Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων...
6
ΜΑΘΗΜΑΤΙΚΑ ΚΑΤΕΥΘΥΝΣΗΣ Γ ΛΥΚΕΙΟΥ ΠΑΝΕΛΛΗΝΙΕΣ ΕΞΕΤΑΣΕΙΣ ΕΤΟΥΣ 2014 ΛΥΣΕΙΣ – ΑΠΑΝΤΗΣΕΙΣ www.lazaridi.info Page 1 of 6 | ΜΑΘΗΜΑΤΙΚΑ ΚΑΤΕΥΘΥΝΣΗΣ Γ ΛΥΚΕΙΟΥ | | Πανελλήνιες Εξετάσεις Γ τάξης Ημερησίου Γενικού Λυκείου 2014 |02 – 06 – 2014 | | ΛΥΣΕΙΣ | |ΘΕΜΑ A Α.1. απόδειξη A.2. ορισμός A.3. ορισμός A.4. Λ / Σ / Σ / Σ / Λ |ΘΕΜΑ Β Β.1. 2 2| | ( ) 4 2 0 z x yi z z zi i =+ ++--=⇒ 2 2 2( ) 2 4 2 0 x y xi i ++--=⇒ 2 2 2( ) 4 2( 1) 0 x y x i +-+-=⇒ 2 2 2 ( 1) 0 x y x i ⇒+-+-=⇒ 2 2 2 0 1 0 x y x +-= ⇒ -= 2 1 1 y x = ⇒ = 1 1 y x =± = Άρα 1 1 z i =+ , 2 1 z i =- Β.2. Είναι 2 1 2 2 1 (1 ) 2 1 1 2 z i i i i z i i + + = = == - - Οπότε () () 39 19 19 39 2 1 2 3 3 3 3 1 3 z w i i i i i z =⋅ =⋅=⋅ ⋅=⋅- ⋅=- Β.3. 1 2 4 u w z z i +=--⇒ ( ) ( ) 3 41 1 u i i i i -=+---⇒ 3 3 4 u i i -=+⇒ 2 2 3 3 4 u i -= +⇒ 3 5 u i ⇒-=⇒ (0 3) 5 u i -+= κύκλος με κέντρο Κ(0, 3), ακτίνα R = 5
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Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων Πανελληνίων 2014
Transcript of Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων...
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2014
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| |
| 2014 |02 06 2014 |
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| A
.1. pi A.2. A.3. A.4. / / / /
|
.1.
22 | | ( ) 4 2 0 z x yiz z z i i = ++ + = 2 22( ) 2 4 2 0x y xi i+ + = 2 22( ) 4 2( 1) 0x y x i+ + =
2 2 2 ( 1) 0x y x i + + = 2 2 2 0
1 0x y
x
+ =
=
2 11
yx
=
=
11
yx
=
=
1 1z i= + , 2 1z i=
.2.
2
12
2
1 (1 ) 21 1 2
z i i i iz i i
+ += = = =
pi ( ) ( )39
19 1939 21
2
3 3 3 3 1 3zw i i i i iz
= = = = =
.3.
1 24u w z z i+ = ( ) ( )3 4 1 1u i i i i = + 3 3 4u i i = + 2 23 3 4u i = +
3 5u i = (0 3 ) 5u i + = (0, 3), R = 5
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2014
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|
.1.
( ) ln( 1),xh x x e x R= + ( ) 111 1
x
x x
eh xe e
= =+ +
[ ( ) 1 01x
h xe
= >+
h R ]
pi ( )h x =( )2
01
x
x
e
e >
+
-
(2 ( ))ln ln1
h h x ee
e
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2014
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.3.
lim ( )x
h x+
= ( )lim ln( 1)xx
x e+
+ = ( )( )lim ln ln 1x xx
e e+
+ = lim ln1
x
xx
e
e+
= +
( )lim ln1 0x+
=
y = 0 ( ) pi hC +
( )lim
x
h xx=
ln( 1)limx
x
x e
x
+=
ln( 1)lim 1x
x
e
x
+ =
1 0 1 R = =
( )( )limx
h x x
= ( )( )limx
h x x
= ( )lim ln( 1)xx
x e x
+ = ( )lim ln( 1)xx
e
+ = 0 =
y x = + y = x pi pi hC
.4.
( )1 *
0
| |E x dx= = ( )1
0
x dx = ( )1
0
( ) ln 2xe h x dx+ = ( ) ( )1
0
( ) ln 2xe h x dx + =
( )1
1
00
( ) ln 2 ( )x xe h x e h x dx = + = ( )1
0
1(1) ln 2 01
x
xe h e dx
e + = +
( )1
0
1 ln( 1) ln 21
x
x
ee e dx
e = + + = +
1
0
21 ln ln( 1)1
xe e
e
+ + = +
[ ]21 ln ln( 1) ln 21
e ee
= + + = +
2 21 ln ln1 1
ee e
+ + = +
2 2ln ln1 1
e ee e
+ + =+
= ( ) 21 ln1
e ee
+ ++
..
*
0 1 (0) ( ) ln 2 ( )h
x h h x h x
( x = 0, ( ) ln 2 (0)h x h= = h x = 0)
( ) ln 2 0h x + ( ) ( )( ) ln 2 0xx e h x = +
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2014
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|
.1.
( ) ( )0 0 0
1lim lim lim 1 01
x xDLH
x x x
e ef x fx
= = = = f xo = 0
0x , ( ) 2 21 ( 1) 1x x x x xe e x e xe ef x
x x x
+ = = =
pi ( ) 1x xA x xe e= +
( ) xA x xe =
( ) 0 0A x x = =
pi ( ) ( )0A x A ( ) 0A x ( x = 0)
0x , ( ) 0f x >
0
( ) (0)lim0x
f x fx
=
0
( ) 1limx
f xx
=
0
1 1lim
x
x
e
x
x
= 20
1limx DLH
x
e x
x
=
0
1lim2
x DLH
x
e
x
=
0
1lim2 2
x
x
e
=
( ) 102
f =
pi f R
x
( )A x
( )A x
0+
( )( )( )
min0, 0
0,0
A
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.2.
)
H f A =R,
( ) ( ) ( )( ) ( )lim , lim 0,x x
f A f x f x +
= = +
( ) 0f x > , x R
-
( ) ( )( )2
1
f xK x f u du
= ( ) ( )( )
( )2 0 1
1 1
0 0f
K f u du f u du
= = =
x = 0 , K
( ) ( )( ) ( )2 2 0K x f f x f x = > ( )2
32 2 2x x xx e xe ef x
x
+ =
-
( )2 1f x = ( ), pi
( ) ( ) ( )1 02
f x f x f = = f 1-1 pi x = 0
)
( ) ( )( ) , 0y t f x t t= ( ) ( ) ( )( )y t f x t x t = pi t = to
( ) ( ) ( )( )o o oy t f x t x t = ( ) ( )2 0o ox t y t = >
( )( ) 12of x t =
( )( ) ( )1 02of x t f = = f 1-1
pi ( ) 0ox t = pi ( ) ( )( ) ( )0 1o oy t f x t f= = = (0, 1)
f
f(x)>0 x
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.3.
( ) 2 2( ) ( 2)xg x e e x= , x > 0 , ( )( )
2( ) ( 2) ( )x x xK x
g x e e x xe e e = 14243
( ) x xK x xe e e= [1, 2] ( ) 21 (2) ( ) ( 2) 0K K e e = < .Bolzano
pi ( )1, 2ox ( ) 0oK x = , pi K 0x >
( ( ) 0, 0xK x xe x = > > )
pi ( ) 0 1, , 2og x x x x x = = = =
pi g
0 1x xe e e e x > > >
( ) ( ) ( ) 0K
o ox x K x K x K x
< < <
( ) ( ) ( )0K
o ox x K x K x K x
< < <
+
( )( )min
1, 1g
x
( )g x
( )g x
1 ox 2
+
( )( )min
2, 2g( )( )max
,o o
x g x
+
+1 ox 2x
( )2 xe e2x
( )K x
( )g x
0
+ +
++ + +
