高考數(shù)學(xué)二輪復(fù)習(xí)核心專(zhuān)題講練：函數(shù)與導(dǎo)數(shù)第4講 函數(shù)與導(dǎo)數(shù)解答題 原卷版

第4講函數(shù)與導(dǎo)數(shù)解答題目錄第一部分：知識(shí)強(qiáng)化第二部分：重難點(diǎn)題型突破突破一：分離變量法與不等式恒（能）成立問(wèn)題突破二：分類(lèi)討論法與不等式恒（能）成立問(wèn)題突破三：同構(gòu)法與不等式恒（能）成立問(wèn)題突破四：端點(diǎn)效應(yīng)與不等式恒（能）成立問(wèn)題突破五：最值定位法解決雙參不等式問(wèn)題突破六：值域法解決雙參等式問(wèn)題突破七：?jiǎn)巫兞坎坏仁阶C明角度1：構(gòu)造函數(shù)，利用單調(diào)性證明不等式角度2：構(gòu)造函數(shù)，利用最值證明不等式角度3：等價(jià)轉(zhuǎn)化與不等式證明角度4：超越放縮與不等式證明突破八：利用導(dǎo)數(shù)證明雙變量不等式角度1：分離雙參，構(gòu)造函數(shù)角度2：糅合雙參（比值糅合）角度3：糅合雙參（差值糅合）角度4：利用指數(shù)（對(duì)數(shù)）平均不等式解決雙變量問(wèn)題第一部分：知識(shí)強(qiáng)化1、不等式恒成立(能成立)求參數(shù)范圍的類(lèi)型與解法（1）分離參數(shù)法用分離參數(shù)法解含參不等式恒成立問(wèn)題，可以根據(jù)不等式的性質(zhì)將參數(shù)分離出來(lái)，得到一個(gè)一端是參數(shù)，另一端是變量表達(dá)式的不等式；步驟：①分類(lèi)參數(shù)（注意分類(lèi)參數(shù)時(shí)自變量SKIPIF1<0的取值范圍是否影響不等式的方向）②轉(zhuǎn)化：若SKIPIF1<0)對(duì)SKIPIF1<0恒成立，則只需SKIPIF1<0；若SKIPIF1<0對(duì)SKIPIF1<0恒成立，則只需SKIPIF1<0．③求最值.（2）分類(lèi)討論法如果無(wú)法分離參數(shù)，可以考慮對(duì)參數(shù)或自變量進(jìn)行分類(lèi)討論求解，如果是二次不等式恒成立的問(wèn)題，可以考慮二次項(xiàng)系數(shù)與判別式的方法(SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0)求解．（3）等價(jià)轉(zhuǎn)化法當(dāng)遇到SKIPIF1<0型的不等式恒成立問(wèn)題時(shí)，一般采用作差法，構(gòu)造“左減右”的函數(shù)SKIPIF1<0或者“右減左”的函數(shù)SKIPIF1<0，進(jìn)而只需滿足SKIPIF1<0，或者SKIPIF1<0，將比較法的思想融入函數(shù)中，轉(zhuǎn)化為求解函數(shù)的最值的問(wèn)題.2、最值定位法解決雙參不等式問(wèn)題（1）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<0（2）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<0（3）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<0（4）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立SKIPIF1<0SKIPIF1<03、值域法解決雙參等式問(wèn)題（任意——存在型）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立①SKIPIF1<0，求出SKIPIF1<0的值域，記為SKIPIF1<0②SKIPIF1<0求出SKIPIF1<0的值域，記為SKIPIF1<0③則SKIPIF1<0,求出參數(shù)取值范圍.4、值域法解決雙參等式問(wèn)題（存在——存在型）SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0成立①SKIPIF1<0，求出SKIPIF1<0的值域，記為SKIPIF1<0②SKIPIF1<0求出SKIPIF1<0的值域，記為SKIPIF1<0③則SKIPIF1<0,求出參數(shù)取值范圍.5、兩個(gè)超越不等式：（注意解答題需先證明后使用）（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③（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④6、指數(shù)不等式法在對(duì)數(shù)均值不等式中，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，根據(jù)對(duì)數(shù)均值不等式有如下關(guān)系：SKIPIF1<07、對(duì)數(shù)均值不等式法兩個(gè)正數(shù)SKIPIF1<0和SKIPIF1<0的對(duì)數(shù)平均定義：SKIPIF1<0對(duì)數(shù)平均與算術(shù)平均、幾何平均的大小關(guān)系：SKIPIF1<0（此式記為對(duì)數(shù)平均不等式）取等條件：當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立．第二部分：重難點(diǎn)題型突破突破一：分離變量法與不等式恒（能）成立問(wèn)題1．（2022·湖南·寧鄉(xiāng)市教育研究中心模擬預(yù)測(cè)）已知SKIPIF1<0，若對(duì)任意的SKIPIF1<0不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的最小值為_(kāi)______.2．（2022·黑龍江·高二期中）已知SKIPIF1<0，若存在SKIPIF1<0，使不等式SKIPIF1<0，對(duì)于SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是______．3．（2022·貴州畢節(jié)·三模（文））函數(shù)SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0零點(diǎn)的個(gè)數(shù)；(2)若對(duì)SKIPIF1<0，SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．4．（2022·廣東·潮州市瓷都中學(xué)三模）已知函數(shù)SKIPIF1<0.(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上恒成立，求整數(shù)SKIPIF1<0的最大值.5．（2022·江蘇·無(wú)錫市第一中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0）.(1)當(dāng)SKIPIF1<0時(shí)，對(duì)于函數(shù)SKIPIF1<0，存在SKIPIF1<0，使得SKIPIF1<0成立，求滿足條件的最大整數(shù)SKIPIF1<0；（SKIPIF1<0）(2)設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.6．（2022·天津市寧河區(qū)蘆臺(tái)第一中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的極值；(2)令SKIPIF1<0是函數(shù)SKIPIF1<0圖像上任意兩點(diǎn)，且滿足SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍；(3)若SKIPIF1<0，使SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的最大值.突破二：分類(lèi)討論法與不等式恒（能）成立問(wèn)題1．（2022·內(nèi)蒙古·霍林郭勒市第一中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，若對(duì)于SKIPIF1<0，SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值集合是_______．2．（2022·天津市瑞景中學(xué)高三期中）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)設(shè)SKIPIF1<0，若對(duì)任意的SKIPIF1<0，SKIPIF1<0恒成立，求SKIPIF1<0的最大值.3．（2022·北京海淀·高三期中）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)，證明：函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有且僅有一個(gè)零點(diǎn)；(3)若對(duì)任意SKIPIF1<0，不等式SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍．4．（2022·福建福州·高二期末）已知函數(shù)SKIPIF1<0(1)當(dāng)SKIPIF1<0時(shí)，求曲線SKIPIF1<0在點(diǎn)（0，f（0））處的切線方程；(2)若存在SKIPIF1<0，使得不等式SKIPIF1<0成立，求m的取值范圍．5．（2022·陜西·咸陽(yáng)市高新一中高二期中（理））已知函數(shù)SKIPIF1<0(1)若曲線SKIPIF1<0在SKIPIF1<0和SKIPIF1<0處的切線互相平行，求SKIPIF1<0的值與函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)設(shè)SKIPIF1<0，若對(duì)任意SKIPIF1<0，均存在SKIPIF1<0，使得SKIPIF1<0，求SKIPIF1<0的取值范圍．6．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的極小值為_(kāi)__________；若函數(shù)SKIPIF1<0，對(duì)于任意的SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.突破三：同構(gòu)法與不等式恒（能）成立問(wèn)題1．（2022·廣西北?！ひ荒＃ɡ恚┮阎瘮?shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求過(guò)點(diǎn)SKIPIF1<0且和曲線SKIPIF1<0相切的直線方程；(2)若對(duì)任意實(shí)數(shù)SKIPIF1<0，不等式SKIPIF1<0恒成立，求實(shí)數(shù)a的取值范圍．2．（2022·江蘇·南京師大蘇州實(shí)驗(yàn)學(xué)校高三階段練習(xí)）已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的極值；(2)當(dāng)SKIPIF1<0時(shí)，是否存在正實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0成立（SKIPIF1<0為自然對(duì)數(shù)的底數(shù)）？若存在，求SKIPIF1<0的取值范圍；若不存在，請(qǐng)說(shuō)明理由．3．（2022·江西·蘆溪中學(xué)高三階段練習(xí)（理））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．突破四：端點(diǎn)效應(yīng)與不等式恒（能）成立問(wèn)題1．設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0．（Ⅰ）當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的零點(diǎn)；（Ⅱ）若對(duì)任意SKIPIF1<0，SKIPIF1<0，恒有SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍．2．（2021·河南大學(xué)附屬中學(xué)高三階段練習(xí)（文））已知函數(shù)f（x）＝ax﹣a+lnx．（1）討論函數(shù)f（x）的單調(diào)性；（2）當(dāng)x∈（1，+∞）時(shí)，曲線y＝f（x）總在曲線y＝a（x2﹣1）的下方，求實(shí)數(shù)a的取值范圍．突破五：最值定位法解決雙參不等式問(wèn)題1．（多選）（2022·福建龍巖·高二期中）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0成立，則a的取值可以是（

）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·江蘇省石莊高級(jí)中學(xué)高二階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，若對(duì)SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0成立，則a的取值范圍是______．3．（2022·陜西·寶雞市金臺(tái)區(qū)教育體育局教研室高二期中（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(shí)，設(shè)SKIPIF1<0，若對(duì)于任意SKIPIF1<0、SKIPIF1<0，均有SKIPIF1<0，求SKIPIF1<0的取值范圍.4．（2022·上海市楊思高級(jí)中學(xué)高三期中）已知函數(shù)SKIPIF1<0.（1）若SKIPIF1<0，求曲線SKIPIF1<0在SKIPIF1<0處切線的方程；（2）求SKIPIF1<0的單調(diào)區(qū)間；（3）設(shè)SKIPIF1<0，若對(duì)任意SKIPIF1<0，均存在SKIPIF1<0，使得SKIPIF1<0，求SKIPIF1<0的取值范圍.5．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值，SKIPIF1<0．（1）求SKIPIF1<0的值與SKIPIF1<0的單調(diào)區(qū)間；（2）設(shè)SKIPIF1<0，已知函數(shù)SKIPIF1<0，若對(duì)于任意SKIPIF1<0、SKIPIF1<0，SKIPIF1<0，都有SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍．6．（2022·遼寧·大連二十四中高一期中）已知函數(shù)SKIPIF1<0是定義域?yàn)镾KIPIF1<0的奇函數(shù)，且SKIPIF1<0.(1)求SKIPIF1<0的值，并判斷SKIPIF1<0的單調(diào)性（不必證明）；(2)設(shè)SKIPIF1<0為正數(shù)，函數(shù)SKIPIF1<0，若對(duì)于任意SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0成立，求SKIPIF1<0的最大值.7．（2022·江蘇省江浦高級(jí)中學(xué)高一期中）已知二次函數(shù)SKIPIF1<0.(1)若關(guān)于SKIPIF1<0的不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍；(2)已知函數(shù)SKIPIF1<0，若對(duì)SKIPIF1<0，SKIPIF1<0，使不等式SKIPIF1<0成立，求SKIPIF1<0的取值范圍.突破六：值域法解決雙參等式問(wèn)題1．（多選）（2022·江蘇·常熟中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若對(duì)任意的SKIPIF1<0，均存在SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的取值可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·山東省實(shí)驗(yàn)中學(xué)高一階段練習(xí)）設(shè)函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0）的定義域的交集為D，集合M是由所有具有性質(zhì)：“對(duì)任意的SKIPIF1<0，都有SKIPIF1<0”的函數(shù)SKIPIF1<0組成的集合.(1)判斷函數(shù)SKIPIF1<0，SKIPIF1<0是不是集合M中的元素？并說(shuō)明理由；(2)設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，若對(duì)任意SKIPIF1<0，總存在SKIPIF1<0，使SKIPIF1<0成立，求實(shí)數(shù)a的取值范圍.3．（2022·河南·新密市第一高級(jí)中學(xué)高一階段練習(xí)）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0．(1)若關(guān)于SKIPIF1<0的方程SKIPIF1<0的兩根滿足一根大于1，另外一根小于1，求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)已知函數(shù)SKIPIF1<0，若對(duì)任意SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．突破七：?jiǎn)巫兞坎坏仁阶C明角度1：構(gòu)造函數(shù)，利用單調(diào)性證明不等式1．（2022·河南焦作·高三期中（理））已知函數(shù)SKIPIF1<0，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線在SKIPIF1<0軸上的截距為SKIPIF1<0．(1)求SKIPIF1<0的最小值；(2)證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．參考數(shù)據(jù)：SKIPIF1<0，SKIPIF1<0．2．（2022·河南·高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0，且曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0．(1)求a，b的值，并求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)證明：SKIPIF1<0．3．（2022·河南宋基信陽(yáng)實(shí)驗(yàn)中學(xué)高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0，(1)若SKIPIF1<0，求SKIPIF1<0的極值；(2)討論SKIPIF1<0的單調(diào)區(qū)間；(3)求證：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.4．（2022·廣東廣州·高二期末）設(shè)x＞0，f(x)=lnx，SKIPIF1<0(1)求證：直線y=x-1與曲線y=f(x)相切；(2)判斷f(x)與g(x)的大小關(guān)系，并加以證明．角度2：構(gòu)造函數(shù)，利用最值證明不等式1．（2022·山西忻州·高三階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，求SKIPIF1<0的取值范圍；(2)若SKIPIF1<0，比較SKIPIF1<0與SKIPIF1<0的大小關(guān)系.2．（2022·云南·巍山彝族回族自治縣第二中學(xué)高二階段練習(xí)）函數(shù)SKIPIF1<0.(1)求證：SKIPIF1<0；(2)求證：SKIPIF1<0.3．（2022·河南·一模（理））已知函數(shù)SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．角度3：等價(jià)轉(zhuǎn)化與不等式證明1．（2022·江西景德鎮(zhèn)·三模（文））已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的極值；(2)已知SKIPIF1<0，求證：當(dāng)SKIPIF1<0時(shí)，總有SKIPIF1<0.2．（2022·全國(guó)·高三專(zhuān)題練習(xí)（文））已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的單調(diào)性；(2)若函數(shù)SKIPIF1<0，證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.3．（2022·山東·菏澤一中高二階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0…為自然對(duì)數(shù)的底數(shù).(1)當(dāng)SKIPIF1<0時(shí)，若過(guò)點(diǎn)SKIPIF1<0與函數(shù)SKIPIF1<0相切的直線有兩條，求SKIPIF1<0的取值范圍；(2)若SKIPIF1<0，SKIPIF1<0，證明：SKIPIF1<0.4．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0．(1)求SKIPIF1<0的極大值點(diǎn)和極小值點(diǎn)；(2)若函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0．5．（2022·廣西·欽州一中高二期中（理））已知函數(shù)SKIPIF1<0.（1）求曲線SKIPIF1<0在SKIPIF1<0處的切線方程；（2）若SKIPIF1<0，證明不等式SKIPIF1<0在SKIPIF1<0上成立.角度4：超越放縮與不等式證明1．（2022·江蘇省包場(chǎng)高級(jí)中學(xué)高三開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0.(1)設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，求SKIPIF1<0的值并討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，證明：SKIPIF1<0.2．（2022·安徽·高二期末）函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的極值；(2)當(dāng)SKIPIF1<0，且SKIPIF1<0.①證明：SKIPIF1<0有兩個(gè)極值點(diǎn)；②證明：對(duì)任意的SKIPIF1<0.3．（2022·四川·成都七中高二期中（理））函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，SKIPIF1<0對(duì)一切SKIPIF1<0恒成立，求a的最大值；(2)證明：SKIPIF1<0，其中e是自然對(duì)數(shù)的底數(shù).4．（2022·四川·成都七中高二期中（文））函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，SKIPIF1<0對(duì)一切SKIPIF1<0恒成立，求a的最大值；(2)證明：SKIPIF1<0，其中e是自然對(duì)數(shù)的底數(shù).突破八：利用導(dǎo)數(shù)證明雙變量不等式角度1：分離雙參，構(gòu)造函數(shù)1．（多選）（2022·全國(guó)·高三專(zhuān)題練習(xí)）若對(duì)任意的SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0，則m的值可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．12．（2022·遼寧·沈陽(yáng)市第三十一中學(xué)高三階段練習(xí)）SKIPIF1<0，均有SKIPIF1<0成立，則SKIPIF1<0的取值范圍為_(kāi)__________.3．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0，SKIPIF1<0（1）SKIPIF1<0處的切線與SKIPIF1<0軸平行．(1)求實(shí)數(shù)SKIPIF1<0的值及SKIPIF1<0的極值；(2)若對(duì)任意SKIPIF1<0，SKIPIF1<0SKIPIF1<0，有SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍．4．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0SKIPIF1<0．(1)若函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)，求SKIPIF1<0的取值范圍；(2)證明：若SKIPIF1<0，則對(duì)于任意的SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0．角度2：糅合雙參（比值糅合）1．（2022·廣東北江實(shí)驗(yàn)學(xué)校模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；(2)任取兩個(gè)正數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，求證：SKIPIF1<0.2．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0，且SKIPIF1<0，證明：SKIPIF1<0.3．（2022