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

Latest revision as of 19:39, 6 May 2007

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

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

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

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

@@ Line 1: / Line 1: @@
 ==Problem==
-We define the operation <math>a*b = \frac{1+a}{1+b^2}</math>, <math>\forall a,b \in \real</math>.
+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
@@ Line 7: / Line 7: @@
 ==Solution==
-<math>\frac{1+\frac{1+2}{1+0^2}}{1+1^2}=\frac{1+3}{2}=2\Rightarrow\mathrm{ A}</math>
+<math>\frac{1+\frac{1+2}{1+0^2}}{1+1^2}=\frac{1+3}{2}=2\Longrightarrow\mathrm{ A}</math>
 ==See also==
-*[[2007 Cyprus MO/Lyceum/Problems]]
+{{CYMO box|year=2007|l=Lyceum|num-b=3|num-a=5}}
-*[[2007 Cyprus MO/Lyceum/Problem 3|Previous Problem]]
+[[Category:Introductory Algebra Problems]]
-*[[2007 Cyprus MO/Lyceum/Problem 5|Next Problem]]