Question

Let \( p(x) \) be a real polynomial of least degree which has a local maximum at \( x = 1 \) and a local minimum at \( x = 3 \). If \( p(1) = 6 \) and \( p(3) = 2 \), then \( p'(0) \) is equal to:

• A. \( 8 \)
• B. \( 9 \)
• C. \( 3 \)
• D. \( 6 \)

Hint:

When a polynomial's derivative has simple roots at given points, assume a factored form for \( p'(x) \) and integrate to find \( p(x) \).

Answer 1

The Correct Option is B

Step 1: Determine conditions.
\n\nSince \( p(x) \) is a polynomial of minimal degree with a local max/min at \( x=1 \) and \( x=3 \), respectively, \( p'(x) \) has roots at \( x=1 \) and \( x=3 \).\n\nTherefore,\n\[\np'(x) = k(x-1)(x-3)\n\]\nwhere \( k \) is a constant.\n\n
Step 2: Calculate \( p(x) \) by integrating \( p'(x) \).
\n\nIntegrate \( p'(x) \):\n\[\np(x) = k\left(\frac{x^3}{3} - 2x^2 + 3x\right) + C\n\]\nwhere \( C \) is the constant of integration.\n\n
Step 3: Apply given conditions.
\n\nGiven:\n\[\np(1) = 6\n\quad \text{and} \quad\np(3) = 2\n\]\n\nSubstitute \( x=1 \) into \( p(x) \):\n\[\nk\left(\frac{1}{3} - 2 + 3\right) + C = 6\n\quad \Rightarrow \quad\nk\left(\frac{4}{3}\right) + C = 6\n\quad \Rightarrow \quad\n\frac{4k}{3} + C = 6 \quad \cdots (1)\n\]\n\nSubstitute \( x=3 \) into \( p(x) \):\n\[\nk\left(\frac{27}{3} - 2(9) + 3(3)\right) + C = 2\n\quad \Rightarrow \quad\nk(9 - 18 + 9) + C = 2\n\quad \Rightarrow \quad\nk(0) + C = 2\n\quad \Rightarrow \quad\nC = 2 \quad \cdots (2)\n\]\n\n
Step 4: Solve for \( k \).
\n\nSubstitute \( C = 2 \) into equation (1):\n\[\n\frac{4k}{3} + 2 = 6\n\quad \Rightarrow \quad\n\frac{4k}{3} = 4\n\quad \Rightarrow \quad\n4k = 12\n\quad \Rightarrow \quad\nk = 3\n\]\n\n
Step 5: Calculate \( p'(0) \).
\n\nNow,\n\[\np'(x) = 3(x-1)(x-3)\n\]\nSubstituting \( x=0 \):\n\[\np'(0) = 3(0-1)(0-3) = 3(-1)(-3) = 9\n\]\n\nThus, \( p'(0) = 9 \).