Explanation

(d)
We have, $2^{x} + 2^{|x|} \geq 2\sqrt{2}$
If $x \geq 0,$then$\ 2^{x} + 2^{x} \geq 2\sqrt{2}$
$\Rightarrow 2^{x} \geq \sqrt{2}\ \Rightarrow x \geq \frac{1}{2}$
and if $x < 0,\$then$\ 2^{x} + 2^{- x} \geq 2\sqrt{2}$
$\Rightarrow t + \frac{1}{t} \geq 2\sqrt{2}\ \ \ \$(where$\ t = 2^{x})$
$\Rightarrow t^{2} - 2\sqrt{2}t + 1 \geq 0$
$\Rightarrow \left( t - \left( \sqrt{2} - 1 \right) \right)\left( t - \left( \sqrt{2} + 1 \right) \right) \geq 0$
$\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 $x \geq \log_{2}\left( \sqrt{2} + 1 \right)$
Which is not possible, because $x > 0$
$\therefore x \in ( - \infty,\log_{2}{(\sqrt{2} - 1)) \cup \left\lbrack \left. \ \frac{1}{2},\infty \right) \right.\ }$