Question

The interval in which the inequality ${\left(\frac{1}{2}\right)}^{\frac{|x+3|}{3-|x|}}>8$ holds true is

• A. $(6,\infty )$
• B. $(-\infty ,3)\cup (6,\infty )$
• C. $(3,6)$
• D. $(-\infty ,3)$

Answer 1

The correct option is C $(3,6)$

${\left(\frac{1}{2}\right)}^{\frac{|x+3|}{3-|x|}}>8$
$\Rightarrow 2^{\frac{-|x+3|}{3-|x|}}>2^3$
$\Rightarrow -\frac{|x+3|}{3-|x|}>3$
$\Rightarrow \frac{|x+3|}{3-|x|}<-3$

Case-1: $x<-3$
$\Rightarrow \frac{-(x+3)}{3+x}<-3$
$\Rightarrow -1<-3$, which is not possible

Case-2: $-3<x<0$
$\Rightarrow \frac{x+3}{3+x}<-3$
$\Rightarrow 1<-3$, which is not possible

Case-3: $x\ge 0$
$\Rightarrow \frac{x+3}{3-x}<-3$
$\Rightarrow \frac{x+3}{3-x}+3<0$
$\Rightarrow \frac{6-x}{3-x}<0$
$\Rightarrow 3<x<6$
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