Difference between revisions of "2023 AMC 12B Problems/Problem 22"
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+ | ==Video Solution 1 by OmegaLearn== | ||
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==See also== | ==See also== |
Revision as of 03:15, 16 November 2023
Contents
Problem
A real-valued function has the property that for all real numbers
and
Which one of the following cannot be the value of
Solution 1
Substituting we get
Substituting
we find
This gives
Plugging in
implies
, so answer choice
is impossible.
~AtharvNaphade
Solution 2
First, we set and
.
Thus, the equation given in the problem becomes
\[
f(0) + f(0) = 2 f(0) \cdot f(0) .
\]
Thus, or 1.
Case 1: .
We set .
Thus, the equation given in the problem becomes
\[
2 f(a) = 0 .
\]
Thus, for all
.
Case 2: .
We set .
Thus, the equation given in the problem becomes
Thus, for any ,
Therefore, an infeasible value of is
\boxed{\textbf{(E) -2}}.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3
The relationship looks suspiciously like a product-to-sum identity. In fact,
which is basically the relation. So we know that
is a valid solution to the function. However, if we define
where
is arbitrary, the above relation should still hold for
so any value in
can be reached, so choices
and
are incorrect.
In addition, using the similar formula for hyperbolic cosine, we know
The range of
is
so choice
is incorrect.
Therefore, the remaining answer is choice
~kxiang
Video Solution 1 by OmegaLearn
See also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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.