Let `f` defined on `[0, 1]` be twice differentiable such that `| f"(x)| <=1` for `x in [0,1]`. if `f(0)=f(1)` then show that `|f'(x)<1` for all `x in [0,1]`.

Explanation

Let `f` be defined on `[0,1]`

`|f''∣<=1` for all `x in [0,1]`

for any `f(x)=x^2`

`f′(x)=2x`

`f''(x)=2`

At `x_1=0`,`f''(0)=2`

At `x^2=1`,`f''(1)=2`

`f(0)=f(1)`

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