📚 Inequalities for AQA A-Level Maths | A-Level AQA 数学不等式考点精讲
Inequalities form a core part of AQA A-Level Mathematics, appearing across pure algebra, graphical methods, and even applied contexts. A solid grasp of techniques – from linear and quadratic inequalities to rational expressions, modulus signs and discriminant conditions – will give you the confidence to tackle any exam question. This guide covers all essential types with clear worked examples, examiner-style warnings and practical advice.
不等式是 AQA A-Level 数学的核心组成部分，出现在纯代数、图形解法甚至应用背景中。牢固掌握求解技巧——从线性不等式、二次不等式到分式不等式、绝对值符号与判别式条件——将使你能够自信地应对任何考试题目。本指南涵盖所有基本类型，并配以清晰的例题、考官式的易错提醒和实用建议。
1. Solving Linear Inequalities | 解线性不等式
Linear inequalities are solved in almost exactly the same way as linear equations. You may add or subtract any number on both sides, and you may multiply or divide by any positive number, without changing the direction of the inequality. The one golden rule is this: if you multiply or divide both sides by a negative number, you must reverse the inequality sign.
线性不等式的解法与线性方程几乎完全相同。两边可以加或减任意数，也可以乘以或除以任意正数，不等号方向都不变。唯一的一条黄金法则是：如果两边同时乘以或除以一个负数，必须反转不等号。
Example 1: Solve 3x − 7 < 5. Add 7 to both sides: 3x < 12. Divide by 3 (positive, so no sign change): x < 4.
示例 1：解 3x − 7 < 5。两边加 7 得 3x < 12。除以 3（正数，不等号不变）得 x < 4。
Example 2: Subtract 5 from both sides: −2x ≥ 6. Divide by −2 (negative, so reverse the sign): x ≤ −3.
示例 2：解 5 − 2x ≥ 11。两边减 5 得 −2x ≥ 6。除以 −2（负数，反转不等号）得 x ≤ −3。
- Step 1: expand any brackets and collect like terms.
第一步：展开所有括号并合并同类项。 - Step 2: move all x-terms to one side and all constants to the other.
第二步：把所有含 x 的项移到一边，常数移到另一边。 - Step 3: divide by the coefficient of x; reverse the sign if that coefficient is negative.
第三步：除以 x 的系数；若该系数为负，则反转不等号。 - Step 4: check your answer by substituting a value from your solution set.
第四步：从解集中取一个值代入原不等式进行检验。
2. Double Inequalities and Compound Inequalities | 双重不等式与复合不等式
e quantity must be greater than −3 and also less than or equal to 7. You can apply the same operation to all three parts simultaneously, which is usually faster and safer than splitting it up.
像 −3 < 2x + 1 ≤ 7 这样的双重不等式同时表示两个条件：该表达式既要大于 −3，又要小于或等于 7。你可以对三个部分同时进行相同的运算，这通常比拆成两个不等式更快也更安全。
−3 < 2x + 1 ≤ 7 → −4 < 2x ≤ 6 → −2 < x ≤ 3
can be written in set notation as {x : −2 < x ≤ 3} or in interval notation as (−2, 3].
三部分同时减 1，再同时除以 2。答案 −2 < x ≤ 3 可用集合记号写成 {x : −2 < x ≤ 3}，或用区间记号写成 (−2, 3]。
If a question gives two separate inequalities joined by "or", solve them independently and take the union of the two solution sets. If they are joined by "and", take the intersection – the overlap.
如果题目给出两个由 "or"（或）连接的不等式，就分别求解，再取两个解集的并集；如果由 "and"（且）连接，则取交集，即重叠部分。
3. Quadratic Inequalities | 二次不等式
Quadratic inequalities are best solved by factorising first and then using a quick sketch of the parabola. The critical values (the roots) divide the number line into regions, and the shape of the curve tells you where it lies above or below the x-axis.
二次不等式最好的解法是先因式分解，然后画出抛物线草图。临界值（即方程的根）把数轴分成若干区间，而曲线的开口方向告诉你它在哪一段位于 x 轴上方或下方。
Example: of x² is positive, the parabola opens upwards, so the curve is below the x-axis between the roots.
例题：解 x² − 5x + 6 < 0。因式分解得 (x − 2)(x − 3) < 0。临界值为 x = 2 和 x = 3。由于 x² 的系数为正，抛物线开口向上，因此曲线在两根之间位于 x 轴下方。
x² − 5x + 6 < 0 → 2 < x < 3
x² − 5x + 6 > 0 → x < 2 or x > 3
| Inequality / 不等式 | Shape / 开口 | Solution / 解集 |
| (x − a)(x − b) < 0, a < b | U-shaped / 开口向上 | a < x < b（两根之间） |
| (x − a)(x − b) > 0, a < b | U-shaped / 开口向上 | x < a or x > b（两根之外） |
| (x − a)(x − b) ≤ 0 | U-shaped / 开口向上 | a ≤ x ≤ b（含端点） |
If the coefficient of x² is negative, the parabola opens downwards and the regions swap over. Always sketch – a five-second sketch prevents the most common sign error in the whole topic.
如果 x² 的系数为负，抛物线开口向下，两个区间会互换。一定要画草图——一张五秒钟的草图就能避免本专题中最常见的符号错误。
4. Quadratic Inequalities that Do Not Factorise | 无法因式分解的二次不等式
When a quadratic will not factorise with integers, find the critical values with the quadratic formula instead. Keep the answers in surd form unless the question asks for a decimal approximation.
当二次式无法用整数因式分解时，改用求根公式求临界值。除非题目要求小数近似值，否则答案应保留根式形式。
Example: Solve x² − 4x + 1 < 0. Using the formula with a = 1, b = −4, c = 1:
例题：解 x² − 4x + 1 < 0。代入 a = 1，b = −4，c = 1 使用公式：
x = (4 ± √16 − 4) ÷ 2 = (4 ± 2√3) ÷ 2 = 2 ± √3
Since the parabola opens upwards, the solution is 2 − √3 < x < 2 + √3, which is approximately 0.27 < x < 3.73. Remember that √12 simplifies to 2√3.
由于抛物线开口向上，解集为 2 − √3 < x < 2 + √3，约为 0.27 < x < 3.73。记住 √12 可化简为 2√3。
5. The Discriminant and Inequalities | 判别式与不等式
The discriminant of ax² + bx + c is the quantity b² − 4ac. It tells you how many real roots a quadratic has, and it is the standard tool for questions that ask when a quadratic is always positive or always negative.
二次式 ax² + bx + c 的判别式是 b² − 4ac。它告诉你该二次方程有多少个实根，也是回答"二次式何时恒正或恒负"这类问题的标准工具。
Δ = b² − 4ac > 0 → two distinct real roots
Δ = b² − 4ac = 0 → one repeated root
Δ = b² − 4ac < 0 → no real roots
For x² + bx + c to be positive for all real values of x, you need the coefficient of x² to be positive and the discriminant to be negative. The curve then sits entirely above the x-axis.
要使 x² + bx + c 对所有实数 x 都为正，需要 x² 的系数为正，且判别式为负。此时曲线完全位于 x 轴上方。
Example: the discriminant is k² − 36. We need k² − 36 < 0, i.e. k² < 36, giving −6 < k < 6.
例题：求使 x² + kx + 9 > 0 对所有 x 成立的 k 的取值范围。此处 a = 1，b = k，c = 9，判别式为 k² − 36。需要 k² − 36 < 0，即 k² < 36，得 −6 < k < 6。
If the question asks for a repeated root or a tangent line, set the discriminant equal to zero and solve the resulting equation. If it asks for two distinct intersection points, use Δ > 0.
如果题目要求重根或相切，就令判别式等于零并解出方程；如果要求两个不同交点，则用 Δ > 0。
6. Rational and Fractional Inequalities | 分式与有理不等式
You must never multiply a rational inequality by the denominator, because the denominator could be negative or zero. The safe method is to multiply by the square of the denominator, which is always positive (except where the denominator is zero), or to move everything to one side and use a sign diagram.
绝不能用分母去乘有理不等式，因为分母可能是负数或零。安全的方法是乘以分母的平方（恒为正，分母为零处除外），或者把所有项移到一边再用符号表。
Example: Solve (x + 1) ÷ (x − 3) < 2, where x ≠ 3. Multiply both sides by (x − 3)²:
例题：解 (x + 1) ÷ (x − 3) < 2，其中 x ≠ 3。两边乘以 (x − 3)²：
(x + 1)(x − 3) < 2(x − 3)²
x² − 2x − 3 < 2x² − 12x + 18
0 < x² − 10x + 21 → (x − 3)(x − 7) > 0
and x = 3 and x = 7 are the critical values, so the final answer is x < 3 or x > 7.
所以 x < 3 或 x > 7。x = 3 被原式排除，而 3 与 7 是临界值，因此最终答案为 x < 3 或 x > 7。
x = −1 and x = 3. A quick sign diagram over the three regions gives x < −1 or x > 3.
对于更简单的情形，例如 (x + 1) ÷ (x − 3) > 0，只需找出分子和分母各自为零的点：x = −1 和 x = 3。对三个区间画快速符号表，即可得 x < −1 或 x > 3。
7. Modulus (Absolute Value) Inequalities | 绝对值不等式
Modulus inequalities have two standard forms that you should memorise, because they convert the problem into a linear or quadratic one immediately.
绝对值不等式有两种标准形式必须牢记，因为它们能立刻把问题转化为线性或二次问题。
|x| < a ⇔ −a < x < a (a > 0)
|x| > a ⇔ x < −a or x > a
Example 1: Solve |2x − 3| ≤ 5. This becomes −5 ≤ 2x − 3 ≤ 5, then −2 ≤ 2x ≤ 8, so −1 ≤ x ≤ 4.
示例 1：解 |2x − 3| ≤ 5。化为 −5 ≤ 2x − 3 ≤ 5，再得 −2 ≤ 2x ≤ 8，所以 −1 ≤ x ≤ 4。
Example 2: Solve |x − 1| > |x + 3|. Since both sides are non-negative, square both sides: (x − 1)² > (x + 3)², giving x² − 2x + 1 > x² + 6x + 9, so −8x > 8 and therefore x < −1.
示例 2：解 |x − 1| > |x + 3|。由于两边都非负，可同时平方：(x − 1)² > (x + 3)²，得 x² − 2x + 1 > x² + 6x + 9，即 −8x > 8，因此 x < −1。
Squaring is only valid when both sides are non-negative. If one side could be negative, you must solve the inequality graphically or split into cases instead.
只有当两边都非负时平方才成立。如果一边可能为负，就必须用图形法或分类讨论来求解。
8. Graphical Interpretation of Inequalities | 不等式的图形解读
Many exam questions ask you to solve f(x) > g(x) or to state where a curve lies above a line. Sketch both graphs on the same axes, find the x-coordinates of the intersection points, and then read off the regions.
许多考题要求解 f(x) > g(x)，或说明曲线位于直线之上的区间。在同一坐标系中画出两条图形，求出交点横坐标，然后读出相应区间。
- Sketch both graphs accurately enough to see the intersections.
把两条图形画到足以看清交点的精度。 - Solve f(x) = g(x) to find the exact boundary values.
解 f(x) = g(x) 求出精确的边界值。 - Decide which graph is on top in each region by testing one convenient value.
在每个区间取一个方便的值代入，判断哪条图形在上方。 - Write the answer as an inequality or a union of intervals.
把答案写成不等式或区间的并集形式。
For the transformation of graphs, remember that y = |f(x)| -axis upwards, which changes the solution set of inequalities involving modulus functions.
在处理图形变换时，要记住 y = |f(x)| 会把曲线位于 x 轴下方的部分向上翻折，这会改变含绝对值函数的不等式解集。
9. Set Notation, Number Lines and Function Domains | 集合记号、数轴与函数定义域
AQA expects you to present solutions in precise set or interval notation. The same solution can be written in several accepted ways, but mixing them up loses marks.
AQA 要求用精确的集合记号或区间记号给出解答。同一个解集有多种可接受的写法，但混用会导致失分。
| Inequality / 不等式 | Set notation / 集合记号 | Interval / 区间 |
| x ≤ −3 | {x : x ≤ −3} | (−∞, −3] |
| −1 ≤ x < 4 | {x : −1 ≤ x < 4} | [−1, 4) |
| x < 2 or x > 5 | {x : x < 2} ∪ {x : x > 5} | (−∞, 2) ∪ (5, ∞) |
Square brackets include the endpoint; round brackets exclude it. The symbol ∪ means union (or) and ∩ means intersection (and). Be careful with ∞ – it is never included, so it always takes a round bracket.
方括号表示包含端点，圆括号表示不包含。符号 ∪ 表示并集（或），∩ 表示交集（且）。注意 ∞ 永远不被包含，因此总是配圆括号。
Inequalities also appear when finding the domain and range of a function, for example when a square root requires the expression inside to be non-negative, or when a denominator must not be zero.
在求函数定义域和值域时也会用到不等式，例如根号内的表达式必须非负，或者分母不能为零。
10. Inequalities in Proof and Completing the Square | 证明中的不等式与配方法
Completing the square is the cleanest way to prove that a quadratic expression is always positive or always negative. The key fact is that any square is non-negative.
配方法是证明二次式恒正或恒负最简洁的方法。关键事实是：任何平方都非负。
(x + a)² ≥ 0 for all real x
Example: t; 0 for all real x. Complete the square: x² + 6x + 11 = (x + 3)² − 9 + 11 = (x + 3)² + 2.
例题：证明对一切实数 x 都有 x² + 6x + 11 > 0。配方得 x² + 6x + 11 = (x + 3)² − 9 + 11 = (x + 3)² + 2。
which proves the statement. The minimum value of the expression is 2, occurring at x = −3.
因为 (x + 3)² ≥ 0，所以 (x + 3)² + 2 ≥ 2 > 0，命题得证。该表达式的最小值为 2，在 x = −3 处取得。
r to prove a condition such as b² ≥ 4ac by arranging the expression into a perfect square.
你也可能被要求用配方法求函数的值域，或通过把表达式整理成完全平方来证明诸如 b² ≥ 4ac 的条件。
11. Common Exam Mistakes | 常见考试失分点
Most lost marks in inequality questions come from a small number of recurring errors. Check your work against this list before you move on.
不等式题目中失去的分数大多来自少数几个反复出现的错误。在继续下一题之前，请对照以下清单检查你的解答。
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