新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練9.2 利用導(dǎo)數(shù)求單調(diào)性（基礎(chǔ)版）（解析版）

9.2利用導(dǎo)數(shù)求單調(diào)性（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一無(wú)參函數(shù)的單調(diào)區(qū)間【例1】（2022高二下·灤南期末）函數(shù)SKIPIF1<0單調(diào)遞減區(qū)間是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】f′(x)=SKIPIF1<0，令f′(x)<0，解得：1<x<e，故f(x)在(1，e)遞減，故答案為：D.【一隅三反】1．（2022·全國(guó)課時(shí)練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0．故選：D.2（2022·全國(guó)·課時(shí)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】B【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，故函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0．故選：B．3．（2022·全國(guó)·高三專題練習(xí)（文））函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為_(kāi)_________．【答案】SKIPIF1<0【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則其在SKIPIF1<0上遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，綜上，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，故答案為：SKIPIF1<0考點(diǎn)二單調(diào)函數(shù)求參數(shù)【例2-1】（2022高三上·成都開(kāi)學(xué)考）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，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<0SKIPIF1<0由題意，已知條件等價(jià)為SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：B.【例2-2】（2022·浙江）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【一隅三反】1．（2022·新疆）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，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】B【解析】依題意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上單調(diào)遞增，則SKIPIF1<0，所以SKIPIF1<0.故選:B.2．（2022·河南）若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，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】B【解析】SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0在SKIPIF1<0上恒成立，而SKIPIF1<0時(shí)，易見(jiàn)SKIPIF1<0，只需要SKIPIF1<0即可，故SKIPIF1<0.故選：B.3．（2022·惠州模擬）若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則實(shí)數(shù)SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0恒成立，分離參數(shù)可得SKIPIF1<0在SKIPIF1<0恒成立，令SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.考點(diǎn)三非單調(diào)函數(shù)求參數(shù)【例3-1】（2022·黑龍江）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)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0可得：SKIPIF1<0.因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在單調(diào)遞增區(qū)間，所以SKIPIF1<0在SKIPIF1<0上有解，即SKIPIF1<0在SKIPIF1<0上有解.設(shè)SKIPIF1<0，由SKIPIF1<0在SKIPIF1<0上恒成立，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0.所以SKIPIF1<0.故選：D【例3-2】（2022·北京十四中高三開(kāi)學(xué)考試）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實(shí)數(shù)k的取值范圍是（

）A．SKIPIF1<0或SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．不存在這樣的實(shí)數(shù)【答案】B【解析】SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，或SKIPIF1<0，所以當(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上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，即函數(shù)SKIPIF1<0極值點(diǎn)為SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0故選：B．【例3-3】（2022·上海）已知函數(shù)SKIPIF1<0存在三個(gè)單調(diào)區(qū)間，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0存在三個(gè)單調(diào)區(qū)間，可得SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，則滿足SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C.【一隅三反】1．（2022·福建·莆田一中）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(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】A【解析】因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，所以SKIPIF1<0在區(qū)間SKIPIF1<0上有解，即SKIPIF1<0在區(qū)間SKIPIF1<0上有解．令SKIPIF1<0，則SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．又因?yàn)镾KIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0.故選：A2．（2022北京）若函數(shù)SKIPIF1<0存在遞減區(qū)間，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題設(shè)，SKIPIF1<0，由SKIPIF1<0存在遞減區(qū)間，即存在SKIPIF1<0使SKIPIF1<0，∴SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0.故選：B3．（2022·全國(guó)·高三專題練習(xí)（文））若函數(shù)SKIPIF1<0在定義域內(nèi)的一個(gè)子區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實(shí)數(shù)k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．(1，2] D．[1，2)【答案】A【解析】顯然函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0．由SKIPIF1<0，得函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0；由SKIPIF1<0，得函數(shù)SKIPIF1<0單調(diào)遞減區(qū)間為SKIPIF1<0．因?yàn)楹瘮?shù)在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，又因?yàn)镾KIPIF1<0為定義域內(nèi)的一個(gè)子區(qū)間，所以SKIPIF1<0，即SKIPIF1<0．綜上可知實(shí)數(shù)k的取值范圍是SKIPIF1<0．故選：A4．（2022·云南）若函數(shù)SKIPIF1<0恰好有三個(gè)不同的單調(diào)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】由題意得SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0恰好有三個(gè)不同的單調(diào)區(qū)間，SKIPIF1<0有兩個(gè)不同的零點(diǎn)，所以，SKIPIF1<0，解得SKIPIF1<0.因此，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D.考點(diǎn)四單調(diào)性的運(yùn)用【例4-1】（2022·湖北模擬）已知函數(shù)SKIPIF1<0，不等式SKIPIF1<0的解集為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0等價(jià)于SKIPIF1<0，解得SKIPIF1<0，即原不等式的解集為SKIPIF1<0.故答案為：B.【例4-2】（2022·湖北模擬）已知：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0、SKIPIF1<0、SKIPIF1<0大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故答案為：B.【例4-3】．（2022·四川·遂寧中學(xué)外國(guó)語(yǔ)實(shí)驗(yàn)學(xué)校高三開(kāi)學(xué)考試（理））設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，SKIPIF1<0的圖像如圖所示，則SKIPIF1<0的圖像最有可能的是（

）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.【一隅三反】1．（2022·江陰模擬）已知SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】令函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，求導(dǎo)得：SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，顯然SKIPIF1<0，則有SKIPIF1<0，所以SKIPIF1<0.故答案為：C2．（2022·全國(guó)課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0在定義域SKIPIF1<0內(nèi)可導(dǎo)，其圖象如圖所示.記SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】對(duì)于不等式對(duì)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則結(jié)合圖象，知原不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則結(jié)合圖象，知原不等式的解集為SKIPIF1<0．綜上，原不等式的解集為SKIPIF1<0．故選：A3．（2022·湖北模擬）已知定義域?yàn)镽的函數(shù)SKIPIF1<0，有SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的解集為．【答案】SKIPIF1<0【解析】SKIPIF1<0，SKIPIF1<0SKIPIF1<0在SKIPIF1<0為增函數(shù)，又SKIPIF1<0為偶函數(shù)，∴SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，解集為SKIPIF1<0故答案為：SKIPIF1<0．9.2利用導(dǎo)數(shù)求單調(diào)性（精練）（基礎(chǔ)版）題組一題組一無(wú)參函數(shù)求單調(diào)區(qū)間1．（2022·廣東·東莞四中高三階段練習(xí)）函數(shù)SKIPIF1<0，則SKIPIF1<0的單調(diào)增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，求導(dǎo)得：SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的單調(diào)增區(qū)間是SKIPIF1<0.故選：B2．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0解不等式SKIPIF1<0，可得SKIPIF1<0，故函數(shù)SKIPIF1<0的遞減區(qū)間為SKIPIF1<0.故選：B．3．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的單調(diào)減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0,由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.故選：B4．（2022·河北·石家莊二中模擬預(yù)測(cè)）已知函數(shù)f(x)滿足SKIPIF1<0，則f(x)的單調(diào)遞減區(qū)間為（

）A．(-,0) B．(1,+∞) C．(-,1) D．(0,+∞)【答案】A【解析】由題設(shè)SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，而SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0遞增，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即SKIPIF1<0遞減，故SKIPIF1<0遞減區(qū)間為(-,0).故選：A5．（2022·湖北黃岡·高三階段練習(xí)）（多選）下列區(qū)間中能使函數(shù)SKIPIF1<0單調(diào)遞增的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【解析】由SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0即SKIPIF1<0，解得SKIPIF1<0，或SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增；所以SKIPIF1<0在定義域內(nèi)是單調(diào)遞增函數(shù),所以函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增.故選：BD.6．（2022·全國(guó)·高二單元測(cè)試）函數(shù)SKIPIF1<0的單調(diào)減區(qū)間為_(kāi)_________．【答案】SKIPIF1<0【解析】∵SKIPIF1<0，則SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0∴函數(shù)SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0故答案為：SKIPIF1<0.7．（2022·全國(guó)課時(shí)練習(xí)）設(shè)函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，則函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為_(kāi)_________．【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又因?yàn)楹瘮?shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<0.8．（2022·廣西）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是______________.【答案】SKIPIF1<0【解析】SKIPIF1<0的定義域?yàn)镽，且SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，即函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0.故答案為：SKIPIF1<0題組二題組二單調(diào)函數(shù)求參數(shù)1．（2022·四川成都·高三階段練習(xí)（文））若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則實(shí)數(shù)k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題意得，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，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，得SKIPIF1<0，所以SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B2．（2022·河南）已知函數(shù)SKIPIF1<0在SKIPIF1<0上為單調(diào)遞增函數(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】A【解析】SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上為單調(diào)遞增函數(shù)，故SKIPIF1<0在SKIPIF1<0上恒成立，所以SKIPIF1<0即SKIPIF1<0，故選：A.3．（2022·江西）已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】C【解析】因?yàn)楹瘮?shù)SKIPIF1<0在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，由SKIPIF1<0在SKIPIF1<0上單調(diào)遞增知，SKIPIF1<0，所以SKIPIF1<0，故選：C4．（2022·全國(guó)·高三專題練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，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】SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)，SKIPIF1<0在SKIPIF1<0上恒成立，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，即SKIPIF1<0，故選：A．5（2022·廣東東莞）若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則實(shí)數(shù)a的取值范圍是（）A．（-1，1） B．SKIPIF1<0 C．（-1，+∞） D．（-1，0）【答案】B【解析】SKIPIF1<0，由題意得：SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上恒成立，因?yàn)镾KIPIF1<0，所以SKIPIF1<0恒成立，故實(shí)數(shù)a的取值范圍是SKIPIF1<0.故選：B6．（2022·吉林吉林·模擬預(yù)測(cè)（文））若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則實(shí)數(shù)a的取值范圍(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由題可知，SKIPIF1<0恒成立，故SKIPIF1<0，即SKIPIF1<0．故選：A﹒7．（2022·全國(guó)·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.8．（2022·河南宋基信陽(yáng)實(shí)驗(yàn)中學(xué)高二階段練習(xí)（理））若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增，所以有SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，因?yàn)镾KIPIF1<0，所以由SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，于是有SKIPIF1<0，故選：D9．（2022·山東聊城）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則實(shí)數(shù)m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，則SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0恒成立，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，導(dǎo)數(shù)不恒為0，故選：D.10．（2022·廣東順德德勝學(xué)校）函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，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】D【解析】：因?yàn)楹瘮?shù)SKIPIF1<0，所以SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在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，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以實(shí)數(shù)a的取值范圍是SKIPIF1<0，故選：D11．（2022·江西吉安）已知函數(shù)SKIPIF1<0在SKIPIF1<0上為單調(diào)遞增函數(shù)，則實(shí)數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上為單調(diào)遞增函數(shù)，所以SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，要滿足SKIPIF1<0①，或SKIPIF1<0②，由①得：SKIPIF1<0，由②得：SKIPIF1<0，綜上：實(shí)數(shù)m的取值范圍是SKIPIF1<0.故選：D12．（2022·江西）若函數(shù)f(x)＝x2＋ax＋SKIPIF1<0在[SKIPIF1<0，＋∞)上是增函數(shù)，則a的取值范圍是（

）A．[－1，0] B．[－1，＋∞)C．[0，3] D．[3，＋∞)【答案】D【解析】f′(x)＝2x＋a－SKIPIF1<0，由于函數(shù)f(x)在[SKIPIF1<0，＋∞)上是增函數(shù)，故f′(x)≥0在[SKIPIF1<0，＋∞)上恒成立.即a≥SKIPIF1<0－2x在[SKIPIF1<0，＋∞)上恒成立.設(shè)h(x)＝SKIPIF1<0－2x，x∈[SKIPIF1<0，＋∞)，易知h(x)在[SKIPIF1<0，＋∞)上為減，∴h(x)max＝h(SKIPIF1<0)＝3，∴a≥3.故選：D13．（2022·湖北）已知函數(shù)SKIPIF1<0，不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0等價(jià)于SKIPIF1<0，解得SKIPIF1<0，即原不等式的解集為SKIPIF1<0.故選：B.14．（2022·河南·高三階段練習(xí)（理））若SKIPIF1<0是R上的減函數(shù)，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】B【解析】由SKIPIF1<0，得SKIPIF1<0，因?yàn)镾KIPIF1<0是R上的減函數(shù)，所以SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0SKIPIF1<0在SKIPIF1<0上恒成立，由于SKIPIF1<0，所以SKIPIF1<0．故選：B.15．（2022·廣西欽州）函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增的一個(gè)必要不充分條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題得SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增，SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立．SKIPIF1<0，而SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，SKIPIF1<0．選項(xiàng)中只有SKIPIF1<0是SKIPIF1<0的必要不充分條件.選項(xiàng)AC是SKIPIF1<0的充分不必要條件，選項(xiàng)B是充要條件.故選：D16．（2022·全國(guó)·高三專題練習(xí)）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【答案】A【解析】求導(dǎo)SKIPIF1<0，令SKIPIF1<0，由SKIPIF1<0在R上單調(diào)，可知SKIPIF1<0恒成立或SKIPIF1<0恒成立，分類討論：SKIPIF1<0（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，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；SKIPIF1<0，即SKIPIF1<0恒成立，符合題意；（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0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；SKIPIF1<0，即SKIPIF1<0恒成立，符合題意；（3）當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，研究SKIPIF1<0內(nèi)的情況即可：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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得極小值，且滿足SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0取得極小值，且滿足SKIPIF1<0SKIPIF1<0，且SKIPIF1<0同理SKIPIF1<0，且SKIPIF1<0又SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故不符合；所以a的取值范圍是SKIPIF1<0故選：A題組三題組三非單調(diào)函數(shù)求參數(shù)1．（2022·福建）若函數(shù)SKIPIF1<0存在單調(diào)遞減區(qū)間,則實(shí)數(shù)b的取值范圍為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0，可得SKIPIF1<0，由題意可得存在SKIPIF1<0，使得SKIPIF1<0，即存在SKIPIF1<0，使得SKIPIF1<0，等價(jià)于SKIPIF1<0，由對(duì)勾函數(shù)性質(zhì)易得SKIPIF1<0，故選B.2．（2022·陜西）若函數(shù)SKIPIF1<0恰好有三個(gè)單調(diào)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】∵函數(shù)f（x）＝ax3﹣3x2+x+1，∴f′（x）＝3ax2﹣6x+1，由函數(shù)f（x）恰好有三個(gè)單調(diào)區(qū)間，得f′（x）有兩個(gè)不相等的零點(diǎn)，∴3ax2﹣6x+1＝0滿足：a≠0，且△＝36﹣12a＞0，解得a＜3，∴a∈（﹣∞，0）∪（0，3）．故選D．3．（2022·西藏）已知函數(shù)SKIPIF1<0．若SKIPIF1<0在SKIPIF1<0內(nèi)不單調(diào)，則實(shí)數(shù)a的取值范圍是______．【答案】SKIPIF1<0【解析】由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0內(nèi)為減函數(shù)時(shí)，則SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，所以SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，當(dāng)SKIPIF1<0在SKIPIF1<0內(nèi)為增函數(shù)時(shí)，則SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，所以SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，令SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞增，在SKIPIF1<0內(nèi)單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0內(nèi)的值域?yàn)镾KIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)時(shí)，a的取值范圍是SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上不單調(diào)時(shí)，實(shí)數(shù)a的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<0．4．（2022·全國(guó)·高三專題練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間(1，4)上不單調(diào)，則實(shí)數(shù)a的取值范圍是___________.【答案】(4，5)【解析】SKIPIF1<0函數(shù)SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則SKIPIF1<0在SKIPIF1<0上存在變號(hào)零點(diǎn)，由SKIPIF1<0得SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0遞減，在SKIPIF1<0遞增，而SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.5．（2022·陜西）若函數(shù)SKIPIF1<0有三個(gè)單調(diào)區(qū)間，則實(shí)數(shù)a的取值范圍是________．【答案】SKIPIF1<0【解析】SKIPIF1<0，由于函數(shù)SKIPIF1<0有三個(gè)單調(diào)區(qū)間，所以SKIPIF1<0有兩個(gè)不相等的實(shí)數(shù)根，所以SKIPIF1<0.故答案為：SKIPIF1<06．（2022·四川）已知函數(shù)SKIPIF1<0.若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實(shí)數(shù)t的取值范圍為_(kāi)_________．【答案】SKIPIF1<0【解析】求導(dǎo)得SKIPIF1<0，易知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單增；SKIPIF1<0，SKIPIF1<0，SKIPIF1<0單減；若使SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)SKIPIF1<0，只需SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<07．（2022·重慶）已知函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)，則SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0【解析】SKIPIF1<0因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)所以SKIPIF1<0必有解當(dāng)SKIPIF1<0只有一個(gè)解時(shí)，SKIPIF1<0得出函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，與題干矛盾，故SKIPIF1<0必有兩個(gè)不等實(shí)根則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0故答案為SKIPIF1<0題組四題組四單調(diào)性的運(yùn)用1．（2022·贛州模擬）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得極大值，則SKIPIF1<0，SKIPIF