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:

For least degree polynomial with extrema: \begin{itemize} \item Two turning points ⇒ cubic. \item Assume derivative as product of linear factors. \item Integrate and use given values. \end{itemize}

Answer 1

The Correct Option is B

Concept: If a polynomial has a local max at \( x=1 \) and local min at \( x=3 \), then: \[ p'(1)=0, \quad p'(3)=0 \] Least degree polynomial with two stationary points → cubic polynomial. Step 1: {\color{red}Assume derivative form.} \[ p'(x) = k(x-1)(x-3) \] Integrate: \[ p(x) = k\int (x^2 - 4x + 3)\,dx \] \[ = k\left(\frac{x^3}{3} - 2x^2 + 3x\right) + C \] Step 2: {\color{red}Use given values.} Using \( p(1)=6 \): \[ k\left(\frac{1}{3} - 2 + 3\right) + C = 6 \] \[ k\left(\frac{4}{3}\right) + C = 6 \quad (1) \] Using \( p(3)=2 \): \[ k\left(9 - 18 + 9\right) + C = 2 \] \[ 0 + C = 2 \Rightarrow C=2 \] Step 3: {\color{red}Find \( k \).} From (1): \[ \frac{4k}{3} + 2 = 6 \] \[ \frac{4k}{3} = 4 \Rightarrow k = 3 \] Step 4: {\color{red}Find \( p'(0) \).} \[ p'(x) = 3(x-1)(x-3) \] \[ p'(0) = 3(-1)(-3) = 9 \]