📚 A-Level Maths: Inequalities – Key Points & Exam Focus | A-Level 数学：不等式 考点精讲
Inequalities form a core part of A-Level algebra, bridging the gap between equation-solving and real-world constraints. They appear across pure mathematics, from quadratic discriminants to calculus applications, and demand careful attention to sign changes, critical values, and valid domains. This revision guide walks through every major type of inequality you will encounter, with worked examples and exam-focused strategies.
不等式是 A-Level 代数部分的核心内容，连接着方程求解与现实约束。它们出现在纯数的各个领域，从二次判别式到微积分应用，需要高度关注符号变化、临界值和有效定义域。本复习指南将带你梳理和攻克考试中每一类重要的不等式，配以典型例题和以考试为导向的解题策略。
1. Understanding Inequalities: Basics & Notation | 不等式基础与符号
Inequalities use the symbols > (greater than), < (less than), ≥ (greater than or equal to) and ≤ (less than or equal to). The most frequent pitfall is reversing the inequality sign when multiplying or dividing by a negative number. For example, -2x > 6 simplifies to x < -3, not x > -3. Always check the sign of the coefficient you are working with.
不等式使用符号 > (大于), < (小于), ≥ (大于或等于) 和 ≤ (小于或等于) 。最常犯的错误是在乘或除以负数时忘记反转不等号。例如 -2x > 6 化简为 x < -3，而不是 x > -3。每次运算时务必检查系数的符号。
Solution sets are expressed using interval notation: parentheses for open ends, brackets for closed ends. For instance, x ≥ 2 is written as [2, ∞), while -1 < x ≤ 4 is (-1, 4]. On a number line, open circles exclude the endpoint and closed circles include it.
解集通常用区间表示：开区间用圆括号，闭区间用方括号。例如 x ≥ 2 表示为 [2, ∞)，而 -1 < x ≤ 4 表示为 (-1, 4]。在数轴上，空心圆圈表示不包含端点，实心圆圈表示包含端点。
2. Linear Inequalities | 线性不等式
Linear inequalities are solved much like linear equations, with the added rule about sign reversal. Consider 5 – 3x ≤ 11. Subtract 5: -3x ≤ 6. Divide by -3: x ≥ -2 (inequality flips). The solution interval is [-2, ∞).
线性不等式的求解与线性方程类似，但需牢记变号规则。例如 5 – 3x ≤ 11，两边减 5 得 -3x ≤ 6，除以 -3 得 x ≥ -2（不等号翻转）。解集区间为 [-2, ∞)。
Double inequalities such as -3 < 2x + 1 < 7 can be treated as two separate inequalities or solved simultaneously. Subtract 1 from all parts: -4 < 2x < 6. Divide by 2: -2 < x < 3. The solution is (-2, 3).
对于 -3 < 2x + 1 < 7 这样的联立不等式，可以拆为两个不等式或同时操作。所有部分减 1：-4 < 2x < 6。再除以 2：-2 < x < 3。解为开区间 (-2, 3)。
3. Quadratic Inequalities | 二次不等式
For a quadratic inequality, first rearrange to make one side zero and factorise the quadratic if possible. Solve x² – 5x + 6 < 0. Factorise: (x - 2)(x - 3) < 0. The critical values are x = 2 and x = 3. Since the coefficient of x² is positive, the parabola opens upwards. The quadratic is negative between the roots, giving the solution 2 < x < 3, or (2, 3).
处理二次不等式，先移项使一边为零，并尽可能进行因式分解。解 x² – 5x + 6 < 0。因式分解为 (x - 2)(x - 3) < 0。临界值为 x = 2 和 x = 3。因为 x² 的系数为正，抛物线开口向上。二次式在两根之间取负值，因此解为 2 < x < 3，即开区间 (2, 3)。
If the inequality were x² – x – 12 ≥ 0, we factorise as (x – 4)(x + 3) ≥ 0. The parabola opens upward, so the quadratic is non-negative outside the roots: x ≤ -3 or x ≥ 4, written as (-∞, -3] ∪ [4, ∞). Always sketch a quick graph or sign chart to avoid guessing.
如果不等式是 x² – x – 12 ≥ 0，因式分解得 (x – 4)(x + 3) ≥ 0。抛物线开口向上，因此二次式在两根之外非负：x ≤ -3 或 x ≥ 4，即 (-∞, -3] ∪ [4, ∞)。务必快速画出草图或符号表，避免凭感觉写答案。
4. Discriminant and the Sign of a Quadratic | 判别式与二次函数的符号
Sometimes you need to determine whether a quadratic expression is always positive or always negative. For ax² + bx + c to be always positive, we require a > 0 and the discriminant Δ = b² – 4ac < 0. For it to be always negative, we require a < 0 and Δ < 0.
有时需要判断一个二次式是否恒正或恒负。要使 ax² + bx + c 恒大于零，要求首项系数 a > 0 且判别式 Δ = b² – 4ac < 0。要使二次式恒小于零，则要求 a < 0 且 Δ < 0。
Example: Find the set of values for which x² + 2x + 5 > 0. Here a = 1 > 0, Δ = 4 – 20 = -16 < 0, so the inequality holds for all real x, i.e., (-∞, ∞). Conversely, -x² + 3x - 7 < 0 for all real x if a < 0 and Δ < 0.
例如：求 x² + 2x + 5 > 0 的解集。这里 a = 1 > 0，Δ = 4 – 20 = -16 < 0，因此不等式对所有实数 x 都成立，解集为 (-∞, ∞)。类似地，若 a < 0 且 Δ < 0，则 -x² + 3x - 7 < 0 对所有实数成立。
5. Absolute Value Inequalities | 绝对值不等式
Recall that |x| represents the distance from 0 on the number line. The inequality |x| < a (with a > 0) is equivalent to -a < x < a. The inequality |x| > a means x < -a or x > a. These patterns extend to linear expressions: |2x – 1| ≤ 5 becomes -5 ≤ 2x – 1 ≤ 5, which solves to -2 ≤ x ≤ 3.
绝对值 |x| 表示数轴上到原点的距离。不等式 |x| < a（a > 0）等价于 -a < x < a。不等式 |x| > a 意味着 x < -a 或 x > a。这些模式可以推广到线性表达式：|2x – 1| ≤ 5 转化为 -5 ≤ 2x – 1 ≤ 5，解得 -2 ≤ x ≤ 3。
For more complex inequalities like |x + 2| > 3, we split into two cases: x + 2 < -3 or x + 2 > 3, giving x < -5 or x > 1. When both sides contain absolute values, e.g. |x + 1| < |x - 3|, it is often safe to square both sides since both are non-negative, leading to (x+1)² < (x-3)², expand and solve: x² + 2x + 1 < x² - 6x + 9 ⇒ 8x < 8 ⇒ x < 1.
对于较为复杂的不等式如 |x + 2| > 3，可拆成两种情况：x + 2 < -3 或 x + 2 > 3，得到 x < -5 或 x > 1。当两边都有绝对值时，例如 |x + 1| < |x - 3|，因为两边非负，可以安全地两边平方得到 (x+1)² < (x-3)²，展开求解得 8x < 8，即 x < 1。
6. Rational Inequalities | 分式不等式
Never multiply through by the denominator to clear fractions, because the denominator might change sign. Instead, bring all terms to one side, combine into a single fraction, and then find the critical values where the numerator or denominator is zero. Use a sign table to check each interval.
切勿直接乘以分母去分母，因为分母的符号可能不确定。正确的做法是将所有项移到一边，合并为一个分式，然后找出使分子或分母为零的临界值。用符号表检查每个区间。
Example: Solve (x – 1)/(x + 2) ≥ 0. Critical values: numerator zero at x = 1, denominator zero at x = -2. Test intervals: x < -2, fraction is positive/positive? Choose a test value, say x = -3: (-4)/(-1) = 4 > 0. For -2 < x < 1, test x = 0: (-1)/(2) < 0. For x > 1, test x = 2: (1)/(4) > 0. The solution is x < -2 or x ≥ 1. Note x = -2 is excluded since the denominator cannot be zero. Write as (-∞, -2) ∪ [1, ∞).
例如：解 (x – 1)/(x + 2) ≥ 0。临界值：分子零点 x = 1，分母零点 x = -2。检验区间：x < -2 时选 x = -3，得 (-4)/(-1) = 4 > 0；-2 < x < 1 时选 x = 0，得 (-1)/2 < 0；x > 1 时选 x = 2，得 1/4 > 0。解集为 x < -2 或 x ≥ 1。注意 x = -2 被排除，因为分母不能为零。写成 (-∞, -2) ∪ [1, ∞)。
7. Polynomial Inequalities of Higher Degree | 高次多项式不等式
When dealing with cubic or quartic inequalities, first factorise completely. Find all real roots and list them in increasing order. These roots split the real line into intervals where the polynomial does not change sign. Pick test points to determine the sign in each region.
处理三次或四次不等式时，首先完全因式分解。找出所有实根并按顺序排列。这些根把实数轴分成若干个区间，在每个区间上多项式符号不变。选取测试点来确定各区间的符号。
Example: (x + 2)(x – 1)(x – 3) ≤ 0. Roots are -2, 1, 3. Test intervals: x < -2 (e.g., x = -3): (-)(-)(-) = negative; -2 < x < 1 (x=0): (+)(-)(-) = positive; 1 < x < 3 (x=2): (+)(+)(-) = negative; x > 3 (x=4): (+)(+)(+) = positive. We need ≤ 0, so include the roots and the negative intervals: (-∞, -2] ∪ [1, 3].
例如：(x + 2)(x – 1)(x – 3) ≤ 0。根为 -2, 1, 3。检验区间：x < -2 时取 x = -3，符号负；-2 < x < 1 时符号正；1 < x < 3 时符号负；x > 3 时符号正。需要 ≤ 0 的部分，包括根所在的点，解集为 (-∞, -2] ∪ [1, 3]。
8. Inequalities Involving Square Roots | 根式不等式
Before solving an inequality with a square root, determine the domain by setting the radicand ≥ 0. For √(x – 4) < 2, we need x - 4 ≥ 0 ⇒ x ≥ 4. Squaring both sides (both sides are non-negative for x ≥ 4) gives x - 4 < 4 ⇒ x < 8. Intersecting with x ≥ 4 yields 4 ≤ x < 8.
求解含有根号的不等式前，需要先确定定义域，即令被开方数 ≥ 0。对 √(x – 4) < 2，要求 x - 4 ≥ 0 即 x ≥ 4。在满足条件的范围内两边平方（两边非负），得 x - 4 < 4 ⇒ x < 8。与 x ≥ 4 取交集得 4 ≤ x < 8。
If the inequality is √(x + 3) > x + 1, squaring immediately could introduce extraneous solutions because the right-hand side might be negative. Split into cases: if x + 1 < 0, the inequality holds as long as the square root is defined (x ≥ -3) because the left side is non-negative. Thus part of solution is -3 ≤ x < -1. If x + 1 ≥ 0 (x ≥ -1), square both sides: x + 3 > (x+1)² ⇒ x + 3 > x² + 2x + 1 ⇒ 0 > x² + x – 2 ⇒ x² + x – 2 < 0 ⇒ (x+2)(x-1) < 0 ⇒ -2 < x < 1. Intersect with x ≥ -1 gives -1 ≤ x < 1. Combine both cases: final answer is -3 ≤ x < 1.
如果不等式是 √(x + 3) > x + 1，直接平方可能会产生增根，因为右边可能为负。此时要分情况：若 x + 1 < 0，只要平方根有定义（x ≥ -3）则不等式自然成立，因为左边非负。故这一部分解为 -3 ≤ x < -1。若 x + 1 ≥ 0 (x ≥ -1)，两边平方：x + 3 > (x+1)² ⇒ 0 > x² + x – 2 ⇒ (x+2)(x-1) < 0 ⇒ -2 < x < 1。与 x ≥ -1 取交集得 -1 ≤ x < 1。合并两种情况得到最终解集 -3 ≤ x < 1。
9. Graphical Inequalities and Regions | 不等式组与图形区域
Inequalities in two variables describe shaded regions in the coordinate plane. For example, y > x² – 4 represents the region above the parabola y = x² – 4, with a dashed boundary line because the inequality is strict. Similarly, y ≤ 2x + 1 is the region on and below the line.
含有两个变量的不等式在坐标系中描述一块阴影区域。例如 y > x² – 4 表示在抛物线 y = x² – 4 上方的区域，边界用虚线表示，因为不等式是严格的。类似地，y ≤ 2x + 1 表示直线及其下方的区域。
Systems of inequalities, such as those in linear programming, require finding the intersection of several half‑planes or regions. Sketch each boundary line (solid or dashed according to the inequality sign) and shade the region that satisfies all conditions simultaneously. The feasible region is often bounded by the intersection points of the boundary lines.
不等式组（如线性规划中的约束条件）需求出多个半平面或区域的交集。画出每条边界线（根据不等号选择实线或虚线），并标出同时满足所有条件的区域。可行域通常由边界线的交点围成。
10. Inequalities with Parameters & Exam Tips | 含参不等式与应试技巧
Parameter problems ask: for which values of k does an inequality hold for all real x, or have a specific solution set? Example: find k such that (k-1)x² + 2x + k > 0 for all x. If k = 1, the expression is 2x + 1, which is not always positive. For k ≠ 1, we need the quadratic to be always positive: (k-1) > 0 and Δ < 0. Δ = 4 - 4k(k-1) = 4 - 4k² + 4k = 4( -k² + k + 1). Set Δ < 0 ⇒ k² - k - 1 > 0. The roots of k² – k – 1 = 0 are (1 ± √5)/2, so k < (1-√5)/2 or k > (1+√5)/2. Intersecting with k > 1 gives k > (1+√5)/2.
含参问题通常会问：当 k 取何值时不等式对全体实数恒成立，或具有特定的解集？例如，求 k 使 (k-1)x² + 2x + k > 0 对所有 x 成立。若 k = 1，表达式变为 2x + 1，不恒正。当 k ≠ 1 时，需二次开口向上 (k-1 > 0) 且 Δ < 0。计算 Δ = 4 - 4k(k-1) = 4(-k² + k + 1)。令 Δ < 0 得 k² - k - 1 > 0，解出 k < (1-√5)/2 或 k > (1+√5)/2。再与 k > 1 取交集，得 k > (1+√5)/2。
Exam tips: always sketch a quick sign chart or graph for non-linear inequalities. Write down domain restrictions before squaring or removing denominators. Double-check whether boundaries are included. When using a calculator, verify with test points. Common mistakes include forgetting to reverse the inequality when multiplying by negatives, mishandling critical values from denominators, and incorrectly assuming a quadratic is always positive without checking the discriminant.
应试技巧：对于非线性不等式，务必快速画出符号表或草图。进行平方或去分母前先明确定义域。要反复核对边界点是否包含在内。使用计算器时，用测试值进行验证。常见错误包括：乘除以负数时忘记反转不等号，遗漏分母产生的临界值，以及没有检查判别式就断定二次式恒正。
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