新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練8.1 定義域（基礎(chǔ)版）（解析版）

8.1定義域（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一具體函數(shù)求定義域【例1】（1）（2022·山東濟(jì)南·二模）函數(shù)SKIPIF1<0的定義域是（2）（2022.廣東潮州）函數(shù)SKIPIF1<0的定義域【答案】（1）SKIPIF1<0（2）SKIPIF1<0【解析】（1）由SKIPIF1<0，得SKIPIF1<0，且SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域是SKIPIF1<0.故選：A.（2）要使函數(shù)SKIPIF1<0有意義，需滿(mǎn)足SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0故函數(shù)定義域?yàn)镾KIPIF1<0【一隅三反】1．（2022·寧夏·銀川一中）函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意得SKIPIF1<0，解得SKIPIF1<0且SKIPIF1<0，故選：D2．（2022·寧夏·銀川一中一模）設(shè)不等式SKIPIF1<0的解集為SKIPIF1<0，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則SKIPIF1<0為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由于不等式SKIPIF1<0等價(jià)于SKIPIF1<0，解得SKIPIF1<0，故集合SKIPIF1<0函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，滿(mǎn)足SKIPIF1<0，故集合SKIPIF1<0，因此通過(guò)集合的交集的運(yùn)算可知，SKIPIF1<0故選：A.3．（2022·北京·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的定義域是_______．【答案】SKIPIF1<0【解析】由題意可得，SKIPIF1<0，解之得SKIPIF1<0則函數(shù)SKIPIF1<0的定義域是SKIPIF1<0故答案為：SKIPIF1<04．（2021·銀川市·寧夏銀川二十四中）函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)__________.【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0解得SKIPIF1<0，所以函數(shù)的定義域?yàn)镾KIPIF1<0,故答案為：SKIPIF1<05．（2020·甘肅武威市·武威十八中高三月考）函數(shù)SKIPIF1<0的定義域是（）A．[－1，4] B．（－1，4] C．[2，4] D．（2，4]【答案】D【解析】由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0所以函數(shù)的定義域?yàn)镾KIPIF1<0故選：D考點(diǎn)二復(fù)合函數(shù)求定義域【例2-1】（2022·陜西·西安高新第三中學(xué)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】要使函數(shù)SKIPIF1<0SKIPIF1<0有意義，則SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故選D.【例2-2】（2022·廣東·化州市第三中學(xué)）已知函數(shù)y＝f（x＋1）定義域是[－2，3]，則y＝f（x－2）的定義域是（）A．[1，6] B．[－1，4] C．[－3，2] D．[－2，3]【答案】A【解析】由題意知，－2≤x≤3，∴－1≤x＋1≤4，∴－1≤x－2≤4，得1≤x≤6，即y＝f（x－2）的定義域?yàn)閇1，6]；故選：A.1.抽象函數(shù)求定義域解題思路：對(duì)應(yīng)法則不變，括號(hào)內(nèi)等范圍1.抽象函數(shù)求定義域解題思路：對(duì)應(yīng)法則不變，括號(hào)內(nèi)等范圍2.定義域求解口訣定義域是何意，自變量有意義；分式分母不為0，對(duì)數(shù)真數(shù)只取正；偶次根式要非負(fù)，三者高考最?？?；和差積商定義域，不等式組求交集；抽象函數(shù)定義域，對(duì)應(yīng)法則內(nèi)相同。溫馨提示【一隅三反】1．（2022·貴州畢節(jié)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】∵函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，∴SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，∴SKIPIF1<0的定義域是SKIPIF1<0，故選：A2．（2022·重慶巴蜀中學(xué)）已知函數(shù)SKIPIF1<0的定義域?yàn)閇1，10]，則SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題意可知，函數(shù)SKIPIF1<0的定義域?yàn)閇1，10]，則函數(shù)SKIPIF1<0成立需要滿(mǎn)足SKIPIF1<0，解得SKIPIF1<0.故選：B.3．（2022·廣東·普寧市第二中學(xué)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0．則函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．[－1，1] B．[SKIPIF1<0，2] C．[1，2] D．[SKIPIF1<0，4]【答案】D【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，解得：SKIPIF1<0.故選：D4．（2022·黑龍江）已知函數(shù)f(x)的定義域?yàn)閇3，6]，則函數(shù)y＝SKIPIF1<0的定義域?yàn)椋ǎ〢．[SKIPIF1<0，＋∞) B．[SKIPIF1<0，2)C．(SKIPIF1<0，＋∞) D．[SKIPIF1<0，2)【答案】B【解析】要使函數(shù)y＝SKIPIF1<0有意義，需滿(mǎn)足SKIPIF1<0SKIPIF1<0?SKIPIF1<0≤x<2.故選：B.5．（2021·天津市第一中學(xué)濱海學(xué)校）設(shè)SKIPIF1<0，則SKIPIF1<0的定義域?yàn)開(kāi)______.【答案】SKIPIF1<0【解析】由SKIPIF1<0得SKIPIF1<0，故SKIPIF1<0且SKIPIF1<0，SKIPIF1<0,SKIPIF1<0或SKIPIF1<0解得：SKIPIF1<0.故答案為：SKIPIF1<0考點(diǎn)三已知定義域求參數(shù)【例3-1】（2022·全國(guó)·高三專(zhuān)題練習(xí)）若函數(shù)y＝SKIPIF1<0的定義域?yàn)镽，則實(shí)數(shù)m的取值范圍是()A．(0，SKIPIF1<0] B．(0，SKIPIF1<0) C．[0，SKIPIF1<0] D．[0，SKIPIF1<0)【答案】D【解析】因?yàn)閥＝SKIPIF1<0的定義域?yàn)镽，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0選D.【例3-2】（2022·全國(guó)·高三專(zhuān)題練習(xí)）（多選）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有意義，則實(shí)數(shù)SKIPIF1<0可能的取值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】AB【解析】函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有意義，等價(jià)于SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，由SKIPIF1<0得SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，所以SKIPIF1<0，故選：AB．【一隅三反】1．（2022·全國(guó)·高三期末）（多選）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值可能是（

）A．0 B．1 C．2 D．3【答案】ABC【解析】因函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，于是得SKIPIF1<0，不等式SKIPIF1<0成立，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，必有SKIPIF1<0，解得SKIPIF1<0，綜上得：SKIPIF1<0，顯然，選項(xiàng)A，B，C都滿(mǎn)足，選項(xiàng)D不滿(mǎn)足.故選：ABC2．（2022·江西）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則實(shí)數(shù)a的取值范圍是___________.【答案】SKIPIF1<0【解析】因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镽，所以SKIPIF1<0的解為R，即函數(shù)SKIPIF1<0的圖象與x軸沒(méi)有交點(diǎn)，（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0與x軸沒(méi)有交點(diǎn)，故SKIPIF1<0成立；（2）當(dāng)SKIPIF1<0時(shí)，要使函數(shù)SKIPIF1<0的圖象與x軸沒(méi)有交點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0.綜上：實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<03．（2022·湖南·新邵縣教研室）已知SKIPIF1<0的定義域?yàn)镾KIPIF1<0，那么a的取值范圍為_(kāi)________．【答案】SKIPIF1<0【解析】依題可知，SKIPIF1<0的解集為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0．8.1定義域（精練）（基礎(chǔ)版）題組一題組一具體函數(shù)求定義域1．（2022·湖南·寧鄉(xiāng)市教育研究中心模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，故選：C2．（2022·湖南·長(zhǎng)郡中學(xué)）函數(shù)SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】依題意SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)的定義域?yàn)镾KIPIF1<0.故選：B．3．（2022·北京石景山·一模）函數(shù)SKIPIF1<0的定義域是_________．【答案】SKIPIF1<0【解析】由SKIPIF1<0，可得SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，故答案為：SKIPIF1<0.4．（2022·上海閔行）函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)__________.【答案】SKIPIF1<0【解析】依題意，SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以所求定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<05．（2022·上海市奉賢中學(xué)高三階段練習(xí)）函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)__________.【答案】SKIPIF1<0【解析】由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)的定義域?yàn)镾KIPIF1<0，故答案為：SKIPIF1<06．（2022·湖南·課時(shí)練習(xí)）求下列函數(shù)的定義域：(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0；【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)SKIPIF1<0【解析】(1)要使函數(shù)SKIPIF1<0有意義，需滿(mǎn)足SKIPIF1<0，解得SKIPIF1<0故函數(shù)定義域?yàn)镾KIPIF1<0(2)要使函數(shù)SKIPIF1<0有意義，需滿(mǎn)足SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0故函數(shù)定義域?yàn)镾KIPIF1<0(3)要使函數(shù)SKIPIF1<0有意義，需滿(mǎn)足SKIPIF1<0，即SKIPIF1<0故函數(shù)定義域?yàn)镾KIPIF1<0題組二題組二復(fù)合函數(shù)求定義域1．（2022·遼寧）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】依題意函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故選：B2．（2022·全國(guó)·高三）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0的定義域是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意得：SKIPIF1<0，解得SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，故函數(shù)的定義域是SKIPIF1<0

.故選：D3．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)?－2,0)，則SKIPIF1<0的定義域?yàn)椋?/p>

）A．(－1,0) B．(－2,0) C．(0,1) D．SKIPIF1<0【答案】C【解析】由題設(shè)，若SKIPIF1<0，則SKIPIF1<0，∴對(duì)于SKIPIF1<0有SKIPIF1<0，故其定義域?yàn)镾KIPIF1<0.故選：C4．（2022·安徽阜陽(yáng)）已知SKIPIF1<0，則SKIPIF1<0的定義域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題可知：SKIPIF1<0且SKIPIF1<0所以函數(shù)定義域?yàn)镾KIPIF1<0且SKIPIF1<0令SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0且SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0故選：C5．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)椋ǎ〢．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．故選：B．6．（2022·全國(guó)·高三專(zhuān)題練習(xí)）若函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)楹瘮?shù)SKIPIF1<0的定義域是SKIPIF1<0，所以SKIPIF1<0.故選：D.7．（2022·河南南陽(yáng)）若函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)_____．【答案】SKIPIF1<0【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0的定義域?yàn)镾KIPIF1<0故答案為：SKIPIF1<08．（2022·甘肅省會(huì)寧縣第一中學(xué)）已知函數(shù)f(x)的定義域是[-1，1]，則函數(shù)f(log2x)的定義域?yàn)開(kāi)___．【答案】SKIPIF1<0【解析】因函數(shù)f(x)的定義域是[-1，1]，則在f(log2x)中，必有SKIPIF1<0，解不等式可得：SKIPIF1<0，即SKIPIF1<0，所以函數(shù)f(log2x)的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<09．（2022·全國(guó)·高三專(zhuān)題練習(xí)）若函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)開(kāi)__________.【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0，要使SKIPIF1<0有意義，需滿(mǎn)足SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<010．（2021·甘肅張掖市）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)椤敬鸢浮縎KIPIF1<0【解析】由函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，得到SKIPIF1<0，故SKIPIF1<0即SKIPIF1<0解得：SKIPIF1<0；所以原函數(shù)的定義域是：SKIPIF1<0題組三題組三已知定義域求參數(shù)1．（2022·福建·廈門(mén)一中）函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，則實(shí)數(shù)a的取值范圍為_(kāi)_______．【答案】SKIPIF1<0【解析】因?yàn)楹瘮?shù)SKIPIF1<0的定義域是SKIPIF1<0.所以不等式SKIPIF1<0恒成立．所以，當(dāng)SKIPIF1<0時(shí)，不等式等價(jià)于SKIPIF1<0，顯然恒成立；當(dāng)SKIPIF1<0時(shí)，則有SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.綜上，實(shí)數(shù)a的取值范圍為SKIPIF1<0.故答案為:SKIPIF1<02．（2022·上海市控江中學(xué)）函數(shù)SKIPIF1<0定義域?yàn)镽，則實(shí)數(shù)k的取值范圍為_(kāi)_____.【答案】SKIPIF1<0【解析】因?yàn)楹瘮?shù)SKIPIF1<0定義域?yàn)镽，所以SKIPIF1<0在R上恒成立，所以SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.3.（2021年廣東肇慶）已知SKIPIF1<0的定義域?yàn)镽，求實(shí)數(shù)SKIPIF1<0的取值范圍.．【答案】SKIPIF1<0或SKIPIF1<0．【解析】由題設(shè)得：SKIPIF1<0在SKIPIF1<0時(shí)恒成立，當(dāng)SKIPIF1<0時(shí)：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0不恒成立∴SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0綜上所述：實(shí)數(shù)SKIPIF1<0的取值范圍是實(shí)數(shù)SKIPIF1<0或SKIPIF1<0．4（2021年廣東韶光）.函數(shù)SKIPIF1<0.若SKIPIF1<0的定義域?yàn)镾KIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】SKIPIF1<0.【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，符合題意；（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的定義域不為SKIPIF1<0，所以SKIPIF1<0；（3）當(dāng)SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0的定義域?yàn)镾KIPIF1<0知拋物線(xiàn)SKIPIF1<0全部在SKIPIF1<0軸上方（或在上方相切），此時(shí)應(yīng)有，解得SKIPIF1<0；綜合（1），（2），（3）有SKIPIF1<0的取值范圍是SKIPIF1<0.5．（2022·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，則SKIPIF1<0的取值范圍是_________．【答案】SKIPIF1<0【解析】由題意可得SKIPIF1<0在SKIPIF1<0上恒成立．①當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0恒成立，SKIPIF1<0符合題意；②當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，解得SKIPIF1<0．綜上可得SKIPIF1<0，∴實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．故答案為：SKIPIF1<0.6．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镽，則a的范圍是________．【答案】SKIPIF1<0【解析】當(dāng)