新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)訓(xùn)練專題09 利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)（含解析）

專題09利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)1、【2022年新高考2卷】曲線y=ln【答案】

y=1e【解析】因?yàn)閥=ln當(dāng)x>0時(shí)y=lnx，設(shè)切點(diǎn)為x0,lnx0又切線過(guò)坐標(biāo)原點(diǎn)，所以?lnx0=1x0當(dāng)x<0時(shí)y=ln?x，設(shè)切點(diǎn)為x1,ln?x又切線過(guò)坐標(biāo)原點(diǎn)，所以?ln?x1=1x故答案為：y=1e2、【2022年新高考1卷】已知函數(shù)f(x)=xA．f(x)有兩個(gè)極值點(diǎn) B．f(x)有三個(gè)零點(diǎn)C．點(diǎn)(0,1)是曲線y=f(x)的對(duì)稱中心 D．直線y=2x是曲線y=f(x)的切線【答案】AC【解析】由題，f'x=3x2?1，令令f'(x)<0得所以f(x)在(?33,33所以x=±3因f(?33)=1+23所以，函數(shù)fx在?當(dāng)x≥33時(shí)，fx≥f3綜上所述，函數(shù)f(x)有一個(gè)零點(diǎn)，故B錯(cuò)誤；令?(x)=x3?x，該函數(shù)的定義域?yàn)镽則?(x)是奇函數(shù)，(0,0)是?(x)的對(duì)稱中心，將?(x)的圖象向上移動(dòng)一個(gè)單位得到f(x)的圖象，所以點(diǎn)(0,1)是曲線y=f(x)的對(duì)稱中心，故C正確；令f'x=3x2當(dāng)切點(diǎn)為(1,1)時(shí)，切線方程為y=2x?1，當(dāng)切點(diǎn)為(?1,1)時(shí)，切線方程為y=2x+3，故D錯(cuò)誤.故選：AC.

3、【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1e4、（2021年全國(guó)高考乙卷數(shù)學(xué)（文）試題）設(shè)SKIPIF1<0，若SKIPIF1<0為函數(shù)SKIPIF1<0的極大值點(diǎn)，則（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】若SKIPIF1<0，則SKIPIF1<0為單調(diào)函數(shù)，無(wú)極值點(diǎn)，不符合題意，故SKIPIF1<0.依題意，SKIPIF1<0為函數(shù)SKIPIF1<0的極大值點(diǎn)，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，SKIPIF1<0，畫出SKIPIF1<0的圖象如下圖所示：由圖可知SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0時(shí)，SKIPIF1<0，畫出SKIPIF1<0的圖象如下圖所示：由圖可知SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.綜上所述，SKIPIF1<0成立.故選：D5、（2021年全國(guó)新高考Ⅰ卷數(shù)學(xué)試題）若過(guò)點(diǎn)SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線，則（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】在曲線SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，所以，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，由題意可知，點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，可得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞減，所以，SKIPIF1<0，由題意可知，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如下圖所示：由圖可知，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個(gè)交點(diǎn).故選：D.6、（2021年全國(guó)高考甲卷數(shù)學(xué)（理）試題）曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為_(kāi)_________．【答案】SKIPIF1<0【解析】由題，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故點(diǎn)在曲線上．求導(dǎo)得：SKIPIF1<0，所以SKIPIF1<0．故切線方程為SKIPIF1<0．故答案為：SKIPIF1<0．7、（2021年全國(guó)新高考Ⅰ卷數(shù)學(xué)試題）函數(shù)SKIPIF1<0的最小值為_(kāi)_____.【答案】1【解析】由題設(shè)知：SKIPIF1<0定義域?yàn)镾KIPIF1<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，有SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增；又SKIPIF1<0在各分段的界點(diǎn)處連續(xù)，∴綜上有：SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；∴SKIPIF1<0故答案為：1.8、（2020年全國(guó)統(tǒng)一高考數(shù)學(xué)試卷（理科）（新課標(biāo)Ⅰ））函數(shù)SKIPIF1<0的圖像在點(diǎn)SKIPIF1<0處的切線方程為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因此，所求切線的方程為SKIPIF1<0，即SKIPIF1<0.故選：B.9、（2020年全國(guó)統(tǒng)一高考數(shù)學(xué)試卷（理科）（新課標(biāo)Ⅲ））若直線l與曲線y=SKIPIF1<0和x2+y2=SKIPIF1<0都相切，則l的方程為（）A．y=2x+1 B．y=2x+SKIPIF1<0 C．y=SKIPIF1<0x+1 D．y=SKIPIF1<0x+SKIPIF1<0【答案】D【解析】設(shè)直線SKIPIF1<0在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，函數(shù)SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，則直線SKIPIF1<0的斜率SKIPIF1<0，設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，由于直線SKIPIF1<0與圓SKIPIF1<0相切，則SKIPIF1<0，兩邊平方并整理得SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0（舍），則直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.故選：D.10、（2020年全國(guó)統(tǒng)一高考數(shù)學(xué)試卷（文科）（新課標(biāo)Ⅰ））曲線SKIPIF1<0的一條切線的斜率為2，則該切線的方程為_(kāi)_____________.【答案】SKIPIF1<0【解析】設(shè)切線的切點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，所以切點(diǎn)坐標(biāo)為SKIPIF1<0，所求的切線方程為SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.題組一、函數(shù)圖像的切線問(wèn)題1-1、（2022·廣東清遠(yuǎn)·高三期末）已知曲線f(x)=(ax+b)ex在點(diǎn)(0,2)處的切線方程為x+y?2=0，則【答案】SKIPIF1<0【解析】因?yàn)閒(x)=(ax+b)ex，所以f'(x)=(ax+a+b)ex，所以故答案為：SKIPIF1<01-2、（2022·廣東潮州·高三期末）曲線y=lnx+ax與直線y=2x?1相切，則【答案】1【解析】由題意，函數(shù)y=lnx+ax，可得設(shè)切點(diǎn)為Px0,因?yàn)榍€y=lnx+ax與直線y=2x?1相切，可得1x0又由y0=lnx0+a聯(lián)立①②，可得x0故答案為：11-3、（2022·山東臨沂·高三期末）已知函數(shù)f(x)=lnx+x2?1，則SKIPIF1<0在SKIPIF1<0處的切線方程為_(kāi)_______．【答案】y=3x?3【解析】因?yàn)楹瘮?shù)f(x)=lnf(1)=0，則f(x)在x=1處的切線的斜率k=f故切線方程為y?0=3(x?1)，即：y=3x?3.故答案為：y=3x?3.1-4、（2022·江蘇如皋·高三期末）已知函數(shù)f(x)＝x3＋ax2－x的圖象在點(diǎn)A(1，f(1))處的切線方程為y＝4x－3，則函數(shù)y＝f(x)的極大值為（）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．－1【答案】A【解析】由由題意得SKIPIF1<0,故SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0時(shí)取得極大值為SKIPIF1<0，故選：A.題組二、利用導(dǎo)數(shù)研究函數(shù)的最值、極值與零點(diǎn)問(wèn)題2-1、（2022·江蘇蘇州·高三期末）已知函數(shù)SKIPIF1<0，則（）A．SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上均有極值B．SKIPIF1<0，使得函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)極值C．SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上有且僅有一個(gè)零點(diǎn)D．SKIPIF1<0，使得函數(shù)SKIPIF1<0在SKIPIF1<0上有兩個(gè)零點(diǎn)【答案】BC【解析】SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0無(wú)極值，A錯(cuò)，B對(duì)．SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0有且僅有一個(gè)零點(diǎn)．SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0有且僅有一個(gè)零點(diǎn)．SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0或0，SKIPIF1<0在SKIPIF1<0，SKIPIF1<0SKIPIF1<0．SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0有且僅有一個(gè)零點(diǎn)．SKIPIF1<0，SKIPIF1<0有且僅有一個(gè)零點(diǎn)，C對(duì)，D錯(cuò)．故選：BC2-2、（2022·江蘇海門·高三期末）已知函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A．(0，SKIPIF1<0) B．[0，SKIPIF1<0) C．[0，SKIPIF1<0] D．(0，SKIPIF1<0)【答案】A【解析】SKIPIF1<0有三個(gè)零點(diǎn)，即方程SKIPIF1<0有三個(gè)根，不妨令SKIPIF1<0,則SKIPIF1<0SKIPIF1<0SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立.當(dāng)SKIPIF1<0趨近于負(fù)無(wú)窮時(shí)，SKIPIF1<0趨近于正無(wú)窮；SKIPIF1<0趨近于正無(wú)窮時(shí)，SKIPIF1<0趨近于SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，滿足題意.故選：A.2-3、（2022·江蘇蘇州·高三期末）設(shè)點(diǎn)SKIPIF1<0是曲線y=x?32lnx上的任意一點(diǎn)，則SKIPIF1<0到直線y=?x的最小距離是__________．【答案】2【解析】由題意得在SKIPIF1<0點(diǎn)的切線與直線y=?x平行設(shè)曲線y=x?32ln由y=?x的斜率為?1，y則由y'x=x0切點(diǎn)(1,1)到x+y=0的距離d=2．故答案為：2-4、（2022·湖北省鄂州高中高三期末）若不同兩點(diǎn)SKIPIF1<0、SKIPIF1<0均在函數(shù)SKIPIF1<0的圖象上，且點(diǎn)SKIPIF1<0、SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱，則稱SKIPIF1<0是函數(shù)SKIPIF1<0的一個(gè)“匹配點(diǎn)對(duì)”（點(diǎn)對(duì)SKIPIF1<0與SKIPIF1<0視為同一個(gè)“匹配點(diǎn)對(duì)”）.已知SKIPIF1<0恰有兩個(gè)“匹配點(diǎn)對(duì)”，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】函數(shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱的圖象所對(duì)應(yīng)的函數(shù)為SKIPIF1<0，SKIPIF1<0的圖象上恰好有兩個(gè)“匹配點(diǎn)對(duì)”等價(jià)于函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0有兩個(gè)交點(diǎn)，即方程SKIPIF1<0SKIPIF1<0有兩個(gè)不等式的正實(shí)數(shù)根，即SKIPIF1<0SKIPIF1<0有兩個(gè)不等式的正實(shí)數(shù)根，即轉(zhuǎn)化為函數(shù)SKIPIF1<0SKIPIF1<0圖象與函數(shù)SKIPIF1<0圖象有2個(gè)交點(diǎn).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.且SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0所以SKIPIF1<0所以SKIPIF1<0SKIPIF1<0圖象與函數(shù)SKIPIF1<0圖象有2個(gè)交點(diǎn).則SKIPIF1<0，解得SKIPIF1<0.故選：B2-5、（2022·山東省淄博實(shí)驗(yàn)中學(xué)高三期末）已知函數(shù)SKIPIF1<0有三個(gè)不同的零點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的值為（）A．3 B．4 C．9 D．16【答案】C【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0有三個(gè)不同的零點(diǎn)SKIPIF1<0.令SKIPIF1<0，SKIPIF1<0在SKIPIF1<0遞增，在SKIPIF1<0上遞減，SKIPIF1<0.SKIPIF1<0時(shí)，SKIPIF1<0.令SKIPIF1<0，SKIPIF1<0必有兩個(gè)根SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0有一解SKIPIF1<0，SKIPIF1<0有兩解SKIPIF1<0，且SKIPIF1<0，故SKIPIF1<0SKIPIF1<0SKIPIF1<0.故選：C2-6、（2022·山東棗莊·高三期末）已知函數(shù)fx=a?2x+1,x≤1A．?12 B．【答案】BD【解析】因?yàn)楹瘮?shù)y=x3與函數(shù)交于點(diǎn)(1,1)由函數(shù)圖象的性質(zhì)得函數(shù)y=(x?a)3與在(1,+由題意，函數(shù)f(x)={a?|2x+1|,x≤1(x?a)3若時(shí)，y=f(x)?x恰有兩個(gè)零點(diǎn)時(shí)，如圖（1）所示，則滿足{a?|2×(?12)+1|>?若時(shí)，y=f(x)?x恰有一個(gè)零點(diǎn)，在時(shí)，y=f(x)?x恰有一個(gè)零點(diǎn)，則{a?|2×(?12)+1|=?12結(jié)合選項(xiàng)，可得的可能取值為?14和6.題組三、利用導(dǎo)數(shù)研究函數(shù)性質(zhì)的綜合性問(wèn)題3-1、（2022·江蘇通州·高三期末）（多選題）已知函數(shù)f(x)＝ekx，g(x)＝SKIPIF1<0，其中k≠0，則（）A．若點(diǎn)P(a，b)在f(x)的圖象上，則點(diǎn)Q(b，a)在g(x)的圖象上B．當(dāng)k＝e時(shí)，設(shè)點(diǎn)A，B分別在f(x)，g(x)的圖象上，則|AB|的最小值為SKIPIF1<0C．當(dāng)k＝1時(shí)，函數(shù)F(x)＝f(x)－g(x)的最小值小于SKIPIF1<0D．當(dāng)k＝－2e時(shí)，函數(shù)G(x)＝f(x)－g(x)有3個(gè)零點(diǎn)【答案】ACD【解析】由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0是SKIPIF1<0的反函數(shù)，它們的圖象關(guān)于直線SKIPIF1<0對(duì)稱，A正確；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0的與直線SKIPIF1<0平行的切線的切點(diǎn)是SKIPIF1<0，SKIPIF1<0到直線SKIPIF1<0的距離是SKIPIF1<0，所以SKIPIF1<0，B錯(cuò)；SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0是增函數(shù)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0，即在SKIPIF1<0上存在唯一零點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，由對(duì)勾函數(shù)知SKIPIF1<0在SKIPIF1<0上是減函數(shù)，SKIPIF1<0，所以SKIPIF1<0，C正確；SKIPIF1<0時(shí)，SKIPIF1<0是減函數(shù)，SKIPIF1<0也是減函數(shù)，它們互為反函數(shù)，作出它們的圖象，如圖，易知它們有一個(gè)交點(diǎn)在直線SKIPIF1<0上，在右側(cè)，SKIPIF1<0的圖象在SKIPIF1<0軸上方，而SKIPIF1<0的圖象在SKIPIF1<0處穿過(guò)SKIPIF1<0軸過(guò)渡到SKIPIF1<0軸下方，之間它們有一個(gè)交點(diǎn)，根據(jù)對(duì)稱性，在左上方，靠近SKIPIF1<0處也有一個(gè)交點(diǎn)，因此函數(shù)SKIPIF1<0SKIPIF1<0與SKIPIF1<0SKIPIF1<0的圖象有3個(gè)交點(diǎn)，所以SKIPIF1<0有3個(gè)零點(diǎn)，D正確．故選：ACD．3-2、（2022·廣東·鐵一中學(xué)高三期末）已知直線SKIPIF1<0恒在函數(shù)SKIPIF1<0的圖象的上方，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】很明顯SKIPIF1<0，否則SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞減，且SKIPIF1<0時(shí)SKIPIF1<0，而SKIPIF1<0當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，不合題意，SKIPIF1<0時(shí)函數(shù)SKIPIF1<0為常函數(shù)，而SKIPIF1<0當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，不合題意，當(dāng)SKIPIF1<0時(shí)，構(gòu)造函數(shù)SKIPIF1<0，由題意可知SKIPIF1<0恒成立，注意到：SKIPIF1<0，據(jù)此可得，函數(shù)在區(qū)間SKIPIF1<0上的單調(diào)遞減，在區(qū)間SKIPIF1<0上單調(diào)遞增，則：SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，則SKIPIF1<0，還是SKIPIF1<0在SKIPIF1<0處取得極值，結(jié)合題意可知：SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0.故選：A.3-3、（2022·湖南常德·高三期末）若函數(shù)SKIPIF1<0為定義在R上的奇函數(shù)，SKIPIF1<0為SKIPIF1<0的導(dǎo)函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則不等式SKIPIF1<0的解集為（）A．SKIPIF1<0 B．SKIPIF1<0C．（0，2） D．SKIPIF1<0【答案】D【解析】令SKIPIF1<0，則SKIPIF1<0，因?yàn)?，?dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0為定義在R上的奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，所以不等式SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)镾KIPIF1<0為定義在R上的奇函數(shù)，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0不滿足SKIPIF1<0，綜上，不等式的解集為SKIPIF1<0故選：D3-4、（2022·廣東揭陽(yáng)·高三期末）已知函數(shù)SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0可作兩條直線與函數(shù)SKIPIF1<0相切，則下列結(jié)論正確的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0的最大值為2 D．SKIPIF1<0【答案】B【解析】設(shè)切點(diǎn)為SKIPIF1<0，又SKIPIF1<0，則切線的斜率SKIPIF1<0又SKIPIF1<0，即有SKIPIF1<0，整理得SKIPIF1<0，由于過(guò)點(diǎn)SKIPIF1<0可作兩條直線與函數(shù)SKIPIF1<0相切所以關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不同的正根，設(shè)為SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，故B正確，A錯(cuò)誤，對(duì)于C，取SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0的最大值不可能為2，故C錯(cuò)誤，對(duì)于D，取SKIPIF1<0，則SKIPIF1<0，故D錯(cuò)誤.故選：B.3-5、（2022·湖北襄陽(yáng)·高三期末）關(guān)于函數(shù)SKIPIF1<0有下列四個(gè)結(jié)論：①函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱；②函數(shù)SKIPIF1<0在定義域內(nèi)是增函數(shù)；③曲線SKIPIF1<0在SKIPIF1<0處的切線為SKIPIF1<0；④函數(shù)SKIPIF1<0無(wú)零點(diǎn)；其中正確結(jié)論的個(gè)數(shù)為（）A．4 B．3 C．2 D．1【答案】C【解析】對(duì)于函數(shù)SKIPIF1<0，有SKIPIF1<0，故SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱，所以①不正確；SKIPIF1<0，而SKIPIF1<0，當(dāng)且僅當(dāng)x=0時(shí)取等號(hào)，所以SKIPIF1<0，故SKIPIF1<0在定義域內(nèi)是增函數(shù)，故②正確；SKIPIF1<0，故線SKIPIF1<0在SKIPIF1<0處的切線為SKIPIF1<0，即SKIPIF1<0，故③正確；由SKIPIF1<0可知，SKIPIF1<0在（-1,0）之間有零點(diǎn)，故④錯(cuò)誤，故選：C.1、（2022·廣東揭陽(yáng)·高三期末）已知函數(shù)fx=ex?sinx【答案】SKIPIF1<0【解析】對(duì)函數(shù)SKIPIF1<0求導(dǎo)可得f'x=exsinx+cosx，把SKIPIF1<0代入可得f則切線方程的斜率SKIPIF1<0.又因?yàn)镾KIPIF1<0，所以切點(diǎn)為SKIPIF1<0，從而可得切線方程為SKIPIF1<0.故答案為：SKIPIF1<0.2