新高考数学一轮复习考点练习考点01 集合 (含解析)

考点01集合考向一集合的含义与基本关系1.元素与集合个关系SKIPIF1<0则SKIPIF1<0,SKIPIF1<02.集合间的基本关系SKIPIF1<0，SKIPIF1<0可知SKIPIF1<0或SKIPIF1<0，则SKIPIF1<0是SKIPIF1<0的子集；A⊆B，∃x0∈B，x0∉A，则SKIPIF1<0是SKIPIF1<0的真子集；SKIPIF1<0且SKIPIF1<0，则SKIPIF1<0.1.已知集合SKIPIF1<0，SKIPIF1<0，则A∩B中元素的个数为A．2 B．3C．4 D．5【答案】B【解析】由题意，SKIPIF1<0，故SKIPIF1<0中元素的个数为3.故选B.2.已知集合SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0中元素的个数为A．2 B．3C．4 D．6【答案】C【解析】由题意，SKIPIF1<0中的元素满足SKIPIF1<0，且SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以满足SKIPIF1<0的有SKIPIF1<0，故SKIPIF1<0中元素的个数为4.故选C．考向二集合的基本运算集合的基本运算SKIPIF1<0SKIPIF1<0SKIPIF1<01.已知集合SKIPIF1<0则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0，又因为SKIPIF1<0，所以SKIPIF1<0，故选D．2.设全集SKIPIF1<0，集合SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由题意结合补集的定义可知SKIPIF1<0，则SKIPIF1<0.故选C．3.已知集合U={−2，−1，0，1，2，3}，A={−1，0，1}，B={1，2}，则SKIPIF1<0A．{−2，3} B．{−2，2，3}C．{−2，−1，0，3} D．{−2，−1，0，2，3}【答案】A【解析】由题意可得SKIPIF1<0，则SKIPIF1<0.故选A考向三集合与不等式集合的运算中，有时牵扯到不等式的求解，相对来说比较简单。集合SKIPIF1<0与集合SKIPIF1<0集合SKIPIF1<0SKIPIF1<0再求解集合的并集或者交集等等，就简单多了.1.设集合A={x|x2–4≤0}，B={x|2x+a≤0}，且A∩B={x|–2≤x≤1}，则a=A．–4 B．–2C．2 D．4【答案】B【解析】求解二次不等式SKIPIF1<0可得SKIPIF1<0，求解一次不等式SKIPIF1<0可得SKIPIF1<0.由于SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0.故选B．2.设SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由题得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故选B．题组一（真题在线）1.已知集合SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02.已知集合P=SKIPIF1<0，Q=SKIPIF1<0，则PSKIPIF1<0Q=A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03.已知集合SKIPIF1<0，则SKIPIF1<0_____.4.已知集合A={x||x|<3，x∈Z}，B={x||x|>1，x∈Z}，则A∩B=A．SKIPIF1<0 B．{–3，–2，2，3）C．{–2，0，2} D．{–2，2}5.已知集合A={x|x-1≥0},B={0,1,2},则A∩B= ()A．{0} B．{1}C．{1,2} D．{0,1,2}6.设全集为R,集合A={x|0<x<2},B={x|x≥1},则A∩(∁RB)= ()A.{x|0<x≤1} B.{x|0<x<1}C.{x|1≤x<2} D.{x|0<x<2}7.已知集合A={1,2,3},B={2,3},则 ()A．A=B B．A∩B=⌀C．A⫋B D．B⫋A题组二1.设集合SKIPIF1<0,集合SKIPIF1<0,则SKIPIF1<0等于A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02.设集合SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03.已知集合A＝{x|﹣1＜x＜5}，B＝{1，3，5}，则A∩B=A．{1，3} B．{1，3，5}C．{1，2，3，4} D．{0，1，2，3，4，5}4.已知集合SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05.已知集合SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<06.设SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<07.已知集合SKIPIF1<0，集合SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<08.SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09.已知集合SKIPIF1<0，6SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<010.已知集合SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0题组一1.D【解析】SKIPIF1<0，故选D．2.B【解析】SKIPIF1<0.故选B.3.SKIPIF1<0【解析】∵SKIPIF1<0,SKIPIF1<0，∴SKIPIF1<0.故答案为SKIPIF1<0.4.D【解析】因为SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0.故选D．5.C【解析】A={x|x≥1},B={0,1,2},所以A∩B={1,2}.【名师点睛】冠词的运用要把握a一般表示泛指某一类人或物，意为＂任何一个，任一＂。the表示表示特指，或有某类型的修饰。6.B【解析】∁RB={x|x<1},所以A∩(∁RB)={x|0<x<1}.故选B.7.D【解析】由子集的概念知B⫋A,故选D.题组二1.C【解析】集合SKIPIF1<0,集合SKIPIF1<0SKIPIF1<0,则SKIPIF1<0,SKIPIF1<0．故选C．2.D【解析】因为SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.故选D．3.A【解析】∵A＝{x|﹣1＜x＜5}，B＝{1，3，5}，∴A∩B＝{1，3}.故选A．4.B【解析】SKIPIF1<0，∴SKIPIF1<0．故选：B．5.B【解析】因为SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0或SKIPIF1<0.若SKIPIF1<0，则SKIPIF1<0,满足SKIPIF1<0.若SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0