Let `f:(0,oo) in R` be given
`f(x)=overset(x)underset(1//x)int e^-(t+(1)/(t))(1)/(t)dt`, then

Options

A. `f(x)` is monotonically increasing on`[1,oo)`
B. `f(x)` is monotonocally decreasing on`(0,1)`
C. `f(x)+f(1/x)=0`, for all `x epsilon (0,oo)`
D. `f(2^(x))` is an odd function of `x` on `R`

Answer

A, C, D

Explanation

`f(x)=int_(1//x)^(x)e^(-(t+1/t))(dt)/t`
`impliesf'(x)=(e^(-(x+1/x)))/x-(e^(-x+1/x))/(1/x)(-1/(x^(2)))`
`=(2e^(-(x+1/x)))/xgt0` for ` x epsilon[1,oo)`
Therefore `f(x) `is increasing in `[1,oo)`
`f'(x)gt0` for `x epsilon(0,1)`
Hene `f(x)` is increasing
Also `f(x)+f(1/x)=int_(1//x)^(x)e^(-(t+1/t))(dt)/t+int_(-(t+1/t))(dt)/t=0`
`g(x)=f(2^(x))=int_(2^(-x))^(2^(x))(e^(-(t+1/t)))/t dt`
`:. g(-x)=int_(2^(x))^(2^(-x))(e^(-(t+1/t)))/t dt=-g(x)`
Hene `f(2^(x))` is an odd function.