課件教程成果

09級一期B卷參考解一.(每小題6分,12分)求極限⑴lim tanxsin1sin解原式tanxsin1sin limsec2xcosxlim1cos3 lim1cos3xlim3cos2xsinx1 x012x2cos2 1⑵limcosxsin2x1

1lncos limcosxsin2x

limesin2 limlncosx

1lim 1.sinxcosx0sin2 x02sinxcos 2x0sinxcos1 limcosxsin2x1

e e2e (6分,24分)求下列積x2x10

x2x10

2x10

10x102x102 1 2 2 uln解cos(lnx)dx eucosudueudsinueusinusineusinueudcos

eusinueucosueucosx所以cos(lnx)dx

sinucosuC2

sinlnxcoslnx2 1x2ln2d uln d 1x2ln2x02u2 0

sintcos

1

11arctan110s 2s

cos

s

0sintcos 2sin

d2t 2 2

ds

0sinscos 0sintcos 2 cos dt2 sin

2dt0sintcos 0sintcos 于 cos dt 于0sintcos 三.(每小721x2⑴設(shè)z(x,y) x2x2x2

(0,1)x2x2z

x x2x2(x(x2y2)3

z

x (x2y2)3解 x2 x2 于

y

dz

dx

dyy2⑵已知f(x,y,z)lnx 及點A(1,0,1),B(3,2,2),求函數(shù)f(x,y,z)在點A處沿由A到B的方向?qū)?shù),并求此函數(shù)在點A處方向y2解uuurl(2,2,3),故(cos,cos,cos) .,f 2

x y2x y2 yy2x y2yy2x y2y y2z2y2zy2x y2z y2z2y2于

1,

0, 1 2y

1 0(2) 11 2A點的方向?qū)?shù)最大值

z

112021 2xyz 2⑶設(shè)函zz(xy由方xyze給出求xy及x2解令F(x,y,z)ezxy F1,

1,

ez1.于z 2z

,z F ez1 ez z xez (ez1)2 (ez1)3 解根據(jù)向量積的定義，可知ABC的面2SABC|AB||AC|sinA=ABAB(-2，1，1)，AC(1，-3，1)， 5252521SABC

ABACijABACij1k15i5j2ABAC152 xy 的平面方 xy 的平面方程⑵求經(jīng)過直線 且平行于直線 解平面的法向量

2i0jk又顯然所求平面過點(1,-2,-故所求平面方程為2(x-1)+0(y+2)-(z+3)=02x-z-

f(xxxx0的極值 lnx1ln 11ln f(x) (ex)ex xx f(x)0,得駐點x=e.又當0<x<e時 f(x)0,當x>e時 f(x)0,故此為極大值點,極大值

f(e)ee

(x六.(12函數(shù)f(x(x1)2求⑴此函數(shù)的單調(diào)區(qū)間與極值點;⑵此函數(shù)f(x)3(x1)2(x1)22(x1)3(x (x,(x1)2[3(x1)2(x1)](x1)2(x5)(x1)3 (x1)3,f(x)[2(x1)(x5)(x1)2](x1)33(x1)2(x5)(x(x(x1)[(3x9)(x1)3(x1)(x5)]24(x1)(x (xx(-∞,--(-5,-(-1f+0-+0+f----0+f凸極大值-凸凸拐點單調(diào)增加區(qū)間為(-∞,-5),(-1,1)和(1,+∞),單調(diào)減少區(qū)間為(-5,-1)函數(shù)在點顯然曲線有垂直漸近線x=-1,曲線無水平漸近線.由于f (xklim lim xx(x(x 5x22xblim[f(x)kx]lim 2x x(x (x⒈求證不等式sinxtanx 0x2證令f(x)sinxtanx 0x,2

f(x)cosxsec2xsinx(2cos3f(x)sinx2secxsecxtanx cos3x故f(xcosxsec2x2在區(qū)間0x2

內(nèi)單調(diào)增加,而f(00于在區(qū)間0x2

f(x0f(xsinxtanx2x在區(qū)間0x內(nèi)單調(diào)增加2f(x)sinxtanx2xf(0)0.⒉設(shè)函數(shù)f(x)在閉區(qū)間[a,b]上二