2009 AMC 12B Problems/Problem 24
Problem
For how many values of in is ? Note: The functions and denote inverse trigonometric functions.
Solution
First of all, we have to agree on the range of and . This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: and .
Hence we get that , thus our equation simplifies to .
Consider the function . We are looking for roots of on .
By analyzing properties of and (or by computing the derivative of ) one can discover the following properties of :
- .
- is increasing and then decreasing on .
- is decreasing and then increasing on .
- is increasing and then decreasing on .
For we have . Hence has exactly one root on .
For we have . Hence is negative on the entire interval .
Now note that . Hence for we have , and we can easily check that as well.
Thus the only unknown part of is the interval . On this interval, is negative in both endpoints, and we know that it is first increasing and then decreasing. Hence there can be zero, one, or two roots on this interval.
To prove that there are two roots, it is enough to find any from this interval such that .
A good guess is its midpoint, , where the function has its local maximum. We can evaluate: .
Summary: The function has roots on : the first one is , the second one is in , and the last two are in .
Alternate solution
Notice that the graph of the second function is just a straight line that goes up to at and starts at . So the second equation is really just the line on the interval. In addition, notice that the first function has a range of , so we only have to look at the first half of the domain -- for . So we now find values of x such that . Applying sin to both sides, this becomes . Thus, we find values on the domain such that is a reference angle of .
First, of course there is 0. Then, when is in first quadrant, can be in second quadrant, first quadrant (with a revolution), and similarly second quadrant. These are modeled by , , and . So we find answers . So there are four solutions, the answer is .
See Also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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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