Difference between revisions of "2003 AMC 12B Problems/Problem 23"

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==Solution==
 
==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>
+
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>
  
 
~Jske25
 
~Jske25

Latest revision as of 09:17, 1 September 2022

Problem

The number of $x$-intercepts on the graph of $y=\sin(1/x)$ in the interval $(0.0001,0.001)$ is closest to

$\mathrm{(A)}\ 2900 \qquad\mathrm{(B)}\ 3000 \qquad\mathrm{(C)}\ 3100 \qquad\mathrm{(D)}\ 3200 \qquad\mathrm{(E)}\ 3300$

Solution

The function $f(x) = \sin x$ has roots in the form of $\pi n$ for all integers $n$. Therefore, we want $\frac{1}{x} = \pi n$ on $\frac{1}{10000} \le x \le \frac{1}{1000}$, so $1000 \le \frac 1x = \pi n \le 10000$. There are $\frac{10000-1000}{\pi} \approx \boxed{2900} \Rightarrow \mathrm{(A)}$ solutions for $n$ on this interval.

Solution

We know that $x$ belongs to the interval $(0.0001,0.001)$ for $\sin(1/x)$. We see that when we plug in $x$ into $\sin(1/x)$, the argument $(1/x)$ is always from the range $(1000, 10000)$. Therefore, the problem simply asks for all the zeros of $\sin(x)$ with $x$ values between $(1000, 10000)$. We know that the $x$ values of any sine graph is $\pi(n-1)$ so, we see that values of $n$ are any integer value from $320$ to $3184$ and therefore gives us an answer of approximately $2865$ which is answer $\boxed{\text{(A)} 2900}$

~Jske25

See also

2003 AMC 12B (ProblemsAnswer KeyResources)
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

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