Question

Let $A=R-\left\{3\right\},B=R-\left\{1\right\}$ and let $f:A\to B$ defined by $f\left(x\right)=\frac{x-2}{x-3}$, then which among the following options is/are true?

• A. $f$ is many-one function
• B. $f$ is one-one function
• C. $f$ is invertible function
• D. $f$ is into function

Answer 1

Answer: (B), (C)

$f$ is invertible function

Let $x_1,x_2\in A$ and let $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}\Rightarrow x_1{x}_2-3x_1-2x_2+6=x_1{x}_2-2x_1-3x_2+6\Rightarrow x_1=x_2$
$\Rightarrow f$ is one-one function
Range of $f$ can be obtained as
Let $y=f\left(x\right)=\frac{x-2}{x-3}$
$\Rightarrow xy-3y=x-2$
$\Rightarrow x(y-1)=3y-2$
$\Rightarrow x=\frac{3y-2}{y-1}$
Clearly, $x$ is defined if $y\ne 1$ i.e the range of $f$ is $R-\left\{1\right\}$
Hence $f$ is onto function.
So, $f$ being one one and onto function, it has to be invertible.