Algebra and Functions: Key Revision for CCEA A-Level Mathematics | CCEA A-Level 数学：代数和函数考点精讲

📚 Algebra and Functions: Key Revision for CCEA A-Level Mathematics | CCEA A-Level 数学：代数和函数考点精讲

Algebra and Functions form the backbone of CCEA A-Level Mathematics, underpinning topics from quadratics to exponential modelling. A strong command of algebraic manipulation, function theory, and graphical transformations is essential for success across AS and A2 modules. This revision guide consolidates the core concepts, common pitfalls, and exam techniques you will need for CCEA’s algebra and function questions.

代数和函数是 CCEA A-Level 数学的基石，支撑着从二次方程到指数建模的各个主题。熟练掌握代数操作、函数理论以及图像变换对于在 AS 和 A2 模块中取得成功至关重要。本复习指南汇集了 CCEA 代数与函数题所需的核心概念、常见陷阱和考试技巧。

1. Algebraic Simplification and Factorising | 代数化简与因式分解

Mastering simplification is the first step. You must be able to expand brackets, factorise expressions including quadratics, and simplify rational expressions by cancelling common factors. Look for common factors, difference of two squares, and grouping terms.

掌握化简是第一步。你必须能够展开括号、对含二次式的表达式进行因式分解，以及通过约去公因式来化简分式。注意寻找公因式、平方差和分组项。

Always check for hidden common factors. For example, x² – 9 factorises as (x – 3)(x + 3), while 2x² + 8x = 2x(x + 4). When simplifying a rational expression like (x² – 4)/(x – 2), factorise numerator and cancel: (x – 2)(x + 2)/(x – 2) = x + 2, provided x ≠ 2.

时刻检查隐藏的公因式。例如，x² – 9 可分解为 (x – 3)(x + 3)，而 2x² + 8x = 2x(x + 4)。在化简像 (x² – 4)/(x – 2) 这样的分式时，对分子因式分解然后约分：(x – 2)(x + 2)/(x – 2) = x + 2，前提是 x ≠ 2。

2. Polynomials, Factor Theorem and Remainder Theorem | 多项式、因式定理与余数定理

The Factor Theorem states that (x – a) is a factor of polynomial p(x) if and only if p(a) = 0. This is used to factorise cubic and higher-order polynomials by testing possible roots using the constant term’s factors.

因式定理指出，若且唯若 p(a) = 0，则 (x – a) 是多项式 p(x) 的因式。这可用于通过检验常数项的因数来对三次及更高次多项式进行因式分解。

The Remainder Theorem: when p(x) is divided by (x – a), the remainder is p(a). This allows you to find remainders without long division and often appears in CCEA exam questions requiring you to evaluate p(a) for a given a.

余数定理：当 p(x) 除以 (x – a) 时，余数为 p(a)。这使你不必进行长除法即可求得余数，并常在 CCEA 考题中出现，要求你针对给定的 a 计算 p(a)。

For example, for p(x) = x³ – 4x² + x + 6, test x = 2: p(2) = 8 – 16 + 2 + 6 = 0, so (x – 2) is a factor. Then divide or equate coefficients to find the quadratic factor, then factorise further.

例如，对于 p(x) = x³ – 4x² + x + 6，检验 x = 2：p(2) = 8 – 16 + 2 + 6 = 0，因此 (x – 2) 是一个因式。然后进行除法或比较系数以求得二次因式，再进一步分解。

3. Partial Fractions | 分式分解

Partial fractions decompose a rational function into simpler fractions, which is vital for integration and series expansion. CCEA expects you to handle distinct linear factors, repeated linear factors, and irreducible quadratic factors in the denominator.

分式分解将一个有理函数分解为更简单的分式，这对积分和级数展开至关重要。CCEA 要求你掌握分母中不同的线性因式、重复线性因式以及不可约的二次因式的处理。

For distinct linear factors: write (2x + 1)/[(x – 3)(x + 2)] ≡ A/(x – 3) + B/(x + 2). Multiply through by the denominator, substitute suitable x-values to find A and B. Always check for improper fractions first; perform polynomial division if the degree of numerator is equal to or greater than denominator.

对于不同的线性因式：将 (2x + 1)/[(x – 3)(x + 2)] 写成 A/(x – 3) + B/(x + 2)。乘以分母，代入合适的 x 值求出 A 和 B。务必先检查是否为假分式；若分子的次数大于或等于分母，需先进行多项式除法。

4. Indices, Surds and Rationalising | 指数、根式与有理化

Laws of indices (aᵐ × aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁻ⁿ = 1/aⁿ) must be second nature. CCEA questions often combine indices with surds, requiring simplification of expressions like √8 or rationalising denominators such as 1/(√2 – 1).

指数法则（aᵐ × aⁿ = aᵐ⁺ⁿ、(aᵐ)ⁿ = aᵐⁿ、a⁻ⁿ = 1/aⁿ）必须烂熟于心。CCEA 题目常将指数与根式结合，要求化简如 √8 的表达式，或有理化如 1/(√2 – 1) 的分母。

Remember that √a × √b = √(ab) and to rationalise a denominator with a surd, multiply numerator and denominator by the conjugate. For example, 1/(√5 – 2) = (√5 + 2)/( (√5 – 2)(√5 + 2) ) = (√5 + 2)/(5 – 4) = √5 + 2.

记住 √a × √b = √(ab)，而有理化含根式的分母时，将分子和分母同乘以共轭式。例如，1/(√5 – 2) = (√5 + 2)/( (√5 – 2)(√5 + 2) ) = (√5 + 2)/(5 – 4) = √5 + 2。

Fractional indices link to roots: x^(½) = √x, and x^(⅓) = ∛x. Be careful when evaluating negative fractional indices; rewriting as 1/(x^(m/n)) helps avoid errors.

分数指数与根式的关系：x^(½) = √x，x^(⅓) = ∛x。在计算负的分数指数时要小心；将其重写为 1/(x^(m/n)) 有助于避免错误。

5. Quadratic Functions and the Discriminant | 二次函数与判别式

The quadratic formula x = [–b ± √(b² – 4ac)]/(2a) solves ax² + bx + c = 0. The discriminant Δ = b² – 4ac determines the nature of roots: two real distinct roots (Δ > 0), one repeated root (Δ = 0), no real roots (Δ < 0).

二次公式 x = [–b ± √(b² – 4ac)]/(2a) 用于求解 ax² + bx + c = 0。判别式 Δ = b² – 4ac 决定根的性质：两个不等的实根（Δ > 0）、一个重根（Δ = 0）、无实根（Δ < 0）。

Completing the square rewrites a quadratic in the form a(x + p)² + q, revealing the vertex (–p, q). This is essential for sketching graphs and finding maximum/minimum values. For instance, x² + 6x + 5 = (x + 3)² – 4, so the minimum point is (–3, –4).

完成平方将二次式改写为 a(x + p)² + q 的形式，揭示顶点 (–p, q)。这对于绘制图像以及求最大值/最小值至关重要。例如，x² + 6x + 5 = (x + 3)² – 4，因此最小点为 (–3, –4)。

Quadratic inequalities like x² – 3x – 4 > 0 are solved by sketching the parabola and identifying where it is positive. Factorise to (x – 4)(x + 1) > 0, so x < –1 or x > 4.

像 x² – 3x – 4 > 0 这样的二次不等式可通过绘制抛物线并确定其正区间来求解。因式分解为 (x – 4)(x + 1) > 0，因此 x < –1 或 x > 4。

6. Functions: Definitions, Domain and Range | 函数：定义、定义域与值域

A function maps each input (x) to exactly one output (f(x)). CCEA questions frequently ask for the domain and range. Domain is the set of all allowed x-values; range is the set of all possible output values.

函数将每个输入 (x) 映射到唯一的输出 (f(x))。CCEA 题目常要求求定义域和值域。定义域是所有允许的 x 值的集合；值域是所有可能的输出值的集合。

To find the domain, consider restrictions: denominators cannot be zero, square roots require non-negative inside. For f(x) = √(x – 2), the domain is x ≥ 2. Range is found by analysing how the function behaves; for this square root function, the output is ≥ 0, so range is f(x) ≥ 0.

求定义域时需考虑限制条件：分母不能为零，根号下需非负。对于 f(x) = √(x – 2)，定义域为 x ≥ 2。通过分析函数的变化求得值域；对于该平方根函数，输出 ≥ 0，因此值域为 f(x) ≥ 0。

One-to-one functions have exactly one x for each y, which is required for the existence of an inverse function. Use the horizontal line test on the graph to check.

一一对应函数的每个 y 恰对应一个 x，这是反函数存在的必要条件。可用水平线检验法在图像上进行检查。

7. Inverse Functions | 反函数

The inverse function f⁻¹(x) reverses the effect of f. To find it, write y = f(x), swap x and y, then solve for y. For example, if f(x) = 3x – 2, then x = 3y – 2, so y = (x + 2)/3, thus f⁻¹(x) = (x + 2)/3.

反函数 f⁻¹(x) 逆转 f 的效果。求反函数的步骤：令 y = f(x)，交换 x 和 y，然后解出 y。例如，若 f(x) = 3x – 2，则 x = 3y – 2，因此 y = (x + 2)/3，故 f⁻¹(x) = (x + 2)/3。

The domain of f⁻¹ is the range of f. Graphs of f and f⁻¹ are reflections of each other in the line y = x. CCEA often tests your ability to sketch these and state their domains and ranges.

f⁻¹ 的定义域是 f 的值域。f 和 f⁻¹ 的图像关于直线 y = x 对称。CCEA 经常考查你绘制这些图像并写出其定义域和值域的能力。

8. Composite Functions | 复合函数

A composite function combines two functions: fg(x) means first apply g, then apply f. In other words, fg(x) = f(g(x)). Order matters: fg(x) is generally not equal to gf(x).

复合函数由两个函数组合而成：fg(x) 表示先应用 g，再应用 f。换言之，fg(x) = f(g(x))。顺序很重要：fg(x) 一般不等于 gf(x)。

When finding the range of a composite function, you must consider the domain of the inner function and how its output maps through the outer function. For example, if f(x) = √x (x ≥ 0) and g(x) = x – 3, then fg(x) = √(x – 3). The domain requires x – 3 ≥ 0, so x ≥ 3.

求复合函数的值域时，必须考虑内层函数的定义域以及其输出如何通过外层函数映射。例如，若 f(x) = √x（x ≥ 0）且 g(x) = x – 3，则 fg(x) = √(x – 3)。定义域要求 x – 3 ≥ 0，即 x ≥ 3。

Work carefully with functions defined piecewise or with restricted domains, as these appear in CCEA C3 and C4 papers.

处理分段定义的函数或有定义域限制的函数时要非常小心，因为这些会出现在 CCEA C3 和 C4 试卷中。

9. The Modulus Function | 绝对值函数

The modulus function |x| is defined as x if x ≥ 0, and –x if x < 0. Its graph is V-shaped with vertex at origin. CCEA questions involve solving equations such as |2x – 3| = 5 and inequalities like |x + 1| ≤ 4.

绝对值函数 |x| 定义为：若 x ≥ 0 则为 x，若 x < 0 则为 –x。其图像呈 V 形，顶点在原点。CCEA 题目涉及求解方程如 |2x – 3| = 5 和不等式如 |x + 1| ≤ 4。

To solve modulus equations, consider both the positive and negative scenarios: |A| = B means A = B or A = –B. For inequalities, sketch or test intervals. For example, |x – 2| < 3 gives –3 < x – 2 < 3, so –1 < x < 5.

解绝对值方程时，需考虑正负两种情况：|A| = B 意味着 A = B 或 A = –B。对于不等式，可通过画草图或检验区间来求解。例如，|x – 2| < 3 可化为 –3 < x – 2 < 3，因此 –1 < x < 5。

Modulus can also combine with other functions, for instance sketching y = |f(x)| or y = f(|x|). The former reflects negative parts of f(x) in the x-axis, the latter reflects the graph for negative x in the y-axis.

绝对值还可与其他函数结合，例如绘制 y = |f(x)| 或 y = f(|x|) 的图像。前者将 f(x) 的负值部分关于 x 轴反射，后者将负 x 部分的图像关于 y 轴反射。

10. Transformations of Graphs | 图像变换

CCEA requires a solid understanding of graph transformations. The main types are translations, stretches, and reflections. For a function y = f(x), y = f(x – a) translates the graph a units to the right; y = f(x) + a translates it a units up.

CCEA 要求扎实掌握图像变换。主要类型有平移、伸缩和反射。对于函数 y = f(x)，y = f(x – a) 将图像向右平移 a 个单位；y = f(x) + a 将其向上平移 a 个单位。

Stretches: y = a f(x) stretches vertically by factor a; y = f(ax) stretches horizontally by factor 1/a (compression if a > 1). Reflections: y = –f(x) reflects in x-axis; y = f(–x) reflects in y-axis.

伸缩：y = a f(x) 在竖直方向上伸缩 a 倍；y = f(ax) 在水平方向上伸缩因子 1/a（若 a > 1 则为压缩）。反射：y = –f(x) 关于 x 轴反射；y = f(–x) 关于 y 轴反射。

Combining transformations: always apply stretches, reflections, and then translations (or follow the order of operations). CCEA may ask for the exact equation after a sequence of transformations, or to describe the transformations mapping one graph to another.

组合变换：总是先进行伸缩和反射，然后进行平移（或遵循运算顺序）。CCEA 可能要求写出一系列变换后的准确方程，或描述将一个图像映射到另一个图像所经的变换。

11. Exponential and Logarithmic Functions | 指数函数与对数函数

Exponential functions of the form y = aˣ (with a > 0) and the natural exponential y = eˣ are core topics. The graph of y = eˣ passes through (0,1) and grows rapidly. Its inverse is the natural logarithm, y = ln x, defined for x > 0.

形如 y = aˣ（a > 0）的指数函数以及自然指数函数 y = eˣ 是核心主题。y = eˣ 的图像经过 (0,1) 且增长迅速。其反函数是自然对数 y = ln x，定义域为 x > 0。

You must know the laws of logarithms: ln(ab) = ln a + ln b, ln(a/b) = ln a – ln b, ln(aⁿ) = n ln a. These are used to solve exponential equations and in modelling growth and decay.

你必须掌握对数法则：ln(ab) = ln a + ln b，ln(a/b) = ln a – ln b，ln(aⁿ) = n ln a。这些法则用于求解指数方程以及进行增长和衰减建模。

To solve an equation like 5ˣ = 20, take ln of both sides: x ln 5 = ln 20, so x = ln 20 / ln 5. When solving e²ˣ = 6, use the inverse directly: 2x = ln 6, so x = (ln 6)/2.

求解像 5ˣ = 20 这样的方程时，两边取 ln：x ln 5 = ln 20，因此 x = ln 20 / ln 5。在解 e²ˣ = 6 时，直接使用其逆运算：2x = ln 6，所以 x = (ln 6)/2。

Graphing y = aˣ and y = logₐ x shows their inverse relationship. Understand how transformations affect exponential and logarithmic graphs.

绘制 y = aˣ 和 y = logₐ x 的图像可显示它们的互逆关系。理解变换如何影响指数和对数图像。

12. Inequalities with Polynomials and Rational Functions | 多项式与分式函数不等式

CCEA exam questions frequently extend to cubic or rational inequalities. To solve (x – 1)(x + 2)(x – 3) > 0, identify critical values where the expression equals zero: x = –2, 1, 3. Test intervals to determine the sign, resulting in x < –2 or 1 < x < 3.

CCEA 考试题目常扩展到三次不等式或有理不等式。解 (x – 1)(x + 2)(x – 3) > 0 时，找出表达式为零的临界值：x = –2、1、3。检验区间以确定符号，结果 x < –2 或 1 < x < 3。

For rational inequalities like (x + 1)/(x – 2) ≤ 3, bring all terms to one side, combine into a single fraction, find critical values (zeros of numerator and denominator), and test intervals. Remember that the denominator cannot be zero, so exclude x = 2.

对于像 (x + 1)/(x – 2) ≤ 3 这样的分式不等式，将所有项移到一边，合并成一个分式，求出临界值（分子和分母的零点），然后检验区间。记住分母不能为零，因此排除 x = 2。

Always present solutions using inequality notation or interval notation, and be careful with strict versus inclusive inequalities.

始终使用不等式符号或区间符号呈现解集，并注意严格不等式与包含等号的不等式的区别。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Algebra and Functions: Key Revision for CCEA A-Level Mathematics | CCEA A-Level 数学：代数和函数 考点精讲