高考第一輪文科數(shù)學(xué)（人教A版）課時(shí)規(guī)范練14　導(dǎo)數(shù)的概念及運(yùn)算

課時(shí)規(guī)范練14導(dǎo)數(shù)的概念及運(yùn)算基礎(chǔ)鞏固組1.曲線f(x)=x2+2ex在點(diǎn)(0,f(0))處的切線方程為()A.x+2y+2=0 B.2x+y+2=0C.x-2y+2=0 D.2x-y+2=02.已知函數(shù)f(x)=x3-f'(1)x2+2的導(dǎo)數(shù)為f'(x),則f(x)的圖象在點(diǎn)(2,f(2))處的切線的斜率為()A.-8 B.8 C.12 D.163.已知函數(shù)f(x)=x4+ax,若limΔx→0f(A.36 B.12 C.4 D.24.若曲線f(x)=lnx在點(diǎn)(1,0)的切線與曲線g(x)=12x2+mx+72也相切,則m=5.已知函數(shù)f(x)=13x3-4x,x<0,-1x6.曲線f(x)=2x-ex與直線x-y+t=0相切,則t=.

7.已知函數(shù)f(x)=lnxx+a,且f'(1)=1,則a=,曲線y=f(x)在x=e處的切線方程為綜合提升組8.已知函數(shù)f(x)=sinx和g(x)=cosx圖象的一個(gè)公共點(diǎn)為P(x0,y0),現(xiàn)給出以下結(jié)論:①f(x0)=g(x0);②f'(x0)=g'(x0);③f(x)和g(x)的圖象在點(diǎn)P處的切線的傾斜角互補(bǔ);④f(x)和g(x)的圖象在點(diǎn)P處的切線互相垂直.其中正確結(jié)論的序號(hào)是()A.①③ B.②④ C.②③ D.①④9.函數(shù)f(x)=lnx圖象上一點(diǎn)P到直線y=2x的最短距離為()A.2 B.2C.(1+ln2)510.設(shè)曲線y=lnx與y=(x+a)2有一條斜率為1的公切線,則a=()A.-1 B.-34 C.14 11.已知函數(shù)f(x)=ex-lnxx?1x+a,若直線y=0在點(diǎn)(b,f(b))處與曲線y=f(xA.1 B.0 C.-1 D.-1或112.定義:函數(shù)f(x)在(a,b)上的導(dǎo)函數(shù)為f'(x),f'(x)在(a,b)上的導(dǎo)函數(shù)為f″(x),若在(a,b)上f″(x)<0恒成立,則稱函數(shù)f(x)在(a,b)上的“嚴(yán)格凸函數(shù)”,稱區(qū)間(a,b)為函數(shù)f(x)的“嚴(yán)格凸區(qū)間”.則下列正確說(shuō)法的序號(hào)為.(填上所有正確答案的序號(hào))

①函數(shù)f(x)=-x3+3x2+2在(1,+∞)上為“嚴(yán)格凸函數(shù)”;②函數(shù)f(x)=lnxx的“嚴(yán)格凸區(qū)間”為(0,e32);③函數(shù)f(x)=ex-m2x2在(1,4)為“嚴(yán)格凸函數(shù)”,則創(chuàng)新應(yīng)用組13.已知a-lnb=0,c-d=1,則(a-c)2+(b-d)2的最小值是()A.1 B.2 C.2 D.22

參考答案課時(shí)規(guī)范練14導(dǎo)數(shù)的概念及運(yùn)算1.Df(x)=x2+2ex的導(dǎo)數(shù)為f'(x)=2x+2ex,則在點(diǎn)(0,f(0))處的切線的斜率為f'(0)=2,且切點(diǎn)為(0,2),則切線的方程為y=2x+2,即2x-y+2=0.2.B因?yàn)閒'(x)=3x2-2f'(1)x,令x=1,得f'(1)=3-2f'(1),所以f'(1)=1,所以f'(x)=3x2-2x,f(x)的圖象在點(diǎn)(2,f(2))處的切線的斜率為f'(2)=8.3.C根據(jù)題意,f(x)=x4+ax,則f'(x)=4x3+a,則f'(0)=a,若limΔx則lim=3lim=3f'(0)=12,則有3a=12,即a=4.4.-2或4由f(x)=lnx的導(dǎo)數(shù)為f'(x)=1x,可得曲線f(x)=lnx在點(diǎn)(1,0)的切線斜率為1,切線的方程為y=x-1,聯(lián)立y=x-1,y=12x2+mx+72,可得x2+(2m-2)x+9=0,由切線與曲線g(x)=12x2+mx+725.14或-4由題意得f'(x)=x2-4,x<0所以0<a<1,1a2-6.-1∵f(x)=2x-ex,∴f'(x)=2-ex,切線x-y+t=0的斜率為k=1,設(shè)切點(diǎn)P(x0,y0),令f'(x0)=2-ex0=1,解得x0=0,代入f(x)=2x-ex得y0=2×0-e0=-1,∴切點(diǎn)坐標(biāo)為(0,-1),代入切線方程x-y+t=0中得到0+1+t=0,解得t=-7.0y-1e=0由f(x)=lnxx+a,則f'(x)=x+ax-lnx(x+a)2,因?yàn)閒'(1)=1,即11+a=1,解得a=0,所以f(x)=lnxx,f'(x)=1-lnxx8.A對(duì)于①,因?yàn)閒(x0)=y0,g(x0)=y0,則f(x0)=g(x0),故①正確;對(duì)于②,因?yàn)閒(x)和g(x)在點(diǎn)P處的切線不平行且不重合,所以f'(x0)≠g'(x0),故②錯(cuò)誤;對(duì)于③,顯然f'(x0)+g'(x0)=0成立,故③正確;對(duì)于④,假設(shè)f(x)和g(x)的圖象在點(diǎn)P處的切線互相垂直,則有-cosx0sinx0=-1,即sin2x0=2,這與|sin2x0|≤1矛盾,故④錯(cuò)誤.9.C設(shè)與直線y=2x平行且與曲線f(x)=lnx相切的直線的切點(diǎn)坐標(biāo)為(x0,lnx0),因?yàn)閒'(x)=1x,所以1x0=2,解得x0=12,則切點(diǎn)坐標(biāo)為12,-ln2,最短距離為點(diǎn)12,-ln2到直線y=2x的距離,即2×12-(-ln210.B因?yàn)閥=lnx,所以y'=1x.又因?yàn)榍芯€的斜率為1,設(shè)切點(diǎn)為(x0,y0),所以1x0=1,解得x0=1,y0=0,所以切線方程為y=x-1.因?yàn)閥=(x+a)2,設(shè)切點(diǎn)(x,y),所以y'=2x+2a=1,解得x=12-a,代入切線方程得y=-12-a,再將12-a,-12-a代入y=(x+a)211.C由f(x)=ex-lnxx?1x+a可得f'(x)=ex-1-lnxx2+1x2=ex+lnxx2,因?yàn)橹本€y=0在點(diǎn)(b,f(b))處與曲線y=f(x)相切,則f'(b)=0,即eb+lnbb2=0,所以eb·b=-lnbb=1b·ln1b,兩邊同時(shí)取以e為底的對(duì)數(shù),可得ln(eb·b)=ln1b·ln1b,即lneb+lnb=ln1b+lnln1b,所以b+lnb=ln1b+lnln1b,設(shè)g(x)=x+lnx,g'(x)=1+1x>0,函數(shù)在(0,12.①②f(x)=-x3+3x2+2的導(dǎo)函數(shù)f'(x)=-3x2+6x,f″(x)=-6x+6,故f″(x)<0在(1,+∞)上恒成立,所以函數(shù)f(x)=-x3+3x2+2在(1,+∞)上為“嚴(yán)格凸函數(shù)”,所以①正確;f(x)=lnxx的定義域?yàn)?0,+∞)且導(dǎo)函數(shù)f'(x)=1-lnxx2,f″(x)=2lnx-3x3,由f″(x)<0可得2lnx-3<0,解得x∈(0,e32),所以函數(shù)f(x)f(x)=ex-m2x2的導(dǎo)函數(shù)f'(x)=ex-mx,f″(x)=ex-m因?yàn)閒(x)在(1,4)為“嚴(yán)格凸函數(shù)”,故f″(x)<0在(1,4)上恒成立,所以ex-m<0在(1,4)上恒成立,即m>ex在(1,4)上恒成立,故m≥e4,所以③不正確.13.C設(shè)(b,a)是曲線C:y=lnx上的點(diǎn),(d,c)是直線