📚 Inequalities | 不等式
Inequalities form a cornerstone of A-Level Edexcel Mathematics, appearing across pure, statistics, and mechanics. A strong grasp of inequality notation, manipulation, and graphical interpretation not only secures direct marks but also underpins topics such as optimisation, curve sketching, and hypothesis testing. This revision guide distils the essential techniques – from linear to modulus inequalities – with linked English–Chinese explanations to help you master every aspect required by the specification.
不等式是A-Level Edexcel数学的基石，广泛出现在纯数、统计和力学中。熟练掌握不等式的符号、变形与图像解读不仅能直接得分，也为优化、曲线描图和假设检验等主题奠定基础。本复习指南从线性到绝对值不等式提炼关键技巧，通过中英对照的讲解帮助你掌握考试要求的每一个细节。
1. Linear Inequalities | 线性不等式
Linear inequalities involve expressions of the form ax + b > c or ax + b < c. They are solved by isolating x, much like linear equations, but with one crucial rule: when multiplying or dividing by a negative number, the inequality sign reverses. The solution is usually expressed as an interval, such as x > 2 or x ≤ −3, and can be represented on a number line with open or closed circles.
线性不等式涉及 ax + b > c 或 ax + b < c 等形式。求解时如同解线性方程一样将 x 分离，但有一个关键规则：当乘以或除以一个负数时，不等号方向必须反转。解通常表示为区间，例如 x > 2 或 x ≤ −3，并可在数轴上用空心或实心圆点表示。
To solve 3 − 2x ≤ 7, first subtract 3: −2x ≤ 4. Then divide by −2, remembering to flip the inequality: x ≥ −2. The solution set is [−2, ∞). Pay careful attention to the direction change – this is the most common error at A-Level.
求解 3 − 2x ≤ 7，先减去 3：−2x ≤ 4。然后除以 −2，记得反转不等号：x ≥ −2。解集为 [−2, ∞)。要特别注意方向的变化，这是A-Level中最常见的错误。
When a linear inequality contains fractions, multiply every term by a common positive denominator to clear fractions without changing the inequality sign. For instance, (x+1)/2 − (x−3)/3 > 1 can be multiplied by 6 to give 3(x+1) − 2(x−3) > 6, leading to x + 9 > 6, so x > −3.
当线性不等式中含有分式时，每一项乘以公分母（正数）去分母，不等号方向不变。例如 (x+1)/2 − (x−3)/3 > 1 乘以6 得 3(x+1) − 2(x−3) > 6，化简得 x + 9 > 6，故 x > −3。
2. Solving Quadratic Inequalities | 二次不等式求解
Quadratic inequalities, such as x² − 5x + 6 < 0, are best approached by sketching the corresponding parabola. First, find the critical values by solving the quadratic equation x² − 5x + 6 = 0, which gives x = 2 or x = 3. These roots divide the real line into three intervals. Since the coefficient of x² is positive, the graph is ∪-shaped, so the quadratic is negative between the roots. Hence the solution is 2 < x < 3.
二次不等式，如 x² − 5x + 6 < 0，最好通过画对应的抛物线来解决。首先求出临界值，解二次方程 x² − 5x + 6 = 0，得 x = 2 或 x = 3。这些根把实数轴分成三个区间。由于 x² 的系数为正，图像呈∪形，因此二次函数在两根之间为负。故解为 2 < x < 3。
If the quadratic had a negative leading coefficient, say −x² + 4x − 3 ≥ 0, multiply through by −1 (reversing the sign) to obtain x² − 4x + 3 ≤ 0. Factorising gives (x−1)(x−3) ≤ 0. With a positive x² coefficient, the parabola is ∪-shaped, so the expression is ≤ 0 between the roots: 1 ≤ x ≤ 3. Always rearrange to have a positive x² term before sketching, unless you are confident with ∩-shaped graphs.
若二次项系数为负，例如 −x² + 4x − 3 ≥ 0，每一项乘以 −1（不等号反转）得到 x² − 4x + 3 ≤ 0。因式分解得 (x−1)(x−3) ≤ 0。此时 x² 系数为正，抛物线呈∪形，故表达式 ≤ 0 的区域在两根之间：1 ≤ x ≤ 3。除非你对∩形图像很有把握，否则应先将二次项系数化为正再画图。
When the quadratic has no real roots, say x² + x + 1 > 0, the discriminant (b² − 4ac) is negative, meaning the graph never dips below the x-axis. Since the leading coefficient is positive, the quadratic is always positive for all real x. Therefore the solution is all real numbers, ℝ. Conversely, x² + x + 1 < 0 would have no solution.
当二次式没有实根时，例如 x² + x + 1 > 0，判别式 (b² − 4ac) 为负，意味着图像永远不会落在 x 轴下方。由于首项系数为正，该二次式对所有实数 x 恒正。因此解为全体实数 ℝ。相反，x² + x + 1 < 0 则无解。
3. Polynomial Inequalities | 多项式不等式
For cubic or higher-degree polynomial inequalities, such as (x+1)(x−2)(x−4) ≤ 0, draw a sign diagram. First, identify the roots in increasing order: x = −1, 2, 4. These split the real line into intervals. Test a value from each interval to determine the sign of the product. For x < −1, say x = −2, all three factors are negative, so the product is negative. For −1 < x < 2, the factor (x+1) becomes positive, the others negative: product positive. Continue this pattern and shade the regions where the product is ≤ 0. The solution is (−∞, −1] ∪ [2, 4].
对于三次或更高次的多项式不等式，如 (x+1)(x−2)(x−4) ≤ 0，需绘制符号表。首先按从小到大的顺序找出所有根：x = −1, 2, 4。这些根将实数轴分成若干区间。从每个区间取一个测试值，判断乘积的符号。当 x < −1 时，取 x = −2，三个因式皆为负，乘积为负。在 −1 < x < 2 内，(x+1) 为正，其余为负：乘积为正。依此类推，将乘积 ≤ 0 的区间标记出来。解为 (−∞, −1] ∪ [2, 4]。
At a repeated root, the sign may not change. For (x−1)²(x+3) > 0, the root x = 1 is of even multiplicity. The sign diagram shows the expression is positive for x > −3 except at x = 1, where it equals zero. Since the inequality is strict, x = 1 is excluded. The solution is x ∈ (−3, 1) ∪ (1, ∞). Always pay close attention to even multiplicity roots when drawing sign diagrams.
在重根处符号可能不变。对于 (x−1)²(x+3) > 0，x = 1 为偶重根。符号表显示表达式在 x > −3 时为正，但在 x = 1 处为零。由于是不含等号的严格不等式，x = 1 须排除。解为 x ∈ (−3, 1) ∪ (1, ∞)。绘制符号表时一定要特别关注偶重根。
4. Rational Inequalities | 有理不等式
Rational inequalities involve fractions with variables in the denominator, such as (x−2)/(x+1) > 0. The procedure: bring all terms to one side, find critical values by setting both numerator and denominator equal to zero (x = 2 and x = −1), and construct a sign table. The expression changes sign only at these critical points. The fraction is positive when numerator and denominator share the same sign. Testing intervals gives x < −1 or x > 2. Note that x = −1 must be excluded because the denominator cannot be zero.
有理不等式涉及分母含变量的分式，如 (x−2)/(x+1) > 0。解题步骤：将所有项移到一边，令分子与分母分别等于零，求出临界值（x = 2 与 x = −1），然后建立符号表。该分式仅在这些临界点处可能变号。当分子分母同号时分式为正。测试各区间可得 x < −1 或 x > 2。注意 x = −1 必须排除，因为分母不能为零。
If the inequality is (x+3)/(x−1) ≤ 2, first rearrange: (x+3)/(x−1) − 2 ≤ 0 → (x+3 − 2x + 2)/(x−1) ≤ 0 → (−x+5)/(x−1) ≤ 0. Multiply by −1, reversing the inequality: (x−5)/(x−1) ≥ 0. Critical values: x = 1 (excluded) and x = 5 (include if equality holds for ≤ originally; check: at x=5 the original becomes 8/4=2, which satisfies ≤2, so x=5 is included). Sign analysis yields x < 1 or x ≥ 5.
如果不等式为 (x+3)/(x−1) ≤ 2，先移项： (x+3)/(x−1) − 2 ≤ 0 → (x+3 − 2x + 2)/(x−1) ≤ 0 → (−x+5)/(x−1) ≤ 0。乘以 −1，不等号反转：(x−5)/(x−1) ≥ 0。临界值：x = 1（排除）和 x = 5（需检验原不等式等号是否成立；x=5 时原式为 8/4=2，满足 ≤2，故包含 x=5）。符号分析得 x < 1 或 x ≥ 5。
5. Inequalities with Modulus | 绝对值不等式
Modulus inequalities, such as |x − 3| < 5, are interpreted geometrically as the distance from x to 3 on the number line being less than 5. Algebraically, this is equivalent to −5 < x − 3 < 5, leading to −2 < x < 8. For a greater-than inequality like |2x + 1| ≥ 7, it splits into two separate conditions: 2x + 1 ≤ −7 or 2x + 1 ≥ 7, giving x ≤ −4 or x ≥ 3.
绝对值不等式，如 |x − 3| < 5，几何意义是数轴上 x 到 3 的距离小于 5。代数学上等价于 −5 < x − 3 < 5，即 −2 < x < 8。对于大于等于型不等式 |2x + 1| ≥ 7，则拆分为两个独立条件：2x + 1 ≤ −7 或 2x + 1 ≥ 7，得 x ≤ −4 或 x ≥ 3。
More complex cases, like |x − 2| < |x + 4|, are best squared, since both sides are non-negative. Squaring yields (x−2)² < (x+4)², expanding to x² − 4x + 4 < x² + 8x + 16, simplifying to −12x < 12, and dividing by −12 gives x > −1. Always remember that squaring is valid only if both sides are guaranteed non-negative.
更复杂的情形，如 |x − 2| < |x + 4|，最好两边平方，因为两边都是非负的。平方得 (x−2)² < (x+4)²，展开为 x² − 4x + 4 < x² + 8x + 16，化简得 −12x < 12，除以 −12 得 x > −1。一定要记住，只有在两边都确保非负的情况下，平方才是有效的操作。
When solving modulus inequalities like |x + 1| > 2|x − 3|, it is often clearer to consider critical values (where expressions inside modulus equal zero: x = −1 and x = 3) and analyse distinct intervals. Alternatively, squaring both sides gives a quadratic inequality that can be simplified and solved carefully.
求解类似 |x + 1| > 2|x − 3| 的绝对值不等式时，通常更清晰的方法是考虑临界值（使模内表达式为零的点：x = −1 与 x = 3）并分区间讨论。也可以两边平方，得到一个二次不等式再仔细求解。
6. Graphical Representation | 图形表示
Edexcel often asks candidates to represent inequalities graphically on the Cartesian plane. For a linear inequality like y > 2x + 1, first draw the boundary line y = 2x + 1 as a dashed line (since strict inequality). Then shade the region above the line, because y-values are greater than 2x+1. For y ≤ −x + 3, draw a solid line and shade below.
Edexcel 考试常要求考生在平面直角坐标系中表示不等式。对于线性不等式如 y > 2x + 1，首先绘制边界线 y = 2x + 1，用虚线（因为严格不等式）。然后给直线上方区域画上阴影，因为 y 值大于 2x+1。对于 y ≤ −x + 3，画实线并给下方画阴影。
When dealing with quadratic inequalities like y < x² − 4, sketch the parabola y = x² − 4 with a dashed curve, and shade the region below it. If the inequality were y ≥ x² − 4, the parabola would be solid, and shading would be above. A system may combine several boundaries; the feasible region is the intersection of all individual shaded regions.
处理二次不等式如 y < x² − 4 时，画出抛物线 y = x² − 4 的草图，用虚线曲线，然后将曲线下方区域涂上阴影。如果不等式是 y ≥ x² − 4，则抛物线为实线，阴影在上方。多个不等式组合时，可行区域是所有单独阴影区域的交集。
Always label axes and indicate whether boundary lines are solid or dashed. In exam questions, unshaded regions are sometimes left as the required region – read the question carefully. For shading, use pencil in the exam, but in digital formats we describe it as “region R”.
务必给坐标轴标注，并标明边界线是实线还是虚线。考试中有时会要求保留未填充区域作为所求区域——仔细读题。考试中请用铅笔涂阴影，在电子版中我们描述为“区域 R”。
7. System of Inequalities | 不等式组
A system of inequalities asks for the set of points satisfying all given conditions simultaneously. For example, the system x ≥ 0, y ≥ 0, x + y ≤ 4, and y ≥ x − 2 defines a polygonal region. Solve by sketching each boundary and finding the overlapping shaded area. The vertices of this region can be used in linear programming problems, which appear in the Decision Mathematics option but also in some pure contexts.
不等式组要求同时满足所有给定条件的点的集合。例如，不等式组 x ≥ 0, y ≥ 0, x + y ≤ 4, y ≥ x − 2 定义了一个多边形区域。通过画出每条边界并找出重叠的阴影区域来求解。该区域的顶点可用于线性规划问题，这类问题出现在决策数学选科中，也在纯数的某些情境下出现。
To locate the feasible region accurately, find intersection points of the boundary lines by solving pairs of equations. For instance, the intersection of x + y = 4 and y = x − 2 is found by substituting: x + (x − 2) = 4 → 2x = 6 → x = 3, y = 1. Mark these vertices clearly; they are often needed for later calculations.
为了准确定位可行区域，需要联立方程求解边界线的交点。例如，x + y = 4 与 y = x − 2 的交点，代入得 x + (x − 2) = 4 → 2x = 6 → x = 3, y = 1。清楚标出这些顶点；后续计算通常会用到它们。
8. Inequalities and Discriminant | 判别式与不等式
The discriminant Δ = b² − 4ac is closely linked to quadratic inequalities. A quadratic ax² + bx + c > 0 for all real x if and only if a > 0 and Δ < 0. This means the graph is entirely above the x-axis. Similarly, ax² + bx + c < 0 for all x if a < 0 and Δ < 0. These conditions are vital in curve sketching and when proving that a function is always positive or always negative.
判别式 Δ = b² − 4ac 与二次不等式密切相关。二次式 ax² + bx + c > 0 对所有实数 x 成立，当且仅当 a > 0 且 Δ < 0。这意味着图像完全在 x 轴上方。类似地，ax² + bx + c < 0 对所有 x 成立，若 a < 0 且 Δ < 0。这些条件在曲线描图和证明函数恒正或恒负时至关重要。
When a question asks “Find the range of k for which x² + kx + 4 > 0 for all real x”, set a = 1 (>0, satisfied) and Δ = k² − 16 < 0, giving −4 < k < 4. If the quadratic is not monic, ensure a matches the required sign. For example, for kx² + 3x + 1 > 0 ∀ x, we need k > 0 and Δ = 9 − 4k < 0 → k > 9/4.
当题目要求“求 k 的取值范围，使得 x² + kx + 4 > 0 对所有实数 x 成立”，设定 a = 1（>0，已满足），且 Δ = k² − 16 < 0，得 −4 < k < 4。若二次式非首一，需确保 a 符号正确。例如，对于 kx² + 3x + 1 > 0 对所有 x 成立，需 k > 0 且 Δ = 9 − 4k < 0 → k > 9/4。
9. Compound Inequalities | 复合不等式
Compound inequalities combine two or more conditions using ‘and’ (intersection) or ‘or’ (union). A typical form is −3 ≤ 2x − 1 < 5, which means −3 ≤ 2x − 1 AND 2x − 1 < 5. Solve each part separately: add 1 throughout to get −2 ≤ 2x < 6, then divide by 2 to obtain −1 ≤ x < 3. This compact notation is efficient but requires careful handling when negative multiplying is involved.
复合不等式以“且”（交集）或“或”（并集）组合两个或更多条件。常见形式如 −3 ≤ 2x − 1 < 5，表示 −3 ≤ 2x − 1 且 2x − 1 < 5。分别求解两部分：整体加 1 得 −2 ≤ 2x < 6，再除以 2 得到 −1 ≤ x < 3。这种紧凑写法高效，但在涉及乘除负数时必须小心处理。
For ‘or’ compound inequalities such as x < 2 or x ≥ 5, the solution set is the union (−∞, 2) ∪ [5, ∞). No further simplification is possible. These often arise from solved modulus inequalities or from polynomial inequalities where two disjoint intervals satisfy the condition.
对于“或”型的复合不等式，如 x < 2 或 x ≥ 5，解集是并集 (−∞, 2) ∪ [5, ∞)。不能再进一步化简。这类结果常源于绝对值不等式的分解，或多项式不等式中两个不交叠的区间同时满足条件。
Double-sided inequalities like 1 < (3x−2)/2 ≤ 4 are solved by releasing the fraction: multiply all parts by 2 → 2 < 3x − 2 ≤ 8 → add 2 → 4 < 3x ≤ 10 → divide by 3 → 4/3 < x ≤ 10/3. Always check the operations are valid (multiplying by a positive constant).
双侧不等式如 1 < (3x−2)/2 ≤ 4 可先去掉分母：三部分同乘 2 → 2 < 3x − 2 ≤ 8 → 加 2 → 4 < 3x ≤ 10 → 除以 3 → 4/3 < x ≤ 10/3。始终检查运算是否有效（乘以正常数）。
10. Common Mistakes and Tips | 常见错误与技巧
Many marks are lost due to sign reversal errors when multiplying or dividing by negatives. A rigorous habit is to write “×(−1), flip sign” explicitly in your working. When clearing fractions in inequalities, always confirm the multiplier is positive. If the sign of the multiplier is unknown, split into cases instead.
很多失分是因为在乘除负数时忘记反转不等号。养成严谨习惯，在解题过程明确写出“×(−1)，反转符号”。在不等式去分母时，始终确认乘数是正数。若乘数符号未知，应分情况讨论。
Students often mishandle critical values at boundaries, particularly when equalities are involved. Always test whether the endpoints satisfy the original inequality, especially for rational and modulus inequalities. For rational functions, remember that denominator zeros are never included, even if the inequality is non-strict.
学生经常处理不好边界处的临界值，尤其是涉及等号时。始终检验端点是否满足原不等式，对有理不等式和绝对值不等式尤为关键。对于有理函数，记住分母零点永远不能包含在内，即使不等号含等号。
When sketching quadratic inequality solutions, an inaccurate U-shape or upside-down U can lead to reversed answers. Label the roots on the x-axis, note the sign of x², and shade the correct region. Practice sketching quickly but accurately; a table of sign intervals provides a reliable check.
画二次不等式解的草图时，U 形或倒 U 形画错会导致答案完全相反。在 x 轴上标出两根，标明 x² 系数正负，再正确填充区域。练习快速而准确地画图；符号区间表格可提供可靠检查。
In modulus inequalities, avoid blindly squaring without checking whether the expression inside the modulus could be negative. Although squaring is safe when both sides are moduli, verify that the resulting inequality is simpler to solve. Sometimes graphical interpretation or interval splitting is clearer and less error-prone.
解绝对值不等式时，避免不检查模内表达式可能为负的情况下盲目平方。虽然两端均为模时平方是安全的，但要确认平方后不等式更易求解。有时候，几何解释或分区间讨论更清晰，错误更少。
Finally, when expressing final solutions, use correct set notation or interval notation as preferred by Edexcel: e.g., {x: x < 3} or (−∞, 3). For multiple intervals, use the union symbol ∪ and present in increasing order. Precise notation is part of the assessment – don’t lose easy marks.
最后，书写最终答案时使用 Edexcel 偏好的正确集合符号或区间符号，例如 {x: x < 3} 或 (−∞, 3)。对于多个区间，使用并集符号 ∪ 并按从小到大的顺序写出。准确的符号表示是评分的一部分——不要丢掉应得的分数。
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