Difference between revisions of "2003 AMC 12B Problems/Problem 23"
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== Solution == | == Solution == | ||
The function <math>f(x) = \sin x</math> has roots in the form of <math>\pi n</math> for all integers <math>n</math>. Therefore, we want <math>\frac{1}{x} = \pi n</math> on <math>\frac{1}{10000} \le x \le \frac{1}{1000}</math>, so <math>1000 \le \frac 1x = \pi n \le 10000</math>. There are <math>\frac{10000-1000}{\pi} \approx \boxed{2900} \Rightarrow \mathrm{(A)}</math> solutions for <math>n</math> on this interval. | The function <math>f(x) = \sin x</math> has roots in the form of <math>\pi n</math> for all integers <math>n</math>. Therefore, we want <math>\frac{1}{x} = \pi n</math> on <math>\frac{1}{10000} \le x \le \frac{1}{1000}</math>, so <math>1000 \le \frac 1x = \pi n \le 10000</math>. There are <math>\frac{10000-1000}{\pi} \approx \boxed{2900} \Rightarrow \mathrm{(A)}</math> solutions for <math>n</math> on this interval. | ||
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| + | ==Solution== | ||
| + | We know that <math>x</math> belongs to the interval <math>(0.0001,0.001)</math> for <math>\sin(1/x)</math>. We see that when we plug in <math>x</math> into <math>\sin(1/x)</math>, the argument <math>(1/x)</math> is always from the range <math>(1000, 10000)</math>. Therefore, the problem simply asks for all the zeros of <math>\sin(x)</math> with <math>x</math> values between <math>(1000, 10000)</math>. We know that the <math>x</math> values of any sine graph is <math>\pi(n-1)</math> so, we see that values of <math>n</math> are any integer value from <math>320</math> to <math>3184</math> and therefore gives us an answer of approximately <math>2865</math> which is answer <math>\boxed{\text{(A)} 2900}</math> | ||
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| + | ~Jske25 | ||
== See also == | == See also == | ||
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[[Category:Introductory Trigonometry Problems]] | [[Category:Introductory Trigonometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:39, 9 February 2023
Contents
Problem
The number of 
-intercepts on the graph of 
in the interval 
is closest to
Solution
The function 
has roots in the form of 
for all integers 
. Therefore, we want 
on 
, so 
. There are 
solutions for on this interval.
Solution
We know that belongs to the interval for 
. We see that when we plug in into , the argument 
is always from the range 
. Therefore, the problem simply asks for all the zeros of 
with values between . We know that the values of any sine graph is 
so, we see that values of are any integer value from 
to 
and therefore gives us an answer of approximately 
which is answer 
~Jske25
See also
| 2003 AMC 12B (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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
