MCQ
${2^{\sin \theta }} + {2^{\cos \theta }}$ is greater than
- A$\frac{1}{2}$
- B$\sqrt 2 $
- C${2^{\frac{1}{{\sqrt 2 }}}}$
- ✓${2^{\left( {1 - \,\frac{1}{{\sqrt 2 }}} \right)}}$
(${\rm{A}}{\rm{.M}}{\rm{.}} \ge {\rm{G}}{\rm{.M}}{\rm{.}}$)
==> ${2^{\sin \theta }} + {2^{\cos \theta }} \ge {2.2^{(\sin \theta + \cos \theta )/2}}$ .....(i)
Now $(\sin \theta + \cos \theta ) = \sqrt 2 \sin (\theta + \pi /4) \ge - \sqrt 2 $
For all real $\theta$,
${2^{\sin \theta }} + {2^{\cos \theta }} \ge {2.2^{(\sin \theta + \cos \theta )/2}} > 2\,.\,{2^{ - \sqrt 2 /2}}$
==> ${2^{\sin \theta }} + {2^{\cos \theta }} \ge {2^{1 - (1/\sqrt 2 )}}$.
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