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 monotonically increasing on (0,1)
C. f(x) is monotonocally decreasing on (0,1)
D. `f(2^(x))`is an odd function of x on R

Answer

A, C, D

Explanation

We have,
`f(x)=overset(x)underset(1//x)int e^-(t+(1)/(t))(1)/(t)dt`
`rArr f'(x)=(1)/(x)e^-(x+(1)/(x))+(x)/(x^(2))e^-(x-(1)/(x))`
`rArr f'(x)=(2)/(x)e-(x+(1)/(x)) gt 0` for all `x in [1,oo)`
`:.f(x)` is monotonically increasing on `[1,oo)`.
So, option (a) is correct
Again, `f(x)=overset(x)underset(1//x)int e^-(t+(1)/(t))(dt)/(t)`
`rArr f((1)/(x))=overset(1//x)underset(x)int e^-(t+(1)/(t))(dt)/(t)`
`rArr f((1)/(x))=overset(u)underset(1//u)int ue^(-(u+(1)/(u)))((-1)/(u^(2)))du` where `t=(1)/(u)`
`rArr f((1)/(x))=-overset(u)underset(1//u)int e^-(u+(1)/(2))(1)/(u)du=-overset(x)underset(1//x)int e^-(t+(1)/(t))(1)/(t)dt=-f(x)`
`:.f(x)+f((1)/(x))=0`. So, option (c ) is correct.
Finally, `f(x)=overset(x)underset(1//x)int e^-(t+(1)/(t))(1)/(t)dt`
`rArr g(x)f(2^(x))=overset(2^(x))underset(2^(-x))int e^-(t+(1)/(t))(1)/(t)dt`
`rArr g(-x)=overset(2^-(x))underset(2^(x))int e^-(t+(1)/(t))(1)/(t)dt=-overset(2^(x))underset(2^-(x))int e^-(u+(1)/(2))(1)/(u)du=-g(x)`
`g(x)=f(2^(x))` is an odd function of x on R.
So,option(d) is correct.