Competition problem — Algebra — Maths Olympiad Prep

Library / /23 of 520

Algebra Difficulty 2.3 Junior Find the answer

Given the universal set $U=\mathbb{R}$ , the set $M=\{x|x^2-2x-3\leq0\}$ , and $N=\{y|y=x^2+1\}$ , then $M\cap(\complement_{U}N)$ is

Pick one

Solution

Since $M=\{x|(x-3)(x+1)\leq0\}=\{x|-1\leq x\leq3\}$ ,
and $N=\{y|y=x^2+1\}=\{y|y\geq1\}$ , therefore $\complement_{U}N=\{y|y<1\}$ ,
thus $M\cap(\complement_{U}N)=\{x|-1\leq x<1\}$ .
Hence, the correct option is: $\boxed{A}$ .

Analysis: First, simplify the set $M$ , then calculate $M\cap(\complement_{U}N)$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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