📚 A-Level Maths: Algebra and Functions Revision | A-Level 数学：代数和函数考点精讲
This comprehensive revision guide covers the core Algebra and Functions topics in A-Level Mathematics. From simplifying expressions and factorising polynomials to understanding function transformations and solving inequalities, every essential concept is explained with clear examples. Master these techniques to build a solid foundation for calculus, trigonometry, and further pure mathematics.
这份全面的复习指南涵盖了 A-Level 数学中代数和函数的核心主题。从化简表达式和分解多项式，到理解函数变换与解不等式，每一个重要概念都配有清晰的例子。掌握这些技巧，将为微积分、三角学和进阶纯数打下坚实基础。
1. Basic Algebraic Manipulation | 代数基本运算
Simplifying algebraic expressions relies on collecting like terms and applying index laws correctly. For instance, 3a + 5a simplifies to 8a, and x² · x⁴ = x⁶.
代数式的化简依赖于正确合并同类项以及运用指数运算法则。例如，3a + 5a 化简为 8a，而 x² · x⁴ = x⁶。
When expanding brackets, multiply every term inside the bracket by the factor outside. In the expression 4(2x – 3), you obtain 8x – 12.
展开括号时，要将括号外的因子与括号内的每一项相乘。在表达式 4(2x – 3) 中，得到 8x – 12。
Factorising is the reverse process: identifying a common factor. For example, 6ab + 3a² can be written as 3a(2b + a).
因式分解是相反的过程：找出公因子。例如，6ab + 3a² 可以写成 3a(2b + a)。
Negative and fractional indices follow specific rules. The expression x⁻² is equivalent to 1/x², while x^(1/2) represents the square root √x.
负指数和分数指数遵循特定规则。表达式 x⁻² 等价于 1/x²，而 x^(1/2) 表示平方根 √x。
2. Polynomials and Factorisation | 多项式与因式分解
A polynomial in x is an expression of the form aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where n is a non-negative integer. The degree of the polynomial is the highest power of x present.
x 的多项式是形如 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ 的表达式，其中 n 为非负整数。多项式的次数是出现的最高次幂。
Factorising quadratics can be done by inspection when the coefficient of x² is 1. To factorise x² + 5x + 6, find two numbers that multiply to 6 and add to 5: (x + 2)(x + 3).
当 x² 的系数为 1 时，可以通过观察来分解二次式。要分解 x² + 5x + 6，找到乘积为 6、和为 5 的两个数：(x + 2)(x + 3)。
When the coefficient of x² is not 1, factorisation may require splitting the middle term. For 2x² + 7x + 3, rewrite as 2x² + 6x + x + 3, then factor in pairs to get (2x + 1)(x + 3).
当 x² 的系数不是 1 时，分解可能需要拆分中项。对于 2x² + 7x + 3，重写为 2x² + 6x + x + 3，然后分组分解得到 (2x + 1)(x + 3)。
Difference of two squares is a key pattern: a² – b² = (a – b)(a + b). This can be applied to expressions like 9x² – 16, which factorises to (3x – 4)(3x + 4).
平方差是一个重要模式：a² – b² = (a – b)(a + b)。这可用于 9x² – 16 这样的表达式，分解为 (3x – 4)(3x + 4)。
3. Remainder and Factor Theorems | 余数定理与因式定理
The remainder theorem states that if a polynomial f(x) is divided by (x – a), the remainder is f(a). For example, dividing f(x) = 2x³ – 3x + 5 by (x – 2) gives a remainder of f(2) = 2(8) – 6 + 5 = 15.
余数定理指出，如果多项式 f(x) 除以 (x – a)，余数为 f(a)。例如，f(x) = 2x³ – 3x + 5 除以 (x – 2)，余数为 f(2) = 2(8) – 6 + 5 = 15。
The factor theorem is a special case: (x – a) is a factor of f(x) if and only if f(a) = 0. This is used to find one factor, enabling further factorisation of polynomials.
因式定理是特殊情况：(x – a) 是 f(x) 的因式当且仅当 f(a) = 0。这用于找出一个因式，进而分解多项式。
By testing values such as x = 1, -1, 2, -2 using the factor theorem, a factor can be found, and then long division or synthetic division reduces the polynomial degree.
利用因式定理，通过测试 x = 1, -1, 2, -2 等值，可以找到一个因式，然后用长除法或综合除法降低多项式的次数。
Given a cubic with one known factor, dividing yields a quadratic which can be factorised further. This is a common A-Level technique.
对于一个已知一个因式的三次多项式，通过除法得到一个可以继续分解的二次式。这是 A-Level 中的常见技巧。
4. Algebraic Fractions and Partial Fractions | 代数分式与部分分式
Simplifying algebraic fractions involves factorising numerators and denominators and cancelling common factors. For example, (x² – 4)/(x² + 3x + 2) simplifies to (x – 2)/(x + 1) after cancelling (x + 2).
化简代数分式需要将分子和分母分解因式，并约去公因式。例如，(x² – 4)/(x² + 3x + 2) 约去 (x + 2) 后化简为 (x – 2)/(x + 1)。
Partial fractions decompose a single rational expression into a sum of simpler fractions. A proper fraction with distinct linear factors, such as 3/(x(x + 2)), can be written as A/x + B/(x + 2).
部分分式将一个有理表达式分解为几个较简单分式之和。一个具有不同线性因式的真分式，如 3/(x(x + 2))，可以写为 A/x + B/(x + 2)。
Solving for the constants A and B involves multiplying through by the common denominator and equating coefficients or substituting suitable x values.
求解常数 A 和 B 需要乘以公分母，并比较系数或代入合适的 x 值。
For repeated linear factors such as (x – 1)² in the denominator, the decomposition takes the form A/(x – 1) + B/(x – 1)². Irreducible quadratic factors require a numerator of the form Cx + D.
对于重复线性因式，如分母中有 (x – 1)²，分解形式为 A/(x – 1) + B/(x – 1)²。不可约二次因式则需要分子为 Cx + D 的形式。
5. Exponential and Logarithmic Functions | 指数函数与对数函数
Exponential functions are of the form y = aˣ, with a > 0. The special base e ≈ 2.718 gives y = eˣ, whose gradient equals the function value at every point.
指数函数的形式为 y = aˣ，其中 a > 0。以特殊底数 e ≈ 2.718 构成的 y = eˣ，其导数在每一点都等于函数值本身。
Logarithms are the inverse of exponentials: if aˣ = b, then x = logₐb. Key laws include logₐ(xy) = logₐx + logₐy and logₐ(xⁿ) = n logₐx.
对数是指数的逆运算：如果 aˣ = b，那么 x = logₐb。关键法则包括 logₐ(xy) = logₐx + logₐy 和 logₐ(xⁿ) = n logₐx。
The natural logarithm ln x is logₑx. It is used extensively in calculus. The graph of y = ln x has a vertical asymptote at x = 0 and passes through (1, 0).
自然对数 ln x 即 logₑx，在微积分中广泛使用。y = ln x 的图像在 x = 0 处有一条垂直渐近线，且过点 (1, 0)。
Solving exponential equations often involves taking logs on both sides. For 2ˣ = 10, taking ln gives x ln 2 = ln 10, so x = ln 10 / ln 2.
求解指数方程通常需要两边取对数。对于 2ˣ = 10，取 ln 得 x ln 2 = ln 10，因此 x = ln 10 / ln 2。
6. Functions: Domain, Range, and Composition | 函数：定义域、值域与复合
A function f maps each element x from its domain to a unique value f(x). The domain is the set of all possible inputs, and the range is the set of all possible outputs.
一个函数 f 将其定义域中的每一个元素 x 映射到一个唯一的值 f(x)。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。
For f(x) = √(x – 2), the domain is x ≥ 2, and the range is f(x) ≥ 0. Restrictions like denominators cannot be zero must be considered.
对于 f(x) = √(x – 2)，定义域为 x ≥ 2，值域为 f(x) ≥ 0。必须考虑分母不能为零等限制。
Composite functions combine two functions where the output of one becomes the input of the other. For f(x) = 2x and g(x) = x + 3, the composite fg(x) means f(g(x)) = 2(x + 3) = 2x + 6.
复合函数将两个函数结合起来，一个函数的输出成为另一个的输入。对于 f(x) = 2x 和 g(x) = x + 3，复合函数 fg(x) 即 f(g(x)) = 2(x + 3) = 2x + 6。
The order matters: gf(x) = g(f(x)) = 2x + 3, which is different. Also, the domain of the composite must be valid for the inner function.
顺序很重要：gf(x) = g(f(x)) = 2x + 3，两者不同。此外，复合函数的定义域必须对内层函数有效。
7. Inverse Functions | 反函数
An inverse function f⁻¹(x) reverses the mapping of f(x). It exists only if f is one-to-one over its domain. The graph of f⁻¹ is a reflection of f in the line y = x.
反函数 f⁻¹(x) 逆转函数 f(x) 的映射。只有当 f 在其定义域上是一一映射时，反函数才存在。f⁻¹ 的图像是 f 关于直线 y = x 的反射。
To find an inverse, write y = f(x), swap x and y, then solve for y. For f(x) = 2x – 3, set x = 2y – 3, giving y = (x + 3)/2.
要求反函数，写出 y = f(x)，交换 x 和 y，然后解出 y。对于 f(x) = 2x – 3，设 x = 2y – 3，得 y = (x + 3)/2。
The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f. It is important to check the original function’s one-to-one nature.
f⁻¹ 的定义域是 f 的值域，f⁻¹ 的值域是 f 的定义域。检查原函数是否一一对应很重要。
For a quadratic function like f(x) = x² defined for x ≥ 0, the inverse is f⁻¹(x) = √x. Restricting the domain was necessary to make it one-to-one.
对于定义在 x ≥ 0 上的二次函数 f(x) = x²，其反函数为 f⁻¹(x) = √x。为了使其一一映射，限制定义域是必要的。
8. Transformations of Graphs | 函数图像的变换
Translations shift a graph horizontally or vertically. The graph of y = f(x) + c is shifted upwards by c units, while y = f(x + c) shifts it left by c units.
平移将图像水平或垂直移动。y = f(x) + c 的图像向上移动 c 个单位，而 y = f(x + c) 的图像向左移动 c 个单位。
Stretches change the scale. y = kf(x) stretches the graph vertically by factor k, and y = f(kx) compresses it horizontally by factor 1/k (if k > 1).
伸缩改变比例。y = kf(x) 在垂直方向以因子 k 拉伸，y = f(kx) 在水平方向以因子 1/k 压缩（若 k > 1）。
Reflections flip the graph over an axis. y = -f(x) reflects in the x-axis, while y = f(-x) reflects in the y-axis. Combining transformations must follow the correct order.
反射将图像沿坐标轴翻转。y = -f(x) 关于 x 轴反射，y = f(-x) 关于 y 轴反射。组合变换必须遵循正确顺序。
When multiple transformations are applied, it is usually safest to perform horizontal shifts and stretches (affecting x) before vertical ones, but always consider the form y = a f(b(x + c)) + d.
当应用多个变换时，通常先进行影响 x 的水平移动和伸缩，再进行垂直变换最为稳妥，但恒考虑形式 y = a f(b(x + c)) + d。
9. Quadratic Functions and the Discriminant | 二次函数与判别式
A quadratic can be written as y = ax² + bx + c or as y = a(x – p)² + q by completing the square. The vertex of the parabola is at (p, q).
二次函数可以写成 y = ax² + bx + c，或通过配方写成 y = a(x – p)² + q。抛物线的顶点在 (p, q)。
The discriminant Δ = b² – 4ac determines the nature of the roots. If Δ > 0, there are two distinct real roots; if Δ = 0, there is one repeated root; if Δ < 0, there are no real roots.
判别式 Δ = b² – 4ac 决定根的性质。若 Δ > 0，有两个不相等的实根；若 Δ = 0，有一个重根；若 Δ < 0，无实根。
Completing the square is used to find the maximum or minimum value. For a > 0, the parabola opens upward and the vertex is a minimum point; for a < 0, it is a maximum.
配方法用于求最大值或最小值。若 a > 0，抛物线开口向上，顶点为最小值点；若 a < 0，则为最大值点。
Solving quadratic inequalities, such as x² – 5x + 6 > 0, requires finding the roots and sketching the curve to see where the function is positive. The solution is x < 2 or x > 3.
解二次不等式，例如 x² – 5x + 6 > 0，需要求出根并画出图像以确定函数取正值的区间。解为 x < 2 或 x > 3。
10. Inequalities and Modulus | 不等式与模函数
Linear inequalities are solved similarly to equations, but multiplying or dividing by a negative number reverses the inequality sign. For -2x < 4, dividing by -2 gives x > -2.
线性不等式的解法与方程类似，但当乘以或除以负数时，不等号要反向。对于 -2x < 4，除以 -2 得 x > -2。
The modulus function |x| gives the distance from zero. The inequality |x| < a is equivalent to -a < x < a, and |x| > a leads to x < -a or x > a.
模函数 |x| 表示到零的距离。不等式 |x| < a 等价于 -a < x < a，而 |x| > a 则导出 x < -a 或 x > a。
For modulus equations involving expressions, separate into two cases without modulus. Solve |2x – 1| = 5 by setting 2x – 1 = 5 and 2x – 1 = -5, yielding x = 3 and x = -2.
对于包含表达式的模方程，分成无模符号的两种情况。解 |2x – 1| = 5，设 2x – 1 = 5 和 2x – 1 = -5，得 x = 3 和 x = -2。
Inequalities like |3x + 2| ≤ 4 can be solved by writing -4 ≤ 3x + 2 ≤ 4, then simplifying to get the range -2 ≤ x ≤ 2/3.
形如 |3x + 2| ≤ 4 的不等式可写为 -4 ≤ 3x + 2 ≤ 4，然后化简得到范围 -2 ≤ x ≤ 2/3。
11. Sketching Rational Functions | 绘制有理函数的图像
Rational functions take the form f(x) = P(x)/Q(x), where P and Q are polynomials. Key features include intercepts, asymptotes, and the behaviour for large |x|.
有理函数的形式为 f(x) = P(x)/Q(x)，其中 P、Q 为多项式。其主要特征包括截距、渐近线以及 |x| 很大时的行为。
Vertical asymptotes occur where Q(x) = 0 and the function is not defined. After cancelling common factors, the roots of the denominator give vertical lines x = a.
垂直渐近线出现在 Q(x) = 0 且函数无定义处。约去公因式后，分母的根给出垂直直线 x = a。
Horizontal asymptotes are found by considering the limit as x → ±∞. If the degree of P is less than that of Q, y = 0 is the asymptote; if equal, it is y = leading coefficient ratio.
水平渐近线通过求 x → ±∞ 的极限来找到。若 P 的次数小于 Q 的次数，渐近线为 y = 0；若相等，则为 y = 首项系数之比。
To sketch, find any x-intercepts (from P(x) = 0) and the y-intercept (x = 0), then use the asymptotes as guides and plot a few points to determine the curve’s shape.
绘图时，找出 x 截距（由 P(x) = 0）和 y 截距（x = 0），然后以渐近线为引导，标出几个点来确定曲线形状。
12. Algebraic Proof | 代数证明
Algebraic proof involves using logical steps to show that a statement is always true. It often requires expanding, factorising, or applying known identities.
代数证明包括用逻辑步骤来证明一个表述恒为真。通常需要展开、分解或应用已知恒等式。
Prove that the sum of the squares of three consecutive integers is 3 more than three times the square of the middle one. Let integers be n-1, n, n+1; then (n-1)² + n² + (n+1)² = 3n² + 2, while 3n² + 3 = 3n² + 3, thus the difference is constant.
证明三个连续整数的平方和等于中间数的平方的三倍加 3。设整数为 n-1, n, n+1；那么 (n-1)² + n² + (n+1)² = 3n² + 2，而 3n² + 3 = 3n² + 3，因此差为常数。
To prove that the product of two odd numbers is odd, let them be (2m + 1) and (2n + 1). Their product is 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1, which is odd.
要证明两个奇数的积是奇数，设它们为 (2m + 1) 和 (2n + 1)。它们的积为 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1，此为奇数。
Proof by contradiction can be used: assume the opposite and derive an impossibility. This method is especially powerful for irrationality proofs, such as showing √2 is irrational.
反证法可被使用：假设相反情况并推导出不可能的事。这种方法对于无理数证明尤其有力，例如证明 √2 是无理数。
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