新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練4.5 導(dǎo)數(shù)的綜合運(yùn)用（提升版）（解析版）

4.5導(dǎo)數(shù)的綜合運(yùn)用（精講）（提升版）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)一零點(diǎn)的個(gè)數(shù)【例1】（2022·廣東·深圳市光明區(qū)高級(jí)中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的極值點(diǎn)；(2)當(dāng)SKIPIF1<0時(shí)，試討論函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù).【答案】(1)SKIPIF1<0(2)有SKIPIF1<0個(gè)零點(diǎn)【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；綜上，函數(shù)SKIPIF1<0的極值點(diǎn)為SKIPIF1<0.(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的一個(gè)零點(diǎn)，令SKIPIF1<0，可得SKIPIF1<0.因?yàn)镾KIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0，此時(shí)SKIPIF1<0在SKIPIF1<0無(wú)零點(diǎn).②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，此時(shí)SKIPIF1<0在SKIPIF1<0無(wú)零點(diǎn).③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，由零點(diǎn)存在性定理知，存在唯一SKIPIF1<0，使得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增；又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有SKIPIF1<0個(gè)零點(diǎn).綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有SKIPIF1<0個(gè)零點(diǎn).【一隅三反】1．（2022·江蘇·南京市天印高級(jí)中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處的切線方程；(2)判斷函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)，并說(shuō)明理由.【答案】(1)SKIPIF1<0(2)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)零點(diǎn)，理由見解析【解析】(1)SKIPIF1<0，SKIPIF1<0所以函數(shù)SKIPIF1<0的圖象在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0.(2)設(shè)SKIPIF1<0，則SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0單調(diào)遞減；且SKIPIF1<0，SKIPIF1<0，由零點(diǎn)存在定理可知，在區(qū)間SKIPIF1<0存在唯一的SKIPIF1<0，使SKIPIF1<0又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，且SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上沒(méi)有零點(diǎn).

②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0有唯一零點(diǎn)，設(shè)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增；在區(qū)間SKIPIF1<0上SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減，且SKIPIF1<0，故有SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減，且SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以存在SKIPIF1<0，使SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0為SKIPIF1<0的極小值點(diǎn).此時(shí)SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0上沒(méi)有零點(diǎn).③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上沒(méi)有零點(diǎn).綜上SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)零點(diǎn).2．（2022·北京四中三模）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的單調(diào)區(qū)間；(2)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的零點(diǎn)個(gè)數(shù).【答案】(1)單調(diào)增區(qū)間為SKIPIF1<0，單調(diào)減區(qū)間為SKIPIF1<0(2)答案見解析【解析】(1)當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，可得SKIPIF1<0.當(dāng)SKIPIF1<0在區(qū)間SKIPIF1<0上變化時(shí)，SKIPIF1<0，f（x）的變化如下表：x0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<00+0-f（x）極小值1SKIPIF1<0極大值SKIPIF1<0SKIPIF1<0-1所以SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0；SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0.(2)由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立，所以SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.又因?yàn)镾KIPIF1<0，所以f（x）在SKIPIF1<0上有0個(gè)零點(diǎn).當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，可得SKIPIF1<0.由SKIPIF1<0可知存在唯一的SKIPIF1<0使得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上有0個(gè)零點(diǎn).②當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上有1個(gè)零點(diǎn).綜上可得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有2個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有0個(gè)零點(diǎn).3．（2022·云南師大附中高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，求證：SKIPIF1<0在SKIPIF1<0上只有1個(gè)零點(diǎn)【答案】(1)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增(2)證明見解析【解析】(1)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0．令SKIPIF1<0，解得SKIPIF1<0，則有當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．(2)證明：由于SKIPIF1<0，SKIPIF1<0．設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上的零點(diǎn)也是SKIPIF1<0的零點(diǎn)，且有SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，由于SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；又SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0；由于SKIPIF1<0，且SKIPIF1<0，所以存在唯一的SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上有1個(gè)零點(diǎn)．②當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，所以SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．又SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0；所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上無(wú)零點(diǎn)．③當(dāng)SKIPIF1<0時(shí)，由于SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上無(wú)零點(diǎn)．綜合①②③，可知，SKIPIF1<0在SKIPIF1<0上只有1個(gè)零點(diǎn)．考點(diǎn)二已知零點(diǎn)個(gè)數(shù)求參【例2】（2022·全國(guó)·高考真題）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的最大值；(2)若SKIPIF1<0恰有一個(gè)零點(diǎn)，求a的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；所以SKIPIF1<0；(2)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；所以SKIPIF1<0，此時(shí)函數(shù)無(wú)零點(diǎn)，不合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；又SKIPIF1<0，由（1）得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則存在SKIPIF1<0，使得SKIPIF1<0，所以SKIPIF1<0僅在SKIPIF1<0有唯一零點(diǎn)，符合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0有唯一零點(diǎn)，符合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；此時(shí)SKIPIF1<0，由（1）得當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，此時(shí)SKIPIF1<0存在SKIPIF1<0，使得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0有一個(gè)零點(diǎn)，在SKIPIF1<0無(wú)零點(diǎn)，所以SKIPIF1<0有唯一零點(diǎn)，符合題意；綜上，a的取值范圍為SKIPIF1<0.【一隅三反】1．（2022·河南·平頂山市第一高級(jí)中學(xué)模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0有兩個(gè)零點(diǎn)，求實(shí)數(shù)a的取值范圍．【答案】(1)答案見解析；(2)SKIPIF1<0.【解析】(1)由題意知，SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；若SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．(2)因?yàn)镾KIPIF1<0，所以SKIPIF1<0有兩個(gè)零點(diǎn)，即SKIPIF1<0有兩個(gè)零點(diǎn)．若SKIPIF1<0，由（1）知，SKIPIF1<0至多有一個(gè)零點(diǎn)．若SKIPIF1<0，由（1）知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值，最小值為SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，由于SKIPIF1<0，故SKIPIF1<0只有一個(gè)零點(diǎn)：②當(dāng)SKIPIF1<0時(shí)，由于SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0沒(méi)有零點(diǎn)；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0．又SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)．存在SKIPIF1<0，則SKIPIF1<0．又SKIPIF1<0，因此SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)．綜上，實(shí)數(shù)a的取值范圍為SKIPIF1<0．2（2022·全國(guó)·高考真題（理））已知函數(shù)SKIPIF1<0(1)當(dāng)SKIPIF1<0時(shí)，求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)若SKIPIF1<0在區(qū)間SKIPIF1<0各恰有一個(gè)零點(diǎn)，求a的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】(1)SKIPIF1<0的定義域?yàn)镾KIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以切點(diǎn)為SKIPIF1<0SKIPIF1<0,所以切線斜率為2所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0(2)SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0若SKIPIF1<0,當(dāng)SKIPIF1<0,即SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0故SKIPIF1<0在SKIPIF1<0上沒(méi)有零點(diǎn),不合題意SKIPIF1<0若SKIPIF1<0,當(dāng)SKIPIF1<0,則SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增所以SKIPIF1<0,即SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0故SKIPIF1<0在SKIPIF1<0上沒(méi)有零點(diǎn),不合題意SKIPIF1<0若SKIPIF1<0(1)當(dāng)SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增SKIPIF1<0所以存在SKIPIF1<0,使得SKIPIF1<0,即SKIPIF1<0當(dāng)SKIPIF1<0單調(diào)遞減當(dāng)SKIPIF1<0單調(diào)遞增所以當(dāng)SKIPIF1<0當(dāng)SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)又SKIPIF1<0沒(méi)有零點(diǎn),即SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)(2)當(dāng)SKIPIF1<0設(shè)SKIPIF1<0SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增SKIPIF1<0所以存在SKIPIF1<0,使得SKIPIF1<0當(dāng)SKIPIF1<0單調(diào)遞減當(dāng)SKIPIF1<0單調(diào)遞增,SKIPIF1<0又SKIPIF1<0所以存在SKIPIF1<0,使得SKIPIF1<0,即SKIPIF1<0當(dāng)SKIPIF1<0單調(diào)遞增,當(dāng)SKIPIF1<0單調(diào)遞減有SKIPIF1<0而SKIPIF1<0,所以當(dāng)SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn),SKIPIF1<0上無(wú)零點(diǎn)即SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)所以SKIPIF1<0,符合題意所以若SKIPIF1<0在區(qū)間SKIPIF1<0各恰有一個(gè)零點(diǎn),求SKIPIF1<0的取值范圍為SKIPIF1<03．（2022·貴州·貴陽(yáng)一中高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0(1)討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0有三個(gè)零點(diǎn)時(shí)a的取值范圍恰好是SKIPIF1<0求b的值.【答案】(1)答案見解析(2)SKIPIF1<0【解析】(1)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減，若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增，若SKIPIF1<0，則SKIPIF1<0SKIPIF1<0或SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減.(2)可知SKIPIF1<0要有三個(gè)零點(diǎn)，則SKIPIF1<0，且SKIPIF1<0由題意也即是SKIPIF1<0的解集就是SKIPIF1<0，也就是關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集就是SKIPIF1<0，令SKIPIF1<0，時(shí)SKIPIF1<0，所以有SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的解是SKIPIF1<0，滿足條件，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不滿足條件，故SKIPIF1<0，綜合上述SKIPIF1<0.考點(diǎn)三不等式恒（能）成立【例3】（2022·天津市）已知函數(shù)SKIPIF1<0，SKIPIF1<0（SKIPIF1<0），其中e是自然對(duì)數(shù)的底數(shù).(1)當(dāng)SKIPIF1<0時(shí)，（ⅰ）求SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；（ⅱ）求SKIPIF1<0的最小值；(2)討論函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)；(3)若存在SKIPIF1<0，使得SKIPIF1<0成立，求a的取值范圍【答案】(1)（ⅰ）SKIPIF1<0；（ⅱ）SKIPIF1<0(2)答案見解析(3)SKIPIF1<0【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0.（ⅰ）SKIPIF1<0，SKIPIF1<0，∴切線方程為SKIPIF1<0.（ⅱ）SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0.(2)∵SKIPIF1<0（SKIPIF1<0），令SKIPIF1<0得，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0無(wú)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，∴SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)零點(diǎn)，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，∴函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上各有一個(gè)零點(diǎn)，綜上，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上有兩個(gè)零點(diǎn).(3)由SKIPIF1<0得，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上有解，令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不合題意；當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，即a的取值范圍為SKIPIF1<0.【一隅三反】1．（2022·河南安陽(yáng)）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程；(2)對(duì)于SKIPIF1<0，不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，此時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0.(2)SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，其中SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以，存在SKIPIF1<0使得SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，因?yàn)镾KIPIF1<0，所以，SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，所以，SKIPIF1<0，SKIPIF1<0.2．（2022·海南中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，求SKIPIF1<0的單調(diào)區(qū)間；(2)是否存在實(shí)數(shù)a，使SKIPIF1<0對(duì)SKIPIF1<0恒成立，若存在，求出a的值或取值范圍；若不存在，請(qǐng)說(shuō)明理由.【答案】(1)單調(diào)遞減區(qū)間是SKIPIF1<0，單調(diào)遞增區(qū)間是SKIPIF1<0；(2)存在，SKIPIF1<0.【解析】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，因?yàn)镾KIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，單調(diào)遞增區(qū)間是SKIPIF1<0.(2)法一：設(shè)SKIPIF1<0，則SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0不符合題意.②當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.·令SKIPIF1<0，即SKIPIF1<0，取SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0.故SKIPIF1<0不符合題意.③當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞增.因?yàn)镾KIPIF1<0，SKIPIF1<0，所以存在唯一的SKIPIF1<0使得SKIPIF1<0，所以，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增.所以SKIPIF1<0的最小值為SKIPIF1<0，因?yàn)镾KIPIF1<0，即SKIPIF1<0，兩邊取對(duì)數(shù)得SKIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，故當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0恒成立，綜上，存在a符合題意，SKIPIF1<0.法二：設(shè)SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0單調(diào)遞增，①當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以存在唯一SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0.所以當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0單調(diào)遞增.故SKIPIF1<0，即SKIPIF1<0，符合題意.②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以存在唯一SKIPIF1<0，使得SKIPIF1<0，所以當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0單調(diào)遞減，故SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0不符合題意.③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以存在唯一SKIPIF1<0，使得SKIPIF1<0，所以當(dāng)SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，故SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0不符合題意.④當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不符合題意.⑤當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不符合題意.綜上，存在a符合題意，SKIPIF1<0.法三：①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.因?yàn)镾KIPIF1<0在SKIPIF1<0單調(diào)遞增，且SKIPIF1<0，SKIPIF1<0，故存在唯一SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，所以任意SKIPIF1<0，都有SKIPIF1<0.故SKIPIF1<0不符合題意.②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，對(duì)于函數(shù)SKIPIF1<0，SKIPIF1<0.所以SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，故SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0符合題意.③當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，對(duì)于函數(shù)SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0單調(diào)遞增，且SKIPIF1<0，SKIPIF1<0，所以存在SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減.故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，“=”成立.所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0不符合題意.綜上，存在a符合題意，SKIPIF1<0.法四：設(shè)SKIPIF1<0，SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0單調(diào)遞增.又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的值域?yàn)镾KIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的值域?yàn)镾KIPIF1<0.所以SKIPIF1<0的值域?yàn)镾KIPIF1<0.故對(duì)于SKIPIF1<0上任意一個(gè)值SKIPIF1<0，都有唯一的一個(gè)正數(shù)SKIPIF1<0，使得SKIPIF1<0.因?yàn)镾KIPIF1<0，即SKIPIF1<0.設(shè)SKIPIF1<0，SKIPIF1<0，所以要使SKIPIF1<0，只需SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0不符合題意.當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增.所以SKIPIF1<0.設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減.所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立.又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.綜上，存在a符合題意，SKIPIF1<0.3．（2022·青?！ご笸ɑ刈逋磷遄灾慰h教學(xué)研究室三模（理））已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0在SKIPIF1<0上僅有一個(gè)零點(diǎn)，求實(shí)數(shù)a的取值范圍；(2)若對(duì)任意的SKIPIF1<0，SKIPIF1<0恒成立，求實(shí)數(shù)a的取值范圍．【答案】(1)SKIPIF1<0或SKIPIF1<0(2)SKIPIF1<0【解析】(1)SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．又SKIPIF1<0SKIPIF1<0，SKIPIF1<0，所以此時(shí)SKIPIF1<0在SKIPIF1<0上僅有一個(gè)零點(diǎn)，符合題意；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0；令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．要使SKIPIF1<0在SKIPIF1<0上僅有一個(gè)零點(diǎn)，則必有SKIPIF1<0，解得SKIPIF1<0．綜上，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上僅有一個(gè)零點(diǎn)．(2)因?yàn)镾KIPIF1<0，所以對(duì)任意的SKIPIF1<0，SKIPIF1<0恒成立，等價(jià)于SKIPIF1<0在SKIPIF1<0上恒成立．令SKIPIF1<0，則只需SKIPIF1<0即可，則SKIPIF1<0，再令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0有唯一的零點(diǎn)SKIPIF1<0，且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．所以SKIPIF1<0SKIPIF1<0，則有SKIPIF1<0．所以實(shí)數(shù)a的取值范圍為SKIPIF1<0．考點(diǎn)四