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Q.
If $g(x)$ is a polynomial satisfying $g(x) g(y)=g(x)+g(y)+g(x y)-2$ for all real $x$ and $y$ and $g(2)=5$ then $\underset{x \rightarrow 3}{Lt} g(x)$ is
$g(x) \cdot g(y)=g(x)+g(y)+g(x y)-2 \ldots .(1)$
Put $x=1, y=2$, then
$g(1) \cdot g(2)=g(1)+g(2)+g(2)-2$
$5 g(1)=g(1)+5+5-2$
$4 g(1)=8 $
$\therefore g(1)=2$
Put $y=\frac{1}{x}$ in equation (1), we get
$g(x) \cdot g\left(\frac{1}{x}\right)=g(x)+g\left(\frac{1}{x}\right)+g(1)-2$
$g(x) \cdot g\left(\frac{1}{x}\right)=g(x)+g\left(\frac{1}{x}\right)+2-2$
$[\because g(1)=2]$
This is valid only for the polynomial
$\therefore g(x)=1 \pm x^{n} \ldots .(2)$
Now $g(2)=5$ (Given)
$\therefore 1 \pm 2^{n}=5$ [Using equation (2)]
$\pm 2^{n}=4, $
$\Rightarrow 2^{n}=4,-4$
$\therefore 1 \pm 2^{ n }=5$ [Using equation (2)]
$\pm 2^{ n }=4,$
$ \Rightarrow 2^{ n }=4,-4$
Since the value of $2^{ n }$ cannot be $- Ve$.
So, $2^{ n }=4, $
$\Rightarrow n =2$
Now, put $n =2$ in equation (2),
we get $g(x)=1 \pm x^{2}$
$\therefore \underset{x \rightarrow 3}{Lt}(x)=\underset{x \rightarrow 3}{Lt}\left(1 \pm x^{2}\right)=1 \pm(3)^{2}$
$=1 \pm 9=10,-8 $