2003 AMC 12A Problems/Problem 25


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. 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 $\mathbf{(C)}$.

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.

Solution 3 (more basic)

  • When numbers are referred to as positive and negative, it is understood that they are real numbers, as complex numbers don't have a definite positive/negative value.

First, notice that the square root of anything, whether it be a positive number, a negative number, or a complex, cannot yield a negative number. The square root of a positive number is a positive number, and the square root of a negative or complex number will yield a complex number. Complex numbers are a bit harder to address than real numbers, so I won't go into depth on complex domain/range (You don't need to, as this is a 2003 question, and the test writers most likely did not intend for test takers to take complex numbers into consideration. Also, even without considering complex domain/range, we can still get the answer).

Since the range cannot include negative numbers, the domain must be the same. This means that the graph of $f(x)$ must only exist in the real coordinate plane for non-negative x-values. If any negative value of x yields a positive value under the square root, then $f(x)$ will be positive, and then a negative x-value will be part the domain, and that contradicts what the first thing that we found. Now, first consider the graph of a quadratic equation/parabola. We know that a parabola always has a vertex, and always extends until $+$ or $-$ infinity, depending on the $a$ value. Quite conveniently, we have an $a$ value in the problem.

Consider the case where $a$ is positive. Then, the parabola will "start" at the vertex, and extend upwards, hitting the x-axis at any roots. The y-values of a "positive" parabola are $[\infty, 0]$ between x-values of $-\infty$ and $r_1$ (the leftmost root), negative between $r_1$ and $r_2$ (the two roots) and $[0, \infty]$ between $r_2$ and $\infty$. If we take the square root of such a parabola, the new function will exist only from $x=[-\infty, r_1]$ and $x=[r_2, \infty]$. In order words, if $a$ is positive, the graph of $f(x)$ will exist from $-\infty$ to $\infty$ except between the two roots. This means, that regardless of where the roots are, some negative value of $x$ will yield a positive value of $f(x)$. This means a positive value of $a$ will not work.

Now we consider the case where $a$ is negative. The parabola will do the opposite of the above; i.e. extend downwards. Now, either the entire graph of the parabola is below the x-axis (no roots), or some part including the vertex sticks out above it. The graph of the square root of the parabola will either be non-existent, or exist only between the 2 roots, respectively. This could work, as long as both roots are at least 0 (this way we will not have negative x-values in the domain).

Upon examination, we find that $ax^2+bx$ can be rewritten $x(ax+b)$. The roots are then $x=0$ and $x=-\frac{b}{a}$. Since $a$ is negative, and $b$ is positive, $-\frac{b}{a}$ will always be positive. Therefore, the graph of $f(x)$ will only exist from $x=\left[0, -\frac{b}{a}\right]$ (the domain).

Notice that $ax^2+bx$ can also be written in vertex form. The maximum value of $f(x)$ (since $a$ is negative) is the y-coordinate of the vertex, i.e. the range of $f(x)$ is $[0, \text{vertex}]$. Converting $ax^2+bx$ to vertex form yields $f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}$. The y-coordinate of the vertex is then $-\frac{b^2}{4a}$. We want the square root of this, so the range of $f(x)$ is $\left[0, \sqrt{-\frac{b^2}{4a}}\right]$. Since domain and range must be equivalent, we have $\frac{b}{-a} = \sqrt{\frac{b^2}{-4a}} \Rightarrow \frac{b}{-a} = \frac{b}{\sqrt{-4a}} \Rightarrow -a = \sqrt{-4a} \Rightarrow a^2 = -4a \Rightarrow a^2 + 4a = 0 \Rightarrow a(a+4) = 0$. So we have $a= 0, -4$, which is 2 values or $\boxed{\mathbf{(C)}}$

See Also

2003 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last question
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png