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Let $f: R \rightarrow R$ be twice continuously differentiable. Let $f(0)=f(1)=f^{\prime}(0)=0 .$ Then,
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Verified Answer
The correct answers are:
$f^{\prime \prime}(c)=0$ for some $c \in R$
Let function $f(x)=x^{2}(x-1)$
$\Rightarrow \quad f^{\prime}(x)=3 x^{2}-2 x$
and $\quad f^{\prime \prime}(x)=6 x-2$
$\begin{array}{ll}\text { Now, } & f(0)=f(1)=f^{\prime}(0)=0\end{array}$
Then, according to question, at $x=\frac{1}{3}, \quad f^{\prime \prime}(x)=0$
(i.e. $c=1 / 3$ for some $C \in R$ )
$\Rightarrow \quad f^{\prime}(x)=3 x^{2}-2 x$
and $\quad f^{\prime \prime}(x)=6 x-2$
$\begin{array}{ll}\text { Now, } & f(0)=f(1)=f^{\prime}(0)=0\end{array}$
Then, according to question, at $x=\frac{1}{3}, \quad f^{\prime \prime}(x)=0$
(i.e. $c=1 / 3$ for some $C \in R$ )
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