新高考數(shù)學(xué)三輪沖刺 押題卷練習(xí)第17題 導(dǎo)數(shù)綜合應(yīng)用（解答題）（原卷版）

導(dǎo)數(shù)綜合應(yīng)用（解答題）考點(diǎn)4年考題考情分析導(dǎo)數(shù)綜合2023年新高考Ⅰ卷第19題2023年新高考Ⅱ卷第22題2022年新高考Ⅰ卷第22題2022年新高考Ⅱ卷第22題2021年新高考Ⅰ卷第22題2021年新高考Ⅱ卷第22題2020年新高考Ⅰ卷第21題2020年新高考Ⅱ卷第22題導(dǎo)數(shù)大題難度中等或較難，縱觀近幾年的新高考試題，主要求極值最值、用導(dǎo)數(shù)研究函數(shù)單調(diào)性問(wèn)題及參數(shù)范圍求解、不等式證明問(wèn)題、零點(diǎn)及恒成立問(wèn)題等知識(shí)點(diǎn)，同時(shí)也是高考沖刺復(fù)習(xí)的重點(diǎn)復(fù)習(xí)內(nèi)容?？梢灶A(yù)測(cè)2024年新高考命題方向?qū)⒗^續(xù)以導(dǎo)數(shù)綜合問(wèn)題之單調(diào)性、極值最值、求解及證明問(wèn)題為背景展開命題，難度會(huì)降低．1．（2023·新高考Ⅰ卷高考真題第19題）已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．2．（2023·新高考Ⅱ卷高考真題第22題）（1）證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；（2）已知函數(shù)SKIPIF1<0，若SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)，求a的取值范圍．3．（2022·新高考Ⅰ卷高考真題第22題）已知函數(shù)SKIPIF1<0和SKIPIF1<0有相同的最小值．(1)求a；(2)證明：存在直線SKIPIF1<0，其與兩條曲線SKIPIF1<0和SKIPIF1<0共有三個(gè)不同的交點(diǎn)，并且從左到右的三個(gè)交點(diǎn)的橫坐標(biāo)成等差數(shù)列．4．（2022·新高考Ⅱ卷高考真題第22題）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求a的取值范圍；(3)設(shè)SKIPIF1<0，證明：SKIPIF1<0．5．（2021·新高考Ⅰ卷高考真題第22題）已知函數(shù)SKIPIF1<0.（1）討論SKIPIF1<0的單調(diào)性；（2）設(shè)SKIPIF1<0，SKIPIF1<0為兩個(gè)不相等的正數(shù)，且SKIPIF1<0，證明：SKIPIF1<0.6．（2021·新高考Ⅱ卷高考真題第22題）已知函數(shù)SKIPIF1<0．（1）討論SKIPIF1<0的單調(diào)性；（2）從下面兩個(gè)條件中選一個(gè)，證明：SKIPIF1<0只有一個(gè)零點(diǎn)①SKIPIF1<0；②SKIPIF1<0．導(dǎo)函數(shù)與原函數(shù)的關(guān)系SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減極值極值的定義SKIPIF1<0在SKIPIF1<0處先↗后↘，SKIPIF1<0在SKIPIF1<0處取得極大值SKIPIF1<0在SKIPIF1<0處先↘后↗，SKIPIF1<0在SKIPIF1<0處取得極小值兩招破解不等式的恒成立問(wèn)題(1)a≥f(x)恒成立?a≥f(x)max；(2)a≤f(x)恒成立?a≤f(x)min.(1)分離參數(shù)法第一步：將原不等式分離參數(shù)，轉(zhuǎn)化為不含參數(shù)的函數(shù)的最值問(wèn)題；第二步：利用導(dǎo)數(shù)求該函數(shù)的最值；第三步：根據(jù)要求得所求范圍．(2)函數(shù)思想法第一步將不等式轉(zhuǎn)化為含待求參數(shù)的函數(shù)的最值問(wèn)題；第二步：利用導(dǎo)數(shù)求該函數(shù)的極值；第三步：構(gòu)建不等式求解．常用函數(shù)不等式：①SKIPIF1<0，其加強(qiáng)不等式SKIPIF1<0；②SKIPIF1<0，其加強(qiáng)不等式SKIPIF1<0.③SKIPIF1<0，SKIPIF1<0，SKIPIF1<0放縮SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0利用導(dǎo)數(shù)證明不等式問(wèn)題：（1）直接構(gòu)造函數(shù)法：證明不等式SKIPIF1<0（或SKIPIF1<0）轉(zhuǎn)化為證明SKIPIF1<0（或SKIPIF1<0），進(jìn)而構(gòu)造輔助函數(shù)SKIPIF1<0；（2）轉(zhuǎn)化為證不等式SKIPIF1<0（或SKIPIF1<0），進(jìn)而轉(zhuǎn)化為證明SKIPIF1<0（SKIPIF1<0），因此只需在所給區(qū)間內(nèi)判斷SKIPIF1<0的符號(hào)，從而得到函數(shù)SKIPIF1<0的單調(diào)性，并求出函數(shù)SKIPIF1<0的最小值即可.證明極值點(diǎn)偏移的相關(guān)問(wèn)題，一般有以下幾種方法：（1）證明SKIPIF1<0（或SKIPIF1<0）：①首先構(gòu)造函數(shù)SKIPIF1<0，求導(dǎo)，確定函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0的單調(diào)性；②確定兩個(gè)零點(diǎn)SKIPIF1<0，且SKIPIF1<0，由函數(shù)值SKIPIF1<0與SKIPIF1<0的大小關(guān)系，得SKIPIF1<0與零進(jìn)行大小比較；③再由函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性得到SKIPIF1<0與SKIPIF1<0的大小，從而證明相應(yīng)問(wèn)題；（2）證明SKIPIF1<0（或SKIPIF1<0）（SKIPIF1<0、SKIPIF1<0都為正數(shù)）：①首先構(gòu)造函數(shù)SKIPIF1<0，求導(dǎo)，確定函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0的單調(diào)性；②確定兩個(gè)零點(diǎn)SKIPIF1<0，且SKIPIF1<0，由函數(shù)值SKIPIF1<0與SKIPIF1<0的大小關(guān)系，得SKIPIF1<0與零進(jìn)行大小比較；③再由函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性得到SKIPIF1<0與SKIPIF1<0的大小，從而證明相應(yīng)問(wèn)題；（3）應(yīng)用對(duì)數(shù)平均不等式SKIPIF1<0證明極值點(diǎn)偏移：①由題中等式中產(chǎn)生對(duì)數(shù)；②將所得含對(duì)數(shù)的等式進(jìn)行變形得到SKIPIF1<0；③利用對(duì)數(shù)平均不等式來(lái)證明相應(yīng)的問(wèn)題.1．（2024·湖南衡陽(yáng)·二模）已知函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得極值SKIPIF1<0．(1)求SKIPIF1<0的解析式；(2)求SKIPIF1<0在區(qū)間SKIPIF1<0上的最值．2．（2024·河北·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0在SKIPIF1<0處的切線為SKIPIF1<0軸．(1)求SKIPIF1<0的值；(2)求SKIPIF1<0的單調(diào)區(qū)間．3．（2024·廣東韶關(guān)·二模）已知函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線平行于SKIPIF1<0軸．(1)求實(shí)數(shù)SKIPIF1<0；(2)求SKIPIF1<0的單調(diào)區(qū)間和極值．4．（2024·廣東·一模）已知SKIPIF1<0，函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的單調(diào)區(qū)間.(2)討論方程SKIPIF1<0的根的個(gè)數(shù).5．（2024·浙江金華·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0．(1)求函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程；(2)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的最小值．6．（2024·江蘇徐州·一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，求a的取值范圍：(2)若直線SKIPIF1<0與SKIPIF1<0的圖象相切，求a的值．7．（2024·重慶·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0．(1)求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)證明：SKIPIF1<0．8．（2024·遼寧·一模）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SKIPIF1<0的方程；(2)討論SKIPIF1<0的極值.9．（2024·遼寧·二模）已知函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直.(1)求SKIPIF1<0的值；(2)求SKIPIF1<0的單調(diào)區(qū)間和極值.10．（2024·廣東深圳·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求曲線SKIPIF1<0在SKIPIF1<0處的切線方程；(2)求證：SKIPIF1<0的極大值恒為正數(shù)．11．（2024·廣東廣州·一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)求SKIPIF1<0的單調(diào)區(qū)間和極小值；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.12．（2024·湖南·二模）已函數(shù)SKIPIF1<0，其圖象的對(duì)稱中心為SKIPIF1<0.(1)求SKIPIF1<0的值；(2)判斷函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù).13．（2024·湖南邵陽(yáng)·二模）設(shè)函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的極值；(2)若對(duì)任意SKIPIF1<0，有SKIPIF1<0恒成立，求SKIPIF1<0的最大值.14．（2024·山東濟(jì)南·一模）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的單調(diào)區(qū)間；(2)討論SKIPIF1<0極值點(diǎn)的個(gè)數(shù).15．（2024·山東青島·一模）已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為1，求該切線的方程；(2)討論SKIPIF1<0的單調(diào)性．16．（2024·福建漳州·一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0且SKIPIF1<0．(1)證明：曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程過(guò)坐標(biāo)原點(diǎn)．(2)討論函數(shù)SKIPIF1<0的單調(diào)性．17．（2024·江蘇南通·二模）設(shè)函數(shù)SKIPIF1<0.已知SKIPIF1<0的圖象的兩條相鄰對(duì)稱軸間的距離為SKIPIF1<0，且SKIPIF1<0.(1)若SKIPIF1<0在區(qū)間SKIPIF1<0上有最大值無(wú)最小值，求實(shí)數(shù)m的取值范圍；(2)設(shè)l為曲線SKIPIF1<0在SKIPIF1<0處的切線，證明：l與曲線SKIPIF1<0有唯一的公共點(diǎn).18．（2024·重慶·一模）（1）已知函數(shù)SKIPIF1<0，（SKIPIF1<0為自然對(duì)數(shù)的底數(shù)），記SKIPIF1<0的最小值為SKIPIF1<0，求證：SKIPIF1<0；（2）若對(duì)SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．19．（2024·河北唐山·一模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，(1)求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程：(2)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的值域．20．（2024·遼寧大連·一模）已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0恒成立，求a的取值范圍；(2)當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0．21．（2024·河北滄州·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，SKIPIF1<0恒成立，求實(shí)數(shù)a的取值范圍．22．（2024·湖北武漢·模擬預(yù)測(cè)）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是其定義域上的增函數(shù)；(3)若SKIPIF1<0，其中SKIPIF1<0且SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的值．23．（2024·山東棗莊·一模）已知SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，求SKIPIF1<0的取值范圍．24．（2024·福建福州·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)求證：SKIPIF1<0；(3)若SKIPIF1<0且SKIPIF1<0，求證：SKIPIF1<0．25．（2024·浙江·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的值域；(2)若函數(shù)SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍.26．（2024·浙江·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0．(1)如果1和SKIPIF1<0是SKIPIF1<0的兩個(gè)極值點(diǎn)，且SKIPIF1<0的極大值為3，求SKIPIF1<0的極小值；(2)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的單調(diào)性；(3)當(dāng)SKIPIF1<0時(shí)，且函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上最大值為2，最小值為SKIPIF1<0．求SKIPIF