Track / Stage 4 / 62 of 340 #322 of 1964

Problem 322

AMC 12 late, AIME early

Algebra Difficulty 4.6 Multiple choice

Given the sets $P=\left\{x \mid x^{2}=1\right\}$ and $Q=\{x \mid m x=1\}$ . If $Q \subset P$ , then the number of possible values for the real number $m$ is:

Pick one

Official solution

$-1 .(\mathrm{D})$ .
$\because P=\{-1,1\}, Q \subset P \text {, }$
$\therefore Q$ could be $\varnothing,\{-1\},\{1\}$ .
When $Q=\varnothing$ , $m=0$ ;
When $Q=\{-1\}$ , $m=-1$ ;
When $Q=\{1\}$ , $m=1$ .
Thus, $m$ has 3 values: $0,-1,1$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.