高数第2章练习题供参考

|x3讨论函数f(x)x2

第二章总练习题的连续性和可导性 2x

4 13解x1时f(x)可导.f(10)lim x f(10)=lim|x3|2

x1f(10)

4

1,f(1)x23x13

x3 1 4

2

f(1),f

2x设函数f(xAx3Bx2Cx5x

x1试确定常数ABCD的值使f(x)在()可导解f(10lim(2x2)4

f(1)ABC f(1)(2x2) 2f(1)(Ax3Bx2Cx (3Ax22BxC)|x13A2Bf(10)ABCDf(10)f(1)3A2BCABCD3A2BCABCD

{A-9/4,B3/4,C41/4,D13/设函数g(xsin2xf(x其中f(x)在x0连续.问g(x)在x0是否可导若可导求出g(xg(0)2f(xsin2x2f(0)(x0g(0)2f x2sin2 cos2问函数f(x)= 与g(x)= 为什么有相同得导数解因为f(xg(x)5,.设函数f(x)在[1,1]上有定义且满足xf(xx2xx[1,1].证明存在且等于证0f(0)0,f(0)0.xf(x)f(0)f(x) 故f(0)

x x 1( 0),f 1,类似f 设f(x|x24|求f解|x|2时,f(x)x24,f(x)2x.f(2)(x24) 4,f(2)不存在,同理f(1 d2设y1xdx2解y=- 2,dy

d2 , ,1x (1 (1设函数f(x)在(,)上有定义,且满足下列性质任意x()都有f(x)f(0)ff(xxf(x)

f(x)f(x)f(x)ff(x)f(x)f(0)

f(0)f(x)(x0),f(x)f(0)f 设f(x) x1/

(n1,2,);g(x)

x1/

(n1,问f(x)在x0处是否可导g(x)在x0处是否可导f(1/2n)f 1/ 解 0(n1/ 1/ f(x)f(0)00(x1/2n,x0).limf(x)f(0)0,f(0)xg(1/2n)g(0)1/2n11

1/ 1/ 2 g(xg(0)00(x1/2nx0limg(xg(0).g(0)不存在x设

f(x)及yg(x)在[ab]上连续 bf(x)g(x)dx2bf2(x)dxbg b证a[f(xb

2tg(x)]dx

bg2(x)dxt2

bf(x)g(x)dxt

bf2(x)dxbg2x)dx0,则由g的连续性g(x)0,x[ab],不等式两端都a如果bg2(x)dx0,(*)左端的二次函数恒非负,故其判别式非正a 2bf(x)g(x)dx24bg2(x)dxbf2(x)dx bf(x)g(x)dx2bf2(x)dxbg2 求出函f(x)1x1x21 在点x1的导数再将函数f(x)写成f(x)

x/2(x/2)n11x/2

的形式再求f由此证明下列12

2n2证f(x)12xnxn1,f(1)12nf(x)

x/2(x/,1x/

2 f(x)

(1/2(n1)(x/2)n(1/2))(1x/2)(1/2)(x/2(x/(1x/ (1/2(n1)(1/2n1))(1/2)(1/2)(1/21/2n1)1/22(1(n1)/2n)11/2n2n2由类似上题的办法证明1+2x+3x2x

1(n1)xn(1

(x证由等比级数求和xx2xn两端求导得1+2x+3x2

1 (1(n1)xn)(1x)(xxn1)1(n1)xn (1 (1

设yf(x)在[0,1]连续且f(x)>0证

1

dx 0f

10f1证111dx lnxn

dx1f(x)dx11f f 0f 1 1 nn1ln n

(nn

11

1 lnn n1 n n证(1) 11/ ln1111/n 11/n n1 11/ n t

n lnnln2 ln1111111 1 n n lnnln111111

1 1

nln n

en

en1en1|x3讨论函数f(x)x2

第二章总练习题的连续性和可导性 2x

4 13解x1时f(x)可导.f(10)lim x f(10)=lim|x3|2

x1f(10)

4

1,f(1)x23x13

x3 1 4

2

f(1),f

2x设函数f(xAx3Bx2Cx5x

x1试确定常数ABCD的值使f(x)在()可导解f(10lim(2x2)4

f(1)ABC f(1)(2x2) 2f(1)(Ax3Bx2Cx (3Ax22BxC)|x13A2Bf(10)ABCDf(10)f(1)3A2BCABCD3A2BCABCD

{A-9/4,B3/4,C41/4,D13/设函数g(xsin2xf(x其中f(x)在x0连续.问g(x)在x0是否可导若可导求出g(xg(0)2f(xsin2x2f(0)(x0g(0)2f x2sin2 cos2问函数f(x)= 与g(x)= 为什么有相同得导数解因为f(xg(x)5,.设函数f(x)在[1,1]上有定义且满足xf(xx2xx[1,1].证明存在且等于证0f(0)0,f(0)0.xf(x)f(0)f(x) 故f(0)

x x 1( 0),f 1,类似f 设f(x|x24|求f解|x|2时,f(x)x24,f(x)2x.f(2)(x24) 4,f(2)不存在,同理f(1 d2设y1xdx2解y=- 2,dy

d2 , ,1x (1 (1设函数f(x)在(,)上有定义,且满足下列性质任意x()都有f(x)f(0)ff(xxf(x)

f(x)f(x)f(x)ff(x)f(x)f(0)

f(0)f(x)(x0),f(x)f(0)f 设f(x) x1/

(n1,2,);g(x)

x1/

(n1,问f(x)在x0处是否可导g(x)在x0处是否可导f(1/2n)f 1/ 解 0(n1/ 1/ f(x)f(0)00(x1/2n,x0).limf(x)f(0)0,f(0)xg(1/2n)g(0)1/2n11

1/ 1/ 2 g(xg(0)00(x1/2nx0limg(xg(0).g(0)不存在x设

f(x)及yg(x)在[ab]上连续 bf(x)g(x)dx2bf2(x)dxbg b证a[f(xb

2tg(x)]dx

bg2(x)dxt2

bf(x)g(x)dxt

bf2(x)dxbg2x)dx0,则由g的连续性g(x)0,x[ab],不等式两端都a如果bg2(x)dx0,(*)左端的二次函数恒非负,故其判别式非正a 2bf(x)g(x)dx24bg2(x)dxbf2(x)dx bf(x)g(x)dx2bf2(x)dxbg2 求出函f(x)1x1x21 在点x1的导数再将函数f(x)写成f(x)

x/2(x/2)n11x/2

的形式再求f由此证明下列12

2n2证f(x12xnxn1,f(112n 2 由类似上题的x/2(x/f(x)f(x)

1x/ (1/2(n1)(x/2)n(1/2))(1x/2)(1/2)(x/2(x/(1x/ (1/2(n1)(1/2n1))(1/2)(1/2)(1/21/2n1)1/22(1(n1)/2n)11/2n2n21(n1)xnnxn1(1

(x

设

f(x)在[0,1]连续且f(x)>0证

11dx 0f 0f证1=1dx f(x) dx1f(x)dx11ff lnxn

0fn 1 1 n

n1ln

(nn 111lnn111 1 11

1

nen1

e.n证(1) 11/ ln1111/n 11/n n1 11/ n t 1

n (2)lnnln2 ln1111111 lnnln1111lnnln111111 n 111 (c)en1

e.n证(1) 11/ ln1111/n 11/n n1 11/ n t 1

n lnnln2 ln1111111 n111 1 n lnnln 1

nln

1 n

1 n

en1en1 设一物质细杆的长为

量为m(x)2x2(0xl),求比值m,问它的物理意义是什么求极限limm,问它的物理意义是什么x0解(1)m2(xx)22x22(x22xxx22x22(2xxx2m2(2xxx2)2(2xxm是x到xx limmlim2(2xx4xlimm是细杆在点xx0 x0根据定义求下列函数的导函数(1)yax3;(2)y 2px,p0;(3)ysin解(1y

a(xx)3 (x33x2x3xx2x3) alim alim(3x3xxx)3ax

y

2p(xx)2px limxx 2x 2x

2(2(xx x)(xx xx(xx x2x2xx 2p2ylimsin5(xx)sin5x

22xx x2cos5(2xx)sin 2 5(2x 5(2x lim 25limcos lim 25cos 2

求下列曲线yf(x)在指定点M(x0f(x0处的切线方程 (2)yx22,解(1y2xln2y(0)ln2切线方程y1ln2(x0yln2)x(2)y2xy(3)6,切线方程y116(x试求抛物线y22pxp0)上任一点M(xy)(x0,y0) 证y 2px,y p,过点M的切线PMN方程：Yyp(Xx).2222 切线与x轴交点N(X0,0),y2FN22

y(X0x),X0x222

px2pxp2 x2pxp22过M作PQ平行于x轴,则PMQFNM曲线yx22x3上哪一点的切线与直线y4x1平行,并求曲线在该点的切y2x24,x01,y06k

1

切线方程:y64(x1y4x2.法线方程:y6

4(x1),y4x4 离地球中心r处的重力加速度g是r的函数 其表达式g(r)

,r

其中R是地球的半径,M是地球的质量,G是引力常GM,r问g(r)是否为r的连续函数作g(r)的草图g(r)是否是r的可导函数解明显地,rR时g(r)连续.limg(rlimGMrGMr rR limg(r)limGMGMlimg(rg(r)在rRr

rRr r(3)rR时g(r)可导g(RGMg(R)2GMg(Rg(r)在rR不可导 求二次函数P(x),已知:点(13)在曲线yP(x)上,且P(0)3,P(2)abc4abb3,a1,c3(ab)1,P(x)1x23x1 求下列函数的导函数y8x3x7,y24x2y(5x3)(6x22),y5(6x22)12x(5x3)90x236xy(x1)(x1)tanx(x21)tanx,y(2x)tanx(x21)sec2y9xx2,y(92x)(5x6)5(9xx2)5x212x5x (5x (5x y1x1

1

(x1),y (1y (x1),y x3 (x3y

x2x1,y

(2x1)exex(x2x1)e2

x2x yx10x,y10xx10xln1010x(1xyxcosxsinx,ycosxxsinxxcosxsinx yexsinx,yexsinxexcosxex(sinxcos定义若多项式P(x)可表为P(xxx0)mg(xgx0则称x0是P(x)的m重根.已知x0是P(x)的k重根,证明x0是P(x)的 1)重(k证P(x)xx0)kg(xg(x0) P(x)k(xx)k1g(x)(xx)k (xx0)k1(kg(x)(xx0)g(x))(xxh(x0kg(0x0,由定义x0是P(x)的(k1)若f(x)在(aa)中有定义且满足f是偶函数,且f(0)存在试证明f(0

f(x),则称f(x)为偶函数.设f证f(0limf(xf(0)limf(xf(0)limf(xf(0)f(0),f(0) 设f(x)在x处可导证明limf(x0xf(x0x)2f(x limf(x0xf(x0x)1limf(x0xf(x0)f(x0xf(x0) 2x0 1limf(x0x)f(x0)f(x0x)f(x0)2x0 01limf0

x)f(x0)

f(x0x)f(x0)1[f(x)f(x)]f(x2 一质点沿曲线yx2运动且已知时刻t(0t2)P(tx(ty(t))满足直线OP与x轴的夹角恰为t.求时刻t时质点的位置速度及解 x2

tant,y(t) xx(t) (位置(tanttan2v(t)(sec2t,2tantsec2v(t)(2sec2ttant,2sec4t4tan2tsec22sec2t(sec2t,2tan2 ,xf(x)1e1/ x 解f(0)lim1e1/x 1,f(0)lim1e1/x x01e1/ x01e1/设f(x|xa|(x其中(x)在xa处连续且(a0.证明f(x)在xa不可导证f(a)lim(ax)(x)(af(alim(xa)(x)(af x x x) x) x x2(2)[ln(1x)] 1,错.[ln(1x)] 1(1x) 11x21x2

1x

x

x21x2

11x21x2x21x2x21x21

x2 11111111

x2sin2

(14sinxcos x2sin2记f(g(xf(u|u()gx.现设f(xx2求f(x),f(0),f(x2f(sin求df(x2),df(sinx); f(g(x))与f(g(x))是否相同 两者的关系解(1f(x2x,f(00,f(x22x2,f(sinx2sindf(x2)f(x2

x22x22xdf(sinx)

f(sinx)(sinx)2sinxcosxsinf(g(x))与f(g(x))不同,f(g(x))fy

,y

x x3 x3 ysecx,y(cosx)1(cosx)2(cosx)(cosx)2(sinx)tanxsecysin3xcos5x,y3cos3x5sinysin3xcos3x,y3sin2xcosxcos3x3sin3xsin3sin2x(cosxcos3xsinxsin3x)3sin2xcosy1sin2x,y2sinxcosxcosx2(1sin2x)(sinx2cossin2xcosx22x(1sin2x)(sinx2 cos2

cos2y1tan3xtanxx,ytan2xsec2xsec2x1tan2xsec2xtan2xtan2x(sec2x1)tan4yeaxsinbx,yaeaxsinbxbeaxcosbxeax(asinbxbcosy 1x2,y5 1x2(sin1x2 111xsin 1ylntanx,y sec2x 4

4 tan

4

2tanxcos2x 2 4 4 4 4 1 1sec1sin(x

cosy1ln (a0,xa),y1xa(xa)(xa) 2axa (xa)2 x2a2求下列函数的导函数1ax a2ay 0),y1ax a2aay 2

x2 a2a1 a yxarccosx(|x|1),y2xarccosx 1y 1

11x2

1a2 a2y (a x 2a21ax2a21ax a2a2 a2 a2

a2x2 x2x2y ln x2 x 2x22x x2x22x22x x2x2 x2x2 2x2 2x2

x2a2y x2

,x 2(x21)

1 2sgn(1x2y

(x2(x2

x21x2tanx(ax21x2

x2 a2a2

arctan 2y

sec2x1a2b21abtan2a

sec2x ab(ab)tan2 (ab)cos2x(ab)sin2 abcosy x 2x 3x),lny x) 2x) 3x1 x)2 2x)1 x)2 2x)3 3x)yy x xx

3x)3x 2x) yyy

1x2x2,y 21xx2x221xx2a2a2x2,y a2yln(x x2a2),y 1 x x2a2 x2a2 x2y(x1)3(3x1)2(2x).lnyln(x1)2ln(3x1)ln(2 y 1 x 3x 32yy 11. 3x 32xyexeex,yexeexexex(1eexyxaaaxaaax(ayaaxaa1axalna(axa1)aaxlnaaxlnaaxaa1alnaaxaxa1aaxaxln2一的探测器瞄准着一枚安装在发射台上的火箭,它与发射台之间射后10秒钟时,探测器的仰角(t)的变化速率是多少?解(t)1t2t2,tn(t)x(t)t ,2 t

(t) ,(t)

t 2

,(10)

22500.1(弧度 1100 1100 旋转.问:当飞轮的旋转角为2解x(t2cos8t364sin28t16x(t)16sin8t16(t)8t,,t1 ,x 活塞向右移动的速率是16当x0时下列各函数是x的几阶无穷小量yx10x2100x3.1y(x2 2)sinx x2yx(1cosx)x2sin2x22

2阶2已知:当x0时,(x)o(x2试证明(x)证(x)(x)xo(1)x 设(xo(x)(x0(xo(x)(x0).试证明:(x(xo(x)(x证(x(x)(x)(x)o(1)o(1) 上述结果有时可以写成o(x)o(x)计算下列函数在指定点x0处的微

1

1 yxsinx,x/4.ysinxxcosx,

,dy

4 2 4 2 4 y(1x)1,y(0),dyy1x1

1

,y

,dy 2dx(1yxex,yexxexex(1x).dyex(1设y x

(x1),计算当x由3变到3.001时,函数的增量和向相应的微解y=-

,y(3)12y

10.001,dy 试计算532.16的近似值解532.16251.16322(11

5求下列方程所确定的隐函数的导函数1

y(1)x3y3a3(a0).1x3

y3y0,y (2)(xa)2yb)2c2abc为常数).2(xa2yby0,yxa.y

xarctanyx

x2y2y yy,xyyxyy,y y yx1 x

x2 x2 x2 xysinxcos(xy)ysinxycosxsin(xy)(1y)yycosxsin(xy).sin(xy)sinx求下列隐函数在指定的点M的导y22xyx22x40,M(3,2yy2y2xy2x20,yyx1,y(3)7319y 7 e220 5xy0,M10,e2

2 2020exy(yxy)10xy5x2y0,y ,ye xexy 10 e25 设yf(x)由下列参数方程给出,求ydy y3tdy33t23(1t).(t 2 xtlnt

y

lnt11ttx1ttyarcsin 11t

3/ 2(1t dy

2tsgn(t),t2t

y2

1上一点M0x0y0处的切线方程与法线方程.,y2x2 b2,y a2 b2x x y切线方程：yy0 0(xx0),0 a2y0

a2y 法线方程：yy 0(xx),ayxbxy(ab)x0 b2 00 b2焦点F(c,0),F(c,0),cab(ab).设y0.切线斜率k 00 a20MF的斜率 x

MF2的斜率k2y0

x b2 k b2x(xc)a2 2

2 1

ay(xc)bx 1 0

0a2y0x0a2b2 b2(a2cx b2(a2cx (a2b2)xya2cy c2xya2cy cy(a2cx) 0 0 bb k cbk cb2x(xc)a2 1b2 a2yc)b2xtanPMF1 1 0

0a2y0x0 a2b2b2cx

b2(a2

b2(a2

b2 ,故PMFF(a2b2)xy c2xy cy(a2,故PMFF

PMF和FPMF和FMQ都在区

0

22 nyxn,y(n)yex,y(n)

(1)ny (1x)(x1

(1)(11)(1n1)(1 (1 y 11,y(n)(1)nn!1 x(x x

(1

n1 设y(xexcosx证明y2y2y证yexcosxexsinxex(cosxsinyex(cosxsinx)ex(sinxcosx)ex(2siny2y2yex(2sinx)2ex(cosxsinx)2excosx设yx3(x4证明2y2y1)y.x4证yx3 ,y ,y x x (x (x2y2 ,(y1)y 2y2.(x4)4 x4 (x4)3 (x4)4 设y(1x)(2x1)23x1)3求y(6解y(6)6!(108),y(7)要使yex满足方程ypyqy0(其中pq为常数该取哪些值解yexy2exypyqy2pq)ex0,ex该取方程2pq0的根解角速度3t24t3,角加速度6t设f(x)

,其中n为一个正整数,求f(k)(x),k为一个解f(x)

(1x)n,f(k)(x)(n)(n1)(nk1)(1x)nkn(n1)(nk1),f(k)(0)n(n1)(nk1).(1x)nk设yx2ln(1x求y(50).解由Leibniz,y(50)x2ln(1x)(50)50(2x)ln(1x)(49)50492ln(122x2(1x)1(49)50(2x)(1x)1(48)50492(1x)12x2(1)(2)(1491)(1x)50100x(1)(2)(1481)(1x)492450(1)(2)(1471)(1x249!(1 !(1x)49245047!(1 ).证y=(C1eaxC2ebx)C1aeaxC2bebxyC1aeaxC2bebx)C1a2eaxC2b2ebx1证y(C1xC2)eaxC1eaxa(C1xC2)eaxeax(aC1xC1aC2yeaxa(aCxCaC)eax(aC)eax(a2Cxa2C2aC y2aya2eax(a2Cxa2C2aC)2aeax(aCxCaC)a2(CxC)eax1 验证函数yCcost

证yC1sintC2cost,y

列不定积分

b 9C5/x 3C3x2dx x5x

x x)2dx(1 x)dxx4x3/21x2x asec2xdxatanxtan2xdx(sec2x1)dxtanxxcot2d(csc21)dcot21C.x23 1x2dx11x2dxx2arctanx

3

x xx1x1 (1cos2x)sec2xdx(sec2x1)dxtanxx41xdx 4x24x3)dxx4x5/42x3/24x7/44 3

xx2x3dxln|x|

2x

(1dxx

12xx4/ dx

x4/32x1/3x2/33x1/33x2/33x5/35(2coshxsinhx)dx2sinhxcoshx3x2 (x1)2 3/ 1/ 1/2x dx3x2 x 3x12x5/24x3/2

xx

x dx

1 dxcotxtanxsin2xcos2 cos2x 1 11x2x13x dx2392 1x/ln311x/ln223 92

dx1 dx

arctanxx2(1x2 1x2 求解微分方程y(x)abex(ab为常数解y(abex)dxaxbexC(axbexC)dx1ax2bex1CxC 设f(x)满足方程xf(xf(x)x21,求f解(xf(x))x21,xf(x) x31dx1x4x4f(x)1x31C akdxlimkxilimk(ba)k(b0 i(ba)b

a)2 (a

) lim(a(ba) n

a(ba)(ba)2lim(1ni)a(ba)(ba)2lim1n(nnn2 n a(ba)(ba)2lim(11/ (b b2a2

a(ba)

yd及y轴所围的图形的面积又设c0,函数xy)在[cd]上严格递增试求 积分和cy)dya(x)dx其中y(x)是xy)的反函数a(cb ddcab 解cy)dya(x)dxbd 写出函数yx2在区间[0,1]上的Riemann和,其中分割为n等分,中间点为分割小区间的左端点.求出当n时Riemannn1i2 1 1 1 2解sn (n1)n(2n1) 2 (n2i0

6 n n 1求定积分1

= xdx11y2dy112 2

(1sinx)dx 证2(1sinx)dx2(1)dx.2(1sinx)dx2(2)dx 2 11

2xx2dx32证2xx21x)(2x0,x1,x2.当x(,12)时,2xx2递增 1.当x(1/2,)时,2xx2递减, 1

2dx

2xx2dx 21/21/4dx 判断下列各题中两个积分值之大小1 10edx0e0/2x2dx/2(sin0 1

xdx

1x2 设函数yf(x)在[ab]上有定义,并且假定yf(x)在任何闭子区间上有最大值和最小值.对于任意一个分割Tx0ax1x2xn1xnb记mi为f(x)在[xi1,xi]中的最小值,Mi为f(x)在[xi1,xi]中的最 yf(x)在[ab]limmixilimMixi相等

(T)n

(T) 证设yf(x)在[ab]

mixilimf(i)xiaf

(T)0 Mixilimf(i)xiaf

(T)0(T)0 (T)0 limmixilimMixI(T)0 (T)0 mixif(i)xiMix, nlimf(i)xi(T)0 F(x)

,F(x) 1t 1G(x)

21 2sintdt,G(x)