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Question
Let f defined on [0, 1] be twice differentiable such that | f (x) | ≤ 1 for all x ∈ [0, 1]. If f(0) = f(1), then show that | f'(x) | < 1 for all x ∈ [ 0, 1] ?
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Solution
If a function is continuous and differentiable and f(0) = f(1) in given domain x ∈ [0, 1],
then by Rolle's Theorem;
f'(x) = 0 for some x ∈ [0, 1]
Given: | f"(x)| ≤ 1
On integrating both sides we get,
|f'(x)| ≤ x
Now, within interval x ∈ [0, 1]
We get, | f' (x)| < 1.
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