2013秋季试题答案.pdf

E ŒÆêƉÆÆ 2012*2013Æc1 ÆÏÏ"•Á  pêC£e¤A ò땉Y 1. (1)z  x (1, 1) = 3, zxy (1, 1) = −4.  −x + 2y + 2z − 3 = 0 x−1 y−1 z−1 ½ (2) = = .  16 9 −1 2x − 3y + 5z − 4 = 0 (3)2π. (4)π. e + e−1 (5) − 1. 2 (6)2 ž" 2.w,•ŒŠŒ30 ≤ x9y ≤ 0…x − y ≤ 1ž . dzx = zy = 0Œ) µx = y = 0, z(0, 0) = 0; dx = 0µz = y 2 , y = −1žz dy = 0µz = x2 , x = 1žz •ŒŠ1; •ŒŠ1; dy = x − 1, x ∈ [0, 1]µz = x2 − x + 1, x = 1žz •ŒŠ1; ¤±z •ŒŠ•1. f (x) − f (0) − f 0 (0)x f 00 (0) 3.Ï•limx→0 = , x2 2 1 1 |f 00 (0)| + 1 ¤±∃N0 > 0∀n > N0 : |f ( ) − f (0) − f 0 (0) | < ,Ï µ n n n2 (1)f (0) 6= 0ž§?êuѶ (2)f (0) = 0, f 0 (0) 6= 0ž§?ê^‡Âñ¶ (3)f (0) = f 0 (0) = 0ž§?êýéÂñ" x 4.S(x) = ln(1 + ), x ∈ (−2, 2]. 2 ∞ x 3 x−1 3 X (−1)n−1 ln(1 + ) = ln + ln(1 + ) = ln + (x − 1)n n 2 2 3 2 n=1 n3 Âñ••x ∈ (−2, 4]( =3(−2, 2]þ†S(x)ƒ ). X 0 1 2 3 1 1 3 1 P 6 2 10 30 Rx 6. f (0) = 1, f 0 (x) = ex + 2f (x) − 0 f (t)dt, f 0 (0) = 3, 5. f 00 (x) − 2f 0 (x) + f (x) = ex 11• 1 ) µf (x) = (1 + 2x + x2 )ex . 2 R1 7(1)p = 0.8 6x(1 − x)dx ≈ 0.104. R1 (2)[0, 0.2], 0.8 (x − 0.8)6x(1 − x)dx ≈ 0.0072. (3)365 ∗ p ≈ 37.96 12•