Question

The set(s) of values of $x$ satisfying the inequality ${\log }_{1/6}\left({\log }_6\left(\frac{x^2+|x|}{|x|+4}\right)\right)>0,$ is/are

• A. $(-\infty ,-2)$
• B. $(-8,-2)$
• C. $(2,8)$
• D. $(2,\infty )$

Answer 1

Answer: (B), (C)

The correct options are
B $(-8,-2)$
C $(2,8)$

${\log }_{1/6}\left({\log }_6\left(\frac{x^2+|x|}{|x|+4}\right)\right)>0$
For the given logarithm to be defined, required conditions are
$\left(i\right){\log }_6\left(\frac{x^2+|x|}{|x|+4}\right)>0\text{and}\left(ii\right)\frac{x^2+|x|}{|x|+4}>0$
$\Rightarrow \frac{x^2+|x|}{|x|+4}>1\text{and}\frac{x^2+|x|}{|x|+4}>0$
$\Rightarrow \frac{x^2+|x|}{|x|+4}>1$
$\Rightarrow x^2+|x|>|x|+4(\because |x|+4\text{is positive})$
$\Rightarrow x^2-4>0$
$\Rightarrow x\in (-\infty ,-2)\cup (2,\infty )...\left(1\right)$

From given inequality,
${\log }_{1/6}\left({\log }_6\left(\frac{x^2+|x|}{|x|+4}\right)\right)>0$
$\Rightarrow {\log }_6\left(\frac{x^2+|x|}{|x|+4}\right)<{\left(\frac{1}{6}\right)}^0$
$\Rightarrow {\log }_6\left(\frac{x^2+|x|}{|x|+4}\right)<1$
$\Rightarrow \frac{x^2+|x|}{|x|+4}<6$
$\Rightarrow x^2+|x|<6|x|+24(\because |x|+4\text{is positive})$
$\Rightarrow x^2-5|x|-24<0$
$\Rightarrow (|x|-8)(|x|+3)<0(\because x^2=|x{|}^2)$
Since, $|x|+3$ is always positive.
$\Rightarrow |x|-8<0$
$\Rightarrow x\in (-8,8)...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right),$
$x\in (-8,-2)\cup (2,8)$