Difference between revisions of "2017 AMC 12A Problems/Problem 15"
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==Solution== | ==Solution== | ||
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: | ||
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Between 0 and 1, <math>f(x)</math> starts at <math>2</math> and increases; clearly there is no zero here. | Between 0 and 1, <math>f(x)</math> starts at <math>2</math> and increases; clearly there is no zero here. | ||
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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 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. | ||
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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. | ||
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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 3 and <math>\pi</math>, <math>f(x)</math> starts at some negative number and increases to -2; there is no zero here either. | ||
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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! | 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! | ||
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Therefore, the answer is <math>\boxed{(D)}</math>. | Therefore, the answer is <math>\boxed{(D)}</math>. | ||
Revision as of 18:51, 8 February 2017
Problem
Let 
, using radian measure for the variable 
. In what interval does the smallest positive value of for which 
lie?
Solution
We must first get an idea of what 
looks like:
Between 0 and 1, starts at 
and increases; clearly there is no zero here.
Between 1 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 3 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 
.
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. 
