Difference between revisions of "1968 AHSME Problems/Problem 16"
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== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | |
| + | Because <math>\frac{1}{x}<2</math>, <math>x>\frac{1}{2}</math> when <math>x</math> is positive. Because <math>\frac{1}{x}>-3</math>, <math>x<\frac{-1}{3}</math> when <math>x</math> is negative. <math>x</math> can never be <math>0</math>, so these two inequalities cover all cases for the value of <math>x</math>. Thus, our answer is <math>\fbox{E}</math>. | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=15|num-a=17}} | + | {{AHSME 35p box|year=1968|num-b=15|num-a=17}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:46, 17 July 2024
Problem
If 
is such that 
and 
, then:
Solution
Because , 
when is positive. Because , 
when is negative. can never be 
, so these two inequalities cover all cases for the value of . Thus, our answer is 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
