Question

Let $A = R - {2}$ and $B = R - {1}$ . Consider the function $f : A ⟶ B$ defined by $f (x) = \frac{x - 1}{x - 2}$ . Show that $f$ is one-one and onto function. Hence, find $f^{- 1}$ .

Accepted Answer

Given, $A = R - {2}$ and $B = R - {1}$ . The function is $f : A ⟶ B$ defined by $f (x) = \frac{x - 1}{x - 2}$ .

First, we shall check if $f$ is a one-one function (i.e. injective function) or not.

For any $x_{1}, x_{2} \in R, f (x_{1}) = f (x_{2}) \Rightarrow \frac{x_{1} - 1}{x_{1} - 2} = \frac{x_{2} - 1}{x_{2} - 2}$

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

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

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

i.e., $f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2}$ for all $x_{1}, x_{2} \in R$ .

Hence, $f$ is a one-one or injective function.

Next, we shall check if $f$ is an onto function or not.

Let $y$ be an arbitrary element of $B$ .

Then, $y = \frac{x - 1}{x - 2}$ and $f (x) = \frac{x - 1}{x - 2} \Rightarrow f (x) = y = \frac{x - 1}{x - 2}$

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

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

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

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

Thus, for all $y \in R - \{1\}$ , there exists a number $x$ , $x \in A$ .

So, each element $R - \{1\}$ in the co-domain has its pre-image in the domain.

Hence, $f$ is an onto or surjective function.

Therefore, the given function is a one-one onto function.

So, $f$ is an invertible function.

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

$\Rightarrow f^{- 1} (y) = \frac{2 y - 1}{y - 1}$ for all $y \in B$ .

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

Let $A = R - {2}$ and $B = R - {1}$ . Consider the function $f : A ⟶ B$ defined by $f (x) = \frac{x - 1}{x - 2}$ . Show that $f$ is one-one and onto function. Hence, find $f^{- 1}$ .

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

Let A=R-{2} and B=R-{1}. Consider the function f:A⟶B defined by f(x)=x-1x-2. Show that f is one-one and onto function. Hence, find f-1.

Important Questions on Relations and Functions

Learn and Prepare for any exam you want !

Let $A = R - {2}$ and $B = R - {1}$ . Consider the function $f : A ⟶ B$ defined by $f (x) = \frac{x - 1}{x - 2}$ . Show that $f$ is one-one and onto function. Hence, find $f^{- 1}$ .