新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專題訓(xùn)練專題10 導(dǎo)數(shù)的綜合運(yùn)用（解析版）

專題10導(dǎo)數(shù)的綜合運(yùn)用1、【2022年全國(guó)乙卷】已知x=x1和x=x2分別是函數(shù)f(x)=2ax?ex【答案】1【解析】解：f'因?yàn)閤1,x所以函數(shù)fx在?∞,x1所以當(dāng)x∈?∞,x1∪x若a>1時(shí)，當(dāng)x<0時(shí)，2lna?a故a>1不符合題意，若0<a<1時(shí)，則方程2lna?a即方程lna?ax即函數(shù)y=lna?a∵0<a<1,∴函數(shù)y=a又∵lna<0,∴y=lna?ax的圖象由指數(shù)函數(shù)設(shè)過(guò)原點(diǎn)且與函數(shù)y=gx的圖象相切的直線的切點(diǎn)為x則切線的斜率為g'故切線方程為y?ln則有?lna?a則切線的斜率為ln2因?yàn)楹瘮?shù)y=lna?a所以eln2a<又0<a<1，所以1e綜上所述，a的范圍為1e2、【2021年新高考2卷】已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0的兩條切線互相垂直，且分別交y軸于M，N兩點(diǎn)，則SKIPIF1<0取值范圍是_______．【答案】SKIPIF1<0【解析】由題意，SKIPIF1<0，則SKIPIF1<0，所以點(diǎn)SKIPIF1<0和點(diǎn)SKIPIF1<0，SKIPIF1<0,所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，同理SKIPIF1<0,所以SKIPIF1<0.故答案為：SKIPIF1<03、（2023年新課標(biāo)全國(guó)Ⅰ卷）1．已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．【詳解】（1）因?yàn)镾KIPIF1<0，定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由于SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得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時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；綜上：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.（2）方法一：由（1）得，SKIPIF1<0，要證SKIPIF1<0，即證SKIPIF1<0，即證SKIPIF1<0恒成立，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0；所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，則SKIPIF1<0恒成立，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，證畢.方法二：令SKIPIF1<0，則SKIPIF1<0，由于SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又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，故SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，所以要證SKIPIF1<0，即證SKIPIF1<0，即證SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0；所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，則SKIPIF1<0恒成立，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，證畢.4、（2023年新課標(biāo)全國(guó)Ⅱ卷）（1）證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；（2）已知函數(shù)SKIPIF1<0，若SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)，求a的取值范圍．【答案】（1）證明見(jiàn)詳解（2）SKIPIF1<0【詳解】（1）構(gòu)建SKIPIF1<0，則SKIPIF1<0對(duì)SKIPIF1<0恒成立，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，所以SKIPIF1<0；構(gòu)建SKIPIF1<0，則SKIPIF1<0，構(gòu)建SKIPIF1<0，則SKIPIF1<0對(duì)SKIPIF1<0恒成立，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，即SKIPIF1<0對(duì)SKIPIF1<0恒成立，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，所以SKIPIF1<0；綜上所述：SKIPIF1<0.（2）令SKIPIF1<0，解得SKIPIF1<0，即函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0在定義域內(nèi)單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0是SKIPIF1<0的極小值點(diǎn)，不合題意，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0因?yàn)镾KIPIF1<0，且SKIPIF1<0，所以函數(shù)SKIPIF1<0在定義域內(nèi)為偶函數(shù)，由題意可得：SKIPIF1<0，（i）當(dāng)SKIPIF1<0時(shí)，取SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，由（1）可得SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，結(jié)合偶函數(shù)的對(duì)稱性可知：SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0是SKIPIF1<0的極小值點(diǎn)，不合題意；（ⅱ）當(dāng)SKIPIF1<0時(shí)，取SKIPIF1<0，則SKIPIF1<0，由（1）可得SKIPIF1<0，構(gòu)建SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0對(duì)SKIPIF1<0恒成立，可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0內(nèi)存在唯一的零點(diǎn)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，結(jié)合偶函數(shù)的對(duì)稱性可知：SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)，符合題意；綜上所述：SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故a的取值范圍為SKIPIF1<05、（2023年全國(guó)乙卷數(shù)學(xué)（理））8．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)是否存在a，b，使得曲線SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，若存在，求a，b的值，若不存在，說(shuō)明理由.(3)若SKIPIF1<0在SKIPIF1<0存在極值，求a的取值范圍.【答案】(1)SKIPIF1<0；(2)存在SKIPIF1<0滿足題意，理由見(jiàn)解析.(3)SKIPIF1<0.【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，據(jù)此可得SKIPIF1<0，函數(shù)在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0.（2）由函數(shù)的解析式可得SKIPIF1<0，函數(shù)的定義域滿足SKIPIF1<0，即函數(shù)的定義域?yàn)镾KIPIF1<0，定義域關(guān)于直線SKIPIF1<0對(duì)稱，由題意可得SKIPIF1<0，由對(duì)稱性可知SKIPIF1<0，取SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，經(jīng)檢驗(yàn)SKIPIF1<0滿足題意，故SKIPIF1<0.即存在SKIPIF1<0滿足題意.（3）由函數(shù)的解析式可得SKIPIF1<0，由SKIPIF1<0在區(qū)間SKIPIF1<0存在極值點(diǎn)，則SKIPIF1<0在區(qū)間SKIPIF1<0上存在變號(hào)零點(diǎn);令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0存在極值點(diǎn)，等價(jià)于SKIPIF1<0在區(qū)間SKIPIF1<0上存在變號(hào)零點(diǎn)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，此時(shí)SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)零點(diǎn)，不合題意;當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，由于SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上無(wú)零點(diǎn)，不符合題意；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得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，故SKIPIF1<0的最小值為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，函數(shù)SKIPIF1<0在定義域內(nèi)單調(diào)遞增，SKIPIF1<0，據(jù)此可得SKIPIF1<0恒成立，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，故SKIPIF1<0，即SKIPIF1<0(取等條件為SKIPIF1<0)，所以SKIPIF1<0，SKIPIF1<0，且注意到SKIPIF1<0，根據(jù)零點(diǎn)存在性定理可知:SKIPIF1<0在區(qū)間SKIPIF1<0上存在唯一零點(diǎn)SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0單調(diào)遞減，注意到SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，從而有SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在變號(hào)零點(diǎn)，符合題意.綜合上面可知:實(shí)數(shù)SKIPIF1<0得取值范圍是SKIPIF1<0.6、【2022年全國(guó)甲卷】已知函數(shù)fx(1)若fx≥0，求(2)證明：若fx有兩個(gè)零點(diǎn)x1,【解析】(1)f(x)的定義域?yàn)?0,+∞f'(x)=(令f(x)=0,得x=1當(dāng)x∈(0,1),f當(dāng)x∈(1,+∞),f若f(x)≥0,則e+1?a≥0,即所以a的取值范圍為(?(2)由題知,f(x)一個(gè)零點(diǎn)小于1,一個(gè)零點(diǎn)大于1不妨設(shè)x要證x1x因?yàn)閤1,因?yàn)閒(x1即證e即證e下面證明x>1時(shí)，e設(shè)g(x)=e則g=(1?設(shè)φ(x)=所以φ(x)>φ(1)=e,而所以exx所以g(x)在(1,+∞即g(x)>g(1)=0,所以e令?(x)=?所以?(x)在(1,+∞即?(x)<?(1)=0,所以lnx?綜上,exx?x7、【2022年全國(guó)乙卷】已知函數(shù)f(1)當(dāng)a=1時(shí)，求曲線y=fx在點(diǎn)0,f(2)若fx在區(qū)間?1,0,0,+【解析】(1)f(x)的定義域?yàn)??1,+當(dāng)a=1時(shí),f(x)=ln(1+x)+xex所以曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=2x(2)f(x)=設(shè)g(x)=1°若a>0,當(dāng)x∈(?1,0),g(x)=e所以f(x)在(?1,0)上單調(diào)遞增,f(x)<f(0)=0故f(x)在(?1,0)上沒(méi)有零點(diǎn),不合題意2°若?1?a?0,當(dāng)x∈(0,+∞所以g(x)在(0,+∞)上單調(diào)遞增所以g(x)>g(0)=1+a?0所以f(x)在(0,+∞)故f(x)在(0,+∞3°若(1)當(dāng)x∈(0,+∞),則g'(x)=ex所以存在m∈(0,1),使得g(m)=0,即f當(dāng)x∈(0,m),f當(dāng)x∈(m,+∞所以當(dāng)x∈(0,m),f(x)<f(0)=0當(dāng)x→+所以f(x)在(m,+∞又(0,m)沒(méi)有零點(diǎn),即f(x)在(0,+∞(2)當(dāng)x∈(?1,0),g(x)=設(shè)?(x)=所以g'(x)在(?1,0)所以存在n∈(?1,0),使得g當(dāng)x∈(?1,n),g當(dāng)x∈(n,0),g'又g(?1)=所以存在t∈(?1,n),使得g(t)=0,即f當(dāng)x∈(?1,t),f(x)單調(diào)遞增,當(dāng)x∈(t,0),f(x)單調(diào)遞減有x→?1,f(x)→?而f(0)=0,所以當(dāng)x∈(t,0),f(x)>0所以f(x)在(?1,t)上有唯一零點(diǎn),(t,0)上無(wú)零點(diǎn)即f(x)在(?1,0)上有唯一零點(diǎn)所以a<?1,符合題意所以若f(x)在區(qū)間(?1,0),(0,+∞)各恰有一個(gè)零點(diǎn),求a題組一、函數(shù)的零點(diǎn)、極值點(diǎn)的綜合性問(wèn)題1-1、（2022·湖北·黃石市有色第一中學(xué)高三期末）（多選題）設(shè)函數(shù)fx=xlnA．不等式gx>0的解集為B．函數(shù)在0,e單調(diào)遞增，在e,+C．當(dāng)x∈1e,1D．若函數(shù)Fx=f【答案】ACD【解析】由題意得f'(x)=對(duì)于A：由g(x)=lnx+1x>0，可得lnx>?1對(duì)于B：g'(x)=1x?x?(所以當(dāng)時(shí)，g'(x)>0，函數(shù)為增函數(shù)，當(dāng)x∈(1,+∞)時(shí)，g'對(duì)于C：當(dāng)x∈1e,1時(shí)，若f所以xlnx?ln令?(x)=x則?'?″當(dāng)x∈1e,1時(shí)，?又?'(1)=0+1?1=0，所以?'所以?(x)=x又?(x)max=?1e所以當(dāng)x∈1e,1對(duì)于D：若函數(shù)Fx則F'(x)=lnx+1?2ax=0有兩個(gè)根，即令m(x)=lnx+1x所以當(dāng)時(shí)，m'(x)>0，函數(shù)m(x)當(dāng)x∈(1,+∞)時(shí)，m'又當(dāng)時(shí)，m(x)→?∞，當(dāng)時(shí)，m(x)→0，m(1)=1，所以2a∈(0,1)，解得a∈0,故選：ACD1-2、（2023·江蘇連云港·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0．(1)求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值；(2)若關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，求實(shí)數(shù)SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0；(2)SKIPIF1<0【分析】（1）利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性，即可求得函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值；（2）由SKIPIF1<0可得出SKIPIF1<0，令SKIPIF1<0，可知直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0的單調(diào)性與極值，數(shù)形結(jié)合可得出實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】（1）解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以，SKIPIF1<0.（2）解：函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由SKIPIF1<0可得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上為減函數(shù)，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以，SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，因?yàn)镾KIPIF1<0，SKIPIF1<0，則存在SKIPIF1<0，使得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.由題意可知，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，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)，故實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.1-3、（2022·河北深州市中學(xué)高三期末）已知函數(shù)SKIPIF1<0．(1)證明：函數(shù)SKIPIF1<0在SKIPIF1<0上存在唯一的零點(diǎn)；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為1，求a的值．【解析】(1)證明：∵SKIPIF1<0，∴SKIPIF1<0．∵SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，∴函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．又SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，故SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上存在唯一的零點(diǎn)．(2)解：由（1）可知存在唯一的SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0（*）．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴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)通增；∴SKIPIF1<0，由（*）式得SKIPIF1<0．∴SKIPIF1<0，顯然SKIPIF1<0是方程的解，又∵SKIPIF1<0是單調(diào)遞減函數(shù)，方程SKIPIF1<0有且僅有唯一的解SKIPIF1<0，把SKIPIF1<0代入（*）式，得SKIPIF1<0，∴SKIPIF1<0，即所求實(shí)數(shù)SKIPIF1<0的值為SKIPIF1<0題組二、利用導(dǎo)數(shù)研究不等式及證明問(wèn)題2-1、（2023·江蘇徐州·徐州市第七中學(xué)?？家荒＃┮阎瘮?shù)SKIPIF1<0.(1)若SKIPIF1<0且函數(shù)SKIPIF1<0在SKIPIF1<0上是單調(diào)遞增函數(shù)，求SKIPIF1<0的取值范圍；(2)設(shè)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0滿足SKIPIF1<0，證明：SKIPIF1<0.【答案】(1)SKIPIF1<0；(2)見(jiàn)解析【分析】(1)由題意可得SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，求導(dǎo)，分SKIPIF1<0、SKIPIF1<0討論SKIPIF1<0在SKIPIF1<0上恒成立即可；(2)由SKIPIF1<0可得SKIPIF1<0，由（1）知SKIPIF1<0，即有SKIPIF1<0，①，令SKIPIF1<0，求導(dǎo)得當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即有SKIPIF1<0，于是得以SKIPIF1<0，代入①式中化簡(jiǎn)即可得證.【詳解】（1）解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上是單調(diào)遞增函數(shù)，所以SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上恒成立，符合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0為單調(diào)遞增函數(shù)，所以存在唯一SKIPIF1<0使得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0遞減，即SKIPIF1<0，SKIPIF1<0，不符合題意；綜上所述SKIPIF1<0；（2）證明：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由（1）可知SKIPIF1<0是增函數(shù)，所以SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，移項(xiàng)得SKIPIF1<0，由（1）知SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，①設(shè)SKIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，代入①式中得到SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，命題得證.2-2、（2023·江蘇南通·統(tǒng)考一模）已知函數(shù)SKIPIF1<0和SKIPIF1<0有相同的最大值.(1)求實(shí)數(shù)SKIPIF1<0；(2)設(shè)直線SKIPIF1<0與兩條曲線SKIPIF1<0和SKIPIF1<0共有四個(gè)不同的交點(diǎn)，其橫坐標(biāo)分別為SKIPIF1<0，證明：SKIPIF1<0.【答案】(1)SKIPIF1<0；(2)見(jiàn)解析【分析】(1)利用導(dǎo)函數(shù)分別討論兩個(gè)函數(shù)的單調(diào)性和最值即可求解；(2)構(gòu)造函數(shù)SKIPIF1<0和SKIPIF1<0，利用導(dǎo)數(shù)和單調(diào)性討論函數(shù)的零點(diǎn)，結(jié)合函數(shù)SKIPIF1<0分類討論對(duì)應(yīng)方程根的個(gè)數(shù)和分布證明.【詳解】（1）SKIPIF1<0，令SKIPIF1<0.SKIPIF1<0有最大值，SKIPIF1<0且SKIPIF1<0在SKIPIF1<0上單調(diào)遞增SKIPIF1<0上單調(diào)遞減，SKIPIF1<0.SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0.（2）由SKIPIF1<0，由SKIPIF1<0，令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)遞增；SKIPIF1<0上單調(diào)遞減，SKIPIF1<0至多兩個(gè)零點(diǎn)，令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)遞增；SKIPIF1<0上單調(diào)遞減；SKIPIF1<0至多兩個(gè)零點(diǎn).令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，設(shè)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，SKIPIF1<0方程無(wú)解，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0方程有唯一解SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，注意到SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0對(duì)SKIPIF1<0恒成立，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上各有一個(gè)零點(diǎn)SKIPIF1<0，SKIPIF1<0示意圖如下注意到SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，因此SKIPIF1<0，即有SKIPIF1<0，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上各有一個(gè)零點(diǎn)SKIPIF1<0.且由SKIPIF1<0，而SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由SKIPIF1<0，由SKIPIF1<0，而SKIPIF1<0而SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，由SKIPIF1<0，于是得SKIPIF1<0，SKIPIF1<0，證畢2-3、（2023·江蘇南通·統(tǒng)考模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若函數(shù)SKIPIF1<0圖象恰與函數(shù)SKIPIF1<0圖象相切，求實(shí)數(shù)SKIPIF1<0的值；(2)若函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0，SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0，SKIPIF1<0，證明：SKIPIF1<0、SKIPIF1<0兩點(diǎn)連線的斜率SKIPIF1<0.【答案】(1)1；(2)證明見(jiàn)解析【分析】（1）設(shè)切點(diǎn)為SKIPIF1<0，結(jié)合導(dǎo)數(shù)的幾何意義求解即可；（2）由SKIPIF1<0有兩個(gè)極值點(diǎn)，可得SKIPIF1<0有兩個(gè)不等的正根SKIPIF1<0，且SKIPIF1<0，可得SKIPIF1<0，要證：SKIPIF1<0，即證SKIPIF1<0.令SKIPIF1<0證SKIPIF1<0，進(jìn)而構(gòu)造函數(shù)，再利用導(dǎo)數(shù)求解即可；【詳解】（1）設(shè)SKIPIF1<0與SKIPIF1<0切于SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.（2）解法一：由SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0有兩個(gè)極值點(diǎn)，SKIPIF1<0，即SKIPIF1<0有兩個(gè)不等的正根SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0SKIPIF1<0，要證：SKIPIF1<0，即證SKIPIF1<0.不妨設(shè)SKIPIF1<0，即證：SKIPIF1<0，即證：SKIPIF1<0，令SKIPIF1<0證SKIPIF1<0令SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上SKIPIF1<0，證畢!解法二：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.所以SKIPIF1<0，所以SKIPIF1<0的斜率SKIPIF1<0SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.不妨設(shè)SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，證畢!2-4、（2023·江蘇南京·南京市秦淮中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求a的取值范圍；(3)設(shè)SKIPIF1<0，證明：SKIPIF1<0．【答案】(1)SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0；(2)SKIPIF1<0；(3)見(jiàn)解析【分析】（1）求出SKIPIF1<0，討論其符號(hào)后可得SKIPIF1<0的單調(diào)性.（2）設(shè)SKIPIF1<0，求出SKIPIF1<0，先討論SKIPIF1<0時(shí)題設(shè)中的不等式不成立，再就SKIPIF1<0結(jié)合放縮法討論SKIPIF1<0符號(hào)，最后就SKIPIF1<0結(jié)合放縮法討論SKIPIF1<0的范圍后可得參數(shù)的取值范圍.（3）由（2）可得SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，從而可得SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，結(jié)合裂項(xiàng)相消法可證題設(shè)中的不等式.【詳解】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0.（2）設(shè)SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0為連續(xù)不間斷函數(shù)，故存在SKIPIF1<0，使得SKIPIF1<0，總有SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0為增函數(shù)，故SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0為增函數(shù)，故SKIPIF1<0，與題設(shè)矛盾.若SKIPIF1<0，則SKIPIF1<0，下證：對(duì)任意SKIPIF1<0，總有SKIPIF1<0成立，證明：設(shè)SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為減函數(shù)，故SKIPIF1<0即SKIPIF1<0成立.由上述不等式有SKIPIF1<0，故SKIPIF1<0總成立，即SKIPIF1<0在SKIPIF1<0上為減函數(shù)，所以SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，

所以SKIPIF1<0在SKIPIF1<0上為減函數(shù)，所以SKIPIF1<0.綜上，SKIPIF1<0.（3）取SKIPIF1<0，則SKIPIF1<0，總有SKIPIF1<0成立，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0即SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立.所以對(duì)任意的SKIPIF1<0，有SKIPIF1<0，整理得到：SKIPIF1<0，故SKIPIF1<0SKIPIF1<0，故不等式成立.題組三、利用導(dǎo)數(shù)研究含參問(wèn)題3-1、（2023·江蘇泰州·泰州中學(xué)校考一模）已知函數(shù)SKIPIF1<0（SKIPIF1<0為自然對(duì)數(shù)的底數(shù)）.(1)若不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)若不等式SKIPIF1<0在SKIPIF1<0上恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0【分析】（1）先判斷SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，再利用單調(diào)性解不等式得解；（2）等價(jià)于SKIPIF1<0對(duì)SKIPIF1<0恒成立，令SKIPIF1<0，利用二次求導(dǎo)對(duì)SKIPIF1<0分類討論求函數(shù)SKIPIF1<0的最大值得解.【詳解】（1）解：SKIPIF1<0，由復(fù)合函數(shù)的單調(diào)性原理得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0.（2）解：SKIPIF1<0對(duì)SKIPIF1<0恒成立令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，若SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0符合題意.若SKIPIF1<0，即SKIPIF1<0時(shí)，（i）若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0這與題設(shè)矛盾，舍去.（ii）若SKIPIF1<0，則存在SKIPIF1<0使SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，此時(shí)SKIPIF1<0這與題設(shè)也矛盾，舍去.綜上：實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<03-2、（2023·江蘇南京·?？家荒＃┮阎瘮?shù)SKIPIF1<0(其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)).（1）當(dāng)SKIPIF1<0時(shí)，求證：函數(shù)SKIPIF1<0圖象上任意一點(diǎn)處的切線斜率均大于SKIPIF1<0；（2）若SKIPIF1<0對(duì)于任意的SKIPIF1<0，SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】（1）見(jiàn)解析；（2）SKIPIF1<0.【分析】（1）代入SKIPIF1<0的值，求出函數(shù)的導(dǎo)數(shù)，結(jié)合函數(shù)的單調(diào)性證明即可；（2）求出SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，得到SKIPIF1<0，得到SKIPIF1<0，再根據(jù)SKIPIF1<0得到結(jié)論成立即可確定SKIPIF1<0的取值范圍．【詳解】解：（1）證明：SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，故SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0的最小值是SKIPIF1<0，即SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，故函數(shù)SKIPIF1<0圖象上任意一點(diǎn)處的切線斜率均大于SKIPIF1<0；（2）先證對(duì)任意SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，故SKIPIF1<0在區(qū)間SKIPIF1<0遞增，在SKIPIF1<0遞減，故SKIPIF1<0，故SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，故SKIPIF1<0在區(qū)間SKIPIF1<0遞減，在區(qū)間SKIPIF1<0遞增，故SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0遞增，故SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0對(duì)于任意SKIPIF1<0，SKIPIF1<0恒成立，SKIPIF1<0，故SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，即SKIPIF1<0對(duì)于任意的SKIPIF1<0，SKIPIF1<0恒成立，綜上：SKIPIF1<0的取值范圍是SKIPIF1<0.3-3、（2023·云南曲靖·統(tǒng)考一模）已知函數(shù)SKIPIF1<0的圖像與直線l：SKIPIF1<0相切于點(diǎn)SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的圖像在點(diǎn)SKIPIF1<0處的切線在x軸上的截距；(2)求c與a的函數(shù)關(guān)系SKIPIF1<0；(3)當(dāng)a為函數(shù)g（a）的零點(diǎn)時(shí)，若對(duì)任意SKIPIF1<0，不等式SKIPIF1<0