Question

Let $p\left(x\right)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p\left(1\right)=6\text{and}p\left(3\right)=2$, then $p\text{'}\left(0\right)$ is

• A. $9$
• B. $5$
• C. $13$
• D. $4$

Answer 1

The correct option is A

$9$

Explanation for the correct option:

Finding the value of $p\text{'}\left(0\right)$

Given : $p\left(1\right)=6\text{and}p\left(3\right)=2$,

$\begin{array}{rcl}p\text{'}\left(x\right)& =& k(x-1)(x-3)\\ & =& k(x^2-4x+3)\\ p\left(x\right)& =& k\left(\frac{x^3}{3}-2x^2+3x\right)+c\end{array}$
Given that$p\left(1\right)=6\Rightarrow \frac{4}{3}k+c=6$
Also,$p\left(3\right)=2\Rightarrow c=2$
So, $k=3$
$\therefore p\prime \left(0\right)=3k=9.$

Hence option(A) is correct.