Application of Derivatives 4 Question 40
####42. 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 zeroes 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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Answer:
Correct Answer: 42. $P_{0}=\frac{1}{2}, \frac{5}{4}$ and $Q_{0}=\frac{5}{4}, \frac{1}{2}$
Solution:
- 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 $h^{\prime}(x)=0$. $\Rightarrow \quad g(x)=0 \quad \Rightarrow \quad h(x)=0$
$\Rightarrow f(x)=0$ or $f^{\prime}(x)=0$
If $f(x)=0$ has 4 minimum solutions.
$f^{\prime}(x)=0$ has 3 minimum solutions.
$h(x)=0$ has 7 minimum solutions.
$\Rightarrow h^{\prime}(x)=g(x)=0$ has 6 minimum solutions.
