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 `-oo

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

Rolle's theorem is applicable in case of phi(x) = a^sin x,a>0 in

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

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

Rolle's theorem is applicable in case of phi(x)=a^(sin x),a>0 in

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

Examine if Rolle's theorem is applicable to the following functions : f(x) = sin x on [0,pi]

Consider function f(x)={x{x}+1,\ \ \ \ \ \ \ \ \ \ \ \ 0lt=x<1 2-{x},\ \ \ \ 1lt=xlt=2\ \ \ where {x}: fractional part function of x . Statement-1: Rolles Theorem is not applicable to f(x) in [0,2] Statement-2: f(0)\ !=f(2)

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

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