Statement-1 : Let `f(x) = |x^2-1|, x in [-2, 2] => f(-2) = f(2)` and hence there must be at least one `c in (-2,2)` so that `f'(c) = 0`, Statement 2: `f'(0) = 0`, where f(x) is the function of `S_1`

Let f(x)=x(x-1)(x-2) , x in [0,2] Find f(0) and f(2)

Let f(x) = x(x-1)(x-2),x in [0,2]. Prove that 'f' satisfies the conditions of Rolle's Theorem and there is more than one 'c' in (0,2) such that f'(C ) = 0

IF f(x) =2x^2 +bx +c and f(0) =3 and f (2) =1, then f(1) is equal to

If f is a function such that f(0) = 2, f(1) = 3, f(x+2) = 2f(x) - f(x + 1), then f(5) is

For all x in [1, 2] Let f"(x) of a non-constant function f(x) exist and satisfy |fprimeprime(x)|<=2. If f(1)=f(2) , then (A) There exist some a in (1,2) such that f'(a)=0 (B) f(x) is strictly increasing in (1,2) (C) There exists atleast one c in (1,2) such that f'(c)>0 (D) |f'(x)| lt 2 AA x in [ 1,2]

A function f(x) is such that f(x+1)=sqrt(f(x)+2) , where x in I^+ uu {0} and f(0)=0 then f(x) is a/an