Let  $f\left(x\right)=\underset{n\to \infty }{\lim }\frac{x^{2n-1}+a{x}^3+b{x}^2}{x^{2n}+1}$ is continuous for all  $x\in R$.If point $A(-a,3)$ and  $B\left(\right(b+1),-1)$ are points of relative maximum and minimum of a cubic polynomial  $y=g\left(x\right)$, then the value of g(2) is:

$f\left(c\right)\{\begin{array}{l}\frac{1}{x;}\text{if\hspace{0.17em}}{x}^2>1\Rightarrow x<-1orx>1\\ a{x}^3+b{x}^2;\text{if}0\le x^2<1\Rightarrow -1<x<1\\ \frac{1/x+a{x}^3+b{x}^2}{2};\text{if\hspace{0.17em}}{x}^2=1\end{array}$

$\because f$  is continuous

$$\begin{gathered} \therefore atx=1 \\ and atx=-1 \\ \begin{array}{l}\therefore b=0\frac{\begin{array}{l}1=a+b\\ -1=-a+b\end{array}}{anda=1}\mathrm{.......}\left(1\right)\\ \mathrm{.......}\left(2\right)\end{array} \\ \therefore \text{point\hspace{0.17em}A\hspace{0.17em}and B\hspace{0.17em}are = }(-1,3)\text{and\hspace{0.17em}}(1,1). \\ \therefore g\text{'}\left(x\right)=\lambda (x-1)(x+1) \\ g\left(x\right)=\lambda (\frac{x^3}{3}-x)+c \\ g(-1)=\frac{2\lambda }{3}+c=3 \\ \frac{g\left(1\right)=-\frac{2\lambda }{3}+c=-1}{c=1\text{and}}\lambda =3 ....\left(3\right) \\ \therefore g\left(x\right)=x^3-3x+1 \\ \therefore g\left(2\right)=3 \end{gathered}$$