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Difference between revisions of "1981 AHSME Problems/Problem 18"

(Created page with "==Problem:== 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\te...")
 
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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
  
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<math>\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65</math>  
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<math>\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65</math>
  
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==Solution:==
  
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==Solution:==
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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>
  
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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>
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https://www.desmos.com/calculator/z6edqwu1kx - Graph
  
  
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~Littlefox_AMC (Eric X)
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~Eric X

Latest revision as of 14:01, 10 November 2022

Problem:

The number of real solutions to the equation \[\dfrac{x}{100}=\sin x\] is

$\textbf{(A)}\ 61\qquad\textbf{(B)}\ 62\qquad\textbf{(C)}\ 63\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 65$

Solution:

The answer to this problem is the number of intersections between the graph of $f(x) = \sin x$ and $f(x) = \frac{1}{100}x.$ 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 $f(x) = \frac{1}{100}x$ 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 $2 \cdot 16 = 32.$ 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 $32 \cdot 2 - 1 = \textbf{(C)}\ 63.$

https://www.desmos.com/calculator/z6edqwu1kx - Graph


~Eric X

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