高中考試數(shù)學(xué)特訓(xùn)練習(xí)含答案-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f(x+h)-f(x-h)1.若f'(x0)=-3,則lim00=()h?→0A.-3C.-9B.-6D.-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的切線方程是()A.x+y+1=0B.x+y-1=0C.3x-y-1=0D.3x-y+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)=sinx-xcosx+2sin2xB.設(shè)函數(shù)f(x)=xlnx,若f'(x)=2,則x=e00C.已知函數(shù)f(x)=3x2e2x,則f'(1)=12e94D.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)=-124.(多選)已知函數(shù)f(x)在x=1處的導(dǎo)數(shù)為-,則f(x)的解析式可能為()1212A.f(x)=-x2+lnxB.f(x)=xexπ1?C.f(x)=sin2x+3D.f(x)=+?5.若曲線f(x)=acosx與曲線g(x)=x2+bx+1在交點(diǎn)(0,m)處有公切線,則a+b=()A.-1B.0C.1D.2.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)處的切線互相垂直,則稱y=f(x)具有T性質(zhì).下列函數(shù)中具有T性質(zhì)的是()6A.y=sinxC.y=exB.y=lnxD.y=x37.(2019全國3,文7,理6)已知曲線y=aex+xlnx在點(diǎn)(1,ae)處的切線方程為y=2x+b,則()A.a=e,b=-1B.a=e,b=1C.a=e-1,b=1D.a=e-1,b=-18.(2020河北唐山一模,文14)曲線f(x)=ex+2sinx-1在點(diǎn)(0,f(0))處的切線方程為.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+2e-x,則曲線y=f(x)在(1,f(1))處的切線方程是.0.(2020山東青島二模,15)已知函數(shù)f(x)=ex-ax的圖像恒過定點(diǎn)A,則點(diǎn)A的坐標(biāo)為;若f(x)在點(diǎn)A處的切線方程為y=2x+1,則a=.1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+1-f'(1)x,則函數(shù)f(x)的圖像在點(diǎn)(1,f(1))處的切線斜率為()1212A.B.-1212C.-3eD.3e-12.已知f(x)為偶函數(shù),當(dāng)x>0時(shí),f(x)=lnx-3x,則曲線y=f(x)在點(diǎn)(-1,-3)處的切線與兩坐標(biāo)軸圍成圖形的面積等于()341412A.1B.C.D.113.(2020全國3,理10)若直線l與曲線y=?和圓x2+y2=都相切,則l的方程為()512A.y=2x+1B.y=2x+121212C.y=x+1D.y=x+1414.(2020廣東茂名一模,理15)P為曲線y=2x2+ln(4x+1)-圖像上的一個(gè)動(dòng)點(diǎn),α為曲線在點(diǎn)P?>處的切線的傾斜角,則當(dāng)α取最小值時(shí)x的值為.15.若直線y=kx+b是曲線y=lnx+2的切線,也是曲線y=ln(x+1)的切線,則b=.創(chuàng)新應(yīng)用組1?(?)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))處的切線的斜率為3,數(shù)列的前n項(xiàng)和為Sn,則S2021的值為()2202102220222023A.B.2202002120192020C.D.e2417.(2020江西上饒三模,文12)已知曲線f(x)=ex+1與曲線g(x)=(x2+2x+1)有公切線l:y=kx+b,設(shè)直線l與x軸交于點(diǎn)P(x,0),則x的值為()00A.1C.eB.0D.-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f(x+h)-f(x-h)?(?+?)-?(?)+?(?)-?(?-?)?(?0+?)-?(?)1.Bf'(x0)=-3,則lim00=???0000=lim0+limh???→0h→0?→0-?→0?(x0-?)-?(?)0=2f'(x0)=-6.-?2.B設(shè)x≥0,則-x≤0,則f(-x)=x2-x.因?yàn)楹瘮?shù)y=f(x)為奇函數(shù),所以f(-x)=x2-x=-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).故切線方程為y=-(x-1),即x+y-1=0.3.BD?對(duì)于A,f'(x)=sinx+xcosx-2sin2x,故A錯(cuò)誤;對(duì)于B,f'(x)=lnx+1,若f'(x)=lnx+1=2,則x=e,故0001?94B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)+,則f'(2)=-,故D正確.故選BD.121212?12124.ADA中f'(x)=-x2+lnx'=-x+,f'(1)=-1+=-;B中f'(x)=(xex)'=ex+xex,f'(1)=2e;C中f'(x)=sinπππ121?1?2112122x+3'=2cos2x+3,f'(1)=2cos2+3≠-;D中f'(x)=+?'=-+.f'(1)=-1+=-.2?故選AD..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)?m=f(0)=g(0),即m=a=1,所以a+b=1.故選C..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..D∵y'=aex+lnx+1,∴k=y'|x=1=ae+1=2,∴ae=1,a=e-1.將點(diǎn)(1,1)代入y=2x+b,得2+b=1,∴b=-1.故選D.678911.y=3x由題可得,f'(x)=ex+2cosx,故f'(0)=e0+2cos0=3.又f(0)=e0+2sin0-1=0,故切線方程為y=3x..ex-y-2e=0因?yàn)槠婧瘮?shù)在關(guān)于原點(diǎn)對(duì)稱的兩點(diǎn)處的切線平行,且f'(x)=3ex2-2e-x,x<0.所以f'(1)=f'(-)=e.又因?yàn)閒(1)=-f(-1)=-e,所以切線為y+e=e(x-1),即ex-y-2e=0.0.(0,1)-1?當(dāng)x=0時(shí),f(0)=e0-a×0=1,所以f(x)的圖像恒過定點(diǎn)(0,1).由題意,f'(x)=ex-a,f'(0)=e0-a=1-a,所以1-a=2,a=-1.1211.A∵f(x)=x2lnx+1-f'(1)x,∴f'(x)=2xlnx+x-f'(1),∴f'(1)=1-f'(1),解得f'(1)=,則函數(shù)f(x)的圖像在點(diǎn)12(1,f(1))處的切線斜率為.故選A.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(-1?x)+3x,f'(x)=+3.于是曲線y=f(x)在點(diǎn)(-1,-3)處的切線斜率k=f'(-1)=2.因此切線方程為y+3=2(x+1),即121214y=2x-1.故切線與兩坐標(biāo)軸圍成圖形的面積S=×1×=.故選C.1113.D由y=?得y'=,設(shè)直線l與曲線y=?的切點(diǎn)為(x,?),則直線l的方程為y-?=(x-0002?2?01121x0),即2?0x-y+?=0,由直線l與圓x2+y2=相切,得圓心(0,0)到直線l的距離等于圓的半徑r=5,即05512|?|051212=,解得x=1(負(fù)值已舍去),所以直線l的方程為y=x+.故選D.01+154?0141414.設(shè)切點(diǎn)P(x,y)-,?0>004?+1y'=4x+4.14∵x>-,∴4x+1>0,0044則tanα=4x0+4=4x0+1+-1≥2(4-1=4-1=3,)4?+1×?0+14?0+104?0+1414當(dāng)且僅當(dāng)4x0+1=4,即x=時(shí),等號(hào)成立.0?0+114即當(dāng)x=時(shí),tanα最小,α取最小值.01?1?+115.1-ln2對(duì)函數(shù)y=lnx+2求導(dǎo),得y'=,對(duì)函數(shù)y=ln(x+1)求導(dǎo),得y'=.設(shè)直線y=kx+b與曲線y=lnx+2相切于點(diǎn)P(x,y),與曲線y=ln(x+1)相切于點(diǎn)P(x,y),則y=lnx+2,y=ln(x+1).由點(diǎn)111222112211P(x,y)在切線上,得y-(lnx+2)=(x-x),由點(diǎn)P(x,y)在切線上,得y-ln(x+1)=(x-x2).因?yàn)檫@兩111112222??2+1111?+1,=121?12條直線表示同一條直線,所以ln2.解得x=,所以k==2,b=lnx+2-1=1-?2?2+111()+1,?1ln?+1=ln?+21更多高中備考資料，到公眾號(hào)【廣東小師姐升學(xué)日記】獲取16.A?f'(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))處的切線的斜率為111?1?+11?(?)122+m=3,解得m=1,即f(x)=x2+x,則==―.數(shù)列的前n項(xiàng)和為Sn,則S2021=1-?(?)?2+?121312021202211202120222022+―+…+―=1-=.故選A.17.B?設(shè)曲線f(x)的切線方程的切點(diǎn)為(m,em+1),由f'(x)=ex+1,得f'(m)=em+1,故切線方程為y-e24e24em+1=em+1(x-m).即y=em+1·