2021-2023年高考數(shù)學(xué)真題分類(lèi)匯編專(zhuān)題03 導(dǎo)數(shù)及其應(yīng)用（選擇題、填空題）（文）（解析版）

專(zhuān)題03導(dǎo)數(shù)及其應(yīng)用（選擇題、填空題）（文）知識(shí)點(diǎn)目錄知識(shí)點(diǎn)1：切線(xiàn)問(wèn)題知識(shí)點(diǎn)2：?jiǎn)握{(diào)性、極最值問(wèn)題知識(shí)點(diǎn)3：比較大小問(wèn)題近三年高考真題知識(shí)點(diǎn)1：切線(xiàn)問(wèn)題1．（2023?甲卷（文））曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0，故函數(shù)在點(diǎn)SKIPIF1<0處的切線(xiàn)斜率SKIPIF1<0，切線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0．故選：SKIPIF1<0．2．（2021?新高考Ⅰ）若過(guò)點(diǎn)SKIPIF1<0可以作曲線(xiàn)SKIPIF1<0的兩條切線(xiàn)，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】法一：函數(shù)SKIPIF1<0是增函數(shù)，SKIPIF1<0恒成立，函數(shù)的圖象如圖，SKIPIF1<0，即切點(diǎn)坐標(biāo)在SKIPIF1<0軸上方，如果SKIPIF1<0在SKIPIF1<0軸下方，連線(xiàn)的斜率小于0，不成立．點(diǎn)SKIPIF1<0在SKIPIF1<0軸或下方時(shí)，只有一條切線(xiàn)．如果SKIPIF1<0在曲線(xiàn)上，只有一條切線(xiàn)；SKIPIF1<0在曲線(xiàn)上側(cè)，沒(méi)有切線(xiàn)；由圖象可知SKIPIF1<0在圖象的下方，并且在SKIPIF1<0軸上方時(shí)，有兩條切線(xiàn)，可知SKIPIF1<0．故選：SKIPIF1<0．法二：設(shè)過(guò)點(diǎn)SKIPIF1<0的切線(xiàn)橫坐標(biāo)為SKIPIF1<0，則切線(xiàn)方程為SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是增函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是減函數(shù)，因此當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，上述關(guān)于SKIPIF1<0的方程有兩個(gè)實(shí)數(shù)解，對(duì)應(yīng)兩條切線(xiàn)．故選：SKIPIF1<0．3．（2022?新高考Ⅰ）若曲線(xiàn)SKIPIF1<0有兩條過(guò)坐標(biāo)原點(diǎn)的切線(xiàn)，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【解析】SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0切線(xiàn)的斜率SKIPIF1<0，SKIPIF1<0切線(xiàn)方程為SKIPIF1<0，又SKIPIF1<0切線(xiàn)過(guò)原點(diǎn)，SKIPIF1<0，整理得：SKIPIF1<0，SKIPIF1<0切線(xiàn)存在兩條，SKIPIF1<0方程有兩個(gè)不等實(shí)根，SKIPIF1<0△SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故答案為：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．4．（2022?新高考Ⅱ）曲線(xiàn)SKIPIF1<0過(guò)坐標(biāo)原點(diǎn)的兩條切線(xiàn)的方程為．【答案】SKIPIF1<0，SKIPIF1<0．【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0切線(xiàn)的斜率SKIPIF1<0，SKIPIF1<0切線(xiàn)方程為SKIPIF1<0，又SKIPIF1<0切線(xiàn)過(guò)原點(diǎn)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0切線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，與SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，SKIPIF1<0切線(xiàn)方程也關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，SKIPIF1<0切線(xiàn)方程為SKIPIF1<0，綜上所述，曲線(xiàn)SKIPIF1<0經(jīng)過(guò)坐標(biāo)原點(diǎn)的兩條切線(xiàn)方程分別為SKIPIF1<0，SKIPIF1<0，故答案為：SKIPIF1<0，SKIPIF1<0．知識(shí)點(diǎn)2：?jiǎn)握{(diào)性、極最值問(wèn)題5．（2023?新高考Ⅱ）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的最小值為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】對(duì)函數(shù)SKIPIF1<0求導(dǎo)可得，SKIPIF1<0，依題意，SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，設(shè)SKIPIF1<0，則SKIPIF1<0，易知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0．故選：SKIPIF1<0．6．（2023?乙卷（文））函數(shù)SKIPIF1<0存在3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】SKIPIF1<0，若函數(shù)SKIPIF1<0存在3個(gè)零點(diǎn)，則SKIPIF1<0，有兩個(gè)不同的根，且極大值大于0極小值小于0，即判別式△SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增，由SKIPIF1<0得SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減，即當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得極大值，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得極小值，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0，①，且SKIPIF1<0，②，則①恒成立，由SKIPIF1<0，SKIPIF1<0，平方得SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，綜上SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故選：SKIPIF1<0．7．（2022?乙卷（文））函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0的最小值、最大值分別為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】SKIPIF1<0【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0得，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0，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，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0上的極大值為SKIPIF1<0，極小值為SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，最大值為SKIPIF1<0，故選：SKIPIF1<0．8．（2022?甲卷（文））當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得最大值SKIPIF1<0，則SKIPIF1<0（2）SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】SKIPIF1<0【解析】由題意SKIPIF1<0（1）SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)函數(shù)取得最值，可得SKIPIF1<0也是函數(shù)的一個(gè)極值點(diǎn)，SKIPIF1<0（1）SKIPIF1<0，即SKIPIF1<0．SKIPIF1<0，易得函數(shù)在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0處，函數(shù)取得極大值，也是最大值，則SKIPIF1<0（2）SKIPIF1<0．故選：SKIPIF1<0．9．（2021?乙卷（文））設(shè)SKIPIF1<0，若SKIPIF1<0為函數(shù)SKIPIF1<0的極大值點(diǎn)，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0及SKIPIF1<0是SKIPIF1<0的兩個(gè)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，由三次函數(shù)的性質(zhì)可知，要使SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)，則函數(shù)SKIPIF1<0的大致圖象如下圖所示，則SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由三次函數(shù)的性質(zhì)可知，要使SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)，則函數(shù)SKIPIF1<0的大致圖象如下圖所示，則SKIPIF1<0；綜上，SKIPIF1<0．故選：SKIPIF1<0．10．（多選題）（2023?新高考Ⅱ）若函數(shù)SKIPIF1<0既有極大值也有極小值，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】函數(shù)定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，由題意，方程SKIPIF1<0即SKIPIF1<0有兩個(gè)正根，設(shè)為SKIPIF1<0，SKIPIF1<0，則有SKIPIF1<0，SKIPIF1<0，△SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0．故選：SKIPIF1<0．11．（多選題）（2022?新高考Ⅰ）已知函數(shù)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0有兩個(gè)極值點(diǎn) B．SKIPIF1<0有三個(gè)零點(diǎn) C．點(diǎn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0的對(duì)稱(chēng)中心 D．直線(xiàn)SKIPIF1<0是曲線(xiàn)SKIPIF1<0的切線(xiàn)【答案】SKIPIF1<0【解析】SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，SKIPIF1<0有兩個(gè)極值點(diǎn)，有且僅有一個(gè)零點(diǎn)，故選項(xiàng)SKIPIF1<0正確，選項(xiàng)SKIPIF1<0錯(cuò)誤；又SKIPIF1<0，則SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，故選項(xiàng)SKIPIF1<0正確；假設(shè)SKIPIF1<0是曲線(xiàn)SKIPIF1<0的切線(xiàn)，設(shè)切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，顯然SKIPIF1<0和SKIPIF1<0均不在曲線(xiàn)SKIPIF1<0上，故選項(xiàng)SKIPIF1<0錯(cuò)誤．故選：SKIPIF1<0．知識(shí)點(diǎn)3：比較大小問(wèn)題12．（2022?天津）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0是定義域SKIPIF1<0上的單調(diào)增函數(shù)，所以SKIPIF1<0，即SKIPIF1<0；因?yàn)镾KIPIF1<0是定義域SKIPIF1<0上的單調(diào)減函數(shù)，所以SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0；因?yàn)镾KIPIF1<0是定義域SKIPIF1<0上的單調(diào)增函數(shù)，所以SKIPIF1<0，即SKIPIF1<0；所以SKIPIF1<0．故選：SKIPIF1<0．13．（2022?甲卷（文））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0（8），又因?yàn)镾KIPIF1<0，故SKIPIF1<0，故選：SKIPIF1<0．14．（2022?新高考Ⅰ）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，SKIPIF1<0在SKIPIF1<0處取最小值SKIPIF1<0（1）SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0；設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，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，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．故選：SKIPIF1<0．15．（2023?甲卷（文））已知函數(shù)SKIPIF1<0．記SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0的開(kāi)口向下，對(duì)稱(chēng)軸為SKIPIF1<