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 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==