If f(x)={(x^(alpha)logx , x > 0),(0, x=...

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`

Similar Questions

Explore conceptually related problems

If f(x) = {:{(x^(alpha)logx,","xgt 0),(0,","x=0):} and Rolle' theorem is applicable to f(x) for x in [0,1] then alpha may be equal to

Rolle's theorem is applicable for the function f(x) = |x-1| in [0,2] .

Let f(x)={(x^(a) , x > 0),(0, x=0):} .Rolle's theorem is applicable to f for x in[0,1] ,if a : (A) -2 (B) -1 (C) 0 (D) (1)/(2)

Rolle's theorem is applicable for the function f(x) = |x-1| "in " x in [0, 2]

Let f (x)= [{:(x ^(2alpha+1)ln x ,,, x gt0),(0 ,,, x =0):} If f (x) satisfies rolle's theorem in interval [0,1], then alpha can be:

Define f(x)={{:(x",",0lexle1),(2-x,,1lexle2):} then Rolles theorem is not applicable to f(x) because

If Rolle's theorem is applicable to the function f(x)=a cos(x+b)+c in [(pi)/(8),(pi)/(4)], then b may be equal to (Here a!=0)