Difference between revisions of "2006 AMC 12A Problems/Problem 18"
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== Problem == | == Problem == | ||
− | The function <math> | + | The function <math>f</math> has the property that for each real number <math>x</math> in its domain, <math>1/x</math> is also in its domain and |
<math>f(x)+f\left(\frac{1}{x}\right)=x</math> | <math>f(x)+f\left(\frac{1}{x}\right)=x</math> | ||
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Therefore, the largest [[set]] of [[real number]]s that can be in the [[domain]] of <math>f</math> is <math>\{-1,1\} \Rightarrow E </math> | Therefore, the largest [[set]] of [[real number]]s that can be in the [[domain]] of <math>f</math> is <math>\{-1,1\} \Rightarrow E </math> | ||
+ | |||
+ | == Solution 2== | ||
+ | We know that <math>f(x) + f \left(\frac{1}{x}\right) = x.</math> Plugging in <math>x = \frac{1}{x}</math> we get <cmath>f \left(\frac{1}{x}\right) + f \left(\frac{1}{\frac{1}{x}}\right) = \frac{1}{x}</cmath> <cmath>f \left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.</cmath> | ||
+ | |||
+ | Also notice <cmath>f \left(\frac{1}{x}\right) + f(x) = x</cmath> by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set <math>\frac{1}{x} = x</math> which gives us <math>x = \pm 1</math> which is answer option <math>\boxed{\mathrm{(E) \ } \{-1,1\}}.</math> | ||
== See also == | == See also == | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
− | [[Category:Introductory Algebra Problems]] | + | [[Category: Introductory Algebra Problems]] |
Latest revision as of 18:34, 18 September 2020
Contents
Problem
The function has the property that for each real number in its domain, is also in its domain and
What is the largest set of real numbers that can be in the domain of ?
Solution
Quickly verifying by plugging in values verifies that and are in the domain.
Plugging in into the function:
Since cannot have two values:
Therefore, the largest set of real numbers that can be in the domain of is
Solution 2
We know that Plugging in we get
Also notice by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set which gives us which is answer option
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.