A) \[\frac{e-1}{2\sqrt{e}}\]
B) \[\frac{e+1}{2\sqrt{e}}\]
C) \[\frac{e-1}{\sqrt{e}}\]
D) \[\frac{e+1}{\sqrt{e}}\]
Correct Answer: B
Solution :
\[\frac{{{e}^{x}}+{{e}^{-x}}}{2}=1+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{4}}}{4\,!}+\frac{{{x}^{6}}}{6\,!}+....\infty \] Putting\[x=\frac{1}{2}\], we get \[1+\frac{1}{4\,.\,2\,!}+\frac{1}{16.4\,!}+\frac{1}{64.6!}+...\infty \] \[=\frac{{{e}^{1/2}}+{{e}^{-1/2}}}{2}=\frac{e+1}{2\sqrt{e}}\].
You need to login to perform this action.
You will be redirected in
3 sec
