2023 AMC 12B Problems/Problem 22: Difference between revisions

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)

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2023 AMC 12B Problems/Problem 22: Difference between revisions

Revision as of 17:40, 15 November 2023

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 17: / Line 17: @@
 ~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
+\[
+f(0) + f(0) = 2 f(0) \cdot f(0) .
+\]
+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
+\[
+f(a) = 0 .
+\]
+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
+\boxed{\textbf{(E) -2}}.
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==See also==