If `f(x)=x^(alpha)log x and f(0)=0`, then the value of `'alpha'` for which Roole's theorem can be applied in [0, 1], is

Options

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

Answer

D

Explanation

Clearly, f(x) is continuous and differentiable on (0,1) for `alpha gt0`
For f(x) to be continuous at x=0 we must have
`underset(x to o^(+)) limf(x)=f(0) i.e.,underset(x to o^(+)) lim x^(alpha)log x" " [ 0xx oo "from for " alpha lt0]`
Now,
`underset(x to o^(+)) lim(log x)/((1)/(x^(alpha)))" " [ (oo)/(oo)"form"]`
`underset(x to o^(+)) lim((1)/(x))/((alpha)/(x^(alpha+1)))underset(xto0^(+))lim-(x^(alpha))/(alpha)=0" " [ :.alpha gt0]`
So, f(x) is continuous at `x = 0 "for" alpha gt0`
Hence, Rolle's theorem can be applied on f(x) in [0,1] for all `alpha+0`. So, option(d) is correct.