新高考數(shù)學(xué)一輪復(fù)習(xí)講練測專題4.1導(dǎo)數(shù)的概念、運(yùn)算及導(dǎo)數(shù)的幾何意義（練）解析版

專題4.1導(dǎo)數(shù)的概念、運(yùn)算及導(dǎo)數(shù)的幾何意義練基礎(chǔ)練基礎(chǔ)1.（2021·浙江高三其他模擬）函數(shù)SKIPIF1<0在SKIPIF1<0處的導(dǎo)數(shù)是（）A．SKIPIF1<0 B．SKIPIF1<0 C．6 D．2【答案】A【解析】利用符合函數(shù)的求導(dǎo)法則SKIPIF1<0，求出SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0，代入x=0，即可求出函數(shù)在x=0處的導(dǎo)數(shù).【詳解】SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0,故當(dāng)x=0時，SKIPIF1<0.故選：A2．（2021·黑龍江哈爾濱市·哈師大附中高三月考（文））曲線SKIPIF1<0在SKIPIF1<0處的切線方程為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】先求得導(dǎo)函數(shù),根據(jù)切點(diǎn)求得斜線的斜率,再由點(diǎn)斜式即可求得方程.【詳解】SKIPIF1<0當(dāng)SKIPIF1<0時,SKIPIF1<0所以在點(diǎn)SKIPIF1<0處的切線方程,由點(diǎn)斜式可得SKIPIF1<0化簡可得SKIPIF1<0故選:D3．（2021·全國高三其他模擬（理））曲線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】D【解析】根據(jù)切點(diǎn)和斜率求得切線方程.【詳解】因為SKIPIF1<0，所以SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的斜率SKIPIF1<0，所以所求切線方程為SKIPIF1<0，即SKIPIF1<0.故選：D4．（2021·山西高三三模（理））已知SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線為l，則l過定點(diǎn)（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】根據(jù)導(dǎo)數(shù)幾何意義求出切線方程，化成斜截式，即可求解【詳解】由SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故過SKIPIF1<0處的切線方程為：SKIPIF1<0，故l過定點(diǎn)SKIPIF1<0故選：A5．（2021·云南曲靖一中高三其他模擬（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】D【解析】利用導(dǎo)數(shù)的幾何意義可知SKIPIF1<0，可求得SKIPIF1<0；根據(jù)SKIPIF1<0為兩曲線公共點(diǎn)可構(gòu)造方程求得SKIPIF1<0，代入可得結(jié)果.【詳解】SKIPIF1<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.6.（2021·重慶高三其他模擬）曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則SKIPIF1<0（）A．SKIPIF1<0 B．0 C．1 D．2【答案】D【解析】求得SKIPIF1<0的導(dǎo)數(shù)，可得切線的斜率，由兩直線垂直的條件，可得SKIPIF1<0的方程，解方程可得所求值．【詳解】解：SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，可得在點(diǎn)SKIPIF1<0處的切線的斜率為SKIPIF1<0，由切線與直線SKIPIF1<0垂直，可得SKIPIF1<0，解得SKIPIF1<0，故選：SKIPIF1<0．7．（2021·重慶八中高三其他模擬）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，若曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為2，則SKIPIF1<0（）A．1 B．SKIPIF1<0 C．0 D．2【答案】C【解析】先由換元法求出SKIPIF1<0的解析式，然后求導(dǎo)，利用導(dǎo)數(shù)的幾何意義先求出SKIPIF1<0的值，然后可得出SKIPIF1<0的值.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．由SKIPIF1<0，解得SKIPIF1<0，從而SKIPIF1<0，故選:C．8．（2018·全國高考真題（理））設(shè)函數(shù)fx=x3+a?1xA．y=?2xB．y=?xC．y=2xD．y=x【答案】D【解析】分析：利用奇函數(shù)偶此項系數(shù)為零求得a=1，進(jìn)而得到f(x)的解析式，再對f(x)求導(dǎo)得出切線的斜率k，進(jìn)而求得切線方程.詳解：因為函數(shù)f(x)是奇函數(shù)，所以a?1=0，解得a=1，所以f(x)=x3+x所以f'(0)=1,f(0)=0，所以曲線y=f(x)在點(diǎn)(0,0)處的切線方程為y?f(0)=f'(0)x，化簡可得y=x，故選D.9.（2021·河南洛陽市·高三其他模擬（理））設(shè)曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行，則SKIPIF1<0等于（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】利用導(dǎo)數(shù)求出曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線的斜率，利用兩直線平行可得出實數(shù)SKIPIF1<0的值.【詳解】對函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，由已知條件可得SKIPIF1<0，所以，SKIPIF1<0.故選：B.10．（2020·河北高三其他模擬（文））已知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為2，則SKIPIF1<0___________.【答案】1【解析】求導(dǎo)數(shù)，由導(dǎo)數(shù)的幾何意義，可得切線的斜率，解方程即可求解.【詳解】解：SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，可得曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線斜率為SKIPIF1<0，解得SKIPIF1<0.故答案為：1.練提升TIDHNEG練提升TIDHNEG1．（2021·浙江金華市·高三三模）已知點(diǎn)P在曲線SKIPIF1<0上，SKIPIF1<0為曲線在點(diǎn)P處的切線的傾斜角，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】首先根據(jù)導(dǎo)數(shù)的幾何意義求得切線斜率的取值范圍，再根據(jù)傾斜角與斜率之間的關(guān)系求得傾斜角的取值范圍.【詳解】因為SKIPIF1<0，由于SKIPIF1<0，所以SKIPIF1<0，根據(jù)導(dǎo)數(shù)的幾何意義可知：SKIPIF1<0,所以SKIPIF1<0，故選：D.2．（2021·四川成都市·石室中學(xué)高三三模）已知函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線方程是SKIPIF1<0，那么SKIPIF1<0（）A．2 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】根據(jù)導(dǎo)數(shù)的幾何意義確定斜率與切點(diǎn)即可求解答案.【詳解】因為SKIPIF1<0，所以SKIPIF1<0，因此切線方程的斜率SKIPIF1<0，所以有SKIPIF1<0，得SKIPIF1<0，又切點(diǎn)在切線上，可得切點(diǎn)坐標(biāo)為SKIPIF1<0，將切點(diǎn)代入SKIPIF1<0中，有SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0.故選：D.3．（2021·四川成都市·成都七中高三月考（文））已知直線SKIPIF1<0為曲線SKIPIF1<0在SKIPIF1<0處的切線，則在直線SKIPIF1<0上方的點(diǎn)是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】利用導(dǎo)數(shù)的幾何意義求得切線的方程，進(jìn)而判定點(diǎn)與切線的位置關(guān)系即可.【詳解】SKIPIF1<0,SKIPIF1<0,又SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，所以切線的方程為SKIPIF1<0,對于A,當(dāng)SKIPIF1<0時，SKIPIF1<0，故點(diǎn)SKIPIF1<0在切線上；對于B,當(dāng)SKIPIF1<0時，SKIPIF1<0，故點(diǎn)SKIPIF1<0在切線下方；對于C,當(dāng)SKIPIF1<0時，SKIPIF1<0，故點(diǎn)SKIPIF1<0在切線上方；對于D,當(dāng)SKIPIF1<01時，SKIPIF1<0,故點(diǎn)SKIPIF1<0在切線下方.故選：C.4．（2021·甘肅高三二模（理））已知函數(shù)SKIPIF1<0，SKIPIF1<0SKIPIF1<0，若經(jīng)過點(diǎn)SKIPIF1<0存在一條直線SKIPIF1<0與SKIPIF1<0圖象和SKIPIF1<0圖象都相切，則SKIPIF1<0（）A．0 B．-1 C．3 D．-1或3【答案】D【解析】先求得過SKIPIF1<0且于SKIPIF1<0相切的切線方程，然后與SKIPIF1<0聯(lián)立，由SKIPIF1<0求解.【詳解】設(shè)直線SKIPIF1<0與SKIPIF1<0相切的切點(diǎn)為SKIPIF1<0，由SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，可得切線的斜率為SKIPIF1<0，則切線的方程為SKIPIF1<0，將SKIPIF1<0代入切線的方程可得SKIPIF1<0，解得SKIPIF1<0，則切線SKIPIF1<0的方程為SKIPIF1<0，聯(lián)立SKIPIF1<0，可得SKIPIF1<0，由SKIPIF1<0，解得SKIPIF1<0或3，故選：D.5．（2021·安徽省泗縣第一中學(xué)高三其他模擬（理））若點(diǎn)SKIPIF1<0是曲線SKIPIF1<0上任意一點(diǎn)，則點(diǎn)SKIPIF1<0到直線SKIPIF1<0的最小距離為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由已知可知曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線與直線SKIPIF1<0平行，利用導(dǎo)數(shù)求出點(diǎn)SKIPIF1<0的坐標(biāo)，利用點(diǎn)到直線的距離公式可求得結(jié)果.【詳解】因為點(diǎn)SKIPIF1<0是曲線SKIPIF1<0任意一點(diǎn)，所以當(dāng)點(diǎn)SKIPIF1<0處的切線和直線SKIPIF1<0平行時，點(diǎn)SKIPIF1<0到直線的SKIPIF1<0的距離最小，因為直線SKIPIF1<0的斜率等于SKIPIF1<0，曲線SKIPIF1<0的導(dǎo)數(shù)SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0（舍去），所以在曲線SKIPIF1<0與直線SKIPIF1<0平行的切線經(jīng)過的切點(diǎn)坐標(biāo)為SKIPIF1<0，所以點(diǎn)SKIPIF1<0到直線SKIPIF1<0的最小距離為SKIPIF1<0.故選：C.6．（2021·安徽省舒城中學(xué)高三三模（理））若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有一條公共切線，且該公共切線與直線SKIPIF1<0平行，則實數(shù)SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】設(shè)函數(shù)SKIPIF1<0圖象上切點(diǎn)為SKIPIF1<0，求出函數(shù)的導(dǎo)函數(shù)，根據(jù)SKIPIF1<0求出切點(diǎn)坐標(biāo)與切線方程，設(shè)函數(shù)SKIPIF1<0的圖象上的切點(diǎn)為SKIPIF1<0SKIPIF1<0，根據(jù)SKIPIF1<0，得到SKIPIF1<0，再由SKIPIF1<0，即可求出SKIPIF1<0，從而得解；【詳解】解：設(shè)函數(shù)SKIPIF1<0圖象上切點(diǎn)為SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，所以切線方程為SKIPIF1<0，即SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0的圖象上的切點(diǎn)為SKIPIF1<0SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍），所以SKIPIF1<0．故選：A7．（2021·全國高三其他模擬）已知直線y＝2x與函數(shù)f（x）＝﹣2lnx+xex+m的圖象相切，則m＝_________.【答案】SKIPIF1<0【解析】設(shè)出切點(diǎn)SKIPIF1<0，根據(jù)切線方程的幾何意義，得到SKIPIF1<0，解方程組即可.【詳解】因為SKIPIF1<0，所以SKIPIF1<0設(shè)切點(diǎn)為SKIPIF1<0，所以切線的斜率為SKIPIF1<0又因為切線方程為y＝2x，因此SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0.故答案為：SKIPIF1<0.8．（2021·黑龍江佳木斯市·佳木斯一中高三三模（理））若兩曲線y＝x2+1與y＝alnx+1存在公切線，則正實數(shù)a的取值范圍是_________．【答案】（0，2e]【解析】設(shè)公切線與曲線y＝x2+1和y＝alnx+1的交點(diǎn)分別為（x1，x12+1），（x2，alnx2+1），其中x2＞0，然后分別求出切線方程，對應(yīng)系數(shù)相等，可以得到SKIPIF1<0，然后轉(zhuǎn)化為﹣SKIPIF1<0＝alnx2﹣a，，然后參變分離得到a＝4x2﹣4x2lnx，進(jìn)而構(gòu)造函數(shù)求值域即可.【詳解】解：設(shè)公切線與曲線y＝x2+1和y＝alnx+1的交點(diǎn)分別為（x1，x12+1），（x2，alnx2+1），其中x2＞0，對于y＝x2+1，y′＝2x，所以與曲線y＝x2+1相切的切線方程為：y﹣（x12+1）＝2x1（x﹣x1），即y＝2x1x﹣x12+1，對于y＝alnx+1，y′＝SKIPIF1<0，所以與曲線y＝alnx+1相切的切線方程為y﹣（alnx2+1）＝SKIPIF1<0（x﹣x2），即y＝SKIPIF1<0x﹣a+1+alnx2，所以SKIPIF1<0，即有﹣SKIPIF1<0＝alnx2﹣a，由a＞0，可得a＝4x2﹣4x2lnx，記f(x)＝4x2﹣4x2lnx（x＞0），f′(x)＝8x﹣4x﹣8xlnx＝4x（1﹣2lnx），當(dāng)x＜SKIPIF1<0時，f′(x)＞0，即f(x)在（0，SKIPIF1<0）上單調(diào)遞增，當(dāng)x＞SKIPIF1<0時，f′(x)＜0，即f(x)在（SKIPIF1<0，+∞）上單調(diào)遞減，所以f(x)max＝f（SKIPIF1<0）＝2e，又x→0時，f(x)→0，x→+∞時，f(x)→﹣∞，所以0＜a≤2e．故答案為：（0，2e]．9．（2021·湖南永州市·高三其他模擬）已知函數(shù)SKIPIF1<0，點(diǎn)SKIPIF1<0為函數(shù)SKIPIF1<0圖象上一動點(diǎn)，則SKIPIF1<0到直線SKIPIF1<0距離的最小值為___________.(注SKIPIF1<0)【答案】SKIPIF1<0【解析】求出導(dǎo)函數(shù)，利用導(dǎo)數(shù)的幾何意義求出切線與已知直線平行時切點(diǎn)坐標(biāo)，然后轉(zhuǎn)化為求點(diǎn)到直線的距離即可求解．【詳解】解：SKIPIF1<0，SKIPIF1<0，與直線SKIPIF1<0平行的切線斜率SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，即切點(diǎn)為SKIPIF1<0，此時點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，即切點(diǎn)為SKIPIF1<0，此時點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，故答案為：SKIPIF1<0.10．（2021·湖北荊州市·荊州中學(xué)高三其他模擬）已知SKIPIF1<0，SKIPIF1<0是曲線SKIPIF1<0上的兩點(diǎn)，分別以SKIPIF1<0，SKIPIF1<0為切點(diǎn)作曲線C的切線SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，切線SKIPIF1<0交y軸于A點(diǎn)，切線SKIPIF1<0交y軸于B點(diǎn)，則線段SKIPIF1<0的長度為___________.【答案】SKIPIF1<0【解析】由兩切線垂直可知，SKIPIF1<0，SKIPIF1<0兩點(diǎn)必分別位于該函數(shù)的兩段上，故可設(shè)出切點(diǎn)坐標(biāo)SKIPIF1<0，表示出兩條切線方程，根據(jù)兩切線垂直，可得SKIPIF1<0，又兩切線分別與SKIPIF1<0軸交于SKIPIF1<0，SKIPIF1<0，則可求出SKIPIF1<0.【詳解】曲線SKIPIF1<0，則SKIPIF1<0，設(shè)SKIPIF1<0，兩切線斜率分別為SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，則不妨設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0由SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0.練真題TIDHNEG練真題TIDHNEG1.（2021·全國高考真題）若過點(diǎn)SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線，則（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】解法一：根據(jù)導(dǎo)數(shù)幾何意義求得切線方程，再構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)圖象，結(jié)合圖形確定結(jié)果；解法二：畫出曲線SKIPIF1<0的圖象，根據(jù)直觀即可判定點(diǎn)SKIPIF1<0在曲線下方和SKIPIF1<0軸上方時才可以作出兩條切線.【詳解】在曲線SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，對函數(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，可得SKIPIF1<0，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，當(dāng)SKIPIF1<0時，SKIPIF1<0，此時函數(shù)SKIPIF1<0單調(diào)遞減，所以，SKIPIF1<0，由題意可知，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個交點(diǎn)，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如下圖所示：由圖可知，當(dāng)SKIPIF1