新高考數(shù)學(xué)一輪復(fù)習(xí)第3章 第02講 導(dǎo)數(shù)與函數(shù)的單調(diào)性 精講+精練（教師版）

第02講導(dǎo)數(shù)與函數(shù)的單調(diào)性(精講+精練）目錄第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第二部分：課前自我評(píng)估測(cè)試第三部分：典型例題剖析高頻考點(diǎn)一：利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間（不含參）高頻考點(diǎn)二：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)高頻考點(diǎn)三：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)區(qū)間高頻考點(diǎn)四：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)高頻考點(diǎn)五：函數(shù)單調(diào)性的應(yīng)用①導(dǎo)函數(shù)與原函數(shù)圖象的單調(diào)性②比較大?、蹣?gòu)造函數(shù)解不等式高頻考點(diǎn)六：含參問(wèn)題討論單調(diào)性①導(dǎo)函數(shù)有效部分是一次型（或可化為一次型）②導(dǎo)函數(shù)有效部分是二次型（或可化為二次型）且可因式分解型③導(dǎo)函數(shù)有效部分是二次型（或可化為二次型）且不可因式分解型第四部分：高考真題感悟第五部分：第02講導(dǎo)數(shù)與函數(shù)的單調(diào)性（精練）第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶1、函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系（導(dǎo)函數(shù)看正負(fù)，原函數(shù)看增減）條件恒有結(jié)論函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上可導(dǎo)SKIPIF1<0SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞增SKIPIF1<0SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞減SKIPIF1<0SKIPIF1<0在SKIPIF1<0內(nèi)是常數(shù)函數(shù)2、求已知函數(shù)（不含參）的單調(diào)區(qū)間①求SKIPIF1<0的定義域②求SKIPIF1<0③令SKIPIF1<0，解不等式，求單調(diào)增區(qū)間④令SKIPIF1<0，解不等式，求單調(diào)減區(qū)間注：求單調(diào)區(qū)間時(shí)，令SKIPIF1<0（或SKIPIF1<0）不跟等號(hào).3、由函數(shù)SKIPIF1<0的單調(diào)性求參數(shù)的取值范圍的方法（1）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(jià)條件中的不等式含等號(hào).（2）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<0令SKIPIF1<0，解不等式，求單調(diào)增區(qū)間SKIPIF1<0，則SKIPIF1<0②已知SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)減區(qū)間SKIPIF1<0令SKIPIF1<0，解不等式，求單調(diào)減區(qū)間SKIPIF1<0，則SKIPIF1<0（3）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)SKIPIF1<0SKIPIF1<0，使得SKIPIF1<04、含參問(wèn)題討論單調(diào)性第一步：求SKIPIF1<0的定義域第二步：求SKIPIF1<0（導(dǎo)函數(shù)中有分母通分）第三步：確定導(dǎo)函數(shù)有效部分，記為SKIPIF1<0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）.接下來(lái)就只需考慮導(dǎo)函數(shù)有效部分，只有該部分決定SKIPIF1<0的正負(fù).第四步：確定導(dǎo)函數(shù)有效部分SKIPIF1<0的類型：①SKIPIF1<0為一次型（或可化為一次型）②SKIPIF1<0為二次型（或可化為二次型）第五步：通過(guò)分析導(dǎo)函數(shù)有效部分，討論SKIPIF1<0的單調(diào)性第二部分：課前自我評(píng)估測(cè)試第二部分：課前自我評(píng)估測(cè)試一、判斷題1．（2021·全國(guó)·高二課前預(yù)習(xí)）函數(shù)f(x)在定義域上都有f′(x)<0，則函數(shù)f(x)在定義域上單調(diào)遞減.()【答案】錯(cuò)誤2．（2021·全國(guó)·高二課前預(yù)習(xí)）函數(shù)f(x)在某區(qū)間內(nèi)單調(diào)遞增，則一定有f′(x)>0.()【答案】錯(cuò)誤3．（2021·全國(guó)·高二課前預(yù)習(xí)）函數(shù)y＝x3＋x的單調(diào)遞增區(qū)間為(－∞，＋∞).()【答案】正確二、單選題1．（2022·廣東·佛山市南海區(qū)桂城中學(xué)高二階段練習(xí)）函數(shù)SKIPIF1<0的圖象如圖所示，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0的符號(hào)不確定【答案】B如圖所示，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0故選：B2．（2022·河北·武安市第三中學(xué)高二階段練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D解：函數(shù)的定義域是SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．故選：D．3．（2022·江西南昌·高二期末（理））若函數(shù)SKIPIF1<0，則SKIPIF1<0的單調(diào)增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C解：因?yàn)楹瘮?shù)SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0，故選：C.4．（2022·湖北·華中師大一附中高一期末）“函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)”是：“實(shí)數(shù)SKIPIF1<0”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】BSKIPIF1<0在SKIPIF1<0上恒成立，可得SKIPIF1<0，SKIPIF1<0，所以“函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)”是：“實(shí)數(shù)SKIPIF1<0”的必要不充分條件.故選：B.第三部分：典型例題剖析第三部分：典型例題剖析高頻考點(diǎn)一：利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間（不含參）1．（2022·廣東·深圳市南山區(qū)華僑城中學(xué)高二階段練習(xí)）函數(shù)SKIPIF1<0的單調(diào)減區(qū)間是（

）A．（－∞，SKIPIF1<0] B．（0，SKIPIF1<0） C．SKIPIF1<0和（0，SKIPIF1<0） D．SKIPIF1<0【答案】B函數(shù)定義域是SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0．即減區(qū)間是SKIPIF1<0．故選：B．2．（2022·福建·福鼎市第一中學(xué)高二階段練習(xí)）函數(shù)SKIPIF1<0的減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C∵SKIPIF1<0，∴SKIPIF1<0，由SKIPIF1<0得，SKIPIF1<0，∴函數(shù)的減區(qū)間是SKIPIF1<0.故選：C.3．（2022·重慶八中高三階段練習(xí)）函數(shù)SKIPIF1<0的遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】DSKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0；SKIPIF1<0在SKIPIF1<0上的單調(diào)遞增區(qū)間為SKIPIF1<0.故選：D.4．（2022·全國(guó)·高二課時(shí)練習(xí)）函數(shù)SKIPIF1<0的減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C由題意，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0的減區(qū)間是SKIPIF1<0.故選：C5．（2022·全國(guó)·高三專題練習(xí)）函數(shù)SKIPIF1<0的遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且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.故選：A.高頻考點(diǎn)二：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)1．（2022·黑龍江·鐵人中學(xué)高二開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0單調(diào)遞增，a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B因?yàn)镾KIPIF1<0在SKIPIF1<0單調(diào)遞增，故SKIPIF1<0SKIPIF1<0在區(qū)間SKIPIF1<0恒成立，即SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞增，則SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0的取值范圍為SKIPIF1<0.故選：B.2．（2022·全國(guó)·高三專題練習(xí)）若函數(shù)f(x)＝x3＋bx2＋cx＋d的單調(diào)遞減區(qū)間為(－1，3)，則b＋c＝（

）A．－12 B．－10 C．8 D．10【答案】ASKIPIF1<0＝3x2＋2bx＋c，由題意知，－1<x<3是不等式3x2＋2bx＋c<0的解，∴－1，3是SKIPIF1<0＝0的兩個(gè)根，∴b＝－3，c＝－9，∴b＋c＝－12.故選：A．3．（2022·廣西欽州·高二期末（文））函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增的一個(gè)必要不充分條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D由題得SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增，SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立．SKIPIF1<0，而SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，SKIPIF1<0．選項(xiàng)中只有SKIPIF1<0是SKIPIF1<0的必要不充分條件.選項(xiàng)AC是SKIPIF1<0的充分不必要條件，選項(xiàng)B是充要條件.故選：D4．（2022·全國(guó)·高二課時(shí)練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞減，則實(shí)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.5．（2022·全國(guó)·高二課時(shí)練習(xí)）若函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A∵SKIPIF1<0在SKIPIF1<0上是減函數(shù)，所以SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，故選：A.【點(diǎn)睛】本題主要考查“分離參數(shù)”在解題中的應(yīng)用、函數(shù)的定義域及利用單調(diào)性求參數(shù)的范圍，屬于中檔題.利用單調(diào)性求參數(shù)的范圍的常見(jiàn)方法：①視參數(shù)為已知數(shù)，依據(jù)函數(shù)的圖象或單調(diào)性定義，確定函數(shù)的單調(diào)區(qū)間，與已知單調(diào)區(qū)間比較求參數(shù)需注意若函數(shù)在區(qū)間SKIPIF1<0上是單調(diào)的，則該函數(shù)在此區(qū)間的任意子集上也是單調(diào)的;②利用導(dǎo)數(shù)轉(zhuǎn)化為不等式SKIPIF1<0或SKIPIF1<0恒成立問(wèn)題求參數(shù)范圍.6．（2022·全國(guó)·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因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<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(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故選：A.高頻考點(diǎn)三：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)區(qū)間1．（2022·河北·武安市第三中學(xué)高二階段練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在單調(diào)遞減區(qū)間，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ASKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，不符合題意；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．故選：A．2．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)遞增區(qū)間，則實(shí)數(shù)b的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A∵函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)增區(qū)間，∴函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在子區(qū)間使得不等式SKIPIF1<0成立，SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0；故選：A．3．（2022·重慶市萬(wàn)州第二高級(jí)中學(xué)高二階段練習(xí)）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.4．（2022·全國(guó)·高二）若函數(shù)SKIPIF1<0存在遞減區(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】B由題設(shè)，SKIPIF1<0，由SKIPIF1<0存在遞減區(qū)間，即存在SKIPIF1<0使SKIPIF1<0，∴SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0.故選：B5．（2021·內(nèi)蒙古·赤峰二中高二期末（理））若函數(shù)SKIPIF1<0存在增區(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】C若函數(shù)SKIPIF1<0不存在增區(qū)間，則函數(shù)SKIPIF1<0單調(diào)遞減，此時(shí)SKIPIF1<0在區(qū)間SKIPIF1<0恒成立，可得SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，故函數(shù)存在增區(qū)間時(shí)實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．故選C.高頻考點(diǎn)四：已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)1．（2022·重慶市青木關(guān)中學(xué)校高二階段練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)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<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D由于函數(shù)SKIPIF1<0在SKIPIF1<0不是單調(diào)函數(shù)，則SKIPIF1<0在SKIPIF1<0內(nèi)存在極值點(diǎn)，所以SKIPIF1<0在SKIPIF1<0內(nèi)有解，即SKIPIF1<0在SKIPIF1<0內(nèi)有解，SKIPIF1<0．故選：D2．（2022·河南·高二階段練習(xí)（理））若函數(shù)SKIPIF1<0在定義域內(nèi)的一個(gè)子區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實(shí)數(shù)k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D由題意得，函數(shù)定義域?yàn)镾KIPIF1<0SKIPIF1<0，令SKIPIF1<0，解得在定義域內(nèi)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，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?yàn)镾KIPIF1<0，得SKIPIF1<0，綜上SKIPIF1<0，故選：D．3．（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(shí)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由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在其定義域上不單調(diào)，即SKIPIF1<0有變號(hào)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(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：A.4．（2022·浙江·高二階段練習(xí)）函數(shù)SKIPIF1<0在區(qū)間[-1，2]上不單調(diào)，則實(shí)數(shù)a的取值范圍是（

）A．(-∞，-3] B．(-3，1)C．[1，+∞) D．(-∞，-3]∪[1，+∞)【答案】BSKIPIF1<0SKIPIF1<0，如果函數(shù)SKIPIF1<0在區(qū)間[-1，2]上單調(diào)，那么a-1≥0或SKIPIF1<0，即SKIPIF1<0，解得a≥1或a≤-3，所以當(dāng)函數(shù)SKIPIF1<0在區(qū)間[-1，2]上不單調(diào)時(shí)，SKIPIF1<0.故選：B5．（2022·安徽省太和中學(xué)高二開(kāi)學(xué)考試）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則實(shí)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由SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，不合題意；②當(dāng)SKIPIF1<0時(shí)，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．6．（2022·江蘇·高二）若函數(shù)SKIPIF1<0在其定義域上不單調(diào)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A可得SKIPIF1<0，SKIPIF1<0SKIPIF1<0在其定義域上不單調(diào)等價(jià)于方程SKIPIF1<0有兩個(gè)解，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故選：A.7．（2022·全國(guó)·高二課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上不單調(diào)的一個(gè)充分不必要條件是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】CSKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上不單調(diào)，令SKIPIF1<0，對(duì)稱軸方程為SKIPIF1<0，則函數(shù)SKIPIF1<0與SKIPIF1<0軸在SKIPIF1<0上有交點(diǎn)．當(dāng)SKIPIF1<0時(shí)，顯然不成立；當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0．四個(gè)選項(xiàng)中的范圍，只有SKIPIF1<0為SKIPIF1<0的真子集，∴SKIPIF1<0在SKIPIF1<0上不單調(diào)的一個(gè)充分不必要條件是SKIPIF1<0.故選：C．高頻考點(diǎn)五：函數(shù)單調(diào)性的應(yīng)用①導(dǎo)函數(shù)與原函數(shù)圖象的單調(diào)性1．（2021·廣西河池·高二階段練習(xí)（理））如果函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象如圖所示，則以下關(guān)于函數(shù)SKIPIF1<0的判斷：①在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增；②在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞減；③在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增；④SKIPIF1<0是極小值點(diǎn)；⑤SKIPIF1<0是極大值點(diǎn).其中不正確的是（

）A．③⑤ B．②③ C．①④⑤ D．①②④【答案】D由圖可知，在區(qū)間SKIPIF1<0內(nèi)，SKIPIF1<0有正有負(fù)，①錯(cuò)誤；在區(qū)間SKIPIF1<0內(nèi)，SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增，②錯(cuò)誤；在區(qū)間SKIPIF1<0內(nèi)，SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞增，③正確；不存在SKIPIF1<0，使當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，④錯(cuò)誤；存在SKIPIF1<0，使當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，如SKIPIF1<0，⑤正確故選：D.2．（2021·福建省漳州第一中學(xué)高二階段練習(xí)）SKIPIF1<0是函數(shù)y＝f(x)的導(dǎo)函數(shù)，若y＝SKIPIF1<0的圖象如圖所示，則函數(shù)y＝f(x)的圖象可能是（

）A． B．C． D．【答案】D由導(dǎo)函數(shù)的圖象可知，當(dāng)x＜0時(shí)，SKIPIF1<0＞0，即函數(shù)f(x)為增函數(shù)；當(dāng)0＜x＜2時(shí)，SKIPIF1<0＜0，即f(x)為減函數(shù)；當(dāng)x＞2時(shí)，SKIPIF1<0＞0，即函數(shù)f(x)為增函數(shù).觀察選項(xiàng)易知D正確.故選：D3．（2021·海南·三亞華僑學(xué)校高三階段練習(xí)）已知函數(shù)SKIPIF1<0的圖象如圖所示，則SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】C由函數(shù)SKIPIF1<0的圖象可知：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增．故選：C4．（2021·全國(guó)·高二課時(shí)練習(xí)）如圖為函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象，那么函數(shù)SKIPIF1<0的圖象可能為（

）A． B．C． D．【答案】A由導(dǎo)函數(shù)SKIPIF1<0的圖象，可知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增．綜上，函數(shù)SKIPIF1<0的圖象可能如A中圖所示故選：A5．（2021·江西省銅鼓中學(xué)高二階段練習(xí)（理））設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)數(shù)，SKIPIF1<0的圖象如圖所示，則SKIPIF1<0的圖像最有可能的是（

）．A． B．C． D．【答案】C解：由導(dǎo)函數(shù)SKIPIF1<0的圖象可知：導(dǎo)函數(shù)SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0，則函數(shù)SKIPIF1<0單調(diào)遞增；導(dǎo)函數(shù)SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0，則函數(shù)SKIPIF1<0單調(diào)遞減；導(dǎo)函數(shù)SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0，則函數(shù)SKIPIF1<0單調(diào)遞增；故選：C②比較大小1．（2022·云南省昆明市第十中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BSKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0是增函數(shù)，而SKIPIF1<0，所以SKIPIF1<0．故選：B．2．（2022·重慶市清華中學(xué)校高二階段練習(xí)）若函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】CSKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.所以SKIPIF1<0，故選項(xiàng)A不正確.SKIPIF1<0,故選項(xiàng)B,D不正確，選項(xiàng)C正確.故選：C3．（2022·河南·民權(quán)縣第一高級(jí)中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C因?yàn)镾KIPIF1<0得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，從而SKIPIF1<0．故選：C.4．（2022·四川·成都外國(guó)語(yǔ)學(xué)校高二階段練習(xí)（理））已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B設(shè)SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，于是當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0單調(diào)遞減，注意到SKIPIF1<0，則有SKIPIF1<0，即SKIPIF1<0.故選：B.③構(gòu)造函數(shù)解不等式1．（2022·全國(guó)·高二課時(shí)練習(xí)）已知定義在R上的奇函數(shù)f（x），當(dāng)x＞0時(shí),SKIPIF1<0，且f（3）＝0，則不等式f（x）≥0的解集為（

）A．（﹣∞，﹣3]∪[3，+∞） B．[﹣3，3]C．（﹣∞，﹣3]∪[0，3] D．[﹣3，0]∪[3，+∞）【答案】D設(shè)SKIPIF1<0，（x＞0），則其導(dǎo)數(shù)SKIPIF1<0，而當(dāng)x＞0時(shí)SKIPIF1<0，所以g′（x）＞0，即g（x）在（0，+∞）上為增函數(shù)，又由f（3）＝0，則SKIPIF1<00，所以SKIPIF1<0區(qū)間（0，3）上，g（x）＜0，在區(qū)間（3，+∞）上，g（x）＞0，則在區(qū)間（0，3）上，f（x）＜0，在區(qū)間（3，+∞）上，f（x）＞0，又由f（x）是定義在R上的奇函數(shù)，則f（0）＝0，SKIPIF1<0，且在區(qū)間（﹣∞，﹣3）上，f（x）＜0，在區(qū)間（﹣3，0）上，f（x）＞0，綜合可得：不等式f（x）≥0的解集為[﹣3，0]∪[3，+∞）．故選：D.2．（2022·河南·高二階段練習(xí)（文））已知函數(shù)SKIPIF1<0對(duì)于任意的SKIPIF1<0滿足SKIPIF1<0，其中SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，則下列各式正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0D．SKIPIF1<0【答案】C令SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，所以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單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0.故選：C3．（2022·全國(guó)·高三專題練習(xí)（理））設(shè)函數(shù)SKIPIF1<0是偶函數(shù)SKIPIF1<0（SKIPIF1<0）的導(dǎo)函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則使得SKIPIF1<0成立的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0D．SKIPIF1<0【答案】D令SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上為減函數(shù)，又∵SKIPIF1<0，∴函數(shù)SKIPIF1<0為定義域上的奇函數(shù)，SKIPIF1<0在SKIPIF1<0上為減函數(shù).又∵SKIPIF1<0，∴SKIPIF1<0，∴不等式SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，或SKIPIF1<0，∵SKIPIF1<0成立的x的取值范圍是SKIPIF1<0，故選：D4．（2022·重慶南開(kāi)中學(xué)高二期末）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足：SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A設(shè)SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0為SKIPIF1<0上的增函數(shù)，而SKIPIF1<0可化為SKIPIF1<0即SKIPIF1<0，故SKIPIF1<0即SKIPIF1<0，所以不等式SKIPIF1<0的解集為SKIPIF1<0，故選：A.5．（2022·甘肅·永昌縣第一高級(jí)中學(xué)高二階段練習(xí)（理））已知f(x)為R上的可導(dǎo)函數(shù)，其導(dǎo)函數(shù)為SKIPIF1<0，且對(duì)于任意的x∈R，均有SKIPIF1<0，則（

）A．e－2021f(－2021)>f(0)，e2021f(2021)<f(0) B．e－2021f(－2021)<f(0)，e2021f(2021)<f(0)C．e－2021f(－2021)>f(0)，e2021f(2021)>f(0) D．e－2021f(－2021)<f(0)，e2021f(2021)>f(0)【答案】D構(gòu)造函數(shù)SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，所以SKIPIF1<0，即SKIPIF1<0.故選：D高頻考點(diǎn)六：含參問(wèn)題討論單調(diào)性①導(dǎo)函數(shù)有效部分是一次型（或可化為一次型）1．（2022·廣東·清遠(yuǎn)市博愛(ài)學(xué)校高二階段練習(xí)）已知函數(shù)SKIPIF1<0．（1）討論SKIPIF1<0的單調(diào)性；【答案】（1）答案見(jiàn)解析；（2）SKIPIF1<0.（1）SKIPIF1<0且SKIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞增；當(dāng)SKIPIF1<0時(shí)：若SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞增；∴SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上遞增；SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增；2．（2022·全國(guó)·高三專題練習(xí)（理））設(shè)a為實(shí)數(shù)，函數(shù)f(x)＝SKIPIF1<0－2x＋2a，x∈R．(1)求f(x)的單調(diào)區(qū)間與極值；【答案】(1)f(x)的單調(diào)遞減區(qū)間是(－∞，ln2)，單調(diào)遞增區(qū)間是(ln2，＋∞)，極小值f(ln2)＝2－2ln2＋2a，無(wú)極大值；(1)由f(x)＝SKIPIF1<0－2x＋2a(x∈R)知SKIPIF1<0＝SKIPIF1<0－2．令SKIPIF1<0＝0，得x＝ln2．當(dāng)x＜ln2時(shí)，SKIPIF1<0＜0，故函數(shù)f(x)在區(qū)間(－∞，ln2)上單調(diào)遞減；當(dāng)x＞ln2時(shí)，SKIPIF1<0＞0，故函數(shù)f(x)在區(qū)間(ln2，＋∞)上單調(diào)遞增．∴f(x)的單調(diào)遞減區(qū)間是(－∞，ln2)，單調(diào)遞增區(qū)間是(ln2，＋∞)，f(x)極小值為f(ln2)＝SKIPIF1<0－2ln2＋2a＝2－2ln2＋2a，無(wú)極大值；2．（2022·全國(guó)·高二）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：SKIPIF1<0

SKIPIF1<0

SKIPIF1<0，SKIPIF1<0

SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，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．3．（2022·全國(guó)·高二課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0.討論SKIPIF1<0的單調(diào)性.解：因?yàn)镾KIPIF1<0，所以定義域?yàn)镾KIPIF1<0，所以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)，由SKIPIF1<0，得SKIPIF1<0；由SKIPIF1<0，得SKIPIF1<0.故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.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)遞減，在SKIPIF1<0上單調(diào)遞增.②導(dǎo)函數(shù)有效部分是二次型（或可化為二次型）且可因式分解型1．（2022·江蘇宿遷·高二期末）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)討論SKIPIF1<0的單調(diào)性；(1)因?yàn)镾KIPIF1<0，故可得SKIPIF1<0SKIPIF1<0，令SKIPIF1<0SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，此時(shí)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，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和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和SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減.2．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，討論SKIPIF1<0的單調(diào)性.【答案】當(dāng)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0，則x∈SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0單調(diào)遞增.當(dāng)a＜0，則x∈SKIPIF1<0時(shí)，SKIPIF1<0；x∈SKIPIF1<0時(shí)，SKIPIF1<0故SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減.綜上所述,當(dāng)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.3．（2022·全國(guó)·高二課時(shí)練習(xí)）已知函數(shù)SKIPIF1<0.（1）當(dāng)SKIPIF1<0時(shí)，求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；（2）求SKIPIF1<0的單調(diào)區(qū)間.【答案】（1）SKIPIF1<0