Question

Let $f$ is a non-constant thrice differentiable function defined on $R$ such that $f(x)=f(6-x)$ and $f^{\prime }(0)=f^{\prime }(2)=f^{\prime }(1)=0$. If $n$ is the minimum number of roots of ${\left(f^{\prime \prime }(x)\right)}^2+f^{\prime }(x)f^{\prime \prime \prime }(x)=0$ in the interval $[0,6]$, then $n$ is divisible by

• A. 2
• B. 3
• C. 4
• D. 6

Answer 1

Answer: (A), (B), (C), (D)

$f^{\prime }(x)=-f^{\prime }(6-x)$, Put $x=3$, we get $f^{\prime }(3)=0$ and curve is symmetrical about line $x=3$.
$\&f^{\prime }(0)=f^{\prime }(1)=f^{\prime }(2)=f^{\prime }(3)=f^{\prime }(4)=f^{\prime }(5)=f^{\prime }(6)=0$
$\therefore f^{\prime }(x)$ is zero for 7 different value
$\therefore f^{\prime \prime }(x)$ is zero for 6 different value (by Rolle's theorem)
$\&{\left(f^{\prime \prime }(x)\right)}^2+f^{\prime }(x)f^{\prime \prime \prime }(x)=\frac{d}{dx}\left(f^{\prime }(x)f^{\prime \prime }(x)\right)$ has 12 roots because $F(x)=f^{\prime }(x)f^{\prime \prime }(x)$ have 12 roots \& by Rolle's theorem.
$\therefore$ Number of roots of ${\left(f^{\prime \prime }(x)\right)}^2+f^{\prime }(x)f^{\prime \prime \prime }(x)=0$ is 12$]$