Let $g \left(x\right)$ is a polynomial function satisfying $g\left(x\right)g\left(y\right)=g\left(x\right)+g\left(y\right)+g\left(xy\right)-2$ for all $x, y \in R.$ If $g\left(2\right)=5,$ then $g\left(5\right)=$

Putting $x=2$ and $y=1$ in the given relation, we obtain

$\begin{array}{l} g\left(2\right)g\left(1\right)=g\left(2\right)+g\left(1\right)+g\left(2\right)-2\\ \Rightarrow 5g\left(1\right)=5+g\left(1\right)+5-2\\ \Rightarrow g\left(1\right)=2\end{array}$

Putting $y=\frac{1}{x}$  in the given relation, we get

          $g\left(x\right)g\left(\frac{1}{x}\right)=g\left(x\right)+g\left(\frac{1}{x}\right)+g\left(1\right)-2$

  $\Rightarrow g\left(x\right)g\left(\frac{1}{x}\right)=g\left(x\right)+g\left(\frac{1}{x}\right) [\because g(1)=2]$

$\Rightarrow g\left(x\right)=x^n+1$                 [See Illustration $6$]

$\begin{array}{r}\Rightarrow g\left(2\right)=2^n+1\Rightarrow 5=2^n+1\Rightarrow n=2\\ \therefore g\left(x\right)=x^2+1\Rightarrow g\left(5\right)=25+1=26.\end{array}$