新高考數(shù)學(xué)一輪復(fù)習(xí)提升練習(xí)考點(diǎn)01 集合 (含解析)

考向01集合（2021年全國(guó)新高考Ⅰ卷數(shù)學(xué)試題）設(shè)集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用交集的定義可求SKIPIF1<0.【詳解】由題設(shè)有SKIPIF1<0，故選：B.【點(diǎn)睛】本題考查集合交集，考查基本分析求解能力，屬基礎(chǔ)題.(1)離散型數(shù)集或抽象集合間的運(yùn)算,常借用Venn圖求解.(2)集合中的元素若是連續(xù)的實(shí)數(shù),常借助數(shù)軸求解,但要注意端點(diǎn)值能否取到.(3)根據(jù)集合的運(yùn)算求參數(shù),先把符號(hào)語言譯成文字語言,然后適時(shí)應(yīng)用數(shù)形結(jié)合求解.集合的基本運(yùn)算運(yùn)算自然語言符號(hào)語言Venn圖交集由屬于集合A且屬于集合B的所有元素組成的集合SKIPIF1<0并集由所有屬于集合A或?qū)儆诩螧的元素組成的集合SKIPIF1<0補(bǔ)集由全集U中不屬于集合A的所有元素組成的集合SKIPIF1<0【知識(shí)拓展】（1）集合運(yùn)算的相關(guān)結(jié)論交集SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0并集SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0補(bǔ)集SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（2）SKIPIF1<01．（2021·安徽安慶市·安慶一中高三三模（理））設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0=（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2021·湖南衡陽市八中高三其他模擬）已知集合SKIPIF1<0，集合SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<01．（2021·山東高三其他模擬）已知集合SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a的取值范圍為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·浙江高三其他模擬）已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·浙江高三其他模擬）已知集合SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·安徽高三其他模擬（文））已知集合SKIPIF1<0則ASKIPIF1<0B=A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2021·沈陽市·遼寧實(shí)驗(yàn)中學(xué)高三其他模擬）已知集合SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<06．（2021·全國(guó)高三月考（文））已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0中元素的個(gè)數(shù)為A．2 B．3 C．4 D．57．（2021·陜西高三其他模擬（理））已知集合SKIPIF1<0，則集合SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．（2021·四川德陽市·高三三模（文））已知集合SKIPIF1<0，SKIPIF1<0.則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2021·陜西寶雞市·高三其他模擬（文））已知集合SKIPIF1<0，SKIPIF1<0，那么SKIPIF1<0的子集個(gè)數(shù)為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2021·遼寧高三其他模擬）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則圖示中陰影部分表示的集合為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<01．（2021年全國(guó)高考乙卷數(shù)學(xué)（文）試題）已知全集SKIPIF1<0，集合SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021年全國(guó)高考乙卷數(shù)學(xué)（理）試題）已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021年全國(guó)高考甲卷數(shù)學(xué)（文）試題）設(shè)集合SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021年全國(guó)高考甲卷數(shù)學(xué)（理）試題）設(shè)集合SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2021年浙江省高考數(shù)學(xué)試題）設(shè)集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2020年高考浙江）已知集合P=SKIPIF1<0，Q=SKIPIF1<0，則PSKIPIF1<0Q=A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<07．（2020年高考江蘇）已知集合SKIPIF1<0，則SKIPIF1<0_____.1．【答案】B【分析】解對(duì)數(shù)不等式得SKIPIF1<0,解一元二次不等式得SKIPIF1<0,進(jìn)而求解集合運(yùn)算.【詳解】解：解對(duì)數(shù)不等式SKIPIF1<0得SKIPIF1<0，故SKIPIF1<0，解一元二次不等式SKIPIF1<0得SKIPIF1<0，則SKIPIF1<0SKIPIF1<0.故選：B.2．【答案】B【分析】解分式不等式求得SKIPIF1<0，由此求得SKIPIF1<0.【詳解】SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0.故選：B1．【答案】C【解析】由題意知SKIPIF1<0，由SKIPIF1<0知SKIPIF1<0，故SKIPIF1<0，解得SKIPIF1<0．故選：C．2．【答案】B【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.故選：B．3．【答案】B【解析】根據(jù)真數(shù)大于零，可得SKIPIF1<0，又SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.故選：B.4．【答案】C【解析】由題意得，SKIPIF1<0，故選：C.5．【答案】C【解析】∵SKIPIF1<0，∴SKIPIF1<0又SKIPIF1<0，∴SKIPIF1<0故選：C.6．【答案】C【解析】由題意可知SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，共有4個(gè)元素.故選：C7．【答案】C【解析】由SKIPIF1<0得，SKIPIF1<0.故選：C8．【答案】C【解析】由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，顯然SKIPIF1<0，所以SKIPIF1<0.故選：C.9．【答案】C【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0的子集個(gè)數(shù)為SKIPIF1<0.故選：C.10．【答案】A【解析】由不等式SKIPIF1<0，可化為SKIPIF1<0，解得SKIPIF1<0，即集合SKIPIF1<0，又由SKIPIF1<0，可得陰影部分所表示的集合為SKIPIF1<0.故選：A.1．【答案】A【分析】首先進(jìn)行并集運(yùn)算，然后進(jìn)行補(bǔ)集運(yùn)算即可.【詳解】由題意可得：SKIPIF1<0，則SKIPIF1<0.故選：A.2．【答案】C【分析】分析可得SKIPIF1<0，由此可得出結(jié)論.【詳解】任取SKIPIF1<0，則SKIPIF1<0，其中