Question

If $g(x)$ is a polynomial satisfying $g(x)g(y)=g(x)+g(y)+g(xy)-2$ for all real $x$ and $y$ and $g(2)=5$ then $\underset{x\to 3}{\mathrm{Lt}}g(x)$ is

• A. 9
• B. 10
• C. 25
• D. 20

Answer 1

Answer: (B)

$g(x)\cdot g(y)=g(x)+g(y)+g(xy)-2\dots .(1)$
Put $x=1,y=2$, then
$g(1)\cdot g(2)=g(1)+g(2)+g(2)-2$
$5g(1)=g(1)+5+5-2$
$4g(1)=8$
$\therefore g(1)=2$
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
$\therefore g(x)=1\pm x^n\dots .(2)$
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 $g(x)=1\pm x^2$
$\therefore \underset{x\to 3}{\mathrm{Lt}}(x)=\underset{x\to 3}{\mathrm{Lt}}\left(1\pm x^2\right)=1\pm (3{)}^2$
$=1\pm 9=10,-8$