Question

Let $f\left(x\right)$ be a polynomial of degree $6,$ which satisfies $\underset{x\to 0}{lim}{(1+\frac{f\left(x\right)}{x^3})}^{\frac{1}{x}}=e^2$ has local maximum at $x=1$ and local minimum at $x=0$ and $2,$ then $\frac{5f\left(3\right)}{4}$ is

Answer 1

The given limit exists only if $f\left(x\right)$ will be as following:
$f\left(x\right)=a{x}^4+b{x}^5+c{x}^6$
$\Rightarrow \underset{x\to 0}{lim}{(1+\frac{f\left(x\right)}{x^3})}^{\frac{1}{x}}=e^2$
$\Rightarrow \underset{x\to 0}{lim}{(1+ax+b{x}^2+c{x}^3)}^{\frac{1}{x}}=e^2$
$\Rightarrow \underset{x\to 0}{lim}{(1+ax)}^{\frac{1}{x}}=e^2$
$\Rightarrow e^a=e^2$
$\Rightarrow a=2$
Thus$f\left(x\right)=2x^4+b{x}^5+c{x}^6{f}^{\prime }\left(x\right)=x^3(8+5bx+6c{x}^2)$
Since $f\left(x\right)$ has local extremum at $x=0,1,2.$
So, $f^{\prime }\left(0\right)=f^{\prime }\left(1\right)=f^{\prime }\left(2\right)=0$
$\Rightarrow c=\frac{2}{3},b=\frac{-12}{5}$
Hence $f\left(x\right)=2x^4-\frac{12}{5}{x}^5+\frac{2}{3}{x}^6$
$\Rightarrow f\left(3\right)=\frac{324}{5}$
$\Rightarrow \frac{5f\left(3\right)}{4}=81$