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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
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Verified Answer
The correct answer is:
6
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 $h^{\prime}(x)=0$
$$
\begin{aligned}
& \Rightarrow \quad g(x)=0 \\
& \Rightarrow \quad h(x)=0 \\
& \Rightarrow \quad f(x)=0 \text { or } f^{\prime}(x)=0 \\
& \text { If } f(x)=0 \text { has } 4 \text { minimum solutions, } \\
& f^{\prime}(x)=0 \text { has } 3 \text { minimum solutions, } \\
& h(x)=0 \text { has } 7 \text { minimum solutions, then } \\
& h^{\prime}(x)=g(x)=0 \text { has } 6 \text { minimum solutions. } \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad g(x)=0 \\
& \Rightarrow \quad h(x)=0 \\
& \Rightarrow \quad f(x)=0 \text { or } f^{\prime}(x)=0 \\
& \text { If } f(x)=0 \text { has } 4 \text { minimum solutions, } \\
& f^{\prime}(x)=0 \text { has } 3 \text { minimum solutions, } \\
& h(x)=0 \text { has } 7 \text { minimum solutions, then } \\
& h^{\prime}(x)=g(x)=0 \text { has } 6 \text { minimum solutions. } \\
&
\end{aligned}
$$
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