Question

Consider the function $f:A\to B$ defined by $f\left(x\right)=\frac{x-2}{x-3}$. Find the range of f.

Answer 1

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

$Yx-3Y=x-2$

$Yx-x=3Y-2$

$x(Y-1)=3Y-2$

$x=\frac{3Y-2}{Y-1}$ denominator should not be zero
$Y-1\ne 0$

$Y\ne 1$ Y is range. So Real number-$1$

Range $=R-1$