• # question_answer ${{2}^{\sin \theta }}+{{2}^{\cos \theta }}$ is greater than [AMU 2000] A) $\frac{1}{2}$ B) $\sqrt{2}$ C) ${{2}^{\frac{1}{\sqrt{2}}}}$ D) ${{2}^{\left( 1-\,\frac{1}{\sqrt{2}} \right)}}$

$\frac{1}{2}\left[ {{2}^{\sin \theta }}+{{2}^{\cos \theta }} \right]\ge \sqrt{{{2}^{\sin \theta }}{{2}^{\cos \theta }}}$   ($\because \text{A}\text{.M}\text{.}\ge \text{G}\text{.M}\text{.}$) Þ ${{2}^{\sin \theta }}+{{2}^{\cos \theta }}\ge {{2.2}^{(\sin \theta +\cos \theta )/2}}$             .....(i) Now $(\sin \theta +\cos \theta )=\sqrt{2}\sin (\theta +\pi /4)\ge -\sqrt{2}$ for all real q ${{2}^{\sin \theta }}+{{2}^{\cos \theta }}\ge {{2.2}^{(\sin \theta +\cos \theta )/2}}>2\,.\,{{2}^{-\sqrt{2}/2}}$ Þ ${{2}^{\sin \theta }}+{{2}^{\cos \theta }}\ge {{2}^{1-(1/\sqrt{2})}}$.