📚 A-Level OCR Maths: Inequalities Key Points | A-Level OCR 数学：不等式考点精讲
Inequalities are a fundamental topic in A-Level OCR Mathematics, bridging algebraic manipulation, graphical interpretation, and logical reasoning. This revision guide covers all essential inequality types tested in the OCR specification, from linear and quadratic to rational and modulus inequalities, along with graphical methods and discriminant applications. Mastery of these techniques will equip you to solve inequality problems accurately and efficiently in the exam.
不等式是 A-Level OCR 数学的基础课题，它将代数操作、图像解读与逻辑推理融为一体。本复习指南涵盖了 OCR 考纲中所有关键的不等式类型，包括线性不等式、二次不等式、分式不等式和绝对值不等式，同时还涉及图像法和判别式的应用。掌握这些方法，你将能准确高效地解决考试中的不等式问题。
1. Solving Linear Inequalities | 解线性不等式
Linear inequalities involve expressions like ax + b > cx + d. The principle is identical to solving linear equations, except that multiplying or dividing by a negative number reverses the inequality sign. The solution set is typically expressed using interval notation, e.g., x > 3 is written as (3, ∞).
线性不等式涉及诸如 ax + b > cx + d 的表达式。解法与解线性方程相同，但要注意：当两边乘以或除以负数时，不等号方向必须改变。解集通常用区间表示，如 x > 3 写作 (3, ∞)。
Example: Solve 2x − 5 < 3x + 2. Collect x terms on one side: −5 − 2 < 3x − 2x → −7 < x, so the solution is x > −7. In interval notation, this is (−7, ∞).
例题：解 2x − 5 < 3x + 2。把含 x 的项移到一边：−5 − 2 < 3x − 2x → −7 < x，因此解为 x > −7。用区间表示即为 (−7, ∞)。
Always present your final answer clearly, either as x > a, x < b, or using set notation such as {x : x > −7}. In OCR exams, both interval and set notations are accepted.
最终答案的书写应清晰，可以是 x > a、x < b 的形式，也可以使用集合记法如 {x : x > −7}。OCR 考试中，区间记法和集合记法均被接受。
2. Quadratic Inequalities and Graphs | 二次不等式与图像
When solving a quadratic inequality such as ax² + bx + c > 0, first find the roots by factorising or using the quadratic formula. Sketch the parabola y = ax² + bx + c, noting whether it opens upwards (a > 0) or downwards (a < 0). The sign of the quadratic is positive outside the roots and negative between them for a > 0.
解二次不等式（如 ax² + bx + c > 0）时，首先通过因式分解或求根公式求出根。画出抛物线 y = ax² + bx + c 的草图，注意开口向上（a > 0）还是向下（a < 0）。当 a > 0 时，二次式的值在两根之外为正，在两根之间为负。
For x² − 5x + 6 < 0, factor as (x − 2)(x − 3) < 0. The roots are 2 and 3. With a positive coefficient of x², the graph is a ∪ shape. The inequality is satisfied where the curve lies below the x‑axis, i.e., between the roots: 2 < x < 3.
对于 x² − 5x + 6 < 0，因式分解得 (x − 2)(x − 3) < 0。根为 2 和 3。由于 x² 系数为正，图像呈 ∪ 形。不等式在曲线位于 x 轴下方时成立，即在两根之间：2 < x < 3。
If the quadratic does not factorise, find the roots with the discriminant or completing the square. The same logic applies: identify where the function is above or below zero.
若二次式无法因式分解，可利用判别式或配方求根，再运用同样的逻辑：判断函数值大于零或小于零的区间。
3. Using the Discriminant in Inequalities | 判别式在不等式中的应用
The discriminant Δ = b² − 4ac determines whether a quadratic has real roots. For a quadratic expression to be always positive (or always negative), we require Δ < 0 and the leading coefficient a of appropriate sign. This concept appears frequently in OCR problems asking for the range of a parameter that keeps a quadratic positive for all real x.
判别式 Δ = b² − 4ac 决定了二次式是否有实根。要使一个二次式恒正（或恒负），需要 Δ < 0 且首项系数 a 的符号适当。这一概念常出现在 OCR 题目中：求使二次式对所有实数 x 恒正的参数取值范围。
Example: Find the values of k for which x² + kx + 9 > 0 for all real x. Here a = 1 > 0, so the quadratic is ∪‑shaped. We need Δ = k² − 36 < 0, giving k² < 36, so −6 < k < 6. The inequality is strict, hence k ∈ (−6, 6).
例题：求使 x² + kx + 9 > 0 对所有实数 x 成立的 k 值。这里 a = 1 > 0，图像是 ∪ 形。需要 Δ = k² − 36 < 0，得出 k² < 36，因此 −6 < k < 6。不等式为严格不等号，故 k ∈ (−6, 6)。
When the inequality is non‑strict (e.g., ≥ 0), Δ ≤ 0 may be required, but check boundary values carefully to ensure the quadratic does not become undefined or violate the condition.
当不等式为非严格（如 ≥ 0）时，可能需要 Δ ≤ 0，但务必仔细检验边界值，确保二次式不会出现未定义或违反条件的情况。
4. Polynomial Inequalities of Higher Degree | 高次多项式不等式
For cubic or quartic inequalities, factorise the polynomial as far as possible. Then use a sign table or a “wave” method where you mark the real roots on a number line and test the sign of the product in each interval. The sign alternates only if the root has odd multiplicity.
对于三次或四次不等式，尽可能将多项式因式分解。然后使用符号表或“波浪线”法：在数轴上标出所有实根，并在每个区间内检验乘积的符号。只有当根的重数为奇数时，符号才会在穿过该根时变号。
Example: Solve (x + 1)(x − 2)(x − 4) > 0. Roots are −1, 2, 4 (all multiplicity 1). Test an x‑value from each interval: for x = −2, product is (−)(−)(−) = −; for x = 0, (+)(−)(−) = +; for x = 3, (+)(+)(−) = −; for x = 5, (+)(+)(+) = +. So the solution intervals are (−1, 2) ∪ (4, ∞).
例题：解 (x + 1)(x − 2)(x − 4) > 0。根为 −1、2、4（重数均为 1）。取各区间的测试值：x = −2 时，乘积为 (负)(负)(负) = 负；x = 0 时，(正)(负)(负) = 正；x = 3 时，(正)(正)(负) = 负；x = 5 时，(正)(正)(正) = 正。因此解集区间为 (−1, 2) ∪ (4, ∞)。
If a factor appears an even number of times (e.g., (x−1)²), the sign does not change at that root. Always sketch a quick sign chart to avoid mistakes.
若某个因子出现偶数次（如 (x−1)²），穿过该根时符号不变。始终快速绘制符号图以避免错误。
5. Rational (Fractional) Inequalities | 分式不等式
Rational inequalities involve expressions like (px + q)/(rx + s) > a. Never multiply by the denominator unless you are certain of its sign. Instead, multiply both sides by the square of the denominator (which is non‑negative) or move all terms to one side to create a single fraction compared to 0.
分式不等式涉及形如 (px + q)/(rx + s) > a 的表达式。切勿随意乘以分母，除非你确定它的正负号。正确的做法是两边乘以分母的平方（它非负），或把所有项移到一边，化为单个分式与 0 比较。
Example: Solve (x − 3)/(x + 1) < 2. Bring 2 to the left: (x − 3)/(x + 1) − 2 < 0 → (x − 3 − 2x − 2)/(x + 1) < 0 → (−x − 5)/(x + 1) < 0. Equivalently, (x + 5)/(x + 1) > 0 after multiplying by −1 and reversing the sign. Critical values: x = −5 and x = −1 (excluded). Test intervals: x < −5 gives +/+ = +; −5 < x < −1 gives −/+ = −; x > −1 gives +/+ = +. The inequality (x + 5)/(x + 1) > 0 holds for x < −5 or x > −1. Do not forget to exclude x = −1.
例题：解 (x − 3)/(x + 1) < 2。将 2 移到左边：(x − 3)/(x + 1) − 2 < 0 → (x − 3 − 2x − 2)/(x + 1) < 0 → (−x − 5)/(x + 1) < 0。乘以 −1 并反转不等号得 (x + 5)/(x + 1) > 0。关键值：x = −5 和 x = −1（分母为零点排除）。测试区间：x < −5 时 ++ = +；−5 < x < −1 时 −+ = −；x > −1 时 ++ = +。不等式 (x + 5)/(x + 1) > 0 的解为 x < −5 或 x > −1。别忘了排除 x = −1。
6. Modulus Inequalities | 绝对值（模）不等式
The modulus function gives the distance from zero. Key facts: |f(x)| ≤ a (with a > 0) is equivalent to −a ≤ f(x) ≤ a. For |f(x)| ≥ a, it means f(x) ≤ −a or f(x) ≥ a. More complex modulus inequalities, such as |ax + b| < cx + d, often require squaring both sides (as both sides are non‑negative) or considering separate cases for when the expression inside the modulus is positive or negative.
绝对值函数表示到零的距离。关键等价关系：对于 a > 0，|f(x)| ≤ a 等价于 −a ≤ f(x) ≤ a；|f(x)| ≥ a 等价于 f(x) ≤ −a 或 f(x) ≥ a。更复杂的模不等式，如 |ax + b| < cx + d，通常需要两边平方（因为两边非负）或根据绝对值内部表达式的正负分情况讨论。
Example: Solve |2x − 1| ≤ 3. This gives −3 ≤ 2x − 1 ≤ 3. Add 1: −2 ≤ 2x ≤ 4. Divide by 2: −1 ≤ x ≤ 2. The solution is the closed interval [−1, 2].
例题：解 |2x − 1| ≤ 3。写为 −3 ≤ 2x − 1 ≤ 3。加 1：−2 ≤ 2x ≤ 4。除以 2：−1 ≤ x ≤ 2。解为闭区间 [−1, 2]。
For |x − 4| > 2x + 1, note that the right side may be negative, so extra care is needed. It is best to split into cases: when x ≥ 4, the inequality becomes x − 4 > 2x + 1 → −5 > x, which contradicts x ≥ 4, so no solution from this branch. When x < 4, it becomes −(x − 4) > 2x + 1 → −x + 4 > 2x + 1 → 3 > 3x → x < 1. Combine with x < 4 to get x < 1. Also ensure 2x + 1 is not constraining beyond the inequality. The final solution is x < 1.
对于 |x − 4| > 2x + 1，注意右边可能为负，需格外小心。最佳方法是分情况讨论：当 x ≥ 4 时，不等式变为 x − 4 > 2x + 1 → −5 > x，与 x ≥ 4 矛盾，此情况无解。当 x < 4 时，变为 −(x − 4) > 2x + 1 → −x + 4 > 2x + 1 → 3 > 3x → x < 1。结合条件 x < 4，得到 x < 1。最终解为 x < 1。
7. Compound Inequalities (Simultaneous) | 复合不等式（同时不等式）
A compound inequality like −3 ≤ 2x + 1 < 5 actually represents two inequalities that must be satisfied simultaneously. You can either solve as a chain by performing the same operation on all three parts, or split into two separate inequalities and find their intersection.
形如 −3 ≤ 2x + 1 < 5 的复合不等式实际上表示两个必须同时满足的不等式。你可以对三部分同时执行同一操作进行求解，也可将其拆分为两个独立不等式并求交集。
Chain method: Subtract 1 throughout: −4 ≤ 2x < 4. Divide by 2: −2 ≤ x < 2. The solution is x ∈ [−2, 2). Splitting method confirms the same.
链式解法：整体减 1：−4 ≤ 2x < 4。除以 2：−2 ≤ x < 2。解为 x ∈ [−2, 2)。拆分法可得到相同结果。
When the inequality involves squares or fractions, consider taking intersections carefully, especially if restrictions arise from denominators. For example, 0 < (x + 1)/(x − 2) ≤ 3 requires solving two rational inequalities and intersecting with the denominator restriction x ≠ 2.
当不等式涉及平方或分式时，要仔细求交集，尤其要考虑分母带来的限制。例如，0 < (x + 1)/(x − 2) ≤ 3 需要求解两个分式不等式，并与分母限制 x ≠ 2 求交集。
8. Solving Inequalities Using Graphs | 利用图像解不等式
Graphical methods are powerful for visualising the solution of f(x) < g(x) or f(x) > a. The inequality f(x) < g(x) holds for x‑values where the graph of y = f(x) lies below that of y = g(x). Similarly, f(x) > a corresponds to the portion of the graph above the horizontal line y = a.
图像法是可视化不等式解的有力工具。f(x) < g(x) 的解就是 y = f(x) 图像位于 y = g(x) 图像下方的 x 取值范围。类似地，f(x) > a 对应图像在水平直线 y = a 上方的区域。
This approach is particularly useful when dealing with modulus functions or when a quadratic and a linear function are given. Suppose the graphs of y = |x + 2| and y = 2x − 1 are sketched. To solve |x + 2| < 2x − 1, identify the interval where the V‑shaped absolute graph is below the line. Read the x‑coordinates of the intersection points and deduce the region.
在处理绝对值函数或给定二次函数与线性函数的图像时，这种方法尤为有用。假设已绘出 y = |x + 2| 和 y = 2x − 1 的图像，要解 |x + 2| < 2x − 1，只需找出绝对值 V 形图位于直线下方的区间，读取交点的 x 坐标，进而确定区域。
In the exam, you may be given a graph and asked to write down the solution of an inequality; always check whether endpoints are included based on dashed/solid lines or open/closed circles.
考试中，你可能会被要求根据给出的图像写出不等式的解；务必根据虚线/实线或空心/实心圆点判断端点是否包含在解内。
9. Inequalities in Modelling and Applications | 不等式建模与应用
Inequalities often arise in real‑world contexts modelled by functions, such as profit > cost, temperature within a safe range, or physical dimensions meeting constraints. The process involves formulating an inequality from the problem statement, simplifying it algebraically, and then solving it using the techniques covered above.
不等式常常出现在函数建模的实际问题中，如利润大于成本、温度处于安全区间，或物理尺寸满足约束条件。解题过程包括根据问题叙述建立不等式、代数化简，然后运用上述方法求解。
Example: A company’s revenue is modelled by R(x) = 50x − x² and costs by C(x) = 20x + 100, where x is the number of items (in hundreds). Find the production levels for which the company makes a profit: 50x − x² > 20x + 100 → 0 > x² − 30x + 100 → x² − 30x + 100 < 0. Factor or complete the square: (x − 15)² < 125 → 15 − √125 < x < 15 + √125. Given the context, x must be positive, so the profit interval is (15 − 5√5, 15 + 5√5).
例题：某公司收入模型为 R(x) = 50x − x²，成本为 C(x) = 20x + 100，x 为产品数量（百件）。求公司盈利的生产水平：50x − x² > 20x + 100 → 0 > x² − 30x + 100 → x² − 30x + 100 < 0。因式分解或配方：(x − 15)² < 125 → 15 − √125 < x < 15 + √125。结合实际，x 应为正数，因此盈利区间为 (15 − 5√5, 15 + 5√5)。
Remember to interpret solutions in the context of the problem, discarding extraneous intervals that do not make practical sense (e.g., negative quantities).
记得将解置于实际情境中解释，舍弃无实际意义的多余区间（如产量为负）。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
Avoid these frequent pitfalls: forgetting to reverse the inequality sign when multiplying or dividing by a negative number; incorrectly multiplying an inequality by a denominator without knowing its sign; missing the exclusion of values that make a denominator zero; writing the solution as a simple interval when it should be a union of intervals; and misinterpreting non‑strict inequalities (≤, ≥) with critical points.
避免这些常见错误：乘以或除以负数时忘记反转不等号；在不清楚分母符号的情况下错误地两边同乘分母；忘记排除使分母为零的值；将本应为并集的解集误写为单一区间；对非严格不等式（≤，≥）的关键点处理不当。
In the exam, always double‑check your factorisation and consider the shape of the graph (smiley or frowny) to avoid sign errors. For modulus inequalities, practise both the algebraic and graphical approaches. When solving rational inequalities, never cross‑multiply unless you know the sign, and always present the final solution with proper interval notation, noting any excluded points.
考试中，务必反复检查因式分解并考虑图像形状（开口向上还是向下），以避免符号错误。对于模不等式，要同时练习代数法和图像法。解分式不等式时，除非确定正负号，否则不要交叉相乘，并且始终用正确的区间记法表示最终解，并注明所有排除点。
- Write out the steps clearly; marks are awarded for method even if the final answer is slightly wrong.
- For “always positive” quadratic conditions, remember Δ < 0 and a > 0.
- Use a number line to verify your solution intervals for higher‑degree polynomials.
- 清晰写出步骤；即使最终答案略有偏差，方法正确也能得过程分。
- 对于二次式恒正的条件，记住 Δ < 0 且 a > 0。
- 处理高次多项式时，使用数轴检验解集区间。
Consistent practice using past OCR questions will build the fluency and confidence needed to handle any inequality problem on your A‑Level paper.
通过持续练习 OCR 历年真题，你将培养出处理 A‑Level 试卷中任何不等式问题所需的熟练度与信心。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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