Question

Let $f:A\to B$ be a function defined as $f\left(x\right)=\frac{x-1}{x-2}$ where A=R-{2} and B=R-{1}. Then f is :

• A. invertible and $f^{-1}\left(y\right)=\frac{3y-1}{y-1}$
• B. invertible and $f^{-1}\left(y\right)=\frac{2y-1}{y-1}$
• C. invertible and $f^{-1}\left(y\right)=\frac{2y+1}{y-1}$
• D. not invertible

Answer 1

The correct option is A invertible and $f^{-1}\left(y\right)=\frac{3y-1}{y-1}$

$f:A\to B$

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

Let $f\left(x\right)=y,x=f^{-1}\left(y\right)$

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

$y(x-2)=x-1$

$xy-2y=x-1$

$xy-x=-1+2y$

$x(y-1)=2y-1$

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

Putting the value of $x$ in above equation

$\left|f^{-1}\right(y)=\frac{2y-1}{y-1}|$

option $B:$ invertible and $f^{-1}\left(y\right)=\frac{2y-1}{y-1}$