新高考數(shù)學(xué)一輪復(fù)習(xí)第3章 第13講 拓展六 泰勒展開式與超越不等式在導(dǎo)數(shù)中的應(yīng)用 精講（教師版）

第13講拓展六：泰勒展開式與超越不等式在導(dǎo)數(shù)中的應(yīng)用(精講）目錄第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第二部分：典型例題剖析高頻考點(diǎn)一：利用超越不等式比較大小高頻考點(diǎn)二：利用對(duì)數(shù)型超越放縮證明不等式高頻考點(diǎn)三：利用指數(shù)型超越放縮證明不等式第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶1、泰勒公式形式：泰勒公式是將一個(gè)在SKIPIF1<0處具有SKIPIF1<0階導(dǎo)數(shù)的函數(shù)利用關(guān)于SKIPIF1<0的SKIPIF1<0次多項(xiàng)式來逼近函數(shù)的方法.若函數(shù)SKIPIF1<0在包含SKIPIF1<0的某個(gè)閉區(qū)間SKIPIF1<0上具有SKIPIF1<0階導(dǎo)數(shù)，且在開區(qū)間SKIPIF1<0上具有SKIPIF1<0階導(dǎo)數(shù)，則對(duì)閉區(qū)間SKIPIF1<0上任意一點(diǎn)SKIPIF1<0，成立下式：SKIPIF1<0其中：SKIPIF1<0表示SKIPIF1<0在SKIPIF1<0處的SKIPIF1<0階導(dǎo)數(shù)，等號(hào)后的多項(xiàng)式稱為函數(shù)SKIPIF1<0在SKIPIF1<0處的泰勒展開式，剩余的SKIPIF1<0是泰勒公式的余項(xiàng)，是SKIPIF1<0的高階無窮小量.2、麥克勞林(Maclaurin)公式SKIPIF1<0雖然麥克勞林公式是泰勒中值定理的特殊形式，僅僅是取SKIPIF1<0的特殊結(jié)果，由于麥克勞林公式使用方便，在高考中經(jīng)常會(huì)涉及到.3、常見函數(shù)的麥克勞林展開式：（1）SKIPIF1<0（2）SKIPIF1<0（3）SKIPIF1<0（4）SKIPIF1<0（5）SKIPIF1<0（6）SKIPIF1<04、兩個(gè)超越不等式：（注意解答題需先證明后使用）4.1對(duì)數(shù)型超越放縮：SKIPIF1<0（SKIPIF1<0）SKIPIF1<0上式（1）中等號(hào)右邊只取第一項(xiàng)得：SKIPIF1<0SKIPIF1<0結(jié)論①用SKIPIF1<0替換上式結(jié)論①中的SKIPIF1<0得：SKIPIF1<0SKIPIF1<0結(jié)論②對(duì)于結(jié)論②左右兩邊同乘“SKIPIF1<0”得SKIPIF1<0，用SKIPIF1<0替換“SKIPIF1<0”得：SKIPIF1<0（SKIPIF1<0）SKIPIF1<0結(jié)論③4.2指數(shù)型超越放縮：SKIPIF1<0（SKIPIF1<0）SKIPIF1<0上式（2）中等號(hào)右邊只取前2項(xiàng)得：SKIPIF1<0SKIPIF1<0結(jié)論①用SKIPIF1<0替換上式結(jié)論①中的SKIPIF1<0得：SKIPIF1<0SKIPIF1<0結(jié)論②當(dāng)SKIPIF1<0時(shí)，對(duì)于上式結(jié)論②SKIPIF1<0SKIPIF1<0結(jié)論③當(dāng)SKIPIF1<0時(shí)，對(duì)于上式結(jié)論②SKIPIF1<0SKIPIF1<0結(jié)論④第二部分：典型例題剖析第二部分：典型例題剖析高頻考點(diǎn)一：利用超越不等式比較大小1．（2022·全國(guó)·高三專題練習(xí)（文））已知SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】先用導(dǎo)數(shù)證明這兩個(gè)重要的不等式①SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”SKIPIF1<0SKIPIF1<0SKIPIF1<0，函數(shù)遞減，SKIPIF1<0函數(shù)遞增故SKIPIF1<0時(shí)函數(shù)取得最小值為0故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”②SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”SKIPIF1<0SKIPIF1<0SKIPIF1<0，函數(shù)遞增，SKIPIF1<0函數(shù)遞減，故SKIPIF1<0時(shí)函數(shù)取得最大值為0，故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”故SKIPIF1<0SKIPIF1<0故選：C2．（2021·安徽·毛坦廠中學(xué)高三階段練習(xí)（理））設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，（其中自然對(duì)數(shù)的底數(shù)SKIPIF1<0）則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上SKIPIF1<0遞增，在SKIPIF1<0上SKIPIF1<0遞減，所以SKIPIF1<0，即SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，考慮到SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0等號(hào)當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取到，故SKIPIF1<0時(shí)SKIPIF1<0，排除A，B.下面比較a，b大小，由SKIPIF1<0得，SKIPIF1<0，故SKIPIF1<0.所以SKIPIF1<0.故選:D3．（2022·全國(guó)·高三專題練習(xí)）已知實(shí)數(shù)a，b，c滿足SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，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，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：A.【點(diǎn)睛】關(guān)鍵點(diǎn)睛：解決本題的關(guān)鍵是利用經(jīng)典不等式SKIPIF1<0可得SKIPIF1<0.4．（2022·河南洛陽·高二期末（文））下列結(jié)論中正確的個(gè)數(shù)為（

）①SKIPIF1<0，SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0．A．0 B．1 C．2 D．3【答案】C【詳解】解：令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，故①正確；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)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0恒成立，所以SKIPIF1<0，故②正確；令SKIPIF1<0，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，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故③錯(cuò)誤；故選：C5．（2021·浙江·模擬預(yù)測(cè)）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，給出以下結(jié)論，正確的個(gè)數(shù)是（

）①SKIPIF1<0；②SKIPIF1<0；③存在無窮多個(gè)SKIPIF1<0,使SKIPIF1<0；④SKIPIF1<0A．4 B．3 C．2 D．1【答案】B【詳解】SKIPIF1<0，SKIPIF1<0,SKIPIF1<0，則SKIPIF1<0單調(diào)遞增且大于0，所以SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0故①正確；令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.因?yàn)镾KIPIF1<0，且SKIPIF1<0，SKIPIF1<0，故②正確；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由歸納法可知，SKIPIF1<0，故不存在無窮多個(gè)SKIPIF1<0,使SKIPIF1<0，故③錯(cuò)誤；由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，累加可得：SKIPIF1<0可知④正確.故選：B.7．（2022·安徽·六安一中高二開學(xué)考試）已知SKIPIF1<0成等比數(shù)列，且SKIPIF1<0，若SKIPIF1<0，則A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】設(shè)SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,則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,則SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0,又SKIPIF1<0成等比數(shù)列,且SKIPIF1<0,設(shè)其公比為SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,故選:A.【點(diǎn)睛】本題考查導(dǎo)數(shù)中的不等式在數(shù)列中的應(yīng)用,以及等比數(shù)列的相關(guān)性質(zhì),屬于中檔題.導(dǎo)數(shù)中存在著一些常用的不等式結(jié)論,學(xué)生可以盡可能掌握.高頻考點(diǎn)二：利用對(duì)數(shù)型超越放縮證明不等式1．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)f(x)＝lnx－ax＋1在x＝2處的切線斜率為－SKIPIF1<0.(1)求實(shí)數(shù)a的值及函數(shù)f(x)的單調(diào)區(qū)間；(2)設(shè)g(x)＝SKIPIF1<0，對(duì)?x1SKIPIF1<0(0，＋∞)，?x2SKIPIF1<0(－∞，0)使得f(x1)≤g(x2)成立，求正實(shí)數(shù)k的取值范圍；(3)證明：SKIPIF1<0＋SKIPIF1<0＋…＋SKIPIF1<0(n∈N*，n≥2).【答案】(1)a＝1，增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0(2)SKIPIF1<0(3)證明見解析(1)由已知得f′(x)＝SKIPIF1<0－a，∴f′(2)＝SKIPIF1<0－a＝－SKIPIF1<0，解得a＝1.于是f′(x)＝SKIPIF1<0－1＝SKIPIF1<0，當(dāng)xSKIPIF1<0(0，1)時(shí)，f′(x)>0，f(x)為增函數(shù)，當(dāng)xSKIPIF1<0(1，＋∞)時(shí)，f′(x)<0，f(x)為減函數(shù)，即f(x)的單調(diào)遞增區(qū)間為(0，1)，單調(diào)遞減區(qū)間為(1，＋∞).(2)由(1)知x1SKIPIF1<0(0，＋∞)，f(x1)≤f(1)＝0，即f(x1)的最大值為0，由題意知：對(duì)?x1SKIPIF1<0(0，＋∞)，?x2SKIPIF1<0(－∞，0)使得f(x1)≤g(x2)成立，只需f(x)max≤g(x)max.∵g(x)＝SKIPIF1<0SKIPIF1<0，（SKIPIF1<0等號(hào)成立）∴只需SKIPIF1<0，解得SKIPIF1<0.(3)證明：要證明SKIPIF1<0(nSKIPIF1<0N*，n≥2).只需證SKIPIF1<0，只需證SKIPIF1<0.由(1)當(dāng)xSKIPIF1<0(1，＋∞)時(shí)，f′(x)<0，f(x)為減函數(shù)，f(x)＝lnx－x＋1≤0，即lnx≤x－1，∴當(dāng)n≥2時(shí)，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0＝SKIPIF1<0，∴SKIPIF1<0.2．（2022·河南·林州一中高二期中（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，證明：SKIPIF1<0.【答案】(1)見解析(2)證明見解析(1)SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增(2)由（1）可知，SKIPIF1<0令SKIPIF1<0，SKIPIF1<0即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0故SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0即SKIPIF1<03．（2022·陜西咸陽·二模（文））已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)證明：SKIPIF1<0.【答案】(1)SKIPIF1<0(2)證明見解析【解析】(1)由題意得：SKIPIF1<0定義域?yàn)镾KIPIF1<0；由SKIPIF1<0得：SKIPIF1<0；設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)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，SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.(2)由（1）知：當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，即SKIPIF1<0；SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題考查導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用，涉及到恒成立問題求解和不等式的證明問題；證明不等式的關(guān)鍵是能夠充分利用（1）中的結(jié)論，將所證不等式進(jìn)行放縮，從而結(jié)合等比數(shù)列求和的知識(shí)進(jìn)行證明.4．（2022·陜西咸陽·二模（理））已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0恒成立，求實(shí)數(shù)k的取值范圍；(2)證明：SKIPIF1<0(SKIPIF1<0，SKIPIF1<0).【答案】(1)SKIPIF1<0；(2)證明見解析﹒【解析】(1)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0＝SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)x＞1時(shí)，SKIPIF1<0單調(diào)遞減，∴SKIPIF1<0，∴SKIPIF1<0；(2)由(1)知，SKIPIF1<0時(shí)，有不等式SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，取“＝”號(hào)，∴SKIPIF1<0，SKIPIF1<0恒成立，令SKIPIF1<0(SKIPIF1<0，且SKIPIF1<0)，則SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0，即SKIPIF1<0(SKIPIF1<0，SKIPIF1<0)，∴SKIPIF1<0(SKIPIF1<0，SKIPIF1<0).【點(diǎn)睛】本題關(guān)鍵是利用(1)中的結(jié)論，取k＝1時(shí)得到不等式SKIPIF1<0，從而得到x＞1時(shí)，SKIPIF1<0，令SKIPIF1<0，即可構(gòu)造不等式SKIPIF1<0，從而通過裂項(xiàng)相消法求出SKIPIF1<0的范圍，從而證明結(jié)論.5．（2022·重慶市實(shí)驗(yàn)中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0且SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0；(3)求證：對(duì)任意的SKIPIF1<0且SKIPIF1<0，都有：SKIPIF1<0…SKIPIF1<0．（其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)）【答案】(1)答案見解析；(2)證明見解析；(3)證明見解析.【解析】(1)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.綜上，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上調(diào)遞增；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)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，要證明SKIPIF1<0，即證SKIPIF1<0，即SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0得，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．所以SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0.(3)由（2）可得SKIPIF1<0，（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立），令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0…SKIPIF1<0…SKIPIF1<0…SKIPIF1<0SKIPIF1<0…SKIPIF1<0，即SKIPIF1<0…SKIPIF1<0，故SKIPIF1<0…SKIPIF1<0.【點(diǎn)睛】本題考察利用導(dǎo)數(shù)研究含參函數(shù)單調(diào)性，以及構(gòu)造函數(shù)利用導(dǎo)數(shù)證明不等式，以及數(shù)列和導(dǎo)數(shù)的綜合，屬綜合困難題.6．（2022·內(nèi)蒙古·元寶山平煤高中高二階段練習(xí)（理））已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)證明：SKIPIF1<0.【答案】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，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；(2)證明見解析.【解析】(1)因?yàn)镾KIPIF1<0（SKIPIF1<0），所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0.若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上為增函數(shù)；若SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，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.(2)當(dāng)SKIPIF1<0時(shí)，由上可知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，有SKIPIF1<0在SKIPIF1<0恒成立，且SKIPIF1<0在SKIPIF1<0上是減函數(shù)，即SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0SKIPIF1<0且SKIPIF1<0，SKIPIF1<0SKIPIF1<0，即：SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）成立．7．（2022·河南·林州一中高二期中（理））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)證明：SKIPIF1<0；(3)若SKIPIF1<0且SKIPIF1<0，證明：SKIPIF1<0.【答案】(1)SKIPIF1<0(2)證明見解析(3)證明見解析【解析】(1)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0.(2)由（1）可得SKIPIF1<0SKIPIF1<0即函數(shù)SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0(3)由（2）可得SKIPIF1<0在SKIPIF1<0上恒成立令SKIPIF1<0，則SKIPIF1<0則SKIPIF1<0故SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)睛：解決第三問時(shí)，關(guān)鍵是由導(dǎo)數(shù)得出SKIPIF1<0，進(jìn)而由對(duì)數(shù)的運(yùn)算證明不等式.高頻考點(diǎn)三：利用指數(shù)型超越放縮證明不等式1．（2022·四川·棠湖中學(xué)高二階段練習(xí)（文））已知函數(shù)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)當(dāng)SKIPIF1<0時(shí)，若關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍；(3)當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(3)證明見解析【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，切點(diǎn)為SKIPIF1<0，斜率SKIPIF1<0，.∴曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0.即SKIPIF1<0.(2)由SKIPIF1<0，得SKIPIF1<0恒成立，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0的最小值為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0的取值范圍是SKIPIF1<0；(3)由（2）知SKIPIF1<0時(shí)，有SKIPIF1<0，所以SKIPIF1<0.①要證SKIPIF1<0，可證SKIPIF1<0，只需證SKIPIF1<0.先證SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單減，在SKIPIF1<0上單增，∴SKIPIF1<0，故SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)），從而當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立.②要證SKIPIF1<0，可證SKIPIF1<0.構(gòu)造函數(shù)SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單增，在SKIPIF1<0上單減，故SKIPIF1<0，即SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)），從而當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.由于SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，綜上所述，當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0.【點(diǎn)睛】要證明SKIPIF1<0，可通過證明SKIPIF1<0來證得.在利用導(dǎo)數(shù)證明不等式的過程中，主要利用的是導(dǎo)數(shù)的工具性的作用，也即利用導(dǎo)數(shù)來求單調(diào)區(qū)間、最值等.2．（2022·河南省杞縣高中模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0恒成立，求實(shí)數(shù)a的值；(2)若SKIPIF1<0，求證：SKIPIF1<0．【答案】(1)1(2)證明見解析【解析】(1)設(shè)SKIPIF1<0，則SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，SKIPIF1<0，不滿足SKIPIF1<0恒成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．所以SKIPIF1<0的最小值為SKIPIF1<0．即SKIPIF1<0，即SKIPIF1<0．設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在（0，1）上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，即SKIPIF1<0，故SKIPIF1<0的解只有SKIPIF1<0．綜上，SKIPIF1<0．(2)證明：先證當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立．令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在（0，1）上單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0．所以要證SKIPIF1<0，即證SKIPIF1<0，即證SKIPIF1<0，即證SKIPIF1<0．設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在（0，1）上單調(diào)遞減，所以SKIPIF1<0，即原不等式成立．所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：此題考查導(dǎo)數(shù)的綜合應(yīng)用，考查利用導(dǎo)數(shù)解決不等式恒成立問題，考查利用導(dǎo)數(shù)證明不等式，解題的關(guān)鍵是先證當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，然后將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，即證SKIPIF1<0，再構(gòu)造函數(shù)求出其最小值大于零即可，考查數(shù)學(xué)轉(zhuǎn)化思想，屬于較難題3．（2022·浙江省諸暨市第二高級(jí)中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0(1)當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程；(2)若SKIPIF1<0且SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍：(3)當(dāng)SKIPIF1<0時(shí)，記SKIPIF1<0，SKIPIF1<0（其中SKIPIF1<0）為SKIPIF1<0在SKIPIF1<0上的兩個(gè)零點(diǎn)，證明：SKIPIF1<0.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0；(3)詳見解析.(1)當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0；(2)由題意可知SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0顯然成立，故SKIPIF1<0；當(dāng)SKIPIF1<0