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Q.
If $p ( x )$ 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:
Since $p ( x )$ has realtive extreme at $x=1 \& 2$
So $ p'(x)=0 \text { at } x=1 \& 2 $
$\Rightarrow p'(x)=A(x-1)(x-2) $
$\Rightarrow p(x)=\int A\left(x^{2}-3 x+2\right) d x$
$p(x)=A\left(\frac{x^{3}}{3}-\frac{3 x^{2}}{2}+2 x\right)+C \ldots$(1)
$P(1)=8$
From (1)
$8= A \left(\frac{1}{3}-\frac{3}{2}+2\right)+ C$
$\Rightarrow 8=\frac{5 A }{6}+ C$
$ \Rightarrow 48=5 A +6 C \dots$(3)
$P (2)=4$
$\Rightarrow \quad 4= A \left(\frac{8}{3}-6+4\right)+ C$
$\Rightarrow 4=\frac{2 A }{3}+ C $
$\Rightarrow 12=2 A +3 C \dots$(4)
From $3 \& 4, C =-12$
So $P (0)= C =-12$
