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Let $f: R \rightarrow R$ be a twice continuously differentiable function such that $f(0)=f(1)=f^{\prime}(0)=0 .$ Then
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The correct answer is:
$f^{\prime \prime}=0$ for some $C \in A$
We have.
$$f: \mathbb{R} \rightarrow \mathbb{R} $$ be a twice continuousty differentiable function such that $f(0)=f(1)=f'(0)=0$
Now, for atleast one value of $\epsilon_{1} \in(0,1)$ $f'\left(c_{1}\right)=0 \quad$ (by Rolle's theorem)
Again. $\quad f'(0)=0=f(c_{1})$
$f'(c)=$ ofor some $c \in\left(0, c_{1}\right)$ (by Rolle's theorem)
$$f: \mathbb{R} \rightarrow \mathbb{R} $$ be a twice continuousty differentiable function such that $f(0)=f(1)=f'(0)=0$
Now, for atleast one value of $\epsilon_{1} \in(0,1)$ $f'\left(c_{1}\right)=0 \quad$ (by Rolle's theorem)
Again. $\quad f'(0)=0=f(c_{1})$
$f'(c)=$ ofor some $c \in\left(0, c_{1}\right)$ (by Rolle's theorem)
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