SATHEE: Application of Derivatives 4 Question 62

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 $p^{'} (0)$ is equal to

(2012)

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Solution:

PLAN If $f (x)$ is least degree polynomial having local maximum and local minimum at $α$ and $β$ .

Then, $f^{'} (x) = λ (x - α) (x - β)$

Here, $p^{'} (x) = λ (x - 1) (x - 3) = λ (x^{2} - 4 x + 3)$

On integrating both sides between 1 to 3 , we get

$\begin{aligned} \int_{1}^{3} p^{'} (x) d x & = \int_{1}^{3} λ (x^{2} - 4 x + 3) d x \\ \Rightarrow & (p (x))_{1}^{3} & = λ \frac{x^{3}}{3} - 2 x^{2} + 3 x^{3} \end{aligned}$

$\begin{array}{llrl} \Rightarrow & p (3) - p (1) & = λ (9 - 18 + 9) - \frac{1}{3} - 2 + 3 \\ \Rightarrow & 2 - 6 & = λ \frac{- 4}{3} \\ \Rightarrow & λ & = 3 \\ \Rightarrow & p^{'} (x) & = 3 (x - 1) (x - 3) \\ ∴ & p^{'} (0) & = 9 \end{array}$

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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′(0) is equal to

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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 $p^{'} (0)$ is equal to