Let A = R – {2} and B = R – {1}.

Solution:

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

To show f is one-one:

Let $f\left(x_{1}\right)=f\left(x_{2}\right)$

$\Rightarrow \frac{x_{1}-1}{x_{1}-2}=\frac{x_{2}-1}{x_{2}-2}$

$\Rightarrow\left(x_{1}-1\right)\left(x_{2}-2\right)=\left(x_{2}-1\right)\left(x_{1}-2\right)$

$\Rightarrow x_{1} x_{2}-2 x_{1}-x_{2}+2=x_{1} x_{2}-2 x_{2}-x_{1}+2$

$\Rightarrow-2 x_{1}-x_{2}=-2 x_{2}-x_{1}$

$\Rightarrow-2 x_{1}+x_{1}=-2 x_{2}+x_{2}$

$\Rightarrow-x_{1}=-x_{2}$

$\Rightarrow x_{1}=x_{2}$

Hence, $f$ is one-one.

To show $f$ is onto:

Let $y \in B$

$\therefore y=f(x)$

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

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

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

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

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

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

Thus, for every value of $y$ in $R-\{1\}$, there exists a pre-image $x=\frac{2 y-1}{y-1}$ in $R-\{2\}$.

Hence, $f$ is onto.

Since, $f$ is one-one and onto

Therefore, $f$ is invertible with $f^{-1}(y)=\frac{2 y-1}{y-1}$.

Hence, $f^{-1}(x)=\frac{2 x-1}{x-1}$.