Difference between revisions of "2023 AMC 12B Problems/Problem 22"
(Solution 3) |
Scrabbler94 (talk | contribs) m (→Solution 3: use \cos and \cosh instead of cos and cosh) |
||
Line 64: | Line 64: | ||
The relationship looks suspiciously like a product-to-sum identity. In fact, | The relationship looks suspiciously like a product-to-sum identity. In fact, | ||
− | <cmath>cos(\alpha)cos(\beta) = \frac{1}{2}(cos(\alpha-\beta)+cos(\alpha+\beta))</cmath> | + | <cmath>\cos(\alpha)\cos(\beta) = \frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))</cmath> |
− | which is basically the relation. So we know that <math>f(x) = cos(x)</math> is a valid solution to the function. However, if we define <math>x=ay,</math> where <math>a</math> is arbitrary, the above relation should still hold for <math>f(x) = cos(ay) = cos(a(1))</math> so any value in <math>[-1,1]</math> can be reached, so choices <math>A,B,</math> and <math>C</math> are incorrect. | + | which is basically the relation. So we know that <math>f(x) = \cos(x)</math> is a valid solution to the function. However, if we define <math>x=ay,</math> where <math>a</math> is arbitrary, the above relation should still hold for <math>f(x) = \cos(ay) = \cos(a(1))</math> so any value in <math>[-1,1]</math> can be reached, so choices <math>A,B,</math> and <math>C</math> are incorrect. |
In addition, using the similar formula for hyperbolic cosine, we know | In addition, using the similar formula for hyperbolic cosine, we know | ||
− | <cmath>cosh(\alpha)cosh(\beta) = \frac{1}{2}(cosh(\alpha-\beta)+cosh(\alpha+\beta))</cmath> | + | <cmath>\cosh(\alpha)\cosh(\beta) = \frac{1}{2}(\cosh(\alpha-\beta)+\cosh(\alpha+\beta))</cmath> |
− | The range of <math>cosh(ay)</math> is <math>[1,\infty)</math> so choice <math>D</math> is incorrect. | + | The range of <math>\cosh(ay)</math> is <math>[1,\infty)</math> so choice <math>D</math> is incorrect. |
Therefore, the remaining answer is choice <math>\boxed{(E)}.</math> | Therefore, the remaining answer is choice <math>\boxed{(E)}.</math> | ||
− | ~kxiang | + | ~kxiang |
==See also== | ==See also== |
Revision as of 19:56, 15 November 2023
Contents
[hide]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
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.