Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ is defined by $\mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{x}\right]$, then

• A. ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=-\mathrm{x}$
• B. ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{1}{\left|\mathrm{x}\right|}$
• C. ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)\mathrm{does}\mathrm{not}\mathrm{exist}$
• D. ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}$

Answer 1

The correct option is C

${\mathrm{f}}^{-1}\left(\mathrm{x}\right)\mathrm{does}\mathrm{not}\mathrm{exist}$

Explanation for the correct answer:

For a function $\mathrm{f}\left(\mathrm{x}\right)$ to be invertible, the function must be one-one and onto.

The range of $\mathrm{f}\left(\mathrm{x}\right)=\left|\mathrm{x}\right|$ is $\left(0,\infty \right)$, while the co-domain of $\mathrm{f}\left(\mathrm{x}\right)$ is given as $R.$

Hence, $\mathrm{f}\left(\mathrm{x}\right)$ is not onto.

Also, since $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}(-\mathrm{x})$, $\mathrm{f}\left(\mathrm{x}\right)$ is also not one-one in its domain.

Therefore, $\mathrm{f}\left(\mathrm{x}\right)$ is not invertible, i.e., the function ${\mathrm{f}}^{-1}\left(\mathrm{x}\right)$ does not exist.

Hence, the correct answer is option (C).