新高考數(shù)學(xué)一輪復(fù)習(xí)章節(jié)專(zhuān)題模擬卷第三章 導(dǎo)數(shù)及其應(yīng)用（解析版）

第三章導(dǎo)數(shù)及其應(yīng)用本試卷22小題，滿(mǎn)分150分。考試用時(shí)120分鐘一、單項(xiàng)選擇題：本題共8小題，每小題5分，共40分。在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。1．（2023·重慶·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0在SKIPIF1<0上單調(diào)遞增”的（

）A．充要條件 B．充分不必要條件C．必要不充分條件 D．既不充分也不必要條件【答案】C【解析】由題SKIPIF1<0若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0恒成立，SKIPIF1<0即SKIPIF1<0，故“SKIPIF1<0”是“SKIPIF1<0在SKIPIF1<0上單調(diào)遞增”的必要不充分條件故選:SKIPIF1<0.2．（2023·四川自貢·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．有2個(gè)極大值點(diǎn) B．有1個(gè)極大值點(diǎn)和1個(gè)極小值點(diǎn)C．有2個(gè)極小值點(diǎn) D．有且僅有一個(gè)極值點(diǎn)【答案】D【分析】求導(dǎo)，根據(jù)導(dǎo)函數(shù)的符號(hào)求得函數(shù)的單調(diào)區(qū)間，再根據(jù)極值點(diǎn)的定義即可得解.【詳解】SKIPIF1<0，因?yàn)镾KIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)），則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(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，所以函數(shù)SKIPIF1<0的極小值點(diǎn)為SKIPIF1<0，沒(méi)有極大值點(diǎn)，即函數(shù)SKIPIF1<0有且僅有一個(gè)極值點(diǎn).故選：D.3．（2023·重慶·統(tǒng)考三模）已知直線y＝ax－a與曲線SKIPIF1<0相切，則實(shí)數(shù)a＝（

）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由SKIPIF1<0且x不為0，得SKIPIF1<0設(shè)切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0.故選：C4．（2023·四川成都·統(tǒng)考二模）若函數(shù)SKIPIF1<0在SKIPIF1<0處有極大值，則實(shí)數(shù)SKIPIF1<0的值為（

）A．1 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】函數(shù)SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0處有極大值，可得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0處有極小值，不合題意.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，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，SKIPIF1<0在SKIPIF1<0處有極大值，符合題意.綜上可得，SKIPIF1<0.5．（2023·廣東汕頭·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的大致圖象為（

）A． B．C． D．【答案】C【解析】SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，故選：C6．（2023·重慶·統(tǒng)考一模）已知函數(shù)fx及其導(dǎo)函數(shù)f'(x)的定義域?yàn)镽，記g(x)=f'(x)，A．f(1)=f(2) B．f(1)=f(3) C．f(1)=f(4) D．f(1)=f(5)【答案】D【分析】根據(jù)f2x+1從而求得g(1)=0，g(x【詳解】因?yàn)閒2x+1是偶函數(shù)，所以f(?2x+1)=f兩邊求導(dǎo)得?2f'(?2所以g(2x+1）=?g(?2x+1)令x=1可得g(1)=?g因?yàn)間(x+2)為偶函數(shù)，所以gx+2=g?x+2，即又因g(4?x)=?g(?x+2),所以g(1)=g(3)=0,故選:D.7．（2023·廣東佛山·統(tǒng)考一模）已知函數(shù)fx=logax,0<x<12a?x,x≥12（A．0,e?C．0,e?【答案】D【分析】當(dāng)0<x<12時(shí)，f(x)=logax≥x構(gòu)造函數(shù)g(x)=2lnxx，求導(dǎo)判斷單調(diào)性，從而推出【詳解】當(dāng)0<x<1由圖可知，0<a此時(shí)若對(duì)任意0<x<1只需loga12≥14，即loga當(dāng)x≥12此時(shí)若對(duì)任意x≥12，(∴l(xiāng)n(1a)≥令g(x)=當(dāng)x∈(0,e),g'∴g(x綜上，116故選:D.8．（2023·黑龍江哈爾濱·哈師大附中統(tǒng)考三模）已知函數(shù)SKIPIF1<0，對(duì)任意的SKIPIF1<0，都有SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】令SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0為SKIPIF1<0上的奇函數(shù)，因?yàn)镾KIPIF1<0時(shí)，SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0為單調(diào)遞減函數(shù)，且SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上為單調(diào)遞減函數(shù)，由不等式SKIPIF1<0，可得SKIPIF1<0整理得到SKIPIF1<0，即SKIPIF1<0，可得SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：B.二、多項(xiàng)選擇題：本題共4小題，每小題5分，共20分。在每小題給出的選項(xiàng)中，有多項(xiàng)符合題目要求。全部選對(duì)的得5分，部分選對(duì)的得2分，有選錯(cuò)的得0分。9．（2023·山西運(yùn)城·統(tǒng)考三模）已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．曲線SKIPIF1<0在SKIPIF1<0處的切線與直線SKIPIF1<0垂直B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0的極小值為SKIPIF1<0D．SKIPIF1<0在SKIPIF1<0上的最小值為SKIPIF1<0【答案】BC【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故A錯(cuò)誤；令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故B正確；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，所以SKIPIF1<0的極小值為SKIPIF1<0，故C正確；SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以最小值為SKIPIF1<0，故D錯(cuò)誤；故選：BC10．（浙江省金麗衢十二校、“七彩陽(yáng)光”2023屆高三下學(xué)期3月聯(lián)考數(shù)學(xué)試題）已知SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【分析】對(duì)于A、B選項(xiàng)，利用條件構(gòu)造SKIPIF1<0，比值換元將問(wèn)題轉(zhuǎn)化為單變量函數(shù)求值域問(wèn)題；對(duì)于C、D選項(xiàng)，構(gòu)造函數(shù)SKIPIF1<0通過(guò)分析單調(diào)性判斷即可.【詳解】∵SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0令SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，令SKIPIF1<0，則SKIPIF1<0綜上SKIPIF1<0，SKIPIF1<0，即B正確；又因?yàn)镾KIPIF1<0，所以SKIPIF1<0令SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0）的零點(diǎn)y滿(mǎn)足SKIPIF1<0∴SKIPIF1<0，解得SKIPIF1<0.所以要證SKIPIF1<0，即證SKIPIF1<0因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以即證SKIPIF1<0而SKIPIF1<0所以SKIPIF1<0成立，即SKIPIF1<0成立，C正確因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，AD錯(cuò)誤.故選：B、C．11．（湖南省名校教研聯(lián)盟2023屆高三下學(xué)期4月聯(lián)考數(shù)學(xué)試題）已知SKIPIF1<0和SKIPIF1<0是定義在上SKIPIF1<0的函數(shù)，若存在區(qū)間SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0則稱(chēng)SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上同步.則（

）A．SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上同步B．存在SKIPIF1<0使得SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上同步C．若存在SKIPIF1<0使得SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上同步，則SKIPIF1<0D．存在區(qū)間SKIPIF1<0使得SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上同步【答案】BC【分析】由題意轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上是否至少存在兩個(gè)零點(diǎn)SKIPIF1<0，SKIPIF1<0，結(jié)合零點(diǎn)存在性定理判斷選項(xiàng)A、B；結(jié)合導(dǎo)數(shù)找到函數(shù)的單調(diào)性及最值，再根據(jù)函數(shù)零點(diǎn)情況判斷選項(xiàng)C、D.【詳解】由題知SKIPIF1<0與SKIPIF1<0在上SKIPIF1<0同步，即SKIPIF1<0在SKIPIF1<0上，至少存在兩個(gè)零點(diǎn)SKIPIF1<0，SKIPIF1<0，對(duì)于A，SKIPIF1<0在上SKIPIF1<0單調(diào)遞增，故A錯(cuò)誤.對(duì)于B，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上必有一個(gè)零點(diǎn)，故B正確.對(duì)于C，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，至多有一個(gè)零點(diǎn)，不符合題意，當(dāng)SKIPIF1<0時(shí)，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，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值，若SKIPIF1<0要有兩個(gè)零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0，故C正確.對(duì)于D，SKIPIF1<0，SKIPIF1<0，則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，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0沒(méi)有零點(diǎn)，故D錯(cuò)誤.故選：BC.12.（重慶市2023屆高三三模數(shù)學(xué)試題）函數(shù)SKIPIF1<0是定義在SKIPIF1<0上不恒為零的可導(dǎo)函數(shù)，對(duì)任意的x，SKIPIF1<0均滿(mǎn)足：SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【分析】先得到SKIPIF1<0，再假設(shè)SKIPIF1<0為正整數(shù)，利用累乘法求出SKIPIF1<0的解析式，再驗(yàn)證SKIPIF1<0不為正整數(shù)時(shí)，SKIPIF1<0也符合題意.利用SKIPIF1<0的解析式容易判斷ABC，根據(jù)錯(cuò)位相減法求和可判斷D.【詳解】令SKIPIF1<0，得SKIPIF1<0，代入SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0為正整數(shù)時(shí)，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，代入SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)，也符合題意，所以SKIPIF1<0.當(dāng)SKIPIF1<0不為正整數(shù)時(shí)，經(jīng)驗(yàn)證SKIPIF1<0也滿(mǎn)足SKIPIF1<0，故SKIPIF1<0為任意實(shí)數(shù)時(shí)，都有SKIPIF1<0.所以SKIPIF1<0，故A正確；SKIPIF1<0，故B正確；所以SKIPIF1<0，SKIPIF1<0，故C不正確；所以SKIPIF1<0SKIPIF1<0，令SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故D正確.故選：ABD三、填空題：本大題共4小題，每小題5分，共20分。13．（上海市建平中學(xué)2023屆高三三模數(shù)學(xué)試題）函數(shù)SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0__________.【答案】SKIPIF1<0【分析】利用復(fù)合函數(shù)的求導(dǎo)法則以及商的導(dǎo)數(shù)運(yùn)算法可求得結(jié)果.【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0.14．（2023·安徽安慶·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0，若不等式SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，則SKIPIF1<0的最小值為_(kāi)_____.【答案】SKIPIF1<0【分析】首先求出SKIPIF1<0，則問(wèn)題即為SKIPIF1<0，可同構(gòu)變形為SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，利用導(dǎo)數(shù)說(shuō)明函數(shù)的單調(diào)性，即可得到SKIPIF1<0，參變分離得到SKIPIF1<0，再令SKIPIF1<0，SKIPIF1<0，利用導(dǎo)數(shù)求出函數(shù)SKIPIF1<0的最大值，即可求出參數(shù)的取值范圍，即可得解.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以不等式SKIPIF1<0即為SKIPIF1<0，即SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0即為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，而SKIPIF1<0，SKIPIF1<0，因此由SKIPIF1<0等價(jià)于SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則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)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0，故正實(shí)數(shù)SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<015.（2023·江西南昌·統(tǒng)考二模）潮汐現(xiàn)象是地球上的海水在太陽(yáng)和月球雙重引力作用下產(chǎn)生的全球性的海水的周期性變化，人們可以利用潮汐進(jìn)行港口貨運(yùn).某港口具體時(shí)刻SKIPIF1<0（單位：小時(shí)）與對(duì)應(yīng)水深SKIPIF1<0（單位：米）的函數(shù)關(guān)系式為SKIPIF1<0.某艘大型貨船要進(jìn)港，其相應(yīng)的吃水深度（船底與水面的距離）為7米，船底與海底距離不小于4.5米時(shí)就是安全的，該船于2點(diǎn)開(kāi)始卸貨（一次卸貨最長(zhǎng)時(shí)間不超過(guò)8小時(shí)），同時(shí)吃水深度以0.375米/小時(shí)的速度減少，該船8小時(shí)內(nèi)沒(méi)有卸完貨，要及時(shí)駛?cè)肷钏畢^(qū)域，則該船第一次停止卸貨的時(shí)刻為_(kāi)_____.【答案】6時(shí)【解析】令船底與海底距離為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0，所以當(dāng)SKIPIF1<0或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在SKIPIF1<0上單調(diào)遞減.又因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0所以該船第一次停止卸貨的時(shí)刻為6時(shí).16．（天津市2023屆高三三模數(shù)學(xué)試題）已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0存在_____個(gè)極值點(diǎn)；若方程SKIPIF1<0有兩個(gè)不等實(shí)根，則SKIPIF1<0的取值范圍是___________【答案】4；SKIPIF1<0【分析】利用導(dǎo)數(shù)研究SKIPIF1<0的單調(diào)性和極值，作出SKIPIF1<0的圖象；由關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，得到函數(shù)SKIPIF1<0與SKIPIF1<0有一個(gè)交點(diǎn)，利用圖象法求解．【詳解】對(duì)于函數(shù)SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．令SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0；令SKIPIF1<0，解得：SKIPIF1<0；所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．而SKIPIF1<0，SKIPIF1<0；SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．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)遞減，在SKIPIF1<0上單調(diào)遞增．而SKIPIF1<0；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．作出SKIPIF1<0的圖象如圖所示：所以函數(shù)存在4個(gè)極值點(diǎn).解關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根,即關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，SKIPIF1<0只有一個(gè)實(shí)數(shù)根SKIPIF1<0，所以關(guān)于SKIPIF1<0的方程SKIPIF1<0有一個(gè)非零的實(shí)數(shù)根，即函數(shù)SKIPIF1<0與SKIPIF1<0有一個(gè)交點(diǎn)，橫坐標(biāo)SKIPIF1<0．結(jié)合圖象可得：SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0．故答案為：4；SKIPIF1<0.四、解答題：本大題共6小題，共70分，請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟。17．（2023·山東濰坊二模）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)討論函數(shù)SKIPIF1<0的極值點(diǎn)的個(gè)數(shù).【解析】（1）∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，而SKIPIF1<0，∴函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程SKIPIF1<0，級(jí)SKIPIF1<0.（2）所以SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0存在一個(gè)極值點(diǎn)SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，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，此時(shí)SKIPIF1<0存在兩個(gè)極值點(diǎn)SKIPIF1<0，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，此時(shí)SKIPIF1<0沒(méi)有極值點(diǎn).當(dāng)SKIPIF1<0時(shí)，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，此時(shí)SKIPIF1<0存在兩個(gè)極值點(diǎn)SKIPIF1<0，SKIPIF1<0.綜上所述：當(dāng)SKIPIF1<0或SKIPIF1<0，存在兩個(gè)極值點(diǎn)；當(dāng)SKIPIF1<0時(shí)，存在一個(gè)極值點(diǎn)；當(dāng)SKIPIF1<0時(shí)，沒(méi)有極值點(diǎn).18．（2023·福建莆田·統(tǒng)考二模）已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0的最小值為0，求a；(2)設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0是增函數(shù)，求a的取值范圍．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【分析】（1）利用函數(shù)最值與極值的關(guān)系推得SKIPIF1<0為SKIPIF1<0的一個(gè)極值點(diǎn)，從而求得SKIPIF1<0，再代回檢驗(yàn)是否滿(mǎn)足題意即可得解；（2）先利用同構(gòu)法得到SKIPIF1<0，再構(gòu)造函數(shù)，結(jié)合（1）中結(jié)論證得SKIPIF1<0，從而得到SKIPIF1<0，由此得解.【詳解】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0的最小值為0，所以SKIPIF1<0為SKIPIF1<0的一個(gè)極值點(diǎn)，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，檢驗(yàn)：當(dāng)SKIPIF1<0時(shí)，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，滿(mǎn)足題意，綜上，SKIPIF1<0.（2）因?yàn)楹瘮?shù)SKIPIF1<0是增函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以方程SKIPIF1<0有解，由（1）可知，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的取值范圍為SKIPIF1<0.19.（2023·山東濰坊·統(tǒng)考一模）已知函數(shù)fx(1)討論fx(2)證明：當(dāng)x∈0,2吋，f【答案】(1)函數(shù)fx在0,+∞(2)證明見(jiàn)解析【分析】（1）求導(dǎo)，再根據(jù)導(dǎo)函數(shù)的符號(hào)即可得出答案；（2）當(dāng)x∈0,2吋，fx≤gx，即證lnxelnx≤【詳解】（1）函數(shù)fx的定義域?yàn)?,+∞f'記?x=lnx所以當(dāng)0<x<1時(shí)，?'當(dāng)x>1時(shí)，?'x所以?x所以f'所以函數(shù)fx在0,+∞（2）原不等式為ex?1ln即證lnxeln設(shè)lx=x所以，當(dāng)x<1時(shí)，l'x>0，lx單調(diào)遞增；當(dāng)x所以lx令tx當(dāng)0<x<1時(shí)，t'x>0，tx單調(diào)遞增；當(dāng)所以t(x)且在x∈0,2上有l(wèi)nx<1x所以在x∈0,2時(shí)，有20.（2023·山東泰安·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0.（1）若SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；（2）若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍.【答案】（1）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，無(wú)遞減區(qū)間（2）SKIPIF1<0【解析】【分析】（1）求出函數(shù)SKIPIF1<0的定義域，利用函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系可得出函數(shù)SKIPIF1<0的增區(qū)間和減區(qū)間；（2）設(shè)SKIPIF1<0，可知SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，對(duì)實(shí)數(shù)SKIPIF1<0的取值進(jìn)行分類(lèi)討論，利用導(dǎo)數(shù)分析函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性，驗(yàn)證SKIPIF1<0對(duì)任意的SKIPIF1<0能否恒成立，綜合可得出實(shí)數(shù)SKIPIF1<0的取值范圍.【小問(wèn)1詳解】解：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則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，SKIPIF1<0，SKIPIF1<0單調(diào)遞增.所以，SKIPIF1<0，則SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，所以，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，無(wú)遞減區(qū)間.【小問(wèn)2詳解】解：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立等價(jià)于SKIPIF1<0在SKIPIF1<0上恒成立，設(shè)SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0圖象為開(kāi)口向上，對(duì)稱(chēng)軸為SKIPIF1<0的拋物線的一部分，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，且SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0在SKIPIF1<0恒成立，滿(mǎn)足題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，所以方程SKIPIF1<0有兩相異實(shí)根，設(shè)為SKIPIF1<0、SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又因?yàn)镾KIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以，SKIPIF1<0在SKIPIF1<0上不恒成立，不滿(mǎn)足題意.綜上，SKIPIF1<0的取值范圍為SKIPIF1<0.21．（2023·湖北·校聯(lián)考三模）已知函數(shù)SKIPIF1<0(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0，在SKIPIF1<0內(nèi)存在不等實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0，證明：SKIPIF1<0．【答案】(1)答案見(jiàn)解析(2)證明見(jiàn)解析【詳解】（1）函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0二次函數(shù)SKIPIF1<0的對(duì)稱(chēng)軸是SKIPIF1<0①若SKIPIF1<0時(shí)，在SKIPIF1<0上SKIPIF1<0，從而SKIPIF1<0，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0；②若SKIPIF1<0時(shí)，SKIPIF1<0，∴函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0；單調(diào)遞減區(qū)間是SKIPIF1<0；（2）由對(duì)稱(chēng)性不妨設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0若SKIPIF1<0，由（1）得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，有SKIPIF1<0，SKIPIF1<0與已知條件矛盾；SKIPIF1<0時(shí)，同理可推出矛盾．SKIPIF1<0，SKIPIF1<0，

要證明：SKIPIF1<0，只需證明：SKIPIF1<0SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴只需證明：SKIPIF1<0又SKIPIF1<0，∴只需證明：SKIPIF1<0構(gòu)造函數(shù)SKIPIF1<0，

其中SKIPIF1<0．SKIPIF1<0SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減，SKIPIF1<0SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0成立SKIPIF1<0由SKIPIF1<0在定義域內(nèi)單調(diào)遞增得，SKIPIF1<0，即SKIPIF1<0成立．22.（2023·山東青島·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，圓SKIPIF1<0．（1）若SKIPIF1<0，寫(xiě)出曲線SKIPIF1<0與圓C的一條公切線的方程(無(wú)需證明)；（2）若曲線SKIPIF1<0與圓C恰有三條公切線．(i)求b的取值范圍；(ii)證明：曲線SKIPIF1<0上存在點(diǎn)SKIPIF1<0SKIPIF1<0，對(duì)任意SKIPIF1<0，SKIPIF1<0．【答案】（1）SKIPIF1<0；（2）(i)SKIPIF1<0；(ii)證明見(jiàn)解析．【解析】【分析】(1)根據(jù)導(dǎo)數(shù)的幾何意義表示出f(x)的切線方程，再根據(jù)該切線與圓相切列出方程，取方程的特殊解即可得到一條共切線的方程；(2)(i)設(shè)曲線SKIPIF1<0與圓SKIPIF1<0公切線SKIPIF1<0的方程為SKIPIF1<0，根據(jù)導(dǎo)數(shù)幾何意義求出k和m的關(guān)系，在根據(jù)圓的切線方程的幾何性質(zhì)得到關(guān)于k的方程，問(wèn)題轉(zhuǎn)化為討論該方程有三