x²+y²−6x+4y−12=0 这类方程进行配方，以求出圆心和半径。切线问题通常需要利用半径的垂直斜率。
x² + y² − 6x + 4y − 12 = 0 → centre (3,−2), radius 5
7. Binomial Expansion | 二项式展开
For a positive integer n, (a+b)ⁿ = Σ from r=0 to n of nCr aⁿ⁻ʳ bʳ, where nCr = n!/(r!(n−r)! sitive integer powers, so the expansion is finite and you do not need validity conditions.
对于正整数 n，(a+b)ⁿ=Σ从r=0到n的 nCr aⁿ⁻ʳ bʳ，其中 nCr=n!/(r!(n−r)!)。在AS阶段，展开通常为正整数次幂，因此展开式是有限的，不需要考虑有效范围条件。
e careful with coefficients and signs when expanding (a−b)ⁿ, and remember to raise the whole term, including its coefficient, to the required power.
示例：(1+2x)³=1+6x+12x²+8x³。展开 (a−b)ⁿ 时要注意系数和符号，并记住将整个项（包括其系数）都进行相应的乘方。
(1+2x)³ = 1 + 3(2x) + 3(2x)² + (2x)³ = 1 + 6x + 12x² + 8x³
8. Trigonometric Functions and Equations | 三角函数与方程
Know exact values: sin 30°=½, cos 60°=½, tan 45°=1, sin 45°=√2/2, cos 45°=√2/2, sin 60°=√3/2, cos 30°=√
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