The Modulus Function | 绝对值函数

📚 The Modulus Function | 绝对值函数

The modulus function, denoted by |x|, is a fundamental concept in A-Level Mathematics that measures the distance of a number from zero on the real number line. It appears frequently in equations, inequalities, graph sketching, and transformations, and it is also the key to understanding absolute error in mechanics and statistics. Mastering its properties and applications is essential for success in Edexcel, AQA, OCR and Cambridge examinations.

模函数（绝对值函数），记作 |x|，是 A-Level 数学中的基础概念，用于衡量实数在数轴上到零的距离。它频繁出现在方程、不等式、图像描绘以及变换中，同时也是理解力学与统计中绝对误差的关键工具。掌握其性质和应用，对于在 Edexcel、AQA、OCR 与剑桥考试中取得好成绩至关重要。

This revision guide rebuilds the topic from the definition upwards, then moves through properties, geometry, equations, inequalities, graphs, calculus and exam technique. Every section contains paired English and Chinese explanations, worked results and examiner-style warnings.

本篇复习指南从定义出发逐层递进，依次讲解性质、几何意义、方程、不等式、图像、微积分以及应试技巧。每一小节都提供中英对照讲解、推导结果与考官式提醒。

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1. Definition and Notation | 定义与符号

The modulus of a real number x, written as |x|, is defined as the non-negative value of x. Algebraically, it is given by |x| = x if x ≥ 0 and |x| branches together cover every real number, so the definition is complete and unambiguous.

实数 x 的模，写作 |x|，定义为 x 的非负值。代数上，当 x ≥ 0 时 |x| = x；当 x < 0 时 |x| = −x。这两个分支合起来覆盖了全体实数，因此定义完整且无歧义。

This piecewise definition reflects the geometric idea of "distance": |3| = 3 and |−3| ive, −x is positive, which is exactly why the modulus can never produce a negative output.

这种分段定义体现了“距离”的几何思想：|3| = 3 与 |−3| = 3 都表示到原点的距离为 3 个单位。请注意，−x 并不一定表示负数；当 x 为负时，−x 反而是正数，这正是模永远不会输出负值的原因。

r words, the graph never dips below the x-axis, and every non-negative output is attained.

模函数有时也被称为绝对值函数，其定义域为全体实数，值域为 [0, ∞)。换言之，其图像永远不会落到 x 轴下方，并且每一个非负输出值都能被取到。

Useful evaluation examples: |−7| = 7, |0| = 0, |2.5| = 2.5, |π − 4| = 4 − π, and |1 − √2| = √2 − 1. In each case you first decide the sign of the expression inside, then either keep it or negate it.

常用求值示例：|−7| = 7，|0| = 0，|2.5| = 2.5，|π − 4| = 4 − π，|1 − √2| = √2 − 1。每一步都是先判断内部表达式的正负，再决定保留原式还是取相反数。

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2. Key Properties | 基本性质

Several algebraic properties stem directly from the definition. Understanding them helps to manipulate modulus expressions efficiently and avoids the algebraic traps that examiners deliberately build into questions.

由定义可直接推导出若干代数性质。理解它们有助于高效处理模表达式，并避开考官刻意设置的代数陷阱。

Property 1 - Non-negativity: |x| ≥ 0 for all real x, and |x| = 0 if and only if x = 0. This is why an equation such as |x − 2| = −5 has no solutions at all.

性质 1：非负性：对所有实数 x，|x| ≥ 0，且 |x| = 0 当且仅当 x = 0。因此像 |x − 2| = −5 这样的方程根本无解。

Property 2 - Symmetry: |−x| = |x|. The function is even, so the graph is symmetric about the y-axis. Equivalently, |x| = |a| implies x = a or x = −a.

性质 2：对称性：|−x| = |x|。该函数为偶函数，因此图像关于 y 轴对称。等价地，|x| = |a| 意味着 x = a 或 x = −a。

Property 3 - Multiplicative property: |ab| = |a|·|b| for any real numbers a and b. This extends to products of any length, for example |abc| = |a|·|b|·|c|.

性质 3：乘法性质：对任意实数 a、b，有 |ab| = |a|·|b|。该性质可推广到任意多个因子的乘积，例如 |abc| = |a|·|b|·|c|。

Property 4 - Quotient property: |a ÷ b| = |a| ÷ |b| d b ≠ 0. Division behaves just like multiplication, but the denominator may never be zero.

性质 4：除法性质：当 b ≠ 0 时，|a ÷ b| = |a| ÷ |b|。除法的表现与乘法一致，但分母绝不能为零。

Property 5 - Squaring: |x|² = x², and therefore √x² = |x| s many modulus equations into ordinary quadratics and is the basis of the squaring method.

性质 5：平方关系：|x|² = x²，因此 √x² = |x|。这一恒等式可以把许多模方程转化为普通二次方程，是“平方解法”的理论基础。

Property 6 - Triangle inequality: |a + b| ≤ |a| + |b| equality only when a and b have the same sign or one of them is zero. Note carefully that |a + b| is generally NOT equal to |a| + |b|.

性质 6：三角不等式：|a + b| ≤ |a| + |b|，当且仅当 a 与 b 同号或其中一个为零时取等号。务必注意，一般地 |a + b| 并不等于 |a| + |b|。

Property 7 - Bounding property: |x| ≥ x and |x| ≥ −x for every real x, so −|x| ≤ x ≤ |x|. This double inequality is what justifies rewriting |x| < a as −a < x < a.

性质 7：界性质：对任意实数 x，有 |x| ≥ x 且 |x| ≥ −x，即 −|x| ≤ x ≤ |x|。这个双边不等式正是把 |x| < a 改写为 −a < x < a 的依据。

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3. Modulus as Distance on the Number Line | 数轴上的距离解释

The expression |x − a| has a precise geometric meaning: it is the distance between the point x and the point a on the number line. This interpretation turns abstract algebra into a picture you can read off in seconds.

表达式 |x − a| 有明确的几何含义：它是数轴上点 x 与点 a 之间的距离。这一解释把抽象代数转化为几秒钟就能看懂的图形。

For example, |x − 5| < 2 means "the distance from x to 5 is less than 2", so x must lie strictly between 3 and 5. Similarly, |x + 3| ≥ 4 means the distance from x to −3 is at least 4, giving x ≤ −7 or x ≥ 1.

例如，|x − 5| < 2 表示“x 到 5 的距离小于 2”，因此 x 必须严格落在 3 与 5 之间。同理，|x + 3| ≥ 4 表示 x 到 −3 的距离至少为 4，于是 x ≤ −7 或 x ≥ 1。

distance statements are worth memorising because they appear in almost every paper. First, |x − a| = |x − b| means x is equidistant from a and b, so x is the midpoint (a + b) ÷ 2. Second, |x − a| = k asks for the two points at distance k on either side of a, namely a − k and a + k.

有两个标准距离结论值得背下来，因为它们几乎出现在每一份试卷中。第一，|x − a| = |x − b| 表示 x 到 a 与到 b 的距离相等，故 x 为 a 与 b 的中点 (a + b) ÷ 2。第二，|x − a| = k 求的是位于 a 两侧、距离为 k 的两个点，即 a − k 与 a + k。

In mechanics and statistics, the same idea appears as absolute error. If a measured value is x and the true value is a, then |x − a| is the magnitude of the error, independent of whether the measurement is too large or too small.

在力学与统计中，同一思想以“绝对误差”的形式出现。若测量值为 x、真实值为 a，则 |x − a| 即为误差的大小，与测量值偏大还是偏小无关。

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4. Solving Modulus Equations | 解绝对值方程

The core technique for |f(x)| Solve each case separately, then collect all solutions. If k < 0, the answer is simply "no solutions".

解 |f(x)| = k（k ≥ 0）的核心技巧是分成两种情形：f(x) = k 或 f(x) = −k。分别求解后再汇总所有解。若 k < 0，答案直接为“无解”。

When both sides contain a modulus, use |f(x)| = |g(x)| ⟺ = g(x) or f(x) = −g(x). This is faster than squaring and usually produces cleaner numbers.

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当两边都含模时，使用 |f(x)| = |g(x)| ⟺ f(x) = g(x) 或 f(x) = −g(x)。这比平方更快，且通常得到更整洁的数据。

gitimate here only because both sides are guaranteed non-negative, but you must still verify each root in the original equation.

另一种方法是平方解法：两边平方得到 (f(x))² = (g(x))²，再因式分解为 (f(x) − g(x))(f(x) + g(x)) = 0。此处平方之所以合法，是因为两边都保证非负，但仍需将每个根代回原方程检验。

Worked example: solve |2x − 5| = 7. Case 1: 2x − 5 = 7 gives x = 6. Case 2: 2x − 5 = −7 gives x = −1. Check: |2(6) − 5| = 7 and |2(−1) − 5| = |−7| = 7. So x = 6 or x = −1.

例题：解 |2x − 5| = 7。情形 1：2x − 5 = 7，得 x = 6。情形 2：2x − 5 = −7，得 x = −1。检验：|2(6) − 5| = 7，|2(−1) − 5| = |−7| = 7。故 x = 6 或 x = −1。

When the unknown appears outside the modulus, extra care is needed. For instance, |x − 1| 0, that is x ≥ 1.5, before any case work begins, because the left side is never negative.

当未知数出现在模之外时须格外小心。例如 |x − 1| = 2x − 3 在开始分类讨论前就必须要求 2x − 3 ≥ 0，即 x ≥ 1.5，因为左边永不为负。

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5. Solving Modulus Inequalities | 解绝对值不等式

The two master rules are: |x| < a ⟺ −a < x < a (a "sandwich" or intersection), and |x| > a ⟺ x < −a or x > a (a "split" or union). These hold for a > 0.

两条总纲式规则是：|x| < a ⟺ −a < x < a（“夹逼”或取交集），以及 |x| > a ⟺ x < −a 或 x > a（“分裂”或取并集）。二者均要求 a > 0。

For expressions like |ax + b| c and solve the three-part inequality simultaneously. This keeps the working compact and reduces sign errors.

对于 |ax + b| ≤ c 这类表达式，把规则中的 x 换成 ax + b：写成 −c ≤ ax + b ≤ c，并同步求解这个三部分不等式。这样书写紧凑，也能减少符号错误。

Worked example: solve |3x − 1| 3 or x ≥ 2. On a number line the solution is two rays pointing outwards, which matches the "greater than" shape.

例题：解 |3x − 1| ≥ 5。分裂为 3x − 1 ≤ −5 或 3x − 1 ≥ 5，得 x ≤ −4 ÷ 3 或 x ≥ 2。在数轴上解集是向两侧延伸的两条射线，正好符合“大于型”的形状。

Worked example: solve |x + 4| < 3. Sandwich: −3 < x + 4 < 3, so −7 < x < −1. The solution is a single interval, which matches the "less than" shape.

例题：解 |x + 4| < 3。夹逼：−3 < x + 4 < 3，故 −7 < x < −1。解集为单一区间，正好符合“小于型”的形状。

If the modulus inequality has the unknown on both sides, for example |x − 3| < 2x, the safest route is a critical-value analysis: find where the expression inside the modulus changes sign, then test one point in each interval.

若不等式两边都含未知数，例如 |x − 3| < 2x，最稳妥的做法是临界值分析：先找出模内表达式变号的位置，然后在每个区间中各取一个测试点。

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6. Graphs of Modulus Functions | 模函数的图像

The graph of y = |x| h is the line y = x with gradient +1; for x < 0 it is the line y = −x with gradient −1.

y = |x| 的图像是经典的 V 形，顶点在原点，关于 y 轴对称。当 x ≥ 0 时图像是斜率为 +1 的直线 y = x；当 x < 0 时是斜率为 −1 的直线 y = −x。

Shifting horizontally gives y = |x − a| with vertex at (a, 0), while shifting vertically gives y = |x| + b with vertex at (0, b). Combining reads as y = |x − a| + b with vertex (a, b).

水平平移得 y = |x − a|，顶点为 (a, 0)；竖直平移得 y = |x| + b，顶点为 (0, b)。二者结合写作 y = |x − a| + b，顶点为 (a, b)。

The transformation y = |f(x)| (x) that lies above the x-axis and reflects any part below the x-axis upwards. This is the "modulus outside" rule and it is the one used most often in graph questions.

变换 y = |f(x)| 保留 y = f(x) 在 x 轴以上的部分，并把 x 轴以下的部分翻折向上。这就是“外模”规则，也是图像题中最常用的规则。

The transformation y = f(|x| aph that is always symmetric about the y-axis, whatever the original function looked like.

变换 y = f(|x|) 则不同：保留右侧分支（x ≥ 0），再关于 y 轴反射得到左侧分支。无论原函数是什么形状，所得图像一定关于 y 轴对称。

ertex of the quadratic, draw the parabola, then fold the negative section upwards. For y = |x² − 4| −4); after folding, the minimum points become (0, 4) and the curve touches zero at x = ±2.

描绘技巧：若模内是二次函数，先求其根与顶点，画出抛物线，再把位于 x 轴下方的部分向上翻折。对于 y = |x² − 4|，抛物线在 x = −2 与 x = 2 处与 x 轴相交，顶点为 (0, −4)；翻折后最低点变为 (0, 4)，并在 x = ±2 处取零。

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7. Transformations Involving Modulus | 含模的变换

ify whether the modulus is applied outside or inside. Write the function in the form y = a|b(x − c)| + d to read off the stretches and translations.

模函数图像的变换与普通函数规则一致，关键在于判断模是加在外面还是里面。把函数写成 y = a|b(x − c)| + d 的形式，即可直接读出伸缩与平移量。

a < 0 the whole graph is reflected in the x-axis, which for a modulus function produces a shape opening downwards.

竖直方向伸缩 a 倍会把 V 形的斜率从 ±1 变为 ±a。若 a < 0，整个图像关于 x 轴反射，对模函数而言会得到开口向下的形状。

Compare y = |x − 3| and y = |x| − 3. The first is a translation of y = |x| vertex (0, −3). Students routinely confuse the two, so always state the vertex explicitly.

请比较 y = |x − 3| 与 y = |x| − 3。前者是 y = |x| 向右平移 3 个单位，顶点为 (3, 0)；后者是向下平移 3 个单位，顶点为 (0, −3)。学生常把两者混淆，因此务必明确写出顶点坐标。

Number of solutions from a graph: rewriting an equation as |f(x)| h graphs lets you count intersections instantly. This is often the fastest route through a "show that there are exactly two solutions" question.

用图像数解的个数：把方程改写为 |f(x)| = g(x)，同时画出两条曲线，即可立刻数出交点个数。在“证明恰有两个解”这类题目中，这往往是最快的路径。

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8. Equations with Two Modulus Expressions | 含两个绝对值表达式的方程

Equations such as |x − 1| + |x − 4| unless you first locate the critical values where each modulus changes sign. Those critical values split the number line into intervals.

像 |x − 1| + |x − 4| = 5 这样的方程，必须先找出每个模变号的临界值，再进行分类讨论。这些临界值把数轴分割成若干区间。

For |x − 1| + |x − 4| interval both moduli can be removed with a fixed sign, producing a simple linear equation.

对 |x − 1| + |x − 4| 而言，临界值为 x = 1 与 x = 4，形成三个区间：x < 1、1 ≤ x ≤ 4 与 x > 4。在每个区间内两个模的符号固定，可化为简单的一次方程。

≤ x ≤ 4 the sum is constantly 3, so no solution arises. On x > 4 we get (x − 1) + (x −

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