# 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: Option: 1 6 Option: 2 -12 Option: 3 -24 Option: 4 12

Since p(x) has relative extreme at x = 1 & 2

so p'(x) = 0 at x = 1 & 2

⇒ p'(x) = A(x – 1) (x – 2)

$\\\Rightarrow p(x)=\int A\left(x^{2}-3 x+2\right) d x \\ p(x)=A\left(\frac{x^{3}}{3}-3 x^{2}+2 x\right)+C\;\;\;\;\;\;\;\;\ldots(1)$

P(1) = 8

from (1)

$8=A\left(\frac{1}{3}-\frac{3}{2}+2\right)+C$

$\\\Rightarrow 8=\frac{5 \mathrm{~A}}{6}+\mathrm{C} \Rightarrow 48=5 \mathrm{~A}+5 \mathrm{C} \\ \mathrm{P}(2)=4 \\ \Rightarrow 4=\mathrm{A}\left(\frac{8}{3}-6+4\right)+\mathrm{C} \\ \Rightarrow 4=\frac{2 \mathrm{~A}}{3}+\mathrm{C} \Rightarrow 12=2 \mathrm{~A}+3 \mathrm{C} \\$

From above

C = –12 So P(0) = C = -12

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