📚 IB CCEA Mathematics: Inequalities Exam Focus Guide | IB CCEA 数学：不等式 考点精讲
Inequalities form a fundamental part of the IB and CCEA mathematics curriculum, bridging algebra, functions, and real-world problem-solving. Mastering them is essential for success in both internal assessment and final examinations, as they appear across topics from linear modelling to calculus. This guide unpacks key concepts, graphical interpretations, and strategic approaches to build your confidence in solving any inequality question that may appear on your paper.
不等式是 IB 和 CCEA 数学课程的基本组成部分，连接了代数、函数与现实问题的解决。掌握不等式对于内部评估和期末考试都至关重要，因为它们遍布线性建模到微积分的各个主题。本指南将剖析关键概念、图解解释和解题策略，帮助您建立信心，应对试卷中可能出现的任何不等式问题。
1. Foundations and Inequality Symbols | 基础与不等式符号
Inequalities compare two expressions using symbols: < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to), and ≠ (not equal to). In IB and CCEA exams, you must interpret these precisely, especially when multiplying or dividing by a negative number, which reverses the inequality sign.
不等式使用下列符号比较两个表达式：<（小于）、>（大于）、≤（小于或等于）、≥（大于或等于）和 ≠（不等于）。在 IB 和 CCEA 考试中，您必须精确解读这些符号，特别是当乘以或除以一个负数时，不等号的方向会反转。
For example, solving -2x > 6 yields x < -3, because dividing by -2 flips the sign. Always check whether the variable is isolated correctly and remember that ≤ and ≥ include the boundary point, which matters when representing solutions on a number line or in set notation.
例如，解 -2x > 6 得到 x < -3，因为除以 -2 翻转了符号。始终检查变量是否被正确隔离，并记住 ≤ 和 ≥ 包含边界点，这在数轴或集合表示法中至关重要。
The solution set of an inequality can be expressed in three common ways: inequality notation (e.g., x ≥ 4), number line diagrams with a filled or open circle, and interval notation. Familiarity with all three is expected in examination mark schemes.
不等式的解集可以用三种常见方式表达：不等式表示法（例如 x ≥ 4）、带实心或空心圆的数轴图，以及区间表示法。阅卷方案要求考生熟悉所有这些形式。
2. Linear Inequalities | 线性不等式
Linear inequalities are the simplest type, involving expressions like 3x + 5 ≤ 14. They are solved using the same algebraic steps as linear equations, with the crucial exception of the sign reversal rule when multiplying or dividing by a negative quantity.
线性不等式是最简单的类型，涉及诸如 3x + 5 ≤ 14 的表达式。求解步骤与线性方程相同，关键区别在于乘以或除以负数时需要反转不等号。
In an examination context, always show each step clearly. For the inequality 3x + 5 ≤ 14, subtract 5: 3x ≤ 9, then divide by 3: x ≤ 3. The solution is all real numbers less than or equal to 3. On a number line, place a filled circle at 3 and shade to the left.
在考试中，每一步都要清晰展示。对于不等式 3x + 5 ≤ 14，减去 5：3x ≤ 9，然后除以 3：x ≤ 3。解是所有小于或等于 3 的实数。在数轴上，在 3 处画一个实心圆并向左涂阴影。
When an inequality involves brackets or fractions, expand or clear denominators first. Remember that if you multiply both sides by a variable expression whose sign is unknown, you may need to consider cases. However, in typical linear inequalities you multiply by positive constants.
当不等式包含括号或分数时，首先展开或去分母。请记住，如果两边乘以一个符号未知的变量表达式，可能需要分情况讨论。但在典型线性不等式中，通常乘以正常数。
3. Interval Notation and Number Lines | 区间表示法与数轴
Interval notation provides a concise way to write solution sets. For instance, x > 2 and x ≤ 5 is written as (2, 5]. The round bracket indicates the endpoint is excluded (open circle), while the square bracket indicates inclusion (filled circle). For unbounded intervals, use ∞ or -∞ with round brackets, since infinity is never reached.
区间表示法是一种简洁表示解集的方法。例如，x > 2 且 x ≤ 5 写作 (2, 5]。圆括号表示端点不包含（空心圆），方括号表示包含（实心圆）。对于无界区间，使用 ∞ 或 -∞ 并配以圆括号，因为无穷大永远无法达到。
The table below summarises common interval types that appear in IB and CCEA papers:
下表总结了 IB 和 CCEA 试卷中常见的区间类型：
| Inequality | Interval Notation | Number Line Representation |
|---|---|---|
| x > 3 | (3, ∞) | Open circle at 3, arrow right |
| x ≤ -2 | (-∞, -2] | Filled circle at -2, arrow left |
| -1 < x < 4 | (-1, 4) | Open circles at -1 and 4, line between |
| 2 ≤ x ≤ 5 | [2, 5] | Filled circles at 2 and 5, line between |
Being able to switch fluently between inequality, interval, and graphical representations is a core skill. Many mark schemes allocate marks specifically for the correct use of brackets and shading direction.
能够在不等于、区间和图形表示之间流畅转换是一项核心技能。许多评分方案专门为正确使用括号和阴影方向分配分数。
4. Quadratic Inequalities | 二次不等式
Quadratic inequalities such as x² – 5x + 6 > 0 require a methodical approach. First, treat the related quadratic equation x² – 5x + 6 = 0 to find critical values. Factorising gives (x – 2)(x – 3) = 0, so x = 2 or x = 3. These divide the real number line into three regions: x < 2, 2 < x < 3, and x > 3.
二次不等式如 x² – 5x + 6 > 0 需要系统的方法。首先，处理相关的二次方程 x² – 5x + 6 = 0 以找到临界值。因式分解得到 (x – 2)(x – 3) = 0，所以 x = 2 或 x = 3。这些点将实数轴分成三个区域：x < 2、2 < x < 3 和 x > 3。
Testing a sample point from each region in the original inequality reveals where the expression is positive. For x < 2 (e.g., x=0), (0)² – 5(0) + 6 = 6 > 0, true. For 2 < x < 3 (e.g., x=2.5), (2.5)² – 5(2.5) + 6 = -0.25 < 0, false. For x > 3 (e.g., x=4), 16 – 20 + 6 = 2 > 0, true. Hence the solution is x < 2 or x > 3, written in interval notation as (-∞, 2) ∪ (3, ∞).
在每个区域取一个样本点代入原不等式，可判断表达式何时为正。对于 x < 2（例如 x=0），(0)² – 5(0) + 6 = 6 > 0，成立。对于 2 < x < 3（例如 x=2.5），(2.5)² – 5(2.5) + 6 = -0.25 < 0，不成立。对于 x > 3（例如 x=4），16 – 20 + 6 = 2 > 0，成立。因此解为 x < 2 或 x > 3，写作区间符号 (-∞, 2) ∪ (3, ∞)。
If the inequality had been ≤ 0, the solution would be the interval where the expression is negative or zero: [2, 3]. The shape of the parabola (opening upward because x² coefficient is positive) helps visualise: values above the x-axis satisfy > 0, below satisfy < 0.
如果不等式是 ≤ 0，解将是表达式为负或零的区间：[2, 3]。抛物线的形状（由于 x² 系数为正，开口向上）有助于直观判断：x 轴上方的值满足 > 0，下方的满足 < 0。
5. Polynomial Inequalities of Higher Degree | 高次多项式不等式
For polynomials of degree 3 or higher, such as (x + 1)(x – 2)(x – 4) ≤ 0, the same sign chart method applies. Find all real roots: x = -1, x = 2, x = 4. These are the critical numbers that partition the number line into four intervals. Because each factor is linear with an odd exponent, the sign of the product changes at each root.
对于三次及更高次的多项式不等式，例如 (x + 1)(x – 2)(x – 4) ≤ 0，可采用相同的符号表格法。找出所有实根：x = -1, x = 2, x = 4。这些是临界数，将数轴分成四个区间。因为每个因子都是奇次幂的线性因子，乘积的符号在每个根处都会改变。
Construct a sign table starting from the rightmost interval, x > 4: all factors are positive, product positive. Moving left across x=4, the factor (x-4) changes sign to negative, so product becomes negative for 2 < x < 4. Cross x=2, (x-2) becomes negative, product positive for -1 < x < 2. Cross x=-1, (x+1) becomes negative, product negative for x < -1. Including zeros, the solution to ≤ 0 is (-∞, -1] ∪ [2, 4].
从最右区间 x > 4 开始构建符号表：所有因子为正，乘积为正。向左越过 x=4，因子 (x-4) 变号，所以在 2 < x < 4 乘积为负。越过 x=2，(x-2) 变号，-1 < x < 2 乘积为正。越过 x=-1，(x+1) 变号，x < -1 乘积为负。包含零点，≤ 0 的解为 (-∞, -1] ∪ [2, 4]。
When a factor appears with an even exponent, e.g., (x – 3)², the sign does not change at that root. The CCEA exam often includes such cases to test deeper understanding. Always write the solution in the format required, and double-check boundary inclusions by substituting critical values back into the original inequality.
当因子以偶次幂出现时，例如 (x – 3)²，符号不会在该根处改变。CCEA 考试常常包含此类情形，以考查深刻理解。始终按要求格式书写解，并通过将临界值代回原不等式来仔细检查边界的包含性。
6. Absolute Value Inequalities | 绝对值不等式
Absolute value inequalities like |2x – 1| > 5 are best approached by interpreting the absolute value as distance. The expression |A| > k (with k > 0) means A is more than k units from zero, leading to two separate inequalities: A < -k or A > k. For |A| < k, the distance is less than k, giving -k < A < k.
对于诸如 |2x – 1| > 5 的绝对值不等式，最好将绝对值理解为距离。表达式 |A| > k（k > 0）意味着 A 距离零点超过 k 个单位，从而导出两个独立不等式：A < -k 或 A > k。对于 |A| < k，距离小于 k，得到 -k < A < k。
Applying this to |2x – 1| > 5 gives 2x – 1 < -5 or 2x – 1 > 5. Solve each: 2x < -4 → x < -2; and 2x > 6 → x > 3. The solution set is (-∞, -2) ∪ (3, ∞). Remember to isolate the absolute term first if there are additional constants outside.
将此应用于 |2x – 1| > 5，得到 2x – 1 < -5 或 2x – 1 > 5。分别求解：2x < -4 → x < -2；以及 2x > 6 → x > 3。解集为 (-∞, -2) ∪ (3, ∞)。如果绝对值外部有其他常数，记得首先将绝对值项隔离。
For |ax + b| ≤ c, the solution is a single interval. Graphically, the absolute value function forms a V-shape; the inequality describes the x-values where the V is below a horizontal line. Always check whether the equality is inclusive, as this affects the bracket type.
对于 |ax + b| ≤ c，解是一个单一区间。从图形上看，绝对值函数呈 V 形；不等式描述的是 V 形在水平线下方的 x 值。始终检查等式是否包含，因为这会影响括号类型。
7. Rational Inequalities | 分式不等式
Rational inequalities involve fractions with variables in the denominator, such as (x + 2)/(x – 3) ≥ 0. The critical values are found by setting the numerator and denominator individually to zero: x = -2 and x = 3. Unlike polynomial inequalities, the denominator’s root is never included because division by zero is undefined.
分式不等式涉及分母中含有变量的分数，例如 (x + 2)/(x – 3) ≥ 0。通过分别令分子和分母为零来寻找临界值：x = -2 和 x = 3。与多项式不等式不同，分母的根永远不能包含在内，因为除以零无定义。
Create a sign chart using the intervals (-∞, -2), (-2, 3), and (3, ∞). Test a value in each: for x = -3, (-1)/(-6) > 0, true. For x = 0, (2)/(-3) < 0, false. For x = 4, (6)/(1) > 0, true. Thus the expression is ≥ 0 for x ≤ -2 or x > 3. Note that x = 3 is excluded with a round bracket: solution is (-∞, -2] ∪ (3, ∞).
利用区间 (-∞, -2)、(-2, 3) 和 (3, ∞) 建立符号表。每个区间取一个测试值：x = -3 时，(-1)/(-6) > 0，成立。x = 0 时，(2)/(-3) < 0，不成立。x = 4 时，(6)/(1) > 0，成立。因此当 x ≤ -2 或 x > 3 时表达式 ≥ 0。注意 x = 3 用圆括号排除：解为 (-∞, -2] ∪ (3, ∞)。
Never multiply both sides by the denominator unless you are absolutely certain of its sign, as this can introduce extraneous solutions. The standard method is to make one side zero, combine into a single fraction, and then analyse signs. This is a key concept tested in both IB and CCEA advanced papers.
切勿将两边同乘分母，除非您绝对确定其符号，因为这会引入增根。标准方法是使一边为零，合并成一个分式，然后分析符号。这是 IB 和 CCEA 高级试卷中考查的关键概念。
8. Systems of Linear Inequalities | 线性不等式组
A system of inequalities consists of two or more inequalities that must be satisfied simultaneously. Graphically, the solution is the region where all shadings overlap. For example, y > 2x – 1 and y ≤ -x + 4 represent a half-plane above one line and a half-plane below or on another.
不等式组由两个或更多必须同时满足的不等式组成。从图形上看，解是所有阴影区域重叠的部分。例如，y > 2x – 1 和 y ≤ -x + 4 分别表示一条直线上方的半平面和另一条直线下方或线上的半平面。
To sketch the region, draw the boundary lines. Use a dashed line for strict inequalities (< or >) and a solid line for ≤ or ≥. Shade the appropriate side of each line, and the feasible region is the intersection. Label any vertices of the region, as they are often required in linear programming problems.
绘制区域时，先画出边界线。严格不等式（< 或 >）使用虚线，≤ 或 ≥ 使用实线。对每条线的适当一侧涂阴影，可行区域即为交集。标出区域的任何顶点，因为线性规划问题中经常需要这些点。
CCEA questions frequently combine linear inequalities with constraints from real-life contexts, such as production limits or budget boundaries. You will need to form the inequalities from word descriptions, graph them accurately on a Cartesian plane, and identify the solution set, sometimes using integer coordinates.
CCEA 试题经常将线性不等式与现实情境约束相结合，如生产限制或预算边界。您需要根据文字描述构建不等式，在笛卡尔平面上精确作图，并识别解集，有时还需使用整数坐标。
9. Graphical Representation of Inequalities in Two Variables | 二元不等式的图形表示
Extending to quadratic curves, an inequality like y < x² – 4 defines a region below a parabola. The boundary is the parabola itself, drawn as a dashed curve because the inequality is strict. Choose a test point, often (0,0), to decide which side to shade: 0 < 0² – 4 is false, so shade the region that does not contain the origin.
扩展到二次曲线，诸如 y < x² – 4 的不等式定义了抛物线下方的一个区域。边界为抛物线本身，因不等式严格而画为虚线曲线。选择一个测试点，通常是 (0,0)，以确定哪一侧要涂阴影：0 < 0² – 4 为假，因此对不含原点的区域涂阴影。
When multiple curves are involved, such as y ≥ x² and x² + y² ≤ 9, the solution is the overlap of the region above the parabola and the interior of a circle of radius 3 centred at the origin. Use different shading directions or colours in rough work to avoid confusion, and clearly indicate the final answer region.
当涉及多条曲线时，例如 y ≥ x² 和 x² + y² ≤ 9，解是抛物线上方区域与以原点为圆心、半径为 3 的圆内部的交集。在草稿中使用不同方向的阴影或不同颜色以避免混淆，并清楚地标出最终答案区域。
In IB examinations, you may be asked to write a system of inequalities that describes a given shaded figure. Analyse each boundary line or curve, determine its equation, and test a point in the shaded region to set the inequality sign correctly.
在 IB 考试中，可能会要求写出一组描述给定阴影图形的不等式。分析每条边界线或曲线，确定其方程，并在阴影区域内测试一个点，以正确设置不等号。
10. Inequalities Involving Exponential and Logarithmic Functions | 涉及指数与对数函数的不等式
When inequalities involve exponentials like 2ˣ > 8, express both sides with the same base if possible: 2ˣ > 2³, and since the base is greater than 1, the inequality sign is preserved when comparing exponents: x > 3. For 0 < base < 1, the inequality direction reverses, because the function is decreasing.
当不等式涉及指数如 2ˣ > 8 时，尽可能将两边表示为同底数：2ˣ > 2³，由于底数大于 1，比较指数时不等号方向保持不变：x > 3。当 0 < 底数 < 1 时，由于函数递减，不等号方向反转。
Logarithmic inequalities, such as log₂(x – 1) ≤ 3, first require the argument to be positive: x > 1. Then rewrite in exponential form: x – 1 ≤ 2³ = 8, giving x ≤ 9. Combining, the solution is 1 < x ≤ 9. Always state the domain restrictions explicitly, as marks are allocated for them.
对数不等式如 log₂(x – 1) ≤ 3，首先要求真数为正：x > 1。然后重写为指数形式：x – 1 ≤ 2³ = 8，得到 x ≤ 9。联立得解为 1 < x ≤ 9。务必明确写出定义域限制，因为评分标准中有相应分值。
These transcend inequalities appear less frequently but are highly discriminating. Practice with bases e and 10, and remember that when taking logs of both sides of an inequality, you must ensure both sides are positive. Alternatively, use the monotonicity of the exponential or logarithmic function to justify the step.
此类超越不等式虽出现频率较低，但区分度极高。练习以 e 和 10 为底的不等式，并记住对不等式两边取对数时，必须确保两边均为正。或者，利用指数或对数函数的单调性来证明步骤合理。
11. Common Mistakes and Examination Strategies | 常见错误与考试策略
A frequent error is forgetting to reverse the inequality sign when multiplying or dividing by a negative. Another is squaring both sides of an inequality without considering the signs of both expressions, which can produce extraneous solutions. In rational inequalities, students often include the denominator’s root in the solution set, leading to an undefined expression.
一个常见错误是在乘以或除以负数时忘记反转不等号。另一个错误是在不考虑两边表达式符号的情况下对不等式两边平方，这可能产生增根。在分式不等式中，学生常常将分母的根包含在解集中，从而得到无定义的表达式。
In the exam, always start by clearly defining the domain of the variable if fractions, roots, or logarithms are present. Show your sign charts or test-point reasoning step by step. If asked to represent on a number line, draw it neatly with a ruler, and use the correct open or filled circle. Mismanagement of brackets in interval notation is a common source of lost marks.
在考试中，如果存在分式、根式或对数，务必首先清晰定义变量的定义域。逐步展示您的符号表或试点推理。如果要求在数轴上表示，要用直尺整齐画出，并使用正确的空心或实心圆。区间表示法中括号的错误使用是失分的常见原因。
When graphing inequalities, clearly label intercepts and intersection points. If a question provides a grid, use a pencil and ensure boundaries are accurate. Time management: linear and quadratic inequality questions are generally straightforward; spend more time on rational or absolute value ones, which carry more marks.
当绘制不等式图形时，清楚标注截距和交点。如果题目提供坐标网格，用铅笔绘图并确保边界准确。时间管理：线性和二次不等式问题通常较直接；在分式或绝对值不等式上多花时间，因为它们分值更高。
12. Exam-style Question Walkthrough | 典型考题详解
Question: Solve the inequality (x² – 4)(x + 1) < 0.
题目：解不等式 (x² – 4)(x + 1) < 0。
Step 1: Factorise completely: (x – 2)(x + 2)(x + 1) < 0. Critical values are x = -2, -1, 2. They divide the line into intervals: (-∞, -2), (-2, -1), (-1, 2), (2, ∞).
第 1 步：完全因式分解：(x – 2)(x + 2)(x + 1) < 0。临界值为 x = -2、-1、2。它们将数轴分成区间：(-∞, -2)、(-2, -1)、(-1, 2)、(2, ∞)。
Step 2: Test a point in each interval to find the sign of the product. For x = -3: (-)(-)(-) = -, negative, satisfies < 0. For x = -1.5: (-)(+)(-) = +, does not satisfy. For x = 0: (-)(+)(+) = -, satisfies. For x = 3: (+)(+)(+) = +, does not satisfy.
第 2 步：在每个区间测试一点以确定乘积符号。x = -3 时：(-)(-)(-) = -，负，满足 < 0。x = -1.5 时：(-)(+)(-) = +，不满足。x = 0 时：(-)(+)(+) = -，满足。x = 3 时：(+)(+)(+) = +，不满足。
Step 3: The solution is where the product is negative: (-∞, -2) ∪ (-1, 2). None of the endpoints are included because the inequality is strict. Check: at x = -2, the expression is zero, which is not < 0. Final answer: x < -2 or -1 < x < 2.
第 3 步：解为乘积为负的区间：(-∞, -2) ∪ (-1, 2)。端点均不包含，因为是不严格不等式。检查：x = -2 时表达式为零，不满足 < 0。最终答案：x < -2 或 -1 < x < 2。
Always present your final answer clearly, using the format requested. If the question does not specify, both inequality and interval forms are acceptable, but be consistent. This systematic approach will secure full marks on polynomial inequality questions.
始终按要求格式清楚呈现最终答案。如果题目未指定，不等式和区间两种形式均可接受，但要一致。这种系统方法将确保在多项式不等式题目上获得满分。
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