AMC12 2006 A

AMC12 2006 A · Q18

AMC12 2006 A · Q18. It mainly tests Functions basics, Manipulating equations.

The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and $f(x)+f\left(\frac{1}{x}\right)=x$ What is the largest set of real numbers that can be in the domain of $f$?

函数 $f$ 具有如下性质：对其定义域中的每个实数 $x$，$1/x$ 也在其定义域中，且 $f(x)+f\left(\frac{1}{x}\right)=x$ $f$ 的定义域中可能包含的最大实数集合是什么？

(A) {x \mid x \neq 0} {x \mid x \neq 0}

(B) {x \mid x < 0} {x \mid x < 0}

(D) {x \mid x \neq -1 \text{ and } x \neq 0 \text{ and } x \neq 1} {x \mid x \neq -1 \text{ 且 } x \neq 0 \text{ 且 } x \neq 1}

(E) {-1, 1} {-1, 1}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Quickly verifying by plugging in values verifies that $-1$ and $1$ are in the domain. $f(x)+f\left(\frac{1}{x}\right)=x$ Plugging in $\frac{1}{x}$ into the function: $f\left(\frac{1}{x}\right)+f\left(\frac{1}{\frac{1}{x}}\right)=\frac{1}{x}$ $f\left(\frac{1}{x}\right)+ f(x)= \frac{1}{x}$ Since $f(x) + f\left(\frac{1}{x}\right)$ cannot have two values: $x = \frac{1}{x}$ $x^2 = 1$ $x=\pm 1$ Therefore, the largest set of real numbers that can be in the domain of $f$ is $\{-1,1\} \Rightarrow E$

快速代入一些值可验证 $-1$ 和 $1$ 在定义域中。 $f(x)+f\left(\frac{1}{x}\right)=x$ 将 $\frac{1}{x}$ 代入： $f\left(\frac{1}{x}\right)+f\left(\frac{1}{\frac{1}{x}}\right)=\frac{1}{x}$ $f\left(\frac{1}{x}\right)+ f(x)= \frac{1}{x}$ 由于 $f(x) + f\left(\frac{1}{x}\right)$ 不可能同时取两个值： $x = \frac{1}{x}$ $x^2 = 1$ $x=\pm 1$ 因此，$f$ 的定义域中可能包含的最大实数集合是 $\{-1,1\} \Rightarrow E$。

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