A) \[{{\log }_{e}}2<I<\frac{\pi }{4}\]
B) \[{{\log }_{e}}2>I\]
C) \[I=\frac{\pi }{4}\]
D) \[I={{\log }_{e}}2\]
Correct Answer: A
Solution :
\[{{x}^{2}}<{{x}^{\frac{\pi }{2}}}\,<x\,\,\forall \,\,x\in \,(0,\,\,1)\] \[\Rightarrow \,\frac{1}{1+x}<\frac{1}{1+{{x}^{\frac{\pi }{2}}}}\,<\frac{1}{1+{{x}^{2}}}\] \[\Rightarrow \,(\ln \,(1+x))_{0}^{1}\,<\int\limits_{0}^{1}{\frac{1}{1+{{x}^{\frac{\pi }{2}}}}}\,<({{\tan }^{-1}}\,x)_{0}^{1}\] \[\Rightarrow \,{{\log }_{e}}2<I<\frac{\pi }{4}\] So, \[{{e}^{2}}=16\,{{m}^{2}}-9\] ?(2)
You need to login to perform this action.
You will be redirected in
3 sec
