新高考數(shù)學(xué)一輪復(fù)習(xí)講與練第08講 一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（二）（練）（解析版）

第02講一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（二）1．已知點(diǎn)SKIPIF1<0在曲線SKIPIF1<0上，則曲線在點(diǎn)SKIPIF1<0處的切線方程為_________.【答案】SKIPIF1<0【詳解】因?yàn)辄c(diǎn)SKIPIF1<0在曲線SKIPIF1<0上，SKIPIF1<0，可得SKIPIF1<0，所以，SKIPIF1<0，對函數(shù)求導(dǎo)得SKIPIF1<0，則曲線在點(diǎn)SKIPIF1<0處的切線斜率為SKIPIF1<0，因此，曲線在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.2．過曲線SKIPIF1<0上一點(diǎn)SKIPIF1<0且與曲線在點(diǎn)SKIPIF1<0處的切線垂直的直線的方程為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：∵SKIPIF1<0，∴SKIPIF1<0，曲線在點(diǎn)SKIPIF1<0處的切線斜率是SKIPIF1<0，∴過點(diǎn)SKIPIF1<0且與曲線在點(diǎn)SKIPIF1<0處的切線垂直的直線的斜率為SKIPIF1<0，∴所求直線方程為SKIPIF1<0，即SKIPIF1<0．故選：A.3．已知曲線SKIPIF1<0與直線SKIPIF1<0相切，則實(shí)數(shù)a的值為__________.【答案】2【解答】解：設(shè)切點(diǎn)為SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，則由題意得，SKIPIF1<0，解得SKIPIF1<0，故答案為：24．若曲線SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，則SKIPIF1<0__________【答案】SKIPIF1<0解：將SKIPIF1<0代入SKIPIF1<0，得切點(diǎn)為SKIPIF1<0，SKIPIF1<0SKIPIF1<0①，又SKIPIF1<0，SKIPIF1<0SKIPIF1<0，SKIPIF1<0②.聯(lián)立①②解得：SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.故答案為：SKIPIF1<05、過點(diǎn)SKIPIF1<0作曲線SKIPIF1<0（SKIPIF1<0）的切線，則切點(diǎn)坐標(biāo)為________．【答案】SKIPIF1<0【詳解】由SKIPIF1<0（SKIPIF1<0），則SKIPIF1<0，化簡得SKIPIF1<0，則SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，顯然SKIPIF1<0不在曲線上，則SKIPIF1<0，得SKIPIF1<0，則切點(diǎn)坐標(biāo)為SKIPIF1<0.故答案為：SKIPIF1<0.6．已知函數(shù)SKIPIF1<0存在單調(diào)遞減區(qū)間，且SKIPIF1<0的圖象在SKIPIF1<0處的切線l與曲線SKIPIF1<0相切，符合情況的切線l（）A．有3條 B．有2條 C．有1條 D．不存在【答案】D【解析】試題分析：SKIPIF1<0，依題意，SKIPIF1<0在SKIPIF1<0上有解.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上無解，不符合題意；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0符合題意，故SKIPIF1<0.易知曲線SKIPIF1<0在SKIPIF1<0處的切線為SKIPIF1<0.假設(shè)該直線與SKIPIF1<0相切，設(shè)切點(diǎn)為SKIPIF1<0，即有SKIPIF1<0，消去SKIPIF1<0化簡得SKIPIF1<0，分別畫出SKIPIF1<0的圖像，觀察可知它們交點(diǎn)橫坐標(biāo)SKIPIF1<0，SKIPIF1<0，這與SKIPIF1<0矛盾，故不存在.7．曲線SKIPIF1<0與曲線SKIPIF1<0有（）條公切線.A．1 B．2 C．3 D．4【答案】B【詳解】設(shè)SKIPIF1<0是曲線SKIPIF1<0圖像上任意一點(diǎn)，SKIPIF1<0，所以SKIPIF1<0，所以過點(diǎn)SKIPIF1<0的切線方程為SKIPIF1<0，整理得SKIPIF1<0①.令SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，所以曲線SKIPIF1<0上過點(diǎn)SKIPIF1<0的切線方程為：SKIPIF1<0，整理得SKIPIF1<0②.由于切線①②重合，故SKIPIF1<0，即SKIPIF1<0③.構(gòu)造函數(shù)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)SKIPIF1<0遞減、當(dāng)SKIPIF1<0時(shí)SKIPIF1<0遞增，注意到當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，且SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0遞增，而SKIPIF1<0，根據(jù)零點(diǎn)存在性定理可知在區(qū)間SKIPIF1<0各存在SKIPIF1<0的一個(gè)零點(diǎn)，也即SKIPIF1<0有兩個(gè)零點(diǎn)，也即方程③有兩個(gè)根，也即曲線SKIPIF1<0和曲線SKIPIF1<0有兩條公切線.故選：B8．已知點(diǎn)M在函數(shù)SKIPIF1<0圖象上，點(diǎn)N在函數(shù)SKIPIF1<0圖象上，則SKIPIF1<0的最小值為（）A．1 B．SKIPIF1<0 C．2 D．3【答案】B【分析】根據(jù)函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0互為反函數(shù)，將問題轉(zhuǎn)化為求函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0平行的切線的切點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離的兩倍，利用導(dǎo)數(shù)求出切點(diǎn)坐標(biāo)，根據(jù)點(diǎn)到直線的距離公式可得結(jié)果.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0與函數(shù)SKIPIF1<0互為反函數(shù)，它們的圖象關(guān)于直線SKIPIF1<0對稱，所以SKIPIF1<0的最小值為函數(shù)SKIPIF1<0的圖象上的點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離的2倍，即為函數(shù)SKIPIF1<0的圖象與直線SKIPIF1<0平行的切線的切點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離的兩倍，因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0的圖象上與直線SKIPIF1<0平行的切線的斜率SKIPIF1<0，所以SKIPIF1<0，所以切點(diǎn)為SKIPIF1<0，它到直線SKIPIF1<0的距離SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0.故選：B.9．設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0取得最小值SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的最小值為___________.【答案】10【詳解】解：SKIPIF1<0表示點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0距離的平方，而點(diǎn)SKIPIF1<0是直線SKIPIF1<0上任一點(diǎn)，點(diǎn)SKIPIF1<0是反比例函數(shù)SKIPIF1<0在第四象限上的點(diǎn)，當(dāng)SKIPIF1<0是斜率為SKIPIF1<0的直線與SKIPIF1<0相切的切點(diǎn)時(shí)，點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離即為SKIPIF1<0的最小值，由SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0取等號，所以函數(shù)SKIPIF1<0的最小值為10，故答案為：1010．若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與曲線SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則SKIPIF1<0__________.【答案】SKIPIF1<0【詳解】SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，可得曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，可得曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的方程為SKIPIF1<0，由兩條切線重合的條件，可得SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，即有SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0.故答案為:SKIPIF1<01．已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的傾斜角為SKIPIF1<0，則SKIPIF1<0的值為（）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B解:函數(shù)SKIPIF1<0的導(dǎo)數(shù)SKIPIF1<0,SKIPIF1<0函數(shù)f(x)在x=1處的傾斜角為SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0故選B.2．曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，則切點(diǎn)SKIPIF1<0的坐標(biāo)是____________.【答案】SKIPIF1<0【詳解】由函數(shù)SKIPIF1<0，則SKIPIF1<0，設(shè)切點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，則斜率SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，切點(diǎn)為SKIPIF1<0，此時(shí)切線方程為SKIPIF1<0；當(dāng)SKIPIF1<0，切點(diǎn)為SKIPIF1<0，不滿足題意，綜上可得，切點(diǎn)為SKIPIF1<0.故答案為：SKIPIF1<0.3．已知SKIPIF1<0軸為曲線SKIPIF1<0的切線，則SKIPIF1<0的值為________.【答案】SKIPIF1<0【詳解】由題意SKIPIF1<0，設(shè)SKIPIF1<0軸與曲線SKIPIF1<0的切點(diǎn)為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.4．已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，則（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0SKIPIF1<0，SKIPIF1<0將SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0，故選D．5．已知直線SKIPIF1<0是曲線SKIPIF1<0的切線，則實(shí)數(shù)SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】設(shè)切點(diǎn)為SKIPIF1<0，∴切線方程是SKIPIF1<0，∴SKIPIF1<0，故答案為：C6．已知函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，曲線SKIPIF1<0在點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0處的切線總是平行時(shí)，則由點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的切線的條數(shù)為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．無法確定【答案】C【解析】詳解：由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0曲線SKIPIF1<0在點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0處的切線總是平行，SKIPIF1<0關(guān)于SKIPIF1<0對稱，即SKIPIF1<0，點(diǎn)SKIPIF1<0，即為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0切線的方程為SKIPIF1<0，將點(diǎn)SKIPIF1<0代入切線方程可得SKIPIF1<0，化為SKIPIF1<0，設(shè)SKIPIF1<0SKIPIF1<0令SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，SKIPIF1<0在SKIPIF1<0處有極大值，在SKIPIF1<0處有極小值，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0與SKIPIF1<0有三個(gè)交點(diǎn)，SKIPIF1<0方程SKIPIF1<0有三個(gè)根，即過SKIPIF1<0的切線有SKIPIF1<0條，故答案為SKIPIF1<0.7．若函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0有公切線，則實(shí)數(shù)SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】【詳解】解：SKIPIF1<0，設(shè)公切線與曲線SKIPIF1<0相切的切點(diǎn)為SKIPIF1<0，

則公共切線為SKIPIF1<0，即SKIPIF1<0，其與SKIPIF1<0相切，聯(lián)立消去SKIPIF1<0得：SKIPIF1<0，則SKIPIF1<0有解，即SKIPIF1<0有解，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，則SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的最小值為SKIPIF1<0.故選：A.8．拋物線上的一動(dòng)點(diǎn)到直線距離的最小值是A． B． C． D．【答案】A【詳解】試題分析：對y=x2求導(dǎo)可求與直線x-y-1=0平行且與拋物線y=x2相切的切線方程，然后利用兩平行線的距離公司可得所求的最小距離d．解：（法一）對y=x2求導(dǎo)可得y′=2x，令y′=2x=1可得x=∴與直線x-y-1=0平行且與拋物線y=x2相切的切點(diǎn)（，），切線方程為y-=x-即x-y-=0由兩平行線的距離公司可得所求的最小距離d=，故選A.9．已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為______．【答案】SKIPIF1<0【詳解】SKIPIF1<0可看成點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離，而點(diǎn)SKIPIF1<0的軌跡是直線SKIPIF1<0，點(diǎn)SKIPIF1<0的軌跡是曲線SKIPIF1<0，則所求最小值可轉(zhuǎn)化為曲線SKIPIF1<0上的點(diǎn)到直線SKIPIF1<0距離的最小值，而曲線SKIPIF1<0在直線SKIPIF1<0上方，平移直線SKIPIF1<0使其與曲線SKIPIF1<0相切，則切點(diǎn)到直線SKIPIF1<0距離即為所求，設(shè)切點(diǎn)SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，切點(diǎn)為SKIPIF1<0則SKIPIF1<0到直線SKIPIF1<0距離SKIPIF1<0.故答案為：SKIPIF1<010．已知函數(shù)SKIPIF1<0，SKIPIF1<0，若存在SKIPIF1<0使得SKIPIF1<0，則實(shí)數(shù)a的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0，所以，SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0在SKIPIF1<0有交點(diǎn)，分情況討論：①直線SKIPIF1<0過點(diǎn)SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0；②直線SKIPIF1<0與SKIPIF1<0相切，設(shè)切點(diǎn)為SKIPIF1<0，得SKIPIF1<0SKIPIF1<0SKIPIF1<0，切點(diǎn)為SKIPIF1<0，故實(shí)數(shù)a的取值范圍是SKIPIF1<0故選：B11．關(guān)于SKIPIF1<0的方程SKIPIF1<0在SKIPIF1<0內(nèi)有且僅有SKIPIF1<0個(gè)根，設(shè)最大的根是SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0的大小關(guān)系是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．以上都不對【答案】C【詳解】由題意作出SKIPIF1<0與SKIPIF1<0在SKIPIF1<0的圖象，如圖所示：∵方程SKIPIF1<0在SKIPIF1<0內(nèi)有且僅有5個(gè)根，最大的根是SKIPIF1<0.∴SKIPIF1<0必是SKIPIF1<0與SKIPIF1<0在SKIPIF1<0內(nèi)相切時(shí)切點(diǎn)的橫坐標(biāo)設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，斜率SKIPIF1<0則SKIPIF1<0故選C.1．（2019·全國·高考真題（文））曲線y=2sinx+cosx在點(diǎn)(π，–1)處的切線方程為A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即點(diǎn)SKIPIF1<0在曲線SKIPIF1<0上．SKIPIF1<0SKIPIF1<0則SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0．故選C．2．（2016·四川·高考真題（文））設(shè)直線l1，l2分別是函數(shù)f(x)=SKIPIF1<0圖象上點(diǎn)P1，P-2處的切線，l1與l2垂直相交于點(diǎn)P，且l1，l2分別與y軸相交于點(diǎn)A，B，則△PAB的面積的取值范圍是A．(0,1) B．(0,2) C．(0,+∞) D．(1,+∞)【答案】A【詳解】試題分析：設(shè)SKIPIF1<0（不妨設(shè)SKIPIF1<0），則由導(dǎo)數(shù)的幾何意義易得切線SKIPIF1<0的斜率分別為SKIPIF1<0由已知得SKIPIF1<0切線SKIPIF1<0的方程分別為SKIPIF1<0，切線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0．分別令SKIPIF1<0得SKIPIF1<0又SKIPIF1<0與SKIPIF1<0的交點(diǎn)為SKIPIF1<0，故選A．3．（2022·浙江·高考真題）設(shè)函數(shù)SKIPIF1<0．(1)求SKIPIF1<0的單調(diào)區(qū)間；(2)已知SKIPIF1<0，曲線SKIPIF1<0上不同的三點(diǎn)SKIPIF1<0處的切線都經(jīng)過點(diǎn)SKIPIF1<0．證明：（?。┤鬝KIPIF1<0，則SKIPIF1<0；（ⅱ）若SKIPIF1<0，則SKIPIF1<0．（注：SKIPIF1<0是自然對數(shù)的底數(shù)）【答案】(1)SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0.(2)（?。┮娊馕?；（ⅱ）見解析.【解答】（1）SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的減區(qū)間為SKIPIF1<0，SKIPIF1<0的增區(qū)間為SKIPIF1<0.（2）（?。┮?yàn)檫^SKIPIF1<0有三條不同的切線，設(shè)切點(diǎn)為SKIPIF1<0，故SKIPIF1<0，故方程SKIPIF1<0有3個(gè)不同的根，該方程可整理為SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為減函數(shù)，在SKIPIF1<0上為增函數(shù)，因?yàn)镾KIPIF1<0有3個(gè)不同的零點(diǎn)，故SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0且SKIPIF1<0，整理得到：SKIPIF1<0且SKIPIF1<0，此時(shí)SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0為SKIPIF1<0上的減函數(shù)，故SKIPIF1<0，故SKIPIF1<0.（ⅱ）當(dāng)SKIPIF1<0時(shí)，同（ⅰ）中討論可得：故SKIPIF1<0在SKIPIF1<0上為減函數(shù)，在SKIPIF1<0上為增函數(shù)，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0有3個(gè)不同的零點(diǎn)，故SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0且SKIPIF1<0，整理得到：SKIPIF1<0，因?yàn)镾KIPIF1<0，故SKIPIF1<0，又SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，則方程SKIPIF1<0即為：SKIPIF1<0即為SKIPIF1<0，記SKIPIF1<0則SKIPIF1<0為SKIPIF1<0有三個(gè)不同的根，設(shè)SKIPIF1<0，SKIPIF1<0，要證：SKIPIF1<0，即證SKIPIF1<0，即證：SKIPIF1<0，即證：SKIPIF1<0，即證：SKIPIF1<0，而SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故即證：SKIPIF1<0，即證：SKIPIF1<0即證：SKIPIF1<0，記SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0即SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為增函數(shù)，故SKIPIF1<0，所以SKIPIF1<0，記SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0為增函數(shù)，故SKIPIF1<0，故SKIPIF1<0即SKIPIF1<0，故原不等式得證：4．（2021·全國·高考真題（理））已知拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，且SKIPIF1<0與圓SKIPIF1<0上點(diǎn)的距離的最小值為SKIPIF1<0．（1）求SKIPIF1<0；（2）若點(diǎn)SKIPIF1<0在SKIPIF1<0上，SKIPIF1<0是SKIPIF1<0的兩條切線，SKIPIF1<0是切點(diǎn)，求SKIPIF1<0面積的最大值．【答案】（1）SKIPIF1<0；（2）SKIPIF1<0.【詳解】（1）[方法一]：利用二次函數(shù)性質(zhì)求最小值由題意知，SKIPIF1<0，設(shè)圓M上的點(diǎn)SKIPIF1<0，則SKIPIF1<0．所以SKIPIF1<0．從而有SKIPIF1<0SKIPIF1<0．因?yàn)镾KIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．又SKIPIF1<0，解之得SKIPIF1<0，因此SKIPIF1<0．[方法二]【最優(yōu)解】：利用圓的幾何意義求最小值拋物線SKIPIF1<0的焦點(diǎn)為SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0與圓SKIPIF1<0上點(diǎn)的距離的最小值為SKIPIF1<0，解得SKIPIF1<0；（2）[方法一]：切點(diǎn)弦方程+韋達(dá)定義判別式求弦長求面積法拋物線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，對該函數(shù)求導(dǎo)得SKIPIF1<0，設(shè)點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，同理可知，直線SKIPIF1<0的方程為SKIPIF1<0，由于點(diǎn)SKIPIF1<0為這兩條直線的公共點(diǎn)，則SKIPIF1<0，所以，點(diǎn)A、SKIPIF1<0的坐標(biāo)滿足方程SKIPIF1<0，所以，直線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，可得SKIPIF1<0，由韋達(dá)定理可得SKIPIF1<0，SKIPIF1<0，所以，SKIPIF1<0，點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，所以，SKIPIF1<0，SKIPIF1<0，由已知可得SKIPIF1<0，所以，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的面積取最大值SKIPIF1<0.[方法二]【最優(yōu)解】：切點(diǎn)弦法+分割轉(zhuǎn)化求面積+三角換元求最值同方法一得到SKIPIF1<0．過P作y軸的平行線交SKIPIF1<0于Q，則SKIPIF1<0．SKIPIF1<0．P點(diǎn)在圓M上，則SKIPIF1<0SKIPIF1<0．故當(dāng)SKIPIF1<0時(shí)SKIPIF1<0的面積最大，最大值為SKIPIF1<0．[方法三]：直接設(shè)直線AB方程法設(shè)切點(diǎn)A，B的坐標(biāo)分別為SKIPIF1<0，SKIPIF1<0．設(shè)SKIPIF1<0，聯(lián)立SKIPIF1<0和拋物線C的方程得SKIPIF1<0整理得SKIPIF1<0．判別式SKIPIF1<0，即SKIPIF1<0，且SKIPIF1<0．拋物線C的方程為SKIPIF1<0，即SKIPIF1<0，有SKIPIF1<0．則SKIPIF1<0，整理得SKIPIF1<0，同理可得SKIPIF1<0．聯(lián)立方程SKIPIF1<0可得點(diǎn)P的坐標(biāo)為SKIPIF1<0，即SKIPIF1<0．將點(diǎn)P的坐標(biāo)代入圓M的方程，得SKIPIF1<0，整理得SKIPIF1<0．由弦長公式得SKIPIF1<0SKIPIF1<0．點(diǎn)P到直線SKIPIF1<0的距離為SKIPIF1<0．所以SKIPIF1<0SKIPIF1<0SKIPIF1<0，其中SKIPIF1<0，即SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．5．（2017·山東·高考真題（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0是自然對數(shù)的底數(shù).（Ⅰ）求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；（Ⅱ）令SKIPIF1<0，討論SKIPIF1<0的單調(diào)性并判斷有無極值，有極值時(shí)求出極值.【答案】（Ⅰ）SKIPIF1<0（Ⅱ）見解析【詳解】（Ⅰ）由題意SKIPIF1<0又SKIPIF1<0，所以SKIPIF1<0，因此

曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即

SKIPIF1<0.（Ⅱ）由題意得

SKIPIF1<0，因?yàn)镾KIPIF1<0SKIPIF1<0SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增.因?yàn)镾KIPIF1<0所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<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，所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取得極小值，極小值是SKIPIF1<0；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0由SKIPIF1<0得SKIPIF1<0，SKIPIF1<0①當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；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.所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取得極大值.極大值為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取到極小值，極小值是SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，無極值；③當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；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；所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取得極大值，極大值是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0取得極小值.極小值是SKIPIF1<0.綜上所述：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，函數(shù)SKIPIF1<0有極小值，極小值是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，函數(shù)SKIPIF1<0有極大值，也有極小值，極大值是SKIPIF1<0極小值是SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，無極值；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，函數(shù)SKIPIF1<0有極大值，也有極小值，極大值是SKIPIF1<0；極小值是SKIPIF1<0.6．（2019·全國·高考真題（理））已知函數(shù)SKIPIF1<0.（1）討論f(x)的單調(diào)性，并證明f(x)有且僅有兩個(gè)零點(diǎn)；（2）設(shè)x0是f(x)的一個(gè)零點(diǎn)，證明曲線y=lnx在點(diǎn)A(x0，lnx0)處的切線也是曲線SKIPIF1<0的切線.【答案】（1）函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上是單調(diào)增函數(shù)，證明見解析；（2）證明見解析.【詳解】（1）函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上是單調(diào)增函數(shù)；當(dāng)SKIPIF1<0，時(shí)，SKIPIF1<0，而SKIPIF1<0，顯然當(dāng)SKIPIF1<0，函數(shù)SKIPIF1<0有零點(diǎn)，而函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有唯一的零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0必有一零點(diǎn)，而函數(shù)SKIPIF1<0在SKIPIF1<0上是單調(diào)遞增，故當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0有唯一的零點(diǎn)綜上所述，函數(shù)SKIPIF1<0的定義域SKIPIF1<0內(nèi)有2個(gè)零點(diǎn)；（2）因?yàn)镾KIPIF1<0是SKIPIF1<0的一個(gè)零點(diǎn)，所以SKIPIF1<0SKIPIF1<0，所以曲線SKIPIF1<0在SKIPIF1<0處的切線SKIPIF1<0的斜率SKIPIF1<0，故曲線SKIPIF1<0在SKIPIF1<0處的切線SKIPIF1<0的方程為：SKIPIF1<0而SKIPIF1<0，所以SKIPIF1<0的方程為SKIPIF1<0，它在縱軸的截距為SKIPIF1<0.設(shè)曲線SKIPIF1<0的切點(diǎn)為SKIPIF1<0，過切點(diǎn)為SKIPIF1<0切線SKIPIF1<0，SKIPIF1<0，所以在SKIPIF1<0處的切線SKIPIF1<0的斜率為SKIPIF1<0，因此切線SKIPIF1<0的方程為SKIPIF1<0，當(dāng)切線SKIPIF1<0的斜率SKIPIF1<0等于直線SKIPIF1<0的斜率SKIPIF1<0時(shí)，即SKIPIF1<0，切線SKIPIF1<0在縱軸的截距為SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，直線SKIPIF1<0的斜率相等，在縱軸上的截距也相等，因此直線SKIPIF1<0重合，故曲線SKIPIF1<0在SKIPIF1<0處的切線也是曲線SKIPIF1<0的切線.7．（2010·湖北·高考真題（文））設(shè)函數(shù)SKIPIF1<0，其中a＞0，曲線SKIPIF1<0在點(diǎn)P（0，SKIPIF1<0）處的切線方程為y=1（Ⅰ）確定b、c的值（Ⅱ）設(shè)曲線SKIPIF1<0在點(diǎn)（SKIPIF1<0）及（SKIPIF1<0）處的切線都過點(diǎn)（0,2）證明：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（Ⅲ）若過點(diǎn)（0,2）可作曲線SKIPIF1<0的三條不同切線，求a的取值范圍．【答案】（Ⅰ）b=0，c=1,（Ⅱ）見解析（Ⅲ）SKIPIF1<0．【詳解】（1）∵f（x）SKIPIF1<0x3SKIPIF1<0x2+bx+c，∴f（0）＝c，f′（x）＝x2﹣ax+b，f′（0）＝b；又∵y＝f（x）在點(diǎn)P（0，f（0））處的切線方程為y＝1，∴f（0）＝1，f′（0）＝0．∴b＝0，c＝1．（2）∵b＝0，c＝1時(shí)，SKIPIF1<0，f'（x）＝x2﹣ax．由于點(diǎn)（t，f（t））處的切線方程為y﹣f（t）＝f'（t）（x﹣t），而點(diǎn)（0，2）在切線上，∴2﹣f（t）＝f'（t）（﹣t），化簡得SKIPIF1<0，即t滿足的方程為SKIPIF1<0．下面用反證法證明．假設(shè)f'（x1）＝f'（x2），由于曲線y＝f（x）在點(diǎn)（x1，f（x1））及（x2，f（x2））處的切線都過點(diǎn)（0，2），則下列等式成立：SKIPIF1<0；由③得x1+x2＝a，由①﹣②得SKIPIF1<0x1x2SKIPIF1<0a2④；又SKIPIF1<0x1x2SKIPIF1<0x1x2＝a2﹣x1（a﹣x1）SKIPIF1<0ax1+a2SKIPIF1<0a2SKIPIF1<0a2∴由④得x1SKIPIF1<0，此時(shí)x2SKIPIF1<0，這與x1≠x2矛盾，∴f′（x1）≠f′（x2）．（3）由（2）知，過點(diǎn)（0，2）可作y＝f（x）的三條切線，等價(jià)于方程2﹣f（t）＝f'（t）（0﹣t）有三個(gè)相異的實(shí)根，即等價(jià)于方程SKIPIF1<0有三個(gè)相異的實(shí)根；設(shè)g（t）SKIPIF1<0t3SKIPIF1<0t2+1，∴g′（t）＝2t2﹣at＝2t（tSKIPIF1<0）；∵a＞0，∴有t（﹣∞，0）0SKIPIF1<0SKIPIF1<0SKIPIF1<0g'（t）+0﹣0+g（t）↗極大值1↘極小值SKIPIF1<0↗由g（t）的單調(diào)性知：要使g（t）＝0有三個(gè)相異的實(shí)根，當(dāng)且僅當(dāng)SKIPIF1<00，即SKIPIF1<0．∴a的取值范圍是SKIPIF1<0．8．（2011·陜西·高考真題（理））如圖，從點(diǎn)SKIPIF1<0作SKIPIF1<0軸的垂線交曲線SKIPIF1<0于點(diǎn)SKIPIF1<0，曲線在SKIPIF1<0點(diǎn)處的切線與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，再從SKIPIF1<0作SKIPIF1<0軸的垂線交曲線于點(diǎn)SKIPIF1<0，依次重復(fù)上述過程得到一系列點(diǎn)：SKIPIF1<0，SKIPIF1<0；SKIPIF1<0，SKIPIF1<0；SKIPIF1<0；SKIPIF1<0，SKIPIF1<0記SKIPIF1<0點(diǎn)的坐標(biāo)為SKIPIF1<0（SKIPIF1<0）（1）試求SKIPIF1<0與SKIPIF1<0的關(guān)系（SKIPIF1<0）（2）求SKIPIF1<0【答案】（1）SKIPIF1<0SKIPIF1<0（2）SKIPIF1<0【詳解】（1）設(shè)點(diǎn)SKIPIF1<0的坐標(biāo)是SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0（SKIPIF1<0）．（2）∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，于是有SKIPIF1<0SKIPIF1<0SKIPIF1<0，即SKIPIF1<0SKIPIF1<0．9．（2015·天津·高考真題（理））已知函數(shù)SKIPIF1<0，其中SKIPIF1<0.（Ⅰ）討論SKIPIF1<0的單調(diào)性；（Ⅱ）設(shè)曲線SKIPIF1<0與SKIPIF1<0軸正半軸的交點(diǎn)為P，曲線在點(diǎn)P處的切線方程為SKIPIF1<0,求證：對于任意的正實(shí)數(shù)SKIPIF1<0，都有SKIPIF1<0；（Ⅲ）若關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)正實(shí)根SKIPIF1<0，求證：SKIPIF1<0【答案】（Ⅰ）當(dāng)SKIPIF