If `f(x)={(x^(alpha)logx , x > 0),(0, x=0):}` and Rolle's theorem is applicable to `f(x)` for `x in [0, 1]` then `alpha` may equal to (A) -2 (B) -1 (C) 0 (D) `1/2`

Options

A. -2
B. -1
C. 0
D. `(1)/(2)`

Answer

D

Explanation

To satisfy Rolle's theorem it should be continuous in [0, 1].
RHL = f(0)
`implies" "underset(xto0^(+))lim(logx)/(x^(-alpha))=0`
`implies" "underset(xto0^(+))lim(1//x)/(-alphax^(-alpha-1))=0`
[Using L'Hospital's Rule]
`implies" "underset(xto0^(+))lim(1)/(alphax^(-alpha))=0`
`implies" "underset(xto0^(+))lim-(x^(alpha))/(alpha)=0`
which shows `alpha gt 0` otherwise it would be discontinuous. Also when `alpha gt 0,` f(x) is differentiable in (0, 1) and f(1) = f(0) = 0. Clearly `alpha gt 0` thus `alpha=(1)/(2)` is the possible answer.