g (x). g(y) = g(x) + g (y) + g (xy)-2 ...(1) Put x = 1,y = 2, then g(1).g(2) = g(1) + g(2) + g(2)-2 5g(1) = g(1) + 5 + 5-2 4g(1) = 8 ∴ g (1) = 2 Put y =
1
x
in equation (1), we get g (x) , g (
1
x
) = g (x) + g (
1
x
) g (1) - 2 g (x) . g(
1
x
) = g (x) + g (
1
x
) + 2 - 2 [Since g (1) = 2] This is valid only for the polynomial ∴ g (x) = 1 ± xn ... (2) Now g (2) = 5 (Given) ∴ 1 ± 2n - 5 [Using equation (2)] ± 2n = 4 ⇒ 2n = 4 , - 4 Since the value of 2n cannot be -Ve. So, 2n = 4 ⇒ n = 2 Now, put n = 2 in equation (2), we get g (x) = 1 ± x2 ∴