Let p (x) be a polynomial of degree three that has a local maximum value 8 at x =1. and a local minimum value 4 at x = 2, then p (0) is equal to

Solution:

We find that $p' (x)$ is a quadratic polynomial having $x = 1$ and $x = 2$ as its two zeros.

$∴ p^{'} (x) = λ (x - 1) (x ∣ - 2)$ .where $λ$ is a positive constant.

$\Rightarrow p (x) = λ (\frac{x^{3}}{3} - \frac{3}{2} x^{2} + 2 x) + C$ ..(i)

where $C$ is a constant .

It is given that $p (1) = 8$ and $p (2) = 4$ .

$\begin{array}{l} ∴ λ (\frac{1}{3} - \frac{3}{2} + 2) + C = 8 and λ (\frac{8}{3} - 6 + 4) + C = 4 \\ \Rightarrow \frac{5 λ}{6} + C = 8 and \frac{2 λ}{3} + C \Rightarrow 4 \Rightarrow C = - 12 and λ = 24 \end{array}$

Hence, $p (0) = C = - 12$ .

Let $p (x)$ be a polynomial of degree three that has a local maximum value 8 at $x = 1$ . and a local minimum value 4 at $x = 2$ , then $p (0)$ is equal to

Solution:

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