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

## 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