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"

m ((brb))
m (finish)
Line 7: Line 7:
 
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>.
  
−
The function has two zeroes at <math>x = 0, \frac{-b}{a}</math>, which must be part of the domain. Since the domain and the range are the same set, it follows that <math>0 \le \frac{-b}{a}</math>, which implies that one (but not both) of <math>a,b</math> is non-positive. If <math>a</math> is positive, then <math>\lim_{x \rightarrow -\infty} ax^2 + bx \ge 0</math>, which implies that a negative number falls in the domain of <math>f(x)</math>, contradiction. Thus <math>a</math> must be non-positive.
+
The function has two zeroes at <math>x = 0, \frac{-b}{a}</math>, which must be part of the domain. Since the domain and the range are the same set, it follows that <math>\frac{-b}{a}</math> is in the codomain of <math>f</math>, or <math>0 \le \frac{-b}{a}</math>. This implies that one (but not both) of <math>a,b</math> is non-positive. If <math>a</math> is positive, then <math>\lim_{x \rightarrow -\infty} ax^2 + bx \ge 0</math>, which implies that a negative number falls in the domain of <math>f(x)</math>, contradiction. Thus <math>a</math> must be non-positive, <math>b</math> is non-negative, and the domain of the function occurs when <math>x(ax+b) > 0</math>, or
  
−
[[Completing the square]], <math>f(x) = \sqrt{a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}} \le \sqrt{\frac{-b^2}{4a}}</math> by the [[Trivial Inequality]] (remember that <math>a \le 0</math>). Since <math>f</math> is continuous and assumes this maximal value at <math>x = \frac{-b}{2a}</math>, it follows that the range of <math>f</math> is <math>0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}</math>.
+
<center><math>0 \le x \le \frac{-b}{a}.</math></center>
  
−
{{incomplete|solution}}
+
[[Completing the square]], <math>f(x) = \sqrt{a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a}} \le \sqrt{\frac{-b^2}{4a}}</math> by the [[Trivial Inequality]] (remember that <math>a \le 0</math>). Since <math>f</math> is continuous and assumes this maximal value at <math>x = \frac{-b}{2a}</math>, it follows that the range of <math>f</math> is
 +
 
 +
<center><math>0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}.</math></center>
 +
 
 +
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>.
  
 
==See Also==
 
==See Also==

Revision as of 14:02, 11 August 2008

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

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 (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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_25&oldid=27363"