Difference between revisions of "2011 AMC 12A Problems/Problem 23"
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<cmath>2=|b^2+1|\geq|b^2|-1</cmath> | <cmath>2=|b^2+1|\geq|b^2|-1</cmath> | ||
so, | so, | ||
− | <cmath>|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}</cmath>. Thus <math>1\leq |b|\leq \sqrt{3}</math>. Hence the maximum value for <math>|b|</math> is <math>\sqrt{3}</math> while the minimum is <math>1</math> (which can be achieved for instance when <math>|a|=1,|b|=\sqrt{3}</math> or <math>|a|=1,|b|=1</math> respectively). Therefore the answer is <math>\boxed{\textbf{(C)}\ | + | <cmath>|b^2|\leq 3\Rightarrow0\leq |b|\leq \sqrt{3}</cmath>. Thus <math>1\leq |b|\leq \sqrt{3}</math>. Hence the maximum value for <math>|b|</math> is <math>\sqrt{3}</math> while the minimum is <math>1</math> (which can be achieved for instance when <math>|a|=1,|b|=\sqrt{3}</math> or <math>|a|=1,|b|=1</math> respectively). Therefore the answer is <math>\boxed{\textbf{(C)}\ \sqrt{3}-1}</math>. |
== See also == | == See also == |
Revision as of 21:20, 23 November 2013
Problem
Let and , where and are complex numbers. Suppose that and for all for which is defined. What is the difference between the largest and smallest possible values of ?
Solution
By algebraic manipulations, we obtain where In order for , we must have , , and .
implies or .
implies , , or .
implies or .
Since , in order to satisfy all 3 conditions we must have either or . In the first case .
For the latter case note that and hence, . On the other hand, so, . Thus . Hence the maximum value for is while the minimum is (which can be achieved for instance when or respectively). Therefore the answer is .
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.