Solving Modulus Problems | 解模问题

📚 Solving Modulus Problems | 解模问题

In A-level mathematics, the modulus function, also known as the absolute value, is a fundamental concept that appears in equations, inequalities, and graphs. Mastering modulus problems is essential for success in the Edexcel exams, as these questions test your algebraic manipulation, logical reasoning, and graphical interpretation skills.

在A-level数学中，模函数（即绝对值）是方程、不等式和图形中的基础概念。掌握解模问题对于在Edexcel考试中取得好成绩至关重要，因为这些问题考查你的代数运算、逻辑推理和图形理解能力。

1. Definition of the Modulus Function | 模函数的定义

The modulus of a number x, denoted as |x|, is its distance from zero on the number line, always non-negative. Formally, |x| = x if x ≥ 0, and |x| = -x if x < 0.

一个数x的模，记为|x|，是它在数轴上到零的距离，始终非负。正式定义：当x ≥ 0时，|x| = x；当x < 0时，|x| = -x。

Key properties: |x| ≥ 0 for all x, |x|² = x², and |x| = √(x²). These properties are often used to transform modulus equations into standard algebraic forms.

关键性质：对于所有x，|x| ≥ 0；|x|² = x²；以及|x| = √(x²)。这些性质常用来将模方程转换为标准代数形式。

|x| = x (if x ≥ 0), |x| = -x (if x < 0)

|x| = x（当x ≥ 0），|x| = -x（当x < 0）

2. Solving Basic Modulus Equations | 解基本模方程

For a simple equation |x| = a, where a > 0, the solutions are x = a or x = -a. If a = 0, the only solution is x = 0. If a < 0, there are no solutions because the modulus is always non-negative.

对于简单方程|x| = a，其中a > 0，解为x = a或x = -a。如果a = 0，唯一解是x = 0。如果a < 0，方程无解，因为模始终非负。

When solving |ax + b| = c, you must consider two cases: ax + b = c and ax + b = -c. Always check your solutions in the original equation to avoid extraneous roots.

当解|ax + b| = c时，必须考虑两种情况：ax + b = c和ax + b = -c。务必检查解是否满足原方程，以避免增根。

|ax + b| = c ⇒ ax + b = c or ax + b = -c

|ax + b| = c ⇒ ax + b = c 或 ax + b = -c

3. Solving Modulus Inequalities (|x| < a and |x| > a) | 解模不等式（|x| < a 和 |x| > a）

For an inequality |x| < a with a > 0, the solution is -a < x < a, which represents the interval between -a and a. This is often written as x ∈ (-a, a).

对于不等式|x| < a（其中a > 0），解为-a < x < a，表示介于-a和a之间的区间，常记为x ∈ (-a, a)。

For |x| > a, the solution is x < -a or x > a, meaning x lies outside the interval. Graphically, |x| < a corresponds to the region between the two lines y = -a and y = a on the x-axis.

对于|x| > a，解为x < -a或x > a，表示x位于区间之外。图像上，|x| < a对应于x轴上y = -a和y = a两条线之间的区域。

|x| < a ⇔ -a < x < a; |x| > a ⇔ x < -a or x > a

|x| < a ⇔ -a < x < a；|x| > a ⇔ x < -a 或 x > a

4. Solving |ax + b| = c and |ax + b| < c | 解|ax + b| = c 和 |ax + b| < c

To solve |ax + b| = c, first ensure c ≥ 0, then write two equations: ax + b = c and ax + b = -c, solving each for x. For example, |2x – 3| = 5 gives 2x – 3 = 5 → x = 4, or 2x – 3 = -5 → x = -1.

要解|ax + b| = c，首先确保c ≥ 0，然后写出两个方程：ax + b = c和ax + b = -c，分别解出x。例如，|2x – 3| = 5可得2x – 3 = 5 → x = 4，或2x – 3 = -5 → x = -1。

For |ax + b| < c, where c > 0, use the property -c < ax + b < c. This is equivalent to solving two linear inequalities simultaneously. Then isolate x to find the range.

对于|ax + b| < c（c > 0），利用性质-c < ax + b < c，这等价于同时求解两个线性不等式。然后分离x求出范围。

|ax + b| < c ⇔ -c < ax + b < c

5. Graphical Interpretation of Modulus | 模的图形解释

The graph of y = |f(x)| is obtained by reflecting the part of y = f(x) that lies below the x-axis in the x-axis. For y = |x|, the graph is a V-shape with the vertex at the origin.

函数y = |f(x)|的图像是通过将y = f(x)位于x轴下方的部分关于x轴反射得到的。对于y = |x|，图像是一个顶点在原点的V形。

When solving modulus equations or inequalities graphically, sketch both sides of the equation. The x-coordinates of the intersection points of y = |f(x)| and y = g(x) are the solutions to |f(x)| = g(x).

用图形法解模方程或不等式时，画出方程两边的图像。y = |f(x)|和y = g(x)交点的x坐标就是方程|f(x)| = g(x)的解。

Example: To solve |x – 1| = 2, sketch y = |x – 1| and y = 2. The intersections are at x = -1 and x = 3. This visual method is especially useful for checking answers.

例如：解|x – 1| = 2时，画出y = |x – 1|和y = 2。交点位于x = -1和x = 3。这种视觉方法特别适合用于检验答案。

6. Solving |f(x)| = g(x) | 解|f(x)| = g(x)

To solve |f(x)| = g(x), first note that g(x) must be ≥ 0 for any solution to exist. Then split into two cases: f(x) = g(x) and f(x) = -g(x). Solve each equation and only accept solutions that satisfy the original condition.

要解|f(x)| = g(x)，首先注意g(x)必须≥ 0才可能有解。然后分为两种情况：f(x) = g(x)和f(x) = -g(x)。分别解方程，只接受满足原条件的解。

For example, solve |x + 2| = 3x. First, 3x ≥ 0 ⇒ x ≥ 0. Case 1: x + 2 = 3x → x = 1. Case 2: x + 2 = -3x → x = -0.5. Since x ≥ 0, only x = 1 is valid.

例如，解|x + 2| = 3x。首先，3x ≥ 0 ⇒ x ≥ 0。情况1：x + 2 = 3x → x = 1。情况2：x + 2 = -3x → x = -0.5。由于x ≥ 0，只有x = 1有效。

A common mistake is ignoring the condition on g(x). Always check the domain at the start.

常见错误是忽视g(x)的条件。开始时要始终检查定义域。

7. Solving |f(x)| > g(x) | 解|f(x)| > g(x)

For an inequality of the form |f(x)| > g(x), the solution set is the union of solutions to f(x) > g(x) and f(x) < -g(x). This is because |f(x)| > g(x) means f(x) is either greater than g(x) or less than -g(x).

对于形式为|f(x)| > g(x)的不等式，解集是f(x) > g(x)和f(x) < -g(x)解集的并集。因为|f(x)| > g(x)意味着f(x)要么大于g(x)，要么小于-g(x)。

Unlike equations, here g(x) can be negative. If g(x) < 0, then |f(x)| > g(x) is always true since |f(x)| ≥ 0. In such cases, the solution is all real x.

与方程不同，这里的g(x)可以为负。如果g(x) < 0，则|f(x)| > g(x)总是成立，因为|f(x)| ≥ 0。这种情况下，解为所有实数x。

|f(x)| > g(x) ⇔ f(x) > g(x) or f(x) < -g(x)

|f(x)| > g(x) ⇔ f(x) > g(x) 或 f(x) < -g(x)

8. Solving |f(x)| < g(x) | 解|f(x)| < g(x)

For |f(x)| < g(x), the inequality is equivalent to -g(x) < f(x) < g(x). This is a double inequality that can be solved by splitting into two inequalities: f(x) < g(x) and f(x) > -g(x).

对于|f(x)| < g(x)，不等式等价于-g(x) < f(x) < g(x)。这是一个双重不等式，可以通过将其拆分为两个不等式来求解：f(x) < g(x)和f(x) > -g(x)。

Note that g(x) must be > 0 for this to have solutions. If g(x) ≤ 0, then |f(x)| < g(x) has no solution, because |f(x)| is always non-negative.

注意g(x)必须> 0才可能有解。如果g(x) ≤ 0，则|f(x)| < g(x)无解，因为|f(x)|始终非负。

|f(x)| < g(x) ⇔ -g(x) < f(x) < g(x)

9. Composite Modulus Functions | 复合模函数

Problems with nested modulus, such as | |x| – 1 | = 2, require careful case analysis. First solve the outer modulus by splitting into |x| – 1 = 2 and |x| – 1 = -2.

涉及嵌套模的问题，例如| |x| – 1 | = 2，需要仔细分类讨论。首先通过拆分外层模求解：|x| – 1 = 2和|x| – 1 = -2。

From the first equation, |x| = 3, giving x = 3 or x = -3. From the second, |x| = -1, which has no solution. So the final answer is x = ±3.

由第一个方程，|x| = 3，得到x = 3或x = -3。由第二个方程，|x| = -1，无解。因此最终答案为x = ±3。

For composite inequalities, apply the rules step by step, always maintaining the direction of the inequality. Drawing a number line helps visualise the solution set.

对于复合不等式，逐步应用规则，始终保持不等号方向。绘制数轴有助于可视化解集。

10. Modulus in Parameter Problems | 参数问题中的模

Problems with parameters, such as |2x – 3| = k, ask you to describe the number of solutions based on k. Since |2x – 3| ≥ 0, if k < 0, no solution; if k = 0, one solution (x = 1.5); if k > 0, two solutions.

涉及参数的问题，例如|2x – 3| = k，要求你根据k描述解的个数。因为|2x – 3| ≥ 0，所以若k < 0，无解；若k = 0，一个解（x = 1.5）；若k > 0，两个解。

In general, for |f(x)| = k, the number of solutions depends on the range of f(x). Sketch y = |f(x)| and y = k to see how many intersection points exist for different values of k.

一般来说，对于|f(x)| = k，解的个数取决于f(x)的值域。画出y = |f(x)|和y = k，观察不同k值下的交点个数。

This type of question often appears in exam as a graph question. Practice identifying critical k values where the number of solutions changes.

这类问题在考试中常以图形题出现。练习识别使解个数发生变化的临界k值。

11. Common Mistakes and Tips | 常见错误与技巧

One major mistake is squaring both sides of an equation or inequality without checking extraneous solutions. For example, |x + 1| = 2x, squaring gives (x + 1)² = (2x)², which may produce invalid negative x values.

一个主要错误是不检查增根就直接对方程或不等式两边平方。例如，|x + 1| = 2x，平方得(x + 1)² = (2x)²，可能产生无效的负x值。

Another common error is forgetting that |x| < a has a single interval, while |x| > a has two separate intervals. Be careful when writing the final solution set.

另一个常见错误是忘记|x| < a是一个区间，而|x| > a是两个分离的区间。写最终解集时要小心。

Tips: Always check solutions by substitution; use graphs for verification; and for inequalities, test points in each interval to confirm the solution.

技巧：始终通过代入检验解；使用图形进行验证；对于不等式，在每个区间内测试点以确认解。

12. Practice and Summary | 练习与总结

To master modulus problems, practice with past paper questions and focus on the transition between algebraic and graphical representations. Review the standard cases: |x| = a, |x| < a, |x| > a, and their composite forms.

要掌握模问题，请练习历年真题，并重点关注代数与图形表示之间的转换。复习标准情形：|x| = a、|x| < a、|x| > a及其复合形式。

Remember the core principle: treat the modulus as a piecewise function. Break the problem into intervals based on critical points where the expression inside the modulus changes sign.

记住核心原则：将模视为分段函数。根据模内部表达式符号变化的临界点，将问题分解为多个区间。

With consistent practice, solving modulus problems becomes straightforward and a reliable way to earn marks in Edexcel mathematics.

通过持续练习，解模问题将变得简单直接，成为在Edexcel数学中稳定得分的方式。

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Solving Modulus Problems | 解模问题