Mathematics - Quadratic Equation Question with Solution | TestHub

$2^x+2^{|x|}\ge 2\sqrt{2}$ ...(i)
Case I: $x\ge 0$, then Eq. (i) becomes
$2^x+2^x\ge 2\sqrt{2}$
$\Rightarrow 2^x\ge \sqrt{2}\Rightarrow x\ge \frac{1}{2}$
Case II: $x<0$, then Eq. (i) becomes
$2^x+2^{-x}\ge 2\sqrt{2}$
$\Rightarrow t+\frac{1}{t}\ge 2\sqrt{2}$, where $2^x=t$
$\Rightarrow t^2-2\sqrt{2}t+1\ge 0$
$\Rightarrow x\le {\log }_2(\sqrt{2}-1)$
Also, $0<\sqrt{2}-1<1,{\log }_2(\sqrt{2}-1)<0$.
$\therefore$ The solution is
$\left(-\infty ,{\log }_2(\sqrt{2}-1)\right)]\cup \left[\frac{1}{2},\infty \right)$