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 $\underset{x\to 0}{\lim }\frac{f(x)}{x^3}=1$, then $5\cdot f(2)$ is equal to _____

Answer 1

Answer: 144

Let $f(x)=x^6+a{x}^5+b{x}^4+c{x}^3+d{x}^2+ex+$
as $\underset{x\to 0}{\lim }\frac{f(x)}{x^3}=1$ non-zero finite
So, $d=e=f=0$
and $f(x)=x^3\left(x^3+a{x}^2+bx+c\right)$
Hence, $\underset{x\to 0}{\lim }\frac{f(x)}{x^3}=c=1$
Now, as $f(x)=x^6+a{x}^5+b{x}^4+x^3$
and $f^{\prime }(x)=0$ at $x=1$ and $x=-1$
i.e., $f^{\prime }(x)=6x^5+5a{x}^4+4b{x}^3+3x^2$
$f^{\prime }(1)=0$
$\Rightarrow 6+5a+4b+3=0$
$\Rightarrow 5a+4b=-9$
&$f^{\prime }(-1)=0$
$\Rightarrow -6+5a-4b+3=0$
$\Rightarrow 5a-4b=-9$
Solving both we get,
$a=\frac{-6}{10}=\frac{-3}{5};b=\frac{-3}{2}$
$\therefore f(x)=x^6-\frac{3}{5}{x}^5-\frac{3}{2}{x}^4+x^3$
$\therefore 5f(2)=5\left[64-\frac{3}{5}\cdot 32-\frac{3}{2}\cdot 16+8\right]$
$=320-96-120+40$
$=144$