The true set of real values of x for whi...

The true set of real values of `x` for which the function `f(x)=xlnx-x+1` is positive is

`(1,oo)`

`(1//e,oo)`

`[e,oo)`

`(0,1)cup(1,oo)`

Text Solution

Verified by Experts

The correct Answer is:

We have fX() =`log_(e)x-x+1,xgt0`
`therefore f(x)=1+log_(e)x-1=log_(e)x`
`therefore f(X)lt0 if 0ltxlt1`
`f(X)gt0 if x gt 1`
also `underset(xrarr0^(+))lim (xlog_(e)x-x+1)=1`
So in (0,1) ,f(x) decreases from 1 to 0 and in `(1,oo)` increases from `0 to oo`
Hence `f(X) gt 0 for x in (0,1)cup(1,oo)`

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