2017 AMC 12A Problems/Problem 15
Contents
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 
.
Solution 2
If you quickly take a moment to sketch the graphs of the three functions, you will see that between 0 and pi/2 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 pi, which is around 3.14. Hence the answer is .
-note: solution makes more sense if you just sketch it
Solution by IronicNinja~
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 | |
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