Jee Main 2024 30 01 2024 Shift1 - Question7
Question 7
If $g^{\prime}\left(\frac{3}{2}\right)=g^{\prime}\left(\frac{1}{2}\right)$ and
$f(x)=\frac{1}{2}[g(x)+g(2-x)]$ and $f^{\prime}\left(\frac{3}{2}\right)=f^{\prime}\left(\frac{1}{2}\right)$ then
(1) $f^{\prime \prime}(x)=0$ has exactly one root in $(0,1)$
(2) $f^{\prime \prime}(x)=0$ has no root in $(0,1)$
(3) $f^{\prime \prime}(x)=0$ has at least two roots in $(0,2)$
(4) $f^{\prime \prime}(x)=0$ has 3 roots in $(0,2)$
Show Answer
Answer: (3)
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 $f^{\prime}\left(\frac{1}{2}\right)=\frac{g^{\prime}\left(\frac{1}{2}\right)-g^{\prime}\left(\frac{3}{2}\right)}{2}=0, f^{\prime}(1)=0$
$\Rightarrow f^{\prime}\left(\frac{3}{2}\right)=f^{\prime}\left(\frac{1}{2}\right)=0$
$\Rightarrow$ roots in $\left(\frac{1}{2}, 1\right)$ and $\left(1, \frac{3}{2}\right)$
$\Rightarrow f^{\prime \prime}(x)$ is zero at least twice in $\left(\frac{1}{2}, \frac{3}{2}\right)$
BETA
