Answer :

False

Rolle’s Theorem states that, Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers.Then there exists some c in (a, b) such that f’(c) = 0.


We have, f(x) = |x – 1| in [0, 2].


Since, polynomial and modulus functions are continuous everywhere f(x) is continuous.


Now, x-1=0


x=1


We need to check if f(x) is differentiable at x=1 or not.


We have,


Lf’(1) =


=


= ( f(x) = 1-x, if x< 1)


=


=


Rf’(1) =


=


= ( f(x) = x-1, if 1< x)


=


Lf’(1) ≠ Rf’(1)


f(x) is not differentiable at x=1.


Hence, rolle’s theorem is not applicable on f(x) since it is not differentiable at x=1 (0,2).


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