Question

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

• A. $a<2$
• B. $2\le a\le 3$
• C. $3<a\le 4$
• D. $a>4$

Answer 1

The correct option is A $a<2$

Let $f\left(x\right)=x^2-2ax+a^2+a-3$
Since f(x) has real roots both less than 3.
Therefore, $D\ge 0$ and $f\left(3\right)>0$
$\begin{array}{cc}\Rightarrow & a^2-(a^2+a-3)\ge 0\\ and& a^2-5a+6>0\\ \Rightarrow & a\le 3and(a-2)(a-3)>0\\ \Rightarrow & a\le 3anda<2ora>3\\ \Rightarrow & a<2\end{array}$
Hence, option (a) is correct answer.