新高考數(shù)學(xué)二輪復(fù)習(xí)考點(diǎn)歸納與演練專題7-1 基本不等式和對(duì)鉤函數(shù)（含解析）

專題7-1基本不等式和對(duì)鉤函數(shù)目錄TOC\o"1-1"\h\u專題7-1基本不等式和對(duì)鉤函數(shù) 1 1題型一：直接法 1題型二：湊配法 4題型三：分離法和換元法 7題型四：常數(shù)代換“1”的代換 11題型五：消元法 15題型六：對(duì)鉤函數(shù) 17 22一、單選題 22二、多選題 26三、填空題 28題型一：直接法【典例分析】例題1．（2022·福建·上杭縣第二中學(xué)高三階段練習(xí)）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0（

）A．有最大值SKIPIF1<0 B．有最小值SKIPIF1<0 C．有最大值4 D．有最小值4【答案】A【分析】利用基本不等式可直接得到函數(shù)的最值.詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故選：A例題2．（2022·黑龍江·哈爾濱德強(qiáng)高級(jí)中學(xué)有限公司高一階段練習(xí)）已知SKIPIF1<0，則SKIPIF1<0有（

）A．最大值0 B．最小值0 C．最大值－4 D．最小值－4【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0有最大值SKIPIF1<0，故選：C【提分秘籍】基本不等式（一正，二定，三相等，特別注意“一正”，“三相等”這兩類(lèi)陷阱）①如果SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立．②其中SKIPIF1<0叫做正數(shù)SKIPIF1<0，SKIPIF1<0的幾何平均數(shù)；SKIPIF1<0叫做正數(shù)SKIPIF1<0，SKIPIF1<0的算數(shù)平均數(shù)．【變式演練】1．（2022·江蘇·連云港市贛馬高級(jí)中學(xué)高一期末）函數(shù)SKIPIF1<0的最小值是（

）A．7 B．9 C．12 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0，故選：C.2．（2022·黑龍江·哈爾濱工業(yè)大學(xué)附屬中學(xué)校高二學(xué)業(yè)考試）若SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．0 B．1 C．SKIPIF1<0 D．2【答案】B【詳解】解：若SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立所以SKIPIF1<0的最小值是1.故選：B.3．（2022·上海虹口·一模）對(duì)于正實(shí)數(shù)SKIPIF1<0，代數(shù)式SKIPIF1<0的最小值為_(kāi)_____．【答案】4【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)取等，所以代數(shù)式SKIPIF1<0的最小值為4，故答案為：44．（2022·河北行唐啟明中學(xué)高一階段練習(xí)）（1）已知SKIPIF1<0，求SKIPIF1<0的最大值【答案】（1）-2；【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立.所以SKIPIF1<0，所以，函數(shù)SKIPIF1<0的最大值為SKIPIF1<0.題型二：湊配法【典例分析】例題1．（2022·四川省南充高級(jí)中學(xué)高一期中）（1）已知SKIPIF1<0，求函數(shù)SKIPIF1<0的最大值．（2）已知SKIPIF1<0，求函數(shù)SKIPIF1<0的最大值．【答案】（1）SKIPIF1<0；（2）1；【詳解】（1）SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因?yàn)镾KIPIF1<0所以函數(shù)SKIPIF1<0SKIPIF1<0的最大值為SKIPIF1<0．（2）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故函數(shù)SKIPIF1<0的最大值為1．例題2．（2022·西藏·拉薩市第二高級(jí)中學(xué)高一期中）若SKIPIF1<0，則SKIPIF1<0的最小值為_(kāi)_____，此時(shí)SKIPIF1<0______.【答案】

SKIPIF1<0

SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取到等號(hào).故SKIPIF1<0的最小值為SKIPIF1<0，此時(shí)SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0【提分秘籍】在例題1中使用基本不等式一定要注意，積定，或者和定，否則需要湊配，比如：SKIPIF1<0直接使用基本不等式，則SKIPIF1<0發(fā)現(xiàn)，和不定，無(wú)法直接使用基本不等式，需要湊配位和定：SKIPIF1<0；再如：SKIPIF1<0直接使用基本不等式，則SKIPIF1<0，發(fā)現(xiàn)積不定，則需要湊配為積定：SKIPIF1<0【變式演練】1．（2022·遼寧·大連八中高一階段練習(xí)）已知SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0，取得最小值時(shí)SKIPIF1<0，則SKIPIF1<0______．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)．故SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0．故答案為：SKIPIF1<0.（2022·云南·屏邊苗族自治縣第一中學(xué)高一階段練習(xí)）（若SKIPIF1<0，求：SKIPIF1<0的最小值.【答案】SKIPIF1<0.【詳解】由題意可得，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0時(shí)，等號(hào)成立，此時(shí)SKIPIF1<0的最小值為SKIPIF1<03．（2022·陜西·興平市南郊高級(jí)中學(xué)高一階段練習(xí)）已知SKIPIF1<0，求SKIPIF1<0的最大值.【答案】）SKIPIF1<0【詳解】解：SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，取等號(hào)，故SKIPIF1<0的最大值是SKIPIF1<0.4．（2022·江蘇·揚(yáng)州大學(xué)附屬中學(xué)東部分校高一期中）求下列函數(shù)的最值(1)已知SKIPIF1<0，求SKIPIF1<0的最小值；【答案】(1)SKIPIF1<0【詳解】（1）由題得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取得等號(hào)，所以SKIPIF1<0的最小值為SKIPIF1<0.題型三：分離法和換元法【典例分析】例題1．（2022·安徽·合肥八中教育集團(tuán)銘傳高級(jí)中學(xué)高一期末）已知SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．6 B．8 C．10 D．12【答案】A【詳解】解法一：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)且僅的SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立.故選：A.解法二：換元法：令SKIPIF1<0，代入SKIPIF1<0換元后可化為：SKIPIF1<0，分離后：SKIPIF1<0，當(dāng)且僅SKIPIF1<0時(shí)，等式成立.例題2．（2022·重慶市育才中學(xué)高一期中）若SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．2 B．4 C．5 D．6【答案】B【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由基本不等式得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故SKIPIF1<0的最小值為4.故選：B解法二：換元法：令SKIPIF1<0（SKIPIF1<0），則SKIPIF1<0化為：SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等式成立【提分秘籍】對(duì)于分式型，可將分母看作一個(gè)整體，直接分離，也可采用換元法，對(duì)于分子，分母中次數(shù)低的式子，一次性換元后，再分離.【變式演練】1．（2022·江蘇省高淳高級(jí)中學(xué)高一階段練習(xí)）已知函數(shù)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0（

）A．有最小值1 B．有最大值1C．有最小值3 D．有最大值3【答案】B【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由基本不等式，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)等號(hào)成立，∴SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0最大值為1.故選：B.2．（2022·湖南·安化縣江英高級(jí)中學(xué)有限公司高一階段練習(xí)）已知SKIPIF1<0，則函數(shù)SKIPIF1<0的最小值是______．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立.所以函數(shù)SKIPIF1<0的最小值是SKIPIF1<0故答案為:SKIPIF1<0.3．（2022·四川省瀘縣第四中學(xué)高二期中（文））當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0的最大值為_(kāi)_____.【答案】SKIPIF1<0【詳解】由題意，SKIPIF1<0，故SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立.故答案為：SKIPIF1<04．（2022·河北·任丘市第一中學(xué)高一期中）解答下列問(wèn)題：已知SKIPIF1<0，求函數(shù)SKIPIF1<0最小值．【答案】9.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立.所以函數(shù)SKIPIF1<0的最小值為9.5．（2022·黑龍江·哈九中高一期中）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的最小值；(2)對(duì)任意SKIPIF1<0，SKIPIF1<0恒成立，求a的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立.SKIPIF1<0的最小值為SKIPIF1<0.（2）SKIPIF1<0，即SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，故SKIPIF1<0.題型四：常數(shù)代換“1”的代換【典例分析】例題1．（2022·重慶·高一階段練習(xí)）已知實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，若不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的最大值為（

）A．9 B．25 C．16 D．12【答案】B【詳解】由SKIPIF1<0得SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以實(shí)數(shù)SKIPIF1<0均是正數(shù)，若不等式SKIPIF1<0恒成立，即SKIPIF1<0；SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立；所以，SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的最大值為25.故選：B.例題2．（2022·四川·成都七中一模（理））已知SKIPIF1<0（SKIPIF1<0），則SKIPIF1<0的最小值為_(kāi)__________.【答案】4【詳解】因?yàn)镾KIPIF1<0，故SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào).故SKIPIF1<0的最小值為4.故答案為：4例題3．（2022·山西運(yùn)城·高三期中）已知實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值是___________.【答案】SKIPIF1<0【詳解】因?yàn)閷?shí)數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0SKIPIF1<0SKIPIF1<0.當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，故SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.【提分秘籍】1的代入：例：已知SKIPIF1<0，求SKIPIF1<0的最小值，解析：SKIPIF1<0.其中SKIPIF1<0，也可以改為SKIPIF1<0，“SKIPIF1<0是常數(shù)”【變式演練】1．（2022·遼寧葫蘆島·高一期中）若SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．16 B．8 C．20 D．12【答案】A【詳解】由題意得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0的最小值為16，故選：A2．（2022·浙江·杭州四中高一期中）設(shè)x，y都是正數(shù)，且SKIPIF1<0，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．3 C．SKIPIF1<0 D．2【答案】A【詳解】SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立.故選：A3．（2022·河北·高一期中）已知SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．20 D．4【答案】D【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故SKIPIF1<0的最小值為4.故選：D.4．（多選）（2022·福建龍巖·高三期中）已知SKIPIF1<0，則SKIPIF1<0的值可能是（

）A．1 B．2.5 C．3 D．4.5【答案】CD【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立.故選：CD5．（2022·重慶·高三階段練習(xí)）已知SKIPIF1<0，則SKIPIF1<0的最小值是______．【答案】SKIPIF1<0【詳解】由于SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立.故答案為：SKIPIF1<0題型五：消元法【典例分析】例題1．（2022·湖北武漢·模擬預(yù)測(cè)）已知正實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為(

)A．0 B．2 C．4 D．6【答案】A【詳解】SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)等式不成立，∴a≠1，∴SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故選：A．例題2．（2022·遼寧·高三期中）若正實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．6 B．5 C．4 D．3【答案】D【詳解】因?yàn)閤＋2y＋xy＝7，所以SKIPIF1<0，所以SKIPIF1<0．因?yàn)镾KIPIF1<0，則SKIPIF1<0所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即x＝1，y＝2時(shí)，等號(hào)成立，所以x＋y的最小值為3．故選：D【提分秘籍】在基本不等式中，涉及到SKIPIF1<0的問(wèn)題，可以轉(zhuǎn)化為SKIPIF1<0，在代入目標(biāo)中求解，這樣二元問(wèn)題轉(zhuǎn)化為一元問(wèn)題，進(jìn)而再利用基本不等式求解目標(biāo)【變式演練】1．（2022·黑龍江·牡丹江市第三高級(jí)中學(xué)高三階段練習(xí)）已知SKIPIF1<0為正實(shí)數(shù)且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】D【詳解】解：因?yàn)镾KIPIF1<0為正實(shí)數(shù)且SKIPIF1<0，所以SKIPIF1<0，所以，SKIPIF1<0因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立；所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立；故選：D2．（2022·河北·承德市高新區(qū)第一中學(xué)高一期中）已知二次函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】若SKIPIF1<0，則函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，不合乎題意，因?yàn)槎魏瘮?shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以，SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值為SKIPIF1<0.故選：B.題型六：對(duì)鉤函數(shù)【典例分析】例題1．（2021·全國(guó)·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0的值域?yàn)開(kāi)_____【答案】SKIPIF1<0【詳解】SKIPIF1<0，令SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0所以SKIPIF1<0SKIPIF1<0，即值域?yàn)椋篠KIPIF1<0.故答案為：SKIPIF1<0例題2．（2020·廣東深圳·高一期末）已知命題SKIPIF1<0是假命題，則實(shí)數(shù)SKIPIF1<0的取值范圍是_______.【答案】SKIPIF1<0【詳解】命題SKIPIF1<0為假命題，則“SKIPIF1<0”為真命題，SKIPIF1<0，令SKIPIF1<0，該對(duì)勾函數(shù)在SKIPIF1<0上單調(diào)遞增，所以，SKIPIF1<0的范圍為SKIPIF1<0，而SKIPIF1<0，SKIPIF1<0恒成立，等價(jià)于SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，所以，“SKIPIF1<0”為真命題時(shí)，SKIPIF1<0；故答案為：SKIPIF1<0例題3．（2015·浙江溫州·高二期中（文））若關(guān)于SKIPIF1<0的方程SKIPIF1<0在區(qū)間SKIPIF1<0上有解，則SKIPIF1<0的取值范圍是________．【答案】SKIPIF1<0【詳解】解：SKIPIF1<0在區(qū)間SKIPIF1<0上有解，即SKIPIF1<0在區(qū)間SKIPIF1<0上有解，令SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又由對(duì)勾函數(shù)的性質(zhì)可知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0．【提分秘籍】對(duì)鉤函數(shù)是對(duì)基本不等式的補(bǔ)充對(duì)鉤函數(shù)是一種類(lèi)似于反比例函數(shù)的一般雙曲函數(shù)，是形如：SKIPIF1<0（SKIPIF1<0）的函數(shù).由圖象得名，又被稱為：“雙勾函數(shù)”、“對(duì)號(hào)函數(shù)”、“雙飛燕函數(shù)”、“耐克函數(shù)”等.函數(shù)SKIPIF1<0（SKIPIF1<0）常考對(duì)鉤函數(shù)SKIPIF1<0（SKIPIF1<0）定義域SKIPIF1<0定義域SKIPIF1<0值域SKIPIF1<0值域SKIPIF1<0奇偶性奇函數(shù)奇偶性奇函數(shù)單調(diào)性SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增；在SKIPIF1<0，SKIPIF1<0單調(diào)遞減單調(diào)性SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增；在SKIPIF1<0，SKIPIF1<0單調(diào)遞減【變式演練】1．（2022·江蘇省天一中學(xué)高一期中）命題“SKIPIF1<0，SKIPIF1<0”為真命題的一個(gè)充分不必要條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】對(duì)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，等價(jià)于SKIPIF1<0，令SKIPIF1<0，利用對(duì)勾函數(shù)性質(zhì)知函數(shù)在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，SKIPIF1<0.因?yàn)镾KIPIF1<0，故A為充要條件，D為充分不必要條件.故選：D2．（2022·全國(guó)·高一課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0的值域?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0，定義域?yàn)镾KIPIF1<0，且SKIPIF1<0令SKIPIF1<0，SKIPIF1<0，利用對(duì)勾函數(shù)的性質(zhì)知，當(dāng)SKIPIF1<0時(shí)，函數(shù)單減；當(dāng)SKIPIF1<0時(shí)，函數(shù)單增；SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0又SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0故選：C3．（2022·北京八中高一期中）函數(shù)SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．3 D．SKIPIF1<0【答案】C【詳解】由對(duì)勾函數(shù)的性質(zhì)可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，故選：C4．（2022·吉林一中高二期末）若SKIPIF1<0使關(guān)于SKIPIF1<0的不等式SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是______．【答案】SKIPIF1<0【詳解】解：SKIPIF1<0，使關(guān)于SKIPIF1<0的不等式SKIPIF1<0成立，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則對(duì)勾函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，故SKIPIF1<0故答案為：SKIPIF1<0一、單選題1．（2022·河南南陽(yáng)·高一階段練習(xí)）若兩個(gè)正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，且存在這樣的SKIPIF1<0使不等式SKIPIF1<0有解，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，則使不等式SKIPIF1<0有解，只需滿足SKIPIF1<0即可，解得SKIPIF1<0.故選：C.2．（2022·廣東清遠(yuǎn)·高一期中）設(shè)SKIPIF1<0，SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)m的最小值是（

）A．SKIPIF1<0 B．2 C．1 D．SKIPIF1<0【答案】D【詳解】∵SKIPIF1<0，SKIPIF1<0，不等式SKIPIF1<0恒成立，即SKIPIF1<0恒成立，∴只需SKIPIF1<0,∵SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào).所以SKIPIF1<0，∴SKIPIF1<0，∴m的最小值為SKIPIF1<0，故選：D3．（2022·湖北·海亮教育仙桃市第一中學(xué)高一階段練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象恒過(guò)定點(diǎn)A，若點(diǎn)A的坐標(biāo)滿足關(guān)于x，y的方程SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．8 B．24 C．4 D．6【答案】C【詳解】因?yàn)楹瘮?shù)SKIPIF1<0圖象恒過(guò)定點(diǎn)SKIPIF1<0又點(diǎn)A的坐標(biāo)滿足關(guān)于SKIPIF1<0，SKIPIF1<0的方程SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)取等號(hào)；所以SKIPIF1<0的最小值為4．故選：C．4．（2022·浙江·杭州四中高一期中）已知SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：A.5．（2022·江蘇·揚(yáng)州中學(xué)高一階段練習(xí)）設(shè)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．有最小值為4 B．有最小值為SKIPIF1<0C．有最小值為SKIPIF1<0 D．無(wú)最小值【答案】B【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，故當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0取最小值SKIPIF1<0，故選：B6．（2022·廣東·惠州市華羅庚中學(xué)高一階段練習(xí)）若指數(shù)函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖象恒過(guò)定點(diǎn)SKIPIF1<0，且點(diǎn)SKIPIF1<0在線段SKIPIF1<0上，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．8 D．9【答案】D【詳解】由題意得：SKIPIF1<0，代入直線得SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)故選：D．7．（2022·貴州貴陽(yáng)·高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0的圖像恒過(guò)一點(diǎn)P，且點(diǎn)P在直線SKIPIF1<0的圖像上，則SKIPIF1<0的最小值為（

）A．4 B．6 C．7 D．8【答案】D【詳解】由函數(shù)SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，整理可得SKIPIF1<0，故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故選：D.8．（2022·黑龍江·鶴崗一中高三階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．4 B．6 C．8 D．10【答案】B【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時(shí)等號(hào)成立，即SKIPIF1<0的最小值為6，故選：B.二、多選題9．（2022·山東·利津縣高級(jí)中學(xué)高三階段練習(xí)）在下列函數(shù)中，最小值是4的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0【答案】BD【詳解】對(duì)于A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0，A錯(cuò)誤；對(duì)于B，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0的最小值為4，B正確；對(duì)于C，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由對(duì)勾函數(shù)性質(zhì)可知：SKIPIF1<0，C錯(cuò)誤；對(duì)于D，SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，所以SKIPIF1<0的最小值為4，D正確.故選：BD10．（2022·浙江·杭州市源清中學(xué)高二期中）下列結(jié)論中正確的結(jié)論是（

）A．SKIPIF1<0的最小值是4 B．SKIPIF1<0的最小值是SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0的最小值是SKIPIF1<0 D．SKIPIF1<0的最大值是25【答案】AC【詳解】對(duì)于SKIPIF1<0:應(yīng)用基本不等式，因?yàn)镾KIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故SKIPIF1<0正確;對(duì)于SKIPIF1<0:應(yīng)用基本不等式SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0取等號(hào)，即SKIPIF1<0不成立，最小值取不到，故SKIPIF1<0錯(cuò)誤；對(duì)于SKIPIF1<0:若SKIPIF1<0,SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)取等號(hào)，故SKIPIF1<0正確；對(duì)于SKIPIF1<0:SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0最大值是5.故SKIPIF1<0錯(cuò)誤；故選：SKIPIF1<0