Given that f(x) f(y) + 2 = f(x) + f(y) + f(xy) ....(i)
Putting x = y = 0 in equation (i), we get f(0) f(0) + 2 = f(0) + f(0) + f(0)
or (f(0))2 + 2 = 3f(0) or (f(0) – 2) (f(0) – 1) = 0 or f(0)
(∴ f(0)≠ 2) ....(ii)
Again putting x = y = 1 in equation (i) and repeating the above steps, we get (f(1) – 2) (f(1) – 1) = 0
But f(1)≠ 1 as f(x) is injective.
∴ f(1) = 2 ....(iii)
Now putting y = 1/x in equation (i), we get