Question

The value of a for which one root of the equation $x^2-(a+1)x+a^2+a-8=0$ exceeds 2 and other is less than 2, are given by

• A. 3 < a < 10
• B. $a\ge 10$
• C. $a\le -2$
• D. -2 < a < -1

Answer 1

The correct option is D -2 < a < -1

$f\left(x\right)=x^2-(a+1)x+a^2+a-8=0$ has roots $\alpha$ and $\beta$ such that $\alpha <2$ and $\beta >2$
$\Rightarrow f\left(2\right)<0$ and $\frac{a+1}{2}<0$
$\Rightarrow a^2-a-6<0$ and a < -1
$\Rightarrow (a-3)(a+2)<0$ and a < -1
$\Rightarrow -2<a<-1$