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Difference between revisions of "2003 AMC 12A Problems/Problem 25"

(I got somewhere, someone else go the rest of the way.)
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== Solution==
 
== Solution==
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The domain of this function is the range of the inverse function, and vice versa, so we find the inverse function:
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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>.
  
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<math>y=\sqrt{ax^2+bx}</math>
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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.
  
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<math>y^2=ax^2+bx</math>
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[[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>.
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<math>x=\dfrac{-b\pm\sqrt{b^2+4ay^2}}{2a}</math>
 
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The domain of this is all real <math>y</math> such that <math>4ay^2\geq -b^2</math>
 
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The range of this function is the domain of the other function, which is all <math>x</math> such that <math>ax^2+bx\geq 0</math>. Thus we need to find all real <math>a</math> such that for all <math>x</math>, either both of those are true or neither are.
 
  
 
{{incomplete|solution}}
 
{{incomplete|solution}}
  
 
==See Also==
 
==See Also==
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*[[2003 AMC 12A Problems/Problem 24 | Previous problem]]
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{{AMC12 box|year=2003|ab=A|num-b=24|after=Last question}}
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*[[2003 AMC 12A Problems]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 12:19, 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 $0 \le \frac{-b}{a}$, which 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.

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 $0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}$.

Template:Incomplete

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
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