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数学分析答案 复旦大学数学系 第三版 全册.pdf
1
1 4
1' 4—
1 C…Œ
§1. …ŒVg
1. )e“§¿xx
1
(1) −2 <
(2) (x − 1)(x + 2)(x − 3) < 0
x + 2
(3)
1
x − 1
< a
(4) 0 6 cos x 6 1
2
x2 − 16 < 0
x2 − 2x > 0
(5)
)
(1) x < − 5
2
‰x > − 3
2
(2) 1 < x < 3‰x < −2
b
b
-3
-2
-1
c
-2
-1
0
0
-
x
c
c
-
1
2
3
x
(3) a > 0§x < 1‰x > 1 +
1
a
¶
0
c
c
1
1 +
1
a
a < 0§1 +
1
a
< x < 1
1 +
a = 0§x < 1
c
c
0
1
1
a
c
1
0
-
x
-
x
-
x
2
(4) 2kπ +
π
3
6 x 6 2kπ +
π
2
(5) −4 < x 6 0‰2 6 x < 4
‰2kπ − π
2
6 x 6 2kπ − π
3
0
c
-4
(k ∈ Z)
c
0
2
-
x
4
-
x
2. y†eØ“
(1) |x − y| > ||x| − |y||
(2) |x1 + x2 + x3 + ··· + xn| 6 |x1| + |x2| + ··· + |xn|
(3) |x + x1 + ··· + xn| > |x| − (|x1| + ··· + |xn|)
y†
(1) ˇ|x||y| > xy§K(x − y)2 > (|x| − |y|)2§u·|x − y| > ||x| − |y||
(2) ^Œ˘8B{y†.
(i) n = 2§d|x1 + x2| 6 |x1| + |x2|§(⁄Æ.
(ii) bn = k(⁄Ƨ=k|x1 + x2 + x3 + ··· + xk| 6 |x1| + |x2| + ··· + |xk|.
Kn = k + 1§|x1 + x2 + x3 + ··· + xk+1| 6 |x1 + x2 + x3 + ··· + xk| + |xk+1| 6 |x1| + |x2| +
··· + |xk| + |xk+1|
n§Øg,Œn§|x1 + x2 + x3 + ··· + xn| 6 |x1| + |x2| + ··· + |xn|⁄Æ.
(3) |x + x1 + ··· + xn| > |x| − |x1 + x2 + x3 + ··· + xn| > |x| − (|x1| + ··· + |xn|)
3. )eØ“§¿xx
(1) |x| > |x + 1|
1
|x| < 4
(2) 2 <
(3) |x| > A
(4) |x − a| < η, η~Œ§η > 0
x − 2
x + 1
>
x − 2
x + 1
(5)
(6) 2 <
1
|x + 2| < 3
)
(1) x < − 1
2
(2) − 1
2
< x < − 1
4
‰
1
4
< x <
1
2
b
0
-1
-
x
e e
1
2
-
0
e
e
1
2
-
x
(3) A > 0§x < −A‰x > A
3
e
e
-
-A
0
A
x
A < 0§x ∈ R
(4) a − η < x < a + η
e
e
a − η
a0
a + η
(5) “du
x − 2
x + 1
< 0§K−1 < x < 2
(6) − 5
3
< x < − 3
2
‰− 5
2
< x < − 7
3
b
-1
0
b
-
2
1
-
x
x
ee
-2
ee
-3
-1
0
-
x
4. ƒe…Œ‰´9§3‰:…Œ
(1) y = f (x) = −x +
‰´9f (−1), f (1)f (2)¶
1
x
(2) y = f (x) =
(3) s = s(t) =
1
t
√
a2 − x2‰´9f (0), f (a)f
−t‰´9s(1), s(2)¶
e
− a
− π
2
¶
¶
(4) y = g(α) = α2 tan α‰´9g(0), g
(5) x = x(θ) = sin θ + cos θ‰´9x
(6) y = f (x) =
1
(x − 1)(x + 2)
π
− π
4
, g
4
, x(−π)
‰´9f (0), f (−1)
2
4
)
(2) …Œ‰´X = [−|a|,|a|]§f (0) = |a|, f (a) = 0, f
1
2e2
(1) …Œ‰´X = (−∞, 0)S(0,∞)§f (−1) = 0, f (1) = 0, f (2) = − 3
(3) …Œ‰´(−∞, 0)S(0,∞)§s(1) =
x ∈ R, x 6= kπ +
− π
π
(6) …Œ‰´X = (−∞,−2)S(−2, 1)S(1, +∞)§f (0) = − 1
(4) …Œ‰´
(5) …Œ‰´X = (−∞,∞)§x
§g(0) = 0, g
= −1, x(−π) = −1
− a
√
3
2
, k ∈ Z
, s(2) =
π2
16
n
o
|a|
π
2
1
e
=
2
x
2
4
2
=
, g
, f (−1) = − 1
2
2
− π
4
= − π2
16
5. ƒe…Œ‰´9
√
2 + x − x2
√
cos x
π
x
(1) y =
(2) y =
(3) y = ln
sin
(4) y =
1
sin πx
)
(1) …Œ‰´X = [−1, 2]§
h
1
2kπ − π
2
π
2
(2) …Œ‰´
, 2kπ +
0,
3
2
(k ∈ Z)§[0, 1]
i
(3) …Œ‰´
(4) …Œ‰´(n − 1, n)(n = 0,±1,±2,··· )§(−∞,−1]S[1, +∞)
(k ∈ Z)§(−∞, 0]
2k + 1
1
2k
,
6. f (x) = x + 1, ϕ(x) = x − 2§`)§|f (x) + ϕ(x)| = |f (x) + |ϕ(x)|
)dfi§f (x)ϕ(x) > 0=(x + 1)(x − 2) > 0§Kx > 2‰x 6 −1.
7. f (x) = (|x| + x)(1 − x)§ƒve“x
(1) f (0) = 0
(2) f (x) < 0
)
(1) f (x) = 0§K|x| + x = 0‰1 − x = 0§=x 6 0‰x = 1
(2) ˇ|x| + x > 0§Kf (x) < 0§1 − x < 0=§=x > 1
8. ª1-5L«>‡|V !‰>{R0C>{R|⁄>·.3ªmS§A, B:m>V –w
⁄~.ƒ>6IC>{R…Œ“.
)dfi9n˘£§V = I(R0 + R).
9. 3˛/NS?,«M§T˛/N.»·a§ph§?Mp·x£ª1-6⁄. T
MN¨V xm…Œ’XV = V (x)§¿§‰´.
)dfi§V = πa2x§§‰´[0, h]§[1, πa2h]
10. ,/Y–¡¨·F/§Xª1-7§.2§>45o§CDL«Y¡§ƒ¡ABCD¡
11. kH¶‡§X^»R¯–zƒ¤ωll¶‡SL›§ƒ›.¡/¡
¨SYh…Œ’X.
)dfi9ª§S = h(h + 2).
0,
H
ωt
lsmt…Œ’X£ª1-8⁄.
)dfi9ª§s = H − ωRt
t ∈
1 + x2, x < 0
x − 1,
x > 0
12. y = f (x) =
§ƒf (−2), f (−1), f (0), f (1)f
)dfi§f (−2) = 5, f (−1) = 2, f (0) = −1, f (1) = 0, f
1
.
2
= − 1
2
.
1
2
0,
1 + t2,
t − 10,
0 6 t < 10
10 6 t 6 20
20 < t 6 30
13. x(t) =
§ƒx(0), x(5), x(10), x(15), x(20), x(25), x(30)§¿xø…Œª/.
5
)dޤx(0) = 0, x(5) = 0, x(10) = 101, x(15) = 226, x(20) = 401, x(25) = 15, x(30) = 20
14. e]y·&›x…Œ.Ue5‰§ØuIS†&§U&›§z›20AGe]8'§
v20–20O.&›360–S§`ø…ŒL“§¿x§ª/.
)dޤy = f (x) =
)dfi9ª§u = u(t) =
8,
1.5t,
16,
24,
0 < x 6 20
20 < x 6 40
40 < x 6 60
30 − 1.5t,
0 6 t 6 10
10 < t 6 20
15. u))n¯§¯/Xª1-9§…Œ’Xu = u(t)(0 6 t 6 20).
16. e…Œf ϕ·˜§o”
(1) f (x) =
x
x
, ϕ(x) = 1
√
x2
(2) f (x) = x, ϕ(x) =
(3) f (x) = 1, ϕ(x) = sin2 x + cos2 x
)
(1) ˇf ‰´(−∞, 0)S(0, +∞)§ϕ‰´(−∞, +∞)§ø…Œ.
(2) ˇf (x) = x, ϕ(x) = |x|§ø…Œ…ŒL“§Kø…Œ.
(3) ˇϕ(x) = sin2 x + cos2 x = 1⁄Ƨø…Œ.
17. y†Øu…Œf (x) = ax + b§egCŒx = xn(n = 1, 2,··· )|⁄Œ§KØA…Œ
yn = f (xn)(n = 1, 2,··· )|⁄Œ.
y†xm−1, xm, xm+1·xn¥?¿3Œ(2 6 m 6 n)
K¿§2xm = xm−1 + xm+1
qyn = f (xn) = axn + b§ Kym−1 = axm−1 + b, ym = axm + b, ym+1 = axm+1 + b§ u ·2ym =
2axm + 2b, ym+1 + ym−1 = axm+1 + b + axm−1 + b = 2axm + 2b§l2ym = ym−1 + ym+1
qxm−1, xm, xm+1·xn¥?¿3Œ§Kym−1, ym, ym+1·yn¥?¿3Œ§u·yn = f (xn)(n =
1, 2,··· )|⁄Œ.
18. XJ›y = f (x)?^upu§⁄l£ª1-10⁄§y†“
u⁄kx1, x2(x1 6= x2)⁄Æ£kªA5…Œ…Œ⁄.
y†3›?:A(x1, f (x1)), B(x2, f (x2))§ºAB§¥:C(xC , yC )§Kf (x1) + f (x2) =
2yC , x1 + x2 = 2xC
q›xD =
x1 + x2
گ:pIyD = f
§KxC = xD
x1 + x2
2
2
q›y = f (x)?^upu§⁄lx1, x2ul:§KyC > yD=
f (x1) + f (x2)
2
>
x1 + x2
f
2
Øu⁄kx1, x2(x1 6= x2)⁄Æ.
f (x1) + f (x2)
> f
2
x1 + x2
Ø
2
y
6
f (x)
A
B
C
0
x1
xD
x2
-
x
19. y†e…Œ3⁄««mS·NO\…Œ
(1) y = x2(0 6 x < +∞)
− π
2
6 x 6 π
2
(2) y = sin x
y†
6
(1) 0 6 x1 < x2
Ky2 − y1 = x2
2 − x2
6 x1 < x2 6 π
2
(2) − π
2
1 = (x2 + x1)(x2 − x1) > 0§u·…Œy = x20 6 xNO\.
x2 + x1
Ky2 − y1 = sin x2 − sin x1 = 2 cos
q− π
2
2
0§ly2 − y1 > 0=…Œy = sin x− π
2
6 x1 < x2 6 π
2
§K− π
2
<
2
x1 + x2
sin
<
π
2
6 x 6 π
2
x2 − x1
2
, 0 <
x2
x1
2 6 π
2
§u·cos
x1 + x2
2
> 0, sin
x2 − x1
2
>
NO\.
20. y†e…Œ3⁄««mS·N~…Œ
(1) y = x2(−∞ < x 6 0)
(2) y = cos x(0 6 x 6 π)
y†
(1) 0 6 x1 < x2
Ky2 − y1 = x2
2 − x2
(2) 0 6 x1 < x2 6 π
1 = (x2 + x1)(x2 − x1) < 0§u·…Œy = x2x 6 0N~.
Ky2 − y1 = cos x2 − cos x1 = −2 sin
2
q0 6 x1 < x2 6 π§K0 <
2 6 π
x1 + x2
2
y2 − y1 < 0=…Œy = cos x0 6 x 6 πN~.
< π, 0 <
x2 + x1
x2
x1
sin
2
2
x2 − x1
§u·sin
x1 + x2
2
> 0, sin
x2 − x1
2
> 0§l
21. ?e…Œ5
(1) y = x + x2 − x5
(2) y = a + b cos x
(3) y = x + sin x + ex
(4) y = x sin
1
x
(5) y = sgnx =
1, x > 0
0, x = 0
−1 x < 0
(6) y =
1
2
< x < +∞
2
x2 ,
sin x2, − 1
2
x2, − ∞ < x < − 1
1
2
2
6 x 6 1
2
)
(1) ˇy = f (x) = x + x2 − x5§Kf (−x) = −x + x2 + x5§f (−x) 6= f (x), f (−x) 6= −f (x)§u·d…Œ
·…Œ.
(2) ˇy = f (x) = a + b cos x§Kf (−x) = a + b cos(−x) = a + b cos x = f (x)§u·d…Œ·…Œ.
(3) ˇy = f (x) = x + sin x + ex§Kf (−x) = −x − sin x + e−x§f (−x) 6= f (x), f (−x) 6= −f (x)§u·d
…Œ·…Œ.
(4) ˇy = f (x) = x sin
§Kf (−x) = −x sin
1
−x
= x sin
1
x
= f (x)§u·d…Œ·…Œ.
1
x
1, x > 0
1, − x > 0
0, x = 0
−1 x < 0
0, − x = 0
−1 − x < 0
(5) ˇy = f (x) =
Kf (−x) =
§
=
−1, x > 0
0, x = 0
x < 0
1
= −f (x)§u·d…Œ·…Œ.
(6) ˇy = f (x) =
1
2
< x < +∞
2
x2 ,
sin x2, − 1
2
x2, − ∞ < x < − 1
1
2
2
2
6 x 6 1
2
< −x < +∞
§
1
2
Kf (−x) =
1
x2,
2
sin x2, − 1
2
2
x2 ,
f (−x) 6= f (x), f (−x) 6= −f (x)§u·d…Œ·…Œ.
(−x)2 ,
sin(−x)2, − 1
2
(−x)2, − ∞ < −x < − 1
1
2
2
6 −x 6 1
2
1
2
=
< x < +∞
6 x 6 1
2
− ∞ < x < − 1
2
7
§
22. `y…Œƒ¨·…Œ§…Œƒ¨·…Œ§…Œ…Œƒ¨·…Œ.
y†f1(x), f2(x)‰´3(−a, a)(a > 0)S…Œ§g1(x), g2(x)‰´3(−a, a)(a > 0)S…
ΤF1(x) = f1(x)f2(x), F2(x) = g1(x)g2(x), F3(x) = f1(x)f2(x)
Kf1(−x) = f1(x), f2(−x) = f2(x), g1(x) = −g1(x), g2(−x) = −g2(x)§u·
F1(−x) = f1(−x)f2(−x) = f1(x)f2(x) = F1(x)
F2(−x) = g1(−x)g2(−x) = (−g1(x))(−g2(x)) = g1(x)g2(x) = F2(x)
F3(−x) = f1(−x)g1(−x) = f1(x)(−g1(x)) = −f1(x)g1(x) = −F3(x)
lF1(x)·…Œ¶F2(x)·…Œ¶F3(x)·…Œ.
23. f (x)‰´3(−∞, +∞)S?…Œ§y†F1(x) ≡ f (x) + f (−x)·…Œ§F2(x) ≡ f (x) − f (−x)·
…Œ.ØAue…ŒF1(x), F2(x)
(1) y = ax
(2) y = (1 + x)n
y†ˇF1(−x) = f (−x) + f (x) = F1(x)§KF1(x) = f (x) + f (−x)·…Œ
qF2(−x) = f (−x) − f (x) = −F2(x)§KF2(x) = f (x) − f (−x)·…Œ.
(1) F1(x) = f (x) + f (−x) = ax + a−x, F2(x) = f (x) − f (−x) = ax − a−x
(2) F1(x) = f (x) + f (−x) = (1 + x)n + (1 − x)n, F2(x) = f (x) − f (−x) = (1 + x)n − (1 − x)n
24. ‘†e…Œ=·–ˇ…Œ§¿ƒ–ˇ
(1) y = sin2 x
(2) y = sin x2
(3) y = sin x +
1
2
sin 2x
π
4
x
(4) y = cos
(5) y = | sin x| + | cos x|
√
(6) y =
tan x
(7) y = x − [x]
(8) y = sin nπx
)
(1) ˇy = sin2 x =
1
2
− 1
2
cos 2x§KT =
2π
2
= π
(2) by = sin x2–ˇ…ŒT = ω > 0
–ˇ…Œ‰´§Ø?x ∈ (−∞, +∞)§ksin(x + ω)2 = sin x2§AOØx = 0AT⁄Ƨ
√
√
√
Ksin ω2 = 0§u·ω2 = kπ, ω =
2 + 1)2kπ = nπ(n ∈ Z +)§u
2ω + ω)2 = sin ω2 = 0§K(
2kπ⁄Ƨsin(
qØx =
(k, n ∈ Z +)
√
kπ(k ∈ Z +)
√
2ω =
k
n
√
2 + 1)2 =
·(
√
√
2 ∈ Q−§
2 + 1)2 = 3 + 2
q(
∈ Q+§Kb⁄Ƨ=…Œy = sin x2·–ˇ…Œ.
k
n
(3) ˇy1 = sin xT = 2π¶y2 =
1
2
sin 2xT = π§Ky = sin x +
1
2
sin 2xT = 2π.
(4) T =
= 8
2π
π
4
8
(5) ˇf (x) = | sin x| + | cos x|, f
x +
†§y = | sin x| + | cos x|T =
π
2
sin
x +
+
cos
π
2
x +
= | cos x| + | sin x| = f (x)
π
2
=
π
2
.
√
tan xT = π.
(6) ˇf (x) = tan xT = π§Ky =
(7) ˇy = x − [x] = (x)§Ky = x − [x]T = 1.
(8) T =
2π
nπ
=
2
n
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