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

6

Clearly, $f(x)$ is a continuous function such that $f(b) f(c)<0, f(c) f(d)<0, f(a)=0$ and $f(e)=0$. So, $f(x)$ has at least 4 zeros in $[a, e]$.

∴  $f'(x)$ has minimum 3 zeros in $[a, e]$

Now,

$g(x)=\left\{f'(x)\right\}^2+f''(x) f(x)$

$\Rightarrow g(x) =\frac{d}{d x}\left\{f'(x) f(x)\right\}$

$\Rightarrow g(x)=\frac{d}{d x} h(x)$, where $h(x)=f(x) f'(x)$

Clearly, h(x) has at least 7 zeros in [a, e]. Therefore, by Rolle's theorem $\frac{d}{d x} h(x)$ i.e. g(x) has at least six zeros in [a, e].