Sequence And Series Question 478
Question: $ {2^{\sin \theta }}+{2^{\cos \theta }} $ is greater than
[AMU 2000]
Options:
A) $ \frac{1}{2} $
B) $ \sqrt{2} $
C) $ {2^{\frac{1}{\sqrt{2}}}} $
D) $ {2^{( 1-,\frac{1}{\sqrt{2}} )}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{1}{2}[ {2^{\sin \theta }}+{2^{\cos \theta }} ]\ge \sqrt{{2^{\sin \theta }}{2^{\cos \theta }}} $ ( $ \because A\text{.M}\text{.}\ge 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})}} $ .
BETA
