Q38 of 105 Page 17

Let F defined on [0, 1] be twice differentiable such that | f”(x) ≤ 1 for all x ϵ [0, 1]. If f(0) = f(1), then show that |f’(x) | < 1 for all x ϵ [0, 1] ?

As f(0) = f(1) and f is differentiable, hence by Rolles theorem:

for some c [0,1]


let us now apply LMVT (as function is twice differentiable) for point c and x [0,1],


hence,


f ”(d)


⇒ f ”(d)


⇒ f ”(d)


A given that | f ”(d)| <=1 for x [0,1]


⇒


⇒


Now both x and c lie in [0,1], hence [0,1]


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