@çbznàa
ò‹Ñïå‚@óibïä
ózÑ—Üa
@pbïšbî‹Üa@Zò†b¾a
öa‹èÜa@óábÐ@óîíäbq
óïjî‹vnÜa@ãíÝÉÜaZójÉ“Üa 2005O2006ZóïaŠ‡Üa@óåÜa
1/2
Œb−fia@ò‡à
3H
Þàbɾa
7
(ÂÕä@5)ßìÿa@æî‹ánÜa
u1 = 2 ﻨﻌﺘﺒﺭ ﺍﻝﻤﺘﺘﺎﻝﻴﺔ ) (u nﺍﻝﻤﻌﺭﻓﺔ ﺒﻤﺎ ﻴﻠﻲ: 1 = + 2 u u n +1 3 n -1 0,5ﺍﺤﺴﺏ . u 2 1 1 1,5 1
-2ﺒﻴﻥ ﺃﻥ. ∀n ∈ ℕ ;u n ≺ 3 : *
-3ﺃﺩﺭﺱ ﺭﺘﺎﺒﺔ ﺍﻝﻤﺘﺘﺎﻝﻴﺔ ) (u n -4ﻨﻌﺘﺒﺭ ﺍﻝﻤﺘﺘﺎﻝﻴﺔ ) (v nﺒﺤﻴﺙ.v n = u n − 3 : ﺃ:ﺒﻴﻥ ﺃﻥ ﺍﻝﻤﺘﺘﺎﻝﻴﺔ ) (v nﻫﻨﺩﺴﻴﺔ ﺜﻡ ﺃﺤﺴﺏ v nﺒﺩﻻﻝﺔ . n
ﻭﺍﺴﺘﻨﺘﺞ ﺃﻥ. ∀n ∈ ℕ ;u n ≥ 2 : *
ﺏ:ﺃﺤﺴﺏ s n = v 1 + v 2 + .... + v nﻭﺍﺴﺘﻨﺘﺞ s n
. lim
∞n →+
(óÕä@5,5@)ðäbrÜa@æî‹ánÜa
5z ﻝﻜل ﻋﺩﺩ ﻋﻘﺩﻱ zﺒﺤﻴﺙ z ≠ −i ) :ﻭ ( z ≠ iﻨﻀﻊ: z2 +1 2 -1 0,5ﺃ :ﺃﺤﺴﺏ ) . ( −3 + 4i
= ) . p (z
1 1
ﺏ:ﺤل ﻓﻲ ℂﺍﻝﻤﻌﺎﺩﻝﺔ. p ( z ) = 3 + i : ﺝ:ﺃﻜﺘﺏ ﻋﻠﻰ ﺍﻝﺸﻜل ﺍﻝﻤﺜﻠﺜﻲ ﺤﻠﻲ ﺍﻝﻤﻌﺎﺩﻝﺔ ﺍﻝﺴﺎﺒﻘﺔ.
1
-2ﺃ:ﺒﻴﻥ ﺃﻥ ﻝﻜل zﻤﻥ } − 1 = 0 : ℂ / {−i , i
1
ﺏ:ﺤﺩﺩ ﻓﻲ ﺍﻝﻤﺴﺘﻭﻯ ﺍﻝﻌﻘﺩﻱ ) ( pﺍﻝﻤﻨﺴﻭﺏ ﺇﻝﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﻤﻨﻅﻡ ) (O , e 1 , e 2ﻤﺠﻤﻭﻋﺔ ﺍﻝﻨﻘﻁ ) M ( zﺒﺤﻴﺙ ﻴﻜﻭﻥ ) p ( z
)
2
(z − z )( z
⇔ . p (z ) ∈ ℝ
ﺤﻘﻴﻘﻴﺎ.
1
-3ﻨﻌﺘﺒﺭ ﻓﻲ ) ( pﺍﻝﻨﻘﻁﺘﻴﻥ A (1 − i ) :ﻭ OB ﻭ . OA ,OB ﺃﺤﺴﺏ
OA
)
1+ i 2
.B
(
(ÂÕä 9,5 @)óÜdà ﺍﻝﺠﺯﺀ:A
ﻨﻌﺘﺒﺭﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ gﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞ ]0, +ﺒﻤﺎ ﻴﻠﻲ+ 1 + ln ( x ) : -1 0,5ﺃﺤﺴﺏ ) g ′ ( xﻝﻜل xﻤﻥ [∞. ]0, + -2ﺃﺩﺭﺱ ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ gﺜﻡ ﺍﺴﺘﻨﺘﺞ ﺇﺸﺎﺭﺓ ) g ( xﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞. ]0, + 1 ﺍﻝﺠﺯﺀ:B
ﻨﻌﺘﺒﺭﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ fﺍﻝﻤﻌﺭﻓﺔ ﺒﻤﺎﻴﻠﻲ− 2 x : 0,5
-1ﺃﺤﺴﺏ ) ( x
. lim+ f x →0
2
) ) ( x ) = (1 + ln ( x
.f
( x ) = −x
.g
ﻋﺩﺩﺍ
2O2ózÑ—Üa 2 -2 0,5ﺃ :ﺘﺤﻘﻕ ﻤﻥ ﺃﻥ − 2 : 0,5ﺏ:ﺍﺤﺴﺏ ) . lim f ( x
) (
ln x ) ln ( x 1 . ∀x ≻ 0 : f ( x ) = x +2 + 4 x x x
∞x →+
1
:3ﺃﺩﺭﺱ ﺍﻝﻔﺭﻭﻉ ﺍﻝﻼﻨﻬﺎﺌﻴﺔ ﻝﻠﻤﻨﺤﻨﻰ ) . (C f
ﻭﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ . ( D ) : y = −2 x
-4 0,5ﺃﺩﺭﺱ ﻭﻀﻊ ﻭﺘﻘﺎﻁﻊ ﻤﻨﺤﻨﻰ ﺍﻝﺩﺍﻝﺔ f ) g (x . ∀x ≻ 0 : f ′ ( x ) = 2 -5ﺃ :ﺒﻴﻥ ﺃﻥ: 1 x 0,5ﺏ:ﺍﺴﺘﻨﺘﺞ ﺇﺸﺎﺭﺓ ) f ′ ( xﺜﻡ ﻀﻊ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ . f -6 0,5ﺃ:ﺒﻴﻥ ﺃﻥ ﺍﻝﻤﺴﺘﻘﻴﻡ ) ( Dﻤﻤﺎﺱ ﻝﻤﻨﺤﻨﻰ ﺍﻝﺩﺍﻝﺔ f
ﻓﻲ ﺍﻝﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻷﻓﺼﻭل
0,5ﺏ:ﺒﻴﻥ ﺃﻥ ﺍﻝﻨﻘﻁﺔ ) I (1, −1ﻫﻲ ﻨﻘﻁﺔ ﺍﻨﻌﻁﺎﻑ ل ) . (C f
1 1 -7 0,5ﺒﻴﻥ ﺃﻨﻪ ﻴﻭﺠﺩ ﻋﺩﺩ ﻭﺤﻴﺩ α -8 0,5ﻝﺘﻜﻥ hﻗﺼﻭﺭ ﺍﻝﺩﺍﻝﺔ fﻋﻠﻰ ﺍﻝﻤﺠﺎل [. I = ]0,1 ﻤﻥ 2 , ﺒﺤﻴﺙ e e
ﺒﻴﻥ ﺃﻥ hﺘﻘﺎﺒل ﻤﻥ ﺍﻝﻤﺠﺎل Iﻨﺤﻭ ﻤﺠﺎل J
(α ) = 0
ﻴﻨﺒﻐﻲ ﺘﺤﺩﻴﺩﻩ.
1
e
= .x0
.f
-9 1.5ﺃﻨﺸﺊ ) ( Dﻭ ) (C fﻭ ) (C h −1ﻓﻲ ﻨﻔﺱ ﺍﻝﻤﻌﻠﻡ ﺍﻝﻤﺘﻌﺎﻤﺩ ﺍﻝﻤﻤﻨﻅﻡ . O , i , j
)ﺍﻝﻭﺤﺩﺓ. 2cm:ﻨﺄﺨﺫ≃ 0, 2 :
1
e2
ﻭ ≃ 0, 4
1
e
)
(
(.
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