Question

Sum of the squares of all integral values of $a$ for which the inequality $x^{2} + a x + a^{2} + 6 a < 0$ is satisfied for all $x \in \left(1 , 2\right)$ must be equal to

• A. $90$
• B. $89$
• C. $88$
• D. $91$

Answer 1

Answer: (D)

Let, $ f \left(x\right) = x^{2} + a x + a^{2} + 6 a$
$∴ \, \, $ $f \left(\right. 1 \left.\right) \leq 0$

$\Rightarrow $ $a^{2} + 7 a + 1 < 0$
or $\frac{- 7 - 3 \sqrt{5}}{2} < a < \frac{- 7 + 3 \sqrt{5}}{2}$ ....(i)
$f \left(\right. 2 \left.\right) \leq 0$
$\Rightarrow $ $a^{2} + 8 a + 4 < 0$
or $ - 4 - 2 \sqrt{3} < a < - 4 + 2 \sqrt{3}$ ....(ii)
and $D>0$
$\Rightarrow a^{2}-4\cdot 1\left(a^{2} + 6 a\right)>0$
$\Rightarrow $ $a^{2} + 8 a < 0$
or $- 8 < a < 0$ ....(iii)
From Eqs. (i), (ii) and (iii), we get,
$\frac{- 7 - 3 \sqrt{5}}{2} \leq a \leq - 4 + 2 \sqrt{3}$
Hence, integral values of $a$ are $- 6 , - 5 , ⁡ - 4 , ⁡ - 3 , ⁡ - 2 , ⁡ - 1$
Required Sum
$= \left(- 6\right)^{2} + \left(- 5\right)^{2} + \left(- 4\right)^{2} + \left(- 3\right)^{2} + \left(- 2\right)^{2} + \left(- 1\right)^{2}$
$ \, = \, 91$ .