Let $g: \mathbf{R} \rightarrow \mathbf{R}$ be a non constant twice differentiable function such that $\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $f(x)=\frac{1}{2}[g(x)+g(2-x)]$, then

Question

$f^{\prime \prime}(x)=0$ for atleast two $x$ in $(0,2)$

$f^{\prime}\left(\frac{3}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)=1$

$f^{\prime \prime}(x)=0$ for no $x$ in $(0,1)$

$f^{\prime \prime}(x)=0$ for exactly one $x$ in $(0,1)$

Solution

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