新高考數(shù)學(xué)二輪復(fù)習(xí) 題型歸納演練專題3-8 利用導(dǎo)函數(shù)證明不等式原卷版

專題3-8利用導(dǎo)函數(shù)證明不等式目錄TOC\o"1-1"\h\u專題3-8利用導(dǎo)函數(shù)證明不等式 1 1題型一：作差法構(gòu)造函數(shù)證明不等式 1題型二：放縮法 9題型三：數(shù)列不等式證明 16 22題型一：作差法構(gòu)造函數(shù)證明不等式【典例分析】例題1．（2022·河南·扶溝縣第二高中高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)當(dāng)SKIPIF1<0時，證明：對任意的SKIPIF1<0，SKIPIF1<0．例題2．（2022·四川·射洪中學(xué)高二期中）已知函數(shù)SKIPIF1<0．(1)若函數(shù)SKIPIF1<0的圖象在點SKIPIF1<0處的切線SKIPIF1<0與直線SKIPIF1<0平行，求切線SKIPIF1<0的方程；(2)若函數(shù)SKIPIF1<0，求證：SKIPIF1<0．【提分秘籍】利用導(dǎo)數(shù)證明不等式SKIPIF1<0的基本方法,可作差構(gòu)造函數(shù)SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．同時SKIPIF1<0，即SKIPIF1<0或若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．同時SKIPIF1<0，即SKIPIF1<0．【變式演練】1．（2022·河南南陽·高二階段練習(xí)（理））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0的極值；(2)當(dāng)SKIPIF1<0時，求證：SKIPIF1<0．2．（2022·廣東中山·高二期末）已知函數(shù)SKIPIF1<0在SKIPIF1<0處有極值2．（Ⅰ）求SKIPIF1<0，SKIPIF1<0的值；（Ⅱ）證明：SKIPIF1<0．3．（2022·廣東·中山紀(jì)念中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0的最小值為SKIPIF1<0．(1)求實數(shù)SKIPIF1<0的值；(2)求證：當(dāng)SKIPIF1<0時，SKIPIF1<0．4．（2022·河南安陽·高二階段練習(xí)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若SKIPIF1<0在定義域上單調(diào)遞減，求SKIPIF1<0的取值范圍；(2)若SKIPIF1<0，SKIPIF1<0，證明：當(dāng)SKIPIF1<0時，SKIPIF1<0.題型二：放縮法【典例分析】例題1．（2022·新疆·克拉瑪依市高級中學(xué)高二階段練習(xí)（理））已知函數(shù)SKIPIF1<0(1)當(dāng)SKIPIF1<0時，求SKIPIF1<0的單調(diào)區(qū)間；(2)當(dāng)SKIPIF1<0時，求證：SKIPIF1<0．例題2．（2022·河南駐馬店·高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0在SKIPIF1<0處的切線經(jīng)過點SKIPIF1<0．(1)求SKIPIF1<0的值；(2)證明：當(dāng)SKIPIF1<0時，SKIPIF1<0．【提分秘籍】放縮構(gòu)造法證明不等式導(dǎo)數(shù)的綜合應(yīng)用題中，最常見的就是SKIPIF1<0和SKIPIF1<0與其他代數(shù)式結(jié)合的問題，對于這類問題，可以先對SKIPIF1<0和SKIPIF1<0進(jìn)行放縮，使問題簡化，便于化簡或判斷導(dǎo)數(shù)的正負(fù)．常見的放縮不等式如下：①SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時取等號；②SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時取等號；③當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時取等號；④當(dāng)SKIPIF1<0時，SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時取等號；⑤SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時取等號；⑥當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時取等號．【變式演練】1．（2022·吉林·東北師大附中模擬預(yù)測（文））已知函數(shù)SKIPIF1<0，SKIPIF1<0，(1)判斷函數(shù)SKIPIF1<0的單調(diào)性；(2)證明：SKIPIF1<0.2．（2022·黑龍江·哈爾濱市第六中學(xué)校高二期末）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的單調(diào)區(qū)間；(2)證明：SKIPIF1<0.3．（2022·河北深州市中學(xué)高三階段練習(xí)）在研究函數(shù)問題時，我們經(jīng)常遇到求函數(shù)在某個區(qū)間上值域的問題，但函數(shù)在區(qū)間端點又恰好沒有意義的情況，此時我們就可以用函數(shù)在這點處的極限來刻畫該點附近數(shù)的走勢，從而得到數(shù)在區(qū)間上的值域.求極限我們有多種方法，其中有一種十分簡單且好用的方法——洛必達(dá)法則該法則表述為：“設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0滿足下列條件：①SKIPIF1<0，SKIPIF1<0；②在點a處函數(shù)SKIPIF1<0和SKIPIF1<0的圖像是連續(xù)且光滑的，即函數(shù)SKIPIF1<0和SKIPIF1<0在點a處存在導(dǎo)數(shù)；③SKIPIF1<0，其中A是某固定實數(shù)；則SKIPIF1<0.”那么，假設(shè)有函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若SKIPIF1<0恒成立，求t的取值范圍；(2)證明：SKIPIF1<0.4．（2022·重慶市永川北山中學(xué)校高二期中）已知函數(shù)SKIPIF1<0,SKIPIF1<0.(1)求證：SKIPIF1<0；(2)設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，求實數(shù)SKIPIF1<0的取值范圍．題型三：數(shù)列不等式證明【典例分析】例題1．（2022·河南駐馬店·高三期中（理））已知函數(shù)SKIPIF1<0(1)求SKIPIF1<0的最大值；(2)求證：SKIPIF1<0例題2．（2020·陜西·安康市教學(xué)研究室三模（文））已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的極值；(2)證明：SKIPIF1<0，SKIPIF1<0.【提分秘籍】數(shù)列中不等式的證明本身就是放縮的結(jié)果，在證明過程中，要善于觀察數(shù)列通項的特點，結(jié)合不等式的結(jié)構(gòu)合理地選擇放大與縮小，常見的兩種放縮方式是：①放縮成等比數(shù)列求和形式；②放縮成裂項求和形式.③借助超越不等式放縮【變式演練】1．（2022·湖南·長沙市麓山濱江實驗學(xué)校高三開學(xué)考試）已知函數(shù)SKIPIF1<0．(1)試比較SKIPIF1<0與1的大小；(2)求證：SKIPIF1<0（SKIPIF1<0）．2．（2022·山西·高三期中）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，比較SKIPIF1<0與2的大??；(2)求證：SKIPIF1<0，SKIPIF1<0．3．（2022·安徽省宣城中學(xué)高二期末）已知函數(shù)SKIPIF1<0.證明：(1)當(dāng)SKIPIF1<0，不等式SKIPIF1<0恒成立；(2)對于任意正整數(shù)SKIPIF1<0，不等式SKIPIF1<0恒成立（其中SKIPIF1<0為自然常數(shù)）一、單選題1．（2021·黑龍江·哈爾濱三中高二期末（文））已知命題SKIPIF1<0：SKIPIF1<0，SKIPIF1<0；命題SKIPIF1<0：若SKIPIF1<0，則SKIPIF1<0下列命題為真命題的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·河南·高二階段練習(xí)（理））若當(dāng)SKIPIF1<0時，關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·山西長治·模擬預(yù)測（理））若SKIPIF1<0，滿足SKIPIF1<0，則SKIPIF1<0（

）A．98 B．99 C．100 D．1014．（2022·湖北武漢·高三開學(xué)考試）若SKIPIF1<0，其中SKIPIF1<0，SKIPIF1<0，則下列結(jié)論一定成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·浙江·效實中學(xué)模擬預(yù)測）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0是自然對數(shù)的底數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<06．（2022·全國·高三專題練習(xí)）下列不等式中正確的是①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0.A．①③ B．①②③ C．② D．①②二、多選題7．（2022·河北滄州·二模）已知實數(shù)SKIPIF1<0滿足SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、填空題8．（2022·重慶一中高三期中）已知函數(shù)SKIPIF1<0，關(guān)于SKIPIF1<0的不等式，SKIPIF1<0的解集是SKIPIF1<0，則SKIPIF1<0___________.9．（2022·江蘇南京·高三階段練習(xí)）當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍為__________.四、解答題10．（2022·全國·高三階段練習(xí)）已知函數(shù)SKIPIF1<0．(1)求曲線SKIPIF1<0在SKIPIF1<0處的切線方程；(2)當(dāng)SKIPIF1<0時，求證：SKIPIF1<0．11．（2022·貴州·高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0(1)求SKIPIF1<0的單調(diào)區(qū)間；(2)證明：SKIPIF1<012．（2022·上海·格致中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0