Question

Let $A=R-\left\{3\right\}$ and $B=R-\left\{1\right\}$ . Consider the function $f:A$ defined by $f\left(x\right)=\frac{x-2}{x-3}$. Prove that $f$ is one-one and and on to function.

Answer 1

$f\left(x\right)=\frac{x-2}{x-3}$

To check $f$ is one-one.

$f\left(x_1\right)=\frac{x_1-2}{x_1-3}$

$f\left(x_2\right)=\frac{x_2-2}{x_2-3}$

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

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

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

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

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

$\Rightarrow -3x_1+2x_1=-3x_2+2x_2$

$\Rightarrow x_1=x_2$

Hence,if $f\left(x_1\right)=f\left(x_2\right)$, then $x_1=x_2$

$\therefore f$ is one-one.

To check $f$ is onto.

Let $f\left(x\right)=y$ such tahat $y\in B$ i.e., $y\in R-\left\{1\right\}$

So, $y=\frac{x-2}{x-3}$

$y(x-3)=x-2$

$xy-3y=x-2$

$xy-x=3y-2$

$x(y-1)=3y-1$

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

For $y=1$, $x$ is not defined.

But it is given that $y\in R\left\{1\right\}$

Hence, $X=\frac{3y-2}{y-1}\in R\left\{3\right\}$

Hence, $f$ is onto.