2018高等数学辅导讲义练习题详解第二章解答

第二 一元函数微分1【解】应选（D）limxf

1)

f( 存在，且

01limf()x x

f(0)

x x

x从而

f()fx1x

limxf(x

fxx0x2sin【解】应选（C）f(0) x x0f(x2xsin1cos1,limf(x不存在,故选 【解】应选

f(x)

f(x)

f(x)【解1】由题设知f(0)0,且1lim lim 2 f

x01cos x 12

x lim 2,f(0)2,故应选x 由

f(x)

1知，令f(x2x1x2f(x)2x1x2满足题设条件，但x

1cos f(x2x1x2x0f(0)2,显然可排除（A）(B)（C）,2【解】应选（B）fxx0处的左、右导数都存在可知,fxx0处左连续且右连续，故fx)x0处连续.【解】应选（D）.

lim[f(x)ex]xlim

ln[f(x)ex limln[f(x)ex]lnx 从 limln[f(x)ex]0,limf(x)f(0) xx0ln[f(xexln[1f(xex1]~f(xex limln[f(x)ex]limf(x)ex f(0)1lnx f(0)1ln【解1】直接 fff[f(x)]f0f(0)

x

f[f(x)]f(x)1

f8则f(0) 2】排除法取f(x2xf(0)2,则（A）(B)(D)均不【解】应选（C）.f(x)x1x(ex1sinx2f(xx1,x2【解】应选（C）.f(x

n1|x|nmax(1,xn1|x|n

x1xxfxx1x0不可导，选(1)正确.由于

f

10,且limx2

则limf(x0,

f(x

x

f(0)

f

x

f(x)f

f(x)fx

1且limx0,则limf(xf(0)f(0 x x f 不正确.令f(x) 显然lim 21.则lim x x0f xf(x)

x f(0)limf(x)f(0)lim[23xsin1cos1]不存在

ff(xx0

令f(x

x3

.x3x10x20x30x40xnf(x的符f(xxx1，xx4xx2xx3处取极小值，所以应选2 f(x)2

x(x1)x2(1x),

xxf(x0

2,x x

0,这两个点都是极值点.f(1)x1【解】应选（A）.f(xt2dt2xtsintdtx2sin2tdtf(x)t2dt4xx2

f(x)42xx2,f(x20.x2【解】应选（B）.由limf(xa0,limln(1xx0ln(1 lim[f(x)a]0,又由f(x)二阶可导知f(x)连续所以limf(xf(0)limf(x)alimf(x)f(0)x0ln(1 limf(x)f(0)f(0)x fx12012,且lim(ex21)

ex2则limf(x1]limf(x10limf(x xf(0)limf(x)f(0)limf(x)1x x0f(x的极小值点，选（C）.f(xf(0)0,f(0)0,f(00可知，x0yf(xyf(xxyf(xxyf(xyf(x 可知选（C）.f(xf(x)]3x2知,f(x3[f(x)]2f(xf(x3[f(x)]2f(x6f(xf(x)]22,x0f(0)0,f(0)2则(0,f(0yf(x

x21,x2x2x2

x21

x

xx2则limy x2 1,lim[yx]lim[x21x]xx2则

x

x x

x2x2x

]11

0y

x2x2x2x21x21x2【解】应选（C）.lim xx22limx2xx12x1x【解】应填

.f(1)limf(x)f(1) (x2)L(xn(n x x1(x1)(x2)L(x(1)n1(n1)!(n

n(n20.y曲线在(0,0)处对应的参数t1.dxe(1t2)d20.y 2t3 d

)2t2dt，t1时，dxdt,dy2dt,故 处dx2，所求切y2x

xcosecos 21.【解】应填xye2. ysinesin曲线在e2 2

点

2 的直角坐标为0,e22 y/2 ,

/2

esinesinecosecose2 ye2(x0)xye2f()f【解】应填1.f(e)1, f()f

11 dx 【解】应填1.由f(x) 2xet2dt1知，x1时y f(x)2e4x2,f(x)1(y)f(x)

(y)d dxf

]dx

f(x)[f

1ff(1(1) 1[f()]

3 【解】应填

xln(12x)x[2x

2

(x)]n

(1)n2n

n1. .f(x

x2x

(x2)(x

3x

xf(n)(x)

(1)n[3

(x

(x2)n1

f(x[(x1)(x2)L(xn)]g(x)x1)(x2)L(xn),(x1)(x2)L(xn)g(1)g(2)Lg(n)0,由定理得g(x)至少有n1个零点，又g(x)是g(xn1g(xn1f(x的驻点个数为【解】应填a4.f(xx42x33x24xa0f(x4x36x26x42(x1)(x2)(2x1f(x)0x2,x1x1.f(x)在(,2)上单调减,在(2,1上单调增,在(1,1) 在(1,f(2)f(1)a4,则a4【解】应填13a8;f(x3x48x36x224xa0f(x)12x324x212x2412(x1)(x2)(x1)f(x)0x1,x1,x2.f(x(,1上单调减,在(1,1)上单调增,在(1,2)(2,f(1)a19,f(1)13a,f(218a,由题设f(1)0,f(1)0,f(20,则13ax【解】F(x)0tf(x （令xtux 0(xu)f(u)dux0f(u)du0ufF(x)1

1 limx0f(u)du0uf x

lim0f(u)duxf(x)xf(x) f(x)

bkxkx0bk(k1)xk2 (

ln(1 由

xf

x2知x

f(x)1

则k3,b1f(0)1,f(0) xyey1xyey10，即由x0，y1，得y x

(2y)yey1(2y)yy2ey1yzxd zx f(lnysinx) cosx,

0d2

d

x

( 又 f(lnysinx) cosxf(lnysinx) sinxd

y d2d2d

f(0)(21)】

f(t)tf(t)fdxx(t) f d2y dt (t) 1

f

d2

d

1

1) 2

() (

(t

3f

dy dtt

2,x(0)(6t)|t0t等式tyteu2du0两端对等式t1y(0)e1，y(0)t

1,y2y22t 令tx2u，x2sin(x)1f(u)du1x2sinf(x由已知得(0) x2sin (x)x3 f(u)dux2f(xsinx)(2xsinxxcos x2sin x3

f(u)duf(xsinx)(sinxcosx)xx0(0)lim(x) 3lim1x2sinxf3x0 lim(2xsinxx2cosx)f(x2sin limf(x2sinx)lim2sinxxcosxf(0)

x x2sin 所以(x)

x3

f(u)duf(xsinx)(sinxcosx

xx x2sin lim(x) x0x3 f(u)duf(xsin sinxcos2f(0)3f(0)2(0)故(xx0x0时，(x连续，所以(x 【解】令uxt，得(x)x0f(u)d (x0)，因(x)1xf(x) xf(u)du (xx2 f(0)0(0)01xf(u)d(0)lim(x)(0)limx

f(x)Ax

x1

x x

x

f

xf(u)d0由于lim(x)lim2f(x)0f(u)du x x0x x x AA(0).所以(xx0 3y2y2yyxyyx y0，由（1）yx，将此代入厡方程有2x3x210x1，对（1）(3y22yx)y2(3y1)y22y1

10x1yy(x2d d2 d2【解】（I）由 1d

dx2t3，当t0dx20L对应的参数为t00L在(x0y0 y(4t0t0)

1(xt0x1y0，得t2t20，解得t01或t02（舍去 由t01知，切点为(2,3)yx由t0t4Lx轴交点分别为(1,0)和(17,0)S2(x1)dx2ydx91(4tt2)d(t2 921(4t2t3)dt7 【分析】令f(xx3xsinxf(x是(,上的奇函数，从而，原方程在区间(,0)和(0,)上实根个数相同，因此，只需讨论(0,)上实根个数。 f(0)0,f(x)3x21cosf(0)20,

f(x)f(x)6xsinx xx(0,x0f(x0,f(xf(x0x(x,f(x0,f(x0f(x)0， x

f(x)则原方程在区间(0,x0)上无实根，在区间(x0,)上有唯一实根。故原方程共有三个实【解】令f(x) xet2dtx3x，则f(x)是(,)上的奇函数，从而，原方0在区间(,0)和(0,上实根个数相同，因此，只需讨论(0, f(0)0,f(x)ex23x2f(0)20,

f(x)f(x)2xex26x xx(0,x0时，当x(x0,)

f(x)f(x)f(x0)0，x

f(x)则原方程在区间(0,x0)上无实根，在区间(x0,)上有唯一实根。故原方程共有三个实【解】原方程变形 x2ex1，令f(x)x2ex1 fx)ex(2xx2xex(2x0x0,x2 x(,0)时，f(x)0,f(x)单调减 x(0,2)时，f(x)0,f(x)单调增 x(2,)时，f(x)0,f(x)单调减limf(x)lim(x2ex1) f(0)1a

f(2)4

f(x)

lim(x21)1x

xe e2e则1）当0a 4e2e当a 4e2e当a 4ylnxyky1lnxylnx 当k0或k 时有唯一根；当0ke【证】fxln(1x

时有两个实根；当k 时无实根 f(x) 1x

exxex 1

1ex1x(1x)e

(xf(0)0,fx0,(x0)f(x2sinxexex4xf(x2cosxexexf(x)2sinxexex.f(x)2cosexex0.f(0)f(0)f(0)【证】ylnxaln(axaxlna f(x)(ax)lnaaln(ax)，则f(x)在[0,)内连续、可导，f(x)lna a

f(x在[0,f(0)0f(x0x0aln(ax)(ax)ln x (ax)xaax x【证】只要证a2ablna和ablnab2即ab和alnabln ln

ln ln ln分别证明f(x) ,g(x)xlnx单调增即可ln【证】令F(xeg(xf(xxF(0)eg(0)0,F(1)2eg(1) 又0f(x)dx20f(x)dx0xdx0f(xx]dx则至少存在一点c(0,1),使f(c)c0,因此F(c)0,由连续函数的零点定理知存在两点a(0,c),b(c,1).使F(a)F(b)0,由定理知至少存在一点(abF(0即eg(f(1]eg(g(f(故f(g(f( 0f(x)dx32f(x)dx3f(x)dx2f(x)dx32f0 则3f(x)dx22f(x)dx32f(c)2f c(0,2),(2 f(cf(F(xeg(x)[f(xf显然F(x)在区间[c,]上满 f()g()[f()f(【证】令F(x)f2(x)f2(x). f(0)f(2)0

f af(2)f(0)2

f b|f(a)|1(|f(0)||f(2)|)1,|f(b)|1(|f(0)||f(2)|)1F(a)f2(a)f2(a) F(b)f2(b)f2(b)而F(0f20)f2(0)Fxf2xf2x在[ab(abF(x点取到它在[a,b]上的最大值F(02f(f(f(0但f()0，否则F()f2()1F(0)，与F()是最大 f(2x【证】（1）因为lim 存在，故limf(2xa)0，由f(x)在[a,b]上 x f(a)0f(x0f(x在(a,bf(x)f(a) x（2）设F(x)x2，g(x) xf(t)dt(axb)a则g(x)f(x)0，故F(x)，g(x)满足 中值定理的条件，于是在(a,b)内存在点，F(b)F(a)g(b)

b2 f(t)dtf(t)d

, f f(t)dtxab2即baf(t)d

2f(（3）因f()f()0f()f(a)，在[a,]上应 日中值定理，知在(a,)存在一点，使f(f()(a，从而由（2）b2aaf(t)da

f()( f()(b2a2) bf(t)dx.(a)a e[g()g(【证】只要证f f( e ,ebg(b)eae

e[g()g(