Difference between revisions of "2009 AMC 12B Problems/Problem 24"
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<math>\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7</math> | <math>\textbf{(A)}\ 3\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 7</math> | ||
− | == Solution == | + | == Solution 1 == |
First of all, we have to agree on the range of <math>\sin^{-1}</math> and <math>\cos^{-1}</math>. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: <math>\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2</math> and <math>\forall x: 0\leq \cos^{-1}(x) \leq \pi</math>. | First of all, we have to agree on the range of <math>\sin^{-1}</math> and <math>\cos^{-1}</math>. This should have been a part of the problem statement -- but as it is missing, we will assume the most common definition: <math>\forall x: -\pi/2 \leq \sin^{-1}(x) \leq \pi/2</math> and <math>\forall x: 0\leq \cos^{-1}(x) \leq \pi</math>. | ||
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<math>f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0</math>. | <math>f(x) = \sin^{-1}(\sin 5\pi/2) - 5\pi/12 = \pi/2 - 5\pi/12 = \pi/12 > 0</math>. | ||
− | Summary: The function <math>f</math> has <math>\boxed{4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>. | + | Summary: The function <math>f</math> has <math>\boxed{\textbf{(B) }4}</math> roots on <math>[0,\pi]</math>: the first one is <math>0</math>, the second one is in <math>(0,\pi/6)</math>, and the last two are in <math>(2\pi/6,3\pi/6)</math>. |
− | + | Actual solutions are <math>x=0</math>, <math>x=\pi/7</math>, <math>x=2\pi/5</math>, and <math>x=3\pi/7</math>. | |
− | |||
− | |||
− | + | ==Solution 2== | |
+ | Since <math>0\leq \cos^{-1}(a) \leq \pi</math> for all <math>a</math>, the equation reduces to <math>\sin^{-1}(\sin(6x)) = x</math>. Since <math>-\pi/2 \leq \sin^{-1}(a) \leq \pi/2</math> for all <math>a</math>, <math>0 \leq x \leq \pi/2</math>. To make the problem easier, we will measure angles in degrees. We will consider each sixth of the interval <math>[0, 90]</math>. | ||
+ | |||
+ | For <math>0 \leq x \leq 15</math>, <math>6x</math> is in the first quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 6x</math>. Setting this equal to <math>x</math> yields the solution <math>x = 0</math>. | ||
+ | |||
+ | For <math>15 \leq x \leq 30</math>, <math>6x</math> is in the second quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 180 - 6x</math>. This yields the solution <math>x = \frac{180}7</math>. | ||
+ | |||
+ | For <math>30 \leq x \leq 45</math>, <math>6x</math> is in the third quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 180 - 6x</math>. As <math>\frac{180}{7}</math> is not on the interval <math>30 \leq x \leq 45</math>, this yields no solution. | ||
+ | |||
+ | For <math>45 \leq x \leq 60</math>, <math>6x</math> is in the fourth quadrant. Thus, <math>\sin^{-1}(\sin(6x)) = 6x - 360</math>. As <math>72</math> is not on the interval <math>45 \leq x \leq 60</math>, this yields no solution. | ||
+ | |||
+ | For <math>60 \leq x \leq 75</math>, <math>6x</math> is in the first quadrant plus a full revolution. Thus, <math>\sin^{-1}(\sin(6x)) = 6x - 360</math>. This yields the solution <math>x = 72</math>. | ||
+ | |||
+ | For <math>75 \leq x \leq 90</math>, <math>6x</math> is in the second quadrant plus a full revolution. Thus <math>\sin^{-1}(\sin(6x)) = 540 - 6x</math>. This yields the solution <math>x = \frac{540}7</math>. | ||
+ | |||
+ | There are <math>\boxed{\textbf{(B) }4}</math> solutions, <math>x=0</math>, <math>x=\pi/7</math>, <math>x=2\pi/5</math>, and <math>x=3\pi/7</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Algebraically, the inverse function of a function should just cancel out, leaving <math>6x=x</math>. 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 <math>2\pi</math>. But instead of a smooth wave, these graphs are made up of zigzags with slope <math>1</math> and <math>-1</math>. Trying a few values, we see that | ||
+ | |||
+ | |||
+ | <math>y = \arcsin{\sin{x}}</math> | ||
+ | |||
+ | |||
+ | peaks at <math>\frac{\pi}{2}</math> and | ||
+ | |||
+ | |||
+ | <math>y = \arccos{\cos{x}}</math> | ||
+ | |||
+ | |||
+ | peaks at <math>\pi</math> | ||
+ | |||
+ | |||
+ | But we want the graph of <math>y = \arcsin{(\sin6{x})}</math>, which has a period of <math>\frac{\pi}{3}</math> instead of <math>2\pi</math>. So this means the interval <math>[0, \pi]</math> will show <math>3</math> periods instead of <math>\frac{1}{2}</math> of a period. Visually it would be lines with slopes <math>6</math> and <math>-6</math>. Using the graph paper given to us, we plot out the two equations according to the above and we see that they intersect <math>4</math> times <math>\Rightarrow \boxed{\text{B}}</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | The most conventional domain and range for <math>\cos^{-1} x</math> is <math>x \in [-1, 1]</math> and <math>\cos^{-1} x \in [0, \pi]</math>. For <math>\cos x</math>, <math>x \in [0, \pi]</math> and <math>\cos x \in [-1, 1]</math>. Since the domain of <math>\cos x</math> is equal to the range of <math>\cos^{-1} x</math>, and the range of <math>\cos x</math> is equal to the domain of <math>\cos^{-1} x</math>, <math>\cos^{-1} (\cos x) = x</math>. | ||
+ | |||
+ | Meaning that <math>\sin^{ - 1}(\sin 6x) = x</math>, <math>\sin x = \sin (6x)</math>. | ||
+ | |||
+ | By graphing <math>\sin x</math> and <math>\sin 6x</math> together, it can be seen that there are 7 intersections for <math>x \in [0, \pi]</math>. | ||
+ | |||
+ | [[File:2009AMC12BP24.png|500px|center]] | ||
+ | |||
+ | However, the range of <math>\sin^{-1} x</math> is <math>[-\frac{\pi}{2}, \frac{\pi}{2}]</math>. Therefore <math>x \in [0, \frac{\pi}{2}]</math> for <math>x = \sin^{ - 1}(\sin 6x)</math>. There are <math>\boxed{\textbf{(B) }4}</math> solutions for <math>x \in [0, \frac{\pi}{2}]</math>. | ||
+ | |||
+ | The <math>4</math> solutions are respectively: | ||
+ | |||
+ | <math>x = 6x</math>, <math>x = 0</math> | ||
+ | |||
+ | <math>x + 2 \pi = 6x</math>, <math>x = \frac25 \pi</math> | ||
+ | |||
+ | <math>\pi - x = 6x</math>, <math>x = \frac{\pi}{7}</math> | ||
+ | |||
+ | <math>3 \pi - x = 6x</math>, <math>x = \frac37 \pi</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== See Also == | == See Also == |
Latest revision as of 07:09, 22 December 2022
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 |
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.