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