新高考數(shù)學二輪復習專題2-5 函數(shù)與導數(shù)壓軸小題歸類（原卷版）

專題2-5函數(shù)導數(shù)壓軸小題歸類目錄TOC\o"1-1"\h\u題型01整數(shù)解型 1題型02函數(shù)零點構(gòu)造型 2題型03同構(gòu):方程零點型同構(gòu) 3題型04同構(gòu):不等式型同構(gòu)求參 4題型05恒成立求參：移項討論型 5題型06恒成立求參：虛設(shè)零點型 5題型07“倍縮”型函數(shù)求參數(shù) 6題型08恒成立求參：“等式”型 7題型09雙變量型不等式范圍最值 8題型10雙變量型：凸凹反轉(zhuǎn)型 9題型11多參型：代換型 10題型12多參型：二次構(gòu)造放縮型 10題型13多參型：韋達定理求參型 11題型14多參型：單峰函數(shù)絕對值型 12題型15導數(shù)與三角函數(shù) 12高考練場 13題型01整數(shù)解型【解題攻略】整數(shù)解，屬于導數(shù)研究函數(shù)的性質(zhì)，根據(jù)題意求得整數(shù)型參數(shù)的取值范圍，或者整數(shù)解求參數(shù)范圍等，涉及函數(shù)的零點問題、方程解的個數(shù)問題、函數(shù)圖像交點個數(shù)問題，一般先通過導數(shù)研究函數(shù)的單調(diào)性、最大值、最小值、變化趨勢等，再借助函數(shù)的大致圖象判斷零點、方程根、交點的情況，歸根到底還是研究函數(shù)的性質(zhì)，如單調(diào)性、極值，然后通過數(shù)形結(jié)合的思想找到解題的思路.【典例1-1】（2021·湖南懷化·二模（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，若對任意的SKIPIF1<0，存在實數(shù)SKIPIF1<0滿足SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的最大值是A．3 B．2 C．4 D．5【典例1-2】.（2020·黑龍江實驗中學三模（理））已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在極值點，且SKIPIF1<0恰好有唯一整數(shù)解，則的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】在關(guān)于SKIPIF1<0的不等式SKIPIF1<0（其中SKIPIF1<0SKIPIF1<0為自然對數(shù)的底數(shù)）的解集中，有且僅有兩個大于2的整數(shù)，則實數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（黑龍江省佳木斯市第一中學2021-2022學年高三上學期第四次調(diào)研考試理科數(shù)學試題）已知偶函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當SKIPIF1<0時，SKIPIF1<0，若關(guān)于x的不等式SKIPIF1<0在SKIPIF1<0上有且只有150個整數(shù)解，則實數(shù)t的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（四川省成都石室中學高三下學期考試數(shù)學（理）試題）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的不等式SKIPIF1<0恰有兩個整數(shù)解，則實數(shù)SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0題型02函數(shù)零點構(gòu)造型【解題攻略】函數(shù)零點構(gòu)造型，涉及到函數(shù)的性質(zhì)應(yīng)用：與對稱有關(guān)的常用結(jié)論：①若點SKIPIF1<0，SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，則SKIPIF1<0；②若SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱，則SKIPIF1<0；③若SKIPIF1<0，則SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱；④若SKIPIF1<0，則SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱．數(shù)形結(jié)合法解決零點問題：①零點個數(shù)：幾個零點②幾個零點的和③幾個零點的積.【典例1-1】（2020·黑龍江實驗中學高三階段練習（理））已知函數(shù)SKIPIF1<0，若實數(shù)SKIPIF1<0互不相等，且SKIPIF1<0，則SKIPIF1<0的取值范圍為______．【典例1-2】.（2020·吉林吉林·三模）已知函數(shù)SKIPIF1<0，若實數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍為___________.【變式1-1】（2022·云南省玉溪第一中學高三）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0的取值范圍是______.【變式1-2】．（2022·浙江·高三專題練習）設(shè)函數(shù)SKIPIF1<0已知SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0的最小值為SKIPIF1<0，則a的值為___________．【變式1-3】.（2021·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若方程SKIPIF1<0有4個不同的實根SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍是______．題型03同構(gòu):方程零點型同構(gòu)【解題攻略】對于既含有指數(shù)式又含有對數(shù)式的等式或不等式，直接求導會出現(xiàn)越求導式子越復雜的情況，此時可通過同構(gòu)函數(shù)，再利用函數(shù)的單調(diào)性，把問題轉(zhuǎn)化為較為簡單的函數(shù)的導數(shù)問題．導函數(shù)求解參數(shù)取值范圍，當函數(shù)中同時出現(xiàn)SKIPIF1<0與SKIPIF1<0，通常使用同構(gòu)來進行求解，難點是尋找構(gòu)造突破口。如SKIPIF1<0變形得到SKIPIF1<0，從而構(gòu)造SKIPIF1<0進行求解.常見同構(gòu)：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0④SKIPIF1<0；SKIPIF1<0【典例1-1】（2024·全國·模擬預(yù)測）已知m是方程SKIPIF1<0的一個根，則SKIPIF1<0（

）A．1 B．2 C．3 D．5【典例1-2】（2023·全國·模擬預(yù)測）若方程SKIPIF1<0在SKIPIF1<0上有實根，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2023·全國·模擬預(yù)測）已知SKIPIF1<0是方程SKIPIF1<0的一個根，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．3【變式1-2】（2023上·四川綿陽·高三四川省綿陽實驗高級中學?？茧A段練習）已知SKIPIF1<0且SKIPIF1<0則一定有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2023上·山東日照·高三統(tǒng)考開學考試）已知正實數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．0 B．1 C．2 D．3題型04同構(gòu):不等式型同構(gòu)求參【解題攻略】（1）乘積模型：SKIPIF1<0（2）商式模型：SKIPIF1<0（3）和差模型：SKIPIF1<0【典例1-1】（2023·全國·安陽市第二中學校聯(lián)考模擬預(yù)測）已知關(guān)于x的不等式SKIPIF1<0在SKIPIF1<0上恒成立，則正數(shù)m的最大值為（

）A．SKIPIF1<0 B．0 C．e D．1【典例1-2】（2020上·北京·高三統(tǒng)考階段練習）已知不等式SKIPIF1<0對SKIPIF1<0恒成立，則實數(shù)a的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2022下·河南·高三校聯(lián)考階段練習）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0上恒成立，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2022上·浙江紹興·高三統(tǒng)考期末）已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，其中SKIPIF1<0為自然對數(shù)的底數(shù)，SKIPIF1<0，則（

）A．SKIPIF1<0既有最小值，也有最大值 B．SKIPIF1<0有最小值，沒有最大值C．SKIPIF1<0有最大值，沒有最小值 D．SKIPIF1<0既沒有最小值，也沒有最大值【變式1-3】（2022上·安徽亳州·高三統(tǒng)考期末）已知SKIPIF1<0，若SKIPIF1<0時，SKIPIF1<0恒成立，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型05恒成立求參：移項討論型【解題攻略】一般地，已知函數(shù)SKIPIF1<0，SKIPIF1<0（1）若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立，故SKIPIF1<0；（2）若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立，故SKIPIF1<0；（3）若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立，故SKIPIF1<0；（4）若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立，故SKIPIF1<0；【典例1-1】（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0有唯一零點，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】.（2022·全國·高三專題練習）若對任意SKIPIF1<0，不等式SKIPIF1<0恒成立，則實數(shù)a的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2020·福建省福州第一中學高三階段練習（理））已知SKIPIF1<0，且SKIPIF1<0時，SKIPIF1<0恒成立，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0有最小值，則實數(shù)SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2022上·江蘇揚州·高三統(tǒng)考階段練習）當SKIPIF1<0時，不等式SKIPIF1<0有解，則實數(shù)m的范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型06恒成立求參：虛設(shè)零點型【解題攻略】虛設(shè)零點法：涉及到導函數(shù)有零點但是求解相對比較繁雜甚至無法求解的情形時，可以將這個零點只設(shè)出來而不必求出來，然后尋找一種整體的轉(zhuǎn)換和過度，再結(jié)合其他條件，進行代換變形，從而最重獲得問題的解決（1）、整體代換：把超越式子（多為指數(shù)和對數(shù)式子）轉(zhuǎn)化為普通的（如二次函數(shù)一次哈數(shù)等）可解式子，如比值代換等等。（2）、反代消參：反解參數(shù)代入，構(gòu)造單一變量的函數(shù)。如果要求解（或者要證明）的結(jié)論與參數(shù)無關(guān)，則可以通過反解參數(shù)，用變量（零點）表示參數(shù)，然后把函數(shù)變成關(guān)于零點的單一函數(shù)，再對單一變量求導就可以解決相應(yīng)的問題。（3）留參降次（留參、消去指對等超越項）：如果要求解的與參數(shù)有關(guān)，則可以通過消去超越項，建立含參數(shù)的方程或者不等式。恒等變形或者化簡方向時保留參數(shù)，通過“降次”變換，一直降到不可再降為止，再結(jié)合條件，求解方程或者不等式，解的相應(yīng)的參數(shù)值或者參數(shù)范圍【典例1-1】（四川省內(nèi)江市威遠中學校2022-2023學年高三上學期第三次月考數(shù)學（理）試題）已知不等式SKIPIF1<0對SKIPIF1<0恒成立，則SKIPIF1<0取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（黑龍江省哈爾濱市第六中學校2022-2023學年高三上學期10月月考數(shù)學試題）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0對一切正實數(shù)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<0【變式1-1】設(shè)實數(shù)SKIPIF1<0，若對任意SKIPIF1<0，不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】已知函數(shù)SKIPIF1<0有唯一零點，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】若對任意SKIPIF1<0，不等式SKIPIF1<0恒成立，則實數(shù)a的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型07“倍縮”型函數(shù)求參數(shù)【解題攻略】如果函數(shù)SKIPIF1<0在定義域的某個區(qū)間SKIPIF1<0（SKIPIF1<0）上的值域恰為SKIPIF1<0（SKIPIF1<0），則稱函數(shù)SKIPIF1<0為SKIPIF1<0上的k倍域函數(shù)，SKIPIF1<0稱為函數(shù)SKIPIF1<0的一個k倍域區(qū)間．把函數(shù)SKIPIF1<0存在區(qū)間SKIPIF1<0，使得函數(shù)SKIPIF1<0為SKIPIF1<0上的SKIPIF1<0倍域函數(shù)，結(jié)合函數(shù)的單調(diào)性，轉(zhuǎn)化為SKIPIF1<0是解答的關(guān)鍵.【典例1-1】（陜西省漢中中學2019屆高三上學期第二次月考數(shù)學（理）試卷）設(shè)函數(shù)的定義域為D，若滿足條件：存在SKIPIF1<0，使SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0，則稱SKIPIF1<0為“倍縮函數(shù)”.若函數(shù)SKIPIF1<0為“倍縮函數(shù)”，則實數(shù)t的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（浙江省杭州學軍中學西溪校區(qū)2020-2021學年高三3月數(shù)學試題）設(shè)函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，若函數(shù)SKIPIF1<0滿足條件：存在SKIPIF1<0，使SKIPIF1<0在SKIPIF1<0上的值域是SKIPIF1<0，則SKIPIF1<0稱為“倍縮函數(shù)”，若函數(shù)SKIPIF1<0為“倍縮函數(shù)”，則實數(shù)SKIPIF1<0的取值范圍是________.【變式1-1】（2020年浙江省新高考考前原創(chuàng)沖刺卷（二））設(shè)函數(shù)SKIPIF1<0的定義域為D，若滿足條件：存在SKIPIF1<0，使SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0，則稱SKIPIF1<0為“倍脹函數(shù)”.若函數(shù)SKIPIF1<0為“倍脹函數(shù)”，則實數(shù)t的取值范圍是________.【變式1-2】（河北省邢臺一中2021-2022學年高三下學期模擬數(shù)學（理)試題）.設(shè)函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0在區(qū)間SKIPIF1<0上的值域為SKIPIF1<0，則稱SKIPIF1<0為“SKIPIF1<0倍函數(shù)”.已知函數(shù)SKIPIF1<0為“3倍函數(shù)”，則實數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2022吉林吉林·高三階段練習（理））設(shè)函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，若滿足條件：存在SKIPIF1<0，使SKIPIF1<0在SKIPIF1<0上的值域為SKIPIF1<0（SKIPIF1<0且SKIPIF1<0），則稱SKIPIF1<0為“SKIPIF1<0倍函數(shù)”，若函數(shù)SKIPIF1<0SKIPIF1<0為“3倍函數(shù)”，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型08恒成立求參：“等式”型【解題攻略】一般地，已知函數(shù)SKIPIF1<0，SKIPIF1<0若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0，則SKIPIF1<0的值域是SKIPIF1<0值域的子集．【典例1-1】（2021·四川·綿陽中學模擬預(yù)測（文））已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，使得SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】．（2022·福建·泉州市城東中學高三）已知SKIPIF1<0，SKIPIF1<0是函數(shù)SKIPIF1<0的兩個極值點，且SKIPIF1<0，當SKIPIF1<0時，不等式SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2022·四川成都·高三階段練習（文））設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0．若對任意的正實數(shù)SKIPIF1<0，SKIPIF1<0，不等式SKIPIF1<0恒成立，則a的最小值為（

）A．0 B．1 C．SKIPIF1<0 D．e【變式1-2】（2022·河南安陽·高三階段練習）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0使得SKIPIF1<0成立，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（江蘇省南京航空航天大學附屬高級中學2020-2021學年高三數(shù)學試題）已知函數(shù)SKIPIF1<0，SKIPIF1<0，對任意的SKIPIF1<0，總存在SKIPIF1<0使得SKIPIF1<0成立，則a的范圍為_________．題型09雙變量型不等式范圍最值【解題攻略】一般地，已知函數(shù)SKIPIF1<0，SKIPIF1<0不等關(guān)系(1)若SKIPIF1<0，SKIPIF1<0，總有SKIPIF1<0成立，故SKIPIF1<0；(2)若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立，故SKIPIF1<0；(3)若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立，故SKIPIF1<0；(4)若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0成立，故SKIPIF1<0．【典例1-1】（2023下·四川眉山·高三眉山市彭山區(qū)第一中學校考階段練習）已知函數(shù)SKIPIF1<0有兩個零點SKIPIF1<0，且SKIPIF1<0，則下列說法不正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0有極小值點【典例1-2】（2023下·福建福州·高三福建省福州第一中學?？迹┮阎瘮?shù)SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2019下·河南鶴壁·高三鶴壁高中?？茧A段練習）已知函數(shù)SKIPIF1<0，SKIPIF1<0，曲線SKIPIF1<0上總存在兩點SKIPIF1<0，SKIPIF1<0，使曲線SKIPIF1<0在SKIPIF1<0兩點處的切線互相平行，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2019下·山西長治·高三統(tǒng)考階段練習）若方程x﹣2lnx+a＝0存在兩個不相等的實數(shù)根x1和x2，則（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2021上·高三單元測試）已知直線SKIPIF1<0分別與函數(shù)SKIPIF1<0和SKIPIF1<0的圖象交于點SKIPIF1<0，SKIPIF1<0，則下列結(jié)論錯誤的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型10雙變量型：凸凹反轉(zhuǎn)型【解題攻略】凸凹翻轉(zhuǎn)型常見思路，如下圖轉(zhuǎn)化為兩個函數(shù)的最值問題是關(guān)鍵?！镜淅?-1】（2023·全國·高三專題練習）設(shè)大于1的兩個實數(shù)a，b滿足SKIPIF1<0，則正整數(shù)n的最大值為（

）．A．7 B．9 C．11 D．12【典例1-2】（2023上·江蘇蘇州·高三統(tǒng)考階段練習）已知正數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【變式1-1】.已知實數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的值為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（安徽省六安市第一中學、合肥八中、阜陽一中三校2021-2022學年高三上學期10月聯(lián)考數(shù)學試題）已知函數(shù)SKIPIF1<0有兩個零點，則SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型11多參型：代換型【解題攻略】不等式中，可以借助對數(shù)均值不等式解決，完整的對數(shù)均值不等式為：SKIPIF1<0，可用兩邊同除SKIPIF1<0，令SKIPIF1<0整體換元的思想來構(gòu)造函數(shù)，證明不等式成立求解參數(shù)【典例1-1】（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0，對于正實數(shù)a，若關(guān)于t的方程SKIPIF1<0恰有三個不同的正實數(shù)根，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2020·江蘇·高三專題練習）若對任意正實數(shù)SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是_________【變式1-1】（2020·全國·高三專題練習（文））設(shè)三次函數(shù)SKIPIF1<0，（a,b,c為實數(shù)且SKIPIF1<0）的導數(shù)為SKIPIF1<0，記SKIPIF1<0，若對任意SKIPIF1<0，不等式SKIPIF1<0恒成立，則SKIPIF1<0的最大值為____________【變式1-2】已知存在SKIPIF1<0，若要使等式SKIPIF1<0成立（e=2.71828…），則實數(shù)SKIPIF1<0的可能的取值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．0【變式1-3】（江蘇省揚州中學2022-2023學年高三考試數(shù)學）若正實數(shù)SKIPIF1<0滿足SKIPIF1<0，則函數(shù)SKIPIF1<0的零點的最大值為______.題型12多參型：二次構(gòu)造放縮型【解題攻略】多參數(shù)型求參數(shù)范圍，或者多參型最值，難點是能夠兩次構(gòu)造函數(shù)，利用導數(shù)求出相應(yīng)函數(shù)的最值【典例1-1】（2023·全國·高三專題練習）已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2021·高三單元測試）已知SKIPIF1<0為自然對數(shù)的底數(shù)，SKIPIF1<0為實數(shù)，且不等式SKIPIF1<0對任意的SKIPIF1<0恒成立.則當SKIPIF1<0取最大值時，SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2021·四川成都·統(tǒng)考模擬預(yù)測）設(shè)SKIPIF1<0，SKIPIF1<0，若關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0上恒成立，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2022·四川南充·高三四川省南充高級中學?？迹┮阎瘮?shù)SKIPIF1<0，若SKIPIF1<0時，恒有SKIPIF1<0，則SKIPIF1<0的最大值為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2023·浙江·高三路橋中學校聯(lián)考）已知SKIPIF1<0，SKIPIF1<0，關(guān)于SKIPIF1<0的不等式SKIPIF1<0無實數(shù)解，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型13多參型：韋達定理求參型【典例1-1】（2023上·北京順義·高三北京市順義區(qū)第一中學校考）若函數(shù)SKIPIF1<0既有極大值也有極小值，則錯誤的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】（2023上·江蘇蘇州·高三蘇州中學?？奸_學考試）若函數(shù)SKIPIF1<0既有極大值也有極小值，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-1】（2023·山東煙臺·統(tǒng)考二模）若函數(shù)SKIPIF1<0有兩個極值點SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-2】（2021·浙江·模擬預(yù)測）已知SKIPIF1<0在SKIPIF1<0上恰有兩個極值點SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】（2023·河南開封·高三統(tǒng)考）已知函數(shù)SKIPIF1<0的兩個極值點分別是SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0C．存在實數(shù)a，使得SKIPIF1<0 D．SKIPIF1<0題型14多參型：單峰函數(shù)絕對值型【典例1-1】（安徽省阜陽市太和第一中學2019-2020學年高三數(shù)學試題）若存在實數(shù)SKIPIF1<0，對任意實數(shù)SKIPIF1<0，使不等式SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍為________.【典例1-2】（中學生標準學術(shù)能力診斷性測試2019-2020學年高三1月（一卷）數(shù)學（理）試題）設(shè)函數(shù)SKIPIF1<0，若對任意的實數(shù)SKIPIF1<0和SKIPIF1<0，總存在SKIPIF1<0，使得SKIPIF1<0，則實數(shù)SKIPIF1<0的最大值為__________．【變式1-1】設(shè)函數(shù)SKIPIF1<0，若對任意的實數(shù)SKIPIF1<0，總存在SKIPIF1<0使得SKIPIF1<0成立，則實數(shù)SKIPIF1<0的取值范圍是________．【變式1-2】若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，對任意SKIPIF1<0，總存在唯一的SKIPIF1<0，使得SKIPIF1<0成立，則實數(shù)a的取值范圍____________．【變式1-3】（浙江省溫州市2021-2022學年高三適應(yīng)性測試一模數(shù)學試題）設(shè)函數(shù)SKIPIF1<0.若SKIPIF1<0在SKIPIF1<0上的最大值為2，則實數(shù)a所有可能的取值組成的集合是________.題型15導數(shù)與三角函數(shù)【典例1-1】函數(shù)SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例1-2】已知函數(shù)SKIPIF1<0，若對于任意的SKIPIF1<0，均有SKIPIF1<0成立，則實數(shù)a的最小值為A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．3【變式1-1】函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0圖象的所有交點的橫坐標之和為___________.【變式1-2】已知SKIPIF1<0，且SKIPIF1<0，其中e為自然對數(shù)的底數(shù)，則下列選項中一定成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【變式1-3】已知函數(shù)SKIPIF1<0，若f(x)在R上單調(diào)，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0高考練場1.（黑龍江省實驗校2020屆高三第三次模擬考試數(shù)學（理）試題）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在極值點，且SKIPIF1<0恰好有唯一整數(shù)解，則的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02.（2021·江蘇·高三開學考試）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為___________.3.（2023·廣東梅州·統(tǒng)考三模）已知實數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．1 B．2 C．4 D．84.（2021·廣東深圳·高三練習）設(shè)SKIPIF1<0，若存在正實數(shù)SKIPIF1<0，使得不等式SKIPIF1<0成立，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05.（2021下·四川眉山··高三練習）若SKIPIF1<0，SKIPIF1<0恒