Difference between revisions of "2006 AMC 12A Problems/Problem 18"
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<math> \mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}</math> | <math> \mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}</math> | ||
| − | <math>\mathrm{(C) \ } \{x|x>0\}</math><math>\mathrm{(D) \ } \{x|x\ne -1\;\ | + | <math>\mathrm{(C) \ } \{x|x>0\}</math><math>\mathrm{(D) \ } \{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\}</math> |
| + | |||
| + | <math>\mathrm{(E) \ } \{-1,1\}</math> | ||
== Solution == | == Solution == | ||
| + | Quickly verifying by plugging in values verifies that <math>-1</math> and <math>1</math> are in the domain. | ||
| + | |||
| + | <math>f(x)+f\left(\frac{1}{x}\right)=x</math> | ||
| + | |||
| + | Plugging in <math> \frac{1}{x} </math> into the [[function]]: | ||
| + | |||
| + | <math>f\left(\frac{1}{x}\right)+f\left(\frac{1}{\frac{1}{x}}\right)=\frac{1}{x}</math> | ||
| + | |||
| + | <math>f\left(\frac{1}{x}\right)+ f(x)= \frac{1}{x}</math> | ||
| + | |||
| + | Since <math>f(x) + f\left(\frac{1}{x}\right) </math> cannot have two values: | ||
| + | |||
| + | <math> x = \frac{1}{x} </math> | ||
| + | |||
| + | <math> x^2 = 1 </math> | ||
| + | |||
| + | <math> x=\pm 1</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 == | ||
* [[2006 AMC 12A Problems]] | * [[2006 AMC 12A Problems]] | ||
| + | |||
| + | {{AMC12 box|year=2006|ab=A|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
Latest revision as of 17:34, 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 ![\[f \left(\frac{1}{x}\right) + f \left(\frac{1}{\frac{1}{x}}\right) = \frac{1}{x}\]](http://latex.artofproblemsolving.com/7/7/1/7719eaa8dde229d3b29828b6b5045ea6194488a5.png)
![\[f \left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.\]](http://latex.artofproblemsolving.com/6/9/c/69c45495e2dce6fe7f6e9a0558605d11e88a9c10.png)
Also notice ![\[f \left(\frac{1}{x}\right) + f(x) = x\]](http://latex.artofproblemsolving.com/2/d/6/2d6c78e7baaf12360805502ec97ade5f739e32cc.png)
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. 
