Question

Let $f\left(x\right)$ be a non-constant, thrice differentiable function defined on $R$ such that $f\left(x\right)=f(6-x)$ and $f^{\prime }\left(0\right)=0=f^{\prime }\left(2\right)=f^{\prime }\left(5\right).$ If $n$ is the minimum number of roots of ${\left(f^{\prime }\right(x\left)\right)}^2+f^{\prime }\left(x\right)\cdot f^{\prime \prime \prime }\left(x\right)=0$ in the interval $[0,6]$, then the value of $\frac{n}{2}$ is

Answer 1

Given : $f\left(x\right)=f(6-x)\cdots \left(i\right)$
On differentiating $\left(i\right)$ w.r.t. $x$, we get
$f^{\prime }\left(x\right)=-f^{\prime }(6-x)\cdots \left(ii\right)$
Putting $x=0,2,3,5$ in $\left(ii\right)$, we get
$\Rightarrow f^{\prime }\left(0\right)=-f^{\prime }\left(6\right)=0$
$\Rightarrow f^{\prime }\left(2\right)=-f^{\prime }\left(4\right)=0$
$\Rightarrow f^{\prime }\left(3\right)=0$
$\Rightarrow f^{\prime }\left(5\right)=-f^{\prime }\left(1\right)=0$
$\therefore f^{\prime }\left(0\right)=0=f^{\prime }\left(2\right)=f^{\prime }\left(3\right)=f^{\prime }\left(5\right)=f^{\prime }\left(1\right)=f^{\prime }\left(4\right)=f^{\prime }\left(6\right)$
$\therefore f^{\prime }\left(x\right)=0$ has minimum $7$ roots in $[0,6]$
Now, consider a function $y=f^{\prime }\left(x\right)$
As $f^{\prime }\left(x\right)$ satisfy Rolle's theorem in intervals $[0,1],[1,2]$,
$[2,3],[3,4],[4,5]$ and $[5,6]$ respectively.
So, using Rolle's theorem, the equation $f^{\prime \prime }\left(x\right)=0$ has minimum 6 roots.
Now, $g\left(x\right)={\left(f^{\prime \prime }\right(x\left)\right)}^2+f^{\prime }\left(x\right)f^{\prime \prime \prime }\left(x\right)=h^{\prime }\left(x\right)$,
where $h\left(x\right)=f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)$
Clearly $h\left(x\right)=0$ has minimum 13 roots in $[0,6]$
Hence again by Rolle's theorem, $g\left(x\right)=h^{\prime }\left(x\right)$ has minimum 12 zeroes in $[0,6]$
$\therefore$ the value of $\frac{n}{2}=6$