If $f\left(x\right)f\left(y\right)+2=f\left(x\right)+f\left(y\right)+f\left(xy\right)\forall x,y\in R$ and $f\left(1\right)=f^{\text{'}}\left(1\right)=2$ , then$f\left(2\right)$  is equal to

$f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{f\left(x\left(1+\frac{h}{x}\right)\right)-f\left(1.x\right)}{h}$

$=\underset{h\to 0}{\lim }\frac{f\left(x\right)-f\left(1+\frac{h}{x}\right)+2-f\left(x\right)f\left(1+\frac{h}{x}\right)-f\left(x\right).f\left(1\right)-2+f\left(x\right)+f\left(1\right)}{h}$

$=\underset{h\to 0}{\lim }\left(f\left(x\right)-1\right)\frac{\left(f\left(1+\frac{h}{x}\right)-f\left(1\right)\right)}{\frac{h}{x}.x}$

$=\frac{f\left(x\right)-1}{x}.f^{\text{'}}\left(1\right)$

$f^{\text{'}}\left(x\right)=2\left\{\frac{f\left(x\right)-1}{x}\right\}$

$\Rightarrow \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)-1}=\frac{2}{x}$

$\ell n\left(f\left(x\right)-1\right)=2\ell nx\_\ell nc$

$f\left(x\right)-1=c{x}^2$

$f\left(x\right)=c{x}^2+1$

as$f\left(1\right)=2\Rightarrow 2=c+1\Rightarrow c=1$

$\Rightarrow f\left(x\right)=x^2+1$ .