Question

设集合A=[x kπ+π/4≤x≤kπ+π/2,π∈Z},集合B={x 6+x-x2≥0},求A∩B,

Answer 1

A∩B = [-2,-π/2] ∪ [π/4,π/2].

根据题目可知,集合B={ x | -2≤x≤3} = [-2,3]

当k=-1时：集合A= [ -0.75π,-0.5π] .注意到 -0.75π < -2 ,-0.5π > -2 ,∴[ -0.75π,-0.5π]部分包含于[-2,3]中

当k=0时：集合A= [ 0.25π,0.5π] .注意到0.5π3.∴[ (5/4)π,(3/2)π]跟[-2,3]没有交集.

( k的其他值不需要讨论,因为明显跟B没有公共部分.)

综上,A∩B = [-2,-π/2] ∪ [π/4,π/2].

明白了吗~