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 $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$ then $5.f\left(2\right)$ is equal to

Answer 1

$f\left(x\right)=x^6+a{x}^5+b{x}^4+x^3$
$\therefore f^{\prime }\left(x\right)=6x^5+5a{x}^4+4b{x}^3+3x^2$
Roots are $1$ and $-1$
$\therefore 6+5a+4b+3=0$ & $-6+5a-4b+3=0$ solving
$a=-\frac{3}{5}b=-\frac{3}{2}$
$\therefore f\left(x\right)=x^6-\frac{3}{5}{x}^5-\frac{3}{2}{x}^4+x^3$
$\therefore 5.f\left(2\right)=5[64-\frac{96}{5}-24+8]=144$