Q. Find the sum of all the real solutions of the inequality $\frac{\left(x^{2} + 2\right) \left(\right. \sqrt{x^{2} - 16} \left.\right)}{\left(x^{4} + 2\right) \left(x^{2} - 9\right)}\leq 0$ .
NTA AbhyasNTA Abhyas 2022
Solution:
Given,
$\frac{\left(x^{2} + 2\right) \left(\right. \sqrt{x^{2} - 16} \left.\right)}{\left(x^{4} + 2\right) \left(x^{2} - 9\right)}\leq 0$
Here, $\left(x^{2} + 2\right)$ and $\left(x^{4} + 2\right)$ are positive $\forall x\in R$ .
Also, $\sqrt{x^{2} - 16}\geq 0$
Then, $\left(x^{2} - 9\right) < 0$
$\Rightarrow \left(x - 3\right)\left(x + 3\right) < 0$
$\Rightarrow -3 < x < 3...\left(\right.1\left.\right)$
Now, for $\sqrt{x^{2} - 16}$ to be defined,
$x^{2}-16\geq 0$
$\Rightarrow x\leq -4$ or $x\geq 4...\left(\right.2\left.\right)$
From $\left(\right.1\left.\right)$ and $\left(\right.2\left.\right)$ ,
$x=\phi$
But for equality in the given problem, numerator can be zero,
i.e., $x^{2}-16=0$
$\Rightarrow x=\pm4$
Hence, sum of all the real solution is $4-4=0$ .
$\frac{\left(x^{2} + 2\right) \left(\right. \sqrt{x^{2} - 16} \left.\right)}{\left(x^{4} + 2\right) \left(x^{2} - 9\right)}\leq 0$
Here, $\left(x^{2} + 2\right)$ and $\left(x^{4} + 2\right)$ are positive $\forall x\in R$ .
Also, $\sqrt{x^{2} - 16}\geq 0$
Then, $\left(x^{2} - 9\right) < 0$
$\Rightarrow \left(x - 3\right)\left(x + 3\right) < 0$
$\Rightarrow -3 < x < 3...\left(\right.1\left.\right)$
$x^{2}-16\geq 0$
$\Rightarrow x\leq -4$ or $x\geq 4...\left(\right.2\left.\right)$
From $\left(\right.1\left.\right)$ and $\left(\right.2\left.\right)$ ,
$x=\phi$
But for equality in the given problem, numerator can be zero,
i.e., $x^{2}-16=0$
$\Rightarrow x=\pm4$
Hence, sum of all the real solution is $4-4=0$ .