Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.1 AMC 10/12 Find the answer

Given the sets $A=\{x|a-1<x<a+1\}$ , $B=\{x|0<x<3\}$ .
1. If $a=0$ , find $A \cap B$ ;
2. If $A \subseteq B$ , find the range of the real number $a$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

1. If $a=0$ , then the set $A=\{x|a-1<x<a+1\}=\{x|-1<x<1\}$ , and $B=\{x|0<x<3\}$ .
Therefore, $A \cap B = \{x|-1<x<1\} \cap \{x|0<x<3\} = \{x|0<x<1\}$ ;
2. If $A \subseteq B$ , then $\begin{cases} a-1\geq 0 \\ a-1\leq 3 \end{cases}$ , which means $1 \leq a \leq 2$ ,
Thus, the range of the real number $a$ is $1 \leq a \leq 2$ .

Therefore, the answers are:
1. $A \cap B = \boxed{\{x|0<x<1\}}$ ;
2. The range of $a$ is $\boxed{1 \leq a \leq 2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.