Modulus Functions, Properties of Modulus Function

Functions are one of the basic concepts in mathematics that have numerous applications in the real world. Be it mega skyscrapers or super-fast cars, their modeling requires practical application of functions. Almost all real-world problems are formulated, interpreted, and solved using functions. Modulus functions help in solving modulus inequalities.

In this article, we will cover the concepts of the modulus function. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Modulus Function

The modulus function can be defined as a function $f: R \rightarrow R$ defined by $f(x)=|x|$ for each $x \in R$.

For each non-negative value of $x, f(x)$ is equal to $x$. But for negative values of $x$, the value of $f(x)$ is the negative of the value of $x$

$|\mathrm{x}|, \mathrm{x} \in \mathbb{R}=\left\{\begin{array}{cc}x, & x \geq 0 \\ -x, & x<0\end{array}\right.$

Range $\in[0, \infty)$

What is Modulus Function?

A modulus function gives the magnitude of a number irrespective of its sign.It also gives us the output as negative if the value of function that we input is less that zero. Hence it is also called the absolute value function. In mathematics, the modulus of a real number $x$ is given by the modulus function, denoted by $|x|$.

$|x|$ is the modulus of $x$, where $x$ is a real number. If $x$ is non-negative then $f(x)$ will be of the same value $x$. If $x$ is negative, then $f(x)$ will be the magnitude of $x$, that is, $\mathrm{f}(\mathrm{x})=-\mathrm{x}$ if $x$ is negative.

Modulus Function Graph

We see how to plot the graph for a modulus function. Let us consider $x$ to be a variable, taking values from $-5$ to $5$. Calculating modulus for the positive values of ' $x$ ', the line plotted in the graph is ' $y=x$ ' and for the negative values of ' $x$ ', the line plotted in the graph is ' $y=-x^{\prime}$.

Modulus Equations: Properties
If $a>0$
1. $|x|=a$, then $x=a,-a$
2. $|x|=|-x|$
3. $|x|^2=x^2$
4. If $|\mathrm{x}|=\mathrm{x}$, then $\mathrm{x}>0$ or $\mathrm{x}=0$
5. If $|x|=-x$, then $x<0$ or $x=0$
6. $|f(x)|=|g(x)|$, then $f(x)=g(x)$ or $f(x)=-g(x)$

Domain and Range of Modulus Function

The modulus function $f(x)=|x|$ can be applied to any real number. i.e., its input can be any real number and hence can be said that its domain is the set of all real numbers $(\mathbb{R})$. The output of the modulus function is always a of non-negative real number and

• The domain of the modulus function is $\mathbb{R}$
• The range of the modulus function is $[0, \infty)$.

We should also take care that the domain of a modulus function $f(x)$ $=\mathrm{a}|\mathrm{x}-\mathrm{h}|+\mathrm{k}$ is still $\mathbb{R}$ but its range varies on the values of '$a$' and '$k$'. Its range is

• $y \geq k$, if $a>0$
• $y \leq k$, if $a<0$

Modulus inequalities

These deal with the inequalities ( $<,>, \leq, \geq$ ) on expressions with absolute value sign.

Properties

If $a, b>0$, then

1. $
\begin{aligned}
& |\mathrm{x}| \leq \mathrm{a} \Rightarrow \mathrm{x}^2 \leq \mathrm{a}^2 \\
& \Rightarrow-\mathrm{a} \leq \mathrm{x} \leq \mathrm{a}
\end{aligned}
$

2. $
\begin{aligned}
& |\mathrm{x}| \geq \mathrm{a} \Rightarrow \mathrm{x}^2 \geq \mathrm{a}^2 \\
& \Rightarrow \mathrm{x} \leq-\mathrm{a} \text { or } \mathrm{x} \geq \mathrm{a}
\end{aligned}
$

3. $
\begin{aligned}
& \mathrm{a} \leq|\mathrm{x}| \leq \mathrm{b} \Rightarrow \mathrm{a}^2 \leq \mathrm{x}^2 \leq \mathrm{b}^2 \\
& \Rightarrow \mathrm{x} \in[-\mathrm{b},-\mathrm{a}] \cup[\mathrm{a}, \mathrm{~b}]
\end{aligned}
$

4. $|x+y|=|x|+|y| \Leftrightarrow x y \geq 0$.

5. $|x-y|=|x|-|y| \Rightarrow x \cdot y \geq 0$ and $|x| \geq|y|$

6. $|x \pm y| \leq|x|+|y|$

7. $|x \pm y| \geq||x|-|y||$

Derivative And Integral of Modulus Function

Since we know that a modulus function $f(x)=|x|$ is equal to $x$ if $x>0$ and $-x$ if $x<0$, hence the derivative of modulus function is 1 if $x>0$ and -1 if $x<0$. The derivative of the modulus function is NOT defined for $x=0$. Hence the derivative of modulus function can be written as $\frac{d(|x|)}{d x} = \frac{x}{|x|}$, for all values of $x$ and $x \neq 0$.

Using the formula of the modulus function and integration formulas, the integral of the modulus function is $(\frac{1}{2}) x^2+C$ if $x \geq 0$, and its integral is $-(1 / 2) x^2+C$ if $x<0$. Hence the integration of the modulus function can be clubbed as:

• $\int|x| d x=(\frac{1}{2}) x^2+C$ if $x \geq 0$
• $\int|x| d x=-(\frac{1}{2}) x^2+C$ if $x<0$

Important Notes on Modulus Function

• The modulus function is also known as the absolute value function. It is denoted by $f(x) = |x|$.
• The domain of modulus functions is the set of all real numbers.
• The range of modulus functions is the set of all real numbers greater than or equal to $0$.
• The vertex of the modulus graph $y = |x|$ is $(0,0)$.
• The vertex of the modulus function $y = a |x - h| + k$ is $(h, k)$.

Recommended Video Based on Modulus Functions

Solved Examples Based On the Modulus Functions

Example 1: If $f(x)=3|x|$ . Then the range of $f(x)$ is

1) $[0, \infty)$
2) $[3, \infty)$
3) $R$
4) $[1, \infty)$

Solution:
We can find a range of $|x|$ by simple manipulation
As $0 \leq|x|<\infty$
Multiplying all 3 sides by 3

$0 \leq 3|x|<\approx$

So, the range is the same i.e $[0, \infty)$
Hence, the answer is the option 1.

Example 2: If $f(x)=-|x|+9$. Then the range of $\mathrm{f}(\mathrm{x})$ is
1) $[8, \infty)$
2) $(9, \infty)$
3) $(-\infty, 0)$
4) $(-\infty, 9]$

Solution:

We can find the range of $|x|$ by simple manipulation

$
\text { As } 0 \leq|x|<\infty
$

multiply all 3 sides by $(-1)$

$
-\infty<-|x| \leq 0
$

Now, adding 9 to all 3 sides

$
-\infty<-|x|+9 \leq 9
$

So, the range is $(-\infty, 9]$

Hence, the answer is the option 4.

Example 3: If $f(x)=|x|+x$, then what is the range of $\mathrm{f}(\mathrm{x})$ ?
1) $[0, \infty 0)$
2) $R$
3) $\{0\}$
4) $(-\infty, 0]$

Solution:
Case 1: If $x \geq 0 ; f(x)=x+x=2 x$
In this case, $x \geq 0$
Multiplying both sides by $2,2 x \geq 0$
Case 2: If $x<0 ; f(x)=-x+x=0$
Hence range $=[0, \infty)$
Hence, the answer is the option 1.

Example 4: The number of real solutions of the equation, $x^2-|x|-12=0$ is:
Solution:
Let $|x|=t$
$\Rightarrow t^2-t-12=0$

Example 5: If $\left|x^2-9\right|+\left|x^2-4\right|=5$, then the set of values of x is:
1) $(-\infty,-3) \cup(3, \infty)$
2) $(-\infty,-2) \cup(3, \infty)$
3) $(-\infty, 3)$
4) $[-3,-2] \cup[2,3]$

Solution:

$\begin{aligned}
& \left|x^2-9\right|+\left|x^2-4\right|=5 \\
& \left|x^2-9\right|+\left|x^2-4\right|=\left|\left(x^2-9\right)-\left(x^2-4\right)\right| \\
& \{\because|a|+|b|=|a-b| \Leftrightarrow a . b \leq 0\}
\end{aligned}$

So, $\left(x^2-9\right)\left(x^2-4\right) \leq 0$

$x \in[-3,-2] \cup[2,3]$
$\Rightarrow(t-4)(t+3)=0$
$\Rightarrow t=4$ or $t=-3$
$\Rightarrow|x|=4$ or $|x|=-3$
$\Rightarrow x=4,-4$ or $x \epsilon \phi$
$\Rightarrow x=4,-4$

Hence, the answer is 2.

Hence, the answer is the option (4).