Question

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

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

Answer 1

The correct option is D invertible and $f^{-1}\left(y\right)=\frac{2y-1}{y-1}$

Let $y=f\left(x\right)$

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

$\Rightarrow yx-2y=x-1$

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

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

So on the given domain the function is invertible and its inverse can be computed as shown above.

So, option D is the correct answer.