Difference between revisions of "2020 AMC 12A Problems/Problem 9"
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Draw a graph of | + | Draw a graph of <math>\tan(2x)</math> and <math>\cos(\tfrac{x}{2})</math> |
− | + | <math>\tan(2x)</math> has a period of <math>\tfrac{\pi}{2},</math> asymptotes at <math>x = \tfrac{\pi}{4}+\tfrac{k\pi}{2},</math> and zeroes at <math>\tfrac{k\pi}{2}</math>. It is positive from <math>(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})</math> and negative elsewhere. | |
− | cos<math>(\ | + | cos<math>(\tfrac{x}{2})</math> has a period of <math>4\pi</math> and zeroes at <math>\pi</math>. It is positive from <math>[0,\pi)</math> and negative elsewhere. |
Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~lopkiloinm | Drawing such a graph would get <math>\boxed{\textbf{E) }5}</math> ~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 <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 | |
− | + | ==Remark== | |
+ | 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. | ||
− | + | [[File:2020 AMC 12A Problem 9.png|center]] | |
− | + | Graph in Desmos: https://www.desmos.com/calculator/gfplgzapww | |
− | + | ~MRENTHUSIASM | |
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2020|ab=A|num-b=8|num-a=10}} | {{AMC12 box|year=2020|ab=A|num-b=8|num-a=10}} | ||
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+ | [[Category:Introductory Trigonometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 06:23, 17 April 2021
Contents
Problem
How many solutions does the equation have on the interval
Solution 1
Draw a graph of and
has a period of asymptotes at and zeroes at . It is positive from and negative elsewhere.
cos has a period of and zeroes at . It is positive from and negative elsewhere.
Drawing such a graph would get ~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 we consider the behaviour of on . Then we see that there are five separate continuous parts of the graph splitting the plane into regions. Since is continuous it must intersect each of the five pieces of at least once. But since is increasing and is decreasing on the interval and continuous increasing functions and decreasing functions can intersect at most once, there are intersections. -codecow
Remark
The graphs of (in red) and (in blue) are shown below.
Graph in Desmos: https://www.desmos.com/calculator/gfplgzapww
~MRENTHUSIASM
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.