新高考數(shù)學(xué)二輪復(fù)習(xí)講義專(zhuān)題11 導(dǎo)數(shù)在不等式恒等式和零點(diǎn)問(wèn)題綜合應(yīng)用（講義）（原卷版）

專(zhuān)題11講：導(dǎo)數(shù)在不等式.恒等式和零點(diǎn)問(wèn)題綜合應(yīng)用【考點(diǎn)專(zhuān)題】考點(diǎn)一：利用導(dǎo)數(shù)研究零點(diǎn)問(wèn)題：（1）確定零點(diǎn)的個(gè)數(shù)問(wèn)題：可利用數(shù)形結(jié)合的辦法判斷交點(diǎn)個(gè)數(shù)，如果函數(shù)較為復(fù)雜，可用導(dǎo)數(shù)知識(shí)確定極值點(diǎn)和單調(diào)區(qū)間從而確定其大致圖象；（2）方程的有解問(wèn)題就是判斷是否存在零點(diǎn)的問(wèn)題，可參變分離，轉(zhuǎn)化為求函數(shù)的值域問(wèn)題處理.可以通過(guò)構(gòu)造函數(shù)g（x）的方法，把問(wèn)題轉(zhuǎn)化為研究構(gòu)造的函數(shù)g（x）的零點(diǎn)問(wèn)題；（3）利用導(dǎo)數(shù)研究函數(shù)零點(diǎn)或方程根，通常有三種思路：考點(diǎn)二：利用導(dǎo)數(shù)證明不等式的基本步驟：(1)作差或變形．(2)構(gòu)造新的函數(shù)h(x)．(3)利用導(dǎo)數(shù)研究h(x)的單調(diào)性或最值．(4)根據(jù)單調(diào)性及最值，得到所證不等式．特別地：當(dāng)作差或變形構(gòu)造的新函數(shù)不能利用導(dǎo)數(shù)求解時(shí)，一般轉(zhuǎn)化為分別求左、右兩端兩個(gè)函數(shù)的最值問(wèn)題．【方法技巧】方法點(diǎn)睛：與SKIPIF1<0和SKIPIF1<0相關(guān)的常見(jiàn)同構(gòu)模型：（1）SKIPIF1<0SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0（或SKIPIF1<0SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0）；（2）SKIPIF1<0SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0（或SKIPIF1<0SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0）；（3）SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0（或SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0）.方法點(diǎn)睛:該題考查函數(shù)與導(dǎo)數(shù)的綜合應(yīng)用,屬于難題,主要應(yīng)用的方法有不等式放縮,關(guān)于常見(jiàn)的放縮有:(1)SKIPIF1<0;(2)SKIPIF1<0;(3)SKIPIF1<0;(4)SKIPIF1<0.【核心題型】題型一：導(dǎo)數(shù)在不等式恒成立問(wèn)題1．（2023·全國(guó)·鄭州中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，對(duì)于SKIPIF1<0，SKIPIF1<0恒成立，則滿足題意的SKIPIF1<0的取值集合為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·河南·統(tǒng)考一模）已知SKIPIF1<0，若不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2023·全國(guó)·高三專(zhuān)題練習(xí)）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題型二：利用導(dǎo)數(shù)研究能成立問(wèn)題4．（2023·廣西柳州·二模）設(shè)函數(shù)SKIPIF1<0（SKIPIF1<0，e為自然對(duì)數(shù)的底數(shù)），若存在SKIPIF1<0使SKIPIF1<0成立，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2022·河南駐馬店·河南省駐馬店高級(jí)中學(xué)校考模擬預(yù)測(cè)）已知e是自然對(duì)數(shù)的底數(shù)．若SKIPIF1<0，使SKIPIF1<0，則實(shí)數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·江西新余·統(tǒng)考二模）若存在兩個(gè)正數(shù)SKIPIF1<0，使得不等式SKIPIF1<0成立，其中SKIPIF1<0，SKIPIF1<0為自然對(duì)數(shù)的底數(shù)，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題型三：利用導(dǎo)數(shù)研究函數(shù)零點(diǎn)問(wèn)題7．（2023·四川綿陽(yáng)·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為（

）A．2 B．3 C．4 D．58．（2022·內(nèi)蒙古呼倫貝爾·校考模擬預(yù)測(cè)）已知SKIPIF1<0，若函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象恰有3個(gè)交點(diǎn)，則實(shí)數(shù)k的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型四：利用導(dǎo)數(shù)研究函數(shù)的根問(wèn)題10．（2022·四川南充·統(tǒng)考一模）已知函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0，若SKIPIF1<0，則關(guān)于x的方程SKIPIF1<0的不同實(shí)根個(gè)數(shù)為（

）A．2 B．3 C．4 D．511．（2023秋·河南安陽(yáng)·高三校考期末）已知SKIPIF1<0是函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象交點(diǎn)的橫坐標(biāo)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．212．（2022·吉林長(zhǎng)春·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象恰有6個(gè)不同的公共點(diǎn)，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型五：利用導(dǎo)數(shù)研究函數(shù)的圖像和性質(zhì)問(wèn)題13．（2022·廣東汕頭·統(tǒng)考三模）已知函數(shù)SKIPIF1<0，若關(guān)于x的方程SKIPIF1<0有四個(gè)不同的實(shí)根，則實(shí)數(shù)k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<014．（2022·天津·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0恰有6個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<015．（2023秋·天津北辰·高三?？计谀┮阎瘮?shù)SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象恰有5個(gè)不同公共點(diǎn)，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型六：利用導(dǎo)數(shù)研究雙變量問(wèn)題16．（2022·福建福州·福建省福州格致中學(xué)?？寄M預(yù)測(cè)）已知SKIPIF1<0，若SKIPIF1<0，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<017．（2022·安徽六安·安徽省舒城中學(xué)?？家荒＃┮阎瘮?shù)SKIPIF1<0．若對(duì)任意的SKIPIF1<0，都存在唯一的SKIPIF1<0，使得SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<018．（2020·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0．若SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0成立，則SKIPIF1<0的最大值為（

）A．3 B．4 C．5 D．SKIPIF1<0題型七：利用導(dǎo)數(shù)解決實(shí)際問(wèn)題19．（2022春·山西太原·高三太原五中?？茧A段練習(xí)）如圖，某款酒杯的容器部分為圓錐，且該圓錐的軸截面為面積是SKIPIF1<0的正三角形.若在該酒杯內(nèi)放置一個(gè)圓柱形冰塊，要求冰塊高度不超過(guò)酒杯口高度，則酒杯可放置圓柱形冰塊的最大體積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<020．（2022·全國(guó)·模擬預(yù)測(cè)）如圖所示，在正方體SKIPIF1<0中，SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0分別在棱SKIPIF1<0，SKIPIF1<0，SKIPIF1<0上（不包含端點(diǎn)），且平面SKIPIF1<0平面SKIPIF1<0，點(diǎn)SKIPIF1<0在線段SKIPIF1<0上，且SKIPIF1<0，則三棱錐SKIPIF1<0的體積的最大值是（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．621．（2022·河南開(kāi)封·統(tǒng)考二模）如圖，將一塊直徑為SKIPIF1<0的半球形石材切割成一個(gè)正四棱柱，則正四棱柱的體積取最大值時(shí)，切割掉的廢棄石材的體積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型八：導(dǎo)數(shù)的綜合問(wèn)題22．（2023·浙江·統(tǒng)考一模）設(shè)函數(shù)SKIPIF1<0,SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí),證明:SKIPIF1<0;(2)若SKIPIF1<0,求a的取值范圍．23．（2023·廣東廣州·統(tǒng)考二模）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.24．（2023·四川涼山·統(tǒng)考一模）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)已知SKIPIF1<0，證明：SKIPIF1<0；(3)若SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．【高考必刷】一、單選題25．（2023·內(nèi)蒙古赤峰·統(tǒng)考模擬預(yù)測(cè)）設(shè)命題p：“SKIPIF1<0”是“SKIPIF1<0”成立的必要不充分條件.命題q：若不等式SKIPIF1<0恒成立，則SKIPIF1<0．下列命題是真命題的（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<026．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若函數(shù)SKIPIF1<0恰有兩個(gè)零點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<027．（2023·廣西柳州·統(tǒng)考模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0為自然對(duì)數(shù)的底數(shù)），若曲線SKIPIF1<0上存在點(diǎn)SKIPIF1<0使SKIPIF1<0成立，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<028．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為（

）A．1 B．3 C．4 D．529．（2022·河北衡水·衡水市第二中學(xué)校考一模）某正六棱錐外接球的表面積為SKIPIF1<0，且外接球的球心在正六棱錐內(nèi)部或底面上，底面正六邊形邊長(zhǎng)SKIPIF1<0，則其體積的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<030．（2022·重慶沙坪壩·重慶八中?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有兩個(gè)不等實(shí)根SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<031．（2023·全國(guó)·高三專(zhuān)題練習(xí)）定義：設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0上的導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上也存在導(dǎo)函數(shù)，則稱函數(shù)SKIPIF1<0在SKIPIF1<0上存在二階導(dǎo)函數(shù)，簡(jiǎn)記為SKIPIF1<0.若在區(qū)間SKIPIF1<0上SKIPIF1<0，則稱函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上為“凹函數(shù)”.已知SKIPIF1<0在區(qū)間SKIPIF1<0上為“凹函數(shù)”，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題32．（2023·遼寧·遼寧實(shí)驗(yàn)中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0在SKIPIF1<0處有極值，且極值為8，則（

）A．SKIPIF1<0有三個(gè)零點(diǎn)B．SKIPIF1<0C．曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0D．函數(shù)SKIPIF1<0為奇函數(shù)33．（2023·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，下列正確的是（

）A．若函數(shù)SKIPIF1<0有且只有1個(gè)零點(diǎn)SKIPIF1<0，則SKIPIF1<0B．若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，則SKIPIF1<0C．若函數(shù)SKIPIF1<0有且只有1個(gè)零點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0D．若SKIPIF1<0有兩個(gè)零點(diǎn)，則SKIPIF1<034．（2023·全國(guó)·模擬預(yù)測(cè)）設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0恒成立，則滿足條件的正整數(shù)SKIPIF1<0可以是（

）A．1 B．2 C．3 D．435．（2023·湖南永州·統(tǒng)考二模）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、填空題36．（2023·四川綿陽(yáng)·綿陽(yáng)中學(xué)校考模擬預(yù)測(cè)）關(guān)于x的不等式SKIPIF1<0在SKIPIF1<0上恒成立，則a的取值范圍是______．37．（20