新高考數(shù)學(xué)二輪復(fù)習(xí)考點(diǎn)歸納與演練專題3-3 利用導(dǎo)數(shù)解決單調(diào)性含參討論問題（解答題）（含解析）

專題3-3利用導(dǎo)數(shù)解決單調(diào)性含參討論問題（解答題）目錄TOC\o"1-1"\h\u專題3-3利用導(dǎo)數(shù)解決單調(diào)性含參討論問題（解答題） 1 1題型一：導(dǎo)函數(shù)有效部分是一次型 1題型二：導(dǎo)函數(shù)有效部分可視為一次型 4題型三：導(dǎo)函數(shù)有效部分是二次型（可因式分解） 8題型四：導(dǎo)函數(shù)有效部分是可視為二次型（可因式分解） 13題型五：導(dǎo)函數(shù)有效部分是二次型（不可因式分解） 18題型六：借助二階導(dǎo)函數(shù)討論單調(diào)性 20 22導(dǎo)函數(shù)有效部分對(duì)于SKIPIF1<0進(jìn)行求導(dǎo)得到SKIPIF1<0，對(duì)SKIPIF1<0初步處理（如通分），提出SKIPIF1<0的恒正部分，將該部分省略，留下的部分則為SKIPIF1<0的有效部分（如：SKIPIF1<0，則記SKIPIF1<0為SKIPIF1<0的有效部分）.題型一：導(dǎo)函數(shù)有效部分是一次型【典型例題】例題1．（2022·北京八十中模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程；(2)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；【答案】(1)SKIPIF1<0；(2)答案見解析；（1）由題設(shè)，SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0.（2）由SKIPIF1<0且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即SKIPIF1<0在定義域上遞減；當(dāng)SKIPIF1<0時(shí)，在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞減，在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0遞增，綜上，SKIPIF1<0時(shí)SKIPIF1<0遞減；SKIPIF1<0時(shí)SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0上遞增.例題2．（2022·河北滄州·二模）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的單調(diào)區(qū)間；【答案】(1)答案見解析(1)函數(shù)SKIPIF1<0，定義域?yàn)镾KIPIF1<0，（i）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；（ii）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，無單調(diào)遞減區(qū)間；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0.【提分秘籍】在例題1中，SKIPIF1<0，可提取有效部分為SKIPIF1<0，只要討論有效部分SKIPIF1<0的正負(fù)即可；在例題2中SKIPIF1<0，可提取有效部分為SKIPIF1<0，只要討論有效部分SKIPIF1<0的正負(fù)即可.【變式演練】1．（2022·北京延慶·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0.SKIPIF1<0(1)若SKIPIF1<0，求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)求SKIPIF1<0的極值和單調(diào)區(qū)間；【答案】(1)SKIPIF1<0(2)答案見解析(1)當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，SKIPIF1<0．

所以SKIPIF1<0，SKIPIF1<0．

所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程SKIPIF1<0．(2)函數(shù)SKIPIF1<0定義域SKIPIF1<0．

求導(dǎo)得SKIPIF1<0.

①當(dāng)SKIPIF1<0時(shí)，因?yàn)镾KIPIF1<0，所以SKIPIF1<0．故SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，此時(shí)SKIPIF1<0無極值.

②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0變化時(shí)，SKIPIF1<0變化如下表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0極小值SKIPIF1<0所以SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，單調(diào)遞增區(qū)間是SKIPIF1<0．

此時(shí)函數(shù)SKIPIF1<0的極小值是SKIPIF1<0，無極大值．2．（2022·安徽·碭山中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】(1)答案見解析；【詳解】（1）由題意得，SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，故函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故函數(shù)SKIPIF1<0在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)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；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．題型二：導(dǎo)函數(shù)有效部分可視為一次型【典型例題】例題1．（2022·青?！ず|市第一中學(xué)模擬預(yù)測(cè)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為自然對(duì)數(shù)的底數(shù)．(1)討論SKIPIF1<0的單調(diào)性；【答案】(1)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時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減．（1）SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得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．例題2．（2022·全國(guó)·模擬預(yù)測(cè)（文））設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；【答案】(1)答案見解析；(1)SKIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，則SKIPIF1<0在SKIPIF1<0上為減函數(shù)，當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，綜上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，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.【提分秘籍】在例題1中，SKIPIF1<0，可提取有效部分為SKIPIF1<0，可以看作一次型，類似一次型討論方式討論SKIPIF1<0的正負(fù)；在例題2中SKIPIF1<0，可提取有效部分為SKIPIF1<0，可以看作一次型，只要討論有效部分SKIPIF1<0的正負(fù)即可.【變式演練】1．（2022·浙江·鎮(zhèn)海中學(xué)高二期中）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0.(1)討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】(1)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；【詳解】（1）SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，綜上，SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.2．（2022·全國(guó)·高三階段練習(xí)（文））已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)討論函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性．【答案】(1)SKIPIF1<0；(2)見解析（1）SKIPIF1<0時(shí)，SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0（2）由題意，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，得SKIPIF1<0，①當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；②當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；③當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；綜上，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；題型三：導(dǎo)函數(shù)有效部分是二次型（可因式分解）【典型例題】例題1．（2022·四川省遂寧市教育局模擬預(yù)測(cè)（文））已知函數(shù)SKIPIF1<0(1)討論SKIPIF1<0的單調(diào)性；【答案】(1)答案不唯一，具體見解析；【詳解】（1）函數(shù)SKIPIF1<0定義域R，求導(dǎo)得SKIPIF1<0，若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增；若SKIPIF1<0，恒有SKIPIF1<0．即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，所以當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的遞減區(qū)間是SKIPIF1<0，遞增區(qū)間是SKIPIF1<0和SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的遞減區(qū)間是SKIPIF1<0，遞增區(qū)間是SKIPIF1<0和SKIPIF1<0.例題2．（2022·湖北武漢·高二期末）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)若曲線SKIPIF1<0與曲線SKIPIF1<0在它們的交點(diǎn)SKIPIF1<0處具有公共切線，求SKIPIF1<0的值；(2)當(dāng)SKIPIF1<0，且SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間．【答案】(1)SKIPIF1<0，SKIPIF1<0(2)答案見解析(1)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0與SKIPIF1<0在交點(diǎn)SKIPIF1<0處具有公共切線，SKIPIF1<0；又SKIPIF1<0，SKIPIF1<0由SKIPIF1<0得：SKIPIF1<0.(2)當(dāng)SKIPIF1<0時(shí)，設(shè)SKIPIF1<0，SKIPIF1<0；設(shè)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；SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0；①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0時(shí)，SKIPIF1<0恒成立，即SKIPIF1<0，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，無單調(diào)遞增區(qū)間；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0；單調(diào)遞增區(qū)間為SKIPIF1<0；綜上所述：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，無單調(diào)遞增區(qū)間；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，SKIPIF1<0；單調(diào)遞增區(qū)間為SKIPIF1<0.【提分秘籍】討論含參函數(shù)單調(diào)性問題時(shí)，求完導(dǎo)函數(shù)后，如果導(dǎo)函數(shù)是二次型，優(yōu)先考慮是否可以因式分解，如例題1：SKIPIF1<0，在討論正負(fù)的過程中，遵循三個(gè)原則：準(zhǔn)則1：最高項(xiàng)系數(shù)含參，從參數(shù)為0開始討論；準(zhǔn)則2：兩根大小不確定，從兩根相等開始討論；準(zhǔn)則3：判斷兩根是否在定義域內(nèi).如例題1中從SKIPIF1<0開始討論。例題2中求導(dǎo)后SKIPIF1<0，記有效部分為SKIPIF1<0，由于最高項(xiàng)系數(shù)含參數(shù)SKIPIF1<0，討論時(shí)從SKIPIF1<0開始討論，當(dāng)SKIPIF1<0時(shí)，從SKIPIF1<0開始討論.【變式演練】1．（2022·陜西西安·模擬預(yù)測(cè)（文））已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0的極值；(2)討論函數(shù)SKIPIF1<0的單調(diào)性．【答案】(1)極小值為SKIPIF1<0，無極大值(2)答案見解析【詳解】（1）易知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0．令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0．∴函數(shù)SKIPIF1<0的極小值為SKIPIF1<0，無極大值．（2）SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0．②當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0；令SKIPIF1<0．得SKIPIF1<0．③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立．④當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0．綜上，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0，減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0和SKIPIF1<0，減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0和SKIPIF1<0，減區(qū)間為SKIPIF1<0．2．（2022·湖南·高三開學(xué)考試）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0.(1)若直線SKIPIF1<0是曲線SKIPIF1<0的切線，求負(fù)數(shù)SKIPIF1<0的值；(2)設(shè)SKIPIF1<0.（i）討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】(1)SKIPIF1<0(2)（i）答案見解析；（1）因?yàn)镾KIPIF1<0，所以SKIPIF1<0由直線SKIPIF1<0是曲線SKIPIF1<0的切線可知SKIPIF1<0，即SKIPIF1<0又SKIPIF1<0，所以SKIPIF1<0，則切點(diǎn)坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0故SKIPIF1<0.（2）（i）SKIPIF1<0.①若SKIPIF1<0即SKIPIF1<0的解為SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；②若SKIPIF1<0即SKIPIF1<0的解為SKIPIF1<0或SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減③若SKIPIF1<0即SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；④若SKIPIF1<0即SKIPIF1<0的解為SKIPIF1<0或SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減.綜上所述：若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減；若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；若SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減.題型四：導(dǎo)函數(shù)有效部分是可視為二次型（可因式分解）【典型例題】例題1．（2022·重慶·高三階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；【答案】(1)答案見解析【詳解】（1）SKIPIF1<0，當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0或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；當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，SKIPIF1<0或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；綜上可知：SKIPIF1<0時(shí)，故SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；例題2．（2022·天津市寶坻區(qū)第一中學(xué)二模）已知函數(shù)SKIPIF1<0．(1)求SKIPIF1<0的最小值；(2)若SKIPIF1<0，討論SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性；【答案】(1)SKIPIF1<0(2)分類討論，答案見解析.(1)SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，SKIPIF1<0得SKIPIF1<0，SKIPIF1<0單調(diào)遞減．SKIPIF1<0得SKIPIF1<0，SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0.(2)SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)．SKIPIF1<0，有SKIPIF1<0單調(diào)遞減，SKIPIF1<0，有SKIPIF1<0單調(diào)遞增，綜上所述，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．【提分秘籍】討論含參函數(shù)單調(diào)性問題時(shí)，求完導(dǎo)函數(shù)后，如果導(dǎo)函數(shù)是二次型，優(yōu)先考慮是否可以因式分解，如例題1：SKIPIF1<0，在討論正負(fù)的過程中，SKIPIF1<0的正負(fù)，可以看做SKIPIF1<0的正負(fù)等同，故為可視為二次函數(shù)型.解題時(shí)，依然遵循三個(gè)原則：準(zhǔn)則1：最高項(xiàng)系數(shù)含參，從參數(shù)為0開始討論；準(zhǔn)則2：兩根大小不確定，從兩根相等開始討論；準(zhǔn)則3：判斷兩根是否在定義域內(nèi).【變式演練】1．（2022·廣東·潮州市綿德中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值；(2)討論函數(shù)SKIPIF1<0的單調(diào)性．【答案】(1)SKIPIF1<0(2)當(dāng)SKIPIF1<0時(shí)，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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；單調(diào)遞減區(qū)間為SKIPIF1<0.(1)SKIPIF1<0定義域?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，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；∴SKIPIF1<0在在SKIPIF1<0上的最小值為SKIPIF1<0；(2)SKIPIF1<0定義域?yàn)镾KIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，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；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0得SKIPIF1<0或SKIPIF1<0.②當(dāng)SKIPIF1<0時(shí)，即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；③當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0或SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0單調(diào)遞減；④當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0在定義域上恒成立，SKIPIF1<0單調(diào)遞增；⑤當(dāng)SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0或SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0，SKIPIF1<0單調(diào)遞減；綜上所述：當(dāng)SKIPIF1<0時(shí)，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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；單調(diào)遞減區(qū)間為SKIPIF1<02．（2022·河北·高二期中）已知函數(shù)SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0的圖象在SKIPIF1<0處的切線方程；(2)討論SKIPIF1<0的單調(diào)性．【答案】(1)SKIPIF1<0(2)答案見解析(1)因?yàn)镾KIPIF1<0，所以SKIPIF1<0．則SKIPIF1<0，故SKIPIF1<0的圖象在SKIPIF1<0處的切線方程為SKIPIF1<0．(2)SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0恒成立，所以當(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．若SKIPIF1<0，則由SKIPIF1<0，得SKIPIF1<0．當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0，函數(shù)單調(diào)遞增；若SKIPIF1<0，則SKIPIF1<0，函數(shù)單調(diào)遞減．當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，SKIPIF1<0在R上恒成立，函數(shù)單調(diào)遞增．當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0，函數(shù)單調(diào)遞增；若SKIPIF1<0，則SKIPIF1<0，函數(shù)單調(diào)遞減．綜上所述，當(dāng)SKIPIF1<0時(shí)，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；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0和SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在R上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0和SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0．題型五：導(dǎo)函數(shù)有效部分是二次型（不可因式分解）【典型例題】例題1．（2022·天津河西·一模）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的極值．(2)討論SKIPIF1<0的單調(diào)性；【答案】(1)極大值為SKIPIF1<0，無極小值(2)答案見解析(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0SKIPIF1<02SKIPIF1<0SKIPIF1<0+0-SKIPIF1<0單調(diào)遞增SKIPIF1<0單調(diào)遞減所以SKIPIF1<0的極大值為SKIPIF1<0，無極小值．(2)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，對(duì)于二次方程SKIPIF1<0，有SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有兩根SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；若SKIPIF1<0，SKIPIF1<0在SKIPIF1<0與SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．【提分秘籍】如本例，求導(dǎo)后SKIPIF1<0，記導(dǎo)函數(shù)有效部分為SKIPIF1<0，判斷為不可因式分解的二次型，此類題型的方法主要采用SKIPIF1<0法；分兩類：①SKIPIF1<0；②SKIPIF1<0，利用求根公式求出方程SKIPIF1<0的兩個(gè)根SKIPIF1<0，SKIPIF1<0，然后再討論SKIPIF1<0的正負(fù)，進(jìn)而討論單調(diào)性，同時(shí)也要注意定義域.【變式演練】1．（2022·內(nèi)蒙古包頭·一模（文））已知函數(shù)SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；【答案】(1)答案見解析（1）由題意可知SKIPIF1<0的定義域?yàn)镽，SKIPIF1<0，對(duì)于SKIPIF1<0，SKIPIF1<0.①當(dāng)SKIPIF1<0，即-3≤a≤3時(shí)SKIPIF1<0，SKIPIF1<0在R上單調(diào)遞增；②當(dāng)SKIPIF1<0，即a<-3或a>3時(shí)，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0；令SKIPIF1<0，則SKIPIF1<0；所以SKIPIF1<0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.綜上，當(dāng)-3≤a≤3時(shí)，SKIPIF1<0在R上單調(diào)遞增；當(dāng)a<-3或a>3時(shí)，SKIPIF1<0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.題型六：借助二階導(dǎo)函數(shù)討論單調(diào)性【典型例題】例題1．（2022·全國(guó)·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0(1)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0單調(diào)性；【答案】(1)SKIPIF1<0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)，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，則SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增.綜上，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.例題2．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的單調(diào)區(qū)間；【答案】(1)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增(1)SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0單調(diào)遞增，注意到SKIPIF1<0∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0，SKIPIF1<0單調(diào)遞增∴SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增【提分秘籍】當(dāng)一階導(dǎo)函數(shù)中含有SKIPIF1<0，SKIPIF1<0，而一階導(dǎo)的正負(fù)難以確定時(shí)，可以通過求二階導(dǎo)，從而判斷一階導(dǎo)的單調(diào)性，進(jìn)而判斷一階導(dǎo)的正負(fù)來討論單調(diào)性.【變式演練】1．（2022·河南南陽·高二期中（理））已知函數(shù)SKIPIF1<0．(1)判斷函數(shù)SKIPIF1<0的單調(diào)性．【答案】(1)SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減(1)因?yàn)镾KIPIF1<0，所以SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0．因?yàn)楫?dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減．1．（2022·貴州貴陽·模擬預(yù)測(cè)（理））已知SKIPIF1<0，函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；【答案】(1)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．【詳解】（1）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·江蘇徐州·高三期中）已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.SKIPIF1<0，SKIPIF1<0分別為函數(shù)SKIPIF1<0，SKIPIF1<0的導(dǎo)函數(shù).(1)討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】(1)答案見解析【詳解】（1）SKIPIF1<0，SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)增；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)減；SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)增.綜上所述，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)增；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)減，在SKIPIF1<0上單調(diào)增3．（2022·江蘇·華羅庚中學(xué)三模）已知函數(shù)SKIPIF1<0，SKIPIF1<0（SKIPIF1<0為自然對(duì)數(shù)的底數(shù)，SKIPIF1<0）．(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；【答案】(1)分類討論，答案見解析.(1)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，①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，無增區(qū)間；②當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0可得SKIP