Difference between revisions of "1981 AHSME Problems/Problem 18"
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| − | The answer to this problem is the number of intersections between the graph of f(x) = | + | The answer to this problem is the number of intersections between the graph of f(x) = sin x and f(x) = (1/100)x. 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 f(x) = (1/100)x will consist of 16 cycles plus a little bit extra for x from 1 to 100. However, the extra is not complete enough to have any intersection at all. Thus, the number of intersections is 2 x 16 = 32. 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 32 x 2 - 1 = <math> \textbf{(E)}\ 63.</math> |
~Eric X | ~Eric X | ||
Revision as of 14:37, 27 July 2022
Problem:
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 f(x) = sin x and f(x) = (1/100)x. 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 f(x) = (1/100)x will consist of 16 cycles plus a little bit extra for x from 1 to 100. However, the extra is not complete enough to have any intersection at all. Thus, the number of intersections is 2 x 16 = 32. 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 32 x 2 - 1 = 
~Eric X
