题目
题型:单选题难度:一般来源:泰安二模
A.1 | B.-1 | C.1或-1 | D.0或1或-1 |
答案
M是x-a=0的解集,而x-a=0⇒x=a;
故M={a},
若M∩N=N,则N⊆M,
①N=∅,则a=0;
②N≠∅,则有N={
1 |
a |
必有
1 |
a |
解可得,a=±1;
综合可得,a=0,1,-1;
故选D.
核心考点
举一反三
A.M∪N | B.M∩N | C.(CuM)∪(CuN) | D.(CuM)∩(CuN) |
A.1 | B.-1 | C.1或-1 | D.0或1或-1 |
1 |
a |
1 |
a |
A.M∪N | B.M∩N | C.(CuM)∪(CuN) | D.(CuM)∩(CuN) |