Question

Let $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\left[\mathrm{x}\right]/1+\mathrm{x}-\left[\mathrm{x}\right]$ where $\left[\mathrm{x}\right]$ denotes the greatest integer less than or equal to $\mathrm{x}$ then the range of $\mathrm{f}$ is

• A. $[0,1]$
• B. $[0,1/2)$
• C. $[0,1)$
• D. $[0,1/2]$

Answer 1

The correct option is B

$[0,1/2)$

Determine the range of the function.

We are given with a function $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-\left[\mathrm{x}\right]/1+\mathrm{x}-\left[\mathrm{x}\right]$. Here $\left[\mathrm{x}\right]$ denotes the greatest integer less than or equal to $\mathrm{x}$.

Now we know that $\mathrm{x}=\left[\mathrm{x}\right]+\left\{\mathrm{x}\right\}$ where, $\left\{\mathrm{x}\right\}$ is the fractional part of $\mathrm{x}$. Therefore $\mathrm{x}-\left[\mathrm{x}\right]=\left\{\mathrm{x}\right\}$. Put this in the given function we get,

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-\left[\mathrm{x}\right]}{1+\mathrm{x}-\left[\mathrm{x}\right]} \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\left\{\mathrm{x}\right\}}{1+\left\{\mathrm{x}\right\}}0\le \left\{\mathrm{x}\right\}<1 \end{gathered}$$

When $\mathrm{x}\in \mathrm{I}$, $\mathrm{f}\left(\mathrm{x}\right)$ will be equal to zero. And when the value of $\mathrm{x}$ will lie in between $\text{'}0\text{'}$and $\text{'}1\text{'}$then $\mathrm{f}\left(\mathrm{x}\right)$ will be less than $\frac{1}{2}$.

Therefore the range of $\mathrm{f}\left(\mathrm{x}\right)$ is $[0,\frac{1}{2})$

Hence, option(B) is the correct answer.