Difference between revisions of "2009 AMC 12B Problems/Problem 24"
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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{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>. | ||
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== See Also == | == See Also == |
Revision as of 16:57, 25 August 2019
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
.
how 'bout no
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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