AMC12 2023 B

AMC12 2023 B · Q22

AMC12 2023 B · Q22. It mainly tests Functions basics, Functional equations (rare).

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 $f(1)?$

一个实值函数 $f$ 具有如下性质：对所有实数 $a$ 和 $b$，\[f(a + b) + f(a - b) = 2f(a) f(b).\] 以下哪一个不可能是 $f(1)$ 的值？

(A) 0 0

(B) 1 1

(D) 2 2

(E) -2 -2

Answer

Correct choice: (E)

正确答案：（E）

Solution

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.

代入 $a = b$ 得到 \[f(2a) + f(0) = 2f(a)^2\] 代入 $a= 0$ 得到 \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\] 这给出 \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\] 代入 $a = \frac{1}{2}$ 得出 $f(1) \geq -1$，因此答案选项 $\boxed{\textbf{(E) -2}}$ 是不可能的。

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