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L e tf:[1,oo)->Ra n df(x)=int1^x(e^t)/t ...

`L e tf:[1,oo)->Ra n df(x)=int_1^x(e^t)/t dt-e^xdotT h e n` `f(x)` is an increasing function `lim_(x->oo)f(x)->oo` `f^(prime)(x)` has a maxima at `x=e` `f(x)` is a decreasing function

A

`f(x)` is an increasing function

B

`lim_(x tooo) f(x)to oo`

C

`f'(x)` has a maxima at `x=e`

D

`f(x)` is a decreasing function

Text Solution

Verified by Experts

The correct Answer is:
A, B

`f(x)=x int_(1)^(x) (e^(t))/t dt-e^(x)`
`:.f'(x) = x(e^(x))/x+int_(1)^(x)(e^(t))/t dt-e^(x)`
`=int_(1)^(x)(e^(t))/t dt gt0 [ :' x epsilon[1,oo)]`
Thus `f(x)` is an increasing function.
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