Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων...
4
ΜΑΘΗΜΑΤΙΚΑ ΚΑΤΕΥΘΥΝΣΗΣ Γ ΛΥΚΕΙΟΥ ΠΑΝΕΛΛΗΝΙΕΣ ΕΞΕΤΑΣΕΙΣ ΕΝΙΑΙΟΥ ΛΥΚΕΙΟΥ ΕΤΟΥΣ 2011 ΛΥΣΕΙΣ – ΑΠΑΝΤΗΣΕΙΣ | www.lazaridi.info | tel. 6977.385.358 | www.lazaridi.info Page 1 of 4 | ΜΑΘΗΜΑΤΙΚΑ ΚΑΤΕΥΘΥΝΣΗΣ Γ ΛΥΚΕΙΟΥ | ΕΝΙΑΙΟ 2011 | ΛΥΣΕΙΣ | |ΘΕΜΑ Α Α1. θεωρία Α2. θεωρία Α3. Σ / Σ / Λ / Λ / Σ |ΘΕΜΑ Β Β1. 1 | 3 | 2 | 3 | 2 2 | 3 | | 3 | 2 | 3 | | 3 | 2 | 3 | | 3 | = - ⇒ = - ⇒ = - + - ⇒ = - + - ⇒ = + + - i z i z i z i z i z i z i z i z Οπότε 1 | ) 3 0 ( | 1 | 3 | = + - ⇒ = - i z i z Κύκλος με κέντρο Κ(0,3) και ακτίνα R = 1 Β2. Είναι 1 | 3 | ) 3 ( ) 3 ( ) 3 ( ) 3 ( 2 = - = - ⋅ - = + ⋅ - i z i z i z i z i z λόγω Β1 Οπότε i z i z i z i z 3 1 3 1 ) 3 ( ) 3 ( - = + ⇒ = + ⋅ - Β3. R z z z i z i z i z i z w B ∈ = + = + + - = - + - = ) Re( 2 3 3 3 1 3 2 Ο z κινείται στο κύκλο με κέντρο Κ(0,3) και ακτίνα R = 1 Άρα 2 2 2 ) Re( 2 2 1 ) Re( 1 ≤ ≤ - ⇒ ≤ ≤ - ⇒ ≤ ≤ - w z z Β4. Αν yi x z + = τότε (λόγω Β3) x z w 2 ) Re( 2 = = Οπότε | | ) ( | | | 2 | | | 2 2 2 2 z y x y x yi x x yi x w z = + = + - = + - = - + = - 1 - K 0 1
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Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων Πανελληνίων 2011
Transcript of Μαθηματικά Κατεύθυνσης Απαντήσεις Θέματων...
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2011
| www.lazaridi.info | tel. 6977.385.358 |
www.lazaridi.info Page 1 of 4
| | 2011 | |
|
1. 2. 3. / / / /
|
1.
1|3|2|3|22|3||3|2|3||3|2|3||3| ===+=+=++ iziziziziziziziz pi
1|)30(|1|3| =+= iziz (0,3) R = 1
2.
1|3|)3()3()3()3( 2===+ iziziziziz 1 pi
iziziziz
3131)3()3(
=+=+
3.
Rzzziziziz
izwB
=+=++=
+= )Re(2333
132
z (0,3) R = 1
222)Re(221)Re(1 wzz
4. yixz += ( 3) xzw 2)Re(2 == pi
||)(|||2||| 2222 zyxyxyixxyixwz =+=+=+=+=
1
K
0 1
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2011
| www.lazaridi.info | tel. 6977.385.358 |
www.lazaridi.info Page 2 of 4
|
( ) Rxxfxxfxfxfe x +=+ ),()(1)()( 0)0()0( == ff
1.
( ) ( ) =+=+ )()()()()()( xfxexfexfxxfexfexfe xxxxx
==+==
1)()(1)()()()(1
xxxxc
xxexfxxfexfxexfecxfxexfe
( ) ( ) 011)ln()()ln()(1)(1)(=
+=
=
==
cxx
x
xxx cxexfxexf
xe
exfexexf
)ln()( xexf x = , Rx
2.
xe
exf
x
x
=
1)(
0010)( === xexf x
3.
( )2
1)2()(xe
xexf
x
x
=
( ) 21,.......0120)( ==== xxxexf x
:
( ) 012 = xe x ( f ) 2 21,
( ) 12)( = xexg x ( )xexg x = 1)(
10)( == xxg
)( xg
)( xg
x
+
1
)( xf
)( xf
x
+
0
( )( )0,0
)0(,0min
f
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2011
| www.lazaridi.info | tel. 6977.385.358 |
www.lazaridi.info Page 3 of 4
g )1,(1 =A ( ) ( )1,1)(lim),(lim
11=
=
exgxgAg
xx
( )10 Ag g 1A = 0)(xg ( ) 012 = xe x 1A pi 11 A pi 0)( 1 =g
g ),1[2 +=A ( ) ( ]1,)1(),(lim2 == + egxgAg x
( )20 Ag g 2A = 0)(xg ( ) 012 = xe x 2A pi 22 A pi 0)( 2 =g
pi
( ) ( ) ( ) 011
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2011
| www.lazaridi.info | tel. 6977.385.358 |
www.lazaridi.info Page 4 of 4
|
1.
=
=
+
=
=+ 0 2
22
0 )(2
20
2
2 )(1)(1
)()(1
)()(1
x
u
xxx
xu
x
utxx t
xdu
uge
ee
xfduug
e
e
xfdttxg
e
e
xf
= 0 2
)()(1x
u
duug
exf =
x u
duug
exf
0
2
)()(1 +=x u
duug
exf
0
2
)(1)( 1)0( =f
+=
x t
xdt
txfe
e
xg
0
2
2 )()(1
pipi +=x u
duuf
exg
0
2
)(1)( 1)0( =g
pi
)()(2
xge
xfx
= xexgxf 2)()( = (1) )()(2
xfe
xgx
= xexgxf 2)()( = (2)
pi (1), (2) pipi ( ) ( )
0)(ln)(ln)(ln)(ln)(
)()()()()()()(
=+==
=
=
c
cxgxfxgxfxgxg
xfxf
xgxfxgxf
Rxxgxfxgxf == ),()()(ln)(ln
2. pi (1) 1 pipi
( ) ( ) ==== xxxx exfexfxfexfxfexgxf 22222 )(2)()(2)()()()( x
cx exfcexf 22
022 )()( =+=
= Rxxf > 0)( Rxexf x = ,)(
3.
=
=====
==
=
1limlim1lim
1
limlimlnlimlnlim1
)(lnlim1
/10/10/100
u
u
DLHu
uuuuu
ux
xxxxx
x
xx
e
u
e
uee
u
e
x
e
ex
e
e
xf
xf
4.
==x
tx
dtedttfxF11
2 2)()( 0)1( =F
>= FexF x 0)( 2 0)()1()(1
