新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)專(zhuān)題23 導(dǎo)數(shù)的綜合問(wèn)題（單選+填空）（解析版）

試卷第=page11頁(yè)，共=sectionpages33頁(yè)專(zhuān)題23導(dǎo)數(shù)的綜合問(wèn)題（單選+填空）一、單選題1．（2023秋·湖北·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0及其導(dǎo)函數(shù)SKIPIF1<0的定義域都為SKIPIF1<0，且SKIPIF1<0為偶函數(shù)，SKIPIF1<0為奇函數(shù)，則下列說(shuō)法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)函數(shù)的奇偶性對(duì)稱(chēng)性可得函數(shù)的周期性以及SKIPIF1<0，再利用復(fù)合函數(shù)的導(dǎo)數(shù)推出SKIPIF1<0的周期以及SKIPIF1<0，進(jìn)而可求解.【詳解】因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0，即SKIPIF1<0，即函數(shù)圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0，因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0，即函數(shù)圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以函數(shù)以4為周期，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0，令SKIPIF1<0，則有SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0以4為周期，所以SKIPIF1<0，所以SKIPIF1<0，C正確，對(duì)于其余選項(xiàng)，根據(jù)題意可假設(shè)SKIPIF1<0滿足周期為4，且關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，SKIPIF1<0，故A錯(cuò)誤；SKIPIF1<0，B錯(cuò)誤；SKIPIF1<0，D錯(cuò)誤，故選:C.2．（2023·江蘇泰州·統(tǒng)考一模）若過(guò)點(diǎn)SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線，切點(diǎn)分別為SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】設(shè)切點(diǎn)SKIPIF1<0，根據(jù)導(dǎo)數(shù)的幾何意義求得切線方程，再根據(jù)切線過(guò)點(diǎn)SKIPIF1<0，結(jié)合韋達(dá)定理可得SKIPIF1<0的關(guān)系，進(jìn)而可得SKIPIF1<0的關(guān)系，再利用導(dǎo)數(shù)即可得出答案.【詳解】設(shè)切點(diǎn)SKIPIF1<0，則切線方程為SKIPIF1<0，又切線過(guò)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0有兩個(gè)不相等實(shí)根SKIPIF1<0，其中SKIPIF1<0或SKIPIF1<0，SKIPIF1<0令SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，SKIPIF1<0，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.故選：D.3．（2023春·廣東惠州·高三?？茧A段練習(xí)）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】因?yàn)椴坏仁絊KIPIF1<0等價(jià)于SKIPIF1<0，故考慮構(gòu)造函數(shù)SKIPIF1<0，結(jié)合已知條件證明其單調(diào)性，結(jié)合單調(diào)性解不等式即可.【詳解】令SKIPIF1<0，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0所以，SKIPIF1<0故SKIPIF1<0故SKIPIF1<0在R上單調(diào)遞減，又因?yàn)镾KIPIF1<0所以，SKIPIF1<0，所以不等式SKIPIF1<0可化為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的解集為SKIPIF1<0故選：B.4．（2023·重慶沙坪壩·高三重慶八中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，若對(duì)于定義域內(nèi)的任意實(shí)數(shù)s，總存在實(shí)數(shù)t使得SKIPIF1<0，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)已知條件將問(wèn)題轉(zhuǎn)化為求函數(shù)沒(méi)有最小值問(wèn)題，利用導(dǎo)數(shù)法求函數(shù)的最值的步驟，但要注意對(duì)參數(shù)SKIPIF1<0進(jìn)行分類(lèi)討論即可求解.【詳解】由題意可知，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，因?yàn)閷?duì)于定義域內(nèi)的任意實(shí)數(shù)s，總存在實(shí)數(shù)t使得SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上沒(méi)有最小值，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，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)遞增，在SKIPIF1<0上單調(diào)遞減.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值為SKIPIF1<0，值域?yàn)镾KIPIF1<0，SKIPIF1<0在SKIPIF1<0內(nèi)無(wú)最小值，因此SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，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)遞增，在SKIPIF1<0上單調(diào)遞減.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值為SKIPIF1<0，顯然SKIPIF1<0，即SKIPIF1<0，在同一坐標(biāo)系內(nèi)作出直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象，如圖所示當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有兩個(gè)根SKIPIF1<0，不妨設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增.所以SKIPIF1<0在SKIPIF1<0與SKIPIF1<0處都取得極小值，SKIPIF1<0，不符合題意，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)取到等號(hào)，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.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值為SKIPIF1<0，不符合題意，綜上所述，實(shí)數(shù)a的取值范圍為SKIPIF1<0故選：D.【點(diǎn)睛】解決本題的關(guān)鍵是將問(wèn)題轉(zhuǎn)化為求函數(shù)SKIPIF1<0沒(méi)有最小值，利用導(dǎo)數(shù)法求函數(shù)的最值步驟，但在研究SKIPIF1<0與SKIPIF1<0的大小關(guān)系時(shí)，借住函數(shù)的圖象，得出對(duì)SKIPIF1<0分SKIPIF1<0和SKIPIF1<0兩種情況討論即可求解.5．（2023春·江蘇南京·高三南京師大附中校考開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)，則實(shí)數(shù)a的取值范圍（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用多次求導(dǎo)的方法，列不等式來(lái)求得SKIPIF1<0的取值范圍.【詳解】SKIPIF1<0的定義域是SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0遞減；在區(qū)間SKIPIF1<0遞增.要使SKIPIF1<0有兩個(gè)極值點(diǎn)，則SKIPIF1<0，此時(shí)SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0，所以SKIPIF1<0，所以實(shí)數(shù)a的取值范圍SKIPIF1<0.故選：D【點(diǎn)睛】利用導(dǎo)數(shù)研究函數(shù)的極值點(diǎn)，當(dāng)一次求導(dǎo)無(wú)法求得函數(shù)的單調(diào)性時(shí)，可利用二次求導(dǎo)的方法來(lái)進(jìn)行求解.在求解的過(guò)程中，要注意原函數(shù)和導(dǎo)函數(shù)間的對(duì)應(yīng)關(guān)系.6．（2023·福建漳州·統(tǒng)考三模）已知函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0，具有相同的零點(diǎn)SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)零點(diǎn)定義可整理得到SKIPIF1<0，令SKIPIF1<0，利用導(dǎo)數(shù)，結(jié)合零點(diǎn)存在定理的知識(shí)可確定SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，并得到SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0可確定SKIPIF1<0，由此化簡(jiǎn)所求式子即可得到結(jié)果.【詳解】由題意知：SKIPIF1<0，SKIPIF1<0，聯(lián)立兩式可得：SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上存在唯一零點(diǎn)SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，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)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：C.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題考查函數(shù)零點(diǎn)、利用導(dǎo)數(shù)求解函數(shù)單調(diào)性的相關(guān)問(wèn)題；解題關(guān)鍵是能夠靈活應(yīng)用零點(diǎn)存在定理確定導(dǎo)函數(shù)的正負(fù)，并得到隱零點(diǎn)所滿足的等量關(guān)系式，進(jìn)而利用等量關(guān)系式化簡(jiǎn)最值和所求式子.7．（2023·湖南·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0（e是自然對(duì)數(shù)的底數(shù)），若存在SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由SKIPIF1<0，得到SKIPIF1<0，再研究函數(shù)SKIPIF1<0的單調(diào)性，得到SKIPIF1<0，將SKIPIF1<0表示出來(lái)，然后利用換元法轉(zhuǎn)化為二次函數(shù)求最值即可.【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，SKIPIF1<0在SKIPIF1<0處取得最小值SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最小值SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值0，所以SKIPIF1<0的取值范圍是SKIPIF1<0.故選：A【點(diǎn)睛】方法點(diǎn)睛：對(duì)于利用導(dǎo)數(shù)研究函數(shù)的綜合問(wèn)題的求解策略：1、通常要構(gòu)造新函數(shù)，利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，求出最值，從而求出參數(shù)的取值范圍；2、利用可分離變量，構(gòu)造新函數(shù)，直接把問(wèn)題轉(zhuǎn)化為函數(shù)的最值問(wèn)題．3、根據(jù)恒成立或有解求解參數(shù)的取值時(shí)，一般涉及分離參數(shù)法，但壓軸試題中很少碰到分離參數(shù)后構(gòu)造的新函數(shù)能直接求出最值點(diǎn)的情況，進(jìn)行求解，若參變分離不易求解問(wèn)題，就要考慮利用分類(lèi)討論法和放縮法，注意恒成立與存在性問(wèn)題的區(qū)別．8．（2023·湖南邵陽(yáng)·統(tǒng)考二模）若不等式SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，則正實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】由題意得SKIPIF1<0恒成立，令SKIPIF1<0，則SKIPIF1<0恒成立，利用SKIPIF1<0的單調(diào)性可得SKIPIF1<0在SKIPIF1<0時(shí)恒成立，即SKIPIF1<0恒成立，構(gòu)造函數(shù)SKIPIF1<0，由其單調(diào)性得SKIPIF1<0，即可得出答案.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0恒成立，即SKIPIF1<0恒成立．令SKIPIF1<0，則SKIPIF1<0恒成立．因?yàn)镾KIPIF1<0恒成立，故SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0時(shí)恒成立，∴SKIPIF1<0恒成立．令SKIPIF1<0，SKIPIF1<0SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0SKIPIF1<0∴SKIPIF1<0單調(diào)遞減．∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0單調(diào)遞減，故SKIPIF1<0．則正實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B．【點(diǎn)睛】方法點(diǎn)睛：不等式恒成立問(wèn)題常見(jiàn)方法：①分離參數(shù)法：分離出函數(shù)中的參數(shù)，問(wèn)題轉(zhuǎn)化為求新函數(shù)的最值或范圍．若SKIPIF1<0恒成立，則SKIPIF1<0；若SKIPIF1<0恒成立，則SKIPIF1<0；②最值法：通過(guò)對(duì)函數(shù)最值的討論得出結(jié)果．若SKIPIF1<0恒成立，則SKIPIF1<0；若SKIPIF1<0恒成立，則SKIPIF1<0；③分段討論法：對(duì)變量SKIPIF1<0進(jìn)行分段討論，然后再綜合處理．9．（2023春·廣東·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0，直線SKIPIF1<0，若有且僅有一個(gè)整數(shù)SKIPIF1<0，使得點(diǎn)SKIPIF1<0在直線l上方，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由定義域得SKIPIF1<0為正整數(shù)，由導(dǎo)數(shù)法研究SKIPIF1<0的圖象，直線l過(guò)定點(diǎn)SKIPIF1<0，由數(shù)形結(jié)合可判斷SKIPIF1<0的值，進(jìn)而列不等式組確定參數(shù)范圍.【詳解】點(diǎn)SKIPIF1<0在直線l上方，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0有且僅有一個(gè)正整數(shù)解.設(shè)SKIPIF1<0，則SKIPIF1<0單調(diào)遞增；SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0.又SKIPIF1<0，故可得SKIPIF1<0圖象如下圖，直線SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0有無(wú)數(shù)個(gè)正整數(shù)解，不合題意，故SKIPIF1<0，又SKIPIF1<0有且僅有一個(gè)正整數(shù)解，故2是唯一的正整數(shù)解，即SKIPIF1<0.故選：C.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：直線l過(guò)定點(diǎn)，則原命題可轉(zhuǎn)化為直線l繞定點(diǎn)旋轉(zhuǎn)，從而滿足條件，可由導(dǎo)數(shù)法研究SKIPIF1<0的圖象，由數(shù)形結(jié)合列式求解.10．（2023·廣東湛江·統(tǒng)考一模）已知函數(shù)SKIPIF1<0及其導(dǎo)函數(shù)SKIPIF1<0的定義域均為R，且SKIPIF1<0為奇函數(shù)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．13 B．16 C．25 D．51【答案】C【分析】根據(jù)題意利用賦值法求出SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的值，推出函數(shù)SKIPIF1<0的周期，結(jié)合SKIPIF1<0，每四個(gè)值為一個(gè)循環(huán)，即可求得答案.【詳解】由SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0．由SKIPIF1<0為奇函數(shù)，得SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0①．又SKIPIF1<0②，由①和②得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，③令SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，得SKIPIF1<0．又SKIPIF1<0④，由③-④得SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0是以8為周期的周期函數(shù)，故SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0，故選：C．【點(diǎn)睛】方法點(diǎn)睛：解決此類(lèi)抽象函數(shù)的求值問(wèn)題時(shí)，涉及到函數(shù)的性質(zhì)，比如奇偶性和對(duì)稱(chēng)軸以及周期性等問(wèn)題，綜合性較強(qiáng)，有一定難度，解答時(shí)往往要采用賦值法求得某些特殊值，繼而推出函數(shù)滿足的性質(zhì)，諸如對(duì)稱(chēng)性和周期性等，從而解決問(wèn)題.11．（2023秋·江蘇南京·高三南京市第一中學(xué)?？计谀┰O(shè)SKIPIF1<0，函數(shù)SKIPIF1<0滿足SKIPIF1<0，則α落于區(qū)間（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由題意，確定函數(shù)的最大值，根據(jù)最值和極值的關(guān)系，可得方程，利用零點(diǎn)存在性定理，可得答案.【詳解】由題意，可知函數(shù)SKIPIF1<0在SKIPIF1<0上當(dāng)SKIPIF1<0時(shí)取得最大值，且SKIPIF1<0，由于SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，根據(jù)零點(diǎn)存在性定理，可知SKIPIF1<0，故選：C.二、填空題12．（2023春·浙江·高三開(kāi)學(xué)考試）已知定義在SKIPIF1<0上可導(dǎo)函數(shù)SKIPIF1<0，對(duì)于任意的實(shí)數(shù)x都有SKIPIF1<0成立，且當(dāng)SKIPIF1<0時(shí)，都有SKIPIF1<0成立，若SKIPIF1<0，則實(shí)數(shù)m的取值范圍是__________．【答案】SKIPIF1<0【分析】構(gòu)造函數(shù)SKIPIF1<0，討論奇偶性和單調(diào)性，根據(jù)函數(shù)的單調(diào)性和奇偶性解不等式.【詳解】令SKIPIF1<0，則易得SKIPIF1<0，即SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故在SKIPIF1<0上單調(diào)遞增，由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0為偶函數(shù)得SKIPIF1<0，又在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，故答案為：SKIPIF1<0．13．（2023秋·浙江杭州·高三期末）已知不等式SKIPIF1<0，對(duì)SKIPIF1<0恒成立，則a的取值范圍是__________．【答案】SKIPIF1<0【分析】根據(jù)已知得出SKIPIF1<0，對(duì)SKIPIF1<0恒成立，而在SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，將SKIPIF1<0化為SKIPIF1<0，令SKIPIF1<0，根據(jù)導(dǎo)數(shù)得出其單調(diào)性，則SKIPIF1<0可化為SKIPIF1<0，即可根據(jù)單調(diào)性得出SKIPIF1<0，令SKIPIF1<0，根據(jù)導(dǎo)數(shù)得出SKIPIF1<0，即可得出SKIPIF1<0，即可得出答案.【詳解】SKIPIF1<0，對(duì)SKIPIF1<0恒成立，則SKIPIF1<0，對(duì)SKIPIF1<0恒成立，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則要滿足SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0化為：SKIPIF1<0，兩邊乘SKIPIF1<0得：SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0不等式SKIPIF1<0，對(duì)SKIPIF1<0恒成立，即SKIPIF1<0時(shí)，SKIPIF1<0恒成立，則SKIPIF1<0可化為：SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則根據(jù)單調(diào)性可得SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，令SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，綜上SKIPIF1<0，故答案為SKIPIF1<0.14．（2023·江蘇連云港·統(tǒng)考模擬預(yù)測(cè)）已知定義在R上的函數(shù)SKIPIF1<0，若SKIPIF1<0有解，則實(shí)數(shù)a的取值范圍是______________．【答案】SKIPIF1<0【分析】分析SKIPIF1<0的奇偶性和單調(diào)性，根據(jù)奇偶性和單調(diào)性求解.【詳解】SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，又SKIPIF1<0，SKIPIF1<0在R的范圍內(nèi)是增函數(shù)，SKIPIF1<0有解等價(jià)于SKIPIF1<0，SKIPIF1<0有解，令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是增函數(shù)，當(dāng)x趨于SKIPIF1<0時(shí)，SKIPIF1<0趨于SKIPIF1<0，滿足題意；當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0是增函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是減函數(shù)，SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0是增函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0是減函數(shù)，并且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0滿足題意，所以a的取值范圍是SKIPIF1<0；故答案為：SKIPIF1<0.15．（2023·湖北武漢·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0與SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a＝____________.【答案】4【分析】由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，根據(jù)SKIPIF1<0解方程即可求出結(jié)果．【詳解】因?yàn)楹瘮?shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0與SKIPIF1<0由SKIPIF1<0，則SKIPIF1<0有兩根SKIPIF1<0與SKIPIF1<0所以SKIPIF1<0，得SKIPIF1<0因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0則SKIPIF1<0，所以SKIPIF1<0故答案為：SKIPIF1<016．（2023·湖北·統(tǒng)考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0，若關(guān)于x的不等式SKIPIF1<0的解集為SKIPIF1<0，則實(shí)數(shù)a的取值范圍為_(kāi)_________．【答案】SKIPIF1<0【分析】分SKIPIF1<0和SKIPIF1<0兩種情況討論，當(dāng)SKIPIF1<0時(shí)，根據(jù)二次函數(shù)的圖象得到SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，分SKIPIF1<0和SKIPIF1<0兩種情況討論，SKIPIF1<0時(shí)，將SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，然后借助函數(shù)的單調(diào)性和最值解不等式即可.【詳解】由題意知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，結(jié)合圖象知SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，顯然成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0．綜上，實(shí)數(shù)a的取值范圍為SKIPIF1<0．故答案為：SKIPIF1<0．17．（2023春·湖南·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，且滿足SKIPIF1<0在SKIPIF1<0上恒成立，則不等式SKIPIF1<0SKIPIF1<0的解集是____________．【答案】SKIPIF1<0【分析】構(gòu)造函數(shù)SKIPIF1<0，再將SKIPIF1<0SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，進(jìn)而根據(jù)SKIPIF1<0的單調(diào)性求解即可.【詳解】令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，由SKIPIF1<0SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.所以不等式SKIPIF1<0SKIPIF1<0的解集是SKIPIF1<0.故答案為：SKIPIF1<0.18．（2023春·湖南長(zhǎng)沙·高三雅禮中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍__________.【答案】SKIPIF1<0【分析】令SKIPIF1<0，分SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，利用導(dǎo)數(shù)法討論求解.【詳解】解：令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不符合題意；②SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上恒為負(fù)，在區(qū)間SKIPIF1<0上恒為正，則需SKIPIF1<0在區(qū)間SKIPIF1<0上恒為負(fù)，在區(qū)間SKIPIF1<0上恒為正，因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則需SKIPIF1<0，此時(shí)SKIPIF1<0，符合題意；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上恒為負(fù)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0在SKIPIF1<0時(shí)取得極大值也是最大值，SKIPIF1<0，解得SKIPIF1<0.所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<019．（2023·廣東茂名·統(tǒng)考一模）e是自然對(duì)數(shù)的底數(shù)，SKIPIF1<0的零點(diǎn)為_(kāi)_____.【答案】SKIPIF1<0##SKIPIF1<0【分析】只用求方程SKIPIF1<0的零點(diǎn)，討論左右兩個(gè)函數(shù)的最值即可求解.【詳解】由SKIPIF1<0得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，取等號(hào)，令SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0解得SKIPIF1<0；令SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，所以要使SKIPIF1<0，只能SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0零點(diǎn)為SKIPIF1<0，故答案為:SKIPIF1<0.20．（2023·廣東湛江·統(tǒng)考一模）若函數(shù)SKIPIF1<0存在兩個(gè)極值點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0______．【答案】SKIPIF1<0【分析】求導(dǎo)得到SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，解得答案.【詳解】SKIPIF1<0，定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0；又SKIPIF1<0，所以SKIPIF1<0．又SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)睛：本題考查了函數(shù)的極值點(diǎn)問(wèn)題，意在考查學(xué)生的計(jì)算能力，轉(zhuǎn)化能力和綜合應(yīng)用能力，其中利用消元的思想解方程是解題的關(guān)鍵.21．（2023春·浙江·高三校聯(lián)考開(kāi)學(xué)考試）已知曲線SKIPIF1<0與SKIPIF1<0的兩條公切線的夾角正切值為SKIPIF1<0，則SKIPIF1<0________．【答案】SKIPIF1<0【分析】由兩曲線互為反函數(shù)，結(jié)合反函數(shù)性質(zhì)及正切函數(shù)倍角公式，可求得兩條公切線的夾角一半的正切值，即可求得直線AD的斜率.設(shè)點(diǎn)A的橫坐標(biāo)為SKIPIF1<0，切點(diǎn)D的橫坐標(biāo)為SKIPIF1<0，由導(dǎo)數(shù)法分別就A、D兩點(diǎn)求同一條切線方程，從而建立方程，化簡(jiǎn)求值.【詳解】SKIPIF1<0與SKIPIF1<0互為反函數(shù)，圖像關(guān)于直線SKIPIF1<0對(duì)稱(chēng)，如圖所示，由題意，兩條公切線的夾角正切值為SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0為銳角，所以SKIPIF1<0.由對(duì)稱(chēng)性，不妨取AD直線進(jìn)行研究，則直線AD的傾斜角SKIPIF1<0，SKIPIF1<0.設(shè)點(diǎn)A的橫坐標(biāo)為SKIPIF1<0，切點(diǎn)D的橫坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0.所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0.∴SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.【點(diǎn)睛】方法點(diǎn)睛：公切線問(wèn)題，一般可在兩曲線上設(shè)出切點(diǎn)，分別求出切線，利用兩切線為同一條切線得出方程，從而進(jìn)一步求解.22．（2023秋·江蘇蘇州·高三統(tǒng)考期末）若對(duì)任意SKIPIF1<0，關(guān)于x的不等式SKIPIF1<0恒成立，則實(shí)數(shù)a的最大值為_(kāi)_______．【答案】SKIPIF1<0##0.75【分析】不等式化為SKIPIF1<0恒成立，由于SKIPIF1<0都是任意實(shí)數(shù)，因此不等式右邊相當(dāng)于兩個(gè)函數(shù)相加：SKIPIF1<0和SKIPIF1<0，后者設(shè)SKIPIF1<0，由導(dǎo)數(shù)求得其最小值，前者由二次函數(shù)性質(zhì)得最小值，兩者相加即得最小值，從而得SKIPIF1<0的范圍，得出結(jié)論．【詳解】原不等式化為SKIPIF1<0恒成立，由于SKIPIF1<0是任意實(shí)數(shù)，SKIPIF1<0也是任意實(shí)數(shù)，∴SKIPIF1<0與SKIPIF1<0是任意實(shí)數(shù)，它們之間沒(méi)有任何影響，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0的最小值是1，所以SKIPIF1<0的最小值是SKIPIF1<0，從而SKIPIF1<0，SKIPIF1<0的最大值是SKIPIF1<0．故答案為：SKIPIF1<0．【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：不等式恒成立求參數(shù)范圍問(wèn)題，一般可采用分離參數(shù)法轉(zhuǎn)化為求函數(shù)的最值，本題分離參數(shù)后，關(guān)鍵是對(duì)變量的理解，本題中由于SKIPIF1<0都是任意實(shí)數(shù)，因此題中SKIPIF1<0與SKIPIF1<0可以看作是兩個(gè)不同的變量，因此不等式右邊轉(zhuǎn)化為兩個(gè)函數(shù)的和，分別求出其最小值后得出結(jié)論．23．（2023秋·河北唐山·高三統(tǒng)考期末）函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【分析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不合題意；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，可證得SKIPIF1<0，SKIPIF1<0，滿足題意.【詳解】SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，有SKIPIF1<0.SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0.SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0.則當(dāng)SKIPIF1<0時(shí)，不滿足SKIPIF1<0.若SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0解得SKIPIF1<0，SKIPIF1<0解得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，可得SKIPIF1<0所以若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0若SKIPIF1<0，SKIPIF1<0設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0解得SKIPIF1<0，SKIPIF1<0解得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0所以若SKIPIF1<0，SKIPIF1<0.綜上可得：若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立.則SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0【點(diǎn)睛】方法點(diǎn)睛：1.導(dǎo)函數(shù)中常用的兩種常用的轉(zhuǎn)化方法：一是利用導(dǎo)數(shù)研究含參函數(shù)的單調(diào)性，?；癁椴坏仁胶愠闪?wèn)題．注意分類(lèi)討論與數(shù)形結(jié)合思想的應(yīng)用；二是函數(shù)的零點(diǎn)、不等式證明常轉(zhuǎn)化為函數(shù)的單調(diào)性、極(最)值問(wèn)題處理．2.利用導(dǎo)數(shù)解決含參函數(shù)的單調(diào)性問(wèn)題時(shí)，一般將其轉(zhuǎn)化為不等式恒成立問(wèn)題，解題過(guò)程中要注意分類(lèi)討論和數(shù)形結(jié)合思想的應(yīng)用.3..證明不等式，構(gòu)造一個(gè)適當(dāng)?shù)暮瘮?shù)，利用它的單調(diào)性進(jìn)行解題，是一種常用技巧.許多問(wèn)題，如果運(yùn)用這種思想去解決，往往能獲得簡(jiǎn)潔明快的思路，有著非凡的功效.24．（2023春·河北邢臺(tái)·高三邢臺(tái)市第二中學(xué)?？茧A段練習(xí)）已知實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0（其中SKIPIF1<0為自然對(duì)數(shù)的底數(shù)），則SKIPIF1<0的最小值是_________．【答案】SKIPIF1<0##SKIPIF1<0【分析】變形給定不等式，構(gòu)造函數(shù)并借助函數(shù)的單調(diào)性，求出SKIPIF1<0的關(guān)系，再利用導(dǎo)數(shù)求出函數(shù)的最值作答.【詳解】SKIPIF1<0，令函數(shù)SKIPIF1<0，求導(dǎo)得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，于是SKIPIF1<0，即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，依題意，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，求導(dǎo)得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，從而函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，所以SKIPIF1<0的最小值是SKIPIF1<0.故答案為：SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)睛：涉及不等式恒成立問(wèn)題，將給定不等式等價(jià)轉(zhuǎn)化，構(gòu)造函數(shù)，利用導(dǎo)數(shù)探求函數(shù)單調(diào)性、最值是解決問(wèn)題的關(guān)鍵.25．（2023·福建泉州·統(tǒng)考三模）已知函數(shù)SKIPIF1<0有兩個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍為_(kāi)__________．【答案】SKIPIF1<0【分析】零點(diǎn)問(wèn)題可以轉(zhuǎn)為為圖像交點(diǎn)問(wèn)題，然后討論a的取值范圍即可.【詳