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Letf $: \mathbb{R} \rightarrow \mathbb{R}$, be twice continuously differentiable (or $f$ " exists and is continuous) such that $f(0)=f(1)=f^{\prime}(0)=0$. Then
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The correct answer is:
$\mathrm{f}^{\prime \prime}(\mathrm{c})=0$ for some $\mathrm{c} \in \mathbb{R}$
Hint:
$f(0)=f(1)=0$
By Rolle's theorem $\mathrm{f}^{\prime}(\mathrm{c})=0$ for some $\mathrm{c} \in(0,1)$
Now, $\mathrm{f}^{\prime}(0)=\mathrm{f}^{\prime}(\mathrm{c})=0$. Again by Rolle's theorem $\mathrm{f}^{\prime \prime}\left(\mathrm{c}_{1}\right)=0 \quad \mathrm{c}_{1} \in(0, \mathrm{c})$
$f(0)=f(1)=0$
By Rolle's theorem $\mathrm{f}^{\prime}(\mathrm{c})=0$ for some $\mathrm{c} \in(0,1)$
Now, $\mathrm{f}^{\prime}(0)=\mathrm{f}^{\prime}(\mathrm{c})=0$. Again by Rolle's theorem $\mathrm{f}^{\prime \prime}\left(\mathrm{c}_{1}\right)=0 \quad \mathrm{c}_{1} \in(0, \mathrm{c})$
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