Mathematics - Functions Question with Solution | TestHub

Let $x=y=1$.

$$F(x)+f(y)+f(xy)=2+f(x)\cdot f(y)$$

$$3f(1)=2+(f(1){)}^2⟹f(1)=1,2$$

But given that $f(1)\ne 1$, so $f(1)=2$.

Now put $y=\frac{1}{x}$.

$$f(x)+f\left(\frac{1}{x}\right)+f(1)=2+f(x)\cdot f\left(\frac{1}{x}\right)$$

So $f(x)=\pm x^n+1$. Now $f(4)=17⟹\pm (4{)}^n+1=17⟹n=2$.

$$f(x)=+(x{)}^2+1$$