Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ is defined by $\mathrm{f}\left(\mathrm{x}\right)=\left[2\mathrm{x}\right]-2\left[\mathrm{x}\right],$$\forall \mathrm{x}\in \mathrm{R},$where $\left[\mathrm{x}\right]$ is the greatest integer not exceeding $x$, then the range of $f$ is

• A. $\left\{\mathrm{x}\in \mathrm{R}:0\le \mathrm{x}\le 1\right\}$
• B. $\left\{0,1\right\}$
• C. $\left\{\mathrm{x}\in \mathrm{R}:\mathrm{x}>0\right\}$
• D. $\left\{\mathrm{x}\in \mathrm{R}:\mathrm{x}\le 0\right\}$

Answer 1

The correct option is B

$\left\{0,1\right\}$

Explanation for the correct answer:

Step 1: Given information

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ then $\begin{array}{rcl}\left[2\mathrm{x}\right]& =& \left[\mathrm{x}\right]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]\left(\because \mathrm{integer}\mathrm{valued}\mathrm{function}\left[2\mathrm{x}\right]\right)\end{array}$

$\begin{array}{rcl}\therefore \mathrm{f}\left(\mathrm{x}\right)& =& \left[2\mathrm{x}\right]-2\left[\mathrm{x}\right]\\ & =& \left[\mathrm{x}\right]+\left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-2\left[\mathrm{x}\right]\\ & =& \left[\mathrm{x}+\left(\frac{1}{2}\right)\right]-\left[\mathrm{x}\right]\end{array}$

Step 2: Finding the values

If $\mathrm{x}=1,$

$\begin{array}{rcl}\therefore \mathrm{f}\left(1\right)& =& \left[1+\left(\frac{1}{2}\right)\right]-1\\ & =& \left[1\right]-1\\ & =& 0\end{array}$

If $\mathrm{x}=0.25$,

$\begin{array}{rcl}\therefore \mathrm{f}(0.25)& =& \left[0.25+\left(\frac{1}{2}\right)\right]-0.25\\ & =& 0.75-0.25\\ & =& 0-0\\ & =& 0\end{array}$

If $\mathrm{x}=0.5$,

$\begin{array}{rcl}\therefore \mathrm{f}(0.5)& =& \left[0.5+\left(\frac{1}{2}\right)\right]-0.5\\ & =& 1-0\\ & =& 1\end{array}$

Therefore, the range of $f$ lies between $\left\{0,1\right\}$

Hence, the correct answer is an option (B).