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`

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)

If f(x)=2(x-1)^(2), x in [0,2] then Rolle's theorem satisfies at

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)

If [[x,1]][[1,0],[2,0]]=0 ,then x equals (A) 0 (B) -2 (C) -1 (D) 2

Let f(x)={{:(x^(alpha)sin\ (1/x)sinpix\ \ \ ;\ \ x\ !=0),( 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ x=0):} If Rolles theorem is applicable to f(x) on [0,1] then range of alpha is (a) -oo lt alpha lt -1 (b) alpha=1 (c) -1 lt alpha lt oo (d) alpha ge 0

If rolle's theorem is applicable to the function f(x)=x^(3)-2x^(2)-3x+7 on the interval [0,k] then k is equal

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

Verify Rolles theorem for function f(x)=x(x-1)^(2) on [0,1]