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

$$\begin{gathered} f\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e \\ \underset{x\to 0}{\lim } 1+\frac{f\left(x\right)}{x^2}=3 \\ f\left(x\right) is divisible by x^2\Rightarrow d=e=0 \\ \underset{x\to 0}{\lim } 1+a{x}^2+bx+c=3 \\ 1+c=3 \\ c=2 \\ f\left(x\right)=a{x}^4+b{x}^3+2x^2 \\ f\text{'}\left(x\right)=4a{x}^3+3b{x}^2+4x \\ given extream values at x=1,x=2 \\ f\text{'}\left(1\right)=0= f\text{'}\left(2\right) \\ f\text{'}\left(x\right)=0 roots are x=\frac{1}{2} \\ 4a{x}^2+3bx+4=0 \\ roots are 1,2 \\ 1\times 2=\frac{4}{4a}\Rightarrow a=\frac{1}{2} \\ 1+2=\frac{-3b}{4a}\Rightarrow b=-4a=-2 \\ f\left(x\right)=\frac{1}{2}{x}^4-2x^3+2x^2 \\ f\left(2\right)=8-16+8=0 \end{gathered}$$