复旦大学数学科学学院 2012~2013学年第二学期期末考试 ■高数C(下)A卷参考答案 1.(1)z(1,1)=3,zx/(1,1)=-4 2)2+2y+2-3=0 1 y 2x-3y+5z-4=0 2显然最大值可在0≤x及y0m>M:()-0)-r(02<0+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。 4S(x)=ln(1+),x∈(-2,2 2)≈ln2+l(×x-1 3 收敛域为x∈(-2,4](但仅在(-2,2上与S(x)相等) 30」 6.f(0)=1,f(x)=c+2f(x)-Jf(t)dt,f(0)=3, f"(x)-2f(x)+f(x)=e 第1页
EåÆÍÆâÆÆ 2012*2013Æc1Æœœ"£ pÍC£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.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 ). 5. X 0 1 2 3 P 1 6 1 2 3 10 1 30 6. f(0) = 1, f 0 (x) = e x + 2f(x) − R x 0 f(t)dt, f 0 (0) = 3, f 00(x) − 2f 0 (x) + f(x) = e x 11ê
解得:f(x)=(1+2x+x2)en 7(1)p=/6x(1-)d≈0104 (2)0.0.2,s(x-0.8)6x(1-)dx≈002 (3)365*p≈37.96 第2页
)µf(x) = (1 + 2x + 1 2 x 2 )e x . 7(1)p = R 1 0.8 6x(1 − x)dx ≈ 0.104. (2)[0, 0.2], R 1 0.8 (x − 0.8)6x(1 − x)dx ≈ 0.0072. (3)365 ∗ p ≈ 37.96 12ê