复旦大学：《高等数学》课程往年试题（高等数学A）高等数学A下-2013试卷A答案

复旦大学数学科学学院 2012~2013学年第二学期期末考试 高数A(下)A卷参考答案 1.(1)z(1,1)=3,2x(1,1)=-4. x+2y+2x-3=0 或 y 16 3y+5z-4=0 (3) (5) )2小时 2.(s)=√2 3因为lmx→ f(x)-f(0)-f(0)xf(0) 所以彐N0>0m>N0:|f(-)-f(0)-f(0)-< f"O|+1 n2,因而: (1)f(0)≠0时,级数发散 (2)f(0)=0,f(0)≠0时,级数条件收敛 (3)f(0)=f(0)=0时,级数绝对收敛。 4记r=Vx2+y,则zx=2xf(r2)+2x2f(r2), 2x2=2f(r2)+[2x2+8x2f(r2)+4x2r2r"(r2) 同理:2y=2f(r2)+[2r2+8y2]f(72)+4y2r2"(r2), 所以:zx+20y=4f(72)+12r2f(2)+4r4f"(r2)=0 记y(t)=f(e),则:y+2y/+y=0,y=(C1+C2t)e- C1+CiNc In r 所以:f(x)= ,代入初始条件得:f(x 5记r=√x2+4y2+42,则 (y) ay 所以:x 0. 原式= .rdydz + ydzd zdxdy=3 x2+4y+4z2=1的外侧 1+(-1)n-1 6(1)a0=1,an=0(n≥1),bn 第1页

E￾åÆÍÆâÆÆ 2012*2013Æc1Æœœ"£  pÍA£e§A ÚÎâY 1. (1)zx(1, 1) = 3, zxy(1, 1) = −4. (2)    −x + 2y + 2z − 3 = 0 2x − 3y + 5z − 4 = 0 ½ x − 1 16 = y − 1 9 = z − 1 −1 . (3)2π. (4)4π 3 . (5)2π 3 . (6)2û" 2.z 0 (s) = √ 2 2 . 3.œèlimx→0 f(x) − f(0) − f 0 (0)x x 2 = f 00(0) 2 , §±∃N0 > 0∀n > N0: |f( 1 n ) − f(0) − f 0 (0) 1 n | < |f 00(0)| + 1 n2 , œ µ (1)f(0) 6= 0ûß?Íu—¶ (2)f(0) = 0, f0 (0) 6= 0ûß?Í^á¬Ò¶ (3)f(0) = f 0 (0) = 0ûß?Í˝È¬Ò" 4.Pr = p x 2 + y 2 , Kzx = 2xf(r 2 ) + 2xr2 f 0 (r 2 ), zxx = 2f(r 2 ) + [2r 2 + 8x 2 ]f 0 (r 2 ) + 4x 2 r 2 f 00(r 2 ), ”nµzyy = 2f(r 2 ) + [2r 2 + 8y 2 ]f 0 (r 2 ) + 4y 2 r 2 f 00(r 2 ), §±µzxx + zyy = 4f(r 2 ) + 12r 2 f 0 (r 2 ) + 4r 4 f 00(r 2 ) = 0. Py(t) = f(e t ), Kµy 00 + 2y 0 + y = 0, y = (C1 + C2t)e −t . §±µf(x) = C1 + C2 ln x x , ì\–©^áµf(x) = ln x x . 5.Pr = p x 2 + 4y 2 + 4z 2 , K: ∂ ∂x( x r 3 ) = 1 r 3 − 3 x 2 r 5 , ∂ ∂y ( y r 3 ) = 1 r 3 − 3 4y 2 r 5 , ∂ ∂z ( z r 3 ) = 1 r 3 − 3 4z 2 r 5 . §±µ ∂ ∂x( x r 3 ) + ∂ ∂y ( y r 3 ) + ∂ ∂z ( z r 3 ) = 0. ™= Z Z x2+4y 4+4z 2=1 ˝ xdydz + ydzdx + zdxdy = 3 Z Z Z x2+4y 4+4z 2≤1 dxdydz = π. 6(1)a0 = 1, an = 0(n ≥ 1), bn = 1 + (−1)n−1 nπ , 11ê

r()~()=+∑c2=1msm(2n-1)x n=1 0,S(7r)=S(丌) =/[(r(a)-g()2+g(a)dx=/2(2-a)2+f(a 所以A2 i=0,1,…,10,B1 b 2 又解:由。1=0得:A2=4,=0,1,10 由。I=0得:B≈在 讠=1.....10. 7(1)由。P=Q解得:a=1,入=1 但是,此时沿单位圆x2+y2=1逆时针的积分=2丌, 所以无论a,如何取值,都不能使积分完全与路径无关。 (2)由ap=0 Q解得 4.入=2 此时Pa+Qdy=d=2y,因而积分与路径无关 第2页

f(x) ∼ S(x) = 1 2 + X∞ n=1 2 (2n − 1)π sin(2n − 1)x S( 7π 2 ) = S(− π 2 ) = 0, S(7π) = S(π) = 1 2 . (2) I = Z π −π ￾f(x) − g(x)
2 + g 2 (x) dx = Z π −π 2 ￾f(x) 2 − g(x)
2 + 1 2 f 2 (x) dx §±Ai = ai 2 , i = 0, 1, . . . , 10, Bi = bi 2 , i = 1, . . . , 10. q)µd ∂ ∂Ai I = 0µAi = ai 2 , i = 0, 1, . . . , 10, d ∂ ∂Bi I = 0µBi = bi 2 , i = 1, . . . , 10. 7(1)d ∂ ∂yP = ∂ ∂xQ)µa = 1, λ = 1. ¥ßdû˜¸†x 2 + y 2 = 1_û»©= 2π, §±Ãÿa, λX¤äß—ÿU¶»©Ü¥ªÃ'" (2)d ∂ ∂yP = ∂ ∂xQ)µa = 4, λ = 2. dûPdx + Qdy = d −2y 2 x 2 + y 2 , œ »©Ü¥ªÃ'" 12ê