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: Analyze the given conditions.
Since \( p(x) \) is a polynomial of least degree with a local maximum and minimum at \( x=1 \) and \( x=3 \), respectively, the derivative \( p'(x) \) must have roots at \( x=1 \) and \( x=3 \). Thus, \[ p'(x) = k(x-1)(x-3) \] where \( k \) is a constant.
Step 2: Find \( p(x) \) by integrating \( p'(x) \).
Integrate \( p'(x) \): \[ p(x) = k\left(\frac{x^3}{3} - 2x^2 + 3x\right) + C \] where \( C \) is the constant of integration.
Step 3: Use the given conditions.
Given: \[ p(1) = 6 \quad \text{and} \quad p(3) = 2 \] Substitute \( x=1 \) into \( p(x) \): \[ k\left(\frac{1}{3} - 2 + 3\right) + C = 6 \quad \Rightarrow \quad k\left(\frac{4}{3}\right) + C = 6 \quad \Rightarrow \quad \frac{4k}{3} + C = 6 \quad \cdots (1) \] Substitute \( x=3 \) into \( p(x) \): \[ k\left(\frac{27}{3} - 2(9) + 3(3)\right) + C = 2 \quad \Rightarrow \quad k(9 - 18 + 9) + C = 2 \quad \Rightarrow \quad k(0) + C = 2 \quad \Rightarrow \quad C = 2 \quad \cdots (2) \]
Step 4: Solve for \( k \).
Substituting \( C = 2 \) in equation (1): \[ \frac{4k}{3} + 2 = 6 \quad \Rightarrow \quad \frac{4k}{3} = 4 \quad \Rightarrow \quad 4k = 12 \quad \Rightarrow \quad k = 3 \]
Step 5: Find \( p'(0) \).
Now, \[ p'(x) = 3(x-1)(x-3) \] Substituting \( x=0 \): \[ p'(0) = 3(0-1)(0-3) = 3(-1)(-3) = 9 \] Thus, \( p'(0) = 9 \).