# 2013-2014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 ?

?

?!(40?) U??)‰e ?K.
π

(1) ?
0

dx ; 2 + cos x

) ?? dx = 2 + cos x
2dt 1+t2 ?t2 +1 1+t2

2

??t = tan tan x √ 2 3

x = 2 + C,

t2

2dt +3

2 t 2 = √ arctan √ + C = √ arctan 3 3 3 ¤±
π 0

2 dx = √ arctan 2 + cos x 3

tan x √ 2 3

π 0

2 π ?0 = =√ 3 2

3π . 3

(2) 3-?x = t, y = t2 , z = t3 ???:, ?

T-?3d:

???1u??x + 2y + z = 4;

) ??:(t, t2 , t3 )?-? ???1u??x + 2y + z = 4 …= :(t, t2 , t3 )?-? ???(1, 2t, 3t2 )? ??x + 2y + z = 4 {??(1, 2, 1)R?§¤± 1 · 1 + 2 · 2t + 1 · 3t2 = 0. ) 1 t = ?1?t = ? , 3 ¤? -?? :?(?1, 1, ?1), 1 1 1 ? , ,? . 3 9 27

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ê?‰??

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x2 + y 2 + z 2 = √ ú v ? 2¤(? ?êz = z (x, y )3:(1, 0, ?1)? § ?z ?x ??dz ?

(3) ?d?§xyz + ) Pa =

?z ?z (1, 0, ?1), b = (1, 0, ?1). ?§ü>éx? ?x ?y yz + xy ?z 1 + ?x 2 x2 + y 2 + z 2

2x + 2z

= 0.

1 u?3(1, 0, ?1)??k √ (2 ? 2a) = 0, 2 2 xz + xy

a = 1. ?§ü>éy ? § ?z ?y

1 ?z + ?y 2 x2 + y 2 + z 2

2y + 2z

= 0.

1 u?3(1, 0, ?1)??k?1 + √ (?2b) = 0, 2 2

√ b = ? 2. ?d √ 2dy. √ ? § § ? ?u = x ? 2 y ,

dz (1, 0, ?1) = adx + bdy = dx ?

(4) 3 g C ? ? ? C ? C ? e, òz = z (x, y ) ? § C ? ?w = w(u, v ) ? 2z ? 2 z 1 ?z √ v = x + 2 y , w = z , ?§? 2 ? y 2 ? · = 0. ?x ?y 2 ?y ) dó?{K§k ?w ?u ?w ?v ?w ?w ?z = + = + , ?x ?u ?x ?v ?x ?u ?v ? k

?z ?w ?u ?w ?v 1 = + = ?√ ?y ?u ?y ?v ?y y

?w ?w ? ?u ?v

.

? 2z ? 2 w ?u ? 2 w ?v ? 2 w ?u ? 2 w ?v ? 2w ? 2w ? 2w = + 2 , + + + = + ?x2 ?u2 ?x ?v?u ?x ?u?v ?x ?v 2 ?x ?u2 ?u?v ?v 2 ? 2z 1 = √ 2 ?y 2y y 1 = √ 2y y ò??(J“\ ?w ?w ? ?u ?v ?w ?w ? ?u ?v 1 ? 2 w ?u ? 2 w ?v ? 2 w ?u ? 2 w ?v ?√ + ? ? y ?u2 ?x ?v?u ?x ?u?v ?x ?v 2 ?x 1 ? 2w ? 2w ? 2w + ? 2 + . y ?u2 ?u?v ?v 2 n 4 ? 2w = 0, ?u?v ? 2w = 0. ?u?v

? 2z ? 2 z 1 ?z ? y ? · = 0?§z{ ?x2 ?y 2 2 ?y 12?

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?

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ú v ? f (x, y )?R2 ? ?ê§é???êλ, x, y , ?kf (λx, λy ) = λf (x, y ).

!(20?)

(1) y??é???"?? l , ?? ê

?f

?l (2) y???êf (x, y )3:(0, 0)??? …= y (1) λ = 0, ?f ?l f (0, 0) = 0. = lim +
t→0 →

→ (0, 0)??3?

?3?êa, b, ? f (x, y ) = ax + by .

l = (r cos θ, r sin θ), ??r > 0, K

→ (0, 0)

tf (cos θ, sin θ) ? 0 f (t cos θ, t sin θ) ? f (0, 0) = lim = f (cos θ, sin θ). t→0+ t t ?? ??

(2) /?0. ?3?êa, b, ? f (x, y ) = ax + by §Kf (?x, ?y ) ? f (0, 0) = a?x + b?y . U ?f (x, y )3:(0, 0)???…df (0, 0) = adx + bdy . /?0. f (x, y )3 :(0, 0)? ? ? § K ? 3 ?êa, b, ? x = λ cos θ, y = λ sin θ, Kk f (x, y ) = f (λ cos θ, λ sin θ) = λf (cos θ, sin θ) = λ ?f ?l df (0, 0) = adx + bdy .

(x, y ) = (0, 0)? §

→ (0, 0)

= λ(a cos θ + b sin θ) = ax + by.

(x, y ) = (0, 0)?§df (0, 0) = 0?f (x, y ) = ax + by ?¤á.

é???êx, y , kf (x, y ) = ax + by .

?

n!(10?)

z = z (x, y )?d?§| z = αx + ?(α)y + ψ (α), 0 = x + ? (α)y + ψ (α)

¤(?

g?Y???ê, ?y: zxx zyy ? (zxy )2 = 0.

y 1???§ü>éx? § zx = αx x + α + ? (α)αx y + ψ (α)αx , 2(?1 ??§§ zx = α. 13? 5?

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ú v ?

1???§ü>éy ? § zy = αy x + ? (α)αy y + ?(α) + ψ (α)αy , 2(?1 u? zxx = αx , zxy = αy , zyx = ? (α)αx , zyy = ? (α)αy . ??z = z (x, y ) g?Y??§¤±zxy = zyx , l zxx zyy ? (zxy )2 = zxx zyy ? zxy zyx = αx · ? (α)αy ? αy · ? (α)αx = 0. ??§§ zx = ?(α).

o!(16?) f (x)?(?∞, +∞)? ?ê§f (x)3??4?m???, …é???êx???h = x+h 1 0, ?k f (t)dt = f (x). 2h x?h (1) y??f (x)3(?∞, +∞)?? ; (2) y??f (x)?(?∞, +∞)? ~ê?ê.
x

?

y 1 2

(1) -F (x) =
x+1 0

f (t)dt, K df (x)3 ? ? 4 ? m ? ? ? ?F (x)3(?∞, +∞)? ? Y. u ? df (x) =

1 f (t)dt = [F (x+1)?F (x?1)]?f (x)3(?∞, +∞)??Y§2d???? ?n?F (x)3(?∞, +∞)? 2 x? 1 1 ? §l df (x) = [F (x + 1) ? F (x ? 1)]?f (x)3(?∞, +∞)?? . 2 1 (2) d 2h
x+h x+h

f (t)dt = f (x)
x?h x? h

f (t)dt = 2hf (x), ü>éh? § f (x + h) + f (x ? h) = 2f (x),

??ü>éh? é??y = 0,

§ f (x + h) ? f (x ? h) = 0. y x = h = , “\??§ 2 f (y ) ? f (0) = 0.

?df (x) ≡ f (0), =f (x)?(?∞, +∞)?

~ê?ê. 14? 5?

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ú v ?

? !(8?) A?B ? ?R ? ? f 8. y ? ? X JA?B ? ? ; 8 § @ oA × B = {(x, y ) ∈ R2 |x ∈ A, y ∈ B }?R2 ? ;8.

y ? ?A?B ? ? ; 8 § ¤ ±A?B ? ? ; 8. ? : {(xm , ym )} ? A × B , K{xm } ? A, {ym } ? B . dA ; ?{xm }k ? ? uA? : f {xmk }, lim xmk = a. dB ; ?{ymk }k ? ? uB ? : f
k→∞

{ymkl },

l→∞

lim ymkl = b. u?{(xm , ym )}kf {(xmkl , ymkl )}??u(a, b) ∈ A × B , ;8.

A × B ?R2 ?

;

8§l A × B ?R2 ?

?

8!(6?)

A?B ??R?

k.48§f (x, y )3A × B ??Y§m(x) = sup f (x, y ), x ∈ A. y
y ∈B

??m(x)3A??—?Y. y ??A?B ??R? k.48§¤±A?B ??;8§ d?K?A×B ?R2 ? ;8. ??f (x, y )3A× B ??Y§¤±dx÷ ?n?f (x, y )3A × B ??—?Y§=é??ε > 0, ?3δ > 0, (x1 , y1 ), (x2 , y2 ) ∈ A × B … (x1 ? x2 )2 + (y1 ? y2 )2 < δ ?§?k|f (x1 , y1 ) ? f (x2 , y2 )| < ε. é??x1 , x2 ∈ A, |x1 ? x2 | < δ , k|f (x1 , y ) ? f (x2 , y )| < ε, ?y ∈ B . df (x1 , y )??y ?ê3B ??Y?B ?k.48?f (x1 , y )3B ? ???§ ?3y1 ∈ B , ? f (x1 , y1 ) = m(x1 ). u?k m(x2 ) ?n? m(x1 ) > m(x2 ) ? ε. ?dé??x1 , x2 ∈ A, |x1 ? x2 | < δ , k |m(x1 ) ? m(x2 )| < ε. U???m(x)3A??—?Y. f (x2 , y1 ) > f (x1 , y1 ) ? ε = m(x1 ) ? ε.

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