📚 OxfordAQA MA01 Pure Mathematics 1 Revision Notes | 牛津AQA MA01 纯数学1 知识点精讲
This article covers the essential topics assessed in the Oxford AQA International AS Mathematics MA01 Pure Mathematics 1 exam, based on the June 2023 mark scheme. Each section includes key concepts, typical problem types, and examiner expectations to help you prepare effectively.
本文基于2023年6月牛津AQA国际AS数学MA01纯数学1评分标准，梳理核心知识点。每节包含关键概念、常见题型与评分要求，助你高效备考。
1. Algebraic Expressions and Laws of Indices | 代数表达式与指数法则
ⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ. Rationalise denominators involving surds such as 1/√a.
通过合并同类项与指数法则化简表达式：aᵐ × aⁿ = aᵐ⁺ⁿ，aᵐ ÷ aⁿ = aᵐ⁻ⁿ，(aᵐ)ⁿ = aᵐⁿ。对含有根式的分母进行有理化，如 1/√a。
- Factorise fully: common factors, difference of two squares, and quadratic trinomials.
- 彻底因式分解：提取公因式、平方差公式及二次三项式。
- t between a× and index notation and evaluate negative and fractional powers: a⁻ⁿ = 1/aⁿ, a1/n = ⁿ√a.
- 转换 a× 与指数记号，计算负指数与分数指数：a⁻ⁿ = 1/aⁿ，a1/n = ⁿ√a。
Example: Simplify (3x²y⁻³)² ÷ (x⁻¹y²)³ = 9x⁴y⁻⁶ ÷ x⁻³y⁶ = 9x⁷y⁻¹²
示例：化简 (3x²y⁻³)² ÷ (x⁻¹y²)³ = 9x⁴y⁻⁶ ÷ x⁻³y⁶ = 9x⁷y⁻¹²
2. Quadratic Functions and Equations | 二次函数与方程
Solve quadratic equations by factorising, using the quadratic formula x = [−b ± √b² − 4ac ), or completing the square. The discriminant Δ = b² − 4ac determines the nature of roots.
通过因式分解、求根公式 x = [−b ± √b² − 4ac] / (2a) 或配方法解二次方程。判别式 Δ = b² − 4ac 决定根的性质。
- Δ > 0: two distinct real roots; Δ = 0: one repeated real root; Δ < 0: no real roots.
- Δ > 0：两个不等实根；Δ = 0：一个重根；Δ < 0：无实根。
- Complete the square to find the vertex of a parabola y = a(x − p)² + q.
- 用配方法求抛物线的顶点 y = a(x − p)² + q。
Vertex at (p, q); line of symmetry x = p.
顶点位于 (p, q)，对称轴为 x = p。
3. Simultaneous Equations | 联立方程组
Solve two linear equations by elimination or substitution. For one linear and one quadratic, substitute the linear expression into the quadratic to form an equation in one variable.
用消元法或代入法求解两个线性方程。对于一线性一二次方程组，将线性表达式代入二次方程，转化为一元方程求解。
Always check solutions in the original equations, especially when squaring may introduce extraneous roots.
注意代回原方程检验，尤其平方过程中可能产生增根。
E.g. y = 2x + 1 and x² + y² = 13 → x² + (2x+1)² = 13 → solve for x.
例：y = 2x + 1 与 x² + y² = 13 → x² + (2x+1)² = 13 → 解出 x。
4. Inequalities and Regions | 不等式与区域
Solve linear and quadratic inequalities. When multiplying or dividing by a negative number, reverse the inequality sign. Represent solutions on a number line or using set notation.
求解线性与二次不等式。乘以或除以负数时需反转不等号方向。在数轴或集合记号中表示解集。
For quadratic inequalities, sketch the graph to identify intervals where the quadratic is positive or negative.
对于二次不等式，画图判断二次函数大于零或小于零的区间。
x² − 5x + 6 ≤ 0 → (x−2)(x−3) ≤ 0 → 2 ≤ x ≤ 3.
x² − 5x + 6 ≤ 0 ⇒ (x−2)(x−3) ≤ 0 ⇒ 2 ≤ x ≤ 3。
5. Polynomials and Factor Theorem | 多项式与因式定理
x). Use polynomial division or equating coefficients to factorise cubic and quartic expressions.
若 f(a) = 0，则 (x − a) 是 f(x) 的因式。通过多项式除法或系数比较法分解三次与四次多项式。
The Remainder Theorem states that when f(x) is divided by (x − a), the remainder is f(a).
余式定理：多项式 f(x) 除以 (x − a) 的余式等于 f(a)。
| Given f(x) = 2x³ − 3x² − 3x + 2, find f(2). If f(2)=0 then (x−2) is a factor. | 已知 f(x) = 2x³ − 3x² − 3x + 2，求 f(2)。若 f(2)=0 则 (x−2) 为因式。 |
6. Binomial Expansion | 二项式展开
e integer n using Pascal’s triangle or the binomial theorem: (1 + x)ⁿ = 1 + nx + [n(n−1)/2!] x² + ...
对于正整数 n，利用帕斯卡三角形或二项式定理展开 (a + b)ⁿ：(1 + x)ⁿ = 1 + nx + [n(n−1)/2!] x² + ...
When n is not a positive integer, the expansion is infinite and valid for |x| < 1. Be able to find specific terms or ranges of validity.
当 n 不是正整数时，展开为无穷级数，且要求 |x| < 1。能求特定项或指出收敛域。
Find coefficient of x³ in (2 − 3x)⁵: term = ⁵C₃ (2)² (−3x)³ = 10 × 4 × (−27x³) = −1080 x³.
求 (2 − 3x)⁵ 中 x³ 的系数：项 = ⁵C₃ (2)² (−3x)³ = 10 × 4 × (−27x³) = −1080 x³。
7. Coordinate Geometry of Straight Lines and Circles | 直线与圆的坐标几何
y₁)/(x₂ − x₁). Parallel lines have equal gradients; perpendicular lines satisfy m₁m₂ = −1.
直线方程：y − y₁ = m(x − x₁)，其中 m = (y₂ − y₁)/(x₂ − x₁)。平行线斜率相等；垂直线满足 m₁m₂ = −1。
centre (a, b), radius r. Complete the square to find centre and radius from general form.
圆的方程：(x − a)² + (y − b)² = r²，圆心 (a, b)，半径 r。将一般式配方求圆心与半径。
nt of the intersection of a line and a circle to determine if a line is a tangent (Δ = 0).
利用直线与圆联立方程的判别式判断相切情况 (Δ = 0)。
8. Trigonometric Ratios and Equations | 三角比与方程
for angles 0°, 30°, 45°, 60°, 90°. Use identities: tan θ = sin θ / cos θ, sin² θ + cos² θ ≡ 1.
熟记 0°, 30°, 45°, 60°, 90° 的精确三角函数值。运用恒等式：tan θ = sin θ / cos θ，sin² θ + cos² θ ≡ 1。
Solve trigonometric equations within a given interval, using CAST diagrams or graphs to find all solutions.
在给定区间内求解三角方程，结合CAST图或图像找出所有解。
in 2θ = 0.5 for 0° ≤ θ ≤ 360°. Then 2θ = 30°, 150°, 390°, 510° → θ = 15°, 75°, 195°, 255°.
解 sin 2θ = 0.5，0° ≤ θ ≤ 360°：2θ = 30°, 150°, 390°, 510° → θ = 15°, 75°, 195°, 255°。
9. Differentiation and Applications | 微分及其应用
of x: d/dx (xⁿ) = n xⁿ⁻¹. Find gradients of curves and equations of tangents and normals.
对 x 的幂函数求导：d/dx (xⁿ) = n xⁿ⁻¹。求曲线斜率及切线与法线方程。
Second derivative d²y/dx² determines whether a stationary point is a maximum, minimum, or point of inflection. Use differentiation to solve optimisation problems.
二阶导数 d²y/dx² 用于判断驻点是极值点还是拐点。用微分解决最优化问题。
- Stationary points occur where dy/dx = 0.
- 驻点位于 dy/dx = 0 处。
- Increasing function: dy/dx > 0; decreasing function: dy/dx < 0.
- 增函数：dy/dx > 0；减函数：dy/dx < 0。
For y = x³ − 3x, dy/dx = 3x² − 3 = 0 → x = ±1. d²y/dx² = 6x: at x = 1, min; x = −1, max.
对于 y = x³ − 3x，dy/dx = 3x² − 3 = 0 → x = ±1。d²y/dx² = 6x：x=1 时极小，x=−1 时极大。
10. Integration and Area | 积分与面积
ⁿ⁺¹/(n+1) + C (n ≠ −1). Find the constant of integration using a given point on the curve.
积分 xⁿ：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)。利用曲线上已知点确定积分常数。
a under a curve y = f(x) between limits x = a and x = b. Be careful with areas below the x‑axis – they contribute as negative values, so use absolute value or split into parts.
定积分计算曲线 y = f(x) 与 x 轴在区间 [a, b] 之间的面积。x 轴下方的区域积分值为负，须取绝对值或分段处理。
= x² + 2 and x-axis from x=1 to x=3: ∫₁³ (x²+2) dx = [x³/3 + 2x]₁³ = (9+6) − (⅓+2) = 12⅔.
曲线 y = x² + 2 与 x 轴在 x=1 到 x=3 间的面积：∫₁³ (x²+2) dx = [x³/3 + 2x]₁³ = (9+6) − (⅓+2) = 12⅔。
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