Difference between revisions of "2023 AMC 12B Problems/Problem 22"
(→See also) |
|||
Line 17: | Line 17: | ||
~AtharvNaphade | ~AtharvNaphade | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | First, we set <math>a \leftarrow 0</math> and <math>b \leftarrow 0</math>. | ||
+ | Thus, the equation given in the problem becomes | ||
+ | \[ | ||
+ | f(0) + f(0) = 2 f(0) \cdot f(0) . | ||
+ | \] | ||
+ | |||
+ | Thus, <math>f(0) = 0</math> or 1. | ||
+ | |||
+ | Case 1: <math>f(0) = 0</math>. | ||
+ | |||
+ | We set <math>b \leftarrow 0</math>. | ||
+ | Thus, the equation given in the problem becomes | ||
+ | \[ | ||
+ | 2 f(a) = 0 . | ||
+ | \] | ||
+ | |||
+ | Thus, <math>f(a) = 0</math> for all <math>a</math>. | ||
+ | |||
+ | Case 2: <math>f(0) = 1</math>. | ||
+ | |||
+ | We set <math>b \leftarrow a</math>. | ||
+ | Thus, the equation given in the problem becomes | ||
+ | <cmath> | ||
+ | \[ | ||
+ | f(2a) + 1 = 2 \left( f(a) \right)^2. | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Thus, for any <math>a</math>, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | f(2a) & = -1 + 2 \left( f(a) \right)^2 \ | ||
+ | & \geq -1 . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, an infeasible value of <math>f(1)</math> is | ||
+ | \boxed{\textbf{(E) -2}}. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== |
Revision as of 17:40, 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)
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.