Μαθηματικά Κατεύθυνσης Απαντήσεις Θεμάτων...
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www.methodiko.net & 2, , : 210 99 40 999 . 201, , : 210 96 36 300
: ,
: 25 2015
1. , . . 194
2. , . . 188
3. , . . 259
4. . , . , . , . , .
B
1. : | 4| = 2| 1| | 4|2 = 4| 1|2
( 4)( 4) = 4( 1)( 1)
4 4 + 16 = 4 4 4 + 4
3 = 12 ||2 = 4||0 || = 2
(0,0) = 2 .
2. . |1| = 2 |1|2 = 4 11 = 4 1 =
1
2 =4
2
:
=212+221=24142
+24241
=221+212=
: = = 0 2 () = 0 () = 0
. :
|| = |212+221|
2|1|
|2|+2|2|
|1|
=
2 2
2+2 2
2= 4
|| 4 4 4.
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3. :
= 4 212+221= 4 1
2 + 22 = 212
12 + 2
2 + 212 = 0 (1 + 2)2 = 0 2 = 1 (1)
() = |3 1|
(1)=
|21 + 1| = |1||1 + 2| = 212 + 22 = 25
() = |3 1| = |22 1| = |1||1 + 2| = 2(1)2 + 22 = 25
() = (), .
1. :
() = (
2 + 1)
=(2 + 1) 2
(2 + 1)2=( 1)2
(2 + 1)2> 0
1 = 1, .
, :
( lim
() , lim
()), :
lim
() = lim
2 + 1= 0
: lim
= 0 lim
1
2+1= 0
lim+
() = lim+
2 + 1= lim+
()
(2 + 1)= lim+
2= lim+
2= +
(0, +).
2. 1-1.
:
(3(2 + 1)) =2
5 (3(2 + 1)) = (2)
3
(2 + 1) = 2
3
2=
(2 + 1) () =
3
2 (1)
=3
2> 0 (1)
. 1-1.
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3. :
() = ()
0
, 0
H (0, +) () = () > 0.
> 0 2 < 4.
[2, 4].
H [2, 4] [0,+] (2, 4).
, 1 (2, 4) :
(1) =(4) (2)
4 2
(1) = ()4
0 ()
2
0
2
(1) = ()4
2
2 (2)
, 1 < 4 :
(1) < (4) ()4
2
2< (4) ()
4
2
< 2 (4), > 0
:
0 < 2 < 4 :
() < (4) () < (4)
:
()
4
2
< (4)
4
2
()
4
2
< (4) []24
()
4
2
< (4) (4 2) ()
4
2
< 2 (4)
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4. (0, +) :
() = (1
()
4
2
)
() = 1
2 ()
4
2
+1
( ()
4
2
)
() = 1
2 ()
4
2
+1
((4) (4) (2) (2))
() = 1
2 ()
4
2
+1
(4 (4) 2 (2))
() =1
2(4 (4) 2 (2) ()
4
2
)
() >1
2(4 (4) 2 (2) 2 (4))
() >1
2(2 (4) 2 (2))
() >2
((4) (2))
> 0 :
2 < 4 (4) (2) > 0 () > 0
, (0, +).
= 0 [0, +].
:
lim0+
() = lim0+
()4
2
= lim0+
( ()4
2)
()
= lim0+
((4) (4) (2) (2))
= lim0+
(4(4) 2(2)) = 2(0) = 2 1 = (0)
de l Hospital.
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www.methodiko.net & 2, , : 210 99 40 999 . 201, , : 210 96 36 300
1.
: () [() + ()] = 2
() () + () () = 2
[() ()]= [2] . : () () = 2 +
= 0 = 0 :
() () = 2
() 1
()= 2
(())2 2 () + 2 = 2 + 1
(() )2= 2 + 1
|() | = 2 + 1 (1)
: 2 + 1 0 () = () 0 (0) = 1
() > 0 .
(1) : () = 2 + 1 () = + 2 + 1
() = ln( + 2 + 1)
2. . H :
() =1
+ 2 + 1 (1 +
2 + 1) =
1
+ 2 + 1 + 2 + 1
2 + 1=
1
2 + 1
() = 1
2 + 1 (2 + 1)
=
1
2 + 1
2
22 + 1=
(2 + 1)2 + 1
() = 0 = 0.
:
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0 +
() +
()
H (, 0] [0, +) (0, (0)) = (0,0).
. H
(0, (0)) :
(): (0) = (0)( 0) =
[0, +) =
(0,0).
() () 0 0 .
:
() = | ()| 2
0
= ( ln ( + 2 + 1)) =1
0
= 1
0
()1
0
() = [2
2]0
1
[()]01 +
1
0
()
=1
2 ln(1 + 2) +
1
0
1
2 + 1 =
1
2 ln(1 + 2) +
2
22 + 1
1
0
=1
2 ln(1 + 2) + [2 + 1]
0
1=1
2 ln(1 + 2) + 2 1 = 2
1
2 ln(1 + 2)
3. () > (0) () > 0 > 0 :
lim0+
( 2()
0 1) |()| = lim
0+
1
()(
2()0 1) () () =
= lim0+
1
()(
2()0 1) () () (1)
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lim0+
() () () = 0 = lim0+
() = 0 (
) :
lim0+
() () = lim0+
= lim0+
1
= lim0+
112
= lim0+
() = 0
lim0+
( 2()
0 1)
()
de lHospital : lim0+
2()
0 1 = 0 1 = 0 (0) = 0.
:
lim0+
( 2()
0 1)
()= lim0+
( 2()
0 1)
()= lim0+
2()
0 2()
()= 0
: lim0+
() = (0) = 1 = 0 (0) = 0.
(1) : lim0+
( 2()
0 1) |()| = 0
4.
() = ( 2) [1 3 (2) 2
0
] + ( 3) [8 3 2()
0
] , [2,3]
H G [2,3]. : (2) = 3 2() 2
0 8
[0,2] : () 2() 2, = 0, :
2() 2
0
< 2 2
0
= [3
3]0
2
=8
3
3 2() 2
0
< 8 3 2() 2
0
8 < 0 (2) < 0
(3) = 1 3 (2)1
0
.
[0,1] (2) 2 = 0 .
(2)1
0
< 2 1
0
= [3
3]0
1
=1
3 3 (2)
1
0
< 1
1 3 (2) 1
0
> 0 (3) > 0
Bolzano () = 0 0 (2,3).
: , , , , , , , ,