📚 Inequalities in IB and OCR Mathematics: Key Points | IB OCR 数学：不等式 考点精讲
Inequalities form a fundamental building block in both IB Analysis & Approaches (AA) and OCR A Level Mathematics. They appear in pure mathematics, coordinate geometry, calculus and even proof questions. This article walks you through the essential inequality types, solving techniques and common pitfalls encountered in exam-style problems. Mastering these concepts will sharpen your algebraic fluency and prepare you for high-mark questions on your final papers.
不等式是 IB 分析与方法和 OCR A Level 数学共同的核心模块。从代数、坐标几何到微积分和证明题，不等式无处不在。本文将系统梳理线性、二次、分式、绝对值等各类不等式的求解技巧，深度剖析考试中的易错点与高分策略。掌握这些内容，能够显著提升你的代数推理能力和应试效率。
1. Linear Inequalities and Number Lines | 线性不等式与数轴表示
Solving linear inequalities such as 3x – 5 > 7 works almost identically to solving linear equations, with one critical exception: multiplying or dividing by a negative number reverses the inequality sign. Always show your answer on a number line using open or closed circles, and express the solution in set notation like {x | x > 4} or interval notation (4, ∞).
解 3x – 5 > 7 这类线性不等式与解方程步骤基本相同，但必须牢记：当两边同乘或同除以一个负数时，不等号方向要反转。解答要画在数轴上，用空心圈或实心点表示边界，并用集合记法 {x | x > 4} 或区间记法 (4, ∞) 写出解集。
In IB exams, you may see combined inequalities such as -2 ≤ 3 – 2x < 5. Split these into two separate inequalities but keep the variable in the middle, then solve simultaneously. The final answer must satisfy both conditions. OCR papers often ask for the solution set as a single inequality and on a number line. Double-check the direction of each sign after isolating x, especially when handling negative coefficients.
IB 考试中常出现 -2 ≤ 3 – 2x < 5 这类联合不等式。解法是拆成两个同向不等式并同时求解，最后取交集。最终答案需用一条不等式表达并画在数轴上。OCR 试题同样重视数轴表示，记得在隔离 x 时若系数为负，及时调整不等号方向，避免致命错误。
2. Solving Quadratic Inequalities | 求解二次不等式
Quadratic inequalities like x² – 4x – 5 < 0 rely on sketching the related parabola y = x² – 4x – 5. First find the roots by factorising: (x – 5)(x + 1) = 0 gives x = -1 and x = 5. Since the coefficient of x² is positive, the graph is U-shaped, so the expression is negative between the roots. Hence the solution is -1 < x < 5.
处理 x² – 4x – 5 < 0 这类二次不等式，关键在于绘制对应抛物线 y = x² – 4x – 5 的草图。先求根，因式分解得 (x – 5)(x + 1) = 0，根为 x = -1 和 x = 5。二次项系数为正，图像呈 U 形，因此不等式的负值区间在两根之间，解为 -1 < x < 5。
If the inequality had been x² – 4x – 5 ≥ 0, the solution would be x ≤ -1 or x ≥ 5. A common mistake is forgetting to check whether the roots themselves are included. Always substitute the roots back into the original inequality to verify. For IB Analysis, be ready to deal with discriminant-related cases where the quadratic has no real roots; if a > 0 and the discriminant is negative, the quadratic is always positive, so inequalities like x² + x + 1 > 0 hold for all real x.
若不等式改为 x² – 4x – 5 ≥ 0，解集便是 x ≤ -1 或 x ≥ 5。常犯错误是忘记检验根是否包含在内。必要时将根代回原式确认。IB 分析课程中还会遇到判别式相关情形：当 a > 0 且判别式小于零时，二次函数恒正，此时 x² + x + 1 > 0 对所有实数均成立。
3. Polynomial Inequalities of Higher Degree | 高次多项式不等式
For cubic or quartic inequalities, such as (x + 2)(x – 1)(x – 3) ≤ 0, construct a sign table. Mark the roots -2, 1, 3 on a number line and test the sign of the product in each interval. The inequality includes the equals condition, so the roots are part of the solution set. The final answer appears as x ≤ -2 or 1 ≤ x ≤ 3.
处理三次或四次不等式如 (x + 2)(x – 1)(x – 3) ≤ 0，要使用符号表。在数轴上标出根 -2、1、3，然后在各区间内任选测试点判断乘积符号。因不等号包含等号，根点属于解集。最终解集形式为 x ≤ -2 或 1 ≤ x ≤ 3。
Pay attention to repeated factors: for (x – 1)²(x + 4) > 0, the square factor is always non-negative and does not change sign at x = 1, but it makes the product zero there. Since the inequality is strict, x = 1 must be excluded, while x = -4 is also excluded. The sign change only occurs at -4, so the solution is x > -4, x ≠ 1. OCR and IB both test this subtlety.
注意重因子的情况：在 (x – 1)²(x + 4) > 0 中，平方项恒非负且在 x = 1 处不改变符号，但其值为零。因是严格不等式，须排除 x = 1；x = -4 也被排除。符号只在 x = -4 处变化，故解为 x > -4 且 x ≠ 1。OCR 和 IB 都对这一细节颇为青睐。
4. Rational Inequalities | 分式不等式
Inequalities involving fractions, like (2x – 3)/(x + 1) ≥ 1, require care: never multiply both sides by the denominator unless you know its sign. Instead, bring all terms to one side and form a single fraction: (2x – 3)/(x + 1) – 1 ≥ 0 → (2x – 3 – (x + 1))/(x + 1) ≥ 0 → (x – 4)/(x + 1) ≥ 0. Now use a sign table with critical points x = -1 (excluded because denominator zero) and x = 4.
遇到 (2x – 3)/(x + 1) ≥ 1 之类的分式不等式，切忌直接乘分母，除非你能确定分母符号。正确做法是将项移到一边通分成单一公式：(2x – 3)/(x + 1) – 1 ≥ 0 → (x – 4)/(x + 1) ≥ 0。然后根据临界点 x = -1（分母为零，须排除）和 x = 4 制作符号表。
The sign table reveals that the fraction is ≥ 0 for x < -1 or x ≥ 4. Do not include x = -1. This ‘rewrite as a single fraction’ method is standard in both syllabi. Remember to watch for non-permissible values when stating the final answer, often requiring combined inequalities or set notation with a union symbol.
符号表显示，该分式在 x < -1 或 x ≥ 4 时非负。切勿包含 x = -1。这种“化为单一公式”的方法是两套教材的标准要求。写最终答案时务必剔除非允许值，常用并集符号或分段不等式呈现。
5. Absolute Value Inequalities | 绝对值不等式
Absolute value inequalities come in two main forms. Type 1: |f(x)| < a, where a > 0, is equivalent to -a < f(x) < a. Type 2: |f(x)| > a becomes f(x) < -a or f(x) > a. Graphically, |x – c| represents the distance from x to c on the number line, so |x – 3| ≤ 2 means all points within 2 units of 3, giving 1 ≤ x ≤ 5.
绝对值不等式主要有两种类型。第一类：|f(x)| < a (a > 0) 等价于 -a < f(x) < a。第二类：|f(x)| > a 等价于 f(x) < -a 或 f(x) > a。从几何角度看，|x – 3| ≤ 2 表示数轴上与 3 距离不超过 2 的所有点，解得 1 ≤ x ≤ 5。
More complicated cases like |2x + 1| < 3x – 2 require a case-by-case analysis based on the sign of 3x – 2 and the definition of absolute value. Always check that any solutions obtained actually satisfy the original inequality’s domain restrictions. IB papers sometimes combine absolute values with other functions, while OCR often sets multi-step algebraic manipulations. Practising both algebraic and graphical approaches builds robustness.
对于 |2x + 1| < 3x – 2 等复杂情形，需根据 3x – 2 的正负及绝对值的分段定义进行分类讨论。求得的解必须代回原不等式验证是否满足定义域限制。IB 试卷有时将绝对值与其它函数混合，OCR 则侧重多步骤代数推导。同时练习代数法与图像法能让你临场不乱。
6. Systems of Linear Inequalities | 线性不等式组
When solving systems of linear inequalities such as y ≥ 2x – 1, x + y < 4 and y > -1, the solution region is the intersection of the half-planes represented by each inequality. Sketch the boundary lines, using dashed lines for strict inequalities (< or >) and solid lines for inclusive ones (≤ or ≥). Shade the unwanted region or the feasible region depending on the convention used in your course.
求解 y ≥ 2x – 1, x + y < 4, y > -1 这类线性不等式组时，解区域是各不等式对应半平面的交集。画出边界直线，严格不等式用虚线，带等号的不等式用实线。根据课程习惯，可涂阴“不合理”区域或仅标示可行域。
In IB, this topic often appears in linear programming problems where you must maximise or minimise an objective function over the feasible region. OCR also includes such problems in the Pure & Statistics strand. You need to identify vertices (corner points) and test them in the objective function. Be precise with scale and shading, and always state final coordinates clearly.
IB 中常见线性规划题，要求在可行域上优化目标函数。OCR 的纯数与统计模块同样包含此类问题。解题需找出可行域的所有顶点坐标，将它们代入目标函数比较。作图时注意比例和描影的准确性，最后清晰写出坐标值。
7. Graphical Interpretation of Inequalities | 不等式的图形解读
Interpreting inequalities graphically is a powerful skill. For example, solving f(x) < g(x) is equivalent to finding where the graph of y = f(x) lies below the graph of y = g(x). This is especially useful when dealing with transcendental functions, modulus graphs or circles. In IB, you might be asked to find the set of x-values where a quadratic is above a linear function.
从图形角度理解不等式是一项关键能力。例如，解 f(x) < g(x) 等价于寻找 y = f(x) 图像位于 y = g(x) 图像下方的 x 区间。处理超越函数、绝对值函数的图像或圆与直线的关系时尤其有效。IB 可能要求你求抛物线上方直线的 x 值范围。
Consider (x – 2)² + y² ≤ 9: this describes a filled circle centred at (2,0) radius 3. When combined with y > x + 1, the solution is the part of the circle’s interior that also lies above the line. Plot the circle with solid boundary, the line dashed, and shade the overlapping region. OCR questions frequently ask for such regions to be clearly indicated on coordinate grids. Always label key intersection points.
再看 (x – 2)² + y² ≤ 9，它表示以 (2,0) 为圆心、半径 3 的实心圆。若再联立 y > x + 1，解区域便是圆内且在直线上方的部分。用实线画圆、虚线画直线，标示重叠区。OCR 经常要求在坐标平面上清晰表示这类区域，记得标出关键交点坐标。
8. Inequalities Involving Parameters | 含参数的不等式
Parameter problems test deeper thinking: ‘Find the range of k such that x² + kx + 4 > 0 for all real x.’ For a quadratic to be always positive, its discriminant must be negative: k² – 16 < 0, leading to -4 < k < 4. Such questions appear in both IB AA and OCR as part of discriminant analysis and proof. They require you to connect inequality solving with the conditions for an always-positive quadratic.
含参不等式考查深度思维：“求 k 的取值范围，使得对所有实数 x 均有 x² + kx + 4 > 0。” 二次式恒正要求判别式小于零：k² – 16 < 0，得 -4 < k < 4。这类题在 IB 分析与方法和 OCR 中均以判别式分析题型出现，需要将不等式求解与恒正二次式的条件紧密结合。
Another typical parameter question involves linear inequalities: ‘Given 2x – k < x + 3 < 5x + 2, determine the possible integer values of k if the solution set includes x = 1.’ Substitute x = 1 into all parts and obtain inequalities for k. IB and OCR both enjoy linking parameters with integer solutions or specified intervals. Always work systematically, separating the compound inequality into two, solving each for the variable and then applying given constraints.
另一类参数题涉及线性不等式组：“已知 2x – k < x + 3 < 5x + 2，且解集包含 x = 1，求 k 的整数值。” 将 x = 1 代入原式，得到关于 k 的不等式。IB 和 OCR 都喜欢将参数与整数解或特定区间挂钩。解题时，先将联合不等式拆为两个，分别求解变量，再代入给定限制条件。
9. Inequalities Proofs (AM-GM, Cauchy) | 不等式证明（AM-GM、柯西）
Proofs involving classical inequalities are a hallmark of higher-level IB and feature in OCR’s advanced pure content. The AM-GM inequality states that for non-negative numbers a and b, (a + b)/2 ≥ √(ab), with equality if and only if a = b. This can be extended to three or four variables. Typical questions ask you to prove an algebraic inequality by starting from (√a – √b)² ≥ 0.
经典不等式的证明是 IB 高阶数学的标志，OCR 进阶纯数也有涉及。均值不等式指出，对非负数 a、b，有 (a + b)/2 ≥ √(ab)，等号当且仅当 a = b 时成立。可推广至三个或更多变量。典型考题要求你从 (√a – √b)² ≥ 0 出发，证明某个代数不等式。
Cauchy-Schwarz inequality also appears: (x₁² + x₂²)(y₁² + y₂²) ≥ (x₁y₁ + x₂y₂)². These proofs require you to expand, rearrange and recognize perfect squares. For instance, to prove a² + b² + c² ≥ ab + bc + ca, you can rewrite it as ½[(a – b)² + (b – c)² + (c – a)²] ≥ 0. Both IB and OCR examiners look for clear logical steps and a concluding statement that matches the original inequality.
柯西-许瓦兹不等式也有出现：(x₁² + x₂²)(y₁² + y₂²) ≥ (x₁y₁ + x₂y₂)²。这类证明需要展开、重组并识别完全平方式。例如，要证 a² + b² + c² ≥ ab + bc + ca，可转化为 ½[(a – b)² + (b – c)² + (c – a)²] ≥ 0。IB 和 OCR 阅卷人都看重清晰的逻辑推进和最终回归原不等式的结语。
10. Exam-style Problems and Common Mistakes | 考试题型与常见错误
Let us consolidate a few exam-style pitfalls. (1) Forgetting to flip the inequality sign when multiplying by a negative in a linear inequality – easily avoided by explicitly checking the coefficient sign. (2) Treating rational inequalities as equations by cross-multiplying without considering denominator sign. (3) In quadratic inequalities, writing x < -1 and x > 5 instead of the correct x < -1 or x > 5 for ‘greater than’ cases. The word ‘and’ would imply an impossible intersection.
以下是一些考试阅卷中常见的失分点。(1) 解线性不等式乘以负数时忘记变号——只要明确检查系数正负即可避免。(2) 解分式不等式时直接交叉相乘，导致忽略分母符号。(3) 二次不等式在“大于零”的情况下，错误地写成 x < -1 和 x > 5，而正确的解是 x < -1 或 x > 5。“和”字意味着无解的交集，属于逻辑错误。
Another frequent error is mishandling set notation and interval notation. For IB, you must be comfortable with both, while OCR expects clean number line representations. Always use open intervals ( ) for strict inequalities and closed intervals [ ] when the endpoint is included. When combining intervals, use the union symbol ∪ appropriately. Practise with questions that require changing between {x | x ≤ a} ∪ {x | x > b} and (-∞, a] ∪ (b, ∞).
此外，集合记法与区间记法混淆也是常见问题。IB 要求两种形式熟练转换，OCR 则看重整洁的数轴表示。严格不等式用开区间 ( )，含等号用闭区间 [ ]。合并区间时恰当地使用并集符号 ∪。多做 {x | x ≤ a} ∪ {x | x > b} 与 (-∞, a] ∪ (b, ∞) 的互换练习会很有帮助。
| Mistake | Correction |
|---|---|
| Cross-multiplying in (x-2)/(x+3) < 0 without sign check | Build a single fraction and use a sign table |
| Writing |2x+1| < 5 as 2x+1 < 5 only | Include the other branch: 2x+1 > -5 |
| Not excluding roots of denominator in rational inequalities | Mark them as open circles and remove from solution set |
Finally, when a quadratic has a repeated root, for example (x – 3)² ≤ 0, the only solution is x = 3. Many students incorrectly state x ≤ 3 or something similar. Drill these special cases to avoid dropping marks. Both IB and OCR exams value precise final answers – a small notation error can cost a full mark even if the concept is understood.
最后，当二次式有重根时，比如 (x – 3)² ≤ 0，解只有 x = 3。许多学生错误地写成 x ≤ 3 之类的解集。务必强化这类特殊情况。IB 和 OCR 都强调最终答案的精确性，一个小的符号错误足以扣掉整道题的分，即便你理解相关的概念。
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