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

易錯(cuò)點(diǎn)04導(dǎo)數(shù)及其應(yīng)用易錯(cuò)題【01】不會(huì)利用等價(jià)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(diǎn)A的切線方程”與“在點(diǎn)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(diǎn)就是切點(diǎn),“過”只說明切線過這個(gè)點(diǎn),這個(gè)點(diǎn)不一定是切點(diǎn),求曲線過某點(diǎn)的切線方程一般先設(shè)切點(diǎn)把問題轉(zhuǎn)化為在某點(diǎn)處的切線,求過某點(diǎn)的切線條數(shù)一般也是先設(shè)切點(diǎn),把問題轉(zhuǎn)化為關(guān)于切點(diǎn)橫坐標(biāo)的方程實(shí)根個(gè)數(shù)問題.易錯(cuò)題【02】對(duì)極值概念理解不準(zhǔn)確致對(duì)于可導(dǎo)函數(shù)f(x)：x0是極值點(diǎn)的充要條件是在x0點(diǎn)兩側(cè)導(dǎo)數(shù)異號(hào),即f′(x)在方程f′(x)＝0的根x0的左右的符號(hào)：“左正右負(fù)”?f(x)在x0處取極大值；“左負(fù)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(diǎn)的必要而不充分條件．對(duì)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(yàn)“左正右負(fù)”或“左負(fù)右正”,防止產(chǎn)生增根．易錯(cuò)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(hào)進(jìn)行討論,當(dāng)a的符號(hào)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(shí)在根據(jù)SKIPIF1<0的大小進(jìn)行討論.易錯(cuò)題【04】不會(huì)利用隱零點(diǎn)研究函數(shù)的性質(zhì)函數(shù)零點(diǎn)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(diǎn)”；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(diǎn)”.利用導(dǎo)數(shù)求函數(shù)的最值或單調(diào)區(qū)間,常常會(huì)把最值問題轉(zhuǎn)化為求導(dǎo)函數(shù)的零點(diǎn)問題,若導(dǎo)數(shù)零點(diǎn)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(diǎn)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(diǎn)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(huì)把切線條數(shù)有2條,轉(zhuǎn)化為關(guān)于SKIPIF1<0的方程有2個(gè)實(shí)根.【答案】D【問診】設(shè)過點(diǎn)SKIPIF1<0的切線與曲線SKIPIF1<0切于SKIPIF1<0,對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0,所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0,即SKIPIF1<0,由題意可知,點(diǎn)SKIPIF1<0在直線SKIPIF1<0上,所以SKIPIF1<0,過點(diǎn)SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線,則方程SKIPIF1<0有兩個(gè)不同實(shí)根,令SKIPIF1<0,則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增,且SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,此時(shí)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(gè)交點(diǎn)時(shí),當(dāng)SKIPIF1<0時(shí),直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個(gè)交點(diǎn).故選D.【叮囑】過某點(diǎn)的切線條數(shù)一般也是先設(shè)切點(diǎn),把問題轉(zhuǎn)化為關(guān)于切點(diǎn)橫坐標(biāo)的方程實(shí)根個(gè)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(shí)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?yàn)楹瘮?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?yàn)镾KIPIF1<0,所以SKIPIF1<0.2.(2021屆江蘇蘇州市高三月考)若過點(diǎn)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(diǎn)為SKIPIF1<0,其中SKIPIF1<0,因?yàn)镾KIPIF1<0,則SKIPIF1<0,故切線斜率為SKIPIF1<0,所以,曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0,即SKIPIF1<0,將點(diǎn)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(gè)交點(diǎn).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(shí),SKIPIF1<0,此時(shí)函數(shù)SKIPIF1<0單調(diào)遞減,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,此時(shí)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(jià)性,“f′(1)＝0”是“x＝1為f(x)的極值點(diǎn)”的必要不充分條件．【答案】－7【問診】f′(x)＝3x2＋2ax＋b,由x＝1時(shí),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(shí),f′(x)＝3x2＋8x－11＝(3x＋11)(x－1)在x＝1兩側(cè)的符號(hào)相反,符合題意．當(dāng)a＝－3,b＝3時(shí),f′(x)＝3(x－1)2在x＝1兩側(cè)的符號(hào)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(yàn)在SKIPIF1<0左右兩側(cè)函數(shù)值的符號(hào).(2022全國1卷T12)設(shè)SKIPIF1<0,若SKIPIF1<0為函數(shù)SKIPIF1<0的極大值點(diǎn),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(gè)零點(diǎn),當(dāng)SKIPIF1<0時(shí),由三次函數(shù)的性質(zhì)可知,要使SKIPIF1<0是SKIPIF1<0的極大值點(diǎn),則函數(shù)SKIPIF1<0的大致圖象如下圖所示,則SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),由三次函數(shù)的性質(zhì)可知,要使SKIPIF1<0是SKIPIF1<0的極大值點(diǎn),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(shí),SKIPIF1<0,函數(shù)單調(diào)遞增,無極值點(diǎn),舍去.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,在SKIPIF1<0和SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)單調(diào)遞增；在SKIPIF1<0時(shí),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(shí),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(shí),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(shí),SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時(shí),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(diǎn)個(gè)數(shù)及零點(diǎn)大小進(jìn)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(diǎn)SKIPIF1<0為曲線SKIPIF1<0上一點(diǎn),若該曲線在點(diǎn)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(diǎn)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(diǎn)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?yàn)镾KIPIF1<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(shí),因?yàn)榉匠蘏KIPIF1<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?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上有唯一的極值點(diǎn)SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上存在唯一零點(diǎn),即SKIPIF1<0在SKIPIF1<0上存在唯一零點(diǎn),且在該零點(diǎn)兩側(cè)SKIPIF1<0的符號(hào)不一致.當(dāng)SKIPIF1<0時(shí),由(2)知,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0無極值點(diǎn),當(dāng)SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0的稱軸為直線SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上存在唯一零點(diǎn),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(shí),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(shí),SKIPIF1<0,所以SKIPIF1<0.故當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0為減函數(shù)；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0為增函數(shù).所以當(dāng)x=1時(shí),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(shí),SKIPIF1<0,SKIPIF1<0在SKIPIF1<0為減函數(shù)；②當(dāng)a>0時(shí),SKIPIF1<0時(shí)SKIPIF1<0,故SKIPIF1<0為減函數(shù)；SKIPIF1<0時(shí),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(diǎn),求實(shí)數(shù)SKIPIF1<0；(2)若SKIPIF1<0在SKIPIF1<0上恒成立,求實(shí)數(shù)SKIPIF1<0的范圍.【警示】不會(huì)引入隱零點(diǎn)研究函數(shù)單調(diào)性【問診】因?yàn)镾KIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0.即SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),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(shí),SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0.終上所述,SKIPIF1<0時(shí),SKIPIF1<0為SKIPIF1<0的極值點(diǎn)成立,所以SKIPIF1<0.(2)由(1)知SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,SKIPIF1<0,①SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0,②SKIPIF1<0時(shí),因?yàn)镾KIPIF1<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(shí),SKIPIF1<0,與SKIPIF1<0矛盾.綜上：SKIPIF1<0時(shí),SKIPIF1<0在SKIPIF1<0上恒成立.所以實(shí)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(diǎn),若導(dǎo)函數(shù)有零點(diǎn),但無法具體確定,可引入隱零點(diǎn).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(diǎn),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?yàn)镾KIPIF1<0且SKIPIF1<0；又SKIPIF1<0是SKIPIF1<0的極值點(diǎn),則SKIPIF1<0,解得SKIPIF1<0,此時(shí)SKIPIF1<0：當(dāng)SKIPIF1<0時(shí),SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0；∴易知：SKIPIF1<0是SKIPIF1<0的極小值點(diǎn),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(duì)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(shí)SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí),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(diǎn)；(2)當(dāng)SKIPIF1<0時(shí),求SKIPIF1<0在SKIPIF1<0上的零點(diǎn)個(gè)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(diǎn)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(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0.由于SKIPIF1<0,因此當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0是SKIPIF1<0在SKIPIF1<0上唯一的極值點(diǎn).(2)當(dāng)SKIPIF1<0時(shí),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(diǎn)存在性定理可知,SKIPIF1<0在SKIPIF1<0上存在唯一的零點(diǎn),在SKIPIF1<0上無零點(diǎn)；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,因此函數(shù)SKIPIF1<0在SKIPIF1<0上無零點(diǎn)；當(dāng)SKIPIF1<0時(shí),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(duì)SKIPIF1<0恒成立.因此當(dāng)SKIPIF1<0時(shí)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(diǎn),在SKIPIF1<0上無零點(diǎn).綜上所述,函數(shù)SKIPIF1<0在SKIPIF1<0上共有2個(gè)零點(diǎn).錯(cuò)1．若點(diǎn)SKIPIF1<0不在函數(shù)SKIPIF1<0的圖象上,且過點(diǎn)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(diǎn)SKIPIF1<0不在SKIPIF1<0的圖象上,則SKIPIF1<0,所以SKIPIF1<0,而SKIPIF1<0,設(shè)過點(diǎn)SKIPIF1<0的直線與SKIPIF1<0的圖象切于點(diǎn)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(diǎn)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(gè)零點(diǎn),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(gè)零點(diǎn),所以可知SKIPIF1<0與SKIPIF1<0同號(hào),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(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】C【解析】由題意,函數(shù)SKIPIF1<0的定義域SKIPIF1<0,且SKIPIF1<0,因?yàn)楹瘮?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?yàn)镾KIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,可得SKIPIF1<0,所以SKIPIF1<0,即實(shí)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(diǎn)SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以切點(diǎn)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(diǎn),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(shí)SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)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(shí),SKIPIF1<0,即SKIPIF1<0單調(diào)遞增當(dāng)SKIPIF1<0時(shí),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(diǎn),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(shí)SKIPIF1<0,SKIPIF1<0遞減；當(dāng)SKIPIF1<0時(shí)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(diǎn),求a的值；(2)當(dāng)SKIPIF1<0時(shí),求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(3)當(dāng)SKIPIF1<0時(shí),設(shè)函數(shù)SKIPIF1<0,證明：SKIPIF1<0.【解析】(1)SKIPIF1<0,因?yàn)镾KIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn),所以SKIPIF1<0,解得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),檢驗(yàn)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(shí),令SKIPIF1<0,得SKIPIF1<0或SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0恒成立；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),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(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上單調(diào)遞減；(3)證明：當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,設(shè)SKIPIF1<0,因?yàn)镾KIPIF1<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(shí),SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),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(shí),求SKIPIF1<0的單調(diào)區(qū)間；(2)若函數(shù)SKIPIF1<0不存在極值點(diǎn),求證：SKIPIF1<0.【解析】(1)當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0令SKIPIF1<0得：SKIPIF1<0或SKIPIF1<0令SKIPIF1<0得：SKIPIF1<0