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Difference between revisions of "2023 AMC 12B Problems/Problem 22"
(→‎Solution 2)
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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  
−
\[
+
<math>[
−
f(0) + f(0) = 2 f(0) \cdot f(0) .
+
f(0) + f(0) = 2 f(0) \cdots f(0) .
−
\]
+
]</math>
  
 
Thus, <math>f(0) = 0</math> or 1.
 
Thus, <math>f(0) = 0</math> or 1.
Line 32: Line 32:
 
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>.
Line 43: Line 43:
 
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>
  
Line 57: Line 57:
  
 
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)

Revision as of 02:54, 16 November 2023

Problem

A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$

$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$

Solution 1

Substituting $a = b$ we get \[f(2a) + f(0) = 2f(a)^2\] Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\] This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\] Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$, so answer choice $\boxed{\textbf{(E) }}$ is impossible.

~AtharvNaphade

Solution 2

First, we set $a \leftarrow 0$ and $b \leftarrow 0$. Thus, the equation given in the problem becomes $[ f(0) + f(0) = 2 f(0) \cdots f(0) . ]$

Thus, $f(0) = 0$ or 1.

Case 1: $f(0) = 0$.

We set $b \leftarrow 0$. Thus, the equation given in the problem becomes $[ 2 f(a) = 0 . ]$

Thus, $f(a) = 0$ for all $a$.

Case 2: $f(0) = 1$.

We set $b \leftarrow a$. Thus, the equation given in the problem becomes \[[ f(2a) + 1 = 2 \left( f(a) \right)^2. ]\]

Thus, for any $a$, \begin{align*} f(2a) & = -1 + 2 \left( f(a) \right)^2 \\ & \geq -1 . \end{align*}

Therefore, an infeasible value of $f(1)$ 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, \[\cos(\alpha)\cos(\beta) = \frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))\] which is basically the relation. So we know that $f(x) = \cos(x)$ is a valid solution to the function. However, if we define $x=ay,$ where $a$ is arbitrary, the above relation should still hold for $f(x) = \cos(ay) = \cos(a(1))$ so any value in $[-1,1]$ can be reached, so choices $A,B,$ and $C$ are incorrect.

In addition, using the similar formula for hyperbolic cosine, we know \[\cosh(\alpha)\cosh(\beta) = \frac{1}{2}(\cosh(\alpha-\beta)+\cosh(\alpha+\beta))\] The range of $\cosh(ay)$ is $[1,\infty)$ so choice $D$ is incorrect.

Therefore, the remaining answer is choice $\boxed{(E)}.$

~kxiang

Video Solution 1 by OmegaLearn

https://youtu.be/UML6WmL0mNk


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

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