新高考數(shù)學一輪復習第2章 第02講 函數(shù)的單調(diào)性與最大（小）值精講+精練（學生版）

第02講函數(shù)的單調(diào)性與最大（?。┲?精講+精練）目錄第一部分：知識點精準記憶第二部分：課前自我評估測試第三部分：典型例題剖析高頻考點一：函數(shù)的單調(diào)性①求函數(shù)的單調(diào)區(qū)間②根據(jù)函數(shù)的單調(diào)性求參數(shù)③復合函數(shù)的單調(diào)性④根據(jù)函數(shù)單調(diào)性解不等式高頻考點二：函數(shù)的最大（小）值①利用函數(shù)單調(diào)性求最值②根據(jù)函數(shù)最值求參數(shù)③不等式恒成立問題④不等式有解問題第四部分：高考真題感悟第五部分：第02講函數(shù)的單調(diào)性與最大（?。┲担ň殻┑谝徊糠郑褐R點精準記憶第一部分：知識點精準記憶1、函數(shù)的單調(diào)性（1）單調(diào)性的定義一般地，設函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，如果對于定義域SKIPIF1<0內(nèi)某個區(qū)間SKIPIF1<0上的任意兩個自變量的值SKIPIF1<0，SKIPIF1<0；①當SKIPIF1<0時，都有SKIPIF1<0，那么就說函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)②當SKIPIF1<0時，都有SKIPIF1<0，那么就說函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)（2）單調(diào)性簡圖：（3）單調(diào)區(qū)間（注意先求定義域）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)或減函數(shù)，則稱函數(shù)SKIPIF1<0在這一區(qū)間上具有（嚴格的）單調(diào)性，區(qū)間SKIPIF1<0叫做函數(shù)SKIPIF1<0的單調(diào)區(qū)間．（4）復合函數(shù)的單調(diào)性（同調(diào)增；異調(diào)減）對于函數(shù)SKIPIF1<0和SKIPIF1<0，如果當SKIPIF1<0時，SKIPIF1<0，且SKIPIF1<0在區(qū)間SKIPIF1<0上和SKIPIF1<0在區(qū)間SKIPIF1<0上同時具有單調(diào)性，則復合函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上具有單調(diào)性，并且具有這樣的規(guī)律：增增(或減減)則增，增減(或減增)則減．2、函數(shù)的最值（1）設函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，如果存在實數(shù)SKIPIF1<0滿足①對于任意的SKIPIF1<0，都有SKIPIF1<0；②存在SKIPIF1<0，使得SKIPIF1<0則SKIPIF1<0為最大值（2）設函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，如果存在實數(shù)SKIPIF1<0滿足①對于任意的SKIPIF1<0，都有SKIPIF1<0；②存在SKIPIF1<0，使得SKIPIF1<0則SKIPIF1<0為最小值3、常用高頻結論（1）設SKIPIF1<0，SKIPIF1<0.①若有SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0在閉區(qū)間SKIPIF1<0上是增函數(shù)；②若有SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0在閉區(qū)間SKIPIF1<0上是減函數(shù).此為函數(shù)單調(diào)性定義的等價形式.（2）函數(shù)相加或相減后單調(diào)性：設SKIPIF1<0，兩個函數(shù)SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性如下表，則SKIPIF1<0在SKIPIF1<0上的單調(diào)性遵循（增+增=增；減+減=減）SKIPIF1<0SKIPIF1<0SKIPIF1<0增增增減減減SKIPIF1<0SKIPIF1<0SKIPIF1<0增減增減增減（3）對鉤函數(shù)單調(diào)性：SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）的單調(diào)性：在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減．（4）常見對鉤函數(shù)：SKIPIF1<0（SKIPIF1<0），的單調(diào)性：在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減．第二部分：課前自我評估測試第二部分：課前自我評估測試一、判斷題1．（2021·江西·貴溪市實驗中學高二階段練習）SKIPIF1<0則SKIPIF1<0在R上是增函數(shù)

()2．（2021·全國·高二課前預習）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值與最小值一定在區(qū)間端點處取得.()二、單選題1．（2022·北京市懷柔區(qū)教科研中心高一期末）下列函數(shù)中，在區(qū)間SKIPIF1<0上是減函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高一）若函數(shù)y＝f(x)在R上單調(diào)遞減，且f(2m－3)>f(－m)，則實數(shù)m的取值范圍是（

）A．(－∞，－1) B．(－1，＋∞) C．(1，＋∞) D．(－∞，1)3．（2022·全國·高三專題練習）函數(shù)y＝SKIPIF1<0在[2，3]上的最小值為（

）A．2 B．SKIPIF1<0C．SKIPIF1<0 D．－SKIPIF1<04．（2022·全國·高三專題練習（理））已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值是SKIPIF1<0,最小值是SKIPIF1<0,則實數(shù)SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0第三部分：典型例題剖析第三部分：典型例題剖析高頻考點一：函數(shù)的單調(diào)性①求函數(shù)的單調(diào)區(qū)間1．（2022·全國·高三專題練習）SKIPIF1<0的單調(diào)增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高一課時練習）函數(shù)SKIPIF1<0的圖象如圖所示，其增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·湖北·孝感市孝南區(qū)第二高級中學高一期中）函數(shù)SKIPIF1<0的減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·四川省峨眉第二中學校高一階段練習）已知函數(shù)SKIPIF1<0在R上單調(diào)遞減，則函數(shù)SKIPIF1<0的增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2021·全國·高一專題練習）函數(shù)SKIPIF1<0的增區(qū)間是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0②根據(jù)函數(shù)的單調(diào)性求參數(shù)一、單選題1．（2022·安徽蕪湖·高一期末）已知函數(shù)SKIPIF1<0是R上的單調(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<02．（2022·天津河西·高一期末）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則實數(shù)k的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·河南·南陽中學高一階段練習）已知函數(shù)SKIPIF1<0是R上的增函數(shù)，則a的取值范圍為（

）A．[-4，0) B．[-4，-2] C．SKIPIF1<0 D．SKIPIF1<04．（2022·河南·溫縣第一高級中學高一階段練習）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上為減函數(shù)，則下列選項正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·廣西梧州·高二期末（理））已知函數(shù)SKIPIF1<0，若對任意的SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，總有SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0③復合函數(shù)的單調(diào)性1．（2022·全國·高三專題練習（文））已知函數(shù)SKIPIF1<0，則該函數(shù)的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高三專題練習）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<03．（2022·全國·高三專題練習）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間是（）A．SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·全國·高三專題練習（文））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<05．（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0的圖象如圖所示，則函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0④根據(jù)函數(shù)單調(diào)性解不等式1．（2022·內(nèi)蒙古包頭·一模（文））設函數(shù)SKIPIF1<0，則滿足SKIPIF1<0的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·河北保定·高一期末）已知函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù)（其中SKIPIF1<0且SKIPIF1<0），則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·四川綿陽·高一期末）若SKIPIF1<0，則滿足SKIPIF1<0的SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·甘肅省會寧縣第一中學高一期末）已知函數(shù)SKIPIF1<0關于直線SKIPIF1<0對稱，且當SKIPIF1<0時，SKIPIF1<0恒成立，則滿足SKIPIF1<0的x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2022·浙江·高三專題練習）已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的增函數(shù)，則滿足SKIPIF1<0的實數(shù)SKIPIF1<0的取值范圍（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·陜西陜西·一模（文））已知SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0高頻考點二：函數(shù)的最大（?。┲耽倮煤瘮?shù)單調(diào)性求最值1．（2022·全國·高三專題練習（理））已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0是單調(diào)遞增函數(shù) B．SKIPIF1<0是奇函數(shù)C．函數(shù)SKIPIF1<0的最大值為SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高三專題練習）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為A．72 B．36 C．12 D．03．（2022·全國·高三專題練習）設函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的增函數(shù)，實數(shù)SKIPIF1<0使得SKIPIF1<0對于任意SKIPIF1<0都成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·全國·高一專題練習）已知SKIPIF1<0，SKIPIF1<0，若對SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0②根據(jù)函數(shù)最值求參數(shù)1．（2021·福建·永安市第三中學高中校高三期中）函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·全國·高一單元測試）設函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值為7，則SKIPIF1<0在SKIPIF1<0上的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·浙江·高一單元測試）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值是4，則實數(shù)SKIPIF1<0的值為（

）A．-1 B．1 C．3 D．1或34．（2019·貴州·興仁市鳳凰中學高一階段練習）已知函數(shù)SKIPIF1<0，SKIPIF1<0，并且函數(shù)SKIPIF1<0的最小值為SKIPIF1<0,則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2021·上?！じ咭粏卧獪y試）一次函數(shù)SKIPIF1<0，在[﹣2，3]上的最大值是SKIPIF1<0，則實數(shù)a的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2021·廣東·廣州四十七中高一期中）己知函數(shù)SKIPIF1<0有最小值，則a的的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<07．（2021·全國·高一課時練習）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為4，則實數(shù)SKIPIF1<0的取值集合為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0③不等式恒成立問題1．（2022·黑龍江·鶴崗一中高三期末（文））已知SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·甘肅武威·高一期末）對SKIPIF1<0，不等式SKIPIF1<0恒成立，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<03．（2022·四川·遂寧中學高一開學考試）對于SKIPIF1<0，不等式SKIPIF1<0恒成立，則實數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·江西·模擬預測（文））已知函數(shù)SKIPIF1<0，當SKIPIF1<0時，不等SKIPIF1<0恒成立，則實數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·廣西梧州·高二期末（理））已知函數(shù)f(x)＝xSKIPIF1<0，g(x)＝2x+a，若?x1∈[SKIPIF1<0，1]，?x2∈[2，3]，使得f(x1)≥g(x2)，則實數(shù)a的取值范圍是()A．a(chǎn)≤1 B．a(chǎn)≥1 C．a(chǎn)≤2 D．a(chǎn)≥2④不等式有解問題1．（2022·河南·平頂山市教育局教育教學研究室高二開學考試（文））已知函數(shù)SKIPIF1<0，SKIPIF1<0，對于任意的SKIPIF1<0，存在SKIPIF1<0，使SKIPIF1<0，則實數(shù)a的取值范圍為（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·江西·景德鎮(zhèn)一中高一期末）已知函數(shù)SKIPIF1<0，SKIPIF1<0，對于任意SKIPIF1<0，存在SKIPIF1<0有SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022·浙江·高三專題練習）當SKIPIF1<0時，若關于SKIPIF1<0的不等式SKIPIF1<0有解，則實數(shù)SKIPIF1<0的取值范圍是（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·山東·棗莊市第三中學高一期中）已知SKIPIF1<0，SKIPIF1<0，若對SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2021·全國·高一單元測試）若SKIPIF1<0，使得SKIPIF1<0，則實數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0第四部分：高考真題感悟第四部分：高考真題感悟1．（2021·全國·高考真題（文））下列函數(shù)中是增函數(shù)的為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·北京·高考真題）已知SKIPIF1<0是定義在上SKIPIF1<0的函數(shù)，那么“函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增”是“函數(shù)SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0”的（

）A．充分而不必要條件 B．必要而不充分條件 C．充分必要條件 D．既不充分也不必要條件3．（2020·山東·高考真題）已知函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，若對于任意兩個不相等的實數(shù)SKIPIF1<0，SKIPIF1<0，總有SKIPIF1<0成立，則函數(shù)SKIPIF1<0一定是（

）A．奇函數(shù) B．偶函數(shù) C．增函數(shù) D．減函數(shù)4．（2019·北京·高考真題（文））下列函數(shù)中，在區(qū)間（0，+SKIPIF1<0）上單調(diào)遞增的是A．SKIPIF1<0 B．y=SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2019·浙江·高考真題）已知SKIPIF1<0，函數(shù)SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0，則實數(shù)SKIPIF1<0的最大值是____.第五部分：第五部分：第02講函數(shù)的單調(diào)性與最大（小）值（精練）一、單選題1．（2022·浙江·高三學業(yè)考試）已知函數(shù)SKIPIF1<0在區(qū)間（-∞，1]是減函數(shù)，則實數(shù)a的取值范圍是（

）A．[1，+∞） B．（-∞，1] C．[-1，+∞） D．（-∞，-1]2．（2022·上海·華師大二附中高一期末）已知函數(shù)SKIPIF1<0可表示為SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<01234則下列結論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0的值域是SKIPIF1<0C．SKIPIF1<0的值域是SKIPIF1<0 D．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增3．（2022·安徽蚌埠·高一期末）若函數(shù)SKIPIF1<0在定義域SKIPIF1<0上的值域為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·河南·高二階段練習（理））函數(shù)SKIPIF1<0SKIPIF1<0的最小值是（

）A．3 B．4 C．5 D．65．（2022·浙江杭州·高一期末）已知SKIPIF1<0設SKIPIF1<0SKIPIF1<0，則函數(shù)SKIPIF1<0的最大值是（

）A．SKIPIF1<0 B．1 C．2 D．36．（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0