Difference between revisions of "1993 AHSME Problems/Problem 12"
(→Solution) |
Du.visionary (talk | contribs) (→Solution) |
||
| (4 intermediate revisions by 2 users not shown) | |||
| 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{4}{ | + | 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{4}{2+\frac{x}{2}}</math> which simplifies to <math>\fbox{E}</math>. |
| + | |||
| + | ~ samrocksnature | ||
| + | |||
| + | ~ clarification by Leon | ||
| + | |||
| + | Note: Wait what | ||
== See also == | == See also == | ||
Latest revision as of 11:46, 3 June 2024
Problem
If 
for all 
, then 
Solution
As , we have that 
. This also means that 
which simplifies to 
.
~ samrocksnature
~ clarification by Leon
Note: Wait what
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. 
