零解, 其系数行列式det f ' (x 0) = 0, 与条件矛盾.

零解, 其系数行列式det f ' (x 0) = 0, 与条件矛盾.

5. 设F (t ) = β ⋅ f (a + t (b - a )), 则F 在[0,1]上连续, 在(0,1)内可微, 故∃θ∈(0,1)使F (1) - F (0) = F ' (θ ), 即β ⋅ ( f (b ) - f (a )) = β ⋅ f ' (a + θ (b - a )) (b - a ) = β ⋅ f ' (c ) (b - a ).

6. 在上题中取β = f (b ) - f (a ). 用Cauchy-Schwarz 不等式, 得| β | 2 ≤ | β | | f ' (c ) (b - a )| =

| β | | f ' (c ) | | b - a |.

7. (1) 不存在 : f (2π ) - f (0) = 0 =⎩⎨⎧==⇒⎪⎭⎫ ⎝⎛-,

0cos ,0sin 2 cos sin c c c c π无解. (2) 设β = (β 1 , β 2), 则0 = β ⋅ f ' (c ) (b - a ) = 2π (- β1 sin c + β 2 cos c ) = 0, β1 sin c = β 2 cos c . 因此β 1 = β 2 = 0时c 任意; β 1 ≠0时c = arctan (β 2 / β 1); β 2 ≠0时c = arctan (β 1 / β 2).

8. (1) f (x 1) = f (x 2) ⇒ | f (x 1) - f (x 2) | = 0 ⇒ | x 1 - x 2 | = 0 ⇒ x 1 = x 2 .

(2) ∃ x 0 使| f ' (x 0)| = 0 ⇒ f ' (x 0) = 0 (用定义) ⇒ 0||)()(lim

000=--→x x x f x f x x ⇒||)()(|lim 000x x x f x f x x --→| = 0 ⇒ ∃ δ > 0使| x - x 0 | < δ 时c x x x f x f <--|

||

)()(|00, 与条件矛盾.

9. 易知f 连续. 设0 < r n < 1, r n →1 (n →∞), f n = r n f , 则f n 满足第1题条件, 故存在唯一的x n ∈A 使x n = f n (x n ). 因为A 有界闭, 故{x n }有收敛子列, 设为{k n x }, 且k n x →x 0 (k →∞), 则由k n x =)()(k k k k n n n n x f r x f =及f 连续, k n r →1(k →∞) 得x 0 = f (x 0).

又, 若y 0 = f (y 0), 则| x 0 - y 0 | = | f (x 0) - f (y 0)| < | x 0 - y 0 |, 不可能, 故上述x 0 唯一.

10. 设p = 1, λ = P dx + Q dy + R dz , 则由d λ = 0得R y = Q z , P z = R x , Q x = P y . 令 ω (x , y , z) =⎰⋅1

),,()(dt z y x u f , 其中 f = (P , Q , R ), u = (tx , ty , tz ). ∵x ∂∂( f (u ) ⋅ (x , y , z)) =x ∂∂f (u ) ⋅ (x , y , z) + f (u ) ⋅x ∂∂(x , y , z ) = t (P x (u ), Q x (u ), R x (u )) ⋅ (x , y , z ) + P (u )

= t ( P x (u ), P y (u ), P z (u )) ⋅ (x , y , z ) + P (u ) = t g ' (t ) + g (t ),

其中 g (t ) = P (u ) = P (tx , ty , tz ), ∴⎰∂∂=∂∂10x

x ω( f (u ) ⋅ (x , y , z)) dt =⎰10t g ' (t ) dt +⎰10)(dt t g (分部积分) = g (1) = P (x , y , z ). 同理可证ω y = Q , ω z = R , 故λ = d ω .

设p = 2, λ = P dy ∧dz + Q dz ∧dx + R dx ∧dy , u 同上, ω = E dx + F dy + G dz , 其中

E =⎰10(Q (u ) z - R (u ) y ) t dt ,

F =⎰10(R (u ) x - P (u ) z ) t dt ,

G =⎰10(P (u ) y - Q (u ) x ) t dt .

由d λ = 0 得P x + Q y + R z = 0.

∵F x - E y =⎰10(t R x (u ) x + R (u ) - t P x (u ) z ) t dt -⎰10(t Q y (u ) z - t R y (u ) y - R (u )) t dt

=⎰10(t 2 R x (u ) x + t 2 R y (u ) y + t 2 R z (u ) z + 2 t R (u ) ) dt =⎰∂∂10t

( t 2 R (u )) dt = R (u ) | t =1 = R (x , y , z ), 同理, G y - F z = P , E z - G x = Q , 故λ = d ω .

设p = 3, λ = P dx ∧ dy ∧dz , u 同上, ω = E dy ∧dz + F dz ∧dx + G dx ∧dy , 其中

E =⎰10)(u P t 2 x dt ,

F =⎰10)(u P t 2 y dt ,

G =⎰10)(u P t 2 z dt , ∵ E x =⎰102t (t x P x (u ) + P (u )) dt , F y =⎰102t (t y P y (u ) + P (u )) dt , G z =⎰10

2t (t z P z (u ) + P (u )) dt , ∴ d ω = (E x + F y + G z ) dx ∧ dy ∧ dz =⎰∂∂102))((dt u P t t dx ∧dy ∧dz = P dx ∧dy ∧dz = λ .