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

(1) –24

(2) –12

(3) 6

(4) 12

(1) –24

(2) –12

(3) 6

(4) 12

**Ans. (2)** –12

**Sol.** Clearly P'(x) = λ (x – 1) (x – 2) where λ > 0

$P(x)=\lambda\left[{x^3\over3}-{3x^2\over2}+2x\right]+C$

given $P(1)=8\implies\lambda\left[{1\over3}-{3\over2}+2\right]+C=8$

${5\lambda\over6}+C=8\quad\quad$ ........(i)

also $P(2)=4\implies\lambda\left[{8\over3}-6+4\right]+C=4$

${2\lambda\over3}+C=4\quad\quad$ ........(ii)

By (i) and (ii) $\implies$ C = – 12

$\implies$ P(0) = – 12