新高考數(shù)學(xué)一輪復(fù)習(xí) 函數(shù)專(zhuān)項(xiàng)重難點(diǎn)突破專(zhuān)題06 函數(shù)的單調(diào)性(解析版)

專(zhuān)題06函數(shù)的單調(diào)性真題再現(xiàn)一、單選題1．（2023·北京·統(tǒng)考高考真題）下列函數(shù)中，在區(qū)間SKIPIF1<0上單調(diào)遞增的是（

）A．SKIPIF1<0B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】對(duì)于A，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故A錯(cuò)誤；對(duì)于B，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故B錯(cuò)誤；對(duì)于C，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C正確；對(duì)于D，因?yàn)镾KIPIF1<0，SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上不單調(diào)，D錯(cuò)誤.故選：C.2．（2023·全國(guó)·統(tǒng)考高考真題）設(shè)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】函數(shù)SKIPIF1<0在R上單調(diào)遞增，而函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則有函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，因此SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D3．（2023·全國(guó)·統(tǒng)考高考真題）已知函數(shù)SKIPIF1<0．記SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0開(kāi)口向下，對(duì)稱(chēng)軸為SKIPIF1<0，因?yàn)镾KIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0由二次函數(shù)性質(zhì)知SKIPIF1<0，因?yàn)镾KIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，綜上，SKIPIF1<0，又SKIPIF1<0為增函數(shù)，故SKIPIF1<0，即SKIPIF1<0.故選：A.4．（2022·天津·統(tǒng)考高考真題）函數(shù)SKIPIF1<0的圖像為（

）A． B．C． D．【解析】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，函數(shù)SKIPIF1<0為奇函數(shù)，A選項(xiàng)錯(cuò)誤；又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，C選項(xiàng)錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0函數(shù)單調(diào)遞增，故B選項(xiàng)錯(cuò)誤；故選：D.5．（2021·全國(guó)·高考真題）下列函數(shù)中是增函數(shù)的為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】對(duì)于A，SKIPIF1<0為SKIPIF1<0上的減函數(shù)，不合題意，舍.對(duì)于B，SKIPIF1<0為SKIPIF1<0上的減函數(shù)，不合題意，舍.對(duì)于C，SKIPIF1<0在SKIPIF1<0為減函數(shù)，不合題意，舍.對(duì)于D，SKIPIF1<0為SKIPIF1<0上的增函數(shù)，符合題意，故選：D.考點(diǎn)一判斷或證明函數(shù)的單調(diào)性一、單選題1．下列函數(shù)中，在區(qū)間SKIPIF1<0上為增函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】由于SKIPIF1<0在SKIPIF1<0為單調(diào)遞減函數(shù)，在SKIPIF1<0時(shí)無(wú)意義，A錯(cuò)誤；SKIPIF1<0在SKIPIF1<0為單調(diào)遞增函數(shù)，B正確；SKIPIF1<0定義域?yàn)镾KIPIF1<0，在SKIPIF1<0無(wú)意義，C錯(cuò)誤；SKIPIF1<0在SKIPIF1<0為單調(diào)遞減函數(shù)，D錯(cuò)誤，故選：B2．下列函數(shù)中，是奇函數(shù)且在區(qū)間SKIPIF1<0上單調(diào)遞增的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】對(duì)于A，SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)的，故A錯(cuò)；對(duì)于B，SKIPIF1<0，定義域不關(guān)于原點(diǎn)對(duì)稱(chēng)，不是奇函數(shù)，故B錯(cuò)；對(duì)于C，SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，故C錯(cuò)；對(duì)于D，SKIPIF1<0，是奇函數(shù)且在區(qū)間SKIPIF1<0上單調(diào)遞增，故D對(duì)；故選：D.3．在下列函數(shù)中：①SKIPIF1<0，②SKIPIF1<0，③SKIPIF1<0，④SKIPIF1<0，在SKIPIF1<0上為增函數(shù)的有（

）A．①② B．③④ C．②③ D．①④【解析】因?yàn)镾KIPIF1<0，所以①SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，不符合題意；②SKIPIF1<0在SKIPIF1<0上為常函數(shù)，不符合題意；③SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，符合題意；④SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，符合題意；故符合題意的為③④.故選：B.二、解答題4．已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0是奇函數(shù).(1)求實(shí)數(shù)SKIPIF1<0，SKIPIF1<0的值；(2)判斷SKIPIF1<0在SKIPIF1<0上的單調(diào)性并用定義證明；(3)若對(duì)任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.【解析】（1）由題意，定義域?yàn)镾KIPIF1<0的函數(shù)SKIPIF1<0是奇函數(shù)，得SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，經(jīng)檢驗(yàn)知，SKIPIF1<0是奇函數(shù).故SKIPIF1<0.（2）由（1）知，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù).證明如下：設(shè)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.∴函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù).（3）由SKIPIF1<0，且SKIPIF1<0是奇函數(shù)，得SKIPIF1<0，∵SKIPIF1<0在SKIPIF1<0上是減函數(shù)，所以SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，即SKIPIF1<0對(duì)任意的SKIPIF1<0恒成立，由SKIPIF1<0，得SKIPIF1<0，∴SKIPIF1<0，所以SKIPIF1<0，故得實(shí)數(shù)SKIPIF1<0的取值范圍SKIPIF1<0.5．已知函數(shù)SKIPIF1<0在SKIPIF1<0為奇函數(shù)，且SKIPIF1<0(1)求SKIPIF1<0值；(2)判斷函數(shù)SKIPIF1<0在SKIPIF1<0的單調(diào)性，并用定義證明；(3)解關(guān)于t的不等式SKIPIF1<0【解析】（1）SKIPIF1<0在SKIPIF1<0為奇函數(shù)，SKIPIF1<0，解得：SKIPIF1<0，

又SKIPIF1<0，解得：SKIPIF1<0，故SKIPIF1<0，經(jīng)檢驗(yàn)滿(mǎn)足題設(shè).（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0

SKIPIF1<0當(dāng)SKIPIF1<0時(shí)函數(shù)SKIPIF1<0在SKIPIF1<0為奇函數(shù)，由SKIPIF1<0SKIPIF1<0，判斷函數(shù)SKIPIF1<0在SKIPIF1<0為單調(diào)遞減，證明：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，

SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0為單調(diào)遞減，（3）則SKIPIF1<0，SKIPIF1<0在SKIPIF1<0為奇函數(shù)，SKIPIF1<0，又函數(shù)SKIPIF1<0在SKIPIF1<0為單調(diào)遞減，SKIPIF1<0，SKIPIF1<0t的不等式的解集為SKIPIF1<06．已知函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，滿(mǎn)足SKIPIF1<0，SKIPIF1<0時(shí)SKIPIF1<0，對(duì)任意正實(shí)數(shù)x，y，都有SKIPIF1<0．(1)求SKIPIF1<0的值；(2)證明：函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)；(3)求不等式SKIPIF1<0的解集．【解析】（1）因?yàn)閷?duì)任意正實(shí)數(shù)x，y，都有SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0.（2）由SKIPIF1<0得SKIPIF1<0，任取SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)；（3）由（1）知，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，由（2）知，函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)；所以SKIPIF1<0，解得SKIPIF1<0，故不等式SKIPIF1<0的解集為SKIPIF1<0.7．函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的函數(shù)，滿(mǎn)足下列條件：①SKIPIF1<0；②SKIPIF1<0；③任意SKIPIF1<0，有SKIPIF1<0．(1)求SKIPIF1<0的值；(2)判斷并證明函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性；(3)解不等式SKIPIF1<0．【解析】（1）SKIPIF1<0任意SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0，有SKIPIF1<0，當(dāng)SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0，（2）結(jié)論：SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)．證明：任取SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0任意SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0．SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)．（3）SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0由（2）可知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)，又SKIPIF1<0，可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．SKIPIF1<0不等式SKIPIF1<0的解集為SKIPIF1<0．考點(diǎn)二求單調(diào)性區(qū)間一、單選題1．函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為（

）A．（–∞，2] B．[2，+∞）C．[0，2] D．[0，+∞）【解析】∵SKIPIF1<0，∴函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是（–∞，2]，增區(qū)間為[2，+∞），∴SKIPIF1<0的單調(diào)遞減區(qū)間是[2，+∞），故選：B．2．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【解析】函數(shù)SKIPIF1<0的定義域需要滿(mǎn)足SKIPIF1<0，解得SKIPIF1<0定義域?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故選：D.3．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【解析】因?yàn)镾KIPIF1<0，所以SKIPIF1<0的增區(qū)間為SKIPIF1<0，故選：D.4．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】SKIPIF1<0的定義域?yàn)镾KIPIF1<0，而SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0的增區(qū)間為SKIPIF1<0.故選：A.5．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】令SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∵函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在定義域內(nèi)遞增，∴根據(jù)復(fù)合函數(shù)的單調(diào)性可知，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0故選：C6．已知函數(shù)SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】依題意，SKIPIF1<0解得a＝－1，故SKIPIF1<0，可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增故選：D7．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【解析】由題意知SKIPIF1<0的定義域?yàn)镾KIPIF1<0．令SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上遞增在SKIPIF1<0上遞減．又SKIPIF1<0在其定義域上遞減．故由復(fù)合函數(shù)的單調(diào)性知原函數(shù)的遞增區(qū)間是SKIPIF1<0，故選：B8．函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0和SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0和SKIPIF1<0【解析】SKIPIF1<0，則由二次函數(shù)的性質(zhì)知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，故SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0和SKIPIF1<0.故選：B二、填空題9．函數(shù)SKIPIF1<0的單調(diào)增區(qū)間為_(kāi)__________.【解析】由SKIPIF1<0得，函數(shù)的定義域是R，設(shè)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上是減函數(shù)，在SKIPIF1<0上是增函數(shù)，∵SKIPIF1<0在定義域上減函數(shù)，∴函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是SKIPIF1<010．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是________【解析】SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，而SKIPIF1<0也單調(diào)遞減，所以SKIPIF1<0單調(diào)遞增，故答案為：SKIPIF1<011．若函數(shù)SKIPIF1<0是偶函數(shù)，則SKIPIF1<0的單調(diào)遞增區(qū)間是__________．【解析】∵函數(shù)SKIPIF1<0是偶函數(shù)，∴SKIPIF1<0，∴SKIPIF1<0，化為SKIPIF1<0，此式對(duì)于任意實(shí)數(shù)SKIPIF1<0都成立，SKIPIF1<0．SKIPIF1<0，∴函數(shù)SKIPIF1<0的遞增區(qū)間是SKIPIF1<0．12．函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是________．【解析】由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)的單調(diào)遞減區(qū)間為SKIPIF1<013．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為_(kāi)_________.【解析】設(shè)SKIPIF1<0，則SKIPIF1<0，對(duì)稱(chēng)軸為SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0為減函數(shù)，函數(shù)SKIPIF1<0為增函數(shù)，則SKIPIF1<0為減函數(shù)，即函數(shù)單調(diào)減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0為減函數(shù)，函數(shù)SKIPIF1<0為減函數(shù)，則SKIPIF1<0為增函數(shù)，即函數(shù)單調(diào)增區(qū)間為SKIPIF1<0.三、雙空題14．函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是_______；函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是_______【解析】（1）SKIPIF1<0是開(kāi)口向上，對(duì)稱(chēng)軸為SKIPIF1<0的拋物線(xiàn)，故SKIPIF1<0是其遞增區(qū)間；（2）SKIPIF1<0，記SKIPIF1<0，則由（1）知，SKIPIF1<0時(shí)，SKIPIF1<0關(guān)于SKIPIF1<0遞增，根據(jù)指數(shù)函數(shù)性質(zhì)，SKIPIF1<0顯然是SKIPIF1<0關(guān)于SKIPIF1<0遞增，根據(jù)復(fù)合函數(shù)“同增異減”的原則，SKIPIF1<0的單調(diào)增區(qū)間是SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0四、解答題15．已知函數(shù)SKIPIF1<0為定義在SKIPIF1<0上的偶函數(shù)，其部分圖象如圖所示.(1)請(qǐng)作出函數(shù)SKIPIF1<0在SKIPIF1<0上的圖象；(2)根據(jù)函數(shù)圖象寫(xiě)出函數(shù)SKIPIF1<0的單調(diào)區(qū)間及最值.【解析】（1）畫(huà)圖如圖：（2）根據(jù)函數(shù)圖象，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的最大值為2，SKIPIF1<0的最小值為-2.16．已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，求a的取值范圍．【解析】（1）根據(jù)題意，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，故SKIPIF1<0的定義域?yàn)镾KIPIF1<0,令SKIPIF1<0，則該函數(shù)在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，因?yàn)楹瘮?shù)SKIPIF1<0為減函數(shù)，所以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；（2）令函數(shù)SKIPIF1<0，該函數(shù)在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．①當(dāng)SKIPIF1<0時(shí)，要使SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0恒成立，故SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，要使SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0恒成立，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，故函數(shù)SKIPIF1<0在SKIPIF1<0上不能單調(diào)遞增，此種情況不可能；綜上，SKIPIF1<0的取值范圍為SKIPIF1<0．考點(diǎn)三圖像與單調(diào)性一、單選題1．已知函數(shù)SKIPIF1<0的圖象如圖所示，則下列說(shuō)法錯(cuò)誤的是（

）

A．SKIPIF1<0是函數(shù)SKIPIF1<0的增區(qū)間 B．SKIPIF1<0是函數(shù)SKIPIF1<0的減區(qū)間C．函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù) D．函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)【解析】根據(jù)函數(shù)圖像可知函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，故A，B正確；函數(shù)SKIPIF1<0在SKIPIF1<0上也單調(diào)遞增，但區(qū)間SKIPIF1<0和SKIPIF1<0不是連續(xù)區(qū)間，并且由圖象可知SKIPIF1<0，因此不能說(shuō)函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，C錯(cuò)誤；由于函數(shù)SKIPIF1<0在SKIPIF1<0時(shí)有定義，由圖象可知SKIPIF1<0，則SKIPIF1<0為函數(shù)的一個(gè)單調(diào)遞減區(qū)間，故函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，D正確，故選：C二、多選題2．函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，SKIPIF1<0在SKIPIF1<0上的圖象如圖所示，則函數(shù)SKIPIF1<0的增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】由圖象，可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.因?yàn)楹瘮?shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，所以函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以函數(shù)SKIPIF1<0的增區(qū)間是SKIPIF1<0和SKIPIF1<0.故選：BC.3．奇函數(shù)SKIPIF1<0在SKIPIF1<0的圖像如圖所示，則下列結(jié)論正確的有（

）A．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0B．函數(shù)SKIPIF1<0在SKIPIF1<0上遞減C．SKIPIF1<0D．函數(shù)SKIPIF1<0在SKIPIF1<0上遞增【解析】根據(jù)圖像可知：SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0遞減，在SKIPIF1<0上遞增，所以根據(jù)奇函數(shù)性質(zhì)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，A正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0遞減，在SKIPIF1<0上遞增，故BD正確.由于SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0，故C錯(cuò)誤.故選：ABD4．設(shè)SKIPIF1<0是定義域?yàn)镾KIPIF1<0的偶函數(shù)，其導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0時(shí)，SKIPIF1<0圖像如圖所示，則可以使SKIPIF1<0成立的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】由SKIPIF1<0時(shí)，SKIPIF1<0圖像可得當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞減，SKIPIF1<0，又SKIPIF1<0是定義域?yàn)镾KIPIF1<0的偶函數(shù)，所以SKIPIF1<0所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞減，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0單調(diào)遞增，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故選：ABD.三、解答題5．已知函數(shù)SKIPIF1<0.(1)在下面的平面直角坐標(biāo)系中，作出函數(shù)SKIPIF1<0的圖象，并寫(xiě)出單調(diào)增區(qū)間；(2)方程SKIPIF1<0有四個(gè)不相等的實(shí)數(shù)根，求實(shí)數(shù)SKIPIF1<0的取值范圍.【解析】（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以，SKIPIF1<0.作出函數(shù)SKIPIF1<0的圖象如下圖由圖像可知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.（2）如圖2，作出函數(shù)SKIPIF1<0與直線(xiàn)SKIPIF1<0的圖象.由圖2知，當(dāng)SKIPIF1<0時(shí)，直線(xiàn)SKIPIF1<0與SKIPIF1<0有4個(gè)交點(diǎn)，即方程SKIPIF1<0有四個(gè)不相等的實(shí)數(shù)根，所以，SKIPIF1<0.6．給定函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.(1)在同一直角坐標(biāo)系中畫(huà)出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象；(2)SKIPIF1<0，用SKIPIF1<0表示SKIPIF1<0，SKIPIF1<0中的最大者，記為SKIPIF1<0，試判斷SKIPIF1<0在區(qū)間SKIPIF1<0的單調(diào)性.【解析】（1）SKIPIF1<0，SKIPIF1<0圖象如圖所示，（2）由（1）及SKIPIF1<0的定義得，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，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，在SKIPIF1<0單調(diào)遞減.7．已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.現(xiàn)已畫(huà)出函數(shù)SKIPIF1<0在SKIPIF1<0軸左側(cè)的圖像，如圖所示.(1)畫(huà)出函數(shù)SKIPIF1<0在y軸右側(cè)的圖像，并寫(xiě)出函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)增區(qū)間；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的解析式.(3)結(jié)合圖像分別直接寫(xiě)出：當(dāng)m為何值時(shí)，關(guān)于x的方程SKIPIF1<0有4個(gè)實(shí)根？【解析】（1）因?yàn)楹瘮?shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，所以函數(shù)的圖像關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，即只需把函數(shù)SKIPIF1<0在SKIPIF1<0軸左側(cè)翻折到SKIPIF1<0軸右側(cè)就可以得到函數(shù)SKIPIF1<0在SKIPIF1<0軸右側(cè)的圖像了.圖像如下圖所示：所以，函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)增區(qū)間為SKIPIF1<0和SKIPIF1<0.（2）解：因?yàn)楹瘮?shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0時(shí)，SKIPIF1<0所以當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.所以，SKIPIF1<0.（3）關(guān)于SKIPIF1<0的方程SKIPIF1<0有幾個(gè)實(shí)根等價(jià)于函數(shù)SKIPIF1<0的圖像與直線(xiàn)SKIPIF1<0有幾個(gè)交點(diǎn).如圖所示，當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的圖像與直線(xiàn)SKIPIF1<0有4個(gè)交點(diǎn)，所以，當(dāng)SKIPIF1<0時(shí)，關(guān)于SKIPIF1<0的方程SKIPIF1<0有4個(gè)實(shí)根.考點(diǎn)四根據(jù)單調(diào)性比較大小一、單選題1．已知函數(shù)SKIPIF1<0是區(qū)間SKIPIF1<0內(nèi)的減函數(shù)，則SKIPIF1<0與SKIPIF1<0的大小關(guān)系為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．不確定【解析】因?yàn)镾KIPIF1<0，又SKIPIF1<0是區(qū)間SKIPIF1<0內(nèi)的減函數(shù)，所以SKIPIF1<0.故選：B．2．設(shè)函數(shù)SKIPIF1<0滿(mǎn)足：對(duì)任意的SKIPIF1<0都有SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】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)在實(shí)數(shù)SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0.故選：A3．若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】由題意可得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0，故選：D4．已知SKIPIF1<0是偶函數(shù)，SKIPIF1<0在SKIPIF1<0上是增函數(shù)，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為：（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】因?yàn)镾KIPIF1<0是偶函數(shù)，所以SKIPIF1<0，SKIPIF1<0.因?yàn)镾KIPIF1<0在SKIPIF1<0上是增函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0.故選;D.5．已知SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】由題意，SKIPIF1<0，在SKIPIF1<0中，函數(shù)單調(diào)遞增，且SKIPIF1<0，∴SKIPIF1<0，在SKIPIF1<0中，函數(shù)單調(diào)遞增，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，故選：A.6．已知函數(shù)SKIPIF1<0，設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)，SKIPIF1<0，又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，由SKIPIF1<0以及SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0．故選:D．7．已知函數(shù)SKIPIF1<0在SKIPIF1<0上是遞減函數(shù)，SKIPIF1<0且SKIPIF1<0，則有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】SKIPIF1<0是減函數(shù)，SKIPIF1<0，SKIPIF1<0；故選：D.8．已知函數(shù)SKIPIF1<0．記SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】令SKIPIF1<0，則SKIPIF1<0開(kāi)口向下，對(duì)稱(chēng)軸為SKIPIF1<0，因?yàn)镾KIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0由二次函數(shù)性質(zhì)知SKIPIF1<0，因?yàn)镾KIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，綜上，SKIPIF1<0，又SKIPIF1<0為增函數(shù)，故SKIPIF1<0，即SKIPIF1<0.故選：A.9．設(shè)SKIPIF1<0為定義在SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0在SKIPIF1<0上為增函數(shù)，則SKIPIF1<0的大小順序?yàn)椋?/p>

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】因?yàn)镾KIPIF1<0為定義在SKIPIF1<0上的偶函數(shù)，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0在SKIPIF1<0上為增函數(shù)，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：B.10．已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，且SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0是偶函數(shù)，SKIPIF1<0是奇函數(shù)，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，對(duì)于A，SKIPIF1<0，但無(wú)法判斷SKIPIF1<0的正負(fù)，故A不正確；對(duì)于B，SKIPIF1<0，但無(wú)法判斷SKIPIF1<0的正負(fù)，故B不正確；對(duì)于C，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，故C不正確；對(duì)于D，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，故D正確.故選：D.11．已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，若對(duì)SKIPIF1<0都有SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0與SKIPIF1<0的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【解析】因?yàn)閷?duì)SKIPIF1<0都有SKIPIF1<0，所以SKIPIF1<0又因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，且SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0．故選：A．12．設(shè)SKIPIF1<0，則SKIPIF1<0大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【解析】設(shè)函數(shù)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，故SKIPIF1<0在SKIPIF1<0上都單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIP