Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 12A Problems/Problem 25"
(→‎Solution)
Line 4: Line 4:
 
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }  </math>
 
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ \mathrm{infinitely \ many} }  </math>
  
−
== Solution==
+
== Solution 1==
 
The function <math>f(x) = \sqrt{x(ax+b)}</math> has a [[codomain]] of all non-negative numbers, or <math>0 \le f(x)</math>. Since the domain and the range of <math>f</math> are the same, it follows that the domain of <math>f</math> also satisfies <math>0 \le x</math>.
 
The function <math>f(x) = \sqrt{x(ax+b)}</math> has a [[codomain]] of all non-negative numbers, or <math>0 \le f(x)</math>. Since the domain and the range of <math>f</math> are the same, it follows that the domain of <math>f</math> also satisfies <math>0 \le x</math>.
  
Line 16: Line 16:
  
 
As the domain and the range are the same, we have that <math>\frac{-b}{a} = \sqrt{\frac{-b^2}{4a}} = \frac{b}{2\sqrt{-a}} \Longrightarrow a(a+4) = 0</math> (we can divide through by <math>b</math> since it is given that <math>b</math> is positive). Hence <math>a = 0, -4</math>, which both we can verify work, and the answer is <math>\mathbf{(C)}</math>.
 
As the domain and the range are the same, we have that <math>\frac{-b}{a} = \sqrt{\frac{-b^2}{4a}} = \frac{b}{2\sqrt{-a}} \Longrightarrow a(a+4) = 0</math> (we can divide through by <math>b</math> since it is given that <math>b</math> is positive). Hence <math>a = 0, -4</math>, which both we can verify work, and the answer is <math>\mathbf{(C)}</math>.
 +
 +
== Solution 2 ==
 +
If <math>f(x)=y</math>, then squaring both sides of the given equation and subtracting <math>ax^2</math> and <math>bx</math> yields <math>y^2-ax^2-bx=0</math>. Completing the square, we get <math>(x+\frac{b}{2a})^2-\frac{y^2}{a}=\frac{b^2}{4a^2}</math> where <math>y\geq 0</math>. Divide out by <math>\frac{b^2}{4a^2}</math> to put the equation in the standard form of an ellipse or hyperbola (depending on the sign of <math>a</math>) to get <math>\frac{(x+\frac{b}{2a})^2}{(\frac{b}{2a})^2}-\frac{y^2}{(\frac{b}{2\sqrt {\pm a}})^2}=1</math>.
 +
 +
Before continuing, it is important to note that because <math>f(x)=\sqrt{ax^2+bx}=\sqrt{ax(x+\frac{b}{a})}</math>, <math>f(x)</math> has roots 0 and <math>-\frac{b}{a}</math>. Now, we can use the function we deduced to figure out some of its properties when:
 +
 +
<math>a>0</math>: A semi-hyperbola above or on the x-axis. Therefore, no positive value of <math>a</math> allows the domain and range to be the same set because the range will always be <math>[0, \infty)</math> and the domain will always be undefined on some finite range between some value and zero.
 +
 +
<math>a=0</math>: 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 <math>[0, \infty )</math>. <math>a=0</math> is the only case where this happens.
 +
 +
<math>a<0</math>: A semi-ellipse in quadrant one. Since its roots are 0 and <math>-\frac{b}{a}</math>, its domain must be <math>[0, -\frac{b}{a}]</math>. To make its domain and range equal, the maximum value of the ellipse must then be <math>-\frac{b}{a}</math>. But we have another expression for the maximum value of the ellipse, which is <math>0+\frac{b}{2\sqrt {\pm a}}</math>. Setting these two expressions equal to each other will find us the final value of <math>a</math> that satisfies the question.
 +
 +
<math>\frac{b}{-a}=\frac{b}{2\sqrt {\pm a}}</math>
 +
 +
<math>-a=2\sqrt {\pm a}</math>
 +
 +
<math>a^2=\pm 4a</math>
 +
 +
<math>a=0,\pm 4</math>
 +
 +
We already knew 0 was a solution from earlier, so -4 is our only new solution (we already ruled out any positive value of <math>a</math> as a solution, so 4 does not work). Thus there are <math>\boxed {2\implies C}</math> values of <math>a</math> that make the domain and range of <math>f(x)</math> the same set.
  
 
==See Also==
 
==See Also==

Revision as of 00:51, 28 October 2017

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 (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_25&oldid=87996"