新高考數(shù)學(xué)二輪復(fù)習(xí)考點(diǎn)歸納與演練專題3-2 利用導(dǎo)數(shù)解決單調(diào)性中求參數(shù)問題（選填）（含解析）

專題3-2利用導(dǎo)數(shù)解決單調(diào)性中求參數(shù)問題（選填）TOC\o"1-1"\h\u 1題型一：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào) 1題型二：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)區(qū)間 6題型三：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào) 8題型四：已知函數(shù)SKIPIF1<0的單調(diào)區(qū)間恰為SKIPIF1<0 11題型五：已知函數(shù)SKIPIF1<0有三個單調(diào)區(qū)間 13 16題型一：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)【典型例題】例題1．（2022·重慶市第七中學(xué)校高二階段練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0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【詳解】由題意得，SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，因為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，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0.故選：A例題2．（2022·全國·高二課時練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞減，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由SKIPIF1<0得SKIPIF1<0，由于函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞減，即SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0，即得SKIPIF1<0在SKIPIF1<0恒成立，所以SKIPIF1<0，故選：D.例題3．（2022·陜西咸陽中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，若對SKIPIF1<0，SKIPIF1<0，都有SKIPIF1<0成立，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因為對SKIPIF1<0，SKIPIF1<0，都有SKIPIF1<0成立，所以對SKIPIF1<0，SKIPIF1<0，都有SKIPIF1<0.設(shè)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0為減函數(shù).SKIPIF1<0，等價于SKIPIF1<0，SKIPIF1<0恒成立，即SKIPIF1<0，SKIPIF1<0恒成立.設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為減函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為增函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.故選：C【提分秘籍】已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)①已知SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增SKIPIF1<0SKIPIF1<0，SKIPIF1<0恒成立.②已知SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減SKIPIF1<0SKIPIF1<0，SKIPIF1<0恒成立.注：已知單調(diào)性，等價條件中的不等式含等號.【變式演練】1．（2021·四川·宜賓市敘州區(qū)第一中學(xué)校高二階段練習(xí)（文））若SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由題意可得：當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0.因為SKIPIF1<0和SKIPIF1<0在SKIPIF1<0上單增，所以SKIPIF1<0在SKIPIF1<0上單增，所以SKIPIF1<0，所以SKIPIF1<0.故選：D2．（2022·全國·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，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】B【詳解】解：SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0時恒成立，即SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0SKIPIF1<0在SKIPIF1<0單調(diào)遞減，故SKIPIF1<0，故SKIPIF1<0．故選：B．3．（2022·陜西省寶雞市長嶺中學(xué)高二期中（理））若函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0上是增函數(shù)，所以SKIPIF1<0對SKIPIF1<0恒成立，則SKIPIF1<0對SKIPIF1<0恒成立，所以SKIPIF1<0對SKIPIF1<0恒成立，則SKIPIF1<0，即SKIPIF1<0.故選：A.4．（2022·山西臨汾·高三期中）設(shè)函數(shù)SKIPIF1<0，若對任意SKIPIF1<0，SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題設(shè)SKIPIF1<0，且SKIPIF1<0，令SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上遞減，所以SKIPIF1<0恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，而SKIPIF1<0在SKIPIF1<0上值域為SKIPIF1<0，所以SKIPIF1<0.故選：A題型二：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)區(qū)間【典型例題】例題1．（2022·江西·上高二中高二階段練習(xí)（文））若函數(shù)SKIPIF1<0存在單調(diào)遞減區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，且其導(dǎo)數(shù)為SKIPIF1<0．由SKIPIF1<0存在單調(diào)遞減區(qū)間知SKIPIF1<0在SKIPIF1<0上有解，即SKIPIF1<0有解．因為函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以SKIPIF1<0．要使SKIPIF1<0有解，只需要SKIPIF1<0的最小值小于SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故選:B．例題2．（2022·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0存在單調(diào)遞增區(qū)間，則SKIPIF1<0的取值范圍是(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0，∴SKIPIF1<0在x∈SKIPIF1<0上有解，即ax+SKIPIF1<00在x∈SKIPIF1<0上有解，即aSKIPIF1<0在x∈SKIPIF1<0上有解．令g(x)SKIPIF1<0，則g′(x)SKIPIF1<0，∴g(x)SKIPIF1<0在(0，e)上單調(diào)遞減，在(e，+∞)上單調(diào)遞增，∴g(x)SKIPIF1<0的最小值為g(e)=SKIPIF1<0，∴a＞SKIPIF1<0．故選：B．【提分秘籍】已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)區(qū)間①已知SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)增區(qū)間SKIPIF1<0SKIPIF1<0，SKIPIF1<0有解.②已知SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)減區(qū)間SKIPIF1<0SKIPIF1<0，SKIPIF1<0有解.【變式演練】1．（2022·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0在SKIPIF1<0存在單調(diào)遞減區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因為函數(shù)SKIPIF1<0在SKIPIF1<0存在單調(diào)遞減區(qū)間，故SKIPIF1<0在區(qū)間SKIPIF1<0上有解.即SKIPIF1<0在區(qū)間SKIPIF1<0有解.即存在SKIPIF1<0，使得SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增.且SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0時SKIPIF1<0；SKIPIF1<0時，SKIPIF1<0，故要滿足題意，只需SKIPIF1<0即可，解得SKIPIF1<0.故選：SKIPIF1<0.2．（2022·福建·福州黎明中學(xué)高三階段練習(xí)）若函數(shù)f(x)＝x2－4ex－ax在R上存在單調(diào)遞增區(qū)間，則實數(shù)a的取值范圍為__________．【答案】SKIPIF1<0【詳解】因為f(x)＝x2－4ex－ax，所以f′(x)＝2x－4ex－a.由題意，f′(x)＝2x－4ex－a＞0，即a＜2x－4ex有解．令g(x)＝2x－4ex，則g′(x)＝2－4ex.令g′(x)＝0，解得x＝－ln2.當(dāng)x∈(－∞，－ln2)時，函數(shù)g(x)＝2x－4ex單調(diào)遞增；當(dāng)x∈(－ln2，＋∞)時，函數(shù)g(x)＝2x－4ex單調(diào)遞減．所以當(dāng)x＝－ln2時，g(x)＝2x－4ex取得最大值－2－2ln2，所以a＜－2－2ln2.題型三：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)【典型例題】例題1．（2022·陜西·蒲城縣蒲城中學(xué)高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】依題意SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上有零點(diǎn)，令SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0單調(diào)遞增，又由SKIPIF1<0，得SKIPIF1<0，故SKIPIF1<0，所以，SKIPIF1<0的取值范圍SKIPIF1<0故選：A例題2．（2022·廣西河池·高二階段練習(xí)（理））若函數(shù)SKIPIF1<0在定義域內(nèi)的一個子區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由題意得，函數(shù)定義域為SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，解得在定義域內(nèi)SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，函數(shù)在區(qū)間SKIPIF1<0內(nèi)不單調(diào)，所以SKIPIF1<0，解得SKIPIF1<0，又因為SKIPIF1<0，得SKIPIF1<0，綜上SKIPIF1<0，故選：D．【提分秘籍】已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)SKIPIF1<0SKIPIF1<0，使得SKIPIF1<0（其中SKIPIF1<0為變號零點(diǎn)）【變式演練】1．（2022·安徽·合肥一中高二階段練習(xí)）若函數(shù)SKIPIF1<0在其定義域上不單調(diào)，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0，因為函數(shù)SKIPIF1<0在其定義域上不單調(diào)，即SKIPIF1<0有變號零點(diǎn)，結(jié)合二次函數(shù)的性質(zhì)，可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：A.2．（2022·四川省資陽中學(xué)高二期中（理））已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由SKIPIF1<0，①當(dāng)SKIPIF1<0時函數(shù)SKIPIF1<0單調(diào)遞增，不合題意；②當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的極值點(diǎn)為SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0不單調(diào)，必有SKIPIF1<0，解得SKIPIF1<0.故選：B．3．（2022·江西·金溪一中高二階段練習(xí)（理））已知函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)不是單調(diào)函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】∵SKIPIF1<0，SKIPIF1<0在SKIPIF1<0內(nèi)不是單調(diào)函數(shù)，故SKIPIF1<0在SKIPIF1<0存在變號零點(diǎn)，即SKIPIF1<0在SKIPIF1<0存在零點(diǎn)，∴SKIPIF1<0.故選：A.4．（2022·上海大學(xué)市北附屬中學(xué)高一期中）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實數(shù)SKIPIF1<0的取值范圍________.【答案】SKIPIF1<0【詳解】解：因為SKIPIF1<0，所以函數(shù)的對稱軸為SKIPIF1<0，因為函數(shù)在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，即實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0題型四：已知函數(shù)SKIPIF1<0的單調(diào)區(qū)間恰為SKIPIF1<0【典型例題】例題1．（2021·四川省成都市玉林中學(xué)高二期中（文））已知函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，則函數(shù)SKIPIF1<0在SKIPIF1<0的值域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：SKIPIF1<0，∵SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0，SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，SKIPIF1<0符合題意，又SKIPIF1<0，SKIPIF1<0（1）SKIPIF1<0，SKIPIF1<0（2）SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0的值域是SKIPIF1<0，SKIPIF1<0．故選：A．例題2．（2022·全國·高二課時練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0、SKIPIF1<0上為增函數(shù)，在SKIPIF1<0上為減函數(shù)，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】因為SKIPIF1<0，則SKIPIF1<0，由題意可知，SKIPIF1<0有兩個不等的零點(diǎn)，設(shè)為SKIPIF1<0、SKIPIF1<0且SKIPIF1<0，因為函數(shù)SKIPIF1<0在SKIPIF1<0、SKIPIF1<0上為增函數(shù)，在SKIPIF1<0上為減函數(shù)，則SKIPIF1<0、SKIPIF1<0，所以，SKIPIF1<0，解得SKIPIF1<0.故選：B.【變式演練】1．（2022·全國·高二課時練習(xí)）已知函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】解：由題可得SKIPIF1<0，則SKIPIF1<0的解集為SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，∴SKIPIF1<0，故選:C．2．（2022·福建漳州·高二期末）已知函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，則關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集是__________.【答案】SKIPIF1<0【詳解】SKIPIF1<0，SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，則不等式SKIPIF1<0的解集為SKIPIF1<0，所以SKIPIF1<0是SKIPIF1<0的兩根，故SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0；所以不等式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0題型五：已知函數(shù)SKIPIF1<0有三個單調(diào)區(qū)間【典型例題】例題1．（2019·河北省隆化存瑞中學(xué)高三階段練習(xí)（理））若函數(shù)SKIPIF1<0恰好有三個單調(diào)區(qū)間，則實數(shù)SKIPIF1<0的取值范圍為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【詳解】因為函數(shù)SKIPIF1<0恰好有三個單調(diào)區(qū)間，所以SKIPIF1<0有兩個不等零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故選D.例題2．（2019·江蘇鹽城·一模）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0存在三個單調(diào)區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【詳解】SKIPIF1<0函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0存在三個單調(diào)區(qū)間即SKIPIF1<00有兩個不等實根，即SKIPIF1<0有兩個不等實根，轉(zhuǎn)化為y=a與y=SKIPIF1<0的圖像有兩個不同的交點(diǎn)SKIPIF1<0令SKIPIF1<0，即x=SKIPIF1<0，即y=SKIPIF1<0在（0，SKIPIF1<0）上單調(diào)遞減，在（SKIPIF1<0，+∞）上單調(diào)遞增．ymin=-SKIPIF1<0，當(dāng)x∈（0，SKIPIF1<0）時，y<0，所以a的范圍為SKIPIF1<0【提分秘籍】已知函數(shù)SKIPIF1<0有三個單調(diào)區(qū)間SKIPIF1<0SKIPIF1<0有兩個不同的實數(shù)根.【變式演練】1．（2022·寧夏·永寧縣文昌中學(xué)高三期末（文））若函數(shù)SKIPIF1<0有三個單調(diào)區(qū)間，則SKIPIF1<0的取值范圍是________________．【答案】b>0【詳解】試題分析：由已知可得SKIPIF1<0在SKIPIF1<0上有不等實根SKIPIF1<0．2．（2022·江西省信豐中學(xué)高二階段練習(xí)（文））若函數(shù)SKIPIF1<0在定義域SKIPIF1<0上恰有三個單調(diào)區(qū)間，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因為函數(shù)SKIPIF1<0在定義域SKIPIF1<0上恰有三個單調(diào)區(qū)間，所以其導(dǎo)函數(shù)在定義域SKIPIF1<0上有兩個不同的零點(diǎn)，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，所以只需SKIPIF1<0，方程SKIPIF1<0在SKIPIF1<0上有兩個不同的實數(shù)根.故選：A.3．（2016·黑龍江雙鴨山·高二階段練習(xí)）若函數(shù)恰有三個單調(diào)區(qū)間，則實數(shù)的取值范圍為A． B．C． D．【答案】D【詳解】試題分析：由題意得，函數(shù)的導(dǎo)數(shù)為，因為函數(shù)恰有三個單調(diào)區(qū)間，所以且有兩個根，即，解得且，故選D.4．（2020·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0恰有三個單調(diào)區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是__________．【答案】SKIPIF1<0或SKIPIF1<0【詳解】分析：求出函數(shù)的導(dǎo)函數(shù)，利用導(dǎo)數(shù)有兩個不同的零點(diǎn)，說明函數(shù)恰好有三個單調(diào)區(qū)間，從而求出a的取值范圍．詳解：∵函數(shù)SKIPIF1<0，∴f′（x）=3x2+6ax+SKIPIF1<0，由函數(shù)f（x）恰好有三個單調(diào)區(qū)間，得f′（x）有兩個不相等的零點(diǎn)，∴3x2+6ax+SKIPIF1<0=0滿足：△=SKIPIF1<0﹣SKIPIF1<0＞0，解得SKIPIF1<0或SKIPIF1<0，故答案為：SKIPIF1<0或SKIPIF1<0．一、單選題1．（2019·四川自貢·高二期末（理））函數(shù)SKIPIF1<0恰有SKIPIF1<0個單調(diào)區(qū)間的必要不充分條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時，由SKIPIF1<0，解得SKIPIF1<0，函數(shù)SKIPIF1<0有兩個單調(diào)區(qū)間；當(dāng)SKIPIF1<0時，由SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，此時函數(shù)SKIPIF1<0恰有3個單調(diào)區(qū)間；當(dāng)SKIPIF1<0時，SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，此時函數(shù)SKIPIF1<0恰有3個單調(diào)區(qū)間．SKIPIF1<0綜上所述SKIPIF1<0是函數(shù)SKIPIF1<0恰有3個單調(diào)區(qū)間的充要條件，分析可得SKIPIF1<0是其必要不充分條件．故選：SKIPIF1<0．2．（2019·河北省隆化存瑞中學(xué)高三階段練習(xí)（理））若函數(shù)SKIPIF1<0恰好有三個單調(diào)區(qū)間，則實數(shù)SKIPIF1<0的取值范圍為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【詳解】因為函數(shù)SKIPIF1<0恰好有三個單調(diào)區(qū)間，所以SKIPIF1<0有兩個不等零點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故選D.3．（2022·河南·駐馬店市第二高級中學(xué)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SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0,因為函數(shù)SKIPIF1<0在區(qū)間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.因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0的取值范圍為SKIPIF1<0.故選：B4．（2021·江蘇·張家港高級中學(xué)高三期中）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【詳解】∵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.5．（2023·全國·高三專題練習(xí)）若函數(shù)SKIPIF1<0在其定義域的一個子區(qū)間SKIPIF1<0內(nèi)不是單調(diào)函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】因為函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0（舍去），因為SKIPIF1<0在定義域的一個子區(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：D6．（2022·福建福州·高三期中）已知函數(shù)SKIPIF1<0，對任意的實數(shù)SKIPIF1<0，且SKIPIF1<0，不等式SKIPIF1<0恒成立，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】B【詳解】不妨設(shè)SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，所以對任意的實數(shù)SKIPIF1<0時，都有SKIPIF1<0，即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．令SKIPIF1<0．則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0，即實數(shù)a的取值范圍是SKIPIF1<0．故選：B．7．（2023·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)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因為SKIPIF1<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，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0．故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．又因為SKIPIF1<0，且當(dāng)SKIPIF1<0時，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：A8．（2022·安徽·合肥一中高二階段練習(xí)）若函數(shù)SKIPIF1<0在其定義域上不單調(diào)，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0，因為函數(shù)SKIPIF1<0在其定義域上不單調(diào)，即SKIPIF1<0有變號零點(diǎn)，結(jié)合二次函數(shù)的性質(zhì)，可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0