Show that the modulus function $f : R \rightarrow R$ given by $f(x) = |x|$ is neither one-one nor onto,where $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.
(N/A) The function $f: R \rightarrow R$ is defined as $f(x) = |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$.
To check if $f$ is one-one:
Consider $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function $f$ is not one-one.
To check if $f$ is onto:
Consider the codomain $R$. For any negative value,such as $-1 \in R$,there exists no $x \in R$ such that $f(x) = |x| = -1$,because the absolute value of any real number is always non-negative $(|x| \ge 0)$.
Therefore,$f$ is not onto.
Hence,the modulus function is neither one-one nor onto.
To check if $f$ is one-one:
Consider $f(-1) = |-1| = 1$ and $f(1) = |1| = 1$.
Since $f(-1) = f(1)$ but $-1 \neq 1$,the function $f$ is not one-one.
To check if $f$ is onto:
Consider the codomain $R$. For any negative value,such as $-1 \in R$,there exists no $x \in R$ such that $f(x) = |x| = -1$,because the absolute value of any real number is always non-negative $(|x| \ge 0)$.
Therefore,$f$ is not onto.
Hence,the modulus function is neither one-one nor onto.
Explore More
Similar Questions
If $n(A) = 5$ and $n(B) = 8$,how many possible functions can be defined from $A$ to $B$?
Medium
View SolutionLet $A = \{x, y, z, u\}$ and $B = \{a, b\}$. $A$ function $f: A \rightarrow B$ is selected randomly. The probability that the function is an onto function is
If $f : R \rightarrow R$ is defined as $f(x) = x - [x] + 3$,$\forall x \in R$,then $f$ is
The function $f: N \rightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2} & , n \text{ is even} \\ -\left(\frac{n-1}{2}\right) & , n \text{ is odd} \end{cases}$ is . . . . . . .
The function $f: R \rightarrow R$ is defined by $f(x)=3^{-x}$. Observe the following statements about it:
$I$. $f$ is one-one
$II$. $f$ is onto
$III$. $f$ is a decreasing function
Out of these, the true statements are:
$I$. $f$ is one-one
$II$. $f$ is onto
$III$. $f$ is a decreasing function
Out of these, the true statements are:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo