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
Let $R$ be the set of real numbers and $f: R \rightarrow R$ be defined by $f(x) = \frac{\{x\}}{1+[x]^2}$,where $[x]$ is the greatest integer less than or equal to $x$,and $\{x\} = x-[x]$. Which of the following statements are true?
$I.$ The range of $f$ is a closed interval.
$II.$ $f$ is continuous on $R$.
$III.$ $f$ is one-one on $R$.
$I.$ The range of $f$ is a closed interval.
$II.$ $f$ is continuous on $R$.
$III.$ $f$ is one-one on $R$.
Let $a > 1$ and $0 < b < 1$. If $f: R \rightarrow [0, 1]$ is defined by $f(x) = \begin{cases} a^x, & -\infty < x < 0 \\ b^x, & 0 \leq x < \infty \end{cases}$,then $f(x)$ is
Let $A = \{(x, y) : 2x + 3y = 23, x, y \in N\}$ and $B = \{x : (x, y) \in A\}$. Then the number of one-one functions from $A$ to $B$ is equal to ................
If $f: N \rightarrow R$ is defined by $f(1)=-1$ and $f(n+1)=3f(n)+2$ for $n \geq 1$, then $f$ is
Which of the following functions is surjective but not injective?
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