若集合M={-1,0,1},集合N={0,1,2},则M∪N等于( )A.{0,1}B.{-1,0,1}C.{0,1,2}D.{-1,0,1,2}
题型:单选题难度:简单来源:不详
若集合M={-1,0,1},集合N={0,1,2},则M∪N等于( )A.{0,1} | B.{-1,0,1} | C.{0,1,2} | D.{-1,0,1,2} |
|
答案
因为M={-1,0,1},N={0,1,2}, 所以M∪N={-1,0,1}∪{0,1,2}={-1,0,1,2}. 故答案为D. |
举一反三
若集合M={-1,0,1,2},N={x|(x+1)(x-2)<0,且x∈Z},则M∩N=( )A.{-1,0,1,2} | B.{0,1,2} | C.{-1,0,1} | D.{0,1} |
|
已知0<a<1,集合A={x题型:x|<1},B={x|logax>0},则A∩B为( )A.(-1,1) | B.(0,1) | C.(0,a) | D.φ |
|
难度:|
查看答案 设集合A={x|-1≤x≤2},B={x|0≤x≤4},则A∩B=( )A.[0,2] | B.[1,2] | C.[0,4] | D.[1,4] |
|
已知集合A={0,1,2,3,4},集合B={x|x=2n,n∈A},则A∩B=( )A.{0} | B.{0,4} | C.{2,4} | D.{0,2,4} |
|
若不等式x2-2x-3≤0的解集为M,函数f(x)=lg(2-|x|)的定义域为N,则M∩N为( )A.(-2,-1] | B.(-1,2) | C.[-1,2) | D.(-1,2] |
|