Difference between revisions of "1983 AHSME Problems/Problem 21"
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After reaching the two possibilities <math>\sqrt{100}-\sqrt{99}</math> and <math>\sqrt{2601}-\sqrt{2600}</math> as in the first solution, note that these can be written as <math>\sqrt{x}-\sqrt{x-1}</math> for <math>x=100</math> and <math>x=2601</math> respectively. Now, <math>\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{x}-\sqrt{x-1}\right) = \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{x-1}} = \frac{\sqrt{x-1} - \sqrt{x}}{2\sqrt{x}\sqrt{x-1}} < 0</math> (as the numerator is always negative and the denominator is always positive for <math>x \geq 1</math>). Therefore, with its derivative being negative, <math>\sqrt{x}-\sqrt{x-1}</math> is a strictly decreasing function over its domain, so the smaller value will be the one with the larger value of <math>x</math>, namely <math>x = 2600</math>. Therefore the answer is <math>\boxed{\textbf{(E)} \ 10 \sqrt{26} - 51}</math> as before. | After reaching the two possibilities <math>\sqrt{100}-\sqrt{99}</math> and <math>\sqrt{2601}-\sqrt{2600}</math> as in the first solution, note that these can be written as <math>\sqrt{x}-\sqrt{x-1}</math> for <math>x=100</math> and <math>x=2601</math> respectively. Now, <math>\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{x}-\sqrt{x-1}\right) = \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{x-1}} = \frac{\sqrt{x-1} - \sqrt{x}}{2\sqrt{x}\sqrt{x-1}} < 0</math> (as the numerator is always negative and the denominator is always positive for <math>x \geq 1</math>). Therefore, with its derivative being negative, <math>\sqrt{x}-\sqrt{x-1}</math> is a strictly decreasing function over its domain, so the smaller value will be the one with the larger value of <math>x</math>, namely <math>x = 2600</math>. Therefore the answer is <math>\boxed{\textbf{(E)} \ 10 \sqrt{26} - 51}</math> as before. | ||
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| + | ==See Also== | ||
| + | {{AHSME box|year=1983|num-b=20|num-a=22}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 23:58, 19 February 2019
Contents
Problem
Find the smallest positive number from the numbers below.
Solution 1
Notice that 
, 
, and similarly, 
, so these numbers can be excluded immediately as they are negative and we seek the smallest positive number. The remaining choices are 
and 
. Observe that 
, and 
, so the smallest positive number in the list is 
.
Solution 2
After reaching the two possibilities 
and 
as in the first solution, note that these can be written as 
for 
and 
respectively. Now, 
(as the numerator is always negative and the denominator is always positive for 
). Therefore, with its derivative being negative, is a strictly decreasing function over its domain, so the smaller value will be the one with the larger value of 
, namely 
. Therefore the answer is as before.
See Also
| 1983 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
