2008 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2008 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

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Concepts:function

Difficulty rating: 1620

12.

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

[1,1],[1,0][-1, 1], [-1, 0]

[1,1],[0,1][-1, 1], [0, 1]

[0,2],[1,0][0, 2], [-1, 0]

[1,3],[1,0][1, 3], [-1, 0]

[1,3],[0,1][1, 3], [0, 1]

Solution:

The value f(x+1)f(x + 1) is defined when 0x+12,0 \le x + 1 \le 2, that is, 1x1,-1 \le x \le 1, so the domain of gg is [1,1].[-1, 1].

As f(x+1)f(x + 1) ranges over [0,1],[0, 1], the value 1f(x+1)1 - f(x + 1) ranges over [0,1][0, 1] as well, so the range of gg is [0,1].[0, 1].

Thus, B is the correct answer.

Problem 12 in Other Years