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2003 AMC 12A Problems/Problem 25

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Problem

Let f(x)= \sqrt{ax^2+bx}. For how many real values of a is there at least one positive value of b for which the domain of f and the range of f are the same set?

\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{inf...

Solution

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. If a is positive, then \lim_{x \rightarrow -\infty} ax^2 + bx \ge 0, which implies that a negative number falls in the domain of f(x), contradiction. 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)}.

See Also

2003 AMC 12A (ProblemsResources)
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Problem 24
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