Question

Consider the function $f:R-\left\{3\right\}\to R-\left\{1\right\}$ defined by $f\left(x\right)=\frac{x-2}{x-3}.$ Then $f$ is

• A. onto but not one-one
• B. both one-one and onto
• C. neither one-one nor onto
• D. one-one but not onto

Answer 1

The correct option is B both one-one and onto

$f\left(x\right)=\frac{x-2}{x-3}$
$\text{Let}f\left(x_1\right)=f\left(x_2\right),x_1,x_2\in D_f$
$\text{Then}\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$
$\Rightarrow x_1=x_2$
Thus, $f$ is one-one.

$\text{Let}y=\frac{x-2}{x-3}$
$\text{Then,}xy-3y=x-2$
$\Rightarrow x=\frac{3y-2}{y-1}$
Here, $x$ is defined for all $x\in R-\left\{1\right\}$
$\therefore R_f=R-\left\{1\right\}$
$\Rightarrow R_f=\text{Co-domain of}f$
$\Rightarrow f$ is onto.