Let <img alt="A= R-\left \{ 2 \right \}and \: B= R-\left \{ 1 \right \}.\; If\;\; f:A\rightarrow B" src="https://entrancecorner.codecogs.com/gif.latex?A%3D%20R-%5Cleft%20%5C%7B%202%20%5Cright%20%5C%7Dand%20%5C%3A%20B%3D%20R-%5Cleft%20%5C%7B%201%20

Let $A= R-\left \{ 2 \right \}and \: B= R-\left \{ 1 \right \}.\; If\;\; f:A\rightarrow B$ is a function defined by $f\left ( x \right )= \frac{x-1}{x-2},$ show that f is one -one and onto. Hence,find $f^{-1}.$

Answers (1)

$f\left ( x \right )= \frac{x-1}{x-2}$
Let $x_{1},x_{2}\: \epsilon A$ such that $f\left ( x_{1} \right )= f\left ( x_{2} \right )$
That is $\frac{x_{1}-1}{x_{1}-2}= \frac{x_{2}-1}{x_{2}-2}$
$\Rightarrow x_{1}x_{2}-2x_{1}-x_{2}+2= x_{1}x_{2}-x_{1}\; \; 2x_{2}+2$
$\Rightarrow x_{1}= x_{2}$
$\therefore f\left ( x \right )$ is one-one function
$Let\: y= f\left ( x \right ),y\epsilon B$
That is $y= \frac{x-1}{x-2}\Rightarrow xy-2y= x-1$
$\Rightarrow xy-x= 2y-1\; \; \Rightarrow x= \frac{2y-1}{y-1}$
clearly, $y\epsilon B$ for all $x\epsilon A$ That implies range $=$ codomain
so, f is onto.
Also $f^{-1}\left ( y \right )= \frac{2y-1}{y-1}$ i.e $f^{-1}\left ( x \right )= \frac{2x-1}{x-1}$