1.

Let `F (x) = f(x) + f ((1)/(x)),` where `f (x) = int _(1) ^(x ) (log t)/(1+t) dt.` Then F (e) equalsA. `1//2`B. 0C. 1D. 2

Answer» Correct Answer - A
`F(e)=f(e)+f((1)/(e))`
`rArrF(e)=int_(1)^(e)(logt)/(1+t)dt+int_(1)^(1//e)(logt)/(1+t)dt[becausef(x)=int_(1)^(x)(logt)/(1+t)dt]`
On putting t `=(1)/(t)` in second integration , we get
`F(e)=int_(1)^(e)(logt)/(1+t)dt+int_(1)^(e)(logt)/(t(1+t))`
`=int_(1)^(e)(logt)/(t)dt=[((logt)^(2))/(2)]_(1)^(e)`
`=(1)/(2)[(loge)^(2)-(log1)^(2)]=(1)/(2)`


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