Question

Let $P\left(x\right)$ be a polynomial of degree $5$ having extrema at $x=-1,1$ and $\underset{x\to 0}{lim}(\frac{P\left(x\right)}{x^3}-2)=4.$ If $M$ and $m$ are the maximum and minimum values of the function $y=P^{\prime }\left(x\right)$ on the set $A=\{x:x^2+6\le 5x\},$ then the value of $\frac{m}{M}$ is

Answer 1

We have $\underset{x\to 0}{lim}(\frac{P\left(x\right)}{x^3}-2)=4$
$\therefore \underset{x\to 0}{lim}\frac{P\left(x\right)}{x^3}=6$

Consider $P\left(x\right)=a{x}^5+b{x}^4+6x^3$
$\Rightarrow P^{\prime }\left(x\right)=5a{x}^4+4b{x}^3+18x^2$
Given $P^{\prime }(-1)=0\Rightarrow 5a-4b=-18$
and $P^{\prime }\left(1\right)=0\Rightarrow 5a+4b=-18$
On solving, we get
$a=\frac{-18}{5},b=0$
Hence, $P\left(x\right)=\frac{-18}{5}{x}^5+6x^3$

$P^{\prime }\left(x\right)=-18x^4+18x^2=18(x^2-x^4)$
and $P^{\prime \prime }\left(x\right)=18(2x-4x^3)=36x(1-2x^2)$
Also, $A=\{x:x^2+6\le 5x\}$
$\Rightarrow x\in [2,3]$

Clearly, $P^{\prime \prime }\left(x\right)<0\forall x\in [2,3]$
So, $y=P^{\prime }\left(x\right)$ is decreasing function in $[2,3]$
$\therefore M=P_{max}^{\prime }(x=2)=18(4-16)=-18\times 12$
and $m=P_{min}^{\prime }(x=3)=18(9-81)=-18\times 72$
Hence, $\frac{m}{M}=\frac{-18\times 72}{-18\times 12}=6$