Question

Let $A=R-3$ and $B=R-1$. Consider the function $f:A\to B$ defined by

$f\left(x\right)=\left(\frac{x-2}{x-3}\right)$. Is $f$ one-one and onto? Justify your answer

Answer 1

$A=R-3,B=R-1$

$f\left(x\right)=\left(\frac{x-2}{x-3}\right)$
$f:A\to B$ is defined as
Let, $x,y\in A$ such that $f\left(x\right)=f\left(y\right)$
$\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}$
$\Rightarrow x-2y-3=y-2x-3$
$\Rightarrow xy-3x-2y+6=xy-3y-2x+6$
$\Rightarrow -3x-2y=-3y-2x$
$\Rightarrow 3x-2x=3y-2y$
$\Rightarrow x=y$
$\therefore f$ is one-one.
Let $y\in B=R-1$. Then, $y\ne 1$
The function $f$ is onto if there exists $x\in A$

such that $f\left(x\right)=y$

Now,$f\left(x\right)=y$
$\Rightarrow \frac{x-2}{x-3}=y$
$\Rightarrow x-2=xy-3y$
$\Rightarrow x(1-y)=-3y+2$
$\Rightarrow x=\frac{2-3y}{1-y}\in A$ $[y\ne 1]$
Thus, for any $y\in B$, there exists
$\frac{2-3y}{1-y}\in A$
such that $f\left(\frac{2-3y}{1-y}\right)=\frac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}$

$\therefore f$ is onto.

Hence, function $f$ is one-one and onto.