AMC 10 · 2003 · #23
Grade 11 algebracountingPick an answer.
Counting oscillations of sin(1/x) on a tiny interval near zero is hard because the wiggles are crushed together and unevenly spaced. Naming u = 1/x (Tool #4) trades that for a far easier related problem: counting zeros of the ordinary sine on a wide interval where the zeros are evenly spaced (Tool #9). The exchange is only legal if it does not lose or duplicate any intercept, so the map has to be checked to be one-to-one. Then the familiar pattern of sine's zeros — every multiple of π, forever (Tool #5) — turns the question into counting integers in a range, which is an endpoint-careful list (Tool #2). Finally, compare the exact count to the five choices (Tool #3), because the question asks only which is closest.
Say what an intercept means here
An intercept is a value where the output is zero, so the task is a counting problem.
The sine is zero exactly when the thing inside it is zero-worthy, so the whole question is about what values 1/x takes.
9.F-IF.A.2Introduce A VariableSubstitute u = 1/x
The reciprocal substitution is one-to-one and decreasing, turning the range into (1000, 10000).
Reciprocal is its own undo, so no zero is lost and none is counted twice — the interval just gets turned inside out.
11.F-BF.B.4Solve An Easier Related ProblemZeros of sine are the multiples of pi
Sine vanishes exactly at multiples of pi, evenly spaced, so we count integers.
Sine returns to zero every half turn around the unit circle, so its zeros march along in perfectly even steps of π.
Sine returns to zero every half turn, so its zeros march along in perfectly even steps.
▸ Why?
Going once round the circle brings the sine back to where it started, so its behaviour repeats for ever.
▸ Why?
Because the gap between neighbouring zeros never changes, counting them is counting the steps of a fixed-step list.
Pin the two endpoints
Checking all four neighbours by hand pins the range to 319 through 3183.
Multiples of π never land on a whole number, so no endpoint is a tie, but each one is close enough that guessing costs a count.
8.NS.A.2Make A Systematic ListCount and compare
That is 2865 intercepts, nearest to 2900, choice (A).
Counting a block of consecutive whole numbers needs the extra +1, because subtracting alone counts the gaps between them rather than the numbers.
6.NS.C.7Eliminate PossibilitiesTurning 1/x into its own variable straightens out crowded wiggles into evenly spaced ones, and then counting is just counting whole numbers between two bounds.
- Say what an intercept means here
- Substitute u = 1/x
- Zeros of sine are the multiples of pi
- Pin the two endpoints
- Count and compare