📚 A-Level Maths Inequalities Mastery | A-Level数学不等式高分攻略
Inequalities are one of the highest-yield topics in A-Level Mathematics. They rarely appear as an entire long question, yet they hide inside almost every other topic: finding the range of a function, deciding where a curve lies above a line, stating the domain of a square root, or interpreting a discriminant. Master them and you protect marks across the whole paper; ignore them and the same three errors keep pulling your grade down.
不等式是 A-Level 数学中性价比最高的考点之一。它很少单独构成一整道大题，却几乎藏在所有其他考点里：求函数值域、判断曲线何时位于直线上方、写出根号函数的定义域、解释判别式的意义。掌握它，你保护的是整张试卷的分数；掌握不了，同样的几个错误就会反复拖低成绩。
This guide rebuilds the topic from the ground up, using the language of real mark schemes: linear inequalities, quadratic inequalities, inequality systems, integer solutions, number line representation, modulus and rational inequalities, and the marking habits that separate a B from an A.
本文从零开始重建这个考点，并严格使用真实评分标准的语言：线性不等式、二次不等式、不等式组、整数解、数轴表示法、绝对值与分式不等式，以及决定 B 与 A 之分的关键评分习惯。
1. Why Inequalities Cost Marks | 不等式为何是失分重灾区
Most lost marks in inequality questions do not come from hard algebra. They come from three repeat offenders: reversing the inequality sign in the wrong direction, confusing intersection (AND) with union (OR), and skipping the number line or sign diagram that the mark scheme explicitly rewards.
不等式题上的失分，大多不是来自困难的代数运算，而是来自三个反复出现的错误：不等号方向反转错误、把交集（且）与并集（或）搞混、以及省略了评分标准明确给分的数轴或符号图步骤。
A second reason is presentation. An examiner marking a five-mark inequality question is scanning for specific evidence: critical values, correct inequality symbols, correct endpoints, and a stated conclusion. If your working is a single line of algebra with no diagram and no final answer in words, you can be mathematically right and still drop two or three marks.
第二个原因是书写呈现。考官批改一道 5 分的不等式题时，是在扫描特定证据：临界值、正确的不等号、正确的端点、以及明确的结论陈述。如果你的过程只有一行代数、没有图、也没有用文字写出最终答案，那么即使数学上完全正确，也可能丢掉两三分。
| Error / 错误 | Typical cost / 典型代价 | Fix / 解决办法 |
| Sign not flipped when dividing by a negative / 除以负数时未反转符号 | A1 lost, answer wrong / 丢 A1，答案错误 | Write the division step explicitly / 把除法那一步写出来 |
| AND/OR confused / 交集与并集混淆 | Final answer wrong / 最终答案错误 | Sketch every region before writing / 先画区间草图再写答案 |
| No number line or diagram / 无数轴或示意图 | Method mark lost / 丢方法分 | Always draw it — it is evidence / 必画，它是得分的证据 |
2. Solving Linear Inequalities | 线性不等式求解
with a ≠ 0. You solve it exactly like a linear equation: add or subtract the same quantity on both sides, multiply or divide both sides by the same non-zero number. Adding and subtracting never change the direction of the inequality sign.
线性不等式的形式为 ax + b > c、ax + b ≥ c、ax + b < c 或 ax + b ≤ c，其中 a ≠ 0。解法与一元一次方程完全一样：两边同时加减同一个量，两边同时乘除同一个非零数。加减运算永远不改变不等号方向。
lips the inequality sign. So 3 − 2x ≥ 11 becomes −2x ≥ 8 and therefore x ≤ −4, not x ≥ −4.
决定整道题对错的那条规则是：两边同时乘以或除以一个负数时，不等号方向反转。所以 3 − 2x ≥ 11 化为 −2x ≥ 8，因此 x ≤ −4，而不是 x ≥ −4。
For brackets, expand first, then collect like terms: 5(x − 2) < 3x + 4 gives 5x − 10 < 3x + 4, so 2x < 14 and x < 7. If the unknown appears in a denominator, do not multiply across blindly — the sign of the denominator is unknown, so move everything to one side and combine into a single fraction instead.
遇到括号先展开再合并同类项：5(x − 2) < 3x + 4 化为 5x − 10 < 3x + 4，于是 2x < 14，得 x < 7。如果未知数出现在分母上，不要直接两边同乘——分母的正负是未知的，正确做法是把所有项移到一边并通分成单个分式。
3. Quadratic Inequalities and Critical Values | 二次不等式与临界值
The standard method for a quadratic inequality is: rearrange so that one side is zero, factorise (or use the quadratic formula, or complete the square), find the two critical values, then test the sign of the expression in each interval.
二次不等式的标准解法是：先整理成一边为零，再因式分解（或用求根公式、配方）求出两个临界值，然后判断表达式在每个区间上的正负。
ards and crosses the x-axis at these two points, so the curve is below the axis strictly between them:
例题：解 4x² − 25 < 0。分解为 (2x − 5)(2x + 5) < 0，临界值为 x = −2.5 与 x = 2.5。抛物线 y = 4x² − 25 开口向上，在这两点穿过 x 轴，因此曲线严格位于两交点之间的轴下方：
−2.5 < x < 2.5
Three structures cover almost every case. If a > 0 and there are two distinct roots r₁ < r₂: the expression is negative for r₁ < x < r₂ and positive for x < r₁ or x > r₂. If a < 0, the shape is inverted, so the intervals swap. If the discriminant Δ < 0, there are no real roots and the sign of the expression never changes — for example x² + x + 1 > 0 is satisfied by every real x.
三种结构几乎覆盖全部情形。若 a > 0 且有两个不同实根 r₁ < r₂：表达式在 r₁ < x < r₂ 上为负，在 x < r₁ 或 x > r₂ 上为正。若 a < 0，开口向下，区间互换。若判别式 Δ < 0，则无实根，表达式符号恒定不变——例如 x² + x + 1 > 0 对一切实数 x 成立。
0 the answer is x ≤ −3 or x ≥ 0.5, and it must be written as a union, not as −3 ≤ x ≤ 0.5.
当解集是两段分离区间时，绝不能写成一个双边不等式。对于 2x² + 5x − 3 ≥ 0，答案是 x ≤ −3 或 x ≥ 0.5，必须写成并集形式，不能写成 −3 ≤ x ≤ 0.5。
4. Sign Diagrams and Parabola Sketches | 符号图与抛物线草图
A sign diagram is a number line with the critical values marked, the sign of each factor written above each interval, and the resulting sign of the product written underneath. Examiners look for it, and it protects you from choosing the wrong interval.
符号图就是一条标出临界值的数轴：在每个区间上方写出各因子的正负，在下方写出乘积的正负。考官会专门找这一步，而它能防止你选错区间。
of x² is positive, the parabola opens upwards, so the product is positive outside the roots and negative between them.
以 2x² + 5x − 3 ≥ 0 为例。分解得 (2x − 1)(x + 3) ≥ 0，临界值为 x = −3 与 x = 0.5。由于 x² 的系数为正，抛物线开口向上，因此乘积在两根之外为正、在两根之间为负。
- Interval x < −3: both factors negative, product positive — satisfy the inequality.
区间 x < −3：两个因子都为负，乘积为正——满足不等式。 - Interval −3 < x < 0.5: one factor negative, one positive, product negative — reject.
区间 −3 < x < 0.5：一负一正，乘积为负——舍去。 - Interval x > 0.5: both factors positive, product positive — satisfy the inequality.
区间 x > 0.5：两个因子都为正，乘积为正——满足不等式。 - Endpoints: ≥ means both x = −3 and x = 0.5 are included.
端点：≥ 表示 x = −3 与 x = 0.5 都包含在内。
A quick sketch of the parabola achieves the same thing in five seconds. Draw the U-shape, mark where it crosses the axis, shade the parts of the curve that lie above or below the axis depending on the inequality symbol, then read the x-values off the axis.
快速画出抛物线草图，五秒钟就能达到同样效果。画出 U 形，标出与 x 轴的交点，根据不等号方向把轴上方或下方的那部分曲线涂出来，再从 x 轴上读出对应的取值区间。
5. Inequality Systems and Intersections | 不等式组与交集
When a question gives two or more inequalities, the solution set is the intersection: the values that satisfy all of them at the same time. Solve each inequality separately, then combine the results on one number line.
当题目给出两个或更多不等式时，解集是交集：即同时满足全部条件的那些取值。先分别解出每个不等式，再把结果合并到同一条数轴上。
Example: solve 4x² − 25 < 0 together with 12 − 5x − 3x² > 0. The first gives −2.5 < x < 2.5. For the second, multiply by −1 to make the x² coefficient positive, remembering to flip the sign: 3x² + 5x − 12 < 0, which factorises as (3x − 4)(x + 3) < 0, giving −3 < x < 1.333… (that is, x < 4/3).
例题：解 4x² − 25 < 0 与 12 − 5x − 3x² > 0 联立。第一个给出 −2.5 < x < 2.5。第二个先两边乘 −1 使 x² 系数为正，注意反转不等号：3x² + 5x − 12 < 0，分解为 (3x − 4)(x + 3) < 0，得 −3 < x < 1.333…（即 x < 4/3）。
Now take the overlap. The lower bound is the larger of −2.5 and −3, which is −2.5. The upper bound is the smaller of 2.5 and 4/3, which is 4/3. Only the common region survives:
取公共部分。下界取 −2.5 与 −3 中较大的一个，即 −2.5；上界取 2.5 与 4/3 中较小的一个，即 4/3。最终只保留公共解集：
−2.5 < x < 4/3
Two habits prevent disaster here. First, draw all the regions on one number line, stacked vertically, so the overlap is visible. Second, when the two regions do not overlap at all, the correct answer is "no solution" — write that in words, because a blank answer scores nothing.
两个习惯可以避免灾难性失误。第一，把全部区间画在同一条数轴上、上下分层排列，让重叠部分一眼可见。第二，若两区间完全没有重叠，正确答案就是"无解"，请用文字写出，因为空白答案一分不得。
6. Integer Solution Problems | 整数解问题
Some questions restrict the variable to integers (often written n ∈ ℤ) and ask you to list every possible value. The algebra is the easy part; the marks are won and lost on endpoints.
有些题目把变量限制为整数（常写作 n ∈ ℤ），要求你列出所有可能取值。代数部分是简单的，分数全在端点上决出。
ecause of ≤, the right endpoint is exclusive because of <. So the values are 2, 3 and 4 — not 5, and not 1.
已知 2 ≤ n < 5 且 n 为整数：左端点因 ≤ 而包含，右端点因 < 而不包含。因此可能值为 2、3、4——既不是 5，也不是 1。
- Inclusive endpoint (≤ or ≥): the number itself is a solution.
包含端点（≤ 或 ≥）：端点数值本身是解。 - Exclusive endpoint (< or >): the number itself is not a solution.
不包含端点（< 或 >）：端点数值本身不是解。 - Questions asking for the greatest or least integer value need the answer stated as a single number, not an interval.
问"最大整数解"或"最小整数解"的题目，答案必须是单个数字，而不是区间。 - If the inequality involves a fraction such as x < 4/3, the largest integer is 1.
若不等式涉及分数如 x < 4/3，最大整数为 1。
A fast check: count how many integers you have listed and test the two end values directly in the original inequality. If either fails, you have misread the endpoint.
快速自检：数一数你列出了几个整数，再把两个端点值直接代回原不等式检验。只要有一个不成立，说明你读错了端点的开闭。
7. Number Line Representation | 数轴表示法
Exam questions often require you to represent the solution set on a number line. The convention is simple and strictly marked: an open circle (○) means the endpoint is not included, a closed circle (●) means the endpoint is included. Arrows or thickened segments show the direction of the region.
考试常要求你在数轴上表示解集。约定很简单，但评分很严格：空心圈（○）表示端点不包含，实心圈（●）表示端点包含。箭头或加粗线段表示区间方向。
| Inequality / 不等式 | Circle at endpoint / 端点圆圈 | Interval notation / 区间记号
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