2009 AMC 12B Problems/Problem 24
Problem
For how many values of in is ? Note: The functions and denote inverse trigonometric functions.
Solution 1
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 .
Actual solutions are , , , and .
Solution 2
Since for all , the equation reduces to . Since for all , . To make the problem easier, we will measure angles in degrees. We will consider each sixth of the interval .
For , is in the first quadrant. Thus, . Setting this equal to yields the solution .
For , is in the second quadrant. Thus, . This yields the solution .
For , is in the third quadrant. Thus, . As is not on the interval , this yields no solution.
For , is in the fourth quadrant. Thus, . As is not on the interval , this yields no solution.
For , is in the first quadrant plus a full revolution. Thus, . This yields the solution .
For , is in the second quadrant plus a full revolution. Thus . This yields the solution .
There are solutions, , , , and .
Solution 3
Algebraically, the inverse function of a function should just cancel out, leaving . However, upon inspection we find that the graphs of these "inverse function of the function" equations are also periodic, like their normal trig function counterparts, due to the fact that inverse trig functions will never return any angle value higher than . But instead of a smooth wave, these graphs are made up of zigzags with slope and . Trying a few values, we see that
peaks at and
peaks at
But we want the graph of , which has a period of instead of . So this means the interval will show periods instead of of a period. Visually it would be lines with slopes and . Using the graph paper given to us, we plot out the two equations according to the above and we see that they intersect times .
Solution 4
The most conventional domain and range for is and . For , and . Since the domain of is equal to the range of , and the range of is equal to the domain of , .
Meaning that , .
By graphing and together, it can be seen that there are 7 intersections for .
However, the range of is . Therefore for . There are solutions for .
The solutions are respectively:
,
,
,
,
See Also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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