Find the range of function $f(x) = \frac{x^2-3x+2}{x^2-4x+3}$.

$R - \{\frac{1}{2},1\}$ $R - \{\frac{1}{2}\}$ $R - \{0,1\}$ $R - \{2\}$

$R - \{\frac{1}{2},1\}$

$y = f(x) = \frac{(x-1)(x-2)}{(x-3)(x-1)}=\frac{x-2}{x-3},x≠1, 3$ Now $yx - 3y = x - 2$ $∴x=\frac{3y-2}{y-1}$ Clearly y ≠ 1. Also, when x = 1, y = 1/2 Therefore, range of function is $R - \{\frac{1}{2},1\}$