新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練9.6 導(dǎo)數(shù)的綜合運(yùn)用（基礎(chǔ)版）（解析版）

9.6導(dǎo)數(shù)的綜合運(yùn)用（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)一零點(diǎn)問題【例1】（2022·全國·成都七中）設(shè)函數(shù)SKIPIF1<0?為常數(shù)).(1)討論SKIPIF1<0?的單調(diào)性；(2)討論函數(shù)SKIPIF1<0?的零點(diǎn)個(gè)數(shù).【答案】(1)遞減區(qū)間SKIPIF1<0，遞增區(qū)間SKIPIF1<0；(2)答案見解析.【解析】（1）當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0求導(dǎo)得：SKIPIF1<0，顯然函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，而SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，所以函數(shù)SKIPIF1<0的遞減區(qū)間是SKIPIF1<0，遞增區(qū)間是SKIPIF1<0.（2）由（1）知函數(shù)SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，求導(dǎo)得SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，函數(shù)SKIPIF1<0在SKIPIF1<0上遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取值集合為SKIPIF1<0，函數(shù)SKIPIF1<0取值集合為SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上的函數(shù)值集合為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的取值集合為SKIPIF1<0，函數(shù)SKIPIF1<0取值集合為SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上的函數(shù)值集合為SKIPIF1<0，所以當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0無零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有一個(gè)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn).【一隅三反】1．（2022·全國·興國中學(xué)）已知函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間，(2)若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，求實(shí)數(shù)m的取值范圍．【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)SKIPIF1<0【解析】（1）由題可得SKIPIF1<0，由題意得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，由SKIPIF1<0得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）因?yàn)镾KIPIF1<0，由（1）可知，SKIPIF1<0在SKIPIF1<0處取得極大值，在SKIPIF1<0處取得極小值，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，依題意，要使SKIPIF1<0有三個(gè)零點(diǎn)，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，經(jīng)檢驗(yàn)，SKIPIF1<0，根據(jù)零點(diǎn)存在定理，可以確定函數(shù)有三個(gè)零點(diǎn)，所以m的取值范圍為SKIPIF1<0．2．（2022·黑龍江）已知函數(shù)SKIPIF1<0，SKIPIF1<0，曲線SKIPIF1<0和SKIPIF1<0在原點(diǎn)處有相同的切線.(1)求SKIPIF1<0的值；(2)判斷函數(shù)SKIPIF1<0在SKIPIF1<0上零點(diǎn)的個(gè)數(shù)，并說明理由.【答案】(1)1(2)1個(gè)零點(diǎn)，理由見解析【解析】（1）依題意得：函數(shù)SKIPIF1<0，其導(dǎo)函數(shù)為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．SKIPIF1<0曲線SKIPIF1<0和SKIPIF1<0在原點(diǎn)處有相同的切線.SKIPIF1<0，SKIPIF1<0．（2）由（1）可知，SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0無零點(diǎn)．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0令SKIPIF1<0則SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，所以存在SKIPIF1<0使得SKIPIF1<0，因此可得SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；又SKIPIF1<0，SKIPIF1<0所以存在SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)．綜上，SKIPIF1<0在SKIPIF1<0上有1個(gè)零點(diǎn)．3．（2022·河南）已知SKIPIF1<0.SKIPIF1<0(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0有一個(gè)零點(diǎn)，求k的取值范圍.【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，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.當(dāng)SKIPIF1<0時(shí)，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.綜上可知，SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.（2）SKIPIF1<0有一個(gè)零點(diǎn)，可得SKIPIF1<0有一個(gè)實(shí)根，令SKIPIF1<0，SKIPIF1<0.令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0.∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.∴SKIPIF1<0.又SKIPIF1<0，∴SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0.SKIPIF1<0大致圖象如圖所示，若直線y＝－k與SKIPIF1<0的圖象有一個(gè)交點(diǎn)，則SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0.∴k的取值范圍是SKIPIF1<0.考點(diǎn)二不等式成立【例2】（2022·江西南昌）已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若不等式SKIPIF1<0在區(qū)間SKIPIF1<0上有解，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0(2)SKIPIF1<0【解析】（1）解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0.（2）解：因?yàn)镾KIPIF1<0，則SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，因此函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，不等式SKIPIF1<0在區(qū)間SKIPIF1<0上無解；②當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，因此SKIPIF1<0，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，不等式SKIPIF1<0在區(qū)間SKIPIF1<0上有解．綜上，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．【例2-2】（2022·四川成都）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求證：SKIPIF1<0；(2)當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，求a的取值范圍．【答案】(1)證明見解析(2)SKIPIF1<0【解析】（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，∴SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0，即SKIPIF1<0．（2）由已知得SKIPIF1<0，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，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，∴SKIPIF1<0，由SKIPIF1<0恒成立得SKIPIF1<0，即SKIPIF1<0，取對數(shù)得SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，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；∴SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，所以a的取值范圍為SKIPIF1<0．【一隅三反】1．（2022·甘肅定西）已知函數(shù)SKIPIF1<0，SKIPIF1<0(1)求SKIPIF1<0在SKIPIF1<0處的切線方程(2)若存在SKIPIF1<0時(shí)，使SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】（1）由SKIPIF1<0，可得SKIPIF1<0，所以切線的斜率SKIPIF1<0，SKIPIF1<0．所以SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0；（2）令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0．2．（2022·四川眉山）已知SKIPIF1<0.(1)求SKIPIF1<0的極值點(diǎn)；(2)若不等式SKIPIF1<0存在正數(shù)解，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)極大值點(diǎn)為SKIPIF1<0，極小值點(diǎn)為SKIPIF1<0(2)SKIPIF1<0【解析】（1）解：函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，令SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，列表如下：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0增極大值減極小值增所以，函數(shù)SKIPIF1<0的極大值點(diǎn)為SKIPIF1<0，極小值點(diǎn)為SKIPIF1<0.（2）解：由題意可知，存在SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，所以，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，則SKIPIF1<0，所以，SKIPIF1<0.3．（2022·廣東廣州·一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若函數(shù)SKIPIF1<0只有一個(gè)零點(diǎn)，求實(shí)數(shù)a的取值所構(gòu)成的集合；(2)若函數(shù)SKIPIF1<0恒成立，求實(shí)數(shù)a的取值范圍.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】（1）當(dāng)SKIPIF1<0時(shí)，顯然滿足題意當(dāng)SKIPIF1<0時(shí)，若函數(shù)SKIPIF1<0只有一個(gè)零點(diǎn)，即SKIPIF1<0只有一個(gè)根，因?yàn)?不是方程的根，所以可轉(zhuǎn)化為SKIPIF1<0只有一個(gè)根，即直線SKIPIF1<0與函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的圖像只有一個(gè)交點(diǎn).SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，在SKIPIF1<0和SKIPIF1<0上，SKIPIF1<0，在SKIPIF1<0上，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.在SKIPIF1<0時(shí)有極小值SKIPIF1<0，SKIPIF1<0圖像如圖所示：由圖可知：若要使直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖像只有一個(gè)交點(diǎn)，則SKIPIF1<0或SKIPIF1<0，綜上SKIPIF1<0.（2）SKIPIF1<0恒成立，等價(jià)于SKIPIF1<0，令SKIPIF1<0（SKIPIF1<0），SKIPIF1<0，①若SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，即SKIPIF1<0，滿足SKIPIF1<0，②若SKIPIF1<0時(shí)，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，不成立故SKIPIF1<0不滿足題意.③若SKIPIF1<0時(shí)，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，只需SKIPIF1<0SKIPIF1<0即可，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，即SKIPIF1<0，綜上：SKIPIF1<0考點(diǎn)三雙變量【例33】（2022·全國·成都七中高三開學(xué)考試（理））設(shè)函數(shù)SKIPIF1<0（SKIPIF1<0?為常數(shù)）.(1)討論SKIPIF1<0?的單調(diào)性；(2)若函數(shù)SKIPIF1<0?有兩個(gè)不相同的零點(diǎn)SKIPIF1<0?，證明:SKIPIF1<0?.【答案】(1)在SKIPIF1<0?上單調(diào)遞減，SKIPIF1<0?上單調(diào)遞增.(2)證明見解析【解析】（1）由SKIPIF1<0（SKIPIF1<0），得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<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，SKIPIF1<0?上單調(diào)遞增.（2）由（1）的結(jié)論，不妨設(shè)SKIPIF1<0?.又SKIPIF1<0?均SKIPIF1<0?，只需證SKIPIF1<0?.構(gòu)造函數(shù)SKIPIF1<0?.則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，而SKIPIF1<0，所以取不到等號(hào)，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單遞增，所以SKIPIF1<0，所以SKIPIF1<0?恒成立，結(jié)論得證.【一隅三反】1．（2022·福建泉州·模擬預(yù)測）已知函數(shù)SKIPIF1<0(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0在SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0，求證：SKIPIF1<0．【答案】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增.(2)見解析【解析】（1）由SKIPIF1<0，求導(dǎo)得SKIPIF1<0，易知SKIPIF1<0恒成立，故看SKIPIF1<0的正負(fù)，即由判別式SKIPIF1<0進(jìn)行判斷，①當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；②當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0時(shí)，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增；綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增.（2）SKIPIF1<0在SKIPIF1<0上由兩個(gè)極值點(diǎn)SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，且SKIPIF1<0為方程SKIPIF1<0的兩個(gè)根，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0將SKIPIF1<0，SKIPIF1<0代入上式，可得：SKIPIF1<0SKIPIF1<0，由題意，需證SKIPIF1<0，令SKIPIF1<0，求導(dǎo)得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，即SKIPIF1<0，故SKIPIF1<0.2．（2022·四川·高三開學(xué)考試（理））已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求證：SKIPIF1<0；(2)當(dāng)SKIPIF1<0時(shí)，已知SKIPIF1<0，SKIPIF1<0是兩個(gè)不相等的正數(shù)且SKIPIF1<0，求證：SKIPIF1<0．【答案】(1)證明見解析(2)證明見解析【解析】（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，∴SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0，即SKIPIF1<0．（2）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，∴SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，根據(jù)題意不妨設(shè)SKIPIF1<0，①先證明SKIPIF1<0，即證SKIPIF1<0，∵SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴只需證SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，∴由SKIPIF1<0得SKIPIF1<0，即得證SKIPIF1<0，∴SKIPIF1<0．②再證明SKIPIF1<0，構(gòu)造過SKIPIF1<0函數(shù)SKIPIF1<0的切線SKIPIF1<0，過SKIPIF1<0與SKIPIF1<0兩點(diǎn)函數(shù)SKIPIF1<0的割線SKIPIF1<0，不妨設(shè)SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0單調(diào)遞增，∵SKIPIF1<0得SKIPIF1<0，SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0得SKIPIF1<0．設(shè)SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0（當(dāng)SKIPIF1<0地取等），得SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0得SKIPIF1<0．由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0．綜上得SKIPIF1<0．8．（2022·全國·興國中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求實(shí)數(shù)m的取值范圍；(2)若SKIPIF1<0，使得SKIPIF1<0，求證：SKIPIF1<0．【答案】(1)SKIPIF1<0(2)證明見解析【解析】（1）由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，其中SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上有最大值，SKIPIF1<0，所以m的取值范圍為SKIPIF1<0；（2）由SKIPIF1<0，可得SKIPIF1<0，整理為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，不妨設(shè)SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，下面證明SKIPIF1<0，即證明SKIPIF1<0，令SKIPIF1<0，即證明SKIPIF1<0，其中SKIPIF1<0，只要證明SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．4．（2022·河南·鄭州市第七中學(xué)高三階段練習(xí)（理））巳知函數(shù)SKIPIF1<0.(1)求函數(shù)f（x）的最大值;(2)若關(guān)于x的方程SKIPIF1<0有兩個(gè)不等實(shí)數(shù)根SKIPIF1<0證明:SKIPIF1<0SKIPIF1<0【答案】(1)2(2)證明見詳解【解析】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0.（2）方程SKIPIF1<0可化為SKIPIF1<0SKIPIF1<0.設(shè)SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上是增函數(shù)，又SKIPIF1<0，所以有SKIPIF1<0，即方程SKIPIF1<0有兩個(gè)實(shí)數(shù)根SKIPIF1<0，SKIPIF1<0.由（1）可知SKIPIF1<0，則有SKIPIF1<0，所以SKIPIF1<0的取值范圍為SKIPIF1<0.因?yàn)榉匠蘏KIPIF1<0有兩個(gè)實(shí)數(shù)根SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，要證SKIPIF1<0，即證SKIPIF1<0.SKIPIF1<0SKIPIF1<0，需證SKIPIF1<0.需證SKIPIF1<0.不妨設(shè)SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，即要證SKIPIF1<0.設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上是增函數(shù)，SKIPIF1<0，即SKIPIF1<0成立，故原式成立.9.6導(dǎo)數(shù)的綜合運(yùn)用（精練）（基礎(chǔ)版）題組一題組一零點(diǎn)問題1．（2022·內(nèi)蒙古包頭·高三開學(xué)考試（理））已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0的單調(diào)區(qū)間；(2)討論SKIPIF1<0的零點(diǎn)情況．【答案】(1)遞增區(qū)間為SKIPIF1<0，遞減區(qū)間為SKIPIF1<0(2)答案見解析【解析】（1）解：當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，可得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，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0在SKIPIF1<0單調(diào)遞減．（2）解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0等價(jià)于SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，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單調(diào)遞增；在SKIPIF1<0單調(diào)遞減，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，如圖所示，可得SKIPIF1<0為SKIPIF1<0的極大值，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0只有1個(gè)交點(diǎn)，即SKIPIF1<0只有1個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0有2個(gè)交點(diǎn)，即SKIPIF1<0有2個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0有3個(gè)交點(diǎn)，即SKIPIF1<0有3個(gè)零點(diǎn)．綜上，SKIPIF1<0時(shí)，SKIPIF1<0只有1個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有2個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有3個(gè)零點(diǎn)．2．（2020·陜西·榆林市第十中學(xué)高三期中（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；(2)設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)答案見解析(2)SKIPIF1<0【解析】（1）解：函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，對任意的SKIPIF1<0，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，即當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，此時(shí)，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0.綜上所述，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0.（2）解：由SKIPIF1<0，可得SKIPIF1<0，其中SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，其中SKIPIF1<0，所以，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以，函數(shù)SKIPIF1<0單調(diào)遞減，所以，函數(shù)SKIPIF1<0的極大值為SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，如下圖所示：由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，因此，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.3．（2022·廣東·金山中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，和SKIPIF1<0，(1)若SKIPIF1<0與SKIPIF1<0有相同的最小值，求SKIPIF1<0的值；(2)設(shè)SKIPIF1<0有兩個(gè)零點(diǎn)，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】（1）SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在R上單調(diào)遞減，無最值，舍去當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0∵SKIPIF1<0，則SKIPIF1<0的定義域?yàn)镾KIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0依題SKIPIF1<0

SKIPIF1<0（2）由題意可知：SKIPIF1<0SKIPIF1<0令SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0即SKIPIF1<0，則SKIPIF1<0∵SKIPIF1<0在SKIPIF1<0上單調(diào)遞增則SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上有兩個(gè)零點(diǎn)由（1）可得：SKIPIF1<0，解得：SKIPIF1<0此時(shí)SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)當(dāng)SKIPIF1<0時(shí)，下證SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)取SKIPIF1<0，則SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0單調(diào)遞減，則SKIPIF1<0，即SKIPIF1<0∵SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0又∵SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0即SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)則SKIPIF1<0的取值范圍為SKIPIF1<04．（2022·安徽省定遠(yuǎn)縣第三中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的導(dǎo)數(shù)．(1)判斷并證明SKIPIF1<0在區(qū)間SKIPIF1<0上存在的極大值點(diǎn)個(gè)數(shù)；(2)判斷SKIPIF1<0的零點(diǎn)個(gè)數(shù)．【答案】(1)SKIPIF1<0在區(qū)間SKIPIF1<0上存在的極大值點(diǎn)個(gè)數(shù)為1，理由見解析；(2)2個(gè)零點(diǎn)，理由見解析.【解析】（1）SKIPIF1<0在區(qū)間SKIPIF1<0上存在的極大值點(diǎn)個(gè)數(shù)為1，理由如下：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，SKIPIF1<0，故存在SKIPIF1<0，使得SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0處取得極大值，故SKIPIF1<0在區(qū)間SKIPIF1<0上存在的極大值點(diǎn)個(gè)數(shù)為1；（2）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，由（1）知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，而SKIPIF1<0，所以當(dāng)SK