新高考數(shù)學(xué)二輪復(fù)習(xí)易錯題專練易錯點04 導(dǎo)數(shù)及其應(yīng)用（含解析）

易錯點04導(dǎo)數(shù)及其應(yīng)用易錯題【01】不會利用等價轉(zhuǎn)化思想及導(dǎo)數(shù)的幾何意義研究曲線的切線求曲線的切線方程一定要注意區(qū)分“過點A的切線方程”與“在點A處的切線方程”的不同．雖只有一字之差,意義完全不同,“在”說明這點就是切點,“過”只說明切線過這個點,這個點不一定是切點,求曲線過某點的切線方程一般先設(shè)切點把問題轉(zhuǎn)化為在某點處的切線,求過某點的切線條數(shù)一般也是先設(shè)切點,把問題轉(zhuǎn)化為關(guān)于切點橫坐標(biāo)的方程實根個數(shù)問題.易錯題【02】對極值概念理解不準(zhǔn)確致對于可導(dǎo)函數(shù)f(x)：x0是極值點的充要條件是在x0點兩側(cè)導(dǎo)數(shù)異號,即f′(x)在方程f′(x)＝0的根x0的左右的符號：“左正右負”?f(x)在x0處取極大值；“左負右正”?f(x)在x0處取極小值,而不僅是f′(x0)＝0.f′(x0)＝0是x0為極值點的必要而不充分條件．對于給出函數(shù)極大(小)值的條件,一定要既考慮f′(x0)＝0,又考慮檢驗“左正右負”或“左負右正”,防止產(chǎn)生增根．易錯題【03】研究含有參數(shù)的函數(shù)單調(diào)性分類標(biāo)準(zhǔn)有誤若函數(shù)的單調(diào)性可轉(zhuǎn)化為解不等式SKIPIF1<0求解此類問題,首先根據(jù)a的符號進行討論,當(dāng)a的符號確定后,再根據(jù)SKIPIF1<0是否在定義域內(nèi)討論,當(dāng)SKIPIF1<0都在定義域內(nèi)時在根據(jù)SKIPIF1<0的大小進行討論.易錯題【04】不會利用隱零點研究函數(shù)的性質(zhì)函數(shù)零點按是否可求精確解可以分為兩類：一類是數(shù)值上能精確求解的,稱之為“顯零點”；另一類是能夠判斷其存在但無法直接表示的,稱之為“隱零點”.利用導(dǎo)數(shù)求函數(shù)的最值或單調(diào)區(qū)間,常常會把最值問題轉(zhuǎn)化為求導(dǎo)函數(shù)的零點問題,若導(dǎo)數(shù)零點存在,但無法求出,我們可以設(shè)其為SKIPIF1<0,再利用導(dǎo)函數(shù)的單調(diào)性確定SKIPIF1<0所在區(qū)間,最后根據(jù)SKIPIF1<0,研究SKIPIF1<0,我們把這類問題稱為隱零點問題.注意若SKIPIF1<0中含有參數(shù)a,關(guān)系式SKIPIF1<0是關(guān)于SKIPIF1<0的關(guān)系式,確定SKIPIF1<0的合適范圍,往往和SKIPIF1<0的范圍有關(guān). 01(2022新高考1卷T7)若過點SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線,則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【警示】不會把切線條數(shù)有2條,轉(zhuǎn)化為關(guān)于SKIPIF1<0的方程有2個實根.【答案】D【問診】設(shè)過點SKIPIF1<0的切線與曲線SKIPIF1<0切于SKIPIF1<0,對函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0,所以曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0,即SKIPIF1<0,由題意可知,點SKIPIF1<0在直線SKIPIF1<0上,所以SKIPIF1<0,過點SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線,則方程SKIPIF1<0有兩個不同實根,令SKIPIF1<0,則SKIPIF1<0.當(dāng)SKIPIF1<0時,SKIPIF1<0,此時函數(shù)SKIPIF1<0單調(diào)遞增,且SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,此時函數(shù)SKIPIF1<0單調(diào)遞減,所以,SKIPIF1<0,如圖所示,當(dāng)直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個交點時,當(dāng)SKIPIF1<0時,直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個交點.故選D.【叮囑】過某點的切線條數(shù)一般也是先設(shè)切點,把問題轉(zhuǎn)化為關(guān)于切點橫坐標(biāo)的方程實根個數(shù)問題.1.(2021屆陜西西安中學(xué)高三期中)若函數(shù)SKIPIF1<0存在平行于SKIPIF1<0軸的切線,則實數(shù)SKIPIF1<0取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為函數(shù)SKIPIF1<0存在平行于SKIPIF1<0軸的切線,所以SKIPIF1<0在SKIPIF1<0上有解,即SKIPIF1<0在SKIPIF1<0上有解,因為SKIPIF1<0,所以SKIPIF1<0.2.(2021屆江蘇蘇州市高三月考)若過點SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線,則()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】設(shè)切點為SKIPIF1<0,其中SKIPIF1<0,因為SKIPIF1<0,則SKIPIF1<0,故切線斜率為SKIPIF1<0,所以,曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0,即SKIPIF1<0,將點SKIPIF1<0的坐標(biāo)代入切線方程可得SKIPIF1<0,設(shè)SKIPIF1<0,則直線SKIPIF1<0與曲線SKIPIF1<0有兩個交點.①若SKIPIF1<0,則SKIPIF1<0,即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,不合乎題意；②若SKIPIF1<0,則SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,此時函數(shù)SKIPIF1<0單調(diào)遞減,當(dāng)SKIPIF1<0時,SKIPIF1<0,此時函數(shù)SKIPIF1<0單調(diào)遞增,所以,SKIPIF1<0.由題意可知SKIPIF1<0,即SKIPIF1<0.故選B. 02已知f(x)＝x3＋ax2＋bx＋a2在x＝1處有極值為10,則a＋b＝________.【警示】忽視了條件的等價性,“f′(1)＝0”是“x＝1為f(x)的極值點”的必要不充分條件．【答案】－7【問診】f′(x)＝3x2＋2ax＋b,由x＝1時,函數(shù)取得極值10,得SKIPIF1<0,解得eq\b\lc\{\rc\(\a\vs4\al\co1(a＝4，,b＝－11，))或eq\b\lc\{\rc\(\a\vs4\al\co1(a＝－3，,b＝3.))當(dāng)a＝4,b＝－11時,f′(x)＝3x2＋8x－11＝(3x＋11)(x－1)在x＝1兩側(cè)的符號相反,符合題意．當(dāng)a＝－3,b＝3時,f′(x)＝3(x－1)2在x＝1兩側(cè)的符號相同,所以a＝－3,b＝3不符合題意,舍去．綜上可知a＝4,b＝－11,∴a＋b＝－7.【叮囑】處理可導(dǎo)函數(shù)SKIPIF1<0在SKIPIF1<0有極值問題,除了保證SKIPIF1<0,還要檢驗在SKIPIF1<0左右兩側(cè)函數(shù)值的符號.(2022全國1卷T12)設(shè)SKIPIF1<0,若SKIPIF1<0為函數(shù)SKIPIF1<0的極大值點,則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】令SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,即SKIPIF1<0及SKIPIF1<0是SKIPIF1<0的兩個零點,當(dāng)SKIPIF1<0時,由三次函數(shù)的性質(zhì)可知,要使SKIPIF1<0是SKIPIF1<0的極大值點,則函數(shù)SKIPIF1<0的大致圖象如下圖所示,則SKIPIF1<0；當(dāng)SKIPIF1<0時,由三次函數(shù)的性質(zhì)可知,要使SKIPIF1<0是SKIPIF1<0的極大值點,則函數(shù)SKIPIF1<0的大致圖象如下圖所示,則SKIPIF1<0；綜上,SKIPIF1<0．故選SKIPIF1<0．2．(2021屆山西長治市高三月考)已知函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值0,則SKIPIF1<0()A．2 B．7 C．2或7 D．3或9【答案】B【解析】SKIPIF1<0,SKIPIF1<0,根據(jù)題意：SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,函數(shù)單調(diào)遞增,無極值點,舍去.當(dāng)SKIPIF1<0時,SKIPIF1<0,在SKIPIF1<0和SKIPIF1<0時,SKIPIF1<0,函數(shù)單調(diào)遞增；在SKIPIF1<0時,SKIPIF1<0,函數(shù)單調(diào)遞減,故函數(shù)在SKIPIF1<0出有極小值,滿足條件.綜上所述：SKIPIF1<0.故選B. 03(2021新高考2卷T22(1))已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性；【警示】討論是分類標(biāo)準(zhǔn)不合理導(dǎo)致解題失誤.【問診】(1)由函數(shù)的解析式可得：SKIPIF1<0,當(dāng)SKIPIF1<0時,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞減,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞增,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞減,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞增,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞減,若SKIPIF1<0,則SKIPIF1<0單調(diào)遞增.【叮囑】此類問題通常根據(jù)導(dǎo)函數(shù)零點個數(shù)及零點大小進行分類討論1.(2021屆河南高三月考)已知函數(shù)SKIPIF1<0SKIPIF1<0(1)已知點SKIPIF1<0為曲線SKIPIF1<0上一點,若該曲線在點SKIPIF1<0處的切線方程為SKIPIF1<0(SKIPIF1<0,SKIPIF1<0),求SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的值；(2)討論函數(shù)SKIPIF1<0的單調(diào)性；(3)若SKIPIF1<0在區(qū)間SKIPIF1<0上有唯一的極值點SKIPIF1<0,求SKIPIF1<0的取值范圍.【解析】(1)SKIPIF1<0,由題意知SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,將點SKIPIF1<0代入方程SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.(2)由題意知函數(shù)的定義域為SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上恒成立,所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時,因為方程SKIPIF1<0的判別式SKIPIF1<0,該方程的兩根分別為SKIPIF1<0,SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0,令SKIPIF1<0,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.(3)由(1)知SKIPIF1<0,令SKIPIF1<0,因為SKIPIF1<0在區(qū)間SKIPIF1<0上有唯一的極值點SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上存在唯一零點,即SKIPIF1<0在SKIPIF1<0上存在唯一零點,且在該零點兩側(cè)SKIPIF1<0的符號不一致.當(dāng)SKIPIF1<0時,由(2)知,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0無極值點,當(dāng)SKIPIF1<0,因為SKIPIF1<0,SKIPIF1<0的稱軸為直線SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上存在唯一零點,必有SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0的取值范圍為SKIPIF1<0.2.(2021屆天津市第二十一中學(xué)高三期中)已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時,求函數(shù)SKIPIF1<0的單調(diào)區(qū)間和極值；(2)討論函數(shù)SKIPIF1<0單調(diào)性．【解析】(1)當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0.故當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0為減函數(shù)；當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0為增函數(shù).所以當(dāng)x=1時,SKIPIF1<0極小值=SKIPIF1<0,無極大值.(2)由SKIPIF1<0可得：SKIPIF1<0.①當(dāng)a≤0時,SKIPIF1<0,SKIPIF1<0在SKIPIF1<0為減函數(shù)；②當(dāng)a>0時,SKIPIF1<0時SKIPIF1<0,故SKIPIF1<0為減函數(shù)；SKIPIF1<0時,SKIPIF1<0,故SKIPIF1<0為增函數(shù). 04(2021屆福建省龍巖高三月考)已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0為SKIPIF1<0的極值點,求實數(shù)SKIPIF1<0；(2)若SKIPIF1<0在SKIPIF1<0上恒成立,求實數(shù)SKIPIF1<0的范圍.【警示】不會引入隱零點研究函數(shù)單調(diào)性【問診】因為SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0.即SKIPIF1<0,當(dāng)SKIPIF1<0時,設(shè)SKIPIF1<0,所以SKIPIF1<0,故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,所以SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0.終上所述,SKIPIF1<0時,SKIPIF1<0為SKIPIF1<0的極值點成立,所以SKIPIF1<0.(2)由(1)知SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,SKIPIF1<0,①SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0,②SKIPIF1<0時,因為SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,SKIPIF1<0；SKIPIF1<0,SKIPIF1<0存在SKIPIF1<0使SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,SKIPIF1<0遞減,當(dāng)SKIPIF1<0時,SKIPIF1<0,與SKIPIF1<0矛盾.綜上：SKIPIF1<0時,SKIPIF1<0在SKIPIF1<0上恒成立.所以實數(shù)SKIPIF1<0的范圍是SKIPIF1<0.【叮囑】求解不等式恒成立或證明不等式一般要利用函數(shù)單調(diào)性,研究函數(shù)單調(diào)性要確定導(dǎo)函數(shù)的零點,若導(dǎo)函數(shù)有零點,但無法具體確定,可引入隱零點.1.(2021屆內(nèi)蒙古海拉爾高三期中)已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0是SKIPIF1<0的極值點,求SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0,求證：SKIPIF1<0【解析】(1)由已知,SKIPIF1<0的定義域為SKIPIF1<0且SKIPIF1<0；又SKIPIF1<0是SKIPIF1<0的極值點,則SKIPIF1<0,解得SKIPIF1<0,此時SKIPIF1<0：當(dāng)SKIPIF1<0時,SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0；∴易知：SKIPIF1<0是SKIPIF1<0的極小值點,且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；(2)若SKIPIF1<0有SKIPIF1<0,設(shè)SKIPIF1<0,SKIPIF1<0；∴SKIPIF1<0；令SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0對任意SKIPIF1<0恒成立,∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；又SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,使得SKIPIF1<0,即SKIPIF1<0,則SKIPIF1<0,即SKIPIF1<0；因此,當(dāng)SKIPIF1<0時SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時,SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0單調(diào)遞減；故SKIPIF1<0,即得證.2.已知SKIPIF1<0,函數(shù)SKIPIF1<0.(1)證明：SKIPIF1<0在SKIPIF1<0上有唯一的極值點；(2)當(dāng)SKIPIF1<0時,求SKIPIF1<0在SKIPIF1<0上的零點個數(shù).【解析】(1)證明：SKIPIF1<0,記SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0.由SKIPIF1<0得SKIPIF1<0在SKIPIF1<0上恒成立,從而SKIPIF1<0在SKIPIF1<0上為增函數(shù),并且SKIPIF1<0,SKIPIF1<0.根據(jù)零點存在性定理可知,存在唯一的SKIPIF1<0使得SKIPIF1<0,并且當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0.由于SKIPIF1<0,因此當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0是SKIPIF1<0在SKIPIF1<0上唯一的極值點.(2)當(dāng)SKIPIF1<0時,SKIPIF1<0,并且根據(jù)(1)知存在SKIPIF1<0使得SKIPIF1<0在SKIPIF1<0上為減函數(shù),在SKIPIF1<0上為增函數(shù).由于SKIPIF1<0,從而SKIPIF1<0.由于SKIPIF1<0,SKIPIF1<0,根據(jù)零點存在性定理可知,SKIPIF1<0在SKIPIF1<0上存在唯一的零點,在SKIPIF1<0上無零點；當(dāng)SKIPIF1<0時,SKIPIF1<0,因此函數(shù)SKIPIF1<0在SKIPIF1<0上無零點；當(dāng)SKIPIF1<0時,記SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上為減函數(shù),所以SKIPIF1<0,即SKIPIF1<0對SKIPIF1<0恒成立.因此當(dāng)SKIPIF1<0時有SKIPIF1<0,因此SKIPIF1<0,結(jié)合SKIPIF1<0知函數(shù)SKIPIF1<0在SKIPIF1<0上存在唯一的零點,在SKIPIF1<0上無零點.綜上所述,函數(shù)SKIPIF1<0在SKIPIF1<0上共有2個零點.錯1．若點SKIPIF1<0不在函數(shù)SKIPIF1<0的圖象上,且過點SKIPIF1<0僅能作一條直線與SKIPIF1<0的圖象相切,則SKIPIF1<0的取值范圍為()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】已知點SKIPIF1<0不在SKIPIF1<0的圖象上,則SKIPIF1<0,所以SKIPIF1<0,而SKIPIF1<0,設(shè)過點SKIPIF1<0的直線與SKIPIF1<0的圖象切于點SKIPIF1<0,則切線的斜率SKIPIF1<0,則SKIPIF1<0,整理得SKIPIF1<0,設(shè)SKIPIF1<0,由于過點SKIPIF1<0僅能作一條直線與SKIPIF1<0的圖象相切,則問題可轉(zhuǎn)化為SKIPIF1<0僅有1個零點,SKIPIF1<0,令SKIPIF1<0,解得：SKIPIF1<0或SKIPIF1<0,令SKIPIF1<0,即SKIPIF1<0,解得：SKIPIF1<0或SKIPIF1<0,令SKIPIF1<0,即SKIPIF1<0,解得：SKIPIF1<0,所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增,在區(qū)間SKIPIF1<0上單調(diào)遞減,可知SKIPIF1<0在區(qū)間SKIPIF1<0或區(qū)間SKIPIF1<0上必有一個零點,所以可知SKIPIF1<0與SKIPIF1<0同號,則SKIPIF1<0,即SKIPIF1<0,解得：SKIPIF1<0或SKIPIF1<0,所以SKIPIF1<0的取值范圍為SKIPIF1<0.故選A.2．(2021屆安徽六安市高三月考)函數(shù)SKIPIF1<0存在與直線SKIPIF1<0平行的切線,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】C【解析】由題意,函數(shù)SKIPIF1<0的定義域SKIPIF1<0,且SKIPIF1<0,因為函數(shù)SKIPIF1<0存在與直線SKIPIF1<0平行的切線,即SKIPIF1<0有解,即SKIPIF1<0在SKIPIF1<0有解,因為SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,可得SKIPIF1<0,所以SKIPIF1<0,即實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選C.3．(2021屆遼寧沈陽市高三月考)若直線SKIPIF1<0與曲線SKIPIF1<0相切,則()A．SKIPIF1<0為定值 B．SKIPIF1<0為定值C．SKIPIF1<0為定值 D．SKIPIF1<0為定值【答案】B【解析】設(shè)直線SKIPIF1<0與曲線SKIPIF1<0切于點SKIPIF1<0,因為SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以切點為SKIPIF1<0,代入直線方程得：SKIPIF1<0,即SKIPIF1<0.故選B.4．(2021屆云南高三月考)已知SKIPIF1<0為函數(shù)SKIPIF1<0的極小值點,則SKIPIF1<0()A．1 B．2 C．3 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0,所以當(dāng)SKIPIF1<0時SKIPIF1<0,當(dāng)SKIPIF1<0時SKIPIF1<0則SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減,故SKIPIF1<0.故選B5．(2021屆河南南陽高三期中)已知函數(shù)SKIPIF1<0在SKIPIF1<0處取得極小值SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0的最大值為()A．2 B．3 C．4 D．6【答案】C【解析】函數(shù)SKIPIF1<0在SKIPIF1<0處取得極小值SKIPIF1<0所以SKIPIF1<0,即SKIPIF1<0,解得：SKIPIF1<0,SKIPIF1<0SKIPIF1<0由SKIPIF1<0得：SKIPIF1<0當(dāng)SKIPIF1<0和SKIPIF1<0時,SKIPIF1<0,即SKIPIF1<0單調(diào)遞增當(dāng)SKIPIF1<0時,SKIPIF1<0,即SKIPIF1<0單調(diào)遞減所以SKIPIF1<0的極大值為SKIPIF1<0,極小值為SKIPIF1<0由SKIPIF1<0得：SKIPIF1<0或SKIPIF1<0由SKIPIF1<0得：SKIPIF1<0或SKIPIF1<0若SKIPIF1<0,SKIPIF1<0,使得SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0,故選C.6．(2021山西太原高三期中)若SKIPIF1<0是函數(shù)SKIPIF1<0的極值點,則函數(shù)()A．有最小值SKIPIF1<0,無最大值 B．有最大值SKIPIF1<0,無最小值C．有最小值SKIPIF1<0,最大值SKIPIF1<0 D．無最大值,無最小值【答案】A【解析】由題設(shè),SKIPIF1<0且SKIPIF1<0,∴SKIPIF1<0,可得SKIPIF1<0.∴SKIPIF1<0且SKIPIF1<0,當(dāng)SKIPIF1<0時SKIPIF1<0,SKIPIF1<0遞減；當(dāng)SKIPIF1<0時SKIPIF1<0,SKIPIF1<0遞增；∴SKIPIF1<0有極小值SKIPIF1<0,無極大值.綜上,有最小值SKIPIF1<0,無最大值.故選A7．(2021北京四中高三期中)設(shè)函數(shù)SKIPIF1<0,其中SKIPIF1<0.(1)若SKIPIF1<0是函數(shù)SKIPIF1<0的極值點,求a的值；(2)當(dāng)SKIPIF1<0時,求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(3)當(dāng)SKIPIF1<0時,設(shè)函數(shù)SKIPIF1<0,證明：SKIPIF1<0.【解析】(1)SKIPIF1<0,因為SKIPIF1<0是函數(shù)SKIPIF1<0的極值點,所以SKIPIF1<0,解得SKIPIF1<0,當(dāng)SKIPIF1<0時,檢驗符合題意,所以a的值為SKIPIF1<0；(2)SKIPIF1<0,SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0時,令SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0恒成立；當(dāng)SKIPIF1<0時,令SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0；綜上,當(dāng)SKIPIF1<0時,SKIPIF1<0在SKIPIF1<0和SKIPIF1<0單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時,SKIPIF1<0在SKIPIF1<0和SKIPIF1<0單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減；(3)證明：當(dāng)SKIPIF1<0時,SKIPIF1<0,設(shè)SKIPIF1<0,因為SKIPIF1<0,SKIPIF1<0,所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,又SKIPIF1<0,所以存在SKIPIF1<0,使SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0；當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,所以函數(shù)SKIPIF1<0的最小值為SKIPIF1<0,所以SKIPIF1<0,從而得證SKIPIF1<0.8．(2021河南南陽高三期中)已知函數(shù)SKIPIF1<0,SKIPIF1<0.(1)當(dāng)SKIPIF1<0時,求SKIPIF1<0的單調(diào)區(qū)間；(2)若函數(shù)SKIPIF1<0不存在極值點,求證：SKIPIF1<0.【解析】(1)當(dāng)SKIPIF1<0時,SKIPIF1<0,則SKIPIF1<0令SKIPIF1<0得：SKIPIF1<0或SKIPIF1<0令SKIPIF1<0得：SKIPIF1<0