Let A = R – {2}, B = R – {1}. Let f : A → B be defined by $f(x)=\frac{x-1}{x-2}$ Then,

f is bijective

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

$f(x_1)=f(x_2)⇒\frac{x_1-1}{x_1-2}=\frac{x_2-1}{x_2-2}$

$⇒x_1x_2-2x_1-x_2+2=x_1x_2-2x_2-x_1+2$

so $x_1=x_2$ (ONE-ONE)

$y=\frac{x-1}{x-2}⇒yx-2y=x-1$

so $yx-x=2y-1$

so $x=\frac{2y-1}{y-1}$

$f^{-1}(x)=\frac{2x-1}{x-1}$  $x≠1$

there exists atleast one x for every y (ONTO)

f is both one-one and onto in the specified domain and range