新高考數(shù)學(xué)二輪復(fù)習(xí)專題培優(yōu)練習(xí)專題10 導(dǎo)數(shù)解答題分類練（解析版）

專題10導(dǎo)數(shù)解答題分類練一、曲線的切線問題1.（2023屆河南省開封市通許縣高三沖刺卷）已知函數(shù)SKIPIF1<0．(1)若函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0相切，求實(shí)數(shù)SKIPIF1<0的值；(2)若函數(shù)SKIPIF1<0有且只有一個(gè)零點(diǎn)，求實(shí)數(shù)SKIPIF1<0的取值范圍．【解析】（1）設(shè)直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象相切于點(diǎn)SKIPIF1<0，因?yàn)镾KIPIF1<0,所以SKIPIF1<0，由②③可得SKIPIF1<0④，易知SKIPIF1<0.由①得SKIPIF1<0，代入④可得SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.故SKIPIF1<0.（2）令SKIPIF1<0，可得SKIPIF1<0，由題意可得SKIPIF1<0只有一個(gè)根.易知SKIPIF1<0不是方程SKIPIF1<0的根，所以SKIPIF1<0，所以由SKIPIF1<0，可得SKIPIF1<0.設(shè)SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的圖象只有一個(gè)交點(diǎn).SKIPIF1<0SKIPIF1<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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增.設(shè)SKIPIF1<0，則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.所以SKIPIF1<0.所以SKIPIF1<0.又SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，畫出函數(shù)SKIPIF1<0的圖象如圖所示：

由圖可知，若SKIPIF1<0與SKIPIF1<0的圖象只有一個(gè)交點(diǎn)，則SKIPIF1<0.所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.2．（2024屆福建省莆田哲理中學(xué)高三上學(xué)期月考）已知函數(shù)SKIPIF1<0，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)試討論函數(shù)SKIPIF1<0的單調(diào)性.【解析】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，切點(diǎn)為SKIPIF1<0又因?yàn)镾KIPIF1<0所以SKIPIF1<0，即SKIPIF1<0所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，即SKIPIF1<0.（2）因?yàn)镾KIPIF1<0，SKIPIF1<0，所以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)，令SKIPIF1<0，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，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，在SKIPIF1<0上單調(diào)遞增3．（2024屆重慶市第一中學(xué)高三上學(xué)期開學(xué)考）已知函數(shù)SKIPIF1<0.(1)設(shè)SKIPIF1<0，經(jīng)過點(diǎn)SKIPIF1<0作函數(shù)SKIPIF1<0圖像的切線，求切線的方程；(2)若函數(shù)SKIPIF1<0有極大值，無最大值，求實(shí)數(shù)SKIPIF1<0的取值范圍.【解析】（1）SKIPIF1<0時(shí)SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，則切線斜率為SKIPIF1<0，切線方程：SKIPIF1<0，將點(diǎn)SKIPIF1<0帶入得：SKIPIF1<0，此時(shí)斜率SKIPIF1<0，所以切線方程為SKIPIF1<0.（2）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，令SKIPIF1<0，則SKIPIF1<0（1）當(dāng)SKIPIF1<0時(shí)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，注意到SKIPIF1<0時(shí)，SKIPIF1<0，注意到SKIPIF1<0時(shí)，SKIPIF1<0，故存在SKIPIF1<0，使得SKIPIF1<0，在SKIPIF1<0時(shí)SKIPIF1<0單調(diào)遞減，在SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，函數(shù)SKIPIF1<0有極小值，無極大值，不符合題意.（2）當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減.當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，無極值和最值.若SKIPIF1<0，即SKIPIF1<0，此時(shí)存在SKIPIF1<0，使得SKIPIF1<0，且在SKIPIF1<0有SKIPIF1<0單調(diào)遞減；在SKIPIF1<0有SKIPIF1<0單調(diào)遞增，此時(shí)SKIPIF1<0為SKIPIF1<0的極大值.注意到SKIPIF1<0時(shí)SKIPIF1<0，要使SKIPIF1<0無最大值，則還應(yīng)滿足SKIPIF1<0，即SKIPIF1<0，同時(shí)SKIPIF1<0，帶入SKIPIF1<0整理得SKIPIF1<0.由于SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0單調(diào)遞減，故SKIPIF1<0，即SKIPIF1<0，綜上實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.4.（2024屆江蘇省南通市高三上學(xué)期質(zhì)量監(jiān)測(cè)）已知函數(shù)SKIPIF1<0的極小值為SKIPIF1<0，其導(dǎo)函數(shù)SKIPIF1<0的圖象經(jīng)過SKIPIF1<0，SKIPIF1<0兩點(diǎn).(1)求SKIPIF1<0的解析式；(2)若曲線SKIPIF1<0恰有三條過點(diǎn)SKIPIF1<0的切線，求實(shí)數(shù)SKIPIF1<0的取值范圍.【解析】（1）SKIPIF1<0，因?yàn)镾KIPIF1<0，且SKIPIF1<0的圖象經(jīng)過SKIPIF1<0，SKIPIF1<0兩點(diǎn).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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增.所以SKIPIF1<0在SKIPIF1<0處取得極小值，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，解方程組SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.（2）設(shè)切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以切線方程為SKIPIF1<0，將SKIPIF1<0代入上式，得SKIPIF1<0.因?yàn)榍€SKIPIF1<0恰有三條過點(diǎn)SKIPIF1<0的切線，所以方程SKIPIF1<0有三個(gè)不同實(shí)數(shù)解.記SKIPIF1<0，則導(dǎo)函數(shù)SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或1.列表：SKIPIF1<0SKIPIF1<00SKIPIF1<01SKIPIF1<0SKIPIF1<0+0-0+SKIPIF1<0↗極大↘極小↗所以SKIPIF1<0的極大值為SKIPIF1<0，SKIPIF1<0的極小值為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故SKIPIF1<0的取值范圍是SKIPIF1<0.5.（2024屆上海市華東師范大學(xué)第二附屬中學(xué)高三上學(xué)期質(zhì)量調(diào)研）設(shè)函數(shù)SKIPIF1<0的定義域?yàn)殚_區(qū)間SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0在SKIPIF1<0處的切線SKIPIF1<0與SKIPIF1<0的圖象只有唯一的公共點(diǎn)，則稱切線SKIPIF1<0是SKIPIF1<0的一條“SKIPIF1<0切線”.(1)判斷函數(shù)SKIPIF1<0是否存在“SKIPIF1<0切線”，若存在，請(qǐng)寫出一條“SKIPIF1<0切線”的方程，若不存在，請(qǐng)說明理由；(2)設(shè)SKIPIF1<0，若對(duì)任意正實(shí)數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0都存在“SKIPIF1<0切線”，求實(shí)數(shù)SKIPIF1<0的取值范圍；(3)已知實(shí)數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0，求證：函數(shù)SKIPIF1<0存在無窮多條“SKIPIF1<0切線”，且至少一條“SKIPIF1<0切線”的切點(diǎn)的橫坐標(biāo)不超過SKIPIF1<0.【解析】（1）記SKIPIF1<0，則SKIPIF1<0取SKIPIF1<0，SKIPIF1<0，切線方程為SKIPIF1<0.與函數(shù)SKIPIF1<0聯(lián)立，得SKIPIF1<0.記SKIPIF1<0，則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上嚴(yán)格增，在SKIPIF1<0上嚴(yán)格減，SKIPIF1<0，故函數(shù)SKIPIF1<0只有一個(gè)零點(diǎn)SKIPIF1<0,故SKIPIF1<0是一條“SKIPIF1<0切線”:（2）SKIPIF1<0設(shè)點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0的圖像上,點(diǎn)SKIPIF1<0處的切線為SKIPIF1<0，與SKIPIF1<0聯(lián)立得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0（*）由題意得直線SKIPIF1<0為“SKIPIF1<0切線”，故方程（*）在SKIPIF1<0上有且僅有一解SKIPIF1<0則SKIPIF1<0或SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0是方程（*）的唯一解（此時(shí)有無數(shù)條“SKIPIF1<0切線”切點(diǎn)橫坐標(biāo)為SKIPIF1<0上的任意值）.若SKIPIF1<0，則SKIPIF1<0（此時(shí)只有一條“SKIPIF1<0切線”，切點(diǎn)的橫坐標(biāo)為SKIPIF1<0）或SKIPIF1<0（此時(shí)有無數(shù)條“SKIPIF1<0切線”,切點(diǎn)橫坐標(biāo)為SKIPIF1<0上的任意值）綜上，SKIPIF1<0.（3）證明：SKIPIF1<0，將點(diǎn)SKIPIF1<0處的切線SKIPIF1<0的方程與SKIPIF1<0聯(lián)立得SKIPIF1<0，記SKIPIF1<0，則直線SKIPIF1<0為“SKIPIF1<0切線”SKIPIF1<0函數(shù)SKIPIF1<0有且僅有一個(gè)零點(diǎn)SKIPIF1<0（此時(shí)，一個(gè)SKIPIF1<0對(duì)應(yīng)一條“SKIPIF1<0切線”），顯然SKIPIF1<0是SKIPIF1<0的零點(diǎn)，故只要SKIPIF1<0沒其他零點(diǎn).此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0即SKIPIF1<0，SKIPIF1<0恒成立，此時(shí)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故此時(shí)SKIPIF1<0為SKIPIF1<0唯一的極小值點(diǎn)（也是最小值點(diǎn)），而SKIPIF1<0，故SKIPIF1<0無其他零點(diǎn)，故直線SKIPIF1<0為“SKIPIF1<0切線”，因SKIPIF1<0的任意性，故函數(shù)SKIPIF1<0存在無窮多條“SKIPIF1<0切線”，有一條“SKIPIF1<0切線”的切點(diǎn)的橫坐標(biāo)為SKIPIF1<0.二、含參函數(shù)的單調(diào)性問題6.（2024屆山東省泰安市肥城市高三上學(xué)期9月月考）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.【解析】（1）函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，可得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，可知SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，可得SKIPIF1<0時(shí)，有SKIPIF1<0，SKIPIF1<0時(shí)，有SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在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時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.（2）證明：當(dāng)SKIPIF1<0時(shí)，要證SKIPIF1<0成立，只需證SKIPIF1<0成立，只需證SKIPIF1<0即可.因?yàn)镾KIPIF1<0，由（1）知，SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0時(shí)，有SKIPIF1<0；SKIPIF1<0時(shí)，有SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，可知SKIPIF1<0，則有SKIPIF1<0，所以有SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立.7.（2024屆江西省豐城厚一學(xué)校高三上學(xué)期9月月考）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，使得SKIPIF1<0.【解析】（1）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，求導(dǎo)得SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，恒有SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0單調(diào)遞減，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0單調(diào)遞減，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0單調(diào)遞增；所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.（2）由（1）知，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值SKIPIF1<0，于是當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，使得SKIPIF1<0成立，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立，令函數(shù)SKIPIF1<0，求導(dǎo)得SKIPIF1<0，令SKIPIF1<0，求導(dǎo)得SKIPIF1<0，于是函數(shù)SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，使得SKIPIF1<0.8.（2024屆四川省仁壽第一中學(xué)校高三上學(xué)期9月月考）已知a為實(shí)常數(shù)，函數(shù)SKIPIF1<0（其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)）(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)設(shè)SKIPIF1<0，函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，求實(shí)數(shù)a的取值范圍．【解析】（1）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時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；綜上：SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上是單調(diào)遞增；當(dāng)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）由（1）得，SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0遞增，不可能有2個(gè)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0遞減，在SKIPIF1<0遞增，函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，∴函數(shù)SKIPIF1<0只有1個(gè)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0遞減，在SKIPIF1<0遞增，SKIPIF1<0為函數(shù)SKIPIF1<0的最小值，令SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0遞增，且SKIPIF1<0，故SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0由于SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有2個(gè)零點(diǎn)．9.（2024屆江蘇省淮安市高三上學(xué)期第一次調(diào)研測(cè)試）已知函數(shù)SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)求證：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．【解析】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；②當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，要證明SKIPIF1<0，只要證SKIPIF1<0，即證SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，列表得aSKIPIF1<01SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0單調(diào)遞減極小值單調(diào)遞增所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．10.（2024屆湖南省長(zhǎng)沙市長(zhǎng)郡中學(xué)高三上學(xué)期月考）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．【解析】（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，所以SKIPIF1<0恒成立，此時(shí)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；綜上所述：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)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．（2）我們先證明引理：SKIPIF1<0，恒有SKIPIF1<0且SKIPIF1<0．引理的證明：設(shè)SKIPIF1<0，SKIPIF1<0．故只需證明SKIPIF1<0，恒有SKIPIF1<0，SKIPIF1<0．由于SKIPIF1<0，知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，恒有SKIPIF1<0．由于SKIPIF1<0，知當(dāng)SKIPIF1<0，均有SKIPIF1<0，所以恒有SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0．所以SKIPIF1<0，恒有SKIPIF1<0．綜上，引理得證．回到原題：由（1）得SKIPIF1<0，故只需證明：對(duì)SKIPIF1<0，恒有SKIPIF1<0，即SKIPIF1<0．由引理得SKIPIF1<0．命題得證．三、函數(shù)零點(diǎn)與方程實(shí)根個(gè)數(shù)問題11.（2024屆江西省全南中學(xué)高三上學(xué)期開學(xué)考試）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的極小值；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有一個(gè)零點(diǎn)，求SKIPIF1<0的取值范圍．【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，解得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，所以函數(shù)SKIPIF1<0的極小值為SKIPIF1<0.（2）解：若SKIPIF1<0時(shí)，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，所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有一個(gè)零點(diǎn).若SKIPIF1<0時(shí)，由SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，①若SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有一個(gè)零點(diǎn)；②若SKIPIF1<0時(shí)，可得SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，又由SKIPIF1<0，只需討論SKIPIF1<0的符號(hào)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有一個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上無零點(diǎn).③若SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，又由SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且只有一個(gè)零點(diǎn)，綜上可得，SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.12.（2023屆海南省?？谑懈呷聦W(xué)期學(xué)生學(xué)科能力診斷）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的極值；(2)若函數(shù)SKIPIF1<0至少有兩個(gè)不同的零點(diǎn)，求實(shí)數(shù)m的最小值.【解析】（1）由題意得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0的極大值為SKIPIF1<0，極小值為SKIPIF1<0.（2）由SKIPIF1<0，即得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0（舍去），令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0的最小值為SKIPIF1<0，又SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，又SKIPIF1<0，故存在唯一的SKIPIF1<0，使SKIPIF1<0，當(dāng)x變化時(shí)，SKIPIF1<0的變化情況如表：xSKIPIF1<0SKIPIF1<0SKIPIF1<01SKIPIF1<0SKIPIF1<0+0-0+SKIPIF1<0+0-0+SKIPIF1<0增極大值減極小值3增當(dāng)SKIPIF1<0且無限趨近于0時(shí)，SKIPIF1<0，由于SKIPIF1<0趨近于負(fù)無窮小，故SKIPIF1<0趨近于負(fù)無窮小，由于SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0，作出SKIPIF1<0的大致圖像如圖：要使函數(shù)SKIPIF1<0至少有兩個(gè)不同的零點(diǎn)，則直線SKIPIF1<0與SKIPIF1<0的圖象至少有兩個(gè)交點(diǎn)，故需使SKIPIF1<0，即實(shí)數(shù)m的最小值為3.13.（2024屆北京市陳經(jīng)綸中學(xué)高三上學(xué)期9月階段性診斷）已知函數(shù)SKIPIF1<0．(1)求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，求實(shí)數(shù)SKIPIF1<0的取值范圍；(3)當(dāng)SKIPIF1<0時(shí)，判斷SKIPIF1<0在SKIPIF1<0零點(diǎn)的個(gè)數(shù)，并說明理由．【解析】（1）由SKIPIF1<0可得SKIPIF1<0，此時(shí)切線斜率為SKIPIF1<0，而SKIPIF1<0；所以切線方程為SKIPIF1<0，即SKIPIF1<0；即曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0；（2）根據(jù)題意，若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，即可得SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0恒成立；令SKIPIF1<0，則SKIPIF1<0；顯然SKIPIF1<0在SKIPIF1<0上滿足SKIPIF1<0，而SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上恒成立；即SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0；所以SKIPIF1<0即可；因此實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.（3）令SKIPIF1<0，即可得SKIPIF1<0；構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，易知SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，如下圖中實(shí)曲線所示：又函數(shù)SKIPIF1<0恒過SKIPIF1<0，且SKIPIF1<0，易知SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0；又SKIPIF1<0，所以SKIPIF1<0（圖中虛線）在SKIPIF1<0范圍內(nèi)恒在SKIPIF1<0（圖中實(shí)直線）的上方；所以由圖易知SKIPIF1<0與SKIPIF1<0在SKIPIF1<0范圍內(nèi)僅有一個(gè)交點(diǎn)，即函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)僅有一個(gè)零點(diǎn).14.（2023屆河南省部分學(xué)校高三押題信息卷）已知函數(shù)SKIPIF1<0．(1)求證：曲線SKIPIF1<0僅有一條過原點(diǎn)的切線；(2)若SKIPIF1<0時(shí)，關(guān)于SKIPIF1<0的方程SKIPIF1<0有唯一解，求實(shí)數(shù)SKIPIF1<0的取值范圍．【解析】（1）SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，設(shè)切點(diǎn)SKIPIF1<0，則切線方程為SKIPIF1<0，當(dāng)切線過原點(diǎn)時(shí)有SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即切點(diǎn)有且只有一個(gè)，則曲線SKIPIF1<0僅有一條過原點(diǎn)的切線，即得證.（2）關(guān)于SKIPIF1<0的方程SKIPIF1<0有唯一解，即方程SKIPIF1<0，SKIPIF1<0有唯一解，令SKIPIF1<0，則SKIPIF1<0.因?yàn)镾KIPIF1<0，故當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.易知SKIPIF1<0的圖象與直線SKIPIF1<0有且僅有一個(gè)交點(diǎn)，滿足題意，此時(shí)SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，設(shè)SKIPIF1<0有兩個(gè)根SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.①若SKIPIF1<0，則當(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，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故要使得SKIPIF1<0有唯一解，則SKIPIF1<0或SKIPIF1<0恒成立.此時(shí)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.則極大值SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故當(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，又SKIPIF1<0恒成立，故SKIPIF1<0，SKIPIF1<0；同理，極小值SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)無最小值，此時(shí)無實(shí)數(shù)SKIPIF1<0使得SKIPIF1<0恒成立.②若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，不滿足SKIPIF1<0；③若SKIPIF1<0，由①可得SKIPIF1<0；故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.綜上所述：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.四、不等式恒成立問題15.（2023屆河南省信陽高級(jí)中學(xué)高三下學(xué)期3月測(cè)試）已知函數(shù)SKIPIF1<0.(1)SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)，求SKIPIF1<0的最小值；(2)證明：對(duì)任意正整數(shù)SKIPIF1<0，都有SKIPIF1<0（其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)）；(3)若SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.【解析】（1）依題意，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上SKIPIF1<0單調(diào)遞減；在區(qū)間SKIPIF1<0上SKIPIF1<0單調(diào)遞增，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取得最小值為SKIPIF1<0.（2）要證明：對(duì)任意正整數(shù)SKIPIF1<0，都有SKIPIF1<0，即證明SKIPIF1<0，即證明SKIPIF1<0，由（1）得SKIPIF1<0，即SKIPIF1<0令SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，所以對(duì)任意正整數(shù)SKIPIF1<0，都有SKIPIF1<0.（3）若不等式SKIPIF1<0恒成立，此時(shí)SKIPIF1<0，則SKIPIF1<0恒成立，令SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0.16.（2024屆河北省保定市唐縣第一中學(xué)高三上學(xué)期9月月考）已知函數(shù)SKIPIF1<0（SKIPIF1<0）．(1)若SKIPIF1<0在SKIPIF1<0上恒成立，求a的取值范圍：(2)設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn)，證明：SKIPIF1<0．【解析】（1）若SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0，令SKIPIF1<0，所以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，所以SKIPIF1<0，所以SKIPIF1<0，即a的取值范圍是SKIPIF1<0．（2）令SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又S