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 $f\left(x\right)$ is $R-Z$

Now, for range
$f\left(x\right)={\log }_{\left\{x\right\}}\left(\frac{x-\left[x\right]}{|x|}\right)\Rightarrow f\left(x\right)={\log }_{\left\{x\right\}}\left(\frac{\left\{x\right\}}{|x|}\right)\Rightarrow f\left(x\right)=1-{\log }_{\left\{x\right\}}|x|(\because \log \frac{m}{n}=\log m-\log n)$

Let $y=f\left(x\right)$
$1-y={\log }_{\left\{x\right\}}|x|$
Now, $0<\left\{x\right\}<1$
Case $1:$ When $|x|>1$
${\log }_{\left\{x\right\}}|x|\in (-\infty ,0)\therefore -\infty <1-y<0\Rightarrow -\infty <1-y<0\Rightarrow 1<y<\infty$

Case $2:$ When $0<x<1$
$\left\{x\right\}=|x|=x\Rightarrow {\log }_{\left\{x\right\}}|x|=1\therefore y=0$

Case $3:$ When $-1<x<0$
$\left\{x\right\}=x+1,|x|=-x$
Now for
$-\frac{1}{2}<x<0\Rightarrow \frac{1}{2}<\left\{x\right\}<1,0<|x|<\frac{1}{2}$
${\log }_{\left\{x\right\}}|x|\in (1,\infty )$
And for
$-1<x<-\frac{1}{2}\Rightarrow 0<\left\{x\right\}<\frac{1}{2},\frac{1}{2}<|x|<1$
${\log }_{\left\{x\right\}}|x|\in (0,1)$
So,
${\log }_{\left\{x\right\}}|x|\in (0,\infty )\Rightarrow 0<1-y<\infty \therefore y\in (-\infty ,1)$

Thus, from above cases we have$B=(-\infty ,1)\cup (1,\infty )$
$\Rightarrow R-B=\left\{1\right\}$

So, the number of integers in $R-B$ is $1.$