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
.
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
.
and
peak at
and
, respectively.
But we want the graph of , which has a period of
instead of $\2\pi$ (Error compiling LaTeX. Unknown error_msg). 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
.
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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