新高考數(shù)學三輪沖刺 押題卷練習第8題 函數(shù)的綜合應(yīng)用（原卷版）

函數(shù)的綜合應(yīng)用考點4年考題考情分析函數(shù)的綜合應(yīng)用2023年新高考Ⅰ卷第11題2023年新高考Ⅱ卷第11題2022年新高考Ⅰ卷第7、10、12題函數(shù)的綜合會以單選題、多選題、填空題、解答題4類題型進行考查，通常伴隨著導數(shù)的考查，在單選題中難度較難，縱觀近幾年的新高考試題，分別以導數(shù)為背景命題考查極值點、零點、函數(shù)值大小比較、函數(shù)的基本性質(zhì)、最值及切線方程等知識點，本內(nèi)容也是新高考沖刺復習的重點復習內(nèi)容?？梢灶A(yù)測2024年新高考命題方向?qū)⒗^續(xù)以導數(shù)綜合應(yīng)用問題展開命題．1．（2023·新高考Ⅰ卷高考真題第11題）已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，則（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0是偶函數(shù) D．SKIPIF1<0為SKIPIF1<0的極小值點2．（2023·新高考Ⅱ卷高考真題第11題）若函數(shù)SKIPIF1<0既有極大值也有極小值，則（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·新高考Ⅰ卷高考真題第7題）設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·新高考Ⅰ卷高考真題第10題）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0有兩個極值點 B．SKIPIF1<0有三個零點C．點SKIPIF1<0是曲線SKIPIF1<0的對稱中心 D．直線SKIPIF1<0是曲線SKIPIF1<0的切線5．（2022·新高考Ⅰ卷高考真題第12題）已知函數(shù)SKIPIF1<0及其導函數(shù)SKIPIF1<0的定義域均為SKIPIF1<0，記SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0均為偶函數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0在定義域內(nèi)，若SKIPIF1<0，其中SKIPIF1<0為奇函數(shù)，SKIPIF1<0為常數(shù)，則最大值SKIPIF1<0，最小值SKIPIF1<0有SKIPIF1<0即SKIPIF1<0倍常數(shù)在定義域內(nèi)，若SKIPIF1<0，其中SKIPIF1<0為奇函數(shù)，SKIPIF1<0為常數(shù)，有SKIPIF1<0即SKIPIF1<0倍常數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0常見函數(shù)的泰勒展開式：結(jié)論1SKIPIF1<0．結(jié)論2SKIPIF1<0．結(jié)論3SKIPIF1<0（SKIPIF1<0）．結(jié)論4SKIPIF1<0．結(jié)論5SKIPIF1<0；SKIPIF1<0；SKIPIF1<0．結(jié)論6SKIPIF1<0；結(jié)論7SKIPIF1<0結(jié)論8SKIPIF1<0．結(jié)論9SKIPIF1<0．放縮程度綜合SKIPIF1<0SKIPIF1<0SKIPIF1<0端點效應(yīng)的類型1.如果函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上,SKIPIF1<0恒成立,則SKIPIF1<0或SKIPIF1<0.2.如果函數(shù)SKIPIF1<0在區(qū)問SKIPIF1<0上,SKIPIF1<0恒成立,且SKIPIF1<0(或SKIPIF1<0),則SKIPIF1<0SKIPIF1<0或SKIPIF1<0.3.如果函數(shù)SKIPIF1<0在區(qū)問SKIPIF1<0上,SKIPIF1<0恒成立,且SKIPIF1<0(或SKIPIF1<0,SKIPIF1<0則SKIPIF1<0或SKIPIF1<0.函數(shù)的凹凸性凹函數(shù)：對于某區(qū)間內(nèi)SKIPIF1<0,都有SKIPIF1<0.凸函數(shù)：對于某區(qū)間內(nèi)SKIPIF1<0,都有SKIPIF1<0.1．（2024·陜西·模擬預(yù)測）設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2024·浙江溫州·二模）已知SKIPIF1<0，則SKIPIF1<0的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2024·廣東佛山·二模）若函數(shù)SKIPIF1<0（SKIPIF1<0）既有極大值也有極小值，則下列結(jié)論一定正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2024·全國·模擬預(yù)測）若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小順序為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2024·全國·模擬預(yù)測）若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2024·遼寧大連·一模）設(shè)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<07．（2024·江蘇·一模）用SKIPIF1<0表示x，y中的最小數(shù)．已知函數(shù)SKIPIF1<0，則SKIPIF1<0的最大值為(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．ln28．（2024·云南·模擬預(yù)測）已知函數(shù)SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0有實數(shù)解，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2024·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0恰有一個零點SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2024·湖南邵陽·二模）已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0為SKIPIF1<0的導函數(shù).若SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上恒成立，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<011．（2024·全國·模擬預(yù)測）設(shè)函數(shù)SKIPIF1<0，記SKIPIF1<0的極小值點為SKIPIF1<0，極大值點為SKIPIF1<0，則SKIPIF1<0（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．（2024·遼寧·模擬預(yù)測）已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，SKIPIF1<0也是定義在SKIPIF1<0上的奇函數(shù)，則關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<013．（2024·全國·模擬預(yù)測）若函數(shù)SKIPIF1<0有兩個零點，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<014．（2024·河南鄭州·模擬預(yù)測）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<015．（2024·浙江·二模）已知函數(shù)SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<016．（2024·山東濟南·一模）若不等式SKIPIF1<0對任意的SKIPIF1<0恒成立，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<017．（2024·福建漳州·一模）已知可導函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0為奇函數(shù)，設(shè)SKIPIF1<0是SKIPIF1<0的導函數(shù)，若SKIPIF1<0為奇函數(shù)，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<018．（2024·湖南邵陽·一模）設(shè)SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<019．（2024·湖南長沙·一模）已知實數(shù)SKIPIF1<0分別滿足SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<020．（2024·貴州貴陽·一模）已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0也是偶函數(shù)，若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<021．（2024·黑龍江哈爾濱·一模）設(shè)SKIPIF1<0且SKIPIF1<0，若函數(shù)SKIPIF1<0有三個極值點，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<022．（2024·遼寧·一模）已知函數(shù)SKIPIF1<0，若SKIPIF1<0成立，則實數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<023．（2024·遼寧·一模）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有五個不等的實數(shù)解，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<024．（2024·全國·模擬預(yù)測）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0內(nèi)有解，則正實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<025．（2024·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0是函數(shù)SKIPIF1<0的導函數(shù)，若不等式SKIPIF1<0對任意的SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<026．（2024·遼寧·二模）若SKIPIF1<0，則（

）A．SKIPIF1<0