Question

If $f\left(x\right)$ is a twice differentiable function such that $f\left(a\right)=0,$ $f\left(b\right)=2,$ $f\left(c\right)=-1,$ $f\left(d\right)=2$ and $f\left(e\right)=0,$ where $a<b<c<d<e,$ then the minimum number of zeros of $g\left(x\right)={f^{\prime }\left(x\right)}^2+f"\left(x\right)\cdot f\left(x\right)$ in the interval $[a,c]$ is

Answer 1

$g\left(x\right)={\left(f^{\prime }\left(x\right)\right)}^2+f"\left(x\right)f\left(x\right)$
$g\left(x\right)=\frac{d}{dx}(f\left(x\right).f^{\prime }\left(x\right))$
Now, Let $h\left(x\right)=f\left(x\right).f^{\prime }\left(x\right)$
Between any two roots of $h\left(x\right)$ there lies one root of $h^{\prime }\left(x\right)=0$
$\Rightarrow g\left(x\right)=0,h\left(x\right)=0$
Now, $f\left(x\right)$ is zero in at least four places, and $f^{\prime }\left(x\right)$ is zero in at least three places.
Now, $h\left(x\right)$ is zero in at least $7$ places, hence $h^{\prime }\left(x\right)$ is zero in atleast $6$ places.