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# 2012-2013第二学期数学分析3-2(基地班)期末考试试题A参考答案

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ú v ?

?

?!(55?§??1(3) K15?§?{z K10?) U??)‰e ?K.
π

(1) ?
0

(x sin x)2 dx;

).
π

(x sin x)2 dx
0 π

= x 6 π3 = 6 π3 = 6 π3 = 6 =
0 3

1 ? cos 2x dx 2 π π 1 π 1 2 ? x sin 2x + x sin 2xdx 4 2 0 0 0 π 1 1 π cos 2xdx ? x cos 2x + 4 4 0 0 π π 1 ? + sin 2x 4 8 0 π ? . 4 x2 · ???

(2) ?E??êw = f (u, v ), u = x + y + z, v = x2 + y 2 + z 2 ). dw =

?f ?f du + dv ?u ?v ?f ?f = (dx + dy + dz ) + (2xdx + 2y dy + 2z dz ) ?u ?v ?f ?f ?f ?f ?f ?f = + 2x dx + + 2y dy + + 2z ?u ?v ?u ?v ?u ?v

dz.

11?

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ê?‰??

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ú v ? x + y = 0, x2 + y 2 = 0,
2 2 →

? 3 3 ? x +y , (3) f (x, y ) = x2 + y 2 ? 0, ?§`?nd; ).

l = (cos θ, sin θ), ?

?f ?l

→ (0, 0),

?

?f (x, y )3 :(0, 0)? ? ? ?

f (t cos θ, t sin θ) ? f (0, 0) t(cos3 θ + sin3 θ) (0 , 0) = lim = lim → t→0+ t→0+ t t ?l = cos3 θ + sin3 θ. ?f f (x, y )3 :(0, 0)? ? ? ?. ? y. ef (x, y )3 :(0, 0)? ? ? § K d 1? ?f f (0, ?y ) ? f (0, 0) (0, 0) = lim = 1? ?y → 0 ?y ?y ?f ?l ? ?f ?l (4) ). 1? x + o(x2 ) cos x 2 = 2 cos y 1 ? y2 + o(y 2 ) x2 y2 = 1? + o(x2 ) · 1 + + o(y 2 ) 2 2 x2 y 2 = 1? + + o(x2 + y 2 ). 2 2
2

?f f (?x, 0) ? f (0, 0) (0, 0) = lim = ?x→0 ?x ?x

→ (0, 0)

=

?f ?f (0, 0) cos θ + (0, 0) sin θ = cos θ + sin θ. ?x ?y

→ (0, 0)

= cos3 θ + sin3 θg?. cos x 3:(0, 0)? cos y V?m?;

??êf (x, y ) =

12?

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ê?‰??

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ú v ? {???-?Γ : x+y+z =3 3:(1, 1, 1)? x2 ? y 2 + 2 z 2 = 2 ?

(5) ?-?S : x2 ? y 2 + 2z 2 = 23:(1, 1, 1)? ??. ).

-f (x, y, z ) = x2 ? y 2 + 2z 2 ? 2, K-?S 3:(1, 1, 1)? {??? ?f (1, 1, 1) = (2x, ?2y, 4z )
(x,y,z )=(1,1,1)

= (2, ?2, 4).

-g (x, y, z ) = x + y + z ? 3, K-?Γ3:(1, 1, 1)? ???? e1 e2 e3 ?f (1, 1, 1) × ?g (1, 1, 1) = 2 ?2 4 = (?6, 2, 4). 1 1 1
→ → →

?

b

!(10?)

f (x)3[a, b]? ? K ? Y … é ? ?x ∈ [a, b], kf (x) < 1, -xn = n
a n→∞

f n (x)dx,

n = 1, 2, · · · . y?? lim xn = 0. y. k

??f (x)3[a, b]??K?Y…é??x ∈ [a, b], kf (x) < 1, ¤±f (x)3[a, b]?k???M < 1. u?
b b

0 q lim nM n (b ? a) = 0,
n→∞

xn = n
a

f (x)dx
n→∞

n

n
a

M n dx = nM n (b ? a).

dü>Y?n? lim xn = 0.

13?

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?

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ú v ?

n!(10?) f (x), g (x)3(?∞, +∞)??Y??§f (0) = 0, f (0) = 0. y???3δ > 0, ? ?§f (x) + tg (x) = 03(?δ, δ )?k???Y)x = x(t), ? x(0) = 0.

y. -?(t, x) = f (x) + tg (x), Kdf (x), g (x)3(?∞, +∞)??Y????(t, x)??x (t, x) = f (x) + tg (x)? 3R2 ? ? Y. q?(0, 0) = f (0) = 0, ?x (0, 0) = f (0) = 0, d ? ? ê ? n ? ? 3δ > 0? ? ? ? ? u(?δ, δ ) ?êx = x(t), ? é??x ∈ (?δ, δ ), ?kf (x(t))+ tg (x(t)) = 0, x(0) = 0, …x = x(t)3(?δ, δ )? ?Y, =?§f (x) + tg (x) = 03(?δ, δ )?k???Y)x = x(t), ? x(0) = 0.

?

o!(10?) 3gC???C? 2 ?z 2 C??w = w(u, v ) =x ?y

C?u =

x , v = x, w = xz ? y e, òz = z (x, y ) y

?§y

? 2z + ?y 2

?§. §dó?{K§k ?w ?u ?w ?v ?z + =x ? 1, u? ?u ?y ?v ?y ?y x ?w ?z ? 2 =x ? 1, y ?u ?y 1 ?w 1 ?z =? 2 + . ?y y ?u x

y.

?w = xz ? y ü>éy ?

??ü>éy ?

§ 2 ?w 1 x ? 2w ? 2z = ? · ? ?y 2 y 3 ?u y2 y 2 ?u2 ? 2z 2 ?w x ? 2w = 3 . + ?y 2 y ?u y 4 ?u2 ,

?d§dy

? 2z ?z 2 +2 = 2 ?y ?y x 2 ?w x ? 2w 2 ?w 2 2 + ? 2 + = . 2 3 2 y ?u y ?u y ?u x x

nz{ x ? 2w = 0, y 3 ?u2 qx = 0, ? 2w = 0. ?u2 14? 5?

ê?‰??
?

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ú v ? f (X )?Rn ?k.8D? ?—?Y?ê. y??f (X )3D?k.. { Xm } ? D , ? lim f (Xm ) = ∞. ? ?D? k . 8 § ¤

?!(10?)

y.

? y. ef (X )3D? ? . § K ? 3 :

m→∞

±{Xm }? k . : § l {Xm }k ? ? f {Xmk }. ? ?f (X )?D? ? — ? Y ? ê § ¤ ± é ? ?ε > 0, ? 3δ > 0, é ? ?X, Y ∈ D, |X ? Y | < δ ? § ? k|f (X ) ? f (Y )| < ε. d{Xmk }? ? ?{Xmk }? … ? § é ? ?δ > 0, ? 3 êK § k, l > K ? § ? k|Xmk ? Xml | < δ . u ? é ? ?ε > 0, ? 3 êK § k, l > K ? § ? k|f (Xmk ) ? f (Xml )| < ε, {f (Xmk )}? … ? § l {f (Xmk )}? ?. ? lim f (Xmk ) = ∞g?.
k→∞

?

8 !(5?)
x

f (x)3[0, 1]? ? Y ? ? §f (0) = 0, é ? ?x ∈ [0, 1], k[f (x)]2 ex .
x

1 +

2
0

|f (t)f (t)|dt. y?: é??x ∈ [0, 1], k|f (x)|
x

y.

-F (x) =
0

|f (t)|dt (x ∈ [0, 1]), K é ? ?x ∈ [0, 1], kF (x) = |f (x)|…|f (x)| =
0

f (t)dt

F (x). u?

x

x

[F (x)]2 = [f (x)]2 ?dF (x) u?k

1+2
0

|f (t)f (t)|dt

1+2
0

F (t)F (t)dt = 1 + [F (x)]2 .

1 + F (x), ?±U ¤ e?x F (x)
x

e?x ,
x

e?t F (t) dt
0 0

e?t dt,

dd e F (x)

?x

1?e , l F ( x) |f (x)| = F (x) ex ? 1, 1 + F (x) ex .

?x

15?

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