Question

If $f\left(x\right)=\frac{x-1}{x+1}$ then $f\left(ax\right)$ in term of $f\left(x\right)$ is equal to

• A. $\frac{f\left(x\right)+a}{1+af\left(x\right)}$
• B. $\frac{(a-1)f\left(x\right)+a+1}{(a+1)f\left(x\right)+a-1}$
• C. $\frac{(a+1)f\left(x\right)+a-1}{(a-1)f\left(x\right)+a+1}$
• D. None of these

Answer 1

The correct option is C $\frac{(a+1)f\left(x\right)+a-1}{(a-1)f\left(x\right)+a+1}$

$f\left(x\right)=\frac{x-1}{x+1}$
$\frac{f\left(x\right)+1}{f\left(x\right)-1}=\frac{x-1+x+1}{x-1-x-1}$
$\frac{f\left(x\right)+1}{f\left(x\right)-1}=\frac{2x}{-2}$
$\frac{1+f\left(x\right)}{1-f\left(x\right)}=x$ ...(1)
$f\left(ax\right)=\frac{ax-1}{ax+1}$
Similarly,

$\frac{1+f\left(ax\right)}{1-f\left(ax\right)}=a\left(x\right)$
$\frac{1+f\left(ax\right)}{1-f\left(ax\right)}=\frac{a+af\left(x\right)}{1-f\left(x\right)}$
$\frac{1+f\left(ax\right)+1-f\left(ax\right)}{1+f\left(ax\right)-1+f\left(ax\right)}=\frac{a+af\left(x\right)+1-f\left(x\right)}{a+af\left(x\right)-1+f\left(x\right)}$
$\frac{1}{f\left(ax\right)}=\frac{a+1+f\left(x\right)(a-1)}{a-1+f\left(x\right)(a+1)}$
$f\left(ax\right)=\frac{a-1+f\left(x\right)(a+1)}{a+1+f\left(x\right)(a-1)}$

Hence, option 'C' is correct.