導(dǎo)數(shù)的概念、意義及運(yùn)算復(fù)習(xí)訓(xùn)練-高三數(shù)學(xué)一輪復(fù)習(xí)

考點(diǎn)規(guī)范練14導(dǎo)數(shù)的概念、意義及運(yùn)算一、基礎(chǔ)鞏固1.已知函數(shù)f(x)=3x+1,則limΔxA.-13 B.13 C.23 2.(多選)下列各式正確的是()A.(x-5)'=-5x-6 B.(cosx)'=sinxC.(sinx)'=cosx D.sinπ3'=3.已知函數(shù)f(x)在R上滿足f(2-x)=2x2-7x+6,則曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程是()A.y=2x-1 B.y=xC.y=3x-2 D.y=-2x+34.已知y=f(x)是可導(dǎo)函數(shù),如圖,直線y=kx+2是曲線y=f(x)在點(diǎn)(3,f(3))處的切線,令g(x)=xf(x),g'(x)是g(x)的導(dǎo)函數(shù),則g'(3)等于()A.-1 B.0 C.2 D.45.已知曲線f(x)=x3-x+3在點(diǎn)P處的切線平行于直線y=2x-1,則點(diǎn)P的坐標(biāo)為()A.(1,3) B.(-1,3)C.(1,3)和(-1,3) D.(1,-3)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=-1 B.a=e,b=1C.a=e-1,b=1 D.a=e-1,b=-17.已知函數(shù)f(x)=-12x2+2xf'(2021)+2021lnx,則f'(2021)=.8.(2020全國Ⅲ,文15)設(shè)函數(shù)f(x)=exx+a.若f'(1)=e49.設(shè)函數(shù)f(x)=g(x)+x2,曲線y=g(x)在點(diǎn)(1,g(1))處的切線方程為y=2x+1,則曲線y=f(x)在點(diǎn)(1,f(1))處切線的斜率為.

10.已知直線ax-by-3=0與曲線f(x)=xex在點(diǎn)P(1,e)處的切線互相垂直,則ab=.11.函數(shù)f(x)=ln(2x+3)-212.若函數(shù)f(x)=12x2-ax+lnx存在垂直于y軸的切線,則實(shí)數(shù)a的取值范圍是.二、綜合應(yīng)用13.若函數(shù)y=f(x),y=g(x)的導(dǎo)函數(shù)的圖象如圖所示,則y=f(x),y=g(x)的圖象可能是()14.若點(diǎn)P是曲線y=x2-lnx上任意一點(diǎn),則點(diǎn)P到直線y=x-2的距離的最小值為()A.1 B.2 C.22 D.15.(多選)若函數(shù)y=f(x)的圖象上存在兩點(diǎn),使得函數(shù)的圖象在這兩點(diǎn)處的切線互相垂直,則稱y=f(x)具有T性質(zhì),則下列函數(shù)中具有T性質(zhì)的是()A.y=cosx B.y=lnxC.y=ex D.y=x216.已知f(x),g(x)分別是定義在R上的偶函數(shù)和奇函數(shù),且f(x)-g(x)=ex+x2+1,則函數(shù)h(x)=2f(x)-g(x)的圖象在點(diǎn)(0,h(0))處的切線方程是.

三、探究創(chuàng)新17.已知直線y=kx是曲線y=ex的切線,也是曲線y=lnx+m的切線,則實(shí)數(shù)k=,實(shí)數(shù)m=.

考點(diǎn)規(guī)范練14導(dǎo)數(shù)的概念、意義及運(yùn)算1.AlimΔx→0f(1-2.AC(x-5)'=-5x-6,A正確;(cosx)'=-sinx,B錯誤;(sinx)'=cosx,C正確;sinπ3'=0,D3.C令2-x=t,可得x=2-t,代入f(2-x)=2x2-7x+6,得f(t)=2(2-t)2-7(2-t)+6,化簡整理得f(t)=2t2-t,即f(x)=2x2-x,則f'(x)=4x-1,可得f(1)=1,f'(1)=3,故所求切線方程為y-1=3(x-1),即y=3x-2.4.B由題圖可知曲線y=f(x)在點(diǎn)(3,f(3))處切線的斜率等于-13,故f'(3)=-∵g(x)=xf(x),∴g'(x)=f(x)+xf'(x),∴g'(3)=f(3)+3f'(3).又由題圖可知f(3)=1,∴g'(3)=1+3×-135.C∵f(x)=x3-x+3,∴f'(x)=3x2-1.設(shè)點(diǎn)P(x,y),則f'(x)=2,即3x2-1=2,解得x=1或x=-1,故P(1,3)或(-1,3).經(jīng)檢驗(yàn),點(diǎn)(1,3),(-1,3)均不在直線y=2x-1上,符合題意.故選C.6.D∵y'=aex+lnx+1,∴k=y'|x=1=ae+1=2,∴ae=1,a=e-1.將(1,1)代入y=2x+b,得2+b=1,解得b=-1.7.2020由題意,得f'(x)=-x+2f'(2021)+2因此有f'(2021)=-2021+2f'(2021)+2所以f'(2021)=2020.8.1對函數(shù)f(x)=exx+a求導(dǎo)得f'(x)=ex(x+a-19.4由導(dǎo)數(shù)的幾何意義及條件,得g'(1)=2.∵函數(shù)f(x)=g(x)+x2,∴f'(x)=g'(x)+2x,∴f'(1)=g'(1)+2=4,∴曲線y=f(x)在點(diǎn)(1,f(1))處切線的斜率為4.10.-12e對函數(shù)f(x)=xex求導(dǎo)可得f'(x)=x'ex+x(ex)'=ex(x+1),則曲線f(x)=xex在點(diǎn)P(1,e)處的切線的斜率為k=f'(1)=e1×(1+1)=2e又直線ax-by-3=0與切線垂直,則有ab=-11.12∵f'(x)=2x∴f'(-1)=-4,∴切線方程為y=-4x-2.∴切線在x軸、y軸上的截距分別為-12,-∴所求三角形的面積為112.[2,+∞)∵f(x)=12x2-ax+lnx,∴f'(x)=x-a+∵f(x)存在垂直于y軸的切線,∴f'(x)存在零點(diǎn),∴x+1x-a=0有解,∴a=x+1x≥當(dāng)且僅當(dāng)x=1時,取等號.13.D由y=f'(x)的圖象知y=f'(x)在區(qū)間(0,+∞)內(nèi)單調(diào)遞減,說明函數(shù)y=f(x)圖象的切線的斜率在區(qū)間(0,+∞)內(nèi)也單調(diào)遞減,故可排除A,C.又由圖象知y=f'(x)與y=g'(x)的圖象在x=x0處相交,說明y=f(x)與y=g(x)的圖象在x=x0處的切線的斜率相同,故可排除B.故選D.14.B因?yàn)楹瘮?shù)y=x2-lnx的定義域?yàn)?0,+∞),所以y'=2x-1x(x>0).令2x-1x=1,解得x=1,則曲線在點(diǎn)P(1,1)處的切線方程為x-y=0,所以兩平行線間的距離為d=215.AD由題意,若y=f(x)具有T性質(zhì),則存在x1,x2,使得f'(x1)f'(x2)=-1.對于選項(xiàng)A,因?yàn)閒'(x)=-sinx,所以存在x1=π2,x2=-π2,使得f'(x1)f'(x2)對于選項(xiàng)B,因?yàn)閒'(x)=1x>0,所以不存在x1,x2,使得f'(x1)f'(x2)=-對于選項(xiàng)C,因?yàn)閒'(x)=ex>0,所以不存在x1,x2,使得f'(x1)f'(x2)=-1;對于選項(xiàng)D,因?yàn)閒'(x)=2x,所以存在x1=1,x2=-14,使得f'(x1)f'(x2)=4x1x2=-1.故選AD16.x-y+4=0∵f(x)-g(x)=ex+x2+1,且f(x)是偶函數(shù),g(x)是奇函數(shù),∴f(-x)-g(-x)=f(x)+g(x)=e-x+x2+1,∴f(x)=ex+e-x+2x2∴h(x)=2f(x)-g(x)=ex+e-x+2x2+2-e-x-ex2=32ex+∴h'(x)=32ex-12e-x+4x,即h'(0)=32又h(0)=4,∴切線方程為x-y+4=0.17.e2對于曲線y=ex,設(shè)切點(diǎn)坐標(biāo)為(x0,ex因?yàn)閥'=