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

Here, A=R-{3},B=R-{1}
and $f:A\to B$ is defined as $f\left(x\right)=\left(\frac{x-2}{x-3}\right)$
Let $x,y\in A$ such that f(x)=f(y)
$\Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3}\Rightarrow (x-2)(y-3)=(y-2)(x-3)$
$\Rightarrow xy-3x-2y+6=xy-3y-2x+6\Rightarrow -3x-2y=-3y-2x\Rightarrow 3x-2x=3y-2y\Rightarrow x=y$
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(x)=y.
Now, f(x)=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

$f\left(\frac{2-3y}{1-y}\right)=\frac{\left(\frac{2-3y}{1-y}\right)-2}{\left(\frac{2-3y}{1-y}\right)-3}=\frac{2-3y-2+2y}{2-3y-3+3y}=\frac{-y}{-1}=y$
Therefore, f is onto. Hence, function f is one-one and onto.