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 $p^{\prime }\left(0\right)$ is

Answer 1

$p\left(x\right)$ will be of degree $3$

$p^{\prime }\left(x\right)=k(x-1)(x-3),k>0$ as 1 is point of maxima and 3 is point of minima.
$=k(x^2-4x+3)$

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

given, $p\left(1\right)=6$

$\Rightarrow 4k+3c=18$ ...(i)

and $p\left(3\right)=2$

$\Rightarrow c=2$

so, $k=3$

$p^{\prime }\left(x\right)=3(x-1)(x-3)$

$p^{\prime }\left(0\right)=3(-1)(-3)=9$