# 2023 AMC 12B Problems/Problem 22

## 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 $\boxed{\textbf{(E) -2}}$ is impossible.

## Solution 2

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

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) kk choudhary

## 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 $\boxed{\textbf{(E) -2}}.$

~kxiang

## Video Solution 2

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)