AMC12 2008 A

AMC12 2008 A · Q12

AMC12 2008 A · Q12. It mainly tests Functions basics, Manipulating equations.

A function $f$ has domain $[0,2]$ and range $[0,1]$. (The notation $[a,b]$ denotes $\{x:a \le x \le b \}$.) What are the domain and range, respectively, of the function $g$ defined by $g(x)=1-f(x+1)$?

函数 $f$ 的定义域是 $[0,2]$，值域是 $[0,1]$。（记号 $[a,b]$ 表示 $\{x:a \le x \le b \}$。）由 $g(x)=1-f(x+1)$ 定义的函数 $g$ 的定义域和值域分别是什么？

(A) [-1, 1], [-1, 0] [-1, 1], [-1, 0]

(B) [-1, 1], [0, 1] [-1, 1], [0, 1]

(D) [1, 3], [-1, 0] [1, 3], [-1, 0]

(E) [1, 3], [0, 1] [1, 3], [0, 1]

Answer

Correct choice: (B)

正确答案：（B）

Solution

$g(x)$ is defined if $f(x + 1)$ is defined. Thus the domain is all $x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]$. Since $f(x + 1) \in [0,1]$, $- f(x + 1) \in [ - 1,0]$. Thus $g(x) = 1 - f(x + 1) \in [0,1]$ is the range of $g(x)$. Thus the answer is $[- 1,1],[0,1] \longrightarrow \boxed{B}$.

当且仅当 $f(x + 1)$ 有定义时，$g(x)$ 才有定义。因此定义域为所有满足 $x + 1 \in [0,2] \rightarrow x \in [ - 1,1]$ 的 $x$。由于 $f(x + 1) \in [0,1]$，所以 $- f(x + 1) \in [ - 1,0]$。因此 $g(x) = 1 - f(x + 1) \in [0,1]$，这就是 $g(x)$ 的值域。因此答案是 $[- 1,1],[0,1] \longrightarrow \boxed{B}$。

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