新高考數(shù)學(xué)一輪復(fù)習(xí)提升訓(xùn)練4.2 利用導(dǎo)數(shù)求單調(diào)性（精講）（解析版）

4.2利用導(dǎo)數(shù)求單調(diào)性（精講）（提升版）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)一單調(diào)區(qū)間（無(wú)參）【例1-1】（2022·新疆）函數(shù)SKIPIF1<0的減區(qū)間是____________.【答案】SKIPIF1<0【解析】由SKIPIF1<0可得SKIPIF1<0所以由SKIPIF1<0可得SKIPIF1<0所以函數(shù)SKIPIF1<0的減區(qū)間是SKIPIF1<0故答案為：SKIPIF1<0【例1-2】（2022·廣東·順德一中）設(shè)曲線SKIPIF1<0在SKIPIF1<0上的單調(diào)遞減區(qū)間是______.【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0所以函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<0.【例1-3】（江蘇省蘇州實(shí)驗(yàn)中學(xué)）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：A【一隅三反】1．函數(shù)f(x)＝x＋2eq\r(1－x)的單調(diào)遞增區(qū)間是()A．(0,1) B．(－∞，1)C．(－∞，0) D．(0，＋∞)【答案】C【解析】f(x)的定義域?yàn)?－∞，1]，f′(x)＝1－eq\f(1,\r(1－x))，令f′(x)＝0，得x＝0.當(dāng)0<x<1時(shí)，f′(x)<0.當(dāng)x<0時(shí)，f′(x)>0.∴f(x)的單調(diào)遞增區(qū)間為(－∞，0)，單調(diào)遞減區(qū)間為(0,1)．2．（皖豫名校聯(lián)盟體2022屆）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<03．已知定義在區(qū)間(0，π)上的函數(shù)f(x)＝x＋2cosx，則f(x)的單調(diào)遞增區(qū)間為．【答案】eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(π,6)))，eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,6)，π))【解析】f′(x)＝1－2sinx，x∈(0，π)．令f′(x)＝0，得x＝eq\f(π,6)或x＝eq\f(5π,6)，當(dāng)0<x<eq\f(π,6)時(shí)，f′(x)>0，當(dāng)eq\f(π,6)<x<eq\f(5π,6)時(shí)，f′(x)<0，當(dāng)eq\f(5π,6)<x<π時(shí)，f′(x)>0，∴f(x)在eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(π,6)))和eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,6)，π))上單調(diào)遞增，在eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)，\f(5π,6)))上單調(diào)遞減．考點(diǎn)二已知單調(diào)性求參數(shù)【例2-1】（2022安徽省“皖東縣中聯(lián)盟）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，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】對(duì)于函數(shù)SKIPIF1<0，導(dǎo)數(shù)SKIPIF1<0.要使函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，只需SKIPIF1<0恒成立.因?yàn)镾KIPIF1<0，只需SKIPIF1<0，只需SKIPIF1<0恒成立.記SKIPIF1<0，只需SKIPIF1<0.SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0.由SKIPIF1<0在SKIPIF1<0上單減，在SKIPIF1<0上單增，且SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0的最大值為1.所以SKIPIF1<0.故選：B【例2-2】（2022.廣東）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)a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∵函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)∴SKIPIF1<0在區(qū)間SKIPIF1<0上有根∴當(dāng)a＝0時(shí)，x＝－1不滿足條件當(dāng)SKIPIF1<0時(shí)，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.故選：D．【一隅三反】1．（2022福建?。┮阎瘮?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：D2．（湖南省三湘名校教育聯(lián)盟2022屆）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.3．（江西省宜春市八校2022屆）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)減區(qū)間，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)遞減區(qū)間，所以存在SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0.故選：A4．（2022·寧夏吳忠）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.考點(diǎn)三單調(diào)性的應(yīng)用之解不等式【例3】（湖南省多所學(xué)校2022屆）已知SKIPIF1<0，則SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】B【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0恒成立，∴SKIPIF1<0在SKIPIF1<0單調(diào)遞增，且SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0是偶函數(shù)，∴SKIPIF1<0的解集是SKIPIF1<0或SKIPIF1<0，故選：B.【一隅三反】1．（陜西省西安地區(qū)八校2022屆）已知函數(shù)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】D【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以不等式SKIPIF1<0等價(jià)于SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以不等式SKIPIF1<0的解集為SKIPIF1<0.故選：D2．（湖北省2022屆）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.3．若函數(shù)f(x)＝lnx＋ex－sinx，則不等式f(x－1)≤f(1)的解集為．【答案】(1,2]【解析】f(x)的定義域?yàn)?0，＋∞)，∴f′(x)＝eq\f(1,x)＋ex－cosx.∵x>0，∴ex>1，∴f′(x)>0，∴f(x)在(0，＋∞)上單調(diào)遞增，又f(x－1)≤f(1)，∴0<x－1≤1，即1<x≤2，原不等式的解集為(1,2]．4．已知函數(shù)f(x)＝xsinx＋cosx＋x2，則不等式f(lnx)＋f

eq\b\lc\(\rc\)(\a\vs4\al\co1(ln\f(1,x)))<2f(1)的解集為．【答案】eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,e)，e))【解析】f(x)＝xsinx＋cosx＋x2是偶函數(shù)，所以f

eq\b\lc\(\rc\)(\a\vs4\al\co1(ln\f(1,x)))＝f(－lnx)＝f(lnx)．則原不等式可變形為f(lnx)<f(1)?f(|lnx|)<f(1)．又f′(x)＝xcosx＋2x＝x(2＋cosx)，由2＋cosx>0，得當(dāng)x>0時(shí)，f′(x)>0.所以f(x)在(0，＋∞)上單調(diào)遞增．∴|lnx|<1?－1<lnx<1?eq\f(1,e)<x<e.考點(diǎn)四單調(diào)性應(yīng)用之比較大小【例4-1】（華大新高考聯(lián)盟名校2022屆）已知實(shí)數(shù)a，b，SKIPIF1<0，e為自然對(duì)數(shù)的底數(shù)，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，求導(dǎo)得SKIPIF1<0，令SKIPIF1<0，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．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0.故選：A．【例4-2】（湖南師范大學(xué)附中2022屆）下列兩數(shù)的大小關(guān)系中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】對(duì)于A，設(shè)SKIPIF1<0，則SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，A錯(cuò)誤；對(duì)于B，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，B正確；對(duì)于C，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，C錯(cuò)誤；對(duì)于D，SKIPIF1<0，D錯(cuò)誤.故選：B.【一隅三反】1．（2022年全國(guó)新高考I卷數(shù)學(xué)試題）設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0,令SKIPIF1<0，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，又SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0故選：C.2．（山東省青州市2022屆）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)單調(diào)遞減,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)單調(diào)遞增,故當(dāng)SKIPIF1<0時(shí),函數(shù)取得最大值SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)單調(diào)遞減,可得SKIPIF1<0,即SKIPIF1<0.故選:C3．（江西省萍鄉(xiāng)市2022屆）設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，可以判斷SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0SKIPIF1<0所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故選：D.4．（湖北省二十一所重點(diǎn)中學(xué)2022屆）已知SKIPIF1<0是自然對(duì)數(shù)的底數(shù)，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，下列說(shuō)法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】根據(jù)題意，設(shè)SKIPIF1<0，易知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0遞減；SKIPIF1<0，即為SKIPIF1<0；SKIPIF1<0，即為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0；SKIPIF1<0，即SKIPIF1<0，故A錯(cuò)，故D錯(cuò)；SKIPIF1<0，即SKIPIF1<0，故B錯(cuò)；構(gòu)造函數(shù)SKIPIF1<0，所以SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0；故選：C.考點(diǎn)五含參函數(shù)的單調(diào)性討論【例5-1】（2022廣西節(jié)選）已知函數(shù)SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；【答案】答案見(jiàn)解析【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0即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)遞減；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【例5-2】（2022安徽）已知函數(shù)SKIPIF1<0，討論f（x）的單調(diào)性；【答案】答案見(jiàn)解析【解析】由題意得：f（x）定義域?yàn)椋?，+∞），SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0在（0，+∞）上恒成立，∴f（x）在（0，+∞）上單調(diào)遞增；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∴f（x）在（0，SKIPIF1<0）上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；綜上所述：當(dāng)SKIPIF1<0時(shí)，f（x）在（0，+∞）上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，f（x）在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.【例5-3】（安徽省江淮名校2022屆）已知函數(shù)SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；【答案】答案見(jiàn)解析【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0．SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減．當(dāng)SKIPIF1<0時(shí)SKIPIF1<0在SKIPIF1<0單調(diào)遞增．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則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．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0．所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減．綜上所述，SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減．SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增．SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減．SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0，SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減．【例5-4】（2022遼寧省沈陽(yáng)市第二中學(xué)）已知函數(shù)SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；【答案】答案見(jiàn)解析【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，對(duì)任意的SKIPIF1<0，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0的減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0在SKIPIF1<0時(shí)的解為SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，此時(shí)，函數(shù)SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0.綜上所述，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0.【一隅三反】1．（2022貴州省貴陽(yáng)市五校）已知SKIPIF1<0，函數(shù)SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；【答案】SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0遞增；SKIPIF1<0時(shí)，SKIPIF1<0的增區(qū)間是SKIPIF1<0，減區(qū)間是SKIPIF1<0．【解析】SKIPIF1<0的定義域是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0遞增，SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0的增區(qū)間是SKIPIF1<0，減區(qū)間是SKIPIF1<0．綜上：SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0遞增；SKIPIF1<0時(shí)，SKIPIF1<0的增區(qū)間是SKIPIF1<0，減區(qū)間是SKIPIF1<0．2．（2022陜西省）已知函數(shù)SKIPIF1<0．討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】答案解析【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0；故SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減．3．（重慶市第八中學(xué)校2022屆高三下學(xué)期適應(yīng)性月考（七）數(shù)學(xué)試題）已知SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；【答案】見(jiàn)解析【解析】SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，則SKIPIF1<0