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09级一期B卷参考解一.(每小题6分,12分)求极限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)求下列积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设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 及点A(1,0,1),B(3,2,2),求函数f(x,y,z)在点A处沿由A到B的方向导数,并求此函数在点A处方向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点的方向导数最大值

z

112021 2xyz 2⑶设函zz(xy由方xyze给出求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 解根据向量积的定义，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求经过直线 且平行于直线 解平面的法向量

2i0jk又显然所求平面过点(1,-2,-故所求平面方程为2(x-1)+0(y+2)-(z+3)=02x-z-

f(xxxx0的极值 lnx1ln 11ln f(x) (ex)ex xx f(x)0,得驻点x=e.又当0<x<e时 f(x)0,当x>e时 f(x)0,故此为极大值点,极大值

f(e)ee

(x六.(12函数f(x(x1)2求⑴此函数的单调区间与极值点;⑵此函数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凸极大值-凸凸拐点单调增加区间为(-∞,-5),(-1,1)和(1,+∞),单调减少区间为(-5,-1)函数在点显然曲线有垂直渐近线x=-1,曲线无水平渐近线.由于f (xklim lim xx(x(x 5x22xblim[f(x)kx]lim 2x x(x (x⒈求证不等式sinxtanx 0x2证令f(x)sinxtanx 0x,2

f(x)cosxsec2xsinx(2cos3f(x)sinx2secxsecxtanx cos3x故f(xcosxsec2x2在区间0x2

内单调增加,而f(00于在区间0x2

f(x0f(xsinxtanx2x在区间0x内单调增加2f(x)sinxtanx2xf(0)0.⒉设函数f(x)在闭区间[a,b]上二