Question

If $f\left(x\right)$ is a twice differentiable function such that $f\left(a\right)=0,f\left(b\right)=2,f\left(c\right)=1,f\left(d\right)=2,f\left(e\right)=0,$ where $a<b<c<d<e,$ then the minimum number of zeroes of $g\left(x\right)=\left(f^{\prime }\right(x){)}^2+f^{\prime \prime }(x\left)f\right(x\left)\right)$ in the interval $[a,e]$ 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 $h^{\prime }\left(x\right)=0$ .. By Rolle's theorem
$\Rightarrow g\left(x\right)=0,h\left(x\right)=0$
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.