Question

Solution of $2^x+2^{|x|}\ge 2\sqrt{2}$ is

• A. $(-\infty ,{\log }_2(\sqrt{2}+1)$
• B. $\left[{\log }_2(\sqrt{2}+1),\infty \right)$
• C. $\left(\frac{1}{2},{\log }_2(\sqrt{2}-1)\right)$
• D. $\left(-\infty ,{\log }_2(\sqrt{2}-1)\right]\cup \left[\frac{1}{2},\infty \right)$

Answer 1

Answer: (D)

We have, $2^x+2^x\ge 2\sqrt{2}$
$(x\ge 0)$
$\Rightarrow 2^x\ge \sqrt{2}$
$\Rightarrow x\ge \frac{1}{2}$
and $2^x+2^{-x}\ge 2\sqrt{2}$
$(x<0)$
$\Rightarrow t+\frac{1}{t}\ge 2\sqrt{2}$
(where $t=2^x$ )
$\Rightarrow t^2-2\sqrt{2}t+1\ge 0$
$\Rightarrow [t-(\sqrt{2}-1)][t-(\sqrt{2}+1)]\ge 0$
$\Rightarrow t\le \sqrt{2}-1$
or $t\ge \sqrt{2}+1$
but $t>0$
$\Rightarrow 02^x\le \sqrt{2}-1$
or$2^x\ge \sqrt{2}+1$
$\Rightarrow -\infty <x\le {\log }_2(\sqrt{2}-1)$
or $x\ge {\log }_2(\sqrt{2}+1)$
(but not acceptable as $x<0$ )
$\therefore x\in \left(-\infty ,{\log }_2(\sqrt{2}-1)\right]\cup \left[\frac{1}{2},\infty \right)$