Question

The number of integral elements in the range of the function $f\left(x\right)=\frac{e^x-\log x}{e^x+\log x};x>1$ is

Answer 1

$y=\frac{e^x-\log x}{e^x+\log x};x>1$
Since, $e^x>\log x,\forall x>1$
$\Rightarrow y>0$
Now, $y{e}^x+y\log x=e^x-\log x$
$\Rightarrow (y-1)e^x=-(1+y)\log x$
$\Rightarrow -\frac{y-1}{1+y}=\frac{\log x}{e^x}>0$
$\Rightarrow \frac{1-y}{1+y}>0\text{and}y>0$
$\Rightarrow y\in (0,1)$
Hence, no integral value of $y$ exists.