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Difference between revisions of "2023 AMC 12B Problems/Problem 22"

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~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
 +
\[
 +
f(0) + f(0) = 2 f(0) \cdot f(0) .
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\]
 +
 +
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>.
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 +
We set <math>b \leftarrow a</math>.
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Thus, the equation given in the problem becomes
 +
<cmath>
 +
\[
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f(2a) + 1 = 2 \left( f(a) \right)^2.
 +
\]
 +
</cmath>
 +
 +
Thus, for any <math>a</math>,
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<cmath>
 +
\begin{align*}
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f(2a) & = -1 + 2 \left( f(a) \right)^2 \\
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& \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

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 \[ f(0) + f(0) = 2 f(0) \cdot f(0) . \]

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)

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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