If `f` is polynomial function satisfying `2+f(x)f(y)=f(x)+f(y)+f(x y)AAx , y in R` and if `f(2)=5,` then find the value of `f(f(2))dot`

Explanation

Put ` y=(1)/(x).` Then
`2+f(x) f((1)/(x))=f(x)+f((1)/(x))+f(1)`
Now, put `x=1`. Then
`2+(f(1))^(2)=3f(1)`
` :. f(1)=1 or 2`
But `f(1) ne 1,` otherwise from the given relation, `2+f(x)f(1)=f(x)+f(1)+f(x) " or " f(x)=1,` which is not possible as given that `f(2)=5`.
Therefore, from (1), we have
`f(x) f((1)/(x))=f(x)+f((1)/(x))`
` or f(x)= +-x^(n)+1`
` :. f(2)=+-2^(n)+1=5`
` or 2^(n)=4 or n=2`
` or f(x)=x^(2)+1`
` or f(f(2))=f(5)=26`