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

Latest revision as of 08:33, 4 August 2024

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.

~AtharvNaphade

Solution 2

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

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 1 by OmegaLearn

https://youtu.be/UML6WmL0mNk

Video Solution 2

https://youtu.be/GK0ZpjxbwLw

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

@@ Line 14: / Line 14: @@
 Substituting <math>a= 0</math> we find <cmath>2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.</cmath>
 This gives <cmath>f(2a) = 2f(a)^2 - f(0) \geq 0-1</cmath>
-Plugging in <math>a = \frac{1}{2}</math> implies <math>f(1) \geq -1</math>, so answer choice <math>\boxed{\textbf{(E) }}</math> is impossible.
+Plugging in <math>a = \frac{1}{2}</math> implies <math>f(1) \geq -1</math>, so answer choice <math>\boxed{\textbf{(E) -2}}</math> is impossible.
 ~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
+<math>[
+f(0) + f(0) = 2 f(0) \times f(0) .
+]</math>
+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
+<math>[
+f(a) = 0 .
+]</math>
+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
+<math>\boxed{\textbf{(E) -2}}.</math>
+~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,
+<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.
+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>
+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{\textbf{(E) -2}}.</math>
+~kxiang
+==Video Solution 1 by OmegaLearn==
+https://youtu.be/UML6WmL0mNk
+==Video Solution 2==
+https://youtu.be/GK0ZpjxbwLw
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==See also==
 {{AMC12 box|ab=B|year=2023|num-b=21|num-a=23}}
-[[Category:Functional equation]]
+[[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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