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

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

Mastering algebra and functions is essential for success in CIE A-Level Mathematics. This revision guide covers polynomial manipulation, factor and remainder theorems, quadratic functions, domains and ranges, composite and inverse functions, graph transformations, and key transcendental functions such as modulus, exponential and logarithmic forms. Each section is structured to mirror the demands of the CIE 9709 syllabus, with techniques clearly explained and illustrated.

掌握代数和函数是取得 CIE A-Level 数学好成绩的关键。本复习指南覆盖多项式的运算、因式与余数定理、二次函数、定义域与值域、复合函数与反函数、图像变换以及模函数、指数函数和对数函数等关键内容。每个部分都根据 CIE 9709 大纲设计，清晰讲解技巧并配有示例。

1. Algebraic Expressions and Polynomials | 代数表达式与多项式

A polynomial is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where the coefficients aᵢ are real numbers and n is a non‑negative integer called the degree. You must be confident in adding, subtracting and multiplying polynomials, and in simplifying by collecting like terms.

多项式是形如 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ 的表达式，其中系数 aᵢ 是实数，n 是非负整数，称为次数。考生必须熟练进行多项式的加减乘运算，并能合并同类项化简。

When multiplying, expand brackets systematically. For example, (2x² + 3x − 1)(x − 4) = 2x³ − 8x² + 3x² − 12x − x + 4 = 2x³ − 5x² − 13x + 4. Always check the degree of the result: it should be the sum of the degrees of the factors.

乘法时，要有条理地展开括号。例如 (2x² + 3x − 1)(x − 4) = 2x³ − 8x² + 3x² − 12x − x + 4 = 2x³ − 5x² − 13x + 4。务必检查结果的次数：它应为各因式次数之和。

Factorisation is equally important. Start by looking for a common factor, then try to factorise trinomials into two binomials. In CIE exams you might need to factorise expressions like 6x² + x − 12 into (2x + 3)(3x − 4). Practice recognising the difference of two squares: a² − b² = (a − b)(a + b).

因式分解同样重要。先寻找公因式，再尝试将三项式分解为两个二项式。在 CIE 考试中，可能需要将 6x² + x − 12 分解为 (2x + 3)(3x − 4)。要熟练识别平方差：a² − b² = (a − b)(a + b)。

2. Factor Theorem and Remainder Theorem | 因式定理与余数定理

The remainder theorem states that when a polynomial f(x) is divided by (x − a), the remainder is f(a). If the remainder is zero, then (x − a) is a factor of f(x) – this is the factor theorem. These tools allow you to factorise higher‑degree polynomials without performing long division each time.

余数定理指出，当多项式 f(x) 除以 (x − a) 时，余数为 f(a)。若余数为零，则 (x − a) 是 f(x) 的一个因式——这就是因式定理。这些工具能让你无需每次都做长除法也能分解高次多项式。

To find a factor of f(x), try small integer values of a (often ±1, ±2, ±3) and evaluate f(a). Once a factor (x − p) is found, divide f(x) by (x − p) using algebraic long division or by equating coefficients. The quotient will be a polynomial of lower degree, often a quadratic that can be factorised further.

要找出 f(x) 的因式，可尝试用较小的整数值 a（通常是 ±1, ±2, ±3）计算 f(a)。一旦找到因式 (x − p)，就用代数长除法或比较系数法将 f(x) 除以 (x − p)。商将是较低次的多项式，通常是可继续分解的二次式。

For example, for f(x) = x³ − 4x² + x + 6, we find f(2) = 0, so (x − 2) is a factor. Dividing gives x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3) = (x − 2)(x − 3)(x + 1).

例如，对于 f(x) = x³ − 4x² + x + 6，我们发现 f(2) = 0，因此 (x − 2) 是因式。相除得 x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3) = (x − 2)(x − 3)(x + 1)。

3. Solving Polynomial Equations | 解多项式方程

Once a polynomial is fully factorised, solving f(x) = 0 becomes straightforward: set each factor equal to zero. For cubics and quartics, the exam will often guide you by giving one factor or one root, allowing you to reduce the equation to a quadratic.

多项式一旦完全分解，求解 f(x) = 0 就很简单：令每个因式等于零。对于三次和四次方程，考题通常会提供一个因式或一个根，使你能够把它降次为二次方程。

If the equation is not already factorised, use the factor theorem to find one rational root, then perform division. Always check your solutions by substituting them back into the original equation, especially when squaring steps were involved.

若方程尚未分解，则利用因式定理找出一个有理根，然后做除法。务必将解代回原方程检验，尤其在涉及平方步骤时更应如此。

For equations that are not polynomials, such as those with square roots or algebraic fractions, you may need to square both sides or clear denominators. After solving, watch out for extraneous roots that do not satisfy the original equation.

对于非多项式的方程，例如含有平方根或代数分式的方程，可能需要两边平方或去分母。求解后要注意舍去不满足原方程的增根。

4. Quadratic Functions: Completing the Square and Vertex | 二次函数：配方法与顶点

A quadratic function can be written as f(x) = ax² + bx + c. Completing the square rewrites it in the form f(x) = a(x − h)² + k, where (h, k) is the vertex of the parabola. This form makes it easy to identify the minimum or maximum value (k), the line of symmetry (x = h), and to sketch the graph.

二次函数可写作 f(x) = ax² + bx + c。配方法将其重写为 f(x) = a(x − h)² + k 的形式，其中 (h, k) 是抛物线的顶点。这种形式便于确定最小值或最大值 (k)、对称轴 (x = h) 并画草图。

To complete the square for 2x² + 8x + 5, first factor out the coefficient of x² from the first two terms: 2(x² + 4x) + 5. Then add and subtract the square of half the coefficient of x inside the bracket: 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3. The vertex is (−2, −3).

对 2x² + 8x + 5 配方，先将前两项的 x² 系数提出：2(x² + 4x) + 5，然后在括号内加上并减去 x 系数一半的平方：2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。顶点为 (−2, −3)。

When a > 0, the parabola opens upwards and the vertex gives the minimum; when a < 0, it opens downwards and the vertex gives the maximum. The y‑intercept is f(0) = c, and the roots (if real) are given by the quadratic formula.

当 a > 0 时，抛物线开口向上，顶点给出最小值；当 a < 0 时，开口向下，顶点给出最大值。y 轴截距为 f(0) = c，实根（若有）由求根公式给出。

5. Discriminant and Quadratic Inequalities | 判别式与二次不等式

The discriminant of the quadratic ax² + bx + c is Δ = b² − 4ac. It determines the nature of the roots without solving the equation. The standard cases are summarised below:

二次式 ax² + bx + c 的判别式为 Δ = b² − 4ac。它能在不解方程的情况下判断根的性质。常见情况总结如下：

Δ = b² − 4ac	Nature of roots	Graph touches x‑axis
Δ > 0	Two distinct real roots	Crosses at two points
Δ = 0	One repeated real root	Touches at one point
Δ < 0	No real roots	Does not meet x‑axis

Quadratic inequalities such as ax² + bx + c > 0 are solved by first finding the critical values where the expression equals zero, then testing intervals on a sign diagram or sketching the graph. Always check whether the inequality is strict or inclusive when writing the final answer.

二次不等式如 ax² + bx + c > 0 的求解方法是先求出表达式等于零的临界值，然后用符号图或草图检验各区间。在写最终答案时，务必注意不等式是严格还是包含等于。

For example, solve 2x² − 3x − 2 ≤ 0. Factorise: (2x + 1)(x − 2) ≤ 0. The critical values are x = −½ and x = 2. A sketch of the upward-opening parabola shows the expression is ≤ 0 between the roots, so the solution is −½ ≤ x ≤ 2.

例如，解 2x² − 3x − 2 ≤ 0。因式分解：(2x + 1)(x − 2) ≤ 0，临界值为 x = −½ 和 x = 2。画出开口向上的抛物线，可知表达式在两根之间 ≤ 0，因此解为 −½ ≤ x ≤ 2。

6. Introduction to Functions: Domain and Range | 函数基础：定义域与值域

A function maps every input value from its domain to exactly one output value in its range. Notation such as f : x ⟼ x² + 1 or f(x) = x² + 1 is used. The domain is the set of possible x‑values, often given explicitly or determined by algebraic restrictions (e.g. denominator ≠ 0, inside a square root ≥ 0).

函数将定义域内的每个输入值都映射为值域内的唯一输出值。常用的表示法有 f : x ⟼ x² + 1 或 f(x) = x² + 1。定义域是可能的 x 值集合，通常由题目明确给出或由代数限制确定（如分母 ≠ 0，平方根内 ≥ 0）。

To find the range, examine the shape of the graph or use completing the square. For a quadratic f(x) = x² − 4x + 3, completing the square gives (x − 2)² − 1, so the minimum value is −1 and the range is f(x) ≥ −1 (for domain all real x).

要求值域，可观察图形形状或使用配方法。对于二次函数 f(x) = x² − 4x + 3，配方得 (x − 2)² − 1，最小值为 −1，因此值域为 f(x) ≥ −1（当定义域为全体实数时）。

Common restrictions: f(x) = 1/(x − 3) has domain x ≠ 3; f(x) = √(2x + 5) has domain x ≥ −2.5. Conversely, the range can be restricted by the given domain; always substitute boundary values and consider the function’s behaviour.

常见限制：f(x) = 1/(x − 3) 的定义域为 x ≠ 3；f(x) = √(2x + 5) 的定义域为 x ≥ −2.5。反之，值域也会受给定定义域的限制；务必代入边界值并考察函数变化趋势。

7. Composite Functions | 复合函数

A composite function combines two functions, applying one after the other. fg(x) is read as ‘f of g of x’ and means f(g(x)). The order matters: fg(x) is generally not the same as gf(x). Work from the inside out: first evaluate g(x), then apply f to the result.

复合函数将两个函数结合，依次作用。fg(x) 读作“f of g of x”，表示 f(g(x))。顺序很重要：fg(x) 通常不同于 gf(x)。运算时由内向外：先计算 g(x)，再对结果施加 f。

For f(x) = 2x + 1 and g(x) = x² − 3, fg(x) = 2(x² − 3) + 1 = 2x² − 5. The domain of fg(x) consists of all x in the domain of g for which g(x) is in the domain of f. Both functions here have domain all real numbers, so fg(x) has domain ℝ.

对于 f(x) = 2x + 1 和 g(x) = x² − 3，fg(x) = 2(x² − 3) + 1 = 2x² − 5。fg(x) 的定义域由所有在 g 的定义域内且使 g(x) 在 f 的定义域内的 x 组成。此处两函数的定义域都是全体实数，故 fg(x) 的定义域为 ℝ。

When a composite function is defined, always check the ‘output of inner must be acceptable for outer’ rule. For instance, if f(x) = √x (domain x ≥ 0) and g(x) = x − 5, then fg(x) = √(x − 5) requires x − 5 ≥ 0, so domain is x ≥ 5.

定义复合函数时，务必检验“内层输出必须在外层定义域内”的规则。例如，若 f(x) = √x（定义域 x ≥ 0），g(x) = x − 5，则 fg(x) = √(x − 5) 要求 x − 5 ≥ 0，因此定义域为 x ≥ 5。

8. Inverse Functions | 反函数

An inverse function, written f⁻¹(x), reverses the effect of f. It exists only if f is one‑to‑one (each y comes from exactly one x). The graph of y = f⁻¹(x) is the reflection of y = f(x) in the line y = x.

反函数记作 f⁻¹(x)，它会逆转 f 的作用。只有当 f 是一一映射（每个 y 只由一个 x 产生）时，反函数才存在。y = f⁻¹(x) 的图像是 y = f(x) 关于直线 y = x 的反射。

To find f⁻¹(x) algebraically, write y = f(x), swap x and y, then solve for y. For f(x) = (2x + 1)/(x − 3) (x ≠ 3), write y = (2x + 1)/(x − 3), swap to x = (2y + 1)/(y − 3), cross‑multiply and rearrange to get y = (3x + 1)/(x − 2). Thus f⁻¹(x) = (3x + 1)/(x − 2), domain x ≠ 2.

代数上求 f⁻¹(x) 的步骤：设 y = f(x)，交换 x 和 y，再解出 y。例如 f(x) = (2x + 1)/(x − 3) (x ≠ 3)，令 y = (2x + 1)/(x − 3)，交换得 x = (2y + 1)/(y − 3)，交叉相乘并整理得 y = (3x + 1)/(x − 2)。因此 f⁻¹(x) = (3x + 1)/(x − 2)，定义域 x ≠ 2。

The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f. This relationship is useful when a function’s domain is restricted to make it one‑to‑one, for example f(x) = x² for x ≥ 0 has inverse f⁻¹(x) = √x.

f⁻¹ 的定义域是 f 的值域，f⁻¹ 的值域是 f 的定义域。当函数被限制定义域以使其一一对应时，这一关系非常有用，例如 f(x) = x² (x ≥ 0) 的反函数是 f⁻¹(x) = √x。

9. Graph Transformations and Solving Equations Graphically | 图像变换与图解方程

Transformations of the graph y = f(x) follow clear rules: y = f(x) + a is a vertical translation by a; y = f(x + a) is a horizontal translation by −a (shift left if a > 0); y = a f(x) is a vertical stretch (scale factor a); y = f(ax) is a horizontal stretch (scale factor 1/a); y = −f(x) is a reflection in the x‑axis; y = f(−x) is a reflection in the y‑axis.

图像 y = f(x) 的变换遵循明确规则：y = f(x) + a 是竖直方向平移 a；y = f(x + a) 是水平方向平移 −a（若 a > 0 则向左移）；y = a f(x) 是竖直拉伸（缩放因子 a）；y = f(ax) 是水平拉伸（缩放因子 1/a）；y = −f(x) 是关于 x 轴反射；y = f(−x) 是关于 y 轴反射。

Using graphs to solve equations is a core CIE skill. To solve f(x) = g(x), plot both y = f(x) and y = g(x) and find the x‑coordinates of their intersections. Alternatively, rearrange the equation to the form h(x) = 0 and find where the graph of y = h(x) crosses the x‑axis.

利用图像解方程是 CIE 的核心技能。要解 f(x) = g(x)，画出 y = f(x) 和 y = g(x)，找出两图像交点的 x 坐标。另一种方法是把方程移项成 h(x) = 0，

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