AQA Maths: Inequalities Essential Revision | AQA 数学：不等式 考点精讲

📚 AQA Maths: Inequalities Essential Revision | AQA 数学：不等式 考点精讲

Inequalities are a fundamental part of the AQA A-level Mathematics specification, appearing in both pure and applied contexts. Understanding how to solve, manipulate, and interpret inequalities is crucial for tackling exam questions effectively. This article distils the key concepts, methods, and common pitfalls you need to master for the AQA examination.

不等式是 AQA A-level 数学大纲中的核心内容，无论在纯数还是应用题中都频繁出现。掌握不等式的求解、变形与解读，对于高效应对考试题目至关重要。本文提炼了 AQA 考试必须掌握的关键概念、解题方法和常见易错点。

1. Linear Inequalities | 线性不等式

Linear inequalities such as 2x + 1 ≤ 5 are solved by isolating the variable x, just like solving linear equations. However, one crucial rule must never be forgotten: when you multiply or divide both sides by a negative number, the inequality sign must be reversed. For instance, solving -3x + 4 > 10: subtract 4 gives -3x > 6, then divide by -3 to obtain x < -2. The solution set can be expressed in set notation {x : x < -2} or in interval notation (-∞, -2).

形如 2x + 1 ≤ 5 的线性不等式通过分离 x 来求解，方法与解方程类似。但有一条关键规则绝对不能忘记：当两边乘以或除以负数时，不等号方向必须反转。例如，解 -3x + 4 > 10：减 4 得 -3x > 6，再除以 -3 得到 x < -2。解集可用集合符号表示为 {x : x < -2}，或用区间符号表示为 (-∞, -2)。

Always present your final answer clearly; AQA accepts both set and interval notations, but the notation should be consistent with the question. For strict inequalities (< or >), use open intervals, and for non-strict inequalities (≤ or ≥), use closed intervals or include the endpoint in set form.

最终答案务必清晰表达；AQA 接受集合符号和区间符号，但应与题目要求一致。对于严格不等式（< 或 >）使用开区间，对于含等号的不等式（≤ 或 ≥）使用闭区间或在集合中包含端点。

2. Quadratic Inequalities | 二次不等式

To solve a quadratic inequality like x² – 5x + 6 > 0, start by finding the critical values from the equation x² – 5x + 6 = 0. These are x = 2 and x = 3. Then sketch the graph of y = x² – 5x + 6, a positive parabola intersecting the x-axis at 2 and 3. The inequality asks where the graph is above the x-axis, i.e. x < 2 or x > 3. A sign table can also be used:

解如 x² – 5x + 6 > 0 的二次不等式，首先由对应方程 x² – 5x + 6 = 0 求出临界值 x = 2 和 x = 3。然后画出 y = x² – 5x + 6 的草图，这是一个开口向上的抛物线，与 x 轴交于 2 和 3。不等式要求图像在 x 轴上方，即 x < 2 或 x > 3。也可使用符号表：

The solution in set notation is {x : x < 2} ∪ {x : x > 3}. For inequalities with ≤ or ≥, the critical values themselves are included. When the quadratic has no real roots (discriminant < 0), it is either always positive or always negative; for example, x² + 2x + 3 < 0 has no solution because the parabola lies entirely above the x-axis.

用集合符号表示的解为 {x : x < 2} ∪ {x : x > 3}。对于带等号的不等式（≤ 或 ≥），临界值本身也包含在解集中。若二次式无实根（判别式 < 0），则它恒正或恒负；例如 x² + 2x + 3 < 0 无解，因为抛物线完全在 x 轴上方。

3. Polynomial Inequalities | 多项式不等式

For polynomial inequalities of degree 3 or higher, fully factorise the expression and determine the sign of each factor on intervals between the real roots. A sign chart is extremely helpful. Consider (x – 1)(x + 2)(x – 4) ≤ 0. The roots are -2, 1, and 4. Test intervals: x < -2, -2 < x < 1, 1 < x < 4, and x > 4. By checking the sign of each factor, you find that the product is negative or zero on the intervals where an odd number of negative factors exist. The solution is x ≤ -2 or 1 ≤ x ≤ 4.

对于三次或更高次的多项式不等式，先完全因式分解，然后在各实根之间的区间内判断每个因式的符号。符号表非常有用。以 (x – 1)(x + 2)(x – 4) ≤ 0 为例，根为 -2, 1, 4。测试区间 x < -2, -2 < x < 1, 1 < x < 4, x > 4。通过判断各因式符号可知，乘积在含有奇数个负因式的区间上为负或零。解为 x ≤ -2 或 1 ≤ x ≤ 4。

If a factor appears an even number of times, the sign does not change at that root (e.g. (x-2)² is always non-negative). For odd multiplicities, the sign changes as you cross the root. Always exclude values that make any undefined expressions invalid, though polynomials are defined everywhere.

若某个因式出现偶数次，则在该根处符号保持不变（例如 (x-2)² 始终非负）。对于奇数次因式，穿过该根时符号改变。多项式处处有定义，无需担心无定义点。

4. Rational Inequalities | 分式不等式

When solving rational inequalities like (x+1)/(x-3) ≥ 2, bring all terms to one side and combine into a single fraction. Never cross-multiply directly by an expression containing x, as its sign is unknown. Rearranging: (x+1)/(x-3) – 2 ≥ 0 → (x+1 – 2(x-3))/(x-3) ≥ 0 → (7 – x)/(x-3) ≥ 0. The critical points are x = 3 (denominator zero) and x = 7 (numerator zero). Using a sign table, the fraction is non-negative when 3 < x ≤ 7. Always exclude x = 3 from the solution. The final answer can be written as {x : 3 < x ≤ 7}.

解分式不等式如 (x+1)/(x-3) ≥ 2 时，应将所有项移到一边，合并成一个分式。绝不能直接乘以含有 x 的表达式，因为它的正负未知。整理得：(x+1)/(x-3) – 2 ≥ 0 → (x+1 – 2x + 6)/(x-3) ≥ 0 → (7 – x)/(x-3) ≥ 0。临界点为 x = 3（分母为零）和 x = 7（分子为零）。通过符号表可知，分式在 3 < x ≤ 7 时非负。务必从解集中排除 x = 3。最终答案可写作 {x : 3 < x ≤ 7}。

Always state the excluded values clearly. A common mistake is to treat the inequality as an equation and simply cross-multiply, losing the sign information. In AQA exams, you must show the rearrangement and sign analysis.

务必清晰注明需排除的数值。一个常见错误是将不等式当作方程处理，直接去分母，从而丢失符号信息。在 AQA 考试中，必须展示移项和符号分析的过程。

5. Manipulating Inequalities | 不等式的变形规则

Basic operations on inequalities follow strict rules. You may add or subtract the same quantity from both sides without affecting the inequality direction. Multiplying or dividing by a positive number preserves the inequality; multiplying or dividing by a negative number reverses the sign. Squaring both sides is only safe when both sides are known to be non-negative; otherwise, extraneous solutions may appear. Taking reciprocals of both sides reverses the inequality if both sides are positive or both are negative, but if they have different signs, the result is not simply a reversal.

不等式的基本运算有严格规则。两边可同时加减同一个量，不等号方向不变。乘以或除以正数方向不变；乘以或除以负数则必须反转不等号。只有当两边均非负时，平方才是安全操作，否则可能产生增根。若两边同号（同正或同负），取倒数会反转不等号；若异号，结果并非简单反转。

For example, if a > b > 0, then 1/a < 1/b. If a > 0 > b, then 1/a > 0 > 1/b, so the original inequality a > b does not translate directly to a simple reciprocal inequality. Always consider the domains when applying functions to both sides of an inequality.

例如，若 a > b > 0，则 1/a < 1/b。若 a > 0 > b，则 1/a > 0 > 1/b，原不等式 a > b 无法直接转变为简单的倒数不等式。在对不等式两边施加函数时，务必考虑定义域。

6. Absolute Value Inequalities | 绝对值不等式

Inequalities involving absolute values, such as |2x – 1| < 5, are solved by rewriting them as compound inequalities. Since |A| < c (with c > 0) means -c < A < c, we get -5 < 2x – 1 < 5, leading to -2 < x < 3. For greater-than inequalities, |x + 3| ≥ 4 splits into two separate inequalities: x + 3 ≥ 4 or x + 3 ≤ -4, giving x &

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