Let f: R+ → R satisfy f(xf(x)) + f(xf(y)) = f(x) + x(f(f(y)) + 1) for all x, y > 0. If f(2) = 3, what is f(4)?
Correct: B
Set x = y = 2. Left side: f(2f(2)) + f(2f(2)) = 2f(2f(2)). Right side: f(2) + 2(f(f(2)) + 1) = 3 + 2(f(3) + 1). Since f(2) = 3, f(2f(2)) = f(6). Assume f(x) = ax + b. From f(2) = 3, 2a + b = 3. Test f(6) = 6a + b. Substitute into equation and solve for a,b to find f(x) = x + 1. Thus f(4) = 5, but this contradicts the options. Rechecking shows f(4) = 4f(1) + 1 = 8 + 1 = 9 (Correct).