Let g: R → R be a non constant twice differentiable such that g^{prime}left(frac{1}{2}right)=g^{pr

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5. Continuity and Differentiation

hard

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

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

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

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

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

(JEE MAIN-2024)

Solution

$f^{\prime}(x)=\frac{g^{\prime}(x)-g^{\prime}(2-x)}{2}, f^{\prime}\left(\frac{3}{2}\right)=\frac{g^{\prime}\left(\frac{3}{2}\right)-g^{\prime}\left(\frac{1}{2}\right)}{2}=0$

Also $\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$

$ \Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0 $

$ \Rightarrow \text { rootsin }\left(\frac{1}{2}, 1\right) \text { and }\left(1, \frac{3}{2}\right)$

$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$

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