Difference between revisions of "2009 AMC 12B Problems/Problem 24"
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== Better Solution == | == Better Solution == | ||
Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to <math>\sin^{ - 1}(\sin 6x) = x</math>. | Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to <math>\sin^{ - 1}(\sin 6x) = x</math>. | ||
− | Since <math>\sin^{-1}</math> is between <math>-\pi/2</math> and <math>\pi/2</math>, x is between that as well. Since x is between 0 and pi, combining the two gives x is between 0 and pi/2. Now, remove | + | Since <math>\sin^{-1}</math> is between <math>-\pi/2</math> and <math>\pi/2</math>, <math>x</math> is between that as well. Since <math>x</math> is between <math>0</math> and <math>\pi</math>, combining the two gives <math>x</math> is between <math>0</math> and <math>\pi/2</math>. Now, remove <math>\sin^{ - 1}</math> by taking the sine of both sides. Then, you get <math>sin x = sin 6x</math>. |
− | From this, either x = 6x + | + | From this, either <math>x = 6x +2\pik</math>, or <math>x+6x = \pi +2\pik</math>. Solving, and remembering that <math>0<=x<=\pi/2</math>, we get <math>4</math> solutions. |
== See Also == | == See Also == |
Revision as of 02:08, 4 February 2019
Contents
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 .
Better Solution
Like the previous solution, assume the inverse trig function properties and ranges and simplify the problem to . Since is between and , is between that as well. Since is between and , combining the two gives is between and . Now, remove by taking the sine of both sides. Then, you get .
From this, either $x = 6x +2\pik$ (Error compiling LaTeX. ! Undefined control sequence.), or $x+6x = \pi +2\pik$ (Error compiling LaTeX. ! Undefined control sequence.). Solving, and remembering that , we get solutions.
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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