If $f(x)=\frac{x-1}{x+1}$, then $f(2x)$ is equal to

$\frac{f(x)+1}{f(x)+3}$ $\frac{3f(x)+1}{f(x)+3}$ $\frac{f(x)+3}{f(x)+1}$ $\frac{f(x)+3}{3f(x)+1}$

$\frac{3f(x)+1}{f(x)+3}$

$f(x)=\frac{x-1}{x+1}⇒f(2x)\frac{2x-1}{2x+1}$   …(i) Also $xf (x)+ f (x) = x −1 ⇒x=\frac{f(x)+1}{1-f(x)}$   …(ii) From equation (i) and (ii)  $⇒f(2x)=\frac{2\left(\frac{f(x)+1}{1-f(x)}\right)-1}{2\left(\frac{f(x)+1}{1-f(x)}\right)+1}⇒f(2x)=\frac{3f(x)+1}{f(x)+3}$