Question

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 :

• A. 6
• B. 12
• C. -12
• D. -24

Hint:

p'(x) has roots 1 and 2, so p'(x) = k(x-1)(x-2); integrate and use p(1) = 8, p(2) = 4.

Answer 1

The Correct Option is C

\(p'(1)=p'(2)=0\) and \(p\) is cubic, so \(p'(x) = k(x-1)(x-2) = k(x^2-3x+2)\).
\(p(x) = k\left(\frac{x^3}{3} - \frac{3x^2}{2} + 2x\right) + C\).
\(p(1) = k\cdot\frac56 + C = 8\) and \(p(2) = k\cdot\frac23 + C = 4\).
Subtracting: \(\frac{k}{6} = 4 \Rightarrow k = 24\), so \(C = 8 - 20 = -12\).
(Check: \(p''(1) = -k \lt 0\), so x = 1 is a local maximum.)
\(p(0) = C = -12\).
Answer: (3) -12