数学分析 第三版 上、下册 课后答案(陈传璋 著) 高等教育出版社

(3) |x + x1 + · · · + xn| |x| − |x1 + x2 + x3 + · · · + xn| |x| − (|x1| + · · · + |xn|)
3. )e ýéŠØ ª§¿xÑx ‰Œµ
(1) |x| > |x + 1|
(2) 2 < 1 < 4 |x|
(3) |x| > A
$
e -A
0
'
e A
Ex
A < 0ž§x ∈ R (4) a − η < x < a + η
' a −eη 0 a
$ a +eη
Ex
(5) ª du x − 2 < 0§K−1 < x < 2 x+1
'
$
-1
0
1
2
Ex
(6)
5 −
< x < − 3 ½− 5
7 <x<−
3
22
3
§¤
§¤
ee
ee
2
2
5. ¦e ¼ê ½Â•9Š•µ
√ (1) y = 2 + x − x2
√ (2) y = cos x
π (3) y = ln sin
x (4) y = 1
sin πx
)µ
(1) ¼ê (2) ¼ê (3) ¼ê (4) ¼ê
½Â••X = [−1, 2]§Š•• 0, 3 2
½Â••
2kπ −
(4) |x − a| < η, η•~ê§η > 0
(5) x − 2 > x − 2 x+1 x+1
1
(6) 2 <
<3
|x + 2|
)µ
1 (1) x < −
2
$
-1 0
Ex
(2)
1 −
< x < −1½1
<x<
1
2
44
2
¨ 1e e 0 2
¨ e 1eபைடு நூலகம்E x
2
3 (3) A 0ž§x < −A½x > A
(1) x < − 5 ½x > − 3
2
2
$' -3 -2 -1 0
Ex
(2) 1 < x < 3½x < −2
$
'
$
-2 -1
0
1
2
3
Ex
(3) a > 0ž§x < 1½x > 1 + 1 ¶ a 0
$' 1 1+ 1
a
a < 0ž§1 + 1 < x < 1 a
' 1+ 1 0
½Â••X = (−∞, 0) (0, ∞)§f (−1) = 0, f (1) = 0, f (2) = − 3
√2
½Â••X = [−|a|, |a|]§f (0) = |a|, f (a) = 0, f − a = 3 |a|
2
2
½Â••(−∞, 0)
(0, ∞)§s(1)
=
1 , s(2) e
=
1 2e2
½Â•• x x ∈ R, x = kπ + π , k ∈ Z
§g(0) = 0, g
π
π2
π
= ,g −
2
4 16
4
½Â••X = (−∞, ∞)§x − π = −1, x(−π) = −1 2
½Â••X = (−∞, −2) (−2, 1) (1, +∞)§f (0) = − 1 , f (−1) = − 1
-3
-2
-1
0
Ex
4. ¦e ¼ê ½Â•9§3‰½:þ ¼êŠµ
(1) y = f (x) = −x + 1 ½Â•9f (−1), f (1)Úf (2)¶ x
√ (2) y = f (x) = a2 − x2
½Â•9f (0), f (a)Úf
−a ¶
2
(3) s = s(t) = 1 e−t ½Â•9s(1), s(2)¶ t
1
1˜Ÿ 4•Ø
1˜Ü© 4•ÐØ
1˜Ù Cþ†¼ê
§1. ¼ê Vg
1. )e Ø ª§¿xÑx ‰Œµ
1 (1) −2 <
x+2 (2) (x − 1)(x + 2)(x − 3) < 0
(3) 1 < a x−1
(4) 0 cos x 1 2
(5)
x2 − 16 < 0 x2 − 2x 0
)µ
a
$ 1
a = 0ž§x < 1
$
0
1
Ex Ex Ex
2
π (4) 2kπ +
x 2kπ + π ½2kπ − π
x
π 2kπ − (k ∈ Z)
3
2
2
3
£
£
£
0
£ Ex
(5) −4 < x 0½2 x < 4
-4
¨ 0
2
4¨ Ex
2. y²e ýéŠØ ªµ
(1) |x − y| ||x| − |y|| (2) |x1 + x2 + x3 + · · · + xn| |x1| + |x2| + · · · + |xn| (3) |x + x1 + · · · + xn| |x| − (|x1| + · · · + |xn|) y²µ (1) Ï|x||y| xy§K(x − y)2 (|x| − |y|)2§u´|x − y| (2) ^êÆ8B{y².
7. f (x) = (|x| + x)(1 − x)§¦÷ve ˆª xŠµ
(4) y = g(α) = α2 tan α ½Â•9g(0), g π , g − π ¶
4
4
(5) x = x(θ) = sin θ + cos θ ½Â•9x − π , x(−π) 2
(6) y = f (x) =
1
½Â•9f (0), f (−1)
(x − 1)(x + 2)
4
)µ
(1) ¼ê (2) ¼ê (3) ¼ê (4) ¼ê (5) ¼ê (6) ¼ê
||x| − |y||
(i) n = 2ž§d|x1 + x2| |x1| + |x2|§ (ؤá. (ii) b n = kž(ؤá§=k|x1 + x2 + x3 + · · · + xk| |x1| + |x2| + · · · + |xk|.
K n = k + 1ž§|x1 + x2 + x3 + · · · + xk+1| |x1 + x2 + x3 + · · · + xk| + |xk+1| |x1| + |x2| + · · · + |xk| + |xk+1| nþŒ•§é˜ƒg,ên§|x1 + x2 + x3 + · · · + xn| |x1| + |x2| + · · · + |xn|þ¤á.
π , 2kπ +
π
(k ∈ Z)§Š••[0, 1]
2
2
½Â••
1
1 ,
(k ∈ Z)§Š••(−∞, 0]
2k + 1 2k
½Â••(n − 1, n)(n = 0, ±1, ±2, · · · )§Š••(−∞, −1]
[1, +∞)
π2 =−
16
6. f (x) = x + 1, ϕ(x) = x − 2§Á)•§|f (x) + ϕ(x)| = |f (x) + |ϕ(x)| )µd®•§ f (x)ϕ(x) 0=(x + 1)(x − 2) 0§Kx 2½x −1.