If $\mathrm{f}\left(\mathrm{x}\right)=\frac{1-\left|\mathrm{x}\right|}{1+\mathrm{x}}, \mathrm{x}\ne -1 \mathrm{then} \mathrm{f}\left(\left[2\mathrm{x}\right]\right), \mathrm{where} \left[.\right]$ represents the greatest integer function is

$\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}1, \mathrm{x}<0\\ =\frac{1-\mathrm{x}}{1+\mathrm{x}}, \mathrm{x}\ge 0\end{array}\right.$
$\mathrm{Then} \mathrm{f}\left(\left[2\mathrm{x}\right]\right)=\left\{\begin{array}{l}1, \left[2\mathrm{x}\right]<0\\ \frac{1-\left[2\mathrm{x}\right]}{1+\left[2\mathrm{x}\right]}, \left[2\mathrm{x}\right]\ge 0 \mathrm{and}\end{array}\right.$
$\mathrm{f}\left(\left[2\mathrm{x}\right]\right)=\left\{\begin{array}{l}\begin{array}{l}1, \mathrm{x}<0,\\ 1 , 0\le \mathrm{x}<\frac{1}{2} \left(i.e\left[2x\right]=0\right)\\ 0, \frac{1}{2}\le \mathrm{x}<1 \left(i.e\left[2x\right]=1\right)\end{array}\\ -1/3, 1\le \mathrm{x}<3/2 ,\left(i.e\left[2x\right]=2\right)\end{array}\right.$