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)
$\&\left(f^{\prime \prime}(x)\right)^2+f^{\prime}(x) f^{\prime \prime \prime}(x)=\frac{d}{d x}\left(f^{\prime}(x) f^{\prime \prime}(x)\right)$ 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$]$