Question:
Let $f(x)$ 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 $\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=1$, then 5. $f(2)$ is equal to
Solution:
$f(x)=x^{6}+a x^{5}+b x^{4}+x^{3}$
$\therefore f^{\prime}(x)=6 x^{5}+5 a x^{4}+4 b x^{3}+3 x^{2}$
Roots $1 \&-1$
$\therefore 6+5 a+4 b+3=0 \&-6+5 a-4 b+3=0$ solving
$a=-\frac{3}{5} \quad b=-\frac{3}{2}$
$\therefore f(x)=x^{6}-\frac{3}{5} x^{5}-\frac{3}{2} x^{4}+x^{3}$
$\therefore 5 \cdot f(2)=5\left[64-\frac{96}{5}-24+8\right]=144$