Difference between revisions of "1965 AHSME Problems/Problem 23"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | |
| + | There are two cases: (i) <math>x-2>0</math> and (ii) <math>x-2<0</math>. In case (i), <math>0<x-2<0.01</math>, so <math>4<x+2<4.01</math>. In case (ii), <math>-0.01<x-2<0</math>, so <math>3.99<x+2<4</math>. Because <math>(x+2)</math> in case (i) can take a larger value, we use it to determine the upper bound for <math>|x^2-4|</math> under the given restrictions. Now, we can see that <math>|x^2-4|=|x-2||x+2|<0.01*4.01=\boxed{0.0401}</math>, which is answer choice <math>\fbox{\textbf{(D)}}</math>. | ||
== See Also == | == See Also == | ||
Latest revision as of 19:04, 18 July 2024
Problem
If we write 
for all 
such that 
, the smallest value we can use for 
is:
Solution
There are two cases: (i) 
and (ii) 
. In case (i), 
, so 
. In case (ii), 
, so 
. Because 
in case (i) can take a larger value, we use it to determine the upper bound for 
under the given restrictions. Now, we can see that 
, which is answer choice 
.
See Also
| 1965 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
