Difference between revisions of "2003 AMC 12A Problems/Problem 25"
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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>. | 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> | + | 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 |
− | + | <center><math>0 \le x \le \frac{-b}{a}.</math></center> | |
− | {{ | + | [[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 |
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+ | <center><math>0 \le f(x) \le \sqrt{\frac{-b^2}{4a}}.</math></center> | ||
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+ | 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 . For how many real values of is there at least one positive value of for which the domain of and the range of are the same set?
Solution
The function has a codomain of all non-negative numbers, or . Since the domain and the range of are the same, it follows that the domain of also satisfies .
The function has two zeroes at , which must be part of the domain. Since the domain and the range are the same set, it follows that is in the codomain of , or . This implies that one (but not both) of is non-positive. If is positive, then , which implies that a negative number falls in the domain of , contradiction. Thus must be non-positive, is non-negative, and the domain of the function occurs when , or
Completing the square, by the Trivial Inequality (remember that ). Since is continuous and assumes this maximal value at , it follows that the range of is
As the domain and the range are the same, we have that (we can divide through by since it is given that is positive). Hence , which both we can verify work, and the answer is .
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 |