2019-2020年秋冬学期微积分期中模拟考试-答案.pdf
2019-2020年秋冬学期微积分期中模拟考试-答案.pdf
2019-2020 学年秋冬学期微积分期中模拟考试答案 命题：丹青学业指导中心 考试时间：2019 年 10 月 26 日 一、 (1) 原式 lim 1 lim 1 2 = ex→0 1−cos x = ex→0 1−cos x ln(ex +ln(1+x2 )) [ln(1+ ln(1+x2 ) )+x2 ] 2 ex ln(1+x2 ) ) 2 x2 ex + 1−cos 1−cos x x) ln(1+ lim ( = ex→0 ln(1+x2 ) 2 lim = ex→0 ex (1−cos x) +2 = e4 (2) n ∑ √ ( 1+ i=1 n i i ∑ i n2 n2 √ √ ; − 1) = <√ n2 1 i +1+1 +1+1 i=1 n2 n i n2 i <√ n ; i 1 +1 1 + n2 + 1 n2 +1 2 而 n ∑ i n 1 1 n(n + 1) 1 1 1 n(n + 1) 1 1 ∑ i √ √ √ √ = → = → ; 2 2 2 2 n 2 n 4 i=1 n 2 n 4 1 1 1 1 +1+1 +1+1 +1+1 +1+1 i=1 1 n n2 n n2 √ 由夹逼定理: lim ( 1 + ni2 − 1) = 14 x→∞ 二、 (1) ′ 2x ln y = x1 ln(1 + x2 ) 求导得 yy = − x12 ln(1 + x2 ) + x1 1+x 2 2 ∴ y ′ = (1 + x2 ) x (− x12 ln(1 + x2 ) + 1+x 2) 1 1 2 dy = (1 + x2 ) x (− x12 ln(1 + x2 ) + 1+x 2 )dx (2) x 不难求得 (sinh x)′ = e +e 2 −x 则 (arcsinh x)′ = (sinh1 y)′ = ey +e1 −y −y 2 2y −2y +e−2y +2 = e +e4 −2 + 1 = x2 + 1 4 ∴ (arcsinh x)′ = √x12 +1 y 又 ( e +e 2 )2 = e 2y 三、 (1) 求导易证 (2) 由 (1) ln(1 + n1 ) ≤ n1 ∴ an = 1 + 12 + · · · + n1 ≥ ln(1 + 1) + ln(1 + 12 ) + · · · + ln(1 + n1 ) = ln(2 · 32 · · · n+1 ) = ln(n + 1) n ∴ {an } 无界 ⇒ {an } 发散 1 2 四、 (1)∀x0 ∈ [1, +∞) |sin √ x − sin √ x0 | = 2|sin √ √ √ √ √ √ x− x0 x− x0 | √0 | · |cos | ≤ | x − x0 | = |√|x−x ≤ 2 2 x+ x0 | 1 |x − x0 | 2 √ √ ∀ϵ > 0, ∃δ = 2ϵ 当 0 < |x − x0 | < δ 时 |sin x − sin x0 | < ϵ √ √ ∴ lim sin x = sin x0 , f (x) 在 [1, +∞) 上连续 x→x0 0 1 1 1 f (x) sin x 2 1 x2 1 2 −m 1 (2) lim+ mx = m =2 m = lim m = lim m xm = lim m x x→0 x→0+ mx x→0+ x→0+ +∞ ∴ 不存在 m ∈ R 使 g(x) 为连续函数 m<2 ̸= g(0) = 13 m = 21 m > 12 五、由条件 |f (0)| ≤ |sin 0| = 0 ⇒ f (0) = 0 (0) 从而 |f ′ (0)| = |lim f (x)−f | = |lim f (x) | = lim | f (x) | ≤ lim | sinx x | = 1 x−0 x x x→0 x→0 x→0 x→0 六、 1 ′′ = y ′ = √1−x 2,y x ′ = 1−x 2y x 3 (1−x2 ) 2 ′ ⇒ (1 − x2 )y ′′ − xy = 0 可以用数学归纳法证明: ∀n, (1 − x2 )y (n+2) − (2n + 1)xy (n+1) − n2 y (n) = 0(用莱布尼茨) n = 1 时,(1 − x2 )y ′′ − xy ′ = 0 求导得 (1 − x2 )y (3) − 3xy (2) − y ′ = 0 设 (1 − x2 )y (k+2) − (2k + 1)xy (k+1) − k 2 y (k) = 0 求导得 (1 − x2 )y (k+3) − 2xy (k+2) − (2k + 1)y (k+1) − (2k + 1)xy k+2 − k 2 y (k+1) = (1 − x2 )y (k+3) − (2k + 3)xy (k+2) − (k + 1)2 y (k+1) = 0 令 x = 0 得:y (n+2) (0) = n2 y (n) (0) 又 y ′ (0) = 1, y ′′ (0) = 0 ∴ y (n) (0) = 0 n = 2k [(n − 2)!!]2 1 n = 2k + 1 (k ∈ N+ ) n=1 七、 f (x)−f (0) = g(x) = g(x)−g(0) x−0 x2 x2 g ′ (x) f (x)−f (0) g ′′ (x) ′ f (0) = lim x−0 = lim g(x)−g(0) = = = = = = = = = lim = lim = 32 2 x 2 x→0 x→0 x→0 2x x→0 L′ Hopitals 八、 ∀x ∈ (a, +∞) 由 Lagrange 中值定理 f (x + 1) − f (x) = f ′ (ξ), ξ ∈ (x, x + 1) x → +∞ 时,ξ → +∞ 从而 lim f ′ (ξ) = B = lim (f (x + 1) − f (x)) = A − A = 0 x→+∞ ξ→+∞ 九、 1 b−a f (a) f (b) a b (b) = bf (a)−af =− b−a ( f (b) f (a) b − a )ab b−a = f (b) f (a) b − a 1 1 b−a 令 F (x) = f (x) , G(x) = x1 由 Cauchy 中值定理 x 1 ∃ξ ∈ (a, b), b−a 十、 f (a) f (b) a b ′ F (b)−F (a) F (ξ) ′ = G(b)−G(a) =G ′ (ξ) = f (ξ) − ξf (ξ) = f (ξ) f ′ (ξ) ξ 1 3 (1) 令 F (x) = f (x + n1 ) − f (x), x ∈ [0, 1], 则 F (0) + F ( n1 ) + · · · + F ( n−1 )=0 n 1∑ k ∴ min F (x) ≤ F ( ) = 0 ≤ max F (x) x∈[0,1] x∈[0,1] n k=0 n n−1 由介值性 ∃ξ ∈ [0, 1] 使 F (ξ) = 0 则 f (ξ) = f (ξ + n1 ) (2) 令 H(x) = f (x) − g(x) 则 H(xn ) = f (xn ) − g(xn ) = f (xn ) − f (xn+1 ) I. 若 ∃n0 ∈ N+ 使得 H(xn0 ) = 0 则结论成立 II. 若 ∃n1 , n2 ∈ N+ 使得 H(xn1 ) > 0, H(xn2 ) < 0 则由介值性结论成立 III. 若 H(xn ) 恒正或恒负, 由 f 单调得 xn 单调, 又因为 {xn } 有界, 则 {xn } 收敛, 则 {f (xn )} 收敛, 则有 lim xn = x0 则结论成立 (∃xf (x) = g(x)) n→∞ 十一、 Fn+1 Fn − Fn+2 Fn−1 = Fn+1 Fn − (Fn+1 + Fn )Fn−1 = Fn (Fn+1 − Fn−1 ) − Fn+1 Fn−1 = Fn2 − Fn+1 Fn−1 Fn2 − Fn+1 Fn−1 = Fn (Fn−1 + Fn−2 ) − (Fn + Fn−1 )Fn−1 2 = Fn Fn−2 − Fn−1 2 = (−1)([)Fn−1 − Fn Fn−2 ) = (−1)[Fn−1 (Fn−2 + Fn−3 ) − (Fn−1 + Fn−2 )Fn−2 ] 2 = (−1)2 (Fn−2 − Fn−1 Fn−3 ) = · · · = (−1)n−2 (F22 − F3 F1 ) = (−1)n 由 Fn+1 Fn − Fn+2 Fn−1 = (−1)n < FF2m+1 ⇒ G2m−1 < G2m+1 令 n = 2m 得 FF2m−1 2m 2m+2 令 n = 2m + 1 得 G2m > G2m+1 n 又 0 < Gn = FFn+1 ≤1 ∴ {G2m } 和 {G2m−1 } 单调有界 ⇒ 收敛 (−1) 2m+1 F2m−1 2m 且 G2m − G2m−1 = FF2m+1 − FF2m−1 = F2m F−F = F2m+1 →0 F2m 2m 2m+1 F2m 2 2m ∴ {Gn } 收敛, 记 ω = lim Gn n→∞ Fn−1 + FFn−2 ⇒ 1 = ω + ω 2 (n → ∞) ⇒ ω = 由 Fn = Fn−1 + Fn−2 ⇒ 1 = FFn−1 n n−1 Fn √ 5−1 2