A) \[{{H}_{2}}O\]
B) 0
C) 1
D) 2
Correct Answer: A
Solution :
\[\because \]\[f(x)=\int_{1}^{x}{\frac{\log t}{1+t}}dt\] and \[F(e)=f(e)+f\left( \frac{1}{e} \right)\] \[\Rightarrow \]\[F(e)=\int_{1}^{e}{\frac{\log t}{1+t}}dt+\int_{1}^{e}{\frac{\log t}{1+t}dt}\] Put\[t=\frac{1}{t}\]in second integration. \[=\int_{1}^{e}{\frac{\log t}{1+t}}dt+\int_{1}^{e}{\frac{\log t}{t(1+t)}dt}\] \[=\int_{1}^{e}{\frac{\log t}{t}}dt=\left[ \frac{{{(\log t)}^{2}}}{2} \right]_{1}^{e}\] \[=\frac{1}{2}[{{(\log e)}^{2}}-{{(\log 1)}^{2}}]=\frac{1}{2}\]
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