Question

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

• A. $\frac{f\left(x\right)+1}{f\left(x\right)+3}$
• B. $\frac{3f\left(x\right)+1}{f\left(x\right)+3}$
• C. $\frac{f\left(x\right)+3}{f\left(x\right)+1}$
• D. $\frac{f\left(x\right)+3}{3f\left(x\right)+1}$

Answer 1

The correct option is B $\frac{3f\left(x\right)+1}{f\left(x\right)+3}$

We have $f\left(x\right)=\frac{x-1}{x+1}...\left(i\right)$
$\Rightarrow \frac{f\left(x\right)+1}{f\left(x\right)-1}=\frac{2x}{-2}$
(applying componendo-dividendo)
$x=\frac{f\left(x\right)+1}{1-f\left(x\right)}....\left(ii\right)$
Now, $f\left(2x\right)=\frac{2x-1}{2x+1}=\frac{2\left(\frac{f\left(x\right)+1}{1-f\left(x\right)}\right)-1}{2\left(\frac{f\left(x\right)+1}{1-f\left(x\right)}\right)+1}=\frac{3f\left(x\right)+1}{f\left(x\right)+3}$