Mathematics - Functions Question with Solution | TestHub

Given $2+f(x)f(y)=f(x)+f(y)+f(xy)$ and $f(2)=5$.

Let $x=1$ and $y=1$, then $2+f(1{)}^2=f(1)+f(1)+f(1)$, so $f(1{)}^2-3f(1)+2=0$.

This factors to $(f(1)-1)(f(1)-2)=0$, so $f(1)=1$ or $f(1)=2$.

Let $x=2$ and $y=1$, then $2+f(2)f(1)=f(2)+f(1)+f(2)$. Since $f(2)=5$, $2+5f(1)=5+f(1)+5$, which simplifies to $4f(1)=8$, so $f(1)=2$.

Therefore, $f(f(1))=f(2)=5$.

Final Answer: The final answer is $\boxed{5}$