新高考數(shù)學(xué)一輪復(fù)習(xí)第3章 第01講 導(dǎo)數(shù)的概念及運(yùn)算 精講+精練（教師版）

第01講導(dǎo)數(shù)的概念及運(yùn)算(精講+精練）目錄第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第二部分：課前自我評(píng)估測(cè)試第三部分：典型例題剖析高頻考點(diǎn)一：導(dǎo)數(shù)的概念高頻考點(diǎn)二：導(dǎo)數(shù)的運(yùn)算高頻考點(diǎn)三：導(dǎo)數(shù)的幾何意義①求切線方程（在型）②求切線方程（過(guò)型）③已知切線方程（或斜率）求參數(shù)④導(dǎo)數(shù)與函數(shù)圖象⑤共切點(diǎn)的公切線問(wèn)題⑥不同切點(diǎn)的公切線問(wèn)題⑦與切線有關(guān)的轉(zhuǎn)化問(wèn)題第四部分：高考真題感悟第五部分：第01講導(dǎo)數(shù)的概念及運(yùn)算（精練）第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶1、平均變化率（1）變化率事物的變化率是相關(guān)的兩個(gè)量的“增量的比值”。如氣球的平均膨脹率是半徑的增量與體積增量的比值.（2）平均變化率一般地，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的平均變化率為：SKIPIF1<0.（3）如何求函數(shù)的平均變化率求函數(shù)的平均變化率通常用“兩步”法：①作差：求出SKIPIF1<0和SKIPIF1<0②作商：對(duì)所求得的差作商，即SKIPIF1<0.2、導(dǎo)數(shù)的概念（1）定義：函數(shù)SKIPIF1<0在SKIPIF1<0處瞬時(shí)變化率是SKIPIF1<0，我們稱它為函數(shù)SKIPIF1<0在SKIPIF1<0處的導(dǎo)數(shù),記作SKIPIF1<0SKIPIF1<0SKIPIF1<0.（2）定義法求導(dǎo)數(shù)步驟：求函數(shù)的增量：SKIPIF1<0；求平均變化率：SKIPIF1<0；求極限，得導(dǎo)數(shù)：SKIPIF1<0.3、導(dǎo)數(shù)的幾何意義函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的導(dǎo)數(shù)的幾何意義，就是曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的斜率SKIPIF1<0，即SKIPIF1<0.4、基本初等函數(shù)的導(dǎo)數(shù)公式基本初等函數(shù)導(dǎo)數(shù)SKIPIF1<0(SKIPIF1<0為常數(shù))SKIPIF1<0SKIPIF1<0(SKIPIF1<0)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0(SKIPIF1<0)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0(SKIPIF1<0，SKIPIF1<0)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<05、導(dǎo)數(shù)的運(yùn)算法則若SKIPIF1<0，SKIPIF1<0存在，則有（1）SKIPIF1<0（2）SKIPIF1<0（3）SKIPIF1<06、復(fù)合函數(shù)求導(dǎo)復(fù)合函數(shù)SKIPIF1<0的導(dǎo)數(shù)和函數(shù)SKIPIF1<0，SKIPIF1<0的導(dǎo)數(shù)間的關(guān)系為SKIPIF1<0，即SKIPIF1<0對(duì)SKIPIF1<0的導(dǎo)數(shù)等于SKIPIF1<0對(duì)SKIPIF1<0的導(dǎo)數(shù)與SKIPIF1<0對(duì)SKIPIF1<0的導(dǎo)數(shù)的乘積．7、曲線的切線問(wèn)題（1）在型求切線方程已知：函數(shù)SKIPIF1<0的解析式.計(jì)算：函數(shù)SKIPIF1<0在SKIPIF1<0或者SKIPIF1<0處的切線方程.步驟：第一步：計(jì)算切點(diǎn)的縱坐標(biāo)SKIPIF1<0（方法：把SKIPIF1<0代入原函數(shù)SKIPIF1<0中），切點(diǎn)SKIPIF1<0.第二步：計(jì)算切線斜率SKIPIF1<0.第三步：計(jì)算切線方程.切線過(guò)切點(diǎn)SKIPIF1<0，切線斜率SKIPIF1<0。根據(jù)直線的點(diǎn)斜式方程得到切線方程：SKIPIF1<0.（2）過(guò)型求切線方程已知：函數(shù)SKIPIF1<0的解析式.計(jì)算：過(guò)點(diǎn)SKIPIF1<0（無(wú)論該點(diǎn)是否在SKIPIF1<0上）的切線方程.步驟：第一步：設(shè)切點(diǎn)SKIPIF1<0第二步：計(jì)算切線斜率SKIPIF1<0；計(jì)算切線斜率SKIPIF1<0；第三步：令：SKIPIF1<0，解出SKIPIF1<0，代入SKIPIF1<0求斜率第三步：計(jì)算切線方程.根據(jù)直線的點(diǎn)斜式方程得到切線方程：SKIPIF1<0.第二部分：課前自我評(píng)估測(cè)試第二部分：課前自我評(píng)估測(cè)試一、判斷題1．（2021·全國(guó)·高二課前預(yù)習(xí)）函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)值就是曲線y＝f(x)在x＝x0處的切線的斜率()【答案】正確函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)值就是曲線y＝f(x)在x＝x0處的切線的斜率.2．（2021·全國(guó)·高二課前預(yù)習(xí)）函數(shù)在x＝x0處的導(dǎo)數(shù)f′(x0)是一個(gè)常數(shù)()【答案】正確函數(shù)在x＝x0處的導(dǎo)數(shù)f′(x0)是一個(gè)常數(shù).3．（2021·全國(guó)·高二課前預(yù)習(xí)）函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)值與Δx的正、負(fù)無(wú)關(guān).()【答案】正確4．（2021·全國(guó)·高二課前預(yù)習(xí)）設(shè)x＝x0＋Δx，則Δx＝x－x0，則Δx趨近于0時(shí)，x趨近于x0，因此，f′(x0)＝SKIPIF1<0SKIPIF1<0＝SKIPIF1<0SKIPIF1<0.()【答案】正確二、單選題1．（2022·河北邢臺(tái)·高二階段練習(xí)）函數(shù)SKIPIF1<0從1到2的平均變化率為（

）A．SKIPIF1<0 B．4 C．SKIPIF1<0 D．6【答案】A函數(shù)SKIPIF1<0從1到2的平均變化率為：SKIPIF1<0．故選：A.2．（2022·四川·攀枝花七中高二階段練習(xí)（理））已知函數(shù)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】DSKIPIF1<0，則SKIPIF1<0SKIPIF1<0故選：D3．（2022·江西九江·二模）曲線SKIPIF1<0在SKIPIF1<0處的切線傾斜角是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B設(shè)曲線SKIPIF1<0在SKIPIF1<0處的切線傾斜角為SKIPIF1<0，因?yàn)镾KIPIF1<0，則SKIPIF1<0，因?yàn)镾KIPIF1<0，因此，SKIPIF1<0.故選：B.4．（2022·安徽滁州·高二階段練習(xí)）曲線SKIPIF1<0在SKIPIF1<0處的切線的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B解：由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以曲線SKIPIF1<0在SKIPIF1<0處的切線的方程為SKIPIF1<0，即SKIPIF1<0．故選：B.第三部分：典型例題剖析第三部分：典型例題剖析高頻考點(diǎn)一：導(dǎo)數(shù)的概念1．（2022·河北邢臺(tái)·高二階段練習(xí)）已知函數(shù)SKIPIF1<0的圖象如圖所示，SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A如圖所示，根據(jù)導(dǎo)數(shù)的幾何意義，可得SKIPIF1<0表示曲線在SKIPIF1<0點(diǎn)處的切線的斜率，即直線SKIPIF1<0的斜率SKIPIF1<0，SKIPIF1<0表示曲線在SKIPIF1<0點(diǎn)處的切線的斜率，即直線SKIPIF1<0的斜率SKIPIF1<0，又由平均變化率的定義，可得SKIPIF1<0表示過(guò)SKIPIF1<0兩點(diǎn)的割線的斜率SKIPIF1<0，結(jié)合圖象，可得SKIPIF1<0，所以SKIPIF1<0.故選：A.2．（2022·安徽·蕪湖一中高二階段練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0處的導(dǎo)數(shù)為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C根據(jù)題意，SKIPIF1<0.故選：C3．（2022·陜西·西安市閻良區(qū)關(guān)山中學(xué)高二階段練習(xí)（理））已知SKIPIF1<0，則SKIPIF1<0________.【答案】SKIPIF1<0SKIPIF1<0.故答案為：SKIPIF1<0.高頻考點(diǎn)二：導(dǎo)數(shù)的運(yùn)算1．（多選）（2022·河北·武安市第三中學(xué)高二階段練習(xí)）下列運(yùn)算正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BDSKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故AD錯(cuò)誤，BC正確.故選：BC.2．（2022·重慶市青木關(guān)中學(xué)校高二階段練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0__________.【答案】4SKIPIF1<0，則SKIPIF1<0故答案為：SKIPIF1<03．（2022·四川·攀枝花七中高二階段練習(xí)（理））求下列函數(shù)的導(dǎo)數(shù)：(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0．【答案】(1)SKIPIF1<0SKIPIF1<0；(2)SKIPIF1<0SKIPIF1<0；(3)SKIPIF1<0SKIPIF1<0.(1)因?yàn)镾KIPIF1<0，故SKIPIF1<0SKIPIF1<0.(2)因?yàn)镾KIPIF1<0，故SKIPIF1<0SKIPIF1<0.(3)因?yàn)镾KIPIF1<0，故SKIPIF1<0SKIPIF1<0.4．（2022·四川·棠湖中學(xué)高二階段練習(xí)（理））求下列函數(shù)的導(dǎo)數(shù)．(1)f（x）=x3e－x；(2)g（x）=cos2x+ln（2x）．【答案】(1)SKIPIF1<0(2)SKIPIF1<0(1)SKIPIF1<0(2)SKIPIF1<05．（2022·甘肅·甘南藏族自治州合作第一中學(xué)高二期末（文））求下列函數(shù)的導(dǎo)數(shù).(1)SKIPIF1<0；(2)SKIPIF1<0.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(1)SKIPIF1<0(2)SKIPIF1<0高頻考點(diǎn)三：導(dǎo)數(shù)的幾何意義①求切線方程（在型）1．（2022·內(nèi)蒙古·赤峰二中高二期末（文））曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線過(guò)點(diǎn)SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0（

）A．SKIPIF1<0 B．0 C．1 D．2【答案】A解：因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以，切線方程為SKIPIF1<0，因?yàn)榍芯€過(guò)點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．故選：A2．（2022·江西·臨川一中高二期末（文））已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】CSKIPIF1<0則SKIPIF1<0，又SKIPIF1<0則函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0故選：C3．（2022·天津市濱海新區(qū)塘沽第一中學(xué)高二階段練習(xí)）曲線SKIPIF1<0在SKIPIF1<0處的切線SKIPIF1<0與坐標(biāo)軸圍成的三角形的面積為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B因?yàn)镾KIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，所以，直線SKIPIF1<0的方程為SKIPIF1<0，直線SKIPIF1<0交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，交SKIPIF1<0軸于點(diǎn)SKIPIF1<0，因此，直線SKIPIF1<0與坐標(biāo)軸圍成的三角形的面積為SKIPIF1<0.故選：B.4．（2022·湖南·一模）若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．2【答案】C由SKIPIF1<0，顯然SKIPIF1<0在曲線SKIPIF1<0上，所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的斜率為SKIPIF1<0，因此切線方程為：SKIPIF1<0，直線SKIPIF1<0的斜率為SKIPIF1<0，因?yàn)榍€SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行，所以SKIPIF1<0，故選：C5．（2022·河南·模擬預(yù)測(cè)（文））函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A依題意，SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，于是有SKIPIF1<0，即SKIPIF1<0，所以所求切線方程為：SKIPIF1<0.故選：A6．（2022·河南·沈丘縣第一高級(jí)中學(xué)高二期末（文））已知函數(shù)SKIPIF1<0，則曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A函數(shù)SKIPIF1<0，求導(dǎo)得：SKIPIF1<0，則SKIPIF1<0，而SKIPIF1<0，于是得：SKIPIF1<0，即SKIPIF1<0，所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0.故選：A②求切線方程（過(guò)型）1．（2022·江西·南昌大學(xué)附屬中學(xué)高二期末（理））曲線y＝lnx在點(diǎn)M處的切線過(guò)原點(diǎn)，則該切線的斜率為（

）A．1 B．e C．－1 D．SKIPIF1<0【答案】D設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，故在SKIPIF1<0點(diǎn)的切線的斜率為SKIPIF1<0，所以SKIPIF1<0，所以切點(diǎn)為SKIPIF1<0，切線的斜率為SKIPIF1<0.故選：D2．（2022·全國(guó)·高三專題練習(xí)）若曲線SKIPIF1<0的一條切線經(jīng)過(guò)點(diǎn)(8，3)，則此切線的斜率為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】C由題意，可設(shè)切點(diǎn)坐標(biāo)為(x0，SKIPIF1<0)，由SKIPIF1<0，得y′＝SKIPIF1<0，切線斜率k＝SKIPIF1<0，由點(diǎn)斜式可得切線方程為y－SKIPIF1<0＝SKIPIF1<0(x－x0)，又切線過(guò)點(diǎn)(8，3)，所以3－SKIPIF1<0＝SKIPIF1<0(8－x0)，整理得x0－6SKIPIF1<0＋8＝0，解得SKIPIF1<0＝4或2，所以切線斜率k＝SKIPIF1<0或SKIPIF1<0.故選：C.3．（2022·江蘇·南京航空航天大學(xué)蘇州附屬中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0，則過(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的切線的條數(shù)為（

）A．0 B．1 C．2 D．3【答案】C解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，設(shè)切點(diǎn)為SKIPIF1<0，所以在切點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0，又SKIPIF1<0在切線上，所以SKIPIF1<0，即SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以過(guò)點(diǎn)SKIPIF1<0可作曲線SKIPIF1<0的切線的條數(shù)為2.故選：C．4．（2022·陜西安康·高三期末（文））曲線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0的切線方程是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B由題意可得點(diǎn)SKIPIF1<0不在曲線SKIPIF1<0上，設(shè)切點(diǎn)為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以所求切線的斜率SKIPIF1<0，所以SKIPIF1<0.因?yàn)辄c(diǎn)SKIPIF1<0是切點(diǎn)，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.設(shè)SKIPIF1<0，明顯SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，所以SKIPIF1<0有唯一解SKIPIF1<0，則所求切線的斜率SKIPIF1<0，故所求切線方程為SKIPIF1<0.故選：B.5．（2022·陜西·西北工業(yè)大學(xué)附屬中學(xué)一模（理））已知SKIPIF1<0，若過(guò)一點(diǎn)SKIPIF1<0可以作出該函數(shù)的兩條切線，則下列選項(xiàng)一定成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A設(shè)切點(diǎn)為SKIPIF1<0，對(duì)函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，則切線斜率為SKIPIF1<0，所以，切線方程為SKIPIF1<0，即SKIPIF1<0，所以，SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，其中SKIPIF1<0，由題意可知，方程SKIPIF1<0有兩個(gè)不等的實(shí)根.SKIPIF1<0.①當(dāng)SKIPIF1<0時(shí)，對(duì)任意的SKIPIF1<0，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則方程SKIPIF1<0至多只有一個(gè)根，不合乎題意；②當(dāng)SKIPIF1<0時(shí)，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.由題意可得SKIPIF1<0，可得SKIPIF1<0.故選：A.6．（2022·江西·模擬預(yù)測(cè)（文））已知曲線SKIPIF1<0與過(guò)點(diǎn)SKIPIF1<0的直線SKIPIF1<0相切，則SKIPIF1<0的斜率為_______．【答案】SKIPIF1<0##SKIPIF1<0解：設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則切線方程為SKIPIF1<0，將點(diǎn)SKIPIF1<0代入得SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0，解得SKIPIF1<0，所以切線的斜率為SKIPIF1<0.故答案為：SKIPIF1<0.7．（2022·全國(guó)·高三專題練習(xí)）已知函數(shù)f(x)＝x＋SKIPIF1<0，若曲線y＝f(x)存在兩條過(guò)(1，0)點(diǎn)的切線，則a的取值范圍是________.【答案】SKIPIF1<0或SKIPIF1<0由題得SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則切線方程為SKIPIF1<0，又切線過(guò)點(diǎn)SKIPIF1<0，可得SKIPIF1<0，整理得SKIPIF1<0，因?yàn)榍€SKIPIF1<0存在兩條切線，故方程有兩個(gè)不等實(shí)根且SKIPIF1<0若SKIPIF1<0，則SKIPIF1<0，為兩個(gè)重根，不成立,即滿足SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．故SKIPIF1<0的取值范圍是SKIPIF1<0或SKIPIF1<0故答案為：SKIPIF1<0或SKIPIF1<0③已知切線方程（或斜率）求參數(shù)1．（2022·北京·北理工附中高二階段練習(xí)）如圖，函數(shù)SKIPIF1<0的圖像在點(diǎn)P處的切線方程是SKIPIF1<0，則SKIPIF1<0（

）A．-2 B．2 C．3 D．無(wú)法確定【答案】B由題圖，SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0.故選：B2．（2022·湖南·長(zhǎng)沙縣實(shí)驗(yàn)中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則SKIPIF1<0（

）A．－2 B．－1 C．2 D．3【答案】B函數(shù)SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，∴SKIPIF1<0，即函數(shù)在SKIPIF1<0處的切線斜率為SKIPIF1<0，由切線與直線SKIPIF1<0垂直，可得SKIPIF1<0，解得SKIPIF1<0.故選：B.3．（2022·吉林白山·一模（理））函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線斜率為1，則SKIPIF1<0（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】A因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0.故選：A．4．（2022·江蘇省蘇州實(shí)驗(yàn)中學(xué)高二階段練習(xí)）已知直線SKIPIF1<0是曲線SKIPIF1<0的切線，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C設(shè)直線SKIPIF1<0與曲線SKIPIF1<0的切線點(diǎn)的橫坐標(biāo)為SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得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，所以SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C.5．（2022·全國(guó)·高三專題練習(xí)）若點(diǎn)P是曲線SKIPIF1<0上任意一點(diǎn)，則點(diǎn)P到直線SKIPIF1<0的距離的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A設(shè)平行于直線SKIPIF1<0且與曲線SKIPIF1<0相切的切線對(duì)應(yīng)切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），故點(diǎn)P的坐標(biāo)為SKIPIF1<0，故點(diǎn)P到直線SKIPIF1<0的最小值為：SKIPIF1<0.故選：A.6．（2022·四川省綿陽(yáng)南山中學(xué)高二階段練習(xí)（理））若曲線SKIPIF1<0存在垂直于y軸的切線，則a的取值范圍是(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C依題意，f(x)存在垂直與y軸的切線，即存在切線斜率SKIPIF1<0的切線，又SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0有正根，即SKIPIF1<0有正根，即函數(shù)y＝－2a與函數(shù)SKIPIF1<0的圖像有交點(diǎn)，令SKIPIF1<0，則g(t)＝SKIPIF1<0，∴g(t)≥g(SKIPIF1<0)＝SKIPIF1<0，∴－2a≥SKIPIF1<0，即a≤SKIPIF1<0.故選：C.7．（2022·全國(guó)·高三專題練習(xí)）點(diǎn)A在直線y＝x上，點(diǎn)B在曲線SKIPIF1<0上，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．2【答案】A設(shè)平行于直線y＝x的直線y＝x＋b與曲線SKIPIF1<0相切，則兩平行線間的距離即為SKIPIF1<0的最小值．設(shè)直線y＝x＋b與曲線SKIPIF1<0的切點(diǎn)為SKIPIF1<0，則由切點(diǎn)還在直線y＝x＋b上可得SKIPIF1<0，由切線斜率等于切點(diǎn)的導(dǎo)數(shù)值可得SKIPIF1<0，聯(lián)立解得m＝1，b＝－1，由平行線間的距離公式可得SKIPIF1<0的最小值為SKIPIF1<0，故選：A.④導(dǎo)數(shù)與函數(shù)圖象1．（2022·北京·北理工附中高二階段練習(xí)）函數(shù)SKIPIF1<0的圖像如圖所示，下列不等關(guān)系正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C如圖所示，根據(jù)導(dǎo)數(shù)的幾何意義，可得SKIPIF1<0表示切線SKIPIF1<0斜率SKIPIF1<0，SKIPIF1<0表示切線SKIPIF1<0斜率SKIPIF1<0，又由平均變化率的定義，可得SKIPIF1<0，表示割線SKIPIF1<0的斜率SKIPIF1<0，結(jié)合圖象，可得SKIPIF1<0，即SKIPIF1<0.故選：C.2．（2022·全國(guó)·高二單元測(cè)試）已知函數(shù)SKIPIF1<0的圖象是下列四個(gè)圖象之一，且其導(dǎo)函數(shù)SKIPIF1<0的圖象如圖所示，則該函數(shù)的圖象是（

）A． B．C． D．【答案】A由函數(shù)f(x)的導(dǎo)函數(shù)y＝f′(x)的圖像自左至右是先減后增，可知函數(shù)y＝f(x)圖像的切線的斜率自左至右先減小后增大，且SKIPIF1<0，在SKIPIF1<0處的切線的斜率為0，故BCD錯(cuò)誤，A正確.故選：A.3．（2022·江蘇·高二）如圖，函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，則SKIPIF1<0（

）.A．1 B．3 C．SKIPIF1<0 D．SKIPIF1<0【答案】D因?yàn)楹瘮?shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，切點(diǎn)的橫坐標(biāo)為SKIPIF1<0，由導(dǎo)數(shù)的幾何意義可得SKIPIF1<0，所以SKIPIF1<0，故選：D.4．（2021·全國(guó)·高二單元測(cè)試）如圖所示，SKIPIF1<0是可導(dǎo)函數(shù)，直線l：y=kx+3是曲線y=f（x）在x=1處的切線，令SKIPIF1<0，SKIPIF1<0是h（x）的導(dǎo)函數(shù)，則SKIPIF1<0的值是（

）A．2 B．1 C．-1 D．-3【答案】D根據(jù)圖象可知SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0.故選：D⑤共切點(diǎn)的公切線問(wèn)題1．（2021·江西·高三階段練習(xí)（理））若曲線SKIPIF1<0與曲線SKIPIF1<0在公共點(diǎn)處有公共切線，則實(shí)數(shù)SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A設(shè)公共點(diǎn)為SKIPIF1<0，SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，曲線SKIPIF1<0在SKIPIF1<0處的切線斜率SKIPIF1<0，SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，曲線SKIPIF1<0在SKIPIF1<0處的切線斜率SKIPIF1<0，因?yàn)閮汕€在公共點(diǎn)SKIPIF1<0處有公共切線，所以SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，故選：A．2．（2021·重慶·高二階段練習(xí)）已知兩曲線SKIPIF1<0和SKIPIF1<0都經(jīng)過(guò)點(diǎn)SKIPIF1<0，且在點(diǎn)SKIPIF1<0處有公切線，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D由題意SKIPIF1<0即SKIPIF1<0設(shè)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，又因?yàn)閮汕€在點(diǎn)P處有公切線，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0所以SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立)故選：D3．（2021·云南·曲靖一中模擬預(yù)測(cè)（理））設(shè)曲線SKIPIF1<0和曲線SKIPIF1<0在它們的公共點(diǎn)SKIPIF1<0處有相同的切線，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】DSKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0為SKIPIF1<0與SKIPIF1<0公共點(diǎn)，SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0.故選：D.4．（2022·全國(guó)·高三專題練習(xí)（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0與SKIPIF1<0在公共點(diǎn)處的切線相同，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0的公共點(diǎn)設(shè)為SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故選：B.5．（2022·全國(guó)·高三專題練習(xí)）若函數(shù)f（x）＝alnx（a∈R）與函數(shù)g（x）SKIPIF1<0在公共點(diǎn)處有共同的切線，則實(shí)數(shù)a的值為（

）A．4 B．SKIPIF1<0 C．SKIPIF1<0 D．e【答案】C由已知得SKIPIF1<0，設(shè)切點(diǎn)橫坐標(biāo)為t，∴SKIPIF1<0，解得SKIPIF1<0.故選：C.⑥不同切點(diǎn)的公切線問(wèn)題1．（2022·河北省唐縣第一中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若直線SKIPIF1<0與函數(shù)SKIPIF1<0，SKIPIF1<0的圖象都相切，則SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B設(shè)直線SKIPIF1<0與函數(shù)SKIPIF1<0，SKIPIF1<0的圖象相切的切點(diǎn)分別為SKIPIF1<0，SKIPIF1<0.由SKIPIF1<0，有SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0.又由SKIPIF1<0，有SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)取“=”.故選：B2．（2022·重慶市育才中學(xué)高三階段練習(xí)）若直線SKIPIF1<0（SKIPIF1<0）為曲線SKIPIF1<0與曲線SKIPIF1<0的公切線，則l的縱截距SKIPIF1<0（

）A．0 B．1 C．e D．SKIPIF1<0【答案】D設(shè)l與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，則由SKIPIF1<0，有SKIPIF1<0.同理，設(shè)l與SKIPIF1<0的切點(diǎn)為SKIPIF1<0，由SKIPIF1<0，有SKIPIF1<0.故SKIPIF1<0解得SKIPIF1<0或SKIPIF1<0則SKIPIF1<0或SKIPIF1<0.因SKIPIF1<0，所以l為SKIPIF1<0時(shí)不成立.故SKIPIF1<0，故選：D.3．（2022·湖北·安陸第一高中高二階段練習(xí)）若存在過(guò)點(diǎn)（0，－2）的直線與曲線SKIPIF1<0和曲線SKIPIF1<0都相切，則實(shí)數(shù)a的值是（

）A．2 B．1 C．0 D．－2【答案】ASKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，若直線與SKIPIF1<0和SKIPIF1<0的切點(diǎn)分別為（SKIPIF1<0，SKIPIF1<0），SKIPIF1<0，∴過(guò)（0，－2）的直線為SKIPIF1<0、SKIPIF1<0，則有SKIPIF1<0，可得SKIPIF1<0．故選：A.4．（2022·湖南永州·二模）若函數(shù)SKIPIF1<0與SKIPIF1<0存在兩條公切線，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】D設(shè)切線與曲線SKIPIF1<0相切于點(diǎn)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，聯(lián)立SKIPIF1<0可得SKIPIF1<0，由題意可得SKIPIF1<0且SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，其中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，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.且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，如下圖所示：由題意可知，直線SKIPIF1<0與曲線SKIPIF1<0有兩個(gè)交點(diǎn)，則SKIPIF1<0，解得SKIPIF1<0.故選：D.5．（2022·山西呂梁·高二期末）若直線SKIPIF1<0是曲線SKIPIF1<0的切線，也是曲線SKIPIF1<0的切線，則SKIPIF1<0__________．【答案】SKIPIF1<0設(shè)曲線SKIPIF1<0的切點(diǎn)為：SKIPIF1<0，由SKIPIF1<0，所以過(guò)該切點(diǎn)的切線斜率為：SKIPIF1<0，于是切線方程為：SKIPIF1<0，因此有：SKIPIF1<0，設(shè)曲線SKIPIF1<0的切點(diǎn)為：SKIPIF1<0，由SKIPIF1<0，所以過(guò)該切點(diǎn)的切線斜率為：SKIPIF1<0，于是切線方程為：SKIPIF1<0，因此有：SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，即SKIPIF1<0，因此SKIPIF1<0，故答案為：SKIPIF1<06．（2022·全國(guó)·高三專題練習(xí)）若曲線SKIPIF1<0與曲線SKIPIF1<0有公切線，則SKIPIF1<0的取值范圍是_____________.【答案】SKIPIF1<0設(shè)SKIPIF1<0是曲線SKIPIF1<0上一點(diǎn)，由SKIPIF1<0，因此過(guò)點(diǎn)SKIPIF1<0的切線的斜率為SKIPIF1<0，所以切線方程為：SKIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，設(shè)SKIPIF1<0是曲線SKIPIF1<0上一點(diǎn)，由SKIPIF1<0，所以過(guò)SKIPIF1<0點(diǎn)的切線的斜率為SKIPIF1<0，所以切線方程為：SKIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，當(dāng)這兩條切線重合時(shí)，就是兩個(gè)曲線的公切線，因此有：SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0設(shè)函數(shù)SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0是減函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因此SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<07．（2022·四川·棠湖中學(xué)高二階段練習(xí)（文））已知SKIPIF1<0（SKIPIF1<0為自然對(duì)數(shù)的成數(shù)），SKIPIF1<0，直線SKIPIF1<0是SKIPIF1<0與SKIPIF1<0的公切線，則直線SKIPIF1<0的方程為________.【答案】SKIPIF1<0或SKIPIF1<0設(shè)公切線SKIPIF1<0與SKIPIF1<0且于點(diǎn)SKIPIF1<0，與曲線SKIPIF1<0切于點(diǎn)SKIPIF1<0，則有SKIPIF1<0SKIPIF1<0②又SKIPIF1<0，∴SKIPIF1<0．∵過(guò)點(diǎn)SKIPIF1<0的直線SKIPIF1<0的斜率為SKIPIF1<0，∴SKIPIF1<0．③由①②③消去SKIPIF1<0整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，直線SKIPIF1<0與曲線SKIPIF1<0的切點(diǎn)為SKIPIF1<0，SKIPIF1<0，此時(shí)切線方程為SKIPIF1<0，即SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，直線SKIPIF1<0與曲線SKIPIF1<0的切點(diǎn)為SKIPIF1<0，SKIPIF1<0，此時(shí)切線方程為SKIPIF1<0，即SKIPIF1<0．故直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0．所以答案為SKIPIF1<0或SKIPIF1<0．⑦與切線有關(guān)的轉(zhuǎn)化問(wèn)題1．（2022·全國(guó)·高三專題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BSKIPIF1<0的最小值可轉(zhuǎn)化為函數(shù)SKIPIF1<0圖像上的點(diǎn)SKIPIF1<0與直線SKIPIF1<0上的點(diǎn)SKIPIF1<0的距離的最小值.【詳解】設(shè)SKIPIF1<0,SKIPIF1<0,點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0上，點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0上，SKIPIF1<0表示曲線SKIPIF1<0上點(diǎn)S