新高考數(shù)學(xué)二輪考點(diǎn)培優(yōu)專題（精講+精練）11 導(dǎo)數(shù)中的不等式證明問(wèn)題（原卷版）

培優(yōu)專題11導(dǎo)數(shù)中的不等式證明問(wèn)題（精講+精練）一、知識(shí)點(diǎn)梳理一、知識(shí)點(diǎn)梳理一、不等式的證明證明不等式的過(guò)程中常使用構(gòu)造法，利用函數(shù)單調(diào)性、極值、最值加以證明．常見(jiàn)的構(gòu)造方法有：(1)直接構(gòu)造法：證明不等式f(x)＞g(x)(f(x)＜g(x))轉(zhuǎn)化為證明f(x)－g(x)＞0(f(x)－g(x)＜0)，進(jìn)而構(gòu)造輔助函數(shù)h(x)＝f(x)－g(x)；(2)適當(dāng)放縮構(gòu)造法：一是根據(jù)已知條件適當(dāng)放縮，二是利用常見(jiàn)的放縮結(jié)論，如①對(duì)數(shù)形式：x≥1＋lnx（x＞0），當(dāng)且僅當(dāng)x＝1時(shí)，等號(hào)成立.②指數(shù)形式：ex≥x＋1（x∈R），當(dāng)且僅當(dāng)x＝0時(shí)，等號(hào)成立.進(jìn)一步可得到一組不等式鏈：ex＞x＋1＞x＞1＋lnx（x＞0，且x≠1）.(3)構(gòu)造“形似”函數(shù)：稍作變形再構(gòu)造，對(duì)原不等式同解變形，如移項(xiàng)、通分、取對(duì)數(shù)，把不等式轉(zhuǎn)化為左、右兩邊是相同結(jié)構(gòu)的式子的形式，根據(jù)“相同結(jié)構(gòu)”構(gòu)造輔助函數(shù)；(4)構(gòu)造雙函數(shù)：若直接構(gòu)造函數(shù)求導(dǎo)難以判斷符號(hào)，導(dǎo)函數(shù)零點(diǎn)也不易求得，因此函數(shù)單調(diào)性與極值點(diǎn)都不易獲得，則可構(gòu)造函數(shù)f(x)和g(x)，利用其最值求解．在證明過(guò)程中，等價(jià)轉(zhuǎn)化是關(guān)鍵，此處f(x)min>g(x)max恒成立．從而f(x)>g(x)，但此處f(x)與g(x)取到最值的條件不是同一個(gè)“x的值”．【常用結(jié)論】1.破解含雙參不等式證明題的3個(gè)關(guān)鍵點(diǎn)（1）轉(zhuǎn)化，即由已知條件入手，尋找雙參所滿足的關(guān)系式，并把含雙參的不等式轉(zhuǎn)化為含單參的不等式.（2）巧構(gòu)造函數(shù)，再借用導(dǎo)數(shù)，判斷函數(shù)的單調(diào)性，從而求其最值.（3）回歸雙參的不等式的證明，把所求的最值應(yīng)用到雙參不等式，即可證得結(jié)果.總結(jié)：雙變量相關(guān)問(wèn)題，解題策略是減少變量，方式為一個(gè)變量用另一個(gè)變量表示，或?qū)勺兞康恼w換元，如下列形式SKIPIF1<0等常見(jiàn)形式2.常見(jiàn)不等式（大題使用需要證明）①SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0②SKIPIF1<0，SKIPIF1<0；SKIPIF1<0；SKIPIF1<0③SKIPIF1<0；SKIPIF1<0；SKIPIF1<0④SKIPIF1<0；SKIPIF1<0⑤SKIPIF1<0；SKIPIF1<0⑥SKIPIF1<0；SKIPIF1<0；SKIPIF1<0，SKIPIF1<0二、題型精講精練二、題型精講精練【典例1】已知函數(shù)SKIPIF1<0.（1）討論SKIPIF1<0的單調(diào)性；（2）當(dāng)SKIPIF1<0時(shí)，證明SKIPIF1<0【典例2】SKIPIF1<0求證：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0【典例3】已知函數(shù)SKIPIF1<0，SKIPIF1<0．（1）討論函數(shù)SKIPIF1<0的單調(diào)性；（2）若SKIPIF1<0、SKIPIF1<0為函數(shù)SKIPIF1<0的兩個(gè)極值點(diǎn)，證明：SKIPIF1<0．【題型訓(xùn)練1-刷真題】一、解答題1．設(shè)函數(shù)SKIPIF1<0，已知SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)．（1）求a；（2）設(shè)函數(shù)SKIPIF1<0．證明:SKIPIF1<0．2．設(shè)a，b為實(shí)數(shù)，且SKIPIF1<0，函數(shù)SKIPIF1<0（1）求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；（2）若對(duì)任意SKIPIF1<0，函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)，求a的取值范圍；（3）當(dāng)SKIPIF1<0時(shí)，證明：對(duì)任意SKIPIF1<0，函數(shù)SKIPIF1<0有兩個(gè)不同的零點(diǎn)SKIPIF1<0，滿足SKIPIF1<0.(注：SKIPIF1<0是自然對(duì)數(shù)的底數(shù))3．已知SKIPIF1<0，函數(shù)SKIPIF1<0，其中e=2.71828…為自然對(duì)數(shù)的底數(shù)．（Ⅰ）證明：函數(shù)SKIPIF1<0在SKIPIF1<0上有唯一零點(diǎn)；（Ⅱ）記x0為函數(shù)SKIPIF1<0在SKIPIF1<0上的零點(diǎn)，證明：（?。㏒KIPIF1<0；（ⅱ）SKIPIF1<0．【題型訓(xùn)練2-刷模擬】1．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<0.2．已知函數(shù)SKIPIF1<0.(1)討論函數(shù)SKIPIF1<0在SKIPIF1<0上的零點(diǎn)個(gè)數(shù)；(2)當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，記SKIPIF1<0，探究SKIPIF1<0與1的大小關(guān)系，并說(shuō)明理由.3．已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.4．已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，求SKIPIF1<0的極值；(2)SKIPIF1<0，若函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0，且SKIPIF1<0，求證：SKIPIF1<0.5．已知函數(shù)SKIPIF1<0.(1)判斷SKIPIF1<0的導(dǎo)函數(shù)在SKIPIF1<0上零點(diǎn)的個(gè)數(shù)，并說(shuō)明理由；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.注：SKIPIF1<0.6．已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)證明：當(dāng)SKIPIF1<0，且SKIPIF1<0時(shí)，SKIPIF1<0．7．已知函數(shù)SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性.(2)若SKIPIF1<0存在兩個(gè)零點(diǎn)SKIPIF1<0，且曲線SKIPIF1<0在SKIPIF1<0和SKIPIF1<0處的切線交于點(diǎn)SKIPIF1<0.①求實(shí)數(shù)SKIPIF1<0的取值范圍；②證明：SKIPIF1<0.8．已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，求實(shí)數(shù)a的取值范圍；(2)當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0，SKIPIF1<0．9．已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0，SKIPIF1<0是方程SKIPIF1<0的兩根，SKIPIF1<0，證明：SKIPIF1<0．10．已知函數(shù)SKIPIF1<0，SKIPIF1<0(1)試判斷函數(shù)SKIPIF1<0在SKIPIF1<0上是否存在極值.若存在，說(shuō)出是極大值還是極小值；若不存在，說(shuō)明理由.(2)設(shè)SKIPIF1<0，若SKIPIF1<0，證明：不等式SKIPIF1<0在SKIPIF1<0上恒成立.11．已知函數(shù)SKIPIF1<0，其中SKIPIF1<0．(1)若SKIPIF1<0有兩個(gè)零點(diǎn)，求SKIPIF