Application Of Derivatives Question 5
Question 5 - 2024 (30 Jan Shift 1)
Let $\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be a non constant twice differentiable such that $\mathrm{g}^{\prime}\left(\frac{1}{2}\right)=\mathrm{g}^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $\mathrm{f}$ is defined as $f(x)=\frac{1}{2}[g(x)+g(2-x)]$, then
(1) $f^{\prime \prime}(x)=0$ for atleast two $x$ in $(0,2)$
(2) $f^{\prime \prime}(x)=0$ for exactly one $x$ in $(0,1)$
(3) $f^{\prime \prime}(x)=0$ for no $x$ in $(0,1)$
(4) $\mathrm{f}^{\prime}\left(\frac{3}{2}\right)+\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=1$
Show Answer
Answer (1)
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 $\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\frac{\mathrm{g}^{\prime}\left(\frac{1}{2}\right)-\mathrm{g}^{\prime}\left(\frac{3}{2}\right)}{2}=0, \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$
$\Rightarrow \mathrm{f}^{\prime}\left(\frac{3}{2}\right)=\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=0$
$\Rightarrow \operatorname{roots} \operatorname{in}\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$
$\Rightarrow \mathrm{f}^{\prime \prime}(\mathrm{x})$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$
