Difference between revisions of "1993 AHSME Problems/Problem 12"
(→Solution) |
(→Solution) |
||
Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
− | As <math>f(2x)=\frac{2}{2+x}</math>, we have that <math>f(x)=\frac{2}{2+\frac{x}{2}}</math>. This also means that <math>2f(x)=\frac{2}{2(2+\frac{x}{2})}</math> which implies that the answer is \fbox{E}$ | + | As <math>f(2x)=\frac{2}{2+x}</math>, we have that <math>f(x)=\frac{2}{2+\frac{x}{2}}</math>. This also means that <math>2f(x)=\frac{2}{2(2+\frac{x}{2})}</math> which implies that the answer is \fbox{E}$. ~ samrocksnature |
== See also == | == See also == |
Revision as of 21:32, 4 April 2021
Problem
If for all , then
Solution
As , we have that . This also means that which implies that the answer is \fbox{E}$. ~ samrocksnature
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.