Question

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

Answer 1

$\underset{x\to 0}{lim}{(1+\frac{f\left(x\right)}{x^3})}^{1/x}=e^2$

If the limit is to exist,$f\left(x\right)$ cannot have terms with degree lesser than 3.

So, let $f\left(x\right)=a{x}^4+b{x}^5+c{x}^6$

Hence, $\underset{x\to 0}{lim}{[1+ax+b{x}^2+c{x}^3]}^{\frac{1}{x}}=e^2$

Applying log on both sides gives

$\underset{x\to 0}{lim}\left(\frac{\log [1+ax+b{x}^2+c{x}^3]}{x}\right)=\log e^2$

using the series $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$

$\underset{x\to 0}{lim}\left(\frac{ax+b{x}^2+c{x}^3}{x}\right)=2$ (higher order terms becomes $0$)

$\Rightarrow$ $a=2$

Therefore, $f\left(x\right)=2x^4+b{x}^5+c{x}^6$

$\Rightarrow$ $f^{\prime }\left(x\right)=8x^3+5b{x}^4+6c{x}^5$

but given that $f\left(x\right)$ has local maximum at $x=1$ and local minimum at $x=0$ and $2$

i.e $f^{\prime }\left(x\right)=0$ when $x=0,1,2$

$\Rightarrow$ $f^{\prime }\left(1\right)=8+5b+6c=0$ and $f^{\prime }\left(2\right)=4+5b+12c=0$

solving above two equations for $b$ and $c$ gives

$b=\frac{-12}{5}$ and $c=\frac{2}{3}$

$\therefore$ $f\left(x\right)=2x^4-\frac{12}{5}{x}^5+\frac{2}{3}{x}^6$

$5f\left(3\right)=5(162-\frac{12}{5}{3}^5+\frac{2}{3}{3}^6)=324$

Hence, $5f\left(3\right)=324$