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Year 12 CIE Mathematics Formula & Theorem Quick Reference | CIE Year 12 数学公式定理速查手册

2026-07-22 18:38 · 本文部分内容由AI辅助创作

📚 Year 12 CIE Mathematics Formula & Theorem Quick Reference | CIE Year 12 数学公式定理速查手册

This quick reference handbook summarises essential formulas and theorems for Year 12 CIE AS Mathematics (9709). It covers Pure Mathematics 1 topics along with key statistics concepts to help you revise efficiently. Keep this guide handy for quick checks and exam preparation.

本速查手册总结了 Year 12 CIE AS 数学 (9709) 的核心公式与定理,涵盖纯数 1 及部分统计内容,帮助高效复习。考试前随时查阅。

1. Quadratics & Discriminant | 二次方程与判别式

The quadratic formula for solving ax² + bx + c = 0 is x = (−b ± √b² − 4ac) / (2a).

二次方程 ax² + bx + c = 0 的求根公式为 x = (−b ± √b² − 4ac) / (2a)。

s: if Δ > 0, two distinct real roots; Δ = 0, one repeated real root; Δ < 0, no real roots.

判别式 Δ = b² − 4ac 决定根的性质:Δ > 0,两个不等实根;Δ = 0,一个重根;Δ < 0,无实根。

e square transforms ax² + bx + c into a(x + p)² + q, where p = b/(2a) and q = c − b²/(4a).

配方法将 ax² + bx + c 转化为 a(x + p)² + q,其中 p = b/(2a),q = c − b²/(4a)。

If α and β are the roots of ax² + bx + c = 0, then sum α + β = −b/a and product αβ = c/a.

若 α、β 为方程 ax² + bx + c = 0 的根,则和 α + β = −b/a,积 αβ = c/a。


2. Equations & Inequalities | 方程与不等式

When multiplying or dividing an inequality by a negative number, reverse the inequality sign: if a < b and c < 0, then ac > bc.

不等式两边同乘或同除负数时,不等号要改变方向:若 a < b 且 c < 0,则 ac > bc。

Quadratic inequalities are solved by factorising, finding critical values, and using a sign diagram to determine intervals where the expression is > 0 or < 0.

解二次不等式需要因式分解,求出临界值,然后用符号图确定表达式大于零或小于零的区间。

For modulus equations |ax + b| = c, rewrite as ax + b = ±c. For inequalities |ax + b| ≤ c, this is equivalent to −c ≤ ax + b ≤ c.

形如 |ax + b| = c 的模方程可化为 ax + b = ±c;对于模不等式 |ax + b| ≤ c,等价于 −c ≤ ax + b ≤ c。

Always check solutions of modulus equations by substitution to avoid extraneous roots.

解模方程后务必代回原方程检验,避免增根。


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3. Functions & Graphs | 函数与图像

The domain of a function f(x) is the set of all possible input values; the range is the set of all output values.

函数 f(x) 的定义域是所有可能输入的集合,值域是所有可能输出的集合。

g(x) = f(g(x)), first apply g, then f. The range of g must be a subset of the domain of f.

复合函数 fg(x) = f(g(x)) 先作用 g 再作用 f;g 的值域必须是 f 定义域的子集。

A function has an inverse f⁻¹ only if it is one‑one. To find f⁻¹, write y = f(x), swap x and y, then solve for y.

仅当函数是一一映射时才存在反函数 f⁻¹。求反函数:令 y = f(x),交换 x 与 y,再解出 y。

= af(x) stretches vertically by factor a; y = f(ax) stretches horizontally by factor 1/a.

图像变换:y = f(x) + a 沿 y 轴平移 a;y = f(x + a) 沿 x 轴平移 −a;y = af(x) 纵向拉伸 a 倍;y = f(ax) 横向压缩 1/a 倍。


4. Coordinate Geometry | 坐标几何

Distance between two points (x₁, y₁) and (x₂, y₂): d = √[(x₂ − x₁)² + (y₂ − y₁)²].

两点 (x₁, y₁) 和 (x₂, y₂) 间的距离为 d = √[(x₂ − x₁)² + (y₂ − y₁)²]。

Midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2).

中点 M = ((x₁ + x₂)/2, (y₁ + y₂)/2)。

m = (y₂ − y₁)/(x₂ − x₁). Two lines are parallel if m₁ = m₂, perpendicular if m₁ × m₂ = −1.

过 (x₁, y₁) 和 (x₂, y₂) 的直线斜率 m = (y₂ − y₁)/(x₂ − x₁)。两直线平行则 m₁ = m₂,垂直则 m₁ × m₂ = −1。

Equation of a straight line: y − y₁ = m(x − x₁) or ax + by + c = 0.

直线方程:y − y₁ = m(x − x

Published by TutorHao | Year 12 Mathematics Revision Series | aleveler.com

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