Difference between revisions of "2003 AMC 12A Problems/Problem 25"
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== Solution== | == Solution== | ||
− | The domain | + | 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>. |
− | <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. |
− | <math> | + | [[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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{{incomplete|solution}} | {{incomplete|solution}} | ||
==See Also== | ==See Also== | ||
− | + | {{AMC12 box|year=2003|ab=A|num-b=24|after=Last question}} | |
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 12:19, 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 , which 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.
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 .
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 |