Question

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

• A. $a<2$
• B. $a>4$
• C. $(-2,0)$
• D. $-2<a<3$

Answer 1

The correct option is A $a<2$

Since, the roots are less than a real number
$D\ge 0$
$(2a{)}^2-4(1\left)\right[a^2+a-3]\ge 0$
$\Rightarrow a\le 3$ .....(1)
Lef $f\left(x\right)=x^2-2ax+a^2+a-3$.
Since, $3$ lies outside the interval $(\alpha ,\beta )$ where $\alpha ,\beta$ are the roots.
$f\left(3\right)>0\Rightarrow a<2$ or $a>3$ ....(2)
Sum of the roots must be less than $6$
$2a<6\Rightarrow a<3$ .....(3)
From (1), (2), (3), we have
$a<2$.