Q.7. If g(x) = 1 + \sqrt{x} and f(g(x)) = 3 + 2\sqrt{x} + x , then f(f(x)) is:
A. x^{2} + 4x + 6
B. x^{4} + x^{2} + 6
C. x^{2} + x + 6
D. x^{4} + 4x^{2} + 6
Answer: D
Explanation:
g(x) = 1 + \sqrt{x} and f(g(x)) = 3 + 2\sqrt{x} + x
f(g(x)) = [(\sqrt{x})^2 + 2\sqrt{x} + 1] + 2 = (\sqrt{x} + 1)^2 + 2 = [g(x)]^2 + 2 \Rightarrow f(x) = x^2 + 2 \Rightarrow f(f(x)) = (x^2 + 2)^2 + 2 = x^4 + 4x^2 + 6