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 $x=3$ . lf $p\left(1\right)=6$ and $p\left(3\right)=2$ , then $p^{\prime }\left(0\right)$ is

• A. $9$
• B. $13$
• C. $11$
• D. $12$

Answer 1

The correct option is A $9$

Given that P(x) is a real polynomial of least degree.
It has two extreme: local maximum at $x=1$ and local minimum at $x=3$.

Hence, $P^{\prime }\left(x\right)=k(x-1)(x-3)$

$=k(x^2-4x+3)$

$\Rightarrow p\left(x\right)=k(\frac{x^3}{3}-2x^2+3x)+c$

Given that $P\left(1\right)=6\Rightarrow \frac{4}{3}k+c=6$

Also, $P\left(3\right)=2\Rightarrow c=2$

So, $k=3$

$\therefore P^{\prime }\left(0\right)=3k=9$.