2003 AMC 12B Problem 23

Attempt Problem 23 of the 2003 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

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23.

The number of xx-intercepts on the graph of y=sin⁡(1x)y = \sin(\frac{1}{x}) in the interval (0.0001,0.001)(0.0001, 0.001) is closest to
29002900
30003000
31003100
32003200
33003300
Answer: A
Concepts:counting integers in a rangefloor and ceiling functionstrigonometry
Small hint:
sin⁡(1x)=0\sin(\frac{1}{x}) = 0 when 1x=kπ\frac{1}{x} = k\pi for a nonzero integer k,k, i.e. x=1kπx = \dfrac{1}{k\pi}
Big hint:
Count integers kk with 1000π<k<10000π\dfrac{1000}{\pi} \lt k \lt \dfrac{10000}{\pi}
Solution:
The intercepts occur where 1x=kπ,\frac{1}{x} = k\pi, that is x=1kπx = \dfrac{1}{k\pi} for a nonzero integer k.k.
The condition 0.0001<1kπ<0.0010.0001 \lt \dfrac{1}{k\pi} \lt 0.001 becomes 1000π<k<10000π. \frac{1000}{\pi} \lt k \lt \frac{10000}{\pi}.
The number of such integers is ⌊10000π⌋−⌊1000π⌋=3183−318=2865, \begin{aligned} &\left\lfloor \frac{10000}{\pi} \right\rfloor - \left\lfloor \frac{1000}{\pi} \right\rfloor \\ &= 3183 - 318 = 2865, \end{aligned} closest to 2900.2900.
Thus, the correct answer is A.
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