期中考試試卷14 15高等數(shù)學(xué)a2試題

(C)-16(x-2)+14(y-0)+11(z-3)=0(D)-16(x+2)+14(y-0)+11(z-3)= 6、曲面x2+y2+z2=a2與x2+y2=2az(a>0)的交線(A)拋物線(B)雙曲線(C)圓(D)x z

7、雙曲線451z

x2+y

y=z =

x y2+z =

(A)y''-2y= (B)5y''-4x= (C)y''-5xy'+4y2= (D)y''-2y'+1=

(x+ z =

x (y+ =2y''2y'3yexsin(2x

y*=ex[Acos(2x)+Bsin(

8、設(shè)f(x,y)= x2+y

y*=xex[Acos(2x)+Bsin(

f(xylimlimf(x,ylimlimf(x,y

y*=Aexcos(y*=Aexsin(

(x,y)fi xfi0yfi yfi0xfi(A)②與③都不存在，但①存在(B)(C)①②③都存在 9、在點(diǎn)(x0,y0)，對函數(shù)f(x,y)下列結(jié)論成立的是

(A)若連續(xù)則偏導(dǎo)數(shù)存 (B)若兩個偏導(dǎo)數(shù)存在則必連(C)兩個偏導(dǎo)數(shù)或都存在，或都不存 (D)兩個偏導(dǎo)數(shù)存在但不一定連程的通解 10、函數(shù)u=xy2+yz3在點(diǎn)(2,-1,1)處的方向?qū)?shù)的最大值

C1y1+C2y2-(1-C1-C2)C1y1+C2y2+

C1y1+C2y2+(1-C1-C2)C1y1+C2y2-(C1+C2)

(A) (B) (D) 11x2cos(xyyzx0在點(diǎn)(0,1,-1)4aba-ba{3-5,8b={-1,1,x}x(A)0(B)1(C)2(D)5P(2,0,3)且與直線

(A)x-y+z=- (B)x+y+z=0(C)x-2y+z=- (D)x-y-z=x-2y+4z-7=0垂直的平面方程 12、設(shè)有三元方程siny+ex-xy-zlnz=0，根據(jù)隱函數(shù)存在定理，存在點(diǎn)(1,0,e)的一…………………裝…………………訂…………………線3x5…………………裝…………………訂…………………線(x2)2y0)4(z3) (B3(x2)5(y0)2(z3) (A)zz(xxxyzzz(xxxyzyy(xxxyzyy(xzzz(x

(A)22p 2p 2p 2 18、設(shè)Wx2y24,z0,z4(x2y2dv

(A)0 dy2(x+

)dz(B)- dy2(x+

4-x413、設(shè)f(xy)為連續(xù)函數(shù)，則使f(xy)dsfyx)4-x4

r

2424dq

r(A)區(qū)域D關(guān)于x軸對 (B)區(qū)域D關(guān)于y軸對

4-19、設(shè)f(x,y)為連續(xù)函數(shù)，則1dxxf(x,y)dy+1 f(x,y)dx(C)區(qū)域D關(guān)于直線y=x對 (D)區(qū)域D關(guān)于直線y=-x對

1dx

f(x,

1dx

f(x, 4- 14f(x,y連續(xù)，則4dqf(rcosqrsinq)rdr

(C)

f(x, (D)

f(x, 22 22

f(x,

f(x,

20f(xy)

-

+2xy-

(A) (B)(0,2)(C) (D)

2 f(x,

2 f(x,

f(xf(x, =

1y''-4y'+3y2e2x f(x, 42 f(rcos,rsin)rdr成立 2x+y

2yexyxe-x 3yC1exC2cos2xC3sin2x（C1C2C3為任意常數(shù)）

f(-x,-y)=f(x,f(-x,y)=f(x,-y)=-f(x,

f(-x,-y)=-f(x,f(-x,y)=f(x,-y)=f(x,

4M(12,-1)xt2,y3t4,zt-1 3(x2+y216、設(shè)W為球體x2+y2+z2￡1與錐體z3(x2+y2W

5、已知abcabc0a?bb?cc?a6xy2z+10與平面2xyz30 (A)dq3sinjcos2jdjr (B)dq6sinjcos2jdjr

0 0 (C)dq3sinjcosjdjr (D)dq6sinjcosjdj0 0

8Lx-1y=1zxy2z-10…………………裝…………………訂…………………線17、圓錐z 在圓柱體x2+…………………裝…………………訂…………………線

9、點(diǎn)(2,1,0)到平面3x4y5z0x2y2z2 21Dx2y2=1,x2y24yxy=3x10、曲線

xOyx+z=

D

ydxdy11f(xyln(1x2y2ln(1+x2+y2

x+y-z-1=22、經(jīng)過直線

xyz012、 2yfixfi0arcsin(x+yfi

x-y+z+1=23、過點(diǎn)(3,-2,1)和點(diǎn)(-1,0,2)13、設(shè)函數(shù)

zf(x2y2x2y2，其中f具有二階連續(xù)偏導(dǎo)數(shù)，則

證明函數(shù)f(x,y)=x-y+x2+ (x,y)?(0,0)在點(diǎn)(0,0)處連續(xù)偏導(dǎo)數(shù)存在但 (x,y)=(0,14、設(shè)方程F(xyz2x)0確定了隱函數(shù)zz(x,y)，其中F具有連續(xù)偏導(dǎo)數(shù)，則x?z-y?z

15、函數(shù)ux2yz在點(diǎn)(1,1,1)l{2-2,1x=t2-16、曲線yt+1在點(diǎn)(0,2,1)z=t 17、2dyysinxdx 2x-18、已知f(u)為連續(xù)函數(shù)，F(xiàn)(tf(x2y2z2dxdydz，其中W為閉區(qū)域Wx2y2z2￡t2t0)F￠(t19、x2+y2

(x2+xyex2+y2)dxdy…………………裝…………………訂…………………線…………………裝…………………訂…………………線20、設(shè)W是由曲面x2y2z2=1W1+x2+y2+z2dxdydz