Question

The range of the function $y=\frac{x-\left[x\right]}{1-\left[x\right]+x}$, where $[.]$ represents the greatest integer function, is

• A. $(0,\frac{1}{2}]$
• B. $R-\left\{0\right\}$
• C. $[0,\frac{1}{2})$
• D. $R-[0,1]$

Answer 1

The correct option is C $[0,\frac{1}{2})$

$y=\frac{x-\left[x\right]}{1-\left[x\right]+x}$
$\Rightarrow y=\frac{\left\{x\right\}}{1+\left\{x\right\}}[\because x=[x]+\{x\left\}\right]$
Domain of $y$ is $R$ as $1+\left\{x\right\}\ne 0$ for any real value of $x$.

$y=\frac{\left\{x\right\}}{1+\left\{x\right\}}$
$\Rightarrow y=\frac{1+\left\{x\right\}-1}{1+\left\{x\right\}}$
$\Rightarrow y=1-\frac{1}{1+\left\{x\right\}}$

Now,
$0\le \left\{x\right\}<1$
$\Rightarrow 1\le 1+\left\{x\right\}<2\Rightarrow \frac{1}{2}<\frac{1}{1+\left\{x\right\}}\le 1$
$\Rightarrow -1\le \frac{-1}{1+\left\{x\right\}}<-\frac{1}{2}$
$\Rightarrow 0\le 1-\frac{1}{1+\left\{x\right\}}<\frac{1}{2}$
$\therefore y\in [0,\frac{1}{2})$