新高考數(shù)學(xué)二輪復(fù)習(xí)強(qiáng)化練習(xí)專題04 導(dǎo)數(shù)的基本應(yīng)用（練）（解析版）

第一篇熱點(diǎn)、難點(diǎn)突破篇專題04導(dǎo)數(shù)的基本應(yīng)用（練）【對(duì)點(diǎn)演練】一、單選題1．（2022·貴州·凱里一中高三階段練習(xí)（文））曲線SKIPIF1<0在SKIPIF1<0點(diǎn)處的切線方程是SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用導(dǎo)數(shù)的幾何意義求解即可.【詳解】SKIPIF1<0，SKIPIF1<0，切點(diǎn)為SKIPIF1<0，切線方程為SKIPIF1<0，∴SKIPIF1<0.故選：A.2．（2022·新疆·伊寧縣第二中學(xué)高三期中（文））設(shè)函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且函數(shù)SKIPIF1<0的部分圖像如圖所示，則（

）A．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 B．函數(shù)SKIPIF1<0在SKIPIF1<0處取得極大值C．函數(shù)SKIPIF1<0在SKIPIF1<0處取得極小值 D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增【答案】D【分析】由導(dǎo)函數(shù)的正負(fù)可得函數(shù)SKIPIF1<0的單調(diào)性，再逐項(xiàng)判斷可得答案.【詳解】由SKIPIF1<0的圖象可得當(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；對(duì)于A，函數(shù)SKIPIF1<0在SKIPIF1<0先遞減，再遞增，故不正確；對(duì)于B，函數(shù)SKIPIF1<0在SKIPIF1<0處取得極小值，故不正確；對(duì)于C，函數(shù)SKIPIF1<0在SKIPIF1<0處取不到極值，故不正確；對(duì)于D，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故正確；故選：D3．（2022·湖北·棗陽(yáng)一中高三期中）已知函數(shù)SKIPIF1<0的圖像在SKIPIF1<0處的切線過點(diǎn)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．2 C．3 D．4【答案】B【分析】結(jié)合導(dǎo)數(shù)求出切線方程，將SKIPIF1<0代入即可求出參數(shù)SKIPIF1<0.【詳解】由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則函數(shù)在SKIPIF1<0處的切線方程為SKIPIF1<0，將SKIPIF1<0代入切線方程可得SKIPIF1<0.故選：B4．（2022·浙江·嘉興一中高三期中）若函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值2，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．2【答案】A【分析】求導(dǎo)，根據(jù)SKIPIF1<0處的極值為2，列方程解方程得到SKIPIF1<0，SKIPIF1<0，即可得到SKIPIF1<0.【詳解】解：SKIPIF1<0，SKIPIF1<0，又函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值2，則SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，經(jīng)檢驗(yàn)滿足要求，所以SKIPIF1<0．故選：A．5．（2020·河南·高三階段練習(xí)（文））函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)SKIPIF1<0在SKIPIF1<0上單調(diào)性求出最值即可【詳解】由SKIPIF1<0可得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0的極小值，也為最小值為SKIPIF1<0，故選：C6．（2023·廣西·模擬預(yù)測(cè)（文））已知函數(shù)SKIPIF1<0存在最大值0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】B【分析】討論SKIPIF1<0與0的大小關(guān)系確定SKIPIF1<0的單調(diào)性，求出SKIPIF1<0的最大值.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，故函數(shù)SKIPIF1<0單調(diào)遞增，不存在最大值；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得出SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞減，所以SKIPIF1<0，解得：SKIPIF1<0SKIPIF1<0.故選：B.二、多選題7．（2022·遼寧葫蘆島·高三階段練習(xí)）已知函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0,SKIPIF1<0,則(

)A．SKIPIF1<0是SKIPIF1<0的極小值點(diǎn) B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【分析】求導(dǎo)，轉(zhuǎn)化為研究二次函數(shù)即可【詳解】SKIPIF1<0因?yàn)镾KIPIF1<0存在兩個(gè)極值點(diǎn)SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0當(dāng)SKIPIF1<0和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是SKIPIF1<0的極大值點(diǎn),且SKIPIF1<0故選：BCD8．（2022·江蘇蘇州·高三期中）已知函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，則（

）A．SKIPIF1<0 B．SKIPIF1<0的最小值是SKIPIF1<0C．SKIPIF1<0圖象與直線SKIPIF1<0相切 D．SKIPIF1<0圖象與直線SKIPIF1<0相切【答案】AD【分析】根據(jù)函數(shù)的對(duì)稱性代入特殊值，求SKIPIF1<0，即可判斷A;利用換元，轉(zhuǎn)化為二次函數(shù)求最值，即可判斷B；聯(lián)立函數(shù)與直線方程，利用方程組的解，判斷交點(diǎn)處的導(dǎo)數(shù)，判斷是否相切，即可判斷C；利用導(dǎo)數(shù)求函數(shù)在SKIPIF1<0處的切線方程，即可判斷D.【詳解】因?yàn)镾KIPIF1<0圖象關(guān)于直線SKIPIF1<0對(duì)稱，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，于是SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，于是SKIPIF1<0，于是SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故A正確；SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0圖象開口向上，對(duì)稱軸是SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0，故B錯(cuò)誤；聯(lián)立方程SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0與直線SKIPIF1<0不能相切，故C不正確；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，故D正確.故選：AD三、填空題9.（2022·黑龍江·哈爾濱七十三中高三階段練習(xí)）函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線方程為_________.【答案】SKIPIF1<0【分析】根據(jù)題意，先求出函數(shù)的導(dǎo)數(shù)，利用導(dǎo)數(shù)的幾何意義，求出切線方程的斜率即可求解.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0，所以SKIPIF1<0，又因?yàn)辄c(diǎn)SKIPIF1<0在函數(shù)圖象上，由導(dǎo)數(shù)的幾何意義可知：切線的斜率SKIPIF1<0，所以所求切線方程為SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，故答案為：SKIPIF1<0或SKIPIF1<0.10．（2022·山東煙臺(tái)·高三期中）若函數(shù)SKIPIF1<0，則SKIPIF1<0的最小值是______．【答案】SKIPIF1<0【分析】因?yàn)槿呛瘮?shù)具有周期性，令SKIPIF1<0，對(duì)函數(shù)求導(dǎo)數(shù)，研究導(dǎo)函數(shù)在區(qū)間內(nèi)的符號(hào)，得到函數(shù)的單調(diào)性，求出最小值.【詳解】不妨設(shè)SKIPIF1<0，SKIPIF1<0則SKIPIF1<0在SKIPIF1<0上的單調(diào)性如下表：x0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0+0-0+SKIPIF1<0SKIPIF1<0極大SKIPIF1<0極小SKIPIF1<0SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以函數(shù)的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.【沖刺提升】一、單選題1．（2022·河南·模擬預(yù)測(cè)（理））如圖是函數(shù)SKIPIF1<0的圖象，則函數(shù)SKIPIF1<0的解析式可以為（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用導(dǎo)數(shù)說明函數(shù)的單調(diào)性，即可判斷.【詳解】解：對(duì)于A：SKIPIF1<0定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，則SKIPIF1<0，即函數(shù)在SKIPIF1<0上單調(diào)遞增，故A錯(cuò)誤；對(duì)于B：SKIPIF1<0定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故B錯(cuò)誤；對(duì)于C：SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0又SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)SKIPIF1<0，即函數(shù)在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故C錯(cuò)誤；對(duì)于D：SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0所以當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即函數(shù)在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，符合題意；故選：D2．（2007·陜西·高考真題（理））SKIPIF1<0是定義在SKIPIF1<0上的非負(fù)可導(dǎo)函數(shù)，且滿足SKIPIF1<0．對(duì)任意正數(shù)a，b，若SKIPIF1<0，則必有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】構(gòu)造函數(shù)SKIPIF1<0，再分類討論即可求解.【詳解】解：令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上為常函數(shù)或遞減,SKIPIF1<0若SKIPIF1<0在SKIPIF1<0上為單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0①，SKIPIF1<0②①②兩式相乘得：所以SKIPIF1<0，SKIPIF1<0若SKIPIF1<0在SKIPIF1<0上為常函數(shù)，且SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0③，SKIPIF1<0④，③④兩式相乘得：所以SKIPIF1<0，綜上所述，SKIPIF1<0故選：A3．（2022·湖北·高三期中）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，結(jié)合函數(shù)的單調(diào)性分別得出SKIPIF1<0，SKIPIF1<0，從而得出答案．【詳解】令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∵SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．綜上，SKIPIF1<0．故選：D．4．（2022·貴州·貴陽(yáng)一中高三階段練習(xí)（文））在給出的①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0三個(gè)不等式中，正確的個(gè)數(shù)為（

）A．0個(gè) B．1個(gè) C．2個(gè) D．3個(gè)【答案】B【分析】構(gòu)造函數(shù)SKIPIF1<0，分析其單調(diào)性可判斷①和②，構(gòu)造函數(shù)SKIPIF1<0，分析其單調(diào)性可判斷③.【詳解】令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0,即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減,可得SKIPIF1<0，即SKIPIF1<0，故①正確；因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故②錯(cuò)誤；再令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0．又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故③錯(cuò)誤，故選：B．5．（2022·浙江·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，對(duì)于任意的SKIPIF1<0、SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，總有SKIPIF1<0成立，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】設(shè)SKIPIF1<0，可知函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù)，可知，對(duì)任意的SKIPIF1<0，SKIPIF1<0，利用導(dǎo)數(shù)求出函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值，即可得出實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】不妨設(shè)SKIPIF1<0，由SKIPIF1<0可得出SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上為增函數(shù)，則SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，所以，SKIPIF1<0，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，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，SKIPIF1<0，所以，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以，SKIPIF1<0，可得SKIPIF1<0，且當(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，所以，SKIPIF1<0.故選：A.【點(diǎn)睛】思路點(diǎn)睛：本題關(guān)鍵點(diǎn)在處理函數(shù)SKIPIF1<0的極值點(diǎn)時(shí)，根據(jù)零點(diǎn)存在定理得出其極值點(diǎn)SKIPIF1<0滿足SKIPIF1<0，通過利用指對(duì)同構(gòu)結(jié)合函數(shù)SKIPIF1<0的單調(diào)性轉(zhuǎn)化為SKIPIF1<0，SKIPIF1<0，利用整體代換法可求得SKIPIF1<0的取值范圍.二、多選題6．（2022·江蘇連云港·高三期中）已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線為SKIPIF1<0，則（

）A．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的極大值為SKIPIF1<0B．若SKIPIF1<0，SKIPIF1<0的斜率為2，則SKIPIF1<0C．若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0D．若存在過點(diǎn)P的直線SKIPIF1<0與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則SKIPIF1<0【答案】AB【分析】當(dāng)SKIPIF1<0時(shí)，求出函數(shù)的導(dǎo)數(shù)，判斷函數(shù)單調(diào)性，求得極值，判斷A;根據(jù)導(dǎo)數(shù)的幾何意義可求得參數(shù)的值，判斷B;利用導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系可得不等式，求得a的范圍，判斷C;根據(jù)導(dǎo)數(shù)的幾何意義，利用斜率關(guān)系，列出相應(yīng)等式，化簡(jiǎn)可得SKIPIF1<0，判斷D.【詳解】當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞減，故SKIPIF1<0時(shí)，取得極大值SKIPIF1<0，A正確；由SKIPIF1<0可知，若SKIPIF1<0，SKIPIF1<0的斜率為2，則SKIPIF1<0，故B正確；若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0恒成立，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，C錯(cuò)誤；若存在過點(diǎn)P的直線SKIPIF1<0與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0的斜率為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0，D錯(cuò)誤，故選：AB.7．（2022·山東·青島超銀高級(jí)中學(xué)高三階段練習(xí)）已知SKIPIF1<0，則（

）A．設(shè)SKIPIF1<0是SKIPIF1<0圖象上的任意一點(diǎn)，SKIPIF1<0是SKIPIF1<0圖象上任一點(diǎn)，則SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0與SKIPIF1<0的圖象有且僅有兩條公切線D．SKIPIF1<0是增函數(shù)【答案】ABC【分析】由導(dǎo)數(shù)的幾何意義可判斷A，由SKIPIF1<0得單調(diào)性可判斷BD，由方程SKIPIF1<0有兩個(gè)解可判斷C.【詳解】在同一坐標(biāo)系上作出SKIPIF1<0的圖象如圖所示：易知SKIPIF1<0和SKIPIF1<0的圖像關(guān)于直線SKIPIF1<0對(duì)稱，作與直線SKIPIF1<0平行且與SKIPIF1<0相切的直線SKIPIF1<0，設(shè)切點(diǎn)SKIPIF1<0，SKIPIF1<0，所以有SKIPIF1<0，解得SKIPIF1<0，即切點(diǎn)為SKIPIF1<0，SKIPIF1<0到直線SKIPIF1<0的距離SKIPIF1<0，即曲線SKIPIF1<0上的動(dòng)點(diǎn)到直線SKIPIF1<0的距離的最小值為SKIPIF1<0，由對(duì)稱性可知:SKIPIF1<0，A正確；設(shè)SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，所以存在SKIPIF1<0，使得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，而SKIPIF1<0，故SKIPIF1<0，而SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，B正確，D錯(cuò)誤.設(shè)SKIPIF1<0與SKIPIF1<0的公切線為SKIPIF1<0，切點(diǎn)分別為SKIPIF1<0SKIPIF1<0，SKIPIF1<0，則有SKIPIF1<0，化簡(jiǎn)得：SKIPIF1<0，即SKIPIF1<0，畫出SKIPIF1<0與SKIPIF1<0的圖像可知：SKIPIF1<0與SKIPIF1<0的圖像有兩個(gè)交點(diǎn)，所以方程SKIPIF1<0有兩個(gè)解，即SKIPIF1<0與SKIPIF1<0的圖象有且僅有兩條公切線，C正確；故選：ABC.【點(diǎn)睛】導(dǎo)函數(shù)中常用的兩種常用的轉(zhuǎn)化方法：一是利用導(dǎo)數(shù)研究含參函數(shù)的單調(diào)性，?；癁椴坏仁胶愠闪栴}．注意分類討論與數(shù)形結(jié)合思想的應(yīng)用；二是函數(shù)的零點(diǎn)、不等式證明常轉(zhuǎn)化為函數(shù)的單調(diào)性、極(最)值問題處理．三、填空題8．（2022·江蘇泰州·高三期中）若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0SKIPIF1<0處的切線也是曲線SKIPIF1<0的切線，則SKIPIF1<0的最小值為_____．【答案】SKIPIF1<0【分析】由兩條曲線的公切線斜率分別等于各曲線上切點(diǎn)處的導(dǎo)數(shù)值，以及各曲線上切點(diǎn)分別滿足切線方程來列方程組，得到SKIPIF1<0與SKIPIF1<0滿足的關(guān)系式，將原式中的SKIPIF1<0替換，再利用基本不等式求最小值即可.【詳解】曲線SKIPIF1<0在點(diǎn)A處的切線可寫作SKIPIF1<0設(shè)該切線在曲線SKIPIF1<0上的切點(diǎn)為SKIPIF1<0，則有SKIPIF1<0，消去t得SKIPIF1<0則SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取得該最小值.故答案為：SKIPIF1<0.9．（2022·遼寧·沈陽(yáng)市第四十中學(xué)高三期中）已知等比數(shù)列SKIPIF1<0的公比SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，則SKIPIF1<0______．【答案】SKIPIF1<0##SKIPIF1<0.【分析】先求出函數(shù)SKIPIF1<0的極值點(diǎn)，從而可得SKIPIF1<0，SKIPIF1<0，再求出公比SKIPIF1<0，進(jìn)而可求出SKIPIF1<0.【詳解】由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以2和3為SKIPIF1<0的極值點(diǎn)，因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.10．（2022·北京大興·高三期中）已知函數(shù)SKIPIF1<0若SKIPIF1<0的值域?yàn)镽，則a的一個(gè)取值為____________；若SKIPIF1<0是R上的增函數(shù)，則實(shí)數(shù)a的取值范圍是____________．【答案】

0（SKIPIF1<0）；

SKIPIF1<0【分析】①SKIPIF1<0的值域?yàn)镽等價(jià)于SKIPIF1<0的值域包含SKIPIF1<0，即SKIPIF1<0，由導(dǎo)數(shù)法，對(duì)分別討論SKIPIF1<0、SKIPIF1<0、SKIPIF1<0下SKIPIF1<0的最大值即可；②SKIPIF1<0是R上的增函數(shù)，則等價(jià)于SKIPIF1<0單調(diào)遞增且SKIPIF1<0，SKIPIF1<0單調(diào)遞增等價(jià)于SKIPIF1<0在SKIPIF1<0恒大于等于0，分別討論SKIPIF1<0、SKIPIF1<0即可【詳解】①SKIPIF1<0值域?yàn)镽等價(jià)于SKIPIF1<0的值域包含SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，即有SKIPIF1<0，故有SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，故當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，即有SKIPIF1<0，故有SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，即有SKIPIF1<0，故有SKIPIF1<0恒成立，故SKIPIF1<0；綜上，SKIPIF1<0的值域?yàn)镽時(shí)，SKIPIF1<0②若SKIPIF1<0是R上的增函數(shù)，等價(jià)于SKIPIF1<0單調(diào)遞增且SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，由SKIPIF1<0單調(diào)遞增即SKIPIF1<0在SKIPIF1<0恒大于等于0得，當(dāng)SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0綜上，SKIPIF1<0或SKIPIF1<0.故答案為：0（SKIPIF1<0）；SKIPIF1<0四、解答題11．（2022·北京·北師大二附中高三期中）已知函數(shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，且在點(diǎn)P處的切線恰好與直線SKIPIF1<0垂直．(1)求函數(shù)SKIPIF1<0的解析式；(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，求實(shí)數(shù)m的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0或SKIPIF1<0【分析】(1)將點(diǎn)SKIPIF1<0坐標(biāo)代入函數(shù)解析式得到關(guān)于SKIPIF1<0的方程，再根據(jù)函數(shù)在切點(diǎn)處的導(dǎo)數(shù)等于切線的斜率再建立關(guān)于SKIPIF1<0的另一個(gè)方程，即可求出SKIPIF1<0，即可確定函數(shù)SKIPIF1<0的解析式;(2)求出函數(shù)的單調(diào)區(qū)間，利用SKIPIF1<0可求解.【詳解】（1）因?yàn)楹瘮?shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0,所以SKIPIF1<0,又因?yàn)镾KIPIF1<0,且SKIPIF1<0點(diǎn)P處的切線恰好與直線SKIPIF1<0垂