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?

$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }$

Solution 1

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 $\mathbf{(C)}$.

Solution 2

If $f(x)=y$, then squaring both sides of the given equation and subtracting $ax^2$ and $bx$ yields $y^2-ax^2-bx=0$. Completing the square, we get $(x+\frac{b}{2a})^2-\frac{y^2}{a}=\frac{b^2}{4a^2}$ where $y\geq 0$. Divide out by $\frac{b^2}{4a^2}$ to put the equation in the standard form of an ellipse or hyperbola (depending on the sign of $a$) to get $\frac{(x+\frac{b}{2a})^2}{(\frac{b}{2a})^2}-\frac{y^2}{(\frac{b}{2\sqrt {\pm a}})^2}=1$.

Before continuing, it is important to note that because $f(x)=\sqrt{ax^2+bx}=\sqrt{ax(x+\frac{b}{a})}$, $f(x)$ has roots 0 and $-\frac{b}{a}$. Now, we can use the function we deduced to figure out some of its properties when:

$a>0$: A semi-hyperbola above or on the x-axis. Therefore, no positive value of $a$ allows the domain and range to be the same set because the range will always be $[0, \infty)$ and the domain will always be undefined on some finite range between some value and zero.

$a=0$: We must refer back to the original function; this results in a horizontal semiparabola in the first quadrant, which satisfies that the domain and range of the function are equal. Specifically, both sets are $[0, \infty )$. $a=0$ is the only case where this happens.

$a<0$: A semi-ellipse in quadrant one. Since its roots are 0 and $-\frac{b}{a}$, its domain must be $[0, -\frac{b}{a}]$. To make its domain and range equal, the maximum value of the ellipse must then be $-\frac{b}{a}$. But we have another expression for the maximum value of the ellipse, which is $0+\frac{b}{2\sqrt {\pm a}}$. Setting these two expressions equal to each other will find us the final value of $a$ that satisfies the question.

$\frac{b}{-a}=\frac{b}{2\sqrt {\pm a}}$

$-a=2\sqrt {\pm a}$

$a^2=\pm 4a$

$a=0,\pm 4$

We already knew 0 was a solution from earlier, so -4 is our only new solution (we already ruled out any positive value of $a$ as a solution, so 4 does not work). Thus there are $\boxed {2\implies C}$ values of $a$ that make the domain and range of $f(x)$ the same set.

See Also

2003 AMC 12A (ProblemsAnswer KeyResources)
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