Question

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

• A. $\left(-\infty, \log _{2}(\sqrt{2}+1)\right.$
• 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} \geq 2 \sqrt{2}$
$(x \geq 0)$
$\Rightarrow 2^{x} \geq \sqrt{2} $
$\Rightarrow x \geq \frac{1}{2}$
and $2^{x}+2^{-x} \geq 2 \sqrt{2}$
$(x < 0)$
$\Rightarrow t+\frac{1}{t} \geq 2 \sqrt{2}$
(where $t=2^{x}$ )
$\Rightarrow t^{2}-2 \sqrt{2} t+1 \geq 0 $
$\Rightarrow {[t-(\sqrt{2}-1)][t-(\sqrt{2}+1)] \geq 0} $
$ \Rightarrow t \leq \sqrt{2}-1 $
or $ t \geq \sqrt{2}+1 $
but $ t > 0 $
$ \Rightarrow 0< 2^{x} \leq \sqrt{2}-1 $
or$ 2^{x} \geq \sqrt{2}+1$
$\Rightarrow -\infty < x \leq \log _{2}(\sqrt{2}-1) $
or $ x \geq \log _{2}(\sqrt{2}+1)$
(but not acceptable as $x < 0$ )
$\therefore x \in\left(-\infty, \log _{2}(\sqrt{2}-1)\right] \cup\left[\frac{1}{2}, \infty\right)$