新高考數(shù)學(xué)一輪復(fù)習(xí)講練測第3章第02講 單調(diào)性問題（講義）（解析版）

第02講單調(diào)性問題知識點(diǎn)一：單調(diào)性基礎(chǔ)問題1、函數(shù)的單調(diào)性函數(shù)單調(diào)性的判定方法：設(shè)函數(shù)SKIPIF1<0在某個(gè)區(qū)間內(nèi)可導(dǎo)，如果SKIPIF1<0，則SKIPIF1<0為增函數(shù)；如果SKIPIF1<0，則SKIPIF1<0為減函數(shù)．2、已知函數(shù)的單調(diào)性問題=1\*GB3①若SKIPIF1<0在某個(gè)區(qū)間上單調(diào)遞增，則在該區(qū)間上有SKIPIF1<0恒成立（但不恒等于0）；反之，要滿足SKIPIF1<0，才能得出SKIPIF1<0在某個(gè)區(qū)間上單調(diào)遞增；=2\*GB3②若SKIPIF1<0在某個(gè)區(qū)間上單調(diào)遞減，則在該區(qū)間上有SKIPIF1<0恒成立（但不恒等于0）；反之，要滿足SKIPIF1<0，才能得出SKIPIF1<0在某個(gè)區(qū)間上單調(diào)遞減．知識點(diǎn)二：討論單調(diào)區(qū)間問題類型一：不含參數(shù)單調(diào)性討論（1）求導(dǎo)化簡定義域（化簡應(yīng)先通分，盡可能因式分解；定義域需要注意是否是連續(xù)的區(qū)間）；（2）變號保留定號去（變號部分：導(dǎo)函數(shù)中未知正負(fù)，需要單獨(dú)討論的部分．定號部分：已知恒正或恒負(fù)，無需單獨(dú)討論的部分）；（3）求根作圖得結(jié)論（如能直接求出導(dǎo)函數(shù)等于0的根，并能做出導(dǎo)函數(shù)與x軸位置關(guān)系圖，則導(dǎo)函數(shù)正負(fù)區(qū)間段已知，可直接得出結(jié)論）；（4）未得結(jié)論斷正負(fù)（若不能通過第三步直接得出結(jié)論，則先觀察導(dǎo)函數(shù)整體的正負(fù)）；（5）正負(fù)未知看零點(diǎn)（若導(dǎo)函數(shù)正負(fù)難判斷，則觀察導(dǎo)函數(shù)零點(diǎn)）；（6）一階復(fù)雜求二階（找到零點(diǎn)后仍難確定正負(fù)區(qū)間段，或一階導(dǎo)函數(shù)無法觀察出零點(diǎn)，則求二階導(dǎo)）；求二階導(dǎo)往往需要構(gòu)造新函數(shù)，令一階導(dǎo)函數(shù)或一階導(dǎo)函數(shù)中變號部分為新函數(shù)，對新函數(shù)再求導(dǎo)．（7）借助二階定區(qū)間（通過二階導(dǎo)正負(fù)判斷一階導(dǎo)函數(shù)的單調(diào)性，進(jìn)而判斷一階導(dǎo)函數(shù)正負(fù)區(qū)間段）；類型二：含參數(shù)單調(diào)性討論（1）求導(dǎo)化簡定義域（化簡應(yīng)先通分，然后能因式分解要進(jìn)行因式分解，定義域需要注意是否是一個(gè)連續(xù)的區(qū)間）；（2）變號保留定號去（變號部分：導(dǎo)函數(shù)中未知正負(fù)，需要單獨(dú)討論的部分．定號部分：已知恒正或恒負(fù)，無需單獨(dú)討論的部分）；（3）恒正恒負(fù)先討論（變號部分因?yàn)閰?shù)的取值恒正恒負(fù)）；然后再求有效根；（4）根的分布來定參（此處需要從兩方面考慮：根是否在定義域內(nèi)和多根之間的大小關(guān)系）；（5）導(dǎo)數(shù)圖像定區(qū)間；【解題方法總結(jié)】1、求可導(dǎo)函數(shù)單調(diào)區(qū)間的一般步驟（1）確定函數(shù)SKIPIF1<0的定義域；（2）求SKIPIF1<0，令SKIPIF1<0，解此方程，求出它在定義域內(nèi)的一切實(shí)數(shù)；（3）把函數(shù)SKIPIF1<0的間斷點(diǎn)（即SKIPIF1<0的無定義點(diǎn)）的橫坐標(biāo)和SKIPIF1<0的各實(shí)根按由小到大的順序排列起來，然后用這些點(diǎn)把函數(shù)SKIPIF1<0的定義域分成若干個(gè)小區(qū)間；（4）確定SKIPIF1<0在各小區(qū)間內(nèi)的符號，根據(jù)SKIPIF1<0的符號判斷函數(shù)SKIPIF1<0在每個(gè)相應(yīng)小區(qū)間內(nèi)的增減性．注：①使SKIPIF1<0的離散點(diǎn)不影響函數(shù)的單調(diào)性，即當(dāng)SKIPIF1<0在某個(gè)區(qū)間內(nèi)離散點(diǎn)處為零，在其余點(diǎn)處均為正（或負(fù)）時(shí)，SKIPIF1<0在這個(gè)區(qū)間上仍舊是單調(diào)遞增（或遞減）的．例如，在SKIPIF1<0上，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，而顯然SKIPIF1<0在SKIPIF1<0上是單調(diào)遞增函數(shù)．②若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0（SKIPIF1<0不恒為0），反之不成立．因?yàn)镾KIPIF1<0，即SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在這個(gè)區(qū)間為常值函數(shù)；同理，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0（SKIPIF1<0不恒為0），反之不成立．這說明在一個(gè)區(qū)間上函數(shù)的導(dǎo)數(shù)大于零，是這個(gè)函數(shù)在該區(qū)間上單調(diào)遞增的充分不必要條件．于是有如下結(jié)論：SKIPIF1<0SKIPIF1<0單調(diào)遞增；SKIPIF1<0單調(diào)遞增SKIPIF1<0；SKIPIF1<0SKIPIF1<0單調(diào)遞減；SKIPIF1<0單調(diào)遞減SKIPIF1<0．題型一：利用導(dǎo)函數(shù)與原函數(shù)的關(guān)系確定原函數(shù)圖像【例1】（2023·全國·高三專題練習(xí)）設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，SKIPIF1<0的圖象如圖所示，則SKIPIF1<0的圖象最有可能的是（

）A． B．C． D．【答案】C【解析】由導(dǎo)函數(shù)的圖象可得當(dāng)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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增.只有C選項(xiàng)的圖象符合.故選：C.【對點(diǎn)訓(xùn)練1】（多選題）（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0且導(dǎo)函數(shù)為SKIPIF1<0，如圖是函數(shù)SKIPIF1<0的圖像，則下列說法正確的是A．函數(shù)SKIPIF1<0的增區(qū)間是SKIPIF1<0B．函數(shù)SKIPIF1<0的增區(qū)間是SKIPIF1<0C．SKIPIF1<0是函數(shù)的極小值點(diǎn)D．SKIPIF1<0是函數(shù)的極小值點(diǎn)【答案】BD【解析】先由題中圖像，確定SKIPIF1<0的正負(fù)，得到函數(shù)SKIPIF1<0的單調(diào)性；從而可得出函數(shù)極大值點(diǎn)與極小值點(diǎn)，進(jìn)而可得出結(jié)果.由題意，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0；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和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，因此函數(shù)SKIPIF1<0在SKIPIF1<0時(shí)取得極小值，在SKIPIF1<0時(shí)取得極大值；故A錯(cuò)，B正確；C錯(cuò)，D正確.故選：BD.【對點(diǎn)訓(xùn)練2】（2023·黑龍江齊齊哈爾·統(tǒng)考二模）已知函數(shù)SKIPIF1<0的圖象如圖所示（其中SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)），下面四個(gè)圖象中可能是SKIPIF1<0圖象的是（

）A． B．C． D．【答案】C【解析】由SKIPIF1<0的圖象知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，等號僅有可能在x＝0處取得，所以SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0單調(diào)遞增，結(jié)合選項(xiàng)只有C符合.故選：C.【對點(diǎn)訓(xùn)練3】（2023·陜西西安·校聯(lián)考一模）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的大致圖像如圖所示，SKIPIF1<0是SKIPIF1<0的導(dǎo)函數(shù)，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】若SKIPIF1<0，則SKIPIF1<0單調(diào)遞減，圖像可知，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞增，由圖像可知SKIPIF1<0，故不等式SKIPIF1<0的解集為SKIPIF1<0．故選：C【解題方法總結(jié)】原函數(shù)的單調(diào)性與導(dǎo)函數(shù)的函數(shù)值的符號的關(guān)系，原函數(shù)SKIPIF1<0單調(diào)遞增SKIPIF1<0導(dǎo)函數(shù)SKIPIF1<0（導(dǎo)函數(shù)等于0，只在離散點(diǎn)成立，其余點(diǎn)滿足SKIPIF1<0）；原函數(shù)單調(diào)遞減SKIPIF1<0導(dǎo)函數(shù)SKIPIF1<0（導(dǎo)函數(shù)等于0，只在離散點(diǎn)成立，其余點(diǎn)滿足SKIPIF1<0）．題型二：求單調(diào)區(qū)間【例2】（2023·江西鷹潭·高三貴溪市實(shí)驗(yàn)中學(xué)校考階段練習(xí)）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】函數(shù)的定義域?yàn)镾KIPIF1<0.SKIPIF1<0，則SKIPIF1<0.令SKIPIF1<0，解得SKIPIF1<0.故選：D【對點(diǎn)訓(xùn)練4】（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0（）A．嚴(yán)格增函數(shù)B．在SKIPIF1<0上是嚴(yán)格增函數(shù)，在SKIPIF1<0上是嚴(yán)格減函數(shù)C．嚴(yán)格減函數(shù)D．在SKIPIF1<0上是嚴(yán)格減函數(shù)，在SKIPIF1<0上是嚴(yán)格增函數(shù)【答案】D【解析】已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以在SKIPIF1<0上是嚴(yán)格減函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以在SKIPIF1<0上是嚴(yán)格增函數(shù)，故選：D.【對點(diǎn)訓(xùn)練5】（2023·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.求導(dǎo)可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由函數(shù)定義域可知，SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0.故選：A.【對點(diǎn)訓(xùn)練6】（2023·高三課時(shí)練習(xí)）函數(shù)SKIPIF1<0（a、b為正數(shù)）的嚴(yán)格減區(qū)間是（

）．A．SKIPIF1<0 B．SKIPIF1<0與SKIPIF1<0C．SKIPIF1<0與SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題得SKIPIF1<0.由SKIPIF1<0，令SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0．所以函數(shù)SKIPIF1<0的嚴(yán)格減區(qū)間是SKIPIF1<0與SKIPIF1<0.選項(xiàng)D，本題的兩個(gè)單調(diào)區(qū)間之間不能用“SKIPIF1<0”連接，所以該選項(xiàng)錯(cuò)誤.故選：C【解題方法總結(jié)】求函數(shù)的單調(diào)區(qū)間的步驟如下：（1）求SKIPIF1<0的定義域（2）求出SKIPIF1<0．（3）令SKIPIF1<0，求出其全部根，把全部的根在SKIPIF1<0軸上標(biāo)出，穿針引線．（4）在定義域內(nèi)，令SKIPIF1<0，解出SKIPIF1<0的取值范圍，得函數(shù)的單調(diào)遞增區(qū)間；令SKIPIF1<0，解出SKIPIF1<0的取值范圍，得函數(shù)的單調(diào)遞減區(qū)間．若一個(gè)函數(shù)具有相同單調(diào)性的區(qū)間不只一個(gè)，則這些單調(diào)區(qū)間不能用“SKIPIF1<0”、“或”連接，而應(yīng)用“和”、“，”隔開．題型三：已知含量參函數(shù)在區(qū)間上單調(diào)或不單調(diào)或存在單調(diào)區(qū)間，求參數(shù)范圍【例3】（2023·寧夏銀川·銀川一中校考三模）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則實(shí)數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．m>1【答案】B【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，所以SKIPIF1<0，解得：SKIPIF1<0故選：B.【對點(diǎn)訓(xùn)練7】（2023·陜西西安·統(tǒng)考三模）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】因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，即SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，又SKIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B【對點(diǎn)訓(xùn)練8】（2023·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0且SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)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】A【解析】令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上遞減，在SKIPIF1<0上遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為增函數(shù)，且函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0，此時(shí)SKIPIF1<0在SKIPIF1<0上遞增，則SKIPIF1<0恒成立，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為減函數(shù)，且函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增，所以SKIPIF1<0，無解，綜上所述，SKIPIF1<0的取值范圍是SKIPIF1<0.故選：A.【對點(diǎn)訓(xùn)練9】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(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<0【答案】B【解析】由題意，SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以在SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0.故選：B【對點(diǎn)訓(xùn)練10】（2023·全國·高三專題練習(xí)）三次函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則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】對函數(shù)SKIPIF1<0求導(dǎo)，得SKIPIF1<0因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0恒成立，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0恒成立；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0；又因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0不是三次函數(shù)，不滿足題意，所以SKIPIF1<0.故選：A．【對點(diǎn)訓(xùn)練11】（2023·青海西寧·高三校考開學(xué)考試）已知函數(shù)SKIPIF1<0.若對任意SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】根據(jù)題意，不妨取SKIPIF1<0，則SKIPIF1<0可轉(zhuǎn)化為SKIPIF1<0，即SKIPIF1<0.令SKIPIF1<0，則對任意SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，即SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立.令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，即實(shí)數(shù)a的取值范圍是SKIPIF1<0，故選:A【對點(diǎn)訓(xùn)練12】（2023·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在單調(diào)遞增區(qū)間，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】D【解析】∵SKIPIF1<0，∴SKIPIF1<0，若SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在單調(diào)遞增區(qū)間，則SKIPIF1<0有解，故SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞增，SKIPIF1<0，故SKIPIF1<0.故選：D.【對點(diǎn)訓(xùn)練13】（2023·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0在其定義域的一個(gè)子區(qū)間SKIPIF1<0內(nèi)不是單調(diào)函數(shù)，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】D【解析】因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0（舍去），因?yàn)镾KIPIF1<0在定義域的一個(gè)子區(qū)間SKIPIF1<0內(nèi)不是單調(diào)函數(shù)，所以SKIPIF1<0，得SKIPIF1<0，綜上，SKIPIF1<0，故選：D【對點(diǎn)訓(xùn)練14】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0）在區(qū)間SKIPIF1<0上存在單調(diào)遞增區(qū)間，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】B【解析】SKIPIF1<0函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)增區(qū)間，SKIPIF1<0函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在子區(qū)間使得不等式SKIPIF1<0成立．SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，得SKIPIF1<0，故選B.考點(diǎn)：導(dǎo)數(shù)的應(yīng)用.【例4】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，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】A【解析】由SKIPIF1<0，得SKIPIF1<0．因?yàn)镾KIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以方程SKIPIF1<0的兩個(gè)根分別位于區(qū)間SKIPIF1<0和SKIPIF1<0上，所以SKIPIF1<0，即SKIPIF1<0解得SKIPIF1<0.故選：A．【對點(diǎn)訓(xùn)練15】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，則SKIPIF1<0（

）A．3 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】B【解析】函數(shù)SKIPIF1<0，則導(dǎo)數(shù)SKIPIF1<0令SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，∴0，4是方程SKIPIF1<0的兩根，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0故選：B.【解題方法總結(jié)】（1）已知函數(shù)在區(qū)間上單調(diào)遞增或單調(diào)遞減，轉(zhuǎn)化為導(dǎo)函數(shù)恒大于等于或恒小于等于零求解，先分析導(dǎo)函數(shù)的形式及圖像特點(diǎn)，如一次函數(shù)最值落在端點(diǎn)，開口向上的拋物線最大值落在端點(diǎn)，開口向下的拋物線最小值落在端點(diǎn)等．（2）已知區(qū)間上函數(shù)不單調(diào)，轉(zhuǎn)化為導(dǎo)數(shù)在區(qū)間內(nèi)存在變號零點(diǎn)，通常用分離變量法求解參變量范圍．（3）已知函數(shù)在區(qū)間上存在單調(diào)遞增或遞減區(qū)間，轉(zhuǎn)化為導(dǎo)函數(shù)在區(qū)間上大于零或小于零有解．題型四：不含參數(shù)單調(diào)性討論【例5】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0.試判斷函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)性并證明你的結(jié)論；【解析】函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù)，證明如下：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù).【對點(diǎn)訓(xùn)練16】（2023·廣東深圳·高三深圳外國語學(xué)校?？茧A段練習(xí)）已知SKIPIF1<0若SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；【解析】若SKIPIF1<0，則SKIPIF1<0，求導(dǎo)得SKIPIF1<0，令SKIPIF1<0可得SKIPIF1<0，令SKIPIF1<0可得SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；在SKIPIF1<0上單調(diào)遞增.【對點(diǎn)訓(xùn)練17】（2023·貴州·校聯(lián)考二模）已知函數(shù)SKIPIF1<0．(1)求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)討論SKIPIF1<0在SKIPIF1<0上的單調(diào)性．【解析】（1）SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，∴曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，即SKIPIF1<0；（2）令SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上遞減，且SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，使SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號成立，顯然，等號不成立，故SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上是減函數(shù)．【對點(diǎn)訓(xùn)練18】（2023·湖南長沙·高三長沙一中?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若SKIPIF1<0，求a的取值范圍；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性；【解析】（1）由題意知SKIPIF1<0的定義域?yàn)镽.①當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，因此SKIPIF1<0.②當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，因?yàn)镾KIPIF1<0，不合題意.所以SKIPIF1<0，此時(shí)SKIPIF1<0恒成立.③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0.綜上可得，a的取值范圍是SKIPIF1<0.（2）設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上恒成立.所以SKIPIF1<0.又由（1）知SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.【對點(diǎn)訓(xùn)練19】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0．判斷SKIPIF1<0的單調(diào)性，并說明理由；【解析】SKIPIF1<0令SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0上遞增，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.【解題方法總結(jié)】確定不含參的函數(shù)的單調(diào)性，按照判斷函數(shù)單調(diào)性的步驟即可，但應(yīng)注意一是不能漏掉求函數(shù)的定義域，二是函數(shù)的單調(diào)區(qū)間不能用并集，要用“逗號”或“和”隔開．題型五：含參數(shù)單調(diào)性討論情形一：函數(shù)為一次函數(shù)【例6】（2023·山東聊城·統(tǒng)考三模）已知函數(shù)SKIPIF1<0．討論SKIPIF1<0的單調(diào)性；【解析】SKIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增．②當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增；在區(qū)間SKIPIF1<0單調(diào)遞減．③當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增．若SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減；在區(qū)間SKIPIF1<0單調(diào)遞增．綜上，SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增；在區(qū)間SKIPIF1<0單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減、在區(qū)間SKIPIF1<0單調(diào)遞增．【對點(diǎn)訓(xùn)練20】（2023·湖北黃岡·黃岡中學(xué)?？级＃┮阎瘮?shù)SKIPIF1<0.討論函數(shù)SKIPIF1<0的單調(diào)性；【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞增；若SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0（舍去）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，【對點(diǎn)訓(xùn)練21】（2023·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0.討論函數(shù)SKIPIF1<0的單調(diào)性；【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0.因?yàn)镾KIPIF1<0，若SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，若SKIPIF1<0，即SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；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.【對點(diǎn)訓(xùn)練22】（2023·福建泉州·泉州五中?？寄M預(yù)測）已知函數(shù)SKIPIF1<0．討論SKIPIF1<0的單調(diào)性；【解析】由函數(shù)SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，可得SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增；②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增.綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增.情形二：函數(shù)為準(zhǔn)一次函數(shù)【對點(diǎn)訓(xùn)練23】（2023·云南師大附中高三階段練習(xí)）已知函數(shù)SKIPIF1<0.討論SKIPIF1<0的單調(diào)性；【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0．令SKIPIF1<0，解得SKIPIF1<0，則有當(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．【對點(diǎn)訓(xùn)練24】（2023·北京·統(tǒng)考模擬預(yù)測）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求曲線SKIPIF1<0在SKIPIF1<0處的切線方程；(2)設(shè)SKIPIF1<0，討論函數(shù)SKIPIF1<0的單調(diào)性；【解析】（1）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0切點(diǎn)坐標(biāo)為SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0切線斜率為SKIPIF1<0，SKIPIF1<0曲線SKIPIF1<0在SKIPIF1<0處切線方程為：SKIPIF1<0.（2）SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0成立，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，無單調(diào)遞增區(qū)間.②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增綜上:SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，無單調(diào)遞增區(qū)間；SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0；【對點(diǎn)訓(xùn)練25】（2023·陜西安康·高三陜西省安康中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0.討論SKIPIF1<0的單調(diào)性；【解析】∵SKIPIF1<0，∴SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，此時(shí)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增.綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，在區(qū)間SKIPIF1<0上單調(diào)遞增.情形三：函數(shù)為二次函數(shù)型方向1、可因式分解【對點(diǎn)訓(xùn)練26】（2023·山東濟(jì)寧·嘉祥縣第一中學(xué)統(tǒng)考三模）已知函數(shù)SKIPIF1<0.討論函數(shù)SKIPIF1<0的單調(diào)性；【解析】因?yàn)镾KIPIF1<0，該函數(shù)的定義域?yàn)镾KIPIF1<0，SKIPIF1<0.因?yàn)镾KIPIF1<0，由SKIPIF1<0得：SKIPIF1<0或SKIPIF1<0.①當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0對任意的SKIPIF1<0恒成立，且SKIPIF1<0不恒為零，此時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0，無減區(qū)間；②當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0；由SKIPIF1<0得SKIPIF1<0.此時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0、SKIPIF1<0，減區(qū)間為SKIPIF1<0；③當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0；由SKIPIF1<0得SKIPIF1<0.此時(shí)函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0、SKIPIF1<0，減區(qū)間為SKIPIF1<0.綜上所述：當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0，無減區(qū)間；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0、SKIPIF1<0，減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0、SKIPIF1<0，減區(qū)間為SKIPIF1<0.【對點(diǎn)訓(xùn)練27】（2023·湖北咸寧·?？寄M預(yù)測）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0.討論函數(shù)SKIPIF1<0的單調(diào)性；【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.①若SKIPIF1<0時(shí)，SKIPIF1<01SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0-0+0-SKIPIF1<0SKIPIF1<0極小值SKIPIF1<0極大值SKIPIF1<0②若SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0單調(diào)遞減，③若SKIPIF1<0時(shí)SKIPIF1<0SKIPIF1<0SKIPIF1<01SKIPIF1<0SKIPIF1<0-0+0-SKIPIF1<0SKIPIF1<0極小值SKIPIF1<0極大值SKIPIF1<0④若SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增.綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增.【對點(diǎn)訓(xùn)練28】（2023·北京海淀·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0．(1)若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與SKIPIF1<0軸平行，求SKIPIF1<0；(2)求SKIPIF1<0的單調(diào)區(qū)間．【解析】（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0SKIPIF1<0.SKIPIF1<0.由題設(shè)知SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.此時(shí)SKIPIF1<0.所以SKIPIF1<0的值為1（2）由（1）得SKIPIF1<0．1）當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的變化情況如下表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0單調(diào)遞增極大值單調(diào)遞減2）當(dāng)SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或2①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0的變化情況如下表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<02SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<00SKIPIF1<0SKIPIF1<0單調(diào)遞減極小值單調(diào)遞增極大值單調(diào)遞減②當(dāng)SKIPIF1<0時(shí)，（ⅰ）當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<02SKIPIF1<0SKIPIF1<0