高考一輪復(fù)習(xí)專(zhuān)項(xiàng)練習(xí)數(shù)學(xué)課時(shí)規(guī)范練14導(dǎo)數(shù)的概念意義及運(yùn)算

課時(shí)規(guī)范練14導(dǎo)數(shù)的概念、意義及運(yùn)算基礎(chǔ)鞏固組1.若f'(x0)=3,則limh→0fA.3 B.6 C.9 D.122.已知奇函數(shù)y=f(x)在區(qū)間(∞,0]上的解析式為f(x)=x2+x,則切點(diǎn)橫坐標(biāo)為1的切線(xiàn)方程是()A.x+y+1=0 B.x+y1=0C.3xy1=0 D.3xy+1=03.(多選)下列結(jié)論正確的有()A.若函數(shù)f(x)=xsinx+cos2x,則f'(x)=sinxxcosx+2sin2xB.設(shè)函數(shù)f(x)=xlnx,若f'(x0)=2,則x0=eC.已知函數(shù)f(x)=3x2e2x,則f'(1)=12eD.設(shè)函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且f(x)=x2+3xf'(2)+lnx,則f'(2)=94.(多選)已知函數(shù)f(x)在x=1處的導(dǎo)數(shù)為12,則f(x)的解析式可能為(A.f(x)=12x2+12lnx B.f(x)=xC.f(x)=sin2x+π3 D.f(x)=1x5.若曲線(xiàn)f(x)=acosx與曲線(xiàn)g(x)=x2+bx+1在交點(diǎn)(0,m)處有公切線(xiàn),則a+b=()A.1 B.0 C.1 D.26.若函數(shù)y=f(x)的圖像上存在兩點(diǎn),使得函數(shù)的圖像在這兩點(diǎn)處的切線(xiàn)互相垂直,則稱(chēng)y=f(x)具有T性質(zhì).下列函數(shù)中具有T性質(zhì)的是()A.y=sinx B.y=lnxC.y=ex D.y=x37.(2019全國(guó)3,文7,理6)已知曲線(xiàn)y=aex+xlnx在點(diǎn)(1,ae)處的切線(xiàn)方程為y=2x+b,則()A.a=e,b=1 B.a=e,b=1C.a=e1,b=1 D.a=e1,b=18.(2020河北唐山一模,文14)曲線(xiàn)f(x)=ex+2sinx1在點(diǎn)(0,f(0))處的切線(xiàn)方程為.

9.(2020山東德州二模,14)已知f(x)為奇函數(shù),當(dāng)x<0時(shí),f(x)=ex3+2ex,則曲線(xiàn)y=f(x)在(1,f(1))處的切線(xiàn)方程是.

10.(2020山東青島二模,15)已知函數(shù)f(x)=exax的圖像恒過(guò)定點(diǎn)A,則點(diǎn)A的坐標(biāo)為;若f(x)在點(diǎn)A處的切線(xiàn)方程為y=2x+1,則a=.

綜合提升組11.(2020陜西西安中學(xué)八模,理5)已知函數(shù)f(x)=x2lnx+1f'(1)x,則函數(shù)f(x)的圖像在點(diǎn)(1,f(1))處的切線(xiàn)斜率為()A.12 B.C.123e D.3e12.已知f(x)為偶函數(shù),當(dāng)x>0時(shí),f(x)=lnx3x,則曲線(xiàn)y=f(x)在點(diǎn)(1,3)處的切線(xiàn)與兩坐標(biāo)軸圍成圖形的面積等于()A.1 B.34 C.1413.(2020全國(guó)3,理10)若直線(xiàn)l與曲線(xiàn)y=x和圓x2+y2=15都相切,則l的方程為(A.y=2x+1 B.y=2x+1C.y=12x+1 D.y=1214.(2020廣東茂名一模,理15)P為曲線(xiàn)y=2x2+ln(4x+1)x>-14圖像上的一個(gè)動(dòng)點(diǎn),α為曲線(xiàn)在點(diǎn)P處的切線(xiàn)的傾斜角,則當(dāng)α取最小值時(shí)15.若直線(xiàn)y=kx+b是曲線(xiàn)y=lnx+2的切線(xiàn),也是曲線(xiàn)y=ln(x+1)的切線(xiàn),則b=.

創(chuàng)新應(yīng)用組16.已知f'(x)=2x+m,且f(0)=0,函數(shù)f(x)的圖像在點(diǎn)A(1,f(1))處的切線(xiàn)的斜率為3,數(shù)列1f(n)的前n項(xiàng)和為Sn,則S2021的值為(A.2021C.202017.(2020江西上饒三模,文12)已知曲線(xiàn)f(x)=ex+1與曲線(xiàn)g(x)=e24(x2+2x+1)有公切線(xiàn)l:y=kx+b,設(shè)直線(xiàn)l與x軸交于點(diǎn)P(x0,0),則x0的值為(A.1 B.0 C.e D.e參考答案課時(shí)規(guī)范練14導(dǎo)數(shù)的概念、意義及運(yùn)算1.Bf'(x0)=3,則limh→0f(x0+h)-f2.B設(shè)x≥0,則x≤0,則f(x)=x2x.因?yàn)楹瘮?shù)y=f(x)為奇函數(shù),所以f(x)=x2x=f(x),即f(x)=x2+x,x≥0.此時(shí)f'(x)=2x+1,x≥0.當(dāng)x=1時(shí),f'(1)=1.又因?yàn)閒(1)=0,所以切點(diǎn)坐標(biāo)為(1,0).故切線(xiàn)方程為y=(x1),即x+y1=0.3.BD對(duì)于A,f'(x)=sinx+xcosx2sin2x,故A錯(cuò)誤;對(duì)于B,f'(x)=lnx+1,若f'(x0)=lnx0+1=2,則x0=e,故B正確;對(duì)于C,f'(x)=6xe2x+6x2e2x,則f'(1)=12e2,故C錯(cuò)誤;對(duì)于D,f'(x)=2x+3f'(2)+1x,則f'(2)=94,故D正確.故選4.ADA中f'(x)=12x2+12lnx'=x+12x,f'(1)=1+12=12;B中f'(x)=(xex)'=ex+xex,f'(1)=2e;C中f'(x)=sin2x+π3'=2cos2x+π3,f'(1)=2cos2+π3≠12;D中f'(x)=1x+x'=1x2+12x.故選AD.5.C依題意得,f'(x)=asinx,g'(x)=2x+b,于是有f'(0)=g'(0),即asin0=2×0+b,則b=0.又因?yàn)閙=f(0)=g(0),即m=a=1,所以a+b=1.故選C.6.A當(dāng)y=sinx時(shí),y'=cosx,因?yàn)閏os0·cosπ=1,所以在函數(shù)y=sinx的圖像上存在兩點(diǎn)x=0,x=π使條件成立,故A正確;函數(shù)y=lnx,y=ex,y=x3的導(dǎo)數(shù)值均非負(fù),不符合題意.故選A.7.D∵y'=aex+lnx+1,∴k=y'|x=1=ae+1=2,∴ae=1,a=e1.將點(diǎn)(1,1)代入y=2x+b,得2+b=1,∴b=1.故選D.8.y=3x由題可得,f'(x)=ex+2cosx,故f'(0)=e0+2cos0=3.又f(0)=e0+2sin01=0,故切線(xiàn)方程為y=3x.9.exy2e=0因?yàn)槠婧瘮?shù)在關(guān)于原點(diǎn)對(duì)稱(chēng)的兩點(diǎn)處的切線(xiàn)平行,且f'(x)=3ex22ex,x<0.所以f'(1)=f'(1)=e.又因?yàn)閒(1)=f(1)=e,所以切線(xiàn)為y+e=e(x1),即exy2e=0.10.(0,1)1當(dāng)x=0時(shí),f(0)=e0a×0=1,所以f(x)的圖像恒過(guò)定點(diǎn)(0,1).由題意,f'(x)=exa,f'(0)=e0a=1a,所以1a=2,a=1.11.A∵f(x)=x2lnx+1f'(1)x,∴f'(x)=2xlnx+xf'(1),∴f'(1)=1f'(1),解得f'(1)=12,則函數(shù)f(x)的圖像在點(diǎn)(1,f(1))處的切線(xiàn)斜率為12.12.C設(shè)x<0,則x>0,于是f(x)=ln(x)3(x)=ln(x)+3x.因?yàn)閒(x)為偶函數(shù),所以當(dāng)x<0時(shí),f(x)=ln(x)+3x,f'(x)=1x+3.于是曲線(xiàn)y=f(x)在點(diǎn)(1,3)處的切線(xiàn)斜率k=f'(1)=2.因此切線(xiàn)方程為y+3=2(x+1),即y=2x1.故切線(xiàn)與兩坐標(biāo)軸圍成圖形的面積S=12×1×13.D由y=x得y'=12x,設(shè)直線(xiàn)l與曲線(xiàn)y=x的切點(diǎn)為(x0,x0),則直線(xiàn)l的方程為yx0=12x0(xx0),即12x0xy+12x0=0,由直線(xiàn)l與圓x2+y2=15相切,得圓心(0,0)到直線(xiàn)l的距離等于圓的半徑r=55,即|114.14設(shè)切點(diǎn)P(x0,y0)y'=4x+4∵x0>14,∴4x0+1>則tanα=4x0+44x0+1=4x0+1+44x0+1當(dāng)且僅當(dāng)4x0+1=44x0+1,即x0=1即當(dāng)x0=14時(shí),tanα最小,α取最小值15.1ln2對(duì)函數(shù)y=lnx+2求導(dǎo),得y'=1x,對(duì)函數(shù)y=ln(x+1)求導(dǎo),得y'=1x+1.設(shè)直線(xiàn)y=kx+b與曲線(xiàn)y=lnx+2相切于點(diǎn)P1(x1,y1),與曲線(xiàn)y=ln(x+1)相切于點(diǎn)P2(x2,y2),則y1=lnx1+2,y2=ln(x2+1).由點(diǎn)P1(x1,y1)在切線(xiàn)上,得y(lnx1+2)=1x1(xx1),由點(diǎn)P2(x2,y2)在切線(xiàn)上,得yln(x2+1)=1x2+1(xx2).因?yàn)檫@兩條直線(xiàn)表示同一條直線(xiàn),所以1x1=1x2+1,ln(x2+1)=ln16.Af'(x)=2x+m,可設(shè)f(x)=x2+mx+c,由f(0)=0,可得c=0.f(x)的圖像在點(diǎn)A(1,f(1))處的切線(xiàn)的斜率為2+m=3,解得m=1,即f(x)=x2+x,則1f(n)=1n2+n=1n-1n+1.數(shù)列1f(n)的前n項(xiàng)和為S17.B設(shè)曲線(xiàn)f(x)的切線(xiàn)方程的切點(diǎn)為(m,em+1),由f'(x)=ex+1,得f'(m)=em+1,