Question

The complete set of values of 'a' such that $x^2+ax+a^2+6a$ < 0 $\forall$ x $ϵ$ [-1, 1] is:

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is D

None of these

f(x) = $x^2+ax+a^2+6a<0$

$\Rightarrow$ f(-1) < 0 , f(1) < 0 , D < 0.

$\Rightarrow$ $a^2-4(a^2+6a)>0,a^2+7a+1<0,a^2+5a+1<0$

-8 < a < 0, $\frac{-7-\sqrt{45}}{2}$ < a < $\frac{-7+\sqrt{45}}{2}$
and $\frac{-5-\sqrt{21}}{2}$ < a <$\frac{-5+\sqrt{21}}{2}$
$\Rightarrow$ $\frac{-5-\sqrt{21}}{2}$ < a < $\frac{-5+\sqrt{21}}{2}$