新高考數(shù)學(xué)一輪復(fù)習(xí)講練測(cè)第3章第02講 單調(diào)性問(wèn)題（練習(xí)）（解析版）

第02講單調(diào)性問(wèn)題（模擬精練+真題演練）1．（2023·全國(guó)·模擬預(yù)測(cè)）已知冪函數(shù)SKIPIF1<0，若SKIPIF1<0，則下列說(shuō)法正確的是（

）A．函數(shù)SKIPIF1<0為奇函數(shù) B．函數(shù)SKIPIF1<0為偶函數(shù)C．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 D．函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減【答案】B【解析】依題意SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0SKIPIF1<0單調(diào)遞減,SKIPIF1<0單調(diào)遞增,SKIPIF1<0知該方程有唯一解SKIPIF1<0，故SKIPIF1<0，易知該函數(shù)為偶函數(shù).故選：B．2．（2023·江西鷹潭·貴溪市實(shí)驗(yàn)中學(xué)校考模擬預(yù)測(cè)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，故選：D3．（2023·廣西玉林·統(tǒng)考模擬預(yù)測(cè)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0，因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，因?yàn)槎魏瘮?shù)SKIPIF1<0的圖象的對(duì)稱(chēng)軸為SKIPIF1<0，且開(kāi)口向上所以SKIPIF1<0的最小值為1，所以SKIPIF1<0.故選：B.4．（2023·甘肅蘭州·校考一模）已知SKIPIF1<0是偶函數(shù)，在(－∞，0)上滿(mǎn)足SKIPIF1<0恒成立，則下列不等式成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0時(shí)，SKIPIF1<0即SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0為偶函數(shù)，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．∴SKIPIF1<0，∴SKIPIF1<0.故選：A．5．（2023·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題意可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，因?yàn)楫?dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因?yàn)镾KIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，故選：A6．（2023·江蘇南京·南京師大附中?？寄M預(yù)測(cè)）已知實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0可得，SKIPIF1<0，即SKIPIF1<0，也即SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0恒成立，所以函數(shù)SKIPIF1<0在定義域SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，故選:B.7．（2023·寧夏銀川·校聯(lián)考二模）已知SKIPIF1<0，SKIPIF1<0，對(duì)SKIPIF1<0，且SKIPIF1<0，恒有SKIPIF1<0，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【答案】A【解析】設(shè)SKIPIF1<0，SKIPIF1<0，對(duì)SKIPIF1<0，且SKIPIF1<0，恒有SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0恒成立，即SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單調(diào)遞減；故SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0.故選：A8．（2023·四川南充·統(tǒng)考三模）已知函數(shù)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【答案】C【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0在定義域上單調(diào)遞增，又SKIPIF1<0使SKIPIF1<0（SKIPIF1<0為常數(shù)）成立，顯然SKIPIF1<0，所以不妨設(shè)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，即函數(shù)SKIPIF1<0在SKIPIF1<0上存在單調(diào)遞增區(qū)間，又SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上有解，則SKIPIF1<0在SKIPIF1<0上有解，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，即常數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：C9．（多選題）（2023·山東濰坊·統(tǒng)考模擬預(yù)測(cè)）下列函數(shù)中，在其定義域內(nèi)既是奇函數(shù)又是增函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】AD【解析】對(duì)于A,SKIPIF1<0,故SKIPIF1<0為奇函數(shù)，SKIPIF1<0，故SKIPIF1<0為定義域內(nèi)的單調(diào)遞增函數(shù)，故A正確，對(duì)于B，SKIPIF1<0，故SKIPIF1<0為非奇非偶函數(shù)，故B錯(cuò)誤，對(duì)于C，SKIPIF1<0在定義域內(nèi)不是單調(diào)增函數(shù)，故C錯(cuò)誤，對(duì)于D，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0定義域內(nèi)既是奇函數(shù)又是增函數(shù)，故D正確，故選：AD10．（多選題）（2023·安徽淮北·統(tǒng)考一模）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0在SKIPIF1<0單調(diào)遞增B．SKIPIF1<0有兩個(gè)零點(diǎn)C．曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處切線(xiàn)的斜率為SKIPIF1<0D．SKIPIF1<0是奇函數(shù)【答案】AC【解析】對(duì)A：SKIPIF1<0，定義域?yàn)镾KIPIF1<0，則SKIPIF1<0SKIPIF1<0，由SKIPIF1<0都在SKIPIF1<0單調(diào)遞增，故SKIPIF1<0SKIPIF1<0也在SKIPIF1<0單調(diào)遞增，又SKIPIF1<0SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0單調(diào)遞增；故A正確；對(duì)B：由A知，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，故SKIPIF1<0只有一個(gè)零點(diǎn)，B錯(cuò)誤；對(duì)C：SKIPIF1<0SKIPIF1<0，根據(jù)導(dǎo)數(shù)幾何意義可知，C正確；對(duì)D：SKIPIF1<0定義域?yàn)镾KIPIF1<0，不關(guān)于原點(diǎn)對(duì)稱(chēng)，故SKIPIF1<0是非奇非偶函數(shù)，D錯(cuò)誤.故選：AC.11．（多選題）（2023·河北·統(tǒng)考模擬預(yù)測(cè)）十六世紀(jì)中葉，英國(guó)數(shù)學(xué)家雷科德在《礪智石》一書(shū)中首先把“=”作為等號(hào)使用，后來(lái)英國(guó)數(shù)學(xué)家哈里奧特首次使用“<”和“>”符號(hào)，不等號(hào)的引入對(duì)不等式的發(fā)展影響深遠(yuǎn)．若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【解析】設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，A正確；由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0單調(diào)遞增，所以SKIPIF1<0，B正確；由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，C錯(cuò)誤；因?yàn)镾KIPIF1<0，所以SKIPIF1<0，D正確．故選：ABD.12．（多選題）（2023·浙江金華·統(tǒng)考模擬預(yù)測(cè)）當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，則自然數(shù)SKIPIF1<0可能為（

）A．0 B．2 C．8 D．12【答案】BC【解析】由于SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0，且SKIPIF1<0時(shí)，故當(dāng)SKIPIF1<0當(dāng)SKIPIF1<0，因此SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最小值SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取最大值SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不妨取SKIPIF1<0，則SKIPIF1<0而SKIPIF1<0，不滿(mǎn)足SKIPIF1<0，故A錯(cuò)誤，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，顯然SKIPIF1<0，故滿(mǎn)足題意，B正確，要使SKIPIF1<0恒成立，則需要SKIPIF1<0，即SKIPIF1<0恒成立即可由于SKIPIF1<0，因此SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，C正確，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不滿(mǎn)足題意，錯(cuò)誤，故選：BC13．（2023·內(nèi)蒙古赤峰·校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的單調(diào)遞減區(qū)間為_(kāi)_____.【答案】SKIPIF1<0【解析】由題得SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<014．（2023·四川雅安·統(tǒng)考模擬預(yù)測(cè)）給出兩個(gè)條件：①SKIPIF1<0，SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（其中SKIPIF1<0為SKIPIF1<0的導(dǎo)函數(shù)）．請(qǐng)寫(xiě)出同時(shí)滿(mǎn)足以上兩個(gè)條件的一個(gè)函數(shù)______．（寫(xiě)出一個(gè)滿(mǎn)足條件的函數(shù)即可）【答案】SKIPIF1<0（答案不唯一）【解析】由SKIPIF1<0，SKIPIF1<0知，函數(shù)SKIPIF1<0可以為指數(shù)函數(shù)，因當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以函數(shù)SKIPIF1<0可以為SKIPIF1<0.故答案為：SKIPIF1<015．（2023·四川·石室中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集為_(kāi)_____________．【答案】SKIPIF1<0【解析】令SKIPIF1<0，定義域?yàn)镽，且SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，SKIPIF1<0變形為SKIPIF1<0，即SKIPIF1<0，其SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，所以SKIPIF1<0在R上單調(diào)遞增，所以SKIPIF1<0，解得：SKIPIF1<0，所以解集為SKIPIF1<0.故答案為：SKIPIF1<016．（2023·寧夏銀川·銀川一中?？既＃┤艉瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則實(shí)數(shù)SKIPIF1<0的取值范圍為_(kāi)_______．【答案】SKIPIF1<0【解析】由SKIPIF1<0可知，其定義域?yàn)镾KIPIF1<0，則SKIPIF1<0，易知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；即函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增；若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則需滿(mǎn)足SKIPIF1<0，解得SKIPIF1<0；所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<017．（2023·天津河西·天津市新華中學(xué)校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0.若函數(shù)SKIPIF1<0為增函數(shù)，求SKIPIF1<0的取值范圍；【解析】∵SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0是增函數(shù)，則SKIPIF1<0，且SKIPIF1<0，可得SKIPIF1<0，故原題意等價(jià)于SKIPIF1<0對(duì)SKIPIF1<0恒成立，構(gòu)建SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0；令SKIPIF1<0，解得SKIPIF1<0；則SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0遞減，故SKIPIF1<0，∴SKIPIF1<0的取值范圍為SKIPIF1<0.18．（2023·四川·校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0若SKIPIF1<0單調(diào)遞增，求a的值；【解析】由SKIPIF1<0可得，SKIPIF1<0，由于函數(shù)SKIPIF1<0單調(diào)遞增，則SKIPIF1<0恒成立，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可知SKIPIF1<0時(shí)，SKIPIF1<0，不滿(mǎn)足題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，又因?yàn)镾KIPIF1<0，即SKIPIF1<0，不滿(mǎn)足題意；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得最小值SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，可知SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，則SKIPIF1<0，由于SKIPIF1<0恒成立，所以，SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故函數(shù)SKIPIF1<0單調(diào)遞增時(shí)，實(shí)數(shù)SKIPIF1<0的值為SKIPIF1<0.19．（2023·貴州貴陽(yáng)·校聯(lián)考模擬預(yù)測(cè)）實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．(1)若SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)討論SKIPIF1<0的單調(diào)性并寫(xiě)出過(guò)程．【解析】（1）由題意得，令SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，由SKIPIF1<0得：SKIPIF1<0．設(shè)SKIPIF1<0，則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)遞增，在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．（2）令SKIPIF1<0，SKIPIF1<0的定義域?yàn)镾KIPIF1<0．SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上是增函數(shù)；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上是減函數(shù)；SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上是增函數(shù)；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上是減函數(shù)；SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上是增函數(shù)；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；④當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上是增函數(shù)，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上是減函數(shù)，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0是增函數(shù)．20．（2023·河南·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．求SKIPIF1<0的單調(diào)區(qū)間；【解析】由已知可得，SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.所以，SKIPIF1<0在SKIPIF1<0處取得唯一極小值，也是最小值SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上恒成立，所以，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．所以，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，無(wú)遞減區(qū)間．21．（2023·湖南·鉛山縣第一中學(xué)校聯(lián)考二模）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù).當(dāng)SKIPIF1<0時(shí)，討論函數(shù)SKIPIF1<0的單調(diào)性；【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0解得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，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得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)遞減，在SKIPIF1<0上單調(diào)遞增；綜上：當(dāng)SKIPIF1<0時(shí)，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在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.22．（2023·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，討論函數(shù)SKIPIF1<0的單調(diào)性；(2)若函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)，求實(shí)數(shù)a的取值范圍．【解析】（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，易知SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減．（2）由題意知SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，不符合題意．②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，不符合題意．③當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．易知SKIPIF1<0，當(dāng)且僅當(dāng)x＝1時(shí)取等號(hào)，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0．所以當(dāng)x＞0時(shí)，SKIPIF1<0．取SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0．又SKIPIF1<0，所以存在SKIPIF1<0，使得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，符合題意．綜上，實(shí)數(shù)a的取值范圍為SKIPIF1<0．1．（2022·全國(guó)·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的單調(diào)性；【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0.2．（2022·北京·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0．(1)求曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程；(2)設(shè)SKIPIF1<0，討論函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性；【解析】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即切點(diǎn)坐標(biāo)為SKIPIF1<0，又SKIPIF1<0，∴切線(xiàn)斜率SKIPIF1<0∴切線(xiàn)方程為：SKIPIF1<0（2）因?yàn)镾KIPIF1<0，

所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上恒成立，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.3．（2022·浙江·統(tǒng)考高考真題）設(shè)函數(shù)SKIPIF1<0．求SKIPIF1<0的單調(diào)區(qū)間；【解析】SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的減區(qū)間為SKIPIF1<0，SKIPIF1<0的增區(qū)間為SKIPIF1<0.4．（2021·全國(guó)·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0．討論SKIPIF1<0的單調(diào)性；【解析】由函數(shù)的解析式可得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞減，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞增，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞減，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞增，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞減，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞增；5．（2021·北京·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0．（1）若SKIPIF1<0，求曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程；（2）若SKIPIF1<0在SKIPIF1<0處取得極值，求SKIPIF1<0的單調(diào)區(qū)間，以及其最大值與最小值．【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，此時(shí)，曲線(xiàn)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0；（2）因?yàn)镾KIPIF1<0，則SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，列表如下：SKIPIF1<0