Let f(x) be a polynomial satisfying $\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \frac{{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)}{2{\mathrm{x}}^5+3}=6$ and $\mathrm{f}\left(1\right)=3,\mathrm{f}\left(3\right)=7 \mathrm{and} \mathrm{f}\left(5\right)=11$ Then the value of $\left|\mathrm{f}\right(0\left)\right|$ is _______ .

Since $\underset{\mathrm{x}\to \mathrm{\infty }}{lim} \frac{{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)}{2{\mathrm{x}}^5+3}=6$ which is finite and non-zero, f(x) must be polynomial of degree '3'.
Also f(1) = 3, f(3) = 7 and f(5) = 11
$\begin{array}{l}\therefore \hspace{0.25em}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\lambda }(\mathrm{x}-1)(\mathrm{x}-3)(\mathrm{x}-5)+(2\mathrm{x}+1)\\ \Rightarrow \underset{\mathrm{x}\to \mathrm{\infty }}{lim} \frac{{\mathrm{x}}^2\left(\mathrm{\lambda }\right(\mathrm{x}-1\left)\right(\mathrm{x}-3\left)\right(\mathrm{x}-5)+2\mathrm{x}+1)}{2{\mathrm{x}}^5+3}=6\\ \Rightarrow \hspace{0.25em}\frac{\mathrm{\lambda }}{2}=6\\ \Rightarrow \mathrm{\lambda }=12\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{f}\left(\mathrm{x}\right)=12(\mathrm{x}-1)(\mathrm{x}-3)(\mathrm{x}-5)+(2\mathrm{x}+1)\end{array}$
So, f(0) = 12(-1)(-3)(-5)+1= 179