Difference between revisions of "2023 AMC 12B Problems/Problem 22"
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Substituting <math>a= 0</math> we find <cmath>2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.</cmath> | Substituting <math>a= 0</math> we find <cmath>2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.</cmath> | ||
This gives <cmath>f(2a) = 2f(a)^2 - f(0) \geq 0-1</cmath> | This gives <cmath>f(2a) = 2f(a)^2 - f(0) \geq 0-1</cmath> | ||
− | Plugging in <math>a = \frac{1}{2}</math> implies <math>f(1) \geq -1</math>, so answer choice <math>\boxed{\textbf{(E) }}</math> is impossible. | + | Plugging in <math>a = \frac{1}{2}</math> implies <math>f(1) \geq -1</math>, so answer choice <math>\boxed{\textbf{(E) -2}}</math> is impossible. |
~AtharvNaphade | ~AtharvNaphade | ||
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First, we set <math>a \leftarrow 0</math> and <math>b \leftarrow 0</math>. | First, we set <math>a \leftarrow 0</math> and <math>b \leftarrow 0</math>. | ||
Thus, the equation given in the problem becomes | Thus, the equation given in the problem becomes | ||
− | + | <math>[ | |
− | f(0) + f(0) = 2 f(0) \ | + | f(0) + f(0) = 2 f(0) \times f(0) . |
− | + | ]</math> | |
Thus, <math>f(0) = 0</math> or 1. | Thus, <math>f(0) = 0</math> or 1. | ||
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We set <math>b \leftarrow 0</math>. | We set <math>b \leftarrow 0</math>. | ||
Thus, the equation given in the problem becomes | Thus, the equation given in the problem becomes | ||
− | + | <math>[ | |
2 f(a) = 0 . | 2 f(a) = 0 . | ||
− | + | ]</math> | |
Thus, <math>f(a) = 0</math> for all <math>a</math>. | Thus, <math>f(a) = 0</math> for all <math>a</math>. | ||
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Thus, the equation given in the problem becomes | Thus, the equation given in the problem becomes | ||
<cmath> | <cmath> | ||
− | + | [ | |
f(2a) + 1 = 2 \left( f(a) \right)^2. | f(2a) + 1 = 2 \left( f(a) \right)^2. | ||
− | + | ] | |
</cmath> | </cmath> | ||
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Therefore, an infeasible value of <math>f(1)</math> is | Therefore, an infeasible value of <math>f(1)</math> is | ||
− | \boxed{\textbf{(E) -2}}. | + | <math>\boxed{\textbf{(E) -2}}.</math> |
− | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) kk choudhary |
==Solution 3== | ==Solution 3== | ||
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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{\textbf{(E) -2}}.</math> |
~kxiang | ~kxiang | ||
Line 77: | Line 77: | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== | ||
https://youtu.be/UML6WmL0mNk | https://youtu.be/UML6WmL0mNk | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | |||
+ | https://youtu.be/GK0ZpjxbwLw | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
Latest revision as of 08:33, 4 August 2024
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
Thus, or 1.
Case 1: .
We set . Thus, the equation given in the problem becomes
Thus, for all .
Case 2: .
We set . Thus, the equation given in the problem becomes
Thus, for any ,
Therefore, an infeasible value of is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) kk choudhary
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
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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.