Difference between revisions of "2020 AMC 12A Problems/Problem 9"

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To find the asymptotes of <math>\tan(2x)</math> we consider the behaviour of <math>\tan(x)</math> on <math>[0,4\pi]</math>. Then we see that there are five separate continuous parts of the graph splitting the plane into regions. Since <math>\cos(\frac{x}{2})</math> is continuous it must intersect each of the five pieces of <math>\tan</math> at least once. But since <math>\tan(2x)</math> is increasing and <math>\cos(\frac{x}{2})</math> is decreasing on the interval and continuous increasing functions and decreasing functions can intersect at most once, there are <math>\boxed{\textbf{E) }5}</math> intersections. -codecow
 
To find the asymptotes of <math>\tan(2x)</math> we consider the behaviour of <math>\tan(x)</math> on <math>[0,4\pi]</math>. Then we see that there are five separate continuous parts of the graph splitting the plane into regions. Since <math>\cos(\frac{x}{2})</math> is continuous it must intersect each of the five pieces of <math>\tan</math> at least once. But since <math>\tan(2x)</math> is increasing and <math>\cos(\frac{x}{2})</math> is decreasing on the interval and continuous increasing functions and decreasing functions can intersect at most once, there are <math>\boxed{\textbf{E) }5}</math> intersections. -codecow
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==Remark==
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The graphs of <math>f(x)=\tan(2x)</math> (in red) and <math>g(x)=\cos\left(\frac{x}{2}\right)</math> (in blue) are shown below.
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[[File:2020 AMC 12A Problem 9.png|center]]
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Graph in Desmos: https://www.desmos.com/calculator/gfplgzapww
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~MRENTHUSIASM
  
 
==See Also==
 
==See Also==

Latest revision as of 06:23, 17 April 2021

Problem

How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1

Draw a graph of $\tan(2x)$ and $\cos(\tfrac{x}{2})$

$\tan(2x)$ has a period of $\tfrac{\pi}{2},$ asymptotes at $x = \tfrac{\pi}{4}+\tfrac{k\pi}{2},$ and zeroes at $\tfrac{k\pi}{2}$. It is positive from $(0,\tfrac{\pi}{4}) \cup (\tfrac{\pi}{2},\tfrac{3\pi}{4}) \cup (\pi,\tfrac{5\pi}{4}) \cup (\tfrac{3\pi}{2},\tfrac{7\pi}{4})$ and negative elsewhere.

cos$(\tfrac{x}{2})$ has a period of $4\pi$ and zeroes at $\pi$. It is positive from $[0,\pi)$ and negative elsewhere.

Drawing such a graph would get $\boxed{\textbf{E) }5}$ ~lopkiloinm

Or you could see the points at which both graphs are positive or both are negative, again yielding 5 such areas. -hi13

edited by - annabelle0913

Solution 2

To find the asymptotes of $\tan(2x)$ we consider the behaviour of $\tan(x)$ on $[0,4\pi]$. Then we see that there are five separate continuous parts of the graph splitting the plane into regions. Since $\cos(\frac{x}{2})$ is continuous it must intersect each of the five pieces of $\tan$ at least once. But since $\tan(2x)$ is increasing and $\cos(\frac{x}{2})$ is decreasing on the interval and continuous increasing functions and decreasing functions can intersect at most once, there are $\boxed{\textbf{E) }5}$ intersections. -codecow

Remark

The graphs of $f(x)=\tan(2x)$ (in red) and $g(x)=\cos\left(\frac{x}{2}\right)$ (in blue) are shown below.

2020 AMC 12A Problem 9.png

Graph in Desmos: https://www.desmos.com/calculator/gfplgzapww

~MRENTHUSIASM

See Also

2020 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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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