Difference between revisions of "1981 AHSME Problems/Problem 18"
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| − | ==Problem | + | ==Problem 18== |
The number of real solutions to the equation <cmath>\dfrac{x}{100}=\sin x</cmath> is | The number of real solutions to the equation <cmath>\dfrac{x}{100}=\sin x</cmath> is | ||
| − | <math>\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65</math> | + | <math>\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65</math> |
| + | ==Solution== | ||
| − | == | + | The answer to this problem is the number of intersections between the graph of <math>f(x) = \sin x</math> and <math>f(x) = \frac{1}{100}x.</math> We can do the right side of the coordinate plane first. Each cycle of the sine wave, consisting of 2π, will have 2 intersections (From the positive part of the sine wave) The line <math>f(x) = \frac{1}{100}x</math> will consist of 16 cycles plus a little bit extra for <math>x</math> from 1 to 100. However, the extra is not complete enough to have any intersection at all. Thus, the number of intersections is <math>2 \cdot 16 = 32.</math> Because of symmetry, we can multiply by two to account for the left side, and subtract one because of the origin. So the answer is <math>32 \cdot 2 - 1 = \textbf{(C)}\ 63.</math> |
| − | + | https://www.desmos.com/calculator/z6edqwu1kx - Graph | |
~Eric X | ~Eric X | ||
| + | |||
| + | ==See also== | ||
| + | |||
| + | {{AHSME box|year=1981|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:12, 28 June 2025
Problem 18
The number of real solutions to the equation ![\[\dfrac{x}{100}=\sin x\]](http://latex.artofproblemsolving.com/f/f/d/ffd7dd7000b0ad3abb2f5ebbaf6410580abd6bc6.png)
is
Solution
The answer to this problem is the number of intersections between the graph of 
and 
We can do the right side of the coordinate plane first. Each cycle of the sine wave, consisting of 2π, will have 2 intersections (From the positive part of the sine wave) The line 
will consist of 16 cycles plus a little bit extra for 
from 1 to 100. However, the extra is not complete enough to have any intersection at all. Thus, the number of intersections is 
Because of symmetry, we can multiply by two to account for the left side, and subtract one because of the origin. So the answer is 
https://www.desmos.com/calculator/z6edqwu1kx - Graph
~Eric X
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
