AMC12 2003 A

The function $f(x) = \sqrt{x(ax+b)}$ has a codomain of all non-negative numbers, or $0 \le f(x)$. Since the domain and the range of $f$ are the same, it follows that the domain of $f$ also satisfies $0 \le x$. The function has two zeroes at $x = 0, \frac{-b}{a}$, which must be part of the domain. Since the domain and the range are the same set, it follows that $\frac{-b}{a}$ is in the codomain of $f$, or $0 \le \frac{-b}{a}$. This implies that one (but not both) of $a,b$ is non-positive. The problem states that there is at least one positive value of b that works, thus $a$ must be non-positive, $b$ is non-negative, and the domain of the function occurs when $x(ax+b) > 0$, or $0 \le x \le \frac{-b}{a}.$ Completing the square, $f(x) = \sqrt{a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}} \le \sqrt{\frac{-b^2}{4a}}$ by the Trivial Inequality (remember that $a \le 0$). Since $f$ is continuous and assumes this maximal value at $x = \frac{-b}{2a}$, it follows that the range of $f$ is $0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}.$ As the domain and the range are the same, we have that $\frac{-b}{a} = \sqrt{\frac{-b^2}{4a}} = \frac{b}{2\sqrt{-a}} \Longrightarrow a(a+4) = 0$ (we can divide through by $b$ since it is given that $b$ is positive). Hence $a = 0, -4$, which both we can verify work, and the answer is $\mathbf{(C)}$.