Question

Let $f:A\to R^{+}$ be a function defined by $f\left(x\right)={\log }_{\left\{x\right\}}\left(\frac{x-\left[x\right]}{|x|}\right)$, where $[.]$ and $\{.\}$ represent greatest integer function and fractional part function respectively. If $B$ is the range of $f$, then the number of integer(s) in $R^{+}-B$ is

Answer 1

$f\left(x\right)={\log }_{\left\{x\right\}}\left(\frac{x-\left[x\right]}{|x|}\right)$
For $f$ to be defined,
$\frac{x-\left[x\right]}{|x|}>0,|x|\ne 0,\left\{x\right\}>0,\left\{x\right\}\ne 1$
$\Rightarrow \frac{\left\{x\right\}}{|x|}>0,x\ne 0,x\notin Z$
$\Rightarrow x\in R-Z$
Hence, domain of $f$ is $R-Z$

Now, for range
$f\left(x\right)={\log }_{\left\{x\right\}}\left(\frac{x-\left[x\right]}{|x|}\right)={\log }_{\left\{x\right\}}\left(\frac{\left\{x\right\}}{|x|}\right)\Rightarrow f\left(x\right)=1-{\log }_{\left\{x\right\}}|x|(\because \log \frac{m}{n}=\log m-\log n)$
Let $y=f\left(x\right)$
$1-y={\log }_{\left\{x\right\}}|x|$
We know that
when $0<a<1,b>1$
${\log }_{a}b\in (-\infty ,0)\therefore -\infty <{\log }_{\left\{x\right\}}|x|<0\Rightarrow -\infty <1-y<0\Rightarrow 1<y<\infty \therefore B=(1,\infty )$
$\Rightarrow R^{+}-B=(0,1]$
The only integer in $(0,1]$ is $1$.