Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 3"
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==Solution== | ==Solution== | ||
| − | <math>2x+4</math> must be non-negative, so <math>x+2</math> must be non-negative. | + | <math>2x+4</math> must be non-negative, so <math>x+2</math> must be non-negative. Therefore, all <math>x\ge-2</math> are in the domain <math>\mathrm{(C)}</math>. |
| − | |||
| − | Therefore, all <math>x\ge-2</math> are in the domain <math>\mathrm{(C)}</math>. | ||
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=2|num-a=4}} | {{CYMO box|year=2006|l=Lyceum|num-b=2|num-a=4}} | ||
Latest revision as of 09:50, 27 April 2008
Problem
is
![$\mathrm{(A)}\ (-2,+\infty)\qquad\mathrm{(B)}\ [0,+\infty)\qquad\mathrm{(C)}\ [-2,+\infty)\qquad\mathrm{(D)}\ [-2,0]\qquad\mathrm{(E)}\ \mathbb{R}$](http://latex.artofproblemsolving.com/e/d/9/ed90c6307eeded48f8d1eaac79fa8343e0b30cfd.png)
Solution
must be non-negative, so 
must be non-negative. Therefore, all 
are in the domain 
.
See also
| 2006 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
