易錯點01集合的概念及運算易錯題【01】對集合中元素的類型理解不到位集合問題是高考必考問題,一般作為容易題出現(xiàn),求解集合問題的關(guān)鍵是理解集合中元素的類型,特別是用描述法表示集合,首先要搞清楚集合中代表元素的含義,再看元素的限制條件,明白集合的類型,是連續(xù)數(shù)集、離散數(shù)集、點集或其他類型的集合.易錯題【02】忽略集合中元素互異性利用元素與集合的關(guān)系或兩集合之間的關(guān)系求參數(shù)的值,集合中元素的互異性常常容易忽略,求解問題時要特別注意,求出以后一定要代入檢驗,看看是否滿足元素的互異性.易錯題【03】忽略空集空集是任何集合的子集,在涉及集合關(guān)系,如根據(jù)SKIPIF1<0求參數(shù)的值或范圍要注意SKIPIF1<0是否可以為SKIPIF1<0,根據(jù)SKIPIF1<0求參數(shù)的值或范圍必須優(yōu)先考慮空集的情況,否則會造成漏解．易錯題【04】忽視集合轉(zhuǎn)化的等價性把用描述法表示的集合轉(zhuǎn)化為用列舉法表述的集合或化簡集合容易忽略等價性,如去分母忽略分母不為零,解含有對數(shù)式的不等式要保證對數(shù)式有意義,要注意集合中的限制條件等. 01(2021年全國統(tǒng)一高考數(shù)學(xué)試卷乙卷理科)已知集合SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【警示】本題出錯的主要原因是不理解集合SKIPIF1<0中元素的類型.【答案】C【問診】誤選A是因為誤認為集合中的元素是代數(shù)式,誤選D是因為只知道SKIPIF1<0是數(shù)集,忽略限制條件,正確求解方法是分別討論當(dāng)SKIPIF1<0是偶數(shù)、奇數(shù)時的集合元素情況,結(jié)合集合的基本運算進行判斷即可,當(dāng)SKIPIF1<0是偶數(shù)時,設(shè)SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0是奇數(shù)時,設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0.【叮囑】求解集合運算問題一定要注意判斷集合中元素的類型,如SKIPIF1<0都是數(shù)集,分別表示方程SKIPIF1<0的解集,函數(shù)SKIPIF1<0的定義域,函數(shù)SKIPIF1<0的值域,SKIPIF1<0表示點集.1.已知集合SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0D．SKIPIF1<0【答案】B【解析】因為SKIPIF1<0,所以SKIPIF1<0,故選B.本題易錯之處是對集合概念理解不準(zhǔn)確,把數(shù)集當(dāng)作點集由SKIPIF1<0解得SKIPIF1<0,誤選D.2．【2017全國卷3理數(shù)】已知集合A=SKIPIF1<0,B=SKIPIF1<0,則ASKIPIF1<0B中元素的個數(shù)為A．3 B．2 C．1 D．0【答案】B【解析】由題意可得,圓SKIPIF1<0與直線SKIPIF1<0相交于兩點SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0中有兩個元素,故選B． 02已知集合A＝{1,x,2},B＝{1,x2},若A∪B＝A,則x的不同取值個數(shù)為()A．1 B．2 C．3 D．4【警示】由x2＝2,解得x1＝eq\r(2),x2＝－eq\r(2).由x2＝x,解得x3＝0,x4＝1.忽略檢驗,誤選D.【答案】C【問診】因為A∪B＝A,所以B?A.所以x2＝2或x2＝x.由x2＝2,解得x＝±eq\r(2),由x2＝x,解得x＝0或x＝1.注意當(dāng)x＝1時,x2＝1,集合A、B中元素不滿足互異性,所以符合題意的x為eq\r(2)或－eq\r(2)或0,故選C.【叮囑】由集合的關(guān)系求集合中的元素,求出以后要代入檢驗,看看是否滿足元素的互異性.1．(2022屆貴州省安順市高三第一次測試)已知集合SKIPIF1<0,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0()A．1 B．0或1或3 C．0或3 D．1或3【答案】C【解析】因為SKIPIF1<0,所以SKIPIF1<0或SKIPIF1<0.①若SKIPIF1<0,則SKIPIF1<0,滿足SKIPIF1<0；②若SKIPIF1<0,則SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,滿足SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0,集合SKIPIF1<0不滿足元素的互異性,不符合題意；綜上所述：SKIPIF1<0或SKIPIF1<0,故選C.2.(2022屆江蘇省高郵市高三10月調(diào)研)已知集合SKIPIF1<0,1,2,3,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0中所含元素的個數(shù)為()A．5 B．6 C．10 D．15【答案】D【解析】因為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以分以下五種情況：①SKIPIF1<0,有四個,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,②SKIPIF1<0,有三個,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,③SKIPIF1<0,有兩個,SKIPIF1<0,SKIPIF1<0,④SKIPIF1<0,有一個,SKIPIF1<0⑤SKIPIF1<0,有五個,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0中所含元素的個數(shù)為15,故選D. 03已知A＝{x∈R|2a≤x≤a＋3}A＝{x∈R|x<－1或x>4},若SKIPIF1<0,則實數(shù)a的取值范圍是________．【警示】忽略A＝?時SKIPIF1<0也成立.【答案】a<－4或a>2【問診】分A＝?和A≠?兩種情況討論：①當(dāng)A≠?時,有eq\b\lc\{\rc\(\a\vs4\al\co1(2a≤a＋3,2a>4或a＋3<－1)),解得a<－4或2<a≤3；②當(dāng)A＝?時,由2a>a＋3,解得a>3.綜上可知,實數(shù)a的取值范圍是a<－4或a>2.【叮囑】造成本題錯誤的根本原因是忽視了“空集是任何集合的子集”這一性質(zhì)．當(dāng)題目中出現(xiàn)A?B,注意對A進行分類討論,即分為A＝?和A≠?兩種情況討論．1．已知集合A＝{x|x2＋(p＋2)x＋1＝0,p∈R},SKIPIF1<0,若A∩B＝?,則實數(shù)p的取值范圍為____________．【答案】(－4,＋∞)【錯解】A∩B＝?,則方程SKIPIF1<0的實根都不是正數(shù),設(shè)其實根分別為SKIPIF1<0,因為SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0,解得SKIPIF1<0.【錯因分析】忽略了SKIPIF1<0的情況.【解析】本題容易略SKIPIF1<0的情況,由題意可得方程有2個負根或無實根,所以SKIPIF1<0或SKIPIF1<0,所以p≤－4.p的取值范圍是(－4,＋∞)．2.設(shè)集合SKIPIF1<0,集合SKIPIF1<0,且SKIPIF1<0,則實數(shù)a的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】集合SKIPIF1<0,因為SKIPIF1<0,所以SKIPIF1<0,當(dāng)集合SKIPIF1<0時,SKIPIF1<0,解得：SKIPIF1<0；當(dāng)集合SKIPIF1<0時,SKIPIF1<0解得：SKIPIF1<0,所以SKIPIF1<0的范圍是SKIPIF1<0,故選D. 04【2019全國卷3理數(shù)3】已知集合SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【警示】求解SKIPIF1<0出錯,誤認為SKIPIF1<0.【答案】A【問診】由題意得,SKIPIF1<0,則SKIPIF1<0．故選A．【叮囑】求解與不等式解集有關(guān)的集合運算問題是高考熱點,解不等式必須是等價轉(zhuǎn)化.1．已知集合SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因為集合SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0或SKIPIF1<0,故選B.2．設(shè)全集SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】B【解析】因為SKIPIF1<0,所以SKIPIF1<0,故選B.錯1．(2022屆山西省呂梁市高三上學(xué)期11月測試)已知集合SKIPIF1<0,集合SKIPIF1<0,則SKIPIF1<0()A．A B．B C．N D．SKIPIF1<0【答案】B【解析】由題設(shè),對于集合SKIPIF1<0：當(dāng)SKIPIF1<0為偶數(shù)時元素屬于集合B,當(dāng)SKIPIF1<0為奇數(shù)時元素不屬于集合B,對于集合B：SKIPIF1<0取任意值其元素都在集合A中,∴SKIPIF1<0.故選B2．(2022屆江西省贛州市2022屆高三上學(xué)期期中)已知SKIPIF1<0、SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0的值為()A．SKIPIF1<0 B．0 C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】C【解析】由SKIPIF1<0且SKIPIF1<0,則SKIPIF1<0,∴SKIPIF1<0,于是SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,根據(jù)集合中元素的互異性可知SKIPIF1<0應(yīng)舍去,因此SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0．故選C.3．(2021屆重慶市西南大學(xué)附屬中學(xué)高三下學(xué)期月考)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【解析】由題意,集合SKIPIF1<0表示數(shù)集,又由集合SKIPIF1<0表示點集,所以SKIPIF1<0.故選B.4．(2022屆湖北省新高考9N聯(lián)盟部分重點中學(xué)高三上學(xué)期11月聯(lián)考)集合SKIPIF1<0,SKIPIF1<0,則A,B間的關(guān)系是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,A錯誤；SKIPIF1<0,B錯誤；SKIPIF1<0,則C錯誤,D正確.故選D.5．(2022屆四川省資陽市高三第一次診斷)已知集合SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0,SKIPIF1<0.故選C.6．(2022屆北京市第十三中學(xué)高三上學(xué)期期中)已知SKIPIF1<0,SKIPIF1<0,則集合SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,由SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0；故選B7．(2022屆河南省湘豫名校聯(lián)盟高三上學(xué)期11月聯(lián)考)已知集合SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】根據(jù)題意,函數(shù)SKIPIF1<0與圓SKIPIF1<0有SKIPIF1<0個交點為SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,故選SKIPIF1<0.8．(2022屆湖北省鄂東南省級示范高中教育教學(xué)改革聯(lián)盟學(xué)校高三上學(xué)期期中聯(lián)考)設(shè)集合SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0得SKIPIF1<0,即SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0,于是得SKIPIF1<0.故選