A) \[{{\log }_{e}}2\]
B) \[{{e}^{2}}\]
C) 0
D) \[\frac{2}{e}\]
Correct Answer: A
Solution :
[a] We have, if \[{{e}^{x}}>2,\frac{2}{{{e}^{x}}}<1\]. Also \[\frac{2}{{{e}^{x}}}>0\] \[\Rightarrow 0<\frac{2}{{{e}^{x}}}<1\therefore Ifx>{{\log }_{e}}2,\left[ \frac{2}{{{e}^{x}}} \right]=0\] Again if \[0<x<{{\log }_{e}}2\] then \[1<{{e}^{x}}<2\] \[\Rightarrow 1>\frac{1}{{{e}^{x}}}>\frac{1}{2}\Rightarrow 2>\frac{2}{{{e}^{x}}}>1\] or \[1<\frac{2}{{{e}^{x}}}<2\] \[\therefore \left[ \frac{2}{{{e}^{x}}} \right]=1\therefore I=\int\limits_{0}^{\infty }{\left[ \frac{2}{{{e}^{x}}} \right]dx=\int\limits_{0}^{\infty }{\left[ 2{{e}^{-x}} \right]}\,dx}\] \[=\int\limits_{0}^{\log \,2}{\left[ 2{{e}^{-x}} \right]dx+\int\limits_{\log \,\,2}^{\infty }{\left[ 2{{e}^{-x}} \right]}dx}\] \[=\int\limits_{0}^{\log \,\,2}{(1)dx+\int\limits_{\log \,\,2}^{\infty }{(0)dx={{\log }_{e}}2}}\]
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