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

Revision as of 19:56, 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)

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

@@ Line 64: / Line 64: @@
 The relationship looks suspiciously like a product-to-sum identity. In fact,
-<cmath>cos(\alpha)cos(\beta) = \frac{1}{2}(cos(\alpha-\beta)+cos(\alpha+\beta))</cmath>
+<cmath>\cos(\alpha)\cos(\beta) = \frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))</cmath>
-which is basically the relation. So we know that <math>f(x) = cos(x)</math> is a valid solution to the function. However, if we define <math>x=ay,</math> where <math>a</math> is arbitrary, the above relation should still hold for <math>f(x) = cos(ay) = cos(a(1))</math> so any value in <math>[-1,1]</math> can be reached, so choices <math>A,B,</math> and <math>C</math> are incorrect.
+which is basically the relation. So we know that <math>f(x) = \cos(x)</math> is a valid solution to the function. However, if we define <math>x=ay,</math> where <math>a</math> is arbitrary, the above relation should still hold for <math>f(x) = \cos(ay) = \cos(a(1))</math> so any value in <math>[-1,1]</math> can be reached, so choices <math>A,B,</math> and <math>C</math> are incorrect.
 In addition, using the similar formula for hyperbolic cosine, we know
-<cmath>cosh(\alpha)cosh(\beta) = \frac{1}{2}(cosh(\alpha-\beta)+cosh(\alpha+\beta))</cmath>
+<cmath>\cosh(\alpha)\cosh(\beta) = \frac{1}{2}(\cosh(\alpha-\beta)+\cosh(\alpha+\beta))</cmath>
-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>
 ~kxiang
 ==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
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