Question

If $R$ denotes the set of all real numbers, then the function $f:R\to R$ defined by $f\left(x\right)=\left|x\right|$ is

• A. only one-one
• B. only onto
• C. both one-one and onto
• D. neither one-one nor onto

Answer 1

The correct option is D

neither one-one nor onto

Checking for the function:

A one-one function is also known as an injective function. One element contains only one element that is a one-one correspondence
Onto function is nothing but a many-one correspondence. It is also known as the surjective function.
A bijective function is a function having both surjective as well as objective functions.

Check the condition $f\left(x\right)=f(-x)$ if this condition satisfies then the given function is neither one-one nor onto.

$f\left(x\right)=\left|x\right|\ge 0$ for all $x$, then the range of $f$ is $(0,\infty ]\Rightarrow$co-domain which implies $f$ is not onto.

Put $x=1$in the function $f\left(x\right)=\left|x\right|$

$\begin{array}{rcl}f\left(1\right)& =& \left|1\right|\\ & =& 1\end{array}$

Put $x=-1$in the function $f\left(x\right)=\left|x\right|$

$\begin{array}{rcl}f(-1)& =& \left|-1\right|\\ & =& 1\end{array}$

Since $\begin{array}{rcl}f\left(1\right)& =& f\left(-1\right)\end{array}$ Which implies $f\left(x\right)=f(-x)$ then given function is neither one-one nor onto

Hence, the correct option is (D).