Question

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

• A. $(-\infty ,lo{g}_2(\sqrt{2}+1\left)\right)$
• B. $(0,\infty )$
• C. $(\frac{1}{2},lo{g}_2(\sqrt{2}-1)$
• D. $(-\infty ,lo{g}_2(\sqrt{2}-1\left)\right]\cup [\frac{1}{2},\infty )$

Answer 1

The correct option is D $(-\infty ,lo{g}_2(\sqrt{2}-1\left)\right]\cup [\frac{1}{2},\infty )$

$2^x+2^{|x|}>2\sqrt{2}$
for x > 0
$2^x+2^x>2^{\frac{3}{2}}$
$\Rightarrow {2.2}^x>2^{\frac{3}{2}}$
$\Rightarrow x+1>,\frac{3}{2}$
$\Rightarrow x>\frac{1}{2}\dots \dots \left(1\right)$
For x < 0
$2^x+2^{-x}>2\sqrt{2}$
$\Rightarrow 2^x+\frac{1}{2^x}>2\sqrt{2}$
=assume $2^x=y$ [y is always greater than 0]
$y+\frac{1}{y}>2\sqrt{2}$
$y^2+1-2\sqrt{2}y>0$
$y^2+1-2\sqrt{2}y=0$
$y=\frac{2\sqrt{2}\pm \sqrt{(2\sqrt{2}{)}^2}-4}{2}$
$=\frac{2\sqrt{2}\pm 2}{2}=\sqrt{2}\pm 1$
$\Rightarrow \frac{[y-\sqrt{2}+1][y-(\sqrt{2}-1\left)\right]}{y}>0$
$\Rightarrow [y-(\sqrt{2}+1\left)\right][y-(\sqrt{2}-1\left)\right]>0[sina,y>0]$
$\Rightarrow yϵ(-\infty ,\sqrt{2}-1)\cup [\sqrt{2}+1,\infty )$
Or $2^xϵ(-\infty ,\sqrt{2}-1)\cup [\sqrt{2}+1,\infty ]$
$\Rightarrow xϵ(-\infty ,lo{g}_2(\sqrt{2}-1\left)\right]\cup \left[lo{g}_2\right(\sqrt{2}+1),\infty )\dots \dots \left(2\right)$
$\sqrt{2}-1=1.414-1=0.414$
$\sqrt{2}-1=1.414+1=2.414$

From the graph, $lo{g}_2(\sqrt{2}-1)=lo{g}_2\left(0.414\right)<0$
And $lo{g}_2(\sqrt{2}+1)=lo{g}_2\left(2.414\right)<0$
Using log tables, we can find $lo{g}_{2}2.414=1.2714$
So, equation (2) can be written as
$xϵ(-\infty ,lo{g}_2(\sqrt{2}-1\left)\right]\cup [1.2714,0)\dots \dots \left(3\right)$
Taking using of (1) and (3), we get
$xϵ(-\infty ,lo{g}_2(\sqrt{2}-1\left)\right]\cup [\frac{1}{2},\infty )$
Since $lo{g}_2(\sqrt{2}+1)>\frac{1}{2}$