📚 Mastering AQA International AS Unit P1: Core Pure Mathematics and Exam Tips | 掌握 AQA 国际 AS 单元 P1：纯数学核心考点与备考要点
AQA International AS Unit P1 Pure Mathematics tests the core tools that underpin the whole A-level Maths course. You need to be fluent in algebra, functions, coordinate geometry, differentiation, integration, and trigonometry. This revision guide walks through the essential topics and the exam techniques that turn knowledge into marks.
AQA 国际 AS 单元 P1 纯数学考查的是支撑整个 A-level 数学课程的核心工具。你需要熟练运用代数、函数、坐标几何、微分、积分和三角学。本复习指南逐一讲解必考主题，以及将知识转化为分数的考试技巧。
1. Algebraic Manipulation and Surds | 代数运算与根式
Algebra is the engine room of P1. You will be expected to simplify expressions involving surds, rationalise denominators such as 1/(2 + √3), and apply the laws of indices confidently. Leaving an answer as (3 + √5)/2 is often acceptable, but you should be able to rewrite it in its simplest exact form.
代数是 P1 的动力核心。你需要化简根式表达式、对如 1/(2 + √3) 的分母有理化，并熟练运用指数律。将答案写为 (3 + √5)/2 通常可以接受，但你应能改写为最简精确形式。
aᵐ × aⁿ = aᵐ⁺ⁿ aᵐ ÷ aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ
For example, √75 simplifies to 5√3 because 75 = 25 × 3. Rationalising 1/√2 gives √2/2. When indices are negative or fractional, remember x⁻¹ = 1/x and x¹/² = √x. These skills appear constantly in later differentiation and integration questions.
例如，√75 化简为 5√3，因为 75 = 25 × 3。1/√2 有理化为 √2/2。当指数为负数或分数时，记住 x⁻¹ = 1/x，x¹/² = √x。这些技能会不断出现在后续的微分和积分题中。
2. Quadratic Functions and the Discriminant | 二次函数与判别式
Quadratics appear in almost every P1 paper. You should be able to solve by factorising, completing the square, and using the quadratic formula. Completing the square also reveals the vertex and helps with sketching the graph.
二次函数几乎出现在每套 P1 试卷中。你需要能够用因式分解、配方法和求根公式求解。配方法还能显示顶点并帮助画图。
x = (−b ± √b² − 4ac) / 2a
nt. If Δ < 0, there are no real roots and the curve is entirely above or below the x-axis.
判别式 Δ = b² − 4ac 是快速判断实根个数的方法。若 Δ > 0，曲线与 x 轴相交两次；若 Δ = 0，x 轴为切线；若 Δ < 0，无实根且曲线完全位于 x 轴上方或下方。
equation x² − 6x + 10 = 0 has no real solutions. The minimum point lies above the x-axis.
例如，y = x² − 6x + 10 的判别式为 Δ = (−6)² − 4 ×
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