Let $f\left(x\right)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2\text{. If }\underset{x\to 0}{lim} \left(1+\frac{f\left(x\right)}{x^2}\right)=3$ then $f\left(2\right)$, is equal to 

It is given that $f\left(x\right)$ is a fourth-degree polynomial such that

$\begin{array}{l}\underset{x\to 0}{lim} \left(1+\frac{f\left(x\right)}{x^2}\right)=3\\ \Rightarrow \underset{x\to 0}{lim} \frac{f\left(x\right)}{x^2}\end{array}$  is finite

 $\Rightarrow f\left(x\right)$ has a repeated root at$x=0$.

Let, $f\left(x\right)=x^2\left(a{x}^2+bx+c\right)$ then.

$\begin{array}{l}\underset{x\to 0}{lim} \left(1+\frac{f\left(x\right)}{x^2}\right)=3\\ \Rightarrow \underset{x\to 0}{lim} \left(1+a{x}^2+bx+c\right)=3\\ \Rightarrow c+1=3\end{array}$

$\Rightarrow c=2$

It is given that $f\left(x\right)$ has extreme values at $x=1$ and $x=2$

$\therefore f^{\mathrm{\prime }}\left(1\right)=0$ and $f^{\mathrm{\prime }}\left(2\right)=0$

$\begin{array}{r}\Rightarrow 4a+3b+2c=032a+12b+4c=0\\ \left[\because f^{\mathrm{\prime }}\left(x\right)=4a{x}^3+3b{x}^2+2cx\right]\end{array}$

$\Rightarrow 4a+3b+4=0$ and $32a+12b+8=0$

$\begin{array}{l}\Rightarrow a=1/2,b=-2\\ \therefore f\left(x\right)=x^2\left(\frac{1}{2}{x}^2-2x+2\right)\\ \Rightarrow f\left(2\right)=4(2-4+2)=0.\end{array}$