If the roots of the equation x2-2ax+a2+a-3=0 are real and less than 3, then

Given equation $x^{2} - 2 ax + a^{2} + a - 3 = 0$

If roots are real, then $D \geq 0$

$\Rightarrow 4 a^{2} - 4 (a^{2} + a - 3) \geq 0 \Rightarrow - a + 3 \geq 0$

$\Rightarrow a - 3 \leq 0 \Rightarrow a \leq 3$

As roots are less than 3, hence $f (x) > 0$

$9 - 6 a + a^{2} + a - 3 > 0 \Rightarrow a^{2} - 5 a + 6 > 0$

$\Rightarrow (a - 2) (a - 3) > 0 \Rightarrow either a < 2 or a > 3$

Hence $a < 2$ satisfy all.

If the roots of the equation $x^{2} - 2 ax + a^{2} + a - 3 = 0$ are real and less than 3, then