Explanation

(b)
Since, $g(x)g(y) = g(x) + g(y) + g(xy) - 2$ ...(i)
Now, at $x = 0,\ \ y = 2$, we get
$g(0)g(2) = g(0) + g(2) + g(0) - 2$
$\Rightarrow \ \ \ \ \ \ 5g(0) = 5 + 2g(0) - 2\ \ \ \ \ \ \ \ \ \ \lbrack\because\ \ g(2) = 5\rbrack$
$\Rightarrow \ \ \ \ \ \ g(0) = 1$
$g(x)$ is given in a polynomial and by the relation given $g(x)$ cannot be linear.
Let $g(x) = x^{2} + k$
$\Rightarrow \ \ \ \ \ \ \ g(x) = x^{2} + 1\ \ \ \ \ \ \ \ \ \ \ \left\lbrack \because\ \ g(0) = 1 \right\rbrack$
$\therefore\ \ \ \ \ \ g(x)$ is satisfied in Eq. (i)
$\therefore\ \ \ \ \lim_{x \rightarrow 3}{g(x)} = g(3) = 3^{2} + 1 = 10$