2^ + 2^ is greater than - Vidyadip

MCQ

${2^{\sin \theta }} + {2^{\cos \theta }}$ is greater than

A
$\frac{1}{2}$
B
$\sqrt 2 $
C
${2^{\frac{1}{{\sqrt 2 }}}}$
✓
${2^{\left( {1 - \,\frac{1}{{\sqrt 2 }}} \right)}}$

✓

Answer

Correct option: D.

${2^{\left( {1 - \,\frac{1}{{\sqrt 2 }}} \right)}}$

d
(d) $\frac{1}{2}\left[ {{2^{\sin \theta }} + {2^{\cos \theta }}} \right] \ge \sqrt {{2^{\sin \theta }}{2^{\cos \theta }}} $

(${\rm{A}}{\rm{.M}}{\rm{.}} \ge {\rm{G}}{\rm{.M}}{\rm{.}}$)

==> ${2^{\sin \theta }} + {2^{\cos \theta }} \ge {2.2^{(\sin \theta + \cos \theta )/2}}$ .....(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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