Mathematics - Functions Question with Solution | TestHub

Putting $x=1,y=2$, then

$$g(1)g(2)=g(1)+g(2)+g(2)-2$$

$$\Rightarrow 5g(1)=8+g(1)$$

$$\therefore g(1)=2$$

Also, replacing $y$ by $\frac{1}{x}$ in the given relation, then

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

or

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

$\Rightarrow g(x)=1\pm x^n$

$\Rightarrow \pm 2^n=2^2$

Taking +ve sign

$$2^n=2^2$$

$$\therefore n=2$$

$$\Rightarrow g(x)=1+x^2$$

$$\therefore g(3)=1+3^2=10$$