Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 4"

 
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==Problem==
 
==Problem==
We define the operation <math>a*b = \frac{1+a}{1+b^2}</math>, <math>\forall a,b \in \real</math>.
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We define the [[operation]] <math>a*b = \frac{1+a}{1+b^2}</math>, <math>\forall a,b \in \real</math>.
  
 
The value of <math>(2*0)*1</math> is
 
The value of <math>(2*0)*1</math> is
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==See also==
 
==See also==
*[[2007 Cyprus MO/Lyceum/Problems]]
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{{CYMO box|year=2007|l=Lyceum|num-b=3|num-a=5}}
  
*[[2007 Cyprus MO/Lyceum/Problem 3|Previous Problem]]
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[[Category:Introductory Algebra Problems]]
 
 
*[[2007 Cyprus MO/Lyceum/Problem 5|Next Problem]]
 

Revision as of 15:19, 6 May 2007

Problem

We define the operation $a*b = \frac{1+a}{1+b^2}$, $\forall a,b \in \real$.

The value of $(2*0)*1$ is

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 0\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{5}{2}$

Solution

$\frac{1+\frac{1+2}{1+0^2}}{1+1^2}=\frac{1+3}{2}=2\Rightarrow\mathrm{ A}$

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 3
Followed by
Problem 5
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