If f: R – {–1} → R – {1} be a function defined by f(x) = \(\frac{x-1}{x+1}\), then

A. f is one-one but not onto

B. f is onto but not one-one

C. f is one-one and onto

D. f^-1(x) = \(\frac{x+1}{x-1}\)

E. (fof)(x) = -\(\frac{1}{x}\); x ≠ 0, -1

Choose the correct answer from the options given below

C, E only

Let x, y be two numbers in set R - {-1} such that f(x) = f(y)

$\frac{x-1}{x+1}=\frac{y-1}{y+1}$ ⇒ Simplifying the equation.

$f(x)=\frac{x-1}{x+1}=\frac{(x+1)}{x+1}-\frac{2}{x+1}=1-\frac{2}{x+1}$

$f'(x)=\frac{2}{(x+1)^2}>0$ ∴ f is one-one

$f(x) = 1-\frac{2}{x+1}∈R$ ...(i)  ∴ f is onto also & $y=\frac{x-1}{x+1}$

$\frac{y+1}{y-1}=\frac{x}{-1}⇒f^{-1}(x)=\frac{x+1}{1-x}⇒\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}=\frac{-2}{2x}=-\frac{1}{x}$