Question

Verify whether the function
$f:$AB where $A=R-\left\{3\right\}$ and $B=R-\left\{1\right\}$, defined by $f\left(x\right)=\frac{x-2}{x-3}$ is one-one and on-to or not. Give reason.

Answer 1

To check one-one

let $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$

$\Rightarrow (x_1-2)(x_2-3)=(x_1-3)(x_2-2)$

$\Rightarrow x_1{x}_2-2x_2-3x_1+6=x_1{x}_2-3x_2-2x_1+6$

$\Rightarrow x_1=x_2$

Thus f is one-one

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

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

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

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

$\Rightarrow x=\frac{3y-2}{y-1}$

here $y\in R-\left\{1\right\}$

thus f is onto.