Question

Let $f(x)$ be a non-constant twice differentiable function on $R$ such that $f(2+x)=f(2-x)$ and $f^{\prime}\left(\frac{1}{2}\right)=f^{\prime}(1)=0$. Then minimum number of roots of the equation $f^{\prime \prime}(x)=0$ in $(0,4)$ are

• A. 2
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (B)

$ f (2+ x )= f (2- x )$
$f^{\prime}(2+x)=-f^{\prime}(2-x)$
Put
$x=0, \,\,\,\, f^{\prime}(2)=0 $
$x=-1 \,\,\,\, f^{\prime}(1)=-f^{\prime}(3)=0 $
$x=\frac{-3}{2}, \,\,\,\, f^{\prime}\left(\frac{1}{2}\right)=-f^{\prime}\left(\frac{7}{2}\right)=0 $
$\therefore f^{\prime}\left(\frac{1}{2}\right)=0=f^{\prime}(1)=f^{\prime}(2)=f^{\prime}(3)=f^{\prime}\left(\frac{7}{2}\right)$
Using Rolle's theorem in $y = f ^{\prime}( x )$.
Minimum number of roots of $f ^{\prime \prime}( x )=0$ are 4 .