新高考數(shù)學(xué)二輪復(fù)習(xí)解答題培優(yōu)練習(xí)專題06 利用導(dǎo)函數(shù)研究能成立（有解）問題(典型題型歸類訓(xùn)練) 原卷版

專題06利用導(dǎo)函數(shù)研究能成立（有解）問題(典型題型歸類訓(xùn)練)目錄TOC\o"1-2"\h\u一、必備秘籍 1二、典型題型 2題型一：單變量有解問題 2題型二：雙變量不等式有解問題 3題型三：雙變量等式有解問題 5三、專項(xiàng)訓(xùn)練 6一、必備秘籍分離參數(shù)法用分離參數(shù)法解含參不等式恒成立問題，可以根據(jù)不等式的性質(zhì)將參數(shù)分離出來，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分類參數(shù)（注意分類參數(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：SKIPIF1<0，使得SKIPIF1<0能成立SKIPIF1<0SKIPIF1<0；SKIPIF1<0，使得SKIPIF1<0能成立SKIPIF1<0SKIPIF1<0．③求最值.二、典型題型題型一：單變量有解問題1．（2023·四川樂山·統(tǒng)考二模）若存在SKIPIF1<0，使不等式SKIPIF1<0成立，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【點(diǎn)睛】思路點(diǎn)睛：構(gòu)造函數(shù)是基本的解題思路，因此觀察題目所給的數(shù)的結(jié)構(gòu)特點(diǎn)，以及數(shù)與數(shù)之間的內(nèi)在聯(lián)系，合理構(gòu)造函數(shù)，利用導(dǎo)數(shù)判斷單調(diào)性是解題的關(guān)鍵.2．（2023·四川成都·石室中學(xué)?？寄M預(yù)測）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0內(nèi)有解，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、填空題3．（2023·河南·校聯(lián)考模擬預(yù)測）已知函數(shù)SKIPIF1<0，若存在唯一的整數(shù)SKIPIF1<0，使得SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛:用導(dǎo)數(shù)求參數(shù)的范圍問題，將題目轉(zhuǎn)化兩個(gè)函數(shù)的交點(diǎn)問題求解是解題的關(guān)鍵.4．（2023·云南·校聯(lián)考三模）設(shè)函數(shù)SKIPIF1<0，若存在唯一整數(shù)SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的取值范圍是.5．（2023·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；(2)若存在SKIPIF1<0，使得SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的最小值．【點(diǎn)睛】關(guān)鍵點(diǎn)睛：本題主要考查了利用導(dǎo)數(shù)解決含參函數(shù)單調(diào)區(qū)間問題，以及不等式能成立問題，難度較難，解答本題的關(guān)鍵在于將不等式問題通過分離參數(shù)法，轉(zhuǎn)化為最值問題，然后構(gòu)造函數(shù)，利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性，解決問題.6．（2023·寧夏銀川·?？寄M預(yù)測）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)如果存在SKIPIF1<0，使得當(dāng)SKIPIF1<0時(shí)，恒有SKIPIF1<0成立，求SKIPIF1<0的取值范圍.【點(diǎn)睛】關(guān)鍵點(diǎn)睛：涉及不等式恒成立問題，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、最值是解決問題的關(guān)鍵.題型二：雙變量不等式有解問題1．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，對(duì)于存在的SKIPIF1<0，存在SKIPIF1<0，使SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2023·四川南充·統(tǒng)考三模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0使SKIPIF1<0（SKIPIF1<0為常數(shù)）成立，則常數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：根據(jù)題意轉(zhuǎn)化為存在SKIPIF1<0，SKIPIF1<0使SKIPIF1<0能成立是其一，其二需要構(gòu)造函數(shù)SKIPIF1<0后分離參數(shù)轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上能成立，再次構(gòu)造函數(shù)SKIPIF1<0，多次利用導(dǎo)數(shù)求其最大值.3．（2023上·廣東中山·高三中山市華僑中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，對(duì)于SKIPIF1<0，都SKIPIF1<0，使SKIPIF1<0，則SKIPIF1<0的取值范圍為.4．（2023下·重慶·高二校聯(lián)考期中）已知函數(shù)SKIPIF1<0，若對(duì)任意SKIPIF1<0都存在SKIPIF1<0，使SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是.5．（2023上·福建莆田·高三莆田一中校考期中）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的最小值；(2)若SKIPIF1<0，且對(duì)SKIPIF1<0，都SKIPIF1<0，使得SKIPIF1<0成立，求實(shí)數(shù)SKIPIF1<0的取值范圍．6．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的單調(diào)性；(2)設(shè)SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，若對(duì)SKIPIF1<0，SKIPIF1<0，使SKIPIF1<0，求實(shí)數(shù)SKIPIF1<0的取值范圍．題型三：雙變量等式有解問題1．（2020·全國·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．

SKIPIF1<0 D．SKIPIF1<02．（2023·河南開封·開封高中?？寄M預(yù)測）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，使得SKIPIF1<0成立，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2023上·北京·高二北京市十一學(xué)校?？计谀┮阎瘮?shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0成立，則n－m的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【點(diǎn)睛】關(guān)鍵點(diǎn)睛：令SKIPIF1<0確定SKIPIF1<0關(guān)于t的函數(shù)式，構(gòu)造函數(shù)并利用導(dǎo)數(shù)求函數(shù)的最小值.4．（2021上·河南商丘·高三睢縣高級(jí)中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題考查函數(shù)中的能成立問題的求解，解題關(guān)鍵是能夠?qū)⒛艹闪⒌臈l件轉(zhuǎn)化為兩個(gè)函數(shù)最值之間大小關(guān)系的比較問題，從而利用導(dǎo)數(shù)、三角函數(shù)知識(shí)求得兩函數(shù)的值域，根據(jù)最值大小關(guān)系構(gòu)造出不等式組.5．（2022下·山東青島·高二山東省萊西市第一中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0.SKIPIF1<0，使得SKIPIF1<0)，求實(shí)數(shù)a的取值范圍.三、專項(xiàng)訓(xùn)練一、單選題1．（2023下·浙江杭州·高二學(xué)軍中學(xué)校考階段練習(xí)）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集中恰有SKIPIF1<0個(gè)整數(shù)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2023下·北京·高二北京市第十二中學(xué)校考期末）已知函數(shù)SKIPIF1<0，若存在SKIPIF1<0，使SKIPIF1<0，則m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2023下·江蘇南通·高二統(tǒng)考階段練習(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)）.若存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022下·天津·高二天津市薊州區(qū)第一中學(xué)校聯(lián)考期中）已知函數(shù)SKIPIF1<0，若對(duì)任意的SKIPIF1<0，存在SKIPIF1<0使得SKIPIF1<0，則實(shí)數(shù)a的取值范圍是（）A．SKIPIF1<0 B．[SKIPIF1<0，4]C．SKIPIF1<0 D．SKIPIF1<05．（2022下·全國·高三校聯(lián)考開學(xué)考試）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0成立，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·全國·高三專題練習(xí)）已知函數(shù)f（x）=SKIPIF1<0，函數(shù)g（x）=asin（SKIPIF1<0x）﹣2a+2（a＞0），若存在x1，x2∈[0，1]，使得f（x1）=g（x2）成立，則實(shí)數(shù)a的取值范圍是（

）A．[﹣SKIPIF1<0，1] B．[SKIPIF1<0，SKIPIF1<0] C．[SKIPIF1<0，SKIPIF1<0] D．[SKIPIF1<0，2]7．（2021上·山西太原·高三太原五中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．若SKIPIF1<0，都SKIPIF1<0，使SKIPIF1<0成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【點(diǎn)睛】結(jié)論點(diǎn)睛：本題考查不等式的恒成立與有解問題，可按如下規(guī)則轉(zhuǎn)化：一般地，已知函數(shù)SKIPIF1<0，SKIPIF1<0（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；（5）若SKIPIF1<0，SKIPIF1<0，有SKIPIF1<0，則SKIPIF1<0的值域是SKIPIF1<0值域的子集．8．（2021下·全國·高三校聯(lián)考專題練習(xí)）設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，若在區(qū)間SKIPIF1<0上存在SKIPIF1<0，使得SKIPIF1<0成立，其中e為自然對(duì)數(shù)的底數(shù)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2018下·四川攀枝花·高三統(tǒng)考階段練習(xí)）已知函數(shù)SKIPIF1<0若對(duì)SKIPIF1<0,使得SKIPIF1<0成立,則實(shí)數(shù)SKIPIF1<0的最小值是A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．3二、填空題10．（2023上·廣東中山·高三中山市華僑中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，對(duì)于SKIPIF1<0，都SKIPIF1<0，使SKIPIF1<0，則SKIPIF1<0的取值范圍為.11．（2021下·四川涼山·高二統(tǒng)考期中）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0，若對(duì)任意的SKIPIF1<0，存在SKIPIF1<0，使得SKIPIF1<0，則實(shí)數(shù)m的取值范圍為．12．（2023下·天津東麗·高二天津市第一百中學(xué)校考階段練習(xí)）已知SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0