Question

Let $f\left(x\right)$ be a polynomial of degree $6$ in $x$, in which the coefficient of $x^6$ is unity and it has extrema at $x=-1$and $x=1$. If $\underset{x\to 0}{\lim }\left(\frac{f\left(x\right)}{x^3}\right)=1$then$5f\left(2\right)=$

Answer 1

Finding the polynomial using the given information:

Since $\underset{x\to 0}{\lim }\left(\frac{f\left(x\right)}{x^3}\right)=1$is defined and finite coefficients of$x^2,x,1$are zero in$f$

let $f\left(x\right)=x^6+a{x}^5+b{x}^4+c{x}^3$

$\underset{x\to 0}{\lim }\left(\frac{x^6+a{x}^5+b{x}^4+c{x}^3}{x^3}\right)=1$

this gives $c=1$

$f\left(x\right)=x^6+a{x}^5+b{x}^4+x^3$

$$\begin{gathered} f\text{'}\left(x\right)=6x^5+5a{x}^4+4b{x}^3+3x^2 \\ f\text{'}\left(1\right)=f\text{'}(-1)=0 \end{gathered}$$

since $1,-1$are critical points

This gives:

$$\begin{gathered} 6+5a+4b+3=0 \\ -6+5a-4b+3=0 \end{gathered}$$

Solving this we get

$a=\frac{-3}{5},b=\frac{-3}{2}$

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

$$\begin{gathered} 5f\left(2\right)=5\left(64-\frac{96}{5}-\frac{48}{2}+8\right) \\ =144 \end{gathered}$$

Hence, the correct answer is $144$