第二章DIERZHANG變化率與導(dǎo)數(shù)§4導(dǎo)數(shù)的四則運算法則課后篇鞏固提升A組1.已知f'(x)是函數(shù)f(x)的導(dǎo)數(shù),f(x)=x2+2xf'(1),則f'(0)等于()A.2 B.2 C.4 D.0解析∵f'(x)=2x+2f'(1),∴f'(1)=2+2f'(1).∴f'(1)=2.∴f'(x)=2x+2×(2)=2x4.∴f'(0)=4.答案C2.下列函數(shù)中,導(dǎo)函數(shù)是偶函數(shù)的是()A.y=sinx B.y=exC.y=lnx D.y=cosx1解析由y=sinx得y'=cosx為偶函數(shù);∵當(dāng)y=ex時,y'=ex為非奇非偶函數(shù),∴B錯;∵y=lnx的定義域為x>0,∴C錯;D中y=cosx12時,y'=sinx為奇函數(shù),故D錯答案A3.已知函數(shù)f(x)=x+sinx+1,其導(dǎo)函數(shù)記為f'(x),則f(2021)+f'(2021)+f(2021)f'(2021)=()A.2021 B.2 C.1 D.0解析因為f'(x)=1+cosx,所以f'(x)為偶函數(shù),所以f'(2021)f'(2021)=f'(2021)f'(2021)=0,所以原式等價于f(2021)+f(2021)=2021+sin2021+1+(2021sin2021+1)=2.故選B.答案B4.曲線y=2x36x上切線平行于x軸的點的坐標(biāo)為()A.(1,4) B.(1,4)C.(1,4)或(1,4) D.(1,4)或(1,4)解析y'=6x26,由y'=0,得x=±1,分別代入y=2x36x,得y=4或y=4,即所求點為(1,4)或(1,4).答案D5.曲線y=3x+sinx在(0,0)點處的切線方程為.

解析對函數(shù)y=3x+sinx求導(dǎo)得y'=3+cosx,則y'|x=0=4,因此,曲線y=3x+sinx在(0,0)點處的切線方程為y=4x,即4xy=0.答案4xy=06.若f(x)=x22x4lnx,則f'(x)>0的解集為.

解析由f(x)=x22x4lnx,得函數(shù)的定義域為(0,+∞),且f'(x)=2x24x=2x2-2x-4x=2(x+1)(x-2)x,由f'答案(2,+∞)7.已知f'(x)是函數(shù)f(x)的導(dǎo)數(shù),f(x)=f'(1)·3x+x2,則f'(2)=.

解析因為f(x)=f'(1)·3x+x2,所以f'(x)=f'(1)·3xln3+2x,得f'(1)=f'(1)·3ln3+2,則f'(1)=21所以f'(2)=21-3ln3×9ln3+4答案6ln3+48.若曲線C:y=x32ax2+2ax上任意一點處的切線的傾斜角都是銳角,則實數(shù)a的取值范圍是.

解析∵曲線在任意一點處的切線的傾斜角都是銳角,∴y'=3x24ax+2a>0恒成立.∴Δ=16a224a<0.∴0<a<32答案0<a<39.求下列函數(shù)的導(dǎo)數(shù):(1)y=x·cosx+x;(2)y=sin4x4+cos4x(3)y=lgx解(1)y'=(x·cosx)'+(x)'=cosxx·sinx+12(2)∵y=sin4x4+cos4=sin2x4+cos2=112sin2x2=112∴y'=34+14cos(3)y'=(=x=1-10.曲線C:y=ax3+bx2+cx+d在點(0,1)處的切線為l1:y=x+1,在點(3,4)處的切線為l2:y=2x+10,求曲線C的方程.解由已知得點(0,1)與點(3,4)均在曲線C上,∴d=1,27a+9b+3c+由導(dǎo)數(shù)的幾何意義得c解得d=1,c=1,a=13,b=1所以曲線C的方程為y=13x3+x2+x+1B組1.直線y=kx+1與曲線y=x3+ax+b相切于點A(1,3),則2a+b的值為()A.2 B.1 C.1 D.2解析由條件可知,點A(1,3)在直線y=kx+1上,則k=2.∵點A在曲線y=x3+ax+b上,∴a+b+1=3,即a+b=2.由y=x3+ax+b,得y'=3x2+a,∴3+a=k=2.∴a=1,b=3.∴2a+b=1.答案C2.(2021全國甲,理13)曲線y=2x-1x+2在點(1,解析由y=2x-1x+2,得y'=5(x+2)2,則在點(1,3)處的切線的斜率為5,所以切線方程為y+3=5(答案5xy+2=03.函數(shù)f(x)=(xa)(xb)(xc)(a,b,c是兩兩互不相等的常數(shù)),則af'(a)解析∵f(x)=x3(a+b+c)x2+(ab+bc+ca)xabc,∴f'(x)=3x22(a+b+c)x+ab+bc+ca.∴f'(a)=(ab)(ac),同理f'(b)=(ba)(bc),f'(c)=(ca)(cb).代入原式,得af'(a答案04.對正整數(shù)n,設(shè)曲線y=xn(1x)在x=2處的切線與y軸交點的縱坐標(biāo)為{an},求數(shù)列ann+1的前解∵y=xn(1x),∴y'=nxn1(1x)xn=nxn1(n+1)xn.∴當(dāng)x=2時,y'=n·2n1(n+1)2n=(n+2)·2n1,f(2)=2n.∴所求的切線方程為y+2n=(n+2)·2n1(x2),令x=0,則y=(n+1)·2n.∴an=(n+1)·2n,ann+1=故數(shù)列ann+1的前n項和為2(1-5.設(shè)函數(shù)f(x)=x3+2ax2+bx+a,g(x)=x23x+2,其中x∈R,a,b為常數(shù),已知曲線y=f(x)與y=g(x)在點(2,0)處有相同的切線,求a,b的值,并