Application of Derivatives 4 Question 62
65. Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$, then $p^{\prime}(0)$ is equal to
(2012)
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Solution:
- PLAN If $f(x)$ is least degree polynomial having local maximum and local minimum at $\alpha$ and $\beta$.
Then, $\quad f^{\prime}(x)=\lambda(x-\alpha)(x-\beta)$
Here, $\quad p^{\prime}(x)=\lambda(x-1)(x-3)=\lambda\left(x^{2}-4 x+3\right)$
On integrating both sides between 1 to 3 , we get
$$ \begin{array}{rlrl} & & \int _1^{3} p^{\prime}(x) d x & =\int _1^{3} \lambda\left(x^{2}-4 x+3\right) d x \\ \Rightarrow \quad & (p(x)) _1^{3} & =\lambda \frac{x^{3}}{3}-2 x^{2}+3 x^{3} \end{array} $$
$$ \begin{array}{llrl} \Rightarrow & & p(3)-p(1) & =\lambda \quad(9-18+9)-\frac{1}{3}-2+3 \\ & \Rightarrow & 2-6 & =\lambda \frac{-4}{3} \\ \Rightarrow & \lambda & =3 \\ \Rightarrow & p^{\prime}(x) & =3(x-1)(x-3) \\ \therefore & & p^{\prime}(0) & =9 \end{array} $$
