A function f : R →R satisfies the equation f(x)f(y) - f(xy) = x + y for all x, y ∈ R and f(1) > 0, then
(a) f(x) = x + 1/2
(b) f(x) = 1/2x + 1
(c) f(x) = x + 1
(d) f(x) = 1/2x - 1
The correct option (c) f(x) = x + 1
Explanation:
Taking x = y = 1 we get
f(1)f(1) – f(1) = 1 + 1 ⇒ f(1)2 – f(1) – 2 = 0
⇒ f(1) = 2 (∵ f(1) > 0)
Taking y = 1 we get
f(x)f(1) – f(x) = x + 1
⇒ 2f(x) – f(x) = x + 1
∴ f(x) = f(x + 1)
