Difference between revisions of "2006 AMC 12A Problems/Problem 18"

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== Problem ==
 
== Problem ==
  
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  
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The function <math>\displaystyle f</math> has the property that for each real number <math>\displaystyle x</math> in its domain, <math>\displaystyle 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>

Revision as of 00:21, 10 August 2006

Problem

The function $\displaystyle f$ has the property that for each real number $\displaystyle x$ in its domain, $\displaystyle 1/x$ is also in its domain and

$f(x)+f\left(\frac{1}{x}\right)=x$

What is the largest set of real numbers that can be in the domain of $f$?

$\mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}$

$\mathrm{(C) \ } \{x|x>0\}$$\mathrm{(D) \ } \{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\}$

$\mathrm{(E) \ }  \{-1,1\}$

Solution

$f(x)+f\left(\frac{1}{x}\right)=x$

Plugging in $\frac{1}{x}$ into the function:

$f\left(\frac{1}{x}\right)+f\left(\frac{1}{\frac{1}{x}}\right)=\frac{1}{x}$

$f\left(\frac{1}{x}\right)+ f(x)= \frac{1}{x}$

Since $f(x) + f\left(\frac{1}{x}\right)$ cannot have two values:

$x = \frac{1}{x}$

$x^2 = 1$

$x=\pm 1$

Therefore, the largest set of real numbers that can be in the domain of $f$ is $\{-1,1\} \Rightarrow E$

See also