Let A = R − {3} and B = R − {1}.


Let $A=R-\{3\}$ and $B=R-\{1\}$. Consider the function $f: A \rightarrow B$ defined by $f(x)=\frac{x-2}{x-3}$. Show that $f$ is one-one and onto and hence find $f^{-1}$.                            [CBSE 2012, 2014]



We have,

$A=R-\{3\}$ and $B=R-\{1\}$

The function $f: A \rightarrow B$ defined by $f(x)=\frac{x-2}{x-3}$

Let $x, y \in A$ such that $f(x)=f(y)$. Then,


$\Rightarrow x y-3 x-2 y+6=x y-2 x-3 y+6$


$\Rightarrow x=y$

$\therefore f$ is one-one.

Let $y \in B$. Then, $y \neq 1$.

The function $f$ is onto if there exists $x \in A$ such that $f(x)=y$.



$\Rightarrow \frac{x-2}{x-3}=y$

$\Rightarrow x-2=x y-3 y$

$\Rightarrow x-x y=2-3 y$

$\Rightarrow x(1-y)=2-3 y$

$\Rightarrow x=\frac{2-3 y}{1-y} \in A \quad[y \neq 1]$

Thus, for any $y \in B$, there exists $\frac{2-3 y}{1-y} \in A$ such that

$f\left(\frac{2-3 y}{1-y}\right)=\frac{\left(\frac{2-y_{g}}{1-y}\right)-2}{\left(\frac{2-3 g}{1-y}\right)-3}=\frac{2-3 y-2+2 y}{2-3 y-3+3 y}=\frac{-y}{-1}=y$

$\therefore f$ is onto.

So, $f$ is one - one and onto fucntion.


As, $x=\left(\frac{2-3 y}{1-y}\right)$

So, $f^{-1}(x)=\left(\frac{2-3 x}{1-x}\right)=\frac{3 x-2}{x-1}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now