If f (x) is a polynomial satisfying f (x) f (1/x) = f (x) + f (1/x) and f (3) = 28, then f (4) is given by
63
65
67
68
Any polynomial satisfying the functional equation f (x) . f (1/x) = f (x) + f (1/x) is of the form xn+1or -xn+1 If 28 =f(3) = -3n+1 then 3n = -27 which is not possible for any n. Hence 28 = f (3) = 3n + 1⇒ 3n= 27 ⇒ n = 3. Thus f (x) = x3 + 1, so f (4) = 43 + 1 = 65.