Question

Let $f\left(x\right)$ be a twice differentiable function with no critical point and $g\left(x\right)=(x+6{)}^{2009}(x+1{)}^{2010}(x+2{)}^{2011}(x-3{)}^{2012}(x-4{)}^{2013}(x-5{)}^{2014}$ be such that $f\left(x\right)+g\left(x\right)f^{\prime }\left(x\right)+f^{\prime \prime }\left(x\right)=0.$ Then $h\left(x\right)=\left(f\right(x){)}^2+(f^{\prime }\left(x\right){)}^2$

• A. is monotonically increasing in $(-2,4)$.
• B. has exactly three points of inflection.
• C. has exactly two points of local maxima.
• D. has a negative point of local minimum.

Answer 1

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

has a negative point of local minimum.

We have, $h\left(x\right)=\left(f\right(x){)}^2+(f^{\prime }\left(x\right){)}^2$
Differentiation w.r.t. $x,$ we get
$h^{\prime }\left(x\right)=2f\left(x\right)f^{\prime }\left(x\right)+2f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)$
$\Rightarrow h^{\prime }\left(x\right)=2f^{\prime }\left(x\right)\left[f\right(x)+f^{\prime \prime }(x\left)\right]$
$\Rightarrow h^{\prime }\left(x\right)=2f^{\prime }\left(x\right)[-g(x\left)f^{\prime }\right(x\left)\right]$
$\Rightarrow h^{\prime }\left(x\right)=-2g\left(x\right)\left(f^{\prime }\right(x){)}^2$
$\Rightarrow h^{\prime }\left(x\right)=-2(x+6{)}^{2009}(x+1{)}^{2010}(x+2{)}^{2011}(x-3{)}^{2012}(x-4{)}^{2013}(x-5{)}^{2014}{\left(f^{\prime }\right(x\left)\right)}^2$

Sign of $h^{\prime }\left(x\right)$


$h^{\prime }\left(x\right)>0$ in $(-2,4)$
$\therefore h\left(x\right)$ is increasing in $(-2,4)$.

By first derivative test,
$x=-1,3$ and $5$ are points of inflection.
$x=-6$ and $x=4$ are points of local maxima.
$x=-2$ is the only the point of local minimum.