复旦大学数学科学学院 2012~2013学年第二学期期末考试 ■高数B(下)A卷参考答案 1.(1)zx(1,1)=3,zxy(1,1)=-4 x+2y+2z-3 或1 16 2x-3y+5z-4= (3)2 e+e (5) (6)2小时。 2显然最大值可在0≤x及y≤0且x-y≤1时取到 由zx=2y=0可解得:r=y=0,2(0,0)=0 由x=0:z=y2,y=-1时z取最大值1 由y=0:z=x2,x=1时z取最大值1 由y=x-1,x∈0,1]:z=x2-x+1,x=1时z取最大值1; 所以z的最大值为1 3因为lmx→0 f(x)-f(0)-f(0)xf"(0) 所以彐N0>0Vn>N:f(2)-f(0)-f(0)-< f"()|+1 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)+2xr2f(r2), 2x=2f(72)+[2n2+8x2]f(r2)+4x2r2f"(r2), 同理:zm=2f(72)+2n2+8y2f(r2)+4g2r2f”(r2) 所以:zx+zy=4f(2)+12r2f(r2)+4r4f"(r2)=0 记y(t)=f(e2),则:y"+2y+y=0.,y=(C1+C2t)et 所以:f(2)≈C1+C2,代入初始条件得:f()≈lx 第1页
EåÆÍÆâÆÆ 2012*2013Æc1Æœœ"£ pÍB£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)π. (5)e + e −1 2 − 1. (6)2û" 2.w,Ååäå30 ≤ x9y ≤ 0Öx − y ≤ 1û. dzx = zy = 0å)µx = y = 0, z(0, 0) = 0; dx = 0µz = y 2 , y = −1ûzÅåä1; dy = 0µz = x 2 , x = 1ûzÅåä1; dy = x − 1, x ∈ [0, 1]µz = x 2 − x + 1, x = 1ûzÅåä1; §±zÅåäè1. 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 . 11ê
5.S(x)=l(1 ∈(-2,2 ln(1+a)=ln+ln(1+-2-)=ln 收敛域为x∈(-2,4(但仅在(-2,2]上与S(x)相等 o e(r2+y)dzdy 由 (=+y)dzdy e-r+y)dzdy x2+y2<1且x,y≥0 x2+y2≤2且x,y≥0 即可证得结论。 7(1)a0=1,an=0(n14、1+(-1 n7 2 f()~S()=5+∑ sin (2n-1) 1 2)=0,5(7m)=S() =/[(f(x)-g(x)2+92(x)dr=/p f(a) g(x)2+产2(x) 所以A=,=0.1,,10,B2=,=1.,10 又解:由。,I=0得:A1=,i=0,1, 由Or=0得:B1=,;=1,10 OB 第2页
5.S(x) = ln(1 + x 2 ), x ∈ (−2, 2]. ln(1 + x 2 ) = ln 3 2 + ln(1 + x − 1 3 ) = ln 3 2 + X∞ n=1 (−1)n−1 n3 n (x − 1)n ¬Òçèx ∈ (−2, 4](=3(−2, 2]˛ÜS(x)É ). 6. Z 1 0 e −x 2 dx 2 = Z 1 0 Z 1 0 e −(x 2+y 2 ) dxdy d Z Z x2+y 2≤1Öx,y≥0 e −(x 2+y 2 ) dxdy < Z 1 0 Z 1 0 e −(x 2+y 2 ) dxdy < Z Z x2+y 2≤2Öx,y≥0 e −(x 2+y 2 ) dxdy =åy(ÿ" 7(1)a0 = 1, an = 0(n ≥ 1), bn = 1 + (−1)n−1 nπ , 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. 12ê