新高考數(shù)學(xué)一輪復(fù)習(xí)過關(guān)訓(xùn)練第17課 導(dǎo)數(shù)與函數(shù)的單調(diào)性（含解析）

試卷第=page11頁，共=sectionpages33頁試卷第=page11頁，共=sectionpages33頁第17課導(dǎo)數(shù)與函數(shù)的單調(diào)性學(xué)校:___________姓名：___________班級：___________考號：___________【基礎(chǔ)鞏固】1．（2022·吉林吉林·模擬預(yù)測）若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則實數(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﹒2．（2022·全國·哈師大附中模擬預(yù)測）已知SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的導(dǎo)函數(shù)，則SKIPIF1<0的圖像大致是（

）A． B．C． D．【答案】B【解析】SKIPIF1<0，SKIPIF1<0為奇函數(shù)，則函數(shù)SKIPIF1<0的圖像關(guān)于原點對稱，排除選項A、D，令SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0遞減，故選B．3．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實數(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，x＝－1不滿足條件當(dāng)SKIPIF1<0時，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0.故選：D．4．（2022·湖北·房縣第一中學(xué)模擬預(yù)測）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】解：因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0等價于SKIPIF1<0，解得SKIPIF1<0，即原不等式的解集為SKIPIF1<0.故選：B.5．（2022·北京·首都師范大學(xué)附屬中學(xué)三模）下列函數(shù)中，既是偶函數(shù)又在SKIPIF1<0上單調(diào)遞減的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】對于A，函數(shù)SKIPIF1<0的定義域為R，關(guān)于原點對稱，且SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，故A不符合題意；對于B，函數(shù)SKIPIF1<0的定義域為R，關(guān)于原點對稱，且SKIPIF1<0，所以函數(shù)SKIPIF1<0為奇函數(shù)，由冪函數(shù)的性質(zhì)知函數(shù)SKIPIF1<0在R上單調(diào)遞增，所以函數(shù)SKIPIF1<0在R上單調(diào)遞減，故B不符合題意；對于C，函數(shù)SKIPIF1<0的定義域為R，關(guān)于原點對稱，且SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時SKIPIF1<0，又SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故C符合題意；對于D，函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，關(guān)于原點對稱，且SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，又SKIPIF1<0，令SKIPIF1<0，令SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故D不符合題意.故選：C.6．（2022·全國·高三專題練習(xí)）函數(shù)SKIPIF1<0的一個單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，則減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】B【解析】函數(shù)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，函數(shù)SKIPIF1<0在其定義域內(nèi)是遞增．當(dāng)SKIPIF1<0時，令SKIPIF1<0，解得：SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0是遞增．SKIPIF1<0函數(shù)SKIPIF1<0的一個單調(diào)遞增區(qū)間為SKIPIF1<0，故得：SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0在SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0是遞減．故選：B．7．（2022·山東·德州市教育科學(xué)研究院二模）已知函數(shù)SKIPIF1<0是偶函數(shù)，其導(dǎo)函數(shù)SKIPIF1<0的圖象見下圖，且SKIPIF1<0對SKIPIF1<0恒成立，則下列說法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0又由導(dǎo)函數(shù)的圖象得，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，SKIPIF1<0SKIPIF1<0故選：D.8．（2022·山東·煙臺二中模擬預(yù)測）已知SKIPIF1<0，p：SKIPIF1<0；q：函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則p是q的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】由SKIPIF1<0可得SKIPIF1<0，又SKIPIF1<0，又SKIPIF1<0，要使函數(shù)在區(qū)間SKIPIF1<0上不單調(diào)，有SKIPIF1<0，解得SKIPIF1<0，顯然SKIPIF1<0，即p是q的充分不必要條件.故選：A.9．（2022·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，若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<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因為SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0的遞增區(qū)間為SKIPIF1<0．由于SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.因此，實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故選：A.10．（多選）（2022·全國·高三專題練習(xí)）對于函數(shù)SKIPIF1<0，下列結(jié)論中正確的是（

）A．SKIPIF1<0在(0，＋∞)上單調(diào)遞增 B．SKIPIF1<0在SKIPIF1<0上單調(diào)遞減C．SKIPIF1<0有最小值 D．SKIPIF1<0有兩個零點【答案】BC【解析】∵SKIPIF1<0，∴SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，所以當(dāng)SKIPIF1<0時，函數(shù)有最小值SKIPIF1<0，即SKIPIF1<0，所以A錯誤，BC正確，D錯誤.故選：BC.11．（多選）（2022·重慶八中高三階段練習(xí)）函數(shù)SKIPIF1<0均是定義在R上的單調(diào)遞增函數(shù)，且SKIPIF1<0，則下列各函數(shù)一定在R上單調(diào)遞增的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【解析】取SKIPIF1<0，故SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，在SKIPIF1<0上，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為減函數(shù)，故A錯誤.而SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，在SKIPIF1<0上，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為減函數(shù)，故D錯誤.設(shè)SKIPIF1<0，SKIPIF1<0，任意SKIPIF1<0，則SKIPIF1<0，因為SKIPIF1<0均是定義在R上的單調(diào)遞增函數(shù)，故SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，故SKIPIF1<0是R上的單調(diào)遞增函數(shù).而SKIPIF1<0因為SKIPIF1<0是定義在R上的單調(diào)遞增函數(shù)，故SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，故SKIPIF1<0是R上的單調(diào)遞增函數(shù).故BC正確.故選：BC12．（多選）（2022·山東·青島二中高三期末）記SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若SKIPIF1<0對任意的正數(shù)都成立，則下列不等式中成立的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【解析】解：因為SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故A錯誤；同理SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，故B正確；因為SKIPIF1<0，所以SKIPIF1<0，構(gòu)造函數(shù)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，化簡得SKIPIF1<0，故C正確；同理SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，化簡得SKIPIF1<0，故D錯誤.故選：BC.13．（2022·湖北·模擬預(yù)測）已知定義域為R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<0．14．（2022·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0有三個單調(diào)區(qū)間，則實數(shù)a的取值范圍是________．【答案】SKIPIF1<0【解析】SKIPIF1<0，由于函數(shù)SKIPIF1<0有三個單調(diào)區(qū)間，所以SKIPIF1<0有兩個不相等的實數(shù)根，所以SKIPIF1<0.故答案為：SKIPIF1<015．（2022·北京·二模）已知奇函數(shù)SKIPIF1<0的定義域為R，且SKIPIF1<0，則SKIPIF1<0的單調(diào)遞減區(qū)間為__________；滿足以上條件的一個函數(shù)是__________．【答案】

SKIPIF1<0

SKIPIF1<0（答案不唯一）【解析】由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，所以當(dāng)SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，所以滿足條件的一個函數(shù)可以為SKIPIF1<0（答案不唯一）故答案為：SKIPIF1<0，SKIPIF1<0（答案不唯一）16．（2022·河北·高三階段練習(xí)）若函數(shù)SKIPIF1<0在SKIPIF1<0上存在單調(diào)遞減區(qū)間，則m的取值范圍是_________．【答案】SKIPIF1<0【解析】SKIPIF1<0，則原向題等價于SKIPIF1<0在SKIPIF1<0上有解，即SKIPIF1<0在SKIPIF1<0上有解，即SKIPIF1<0在SKIPIF1<0上有解，因為SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<017．（2022·山東師范大學(xué)附中高三期中）設(shè)函數(shù)SKIPIF1<0(1)當(dāng)SKIPIF1<0時，求SKIPIF1<0的單調(diào)區(qū)間；(2)任意正實數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時，試判斷SKIPIF1<0與SKIPIF1<0的大小關(guān)系并證明【解】(1)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0令SKIPIF1<0得SKIPIF1<0；令SKIPIF1<0得SKIPIF1<0或SKIPIF1<0故SKIPIF1<0的單增區(qū)間為SKIPIF1<0，單減區(qū)間為SKIPIF1<0，SKIPIF1<0(2)結(jié)論：SKIPIF1<0，證明如下：SKIPIF1<0SKIPIF1<0設(shè)SKIPIF1<0，由SKIPIF1<0均為正數(shù)且SKIPIF1<0得SKIPIF1<0設(shè)SKIPIF1<0，則SKIPIF1<0①當(dāng)SKIPIF1<0時，由SKIPIF1<0得SKIPIF1<0即SKIPIF1<0故SKIPIF1<0單調(diào)遞減，從而SKIPIF1<0而SKIPIF1<0，此時SKIPIF1<0成立②當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增故SKIPIF1<0的最小值為SKIPIF1<0此時只需證SKIPIF1<0，化簡后即證SKIPIF1<0設(shè)SKIPIF1<0，SKIPIF1<0故SKIPIF1<0單調(diào)遞增，從而有SKIPIF1<0，即證SKIPIF1<0綜上：不等式得證．18．（2022·北京·高考真題）已知函數(shù)SKIPIF1<0．(1)求曲線SKIPIF1<0在點SKIPIF1<0處的切線方程；(2)設(shè)SKIPIF1<0，討論函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性；(3)證明：對任意的SKIPIF1<0，有SKIPIF1<0．【解】(1)解：因為SKIPIF1<0，所以SKIPIF1<0，即切點坐標(biāo)為SKIPIF1<0，又SKIPIF1<0，∴切線斜率SKIPIF1<0∴切線方程為：SKIPIF1<0(2)解：因為SKIPIF1<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)解：原不等式等價于SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，即證SKIPIF1<0，∵SKIPIF1<0，SKIPIF1<0，由（2）知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0，∴SKIPIF1<0∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因為SKIPIF1<0，∴SKIPIF1<0，所以命題得證.【素養(yǎng)提升】1．（2022·全國·高考真題）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，因為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時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，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，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．（多選）（2022·福建泉州·模擬預(yù)測）若SKIPIF1<0，則下列式子可能成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【解析】令SKIPIF1<0，SKIPIF1<0則SKIPIF1<0恒成立，所以SKIPIF1<0單調(diào)遞增，其中SKIPIF1<0，SKIPIF1<0，則存在SKIPIF1<0，使得SKIPIF1<0①當(dāng)SKIPIF1<0時，SKIPIF1<0即SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，不滿足SKIPIF1<0，故SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0又因為SKIPIF1<0，所以SKIPIF1<0，D正確；②當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0（1）當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0成立，故SKIPIF1<0，B正確；（2）當(dāng)SKIPIF1<0時，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，因為SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又因為SKIPIF1<0，所以SKIPIF1<0，選項C正確.故選：BCD3．（2022·湖南·長沙一中模擬預(yù)測）已知正實數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的最大值為______．【答案】SKIPIF1<0【解析】由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，設(shè)SKIPIF1<0（SKIPIF1<0），則SKIPIF1<0SKIPIF1<0，SKIPIF1<0遞增，所以由SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0遞增，SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0遞減，所以SKIPIF1<0．故答案為：SKIPIF1<0．4．（2022·江蘇江蘇·三模）設(shè)函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時，討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，求SKIPIF1<0.【解】(1)解：因為SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0所以當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又由SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0