Difference between revisions of "2017 AMC 12A Problems/Problem 15"
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\qquad\textbf{(E)}\ (4,5) </math> | \qquad\textbf{(E)}\ (4,5) </math> | ||
− | ==Solution== | + | ==Solution 1== |
We must first get an idea of what <math>f(x)</math> looks like: | We must first get an idea of what <math>f(x)</math> looks like: | ||
− | Between 0 and 1, <math>f(x)</math> starts at <math>2</math> and increases; clearly there is no zero here. | + | Between <math>0</math> and <math>1</math>, <math>f(x)</math> starts at <math>2</math> and increases; clearly there is no zero here. |
− | Between 1 and <math>\frac{\pi}{2}</math>, <math>f(x)</math> starts at a positive number and increases to <math>\infty</math>; there is no zero here either. | + | Between <math>1</math> and <math>\frac{\pi}{2}</math>, <math>f(x)</math> starts at a positive number and increases to <math>\infty</math>; there is no zero here either. |
Between <math>\frac{\pi}{2}</math> and 3, <math>f(x)</math> starts at <math>-\infty</math> and increases to some negative number; there is no zero here either. | Between <math>\frac{\pi}{2}</math> and 3, <math>f(x)</math> starts at <math>-\infty</math> and increases to some negative number; there is no zero here either. | ||
− | Between 3 and <math>\pi</math>, <math>f(x)</math> starts at some negative number and increases to -2; there is no zero here either. | + | Between <math>3</math> and <math>\pi</math>, <math>f(x)</math> starts at some negative number and increases to -2; there is no zero here either. |
Between <math>\pi</math> and <math>\pi+\frac{\pi}{4} < 4</math>, <math>f(x)</math> starts at -2 and increases to <math>-\frac{\sqrt2}{2} + 2\left(-\frac{\sqrt2}{2}\right) + 3\left(1\right)=3\left(1-\frac{\sqrt2}{2}\right)>0</math>. There is a zero here by the Intermediate Value Theorem. Therefore, the answer is <math>\boxed{\textbf{(D)}}</math>. | Between <math>\pi</math> and <math>\pi+\frac{\pi}{4} < 4</math>, <math>f(x)</math> starts at -2 and increases to <math>-\frac{\sqrt2}{2} + 2\left(-\frac{\sqrt2}{2}\right) + 3\left(1\right)=3\left(1-\frac{\sqrt2}{2}\right)>0</math>. There is a zero here by the Intermediate Value Theorem. Therefore, the answer is <math>\boxed{\textbf{(D)}}</math>. | ||
− | ==Solution 2== | + | ==Solution 2 (Graphing)== |
− | If you quickly take a moment to sketch the graphs of the three functions, you will see that between 0 and pi/ | + | If you quickly take a moment to sketch the graphs of the three functions, you will see that between <math>0</math> and <math>\frac{\pi}{2}</math> everything is positive, while the positive number created by the sin does not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after <math>\pi</math>, which is around <math>3.14</math>. Hence the answer is <math>\boxed{\textbf{(D)}}</math>. |
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Solution by roadchicken~ | Solution by roadchicken~ | ||
− | ( | + | (Not original author) Here is the graph: |
<asy> | <asy> | ||
Label f; | Label f; |
Revision as of 14:26, 19 December 2020
Problem
Let , using radian measure for the variable . In what interval does the smallest positive value of for which lie?
Solution 1
We must first get an idea of what looks like:
Between and , starts at and increases; clearly there is no zero here.
Between and , starts at a positive number and increases to ; there is no zero here either.
Between and 3, starts at and increases to some negative number; there is no zero here either.
Between and , starts at some negative number and increases to -2; there is no zero here either.
Between and , starts at -2 and increases to . There is a zero here by the Intermediate Value Theorem. Therefore, the answer is .
Solution 2 (Graphing)
If you quickly take a moment to sketch the graphs of the three functions, you will see that between and everything is positive, while the positive number created by the sin does not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after , which is around . Hence the answer is .
Solution by roadchicken~
(Not original author) Here is the graph:
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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.