Question

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

• A. $9$
• B. $10$
• C. $25$
• D. $20$

Answer 1

Answer: (B)

$10$

$g\left(x\right).g\left(y\right)=g\left(x\right)+g\left(y\right)+g\left(xy\right)-2....\left(1\right)$
Put $x=1,y=2$, then
$g\left(1\right).g\left(2\right)=g\left(1\right)+g\left(2\right)+g\left(2\right)-2$
$5g\left(1\right)=g\left(1\right)+5+5-2$
$4g\left(1\right)=8\therefore g\left(1\right)=2$
Put $y=\frac{1}{x}$ in equation $\left(1\right)$, we get

$g\left(x\right).g\left(\frac{1}{x}\right)=g\left(x\right)+g\left(\frac{1}{x}\right)+g\left(1\right)-2$
$g\left(x\right).g\left(\frac{1}{x}\right)=g\left(x\right)+g\left(\frac{1}{x}\right)+2-2[\because g(1)=2]$
This is valid only for the polynomial
$\therefore g\left(x\right)=1\pm x^n....\left(2\right)$
Now $g\left(2\right)=5$ (Given)
$\therefore 1\pm 2^n=5$ [Using equation $\left(2\right)$]
$\pm 2^n=4,⟹2^n=4,-4$
Since the value of $2^n$ cannot be $-Ve$.
So, $2^n=4,⟹n=2$
Now, put $n=2$ in equation $\left(2\right)$, we get
$g\left(x\right)=1\pm x^2$
$\therefore \underset{x\to 3}{Lt}g\left(x\right)=\underset{x\to 3}{Lt}(1\pm x^2)=1\pm (3{)}^2$
$=1\pm 9=10,-8$.

Hence, option B is correct.