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

Let, \(g(x)=\frac{d}{d x}\left[f(x) \cdot f^{\prime}(x)\right]\) to get the zero of \(g(x)\) we take function \(h(x)=f(x) \cdot f^{\prime}(x)\) between any two roots of \(h(x)\) there lies atleast one root of \(h^{\prime}(x)=0\)
\[
\begin{aligned}
& \Rightarrow \quad g(x)=0 \\
& \Rightarrow \quad h(x)=0 \\
& \Rightarrow \quad f(x)=0 \text { or } f^{\prime}(x)=0 \\
& \text { If } f(x)=0 \text { has } 4 \text { minimum solutions, } \\
& f^{\prime}(x)=0 \text { has } 3 \text { minimum solutions, } \\
& h(x)=0 \text { has } 7 \text { minimum solutions, then } \\
& h^{\prime}(x)=g(x)=0 \text { has } 6 \text { minimum solutions. } \\
&
\end{aligned}
\]