Difference between revisions of "1990 AHSME Problems/Problem 12"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | If <math>f(x)=-\sqrt2</math>, then <math>ax^2=0\implies x=0</math>. Therefore <math>f(\sqrt2)=0\implies 2a=\sqrt2</math>, so <math>\fbox{D}</math> |
== See also == | == See also == | ||
Latest revision as of 14:47, 3 September 2024
Problem
Let 
be the function defined by 
for some positive 
. If 
then 
Solution
If 
, then 
. Therefore 
, so 
See also
| 1990 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. 
