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corrigé ratt 2007 .pdf



Nom original: corrigé ratt 2007.pdf
Titre: corrigé bac ratrrapage2007
Auteur: Administrateur

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‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ‬
‫ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬
‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬

‫ﺍﻝﺸﻌﺏ‪ :‬ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ ﺍﻷﺼﻴﻠﺔ‬
‫ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ‬
‫ﺍﻝﻌﻠﻭﻡ ﺍﻝﺯﺭﺍﻋﻴﺔ‬

‫ا
ا ول ‪:‬‬
‫ ‬
‫ﻓﻲ ﺍﻝﻔﻀﺎﺀ ﺍﻝﻤﻨﺴﻭﺏ ﺇﻝﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﻤﻨﻅﻡ ) ‪ (o, i , j , k‬ﻝﺩﻴﻨﺎ ﺍﻝﻨﻘﻁ )‪ A(2, 0, −1‬ﻭ )‪B (2, 4, 2‬‬
‫ﻭ )‪ C (3,3, 3‬ﻭ ﺍﻝﻔﻠﻜﺔ ) ‪ ( S‬ﺍﻝﺘﻲ ﻤﻌﺎﺩﻝﺘﻬﺎ ﺍﻝﺩﻴﻜﺎﺭﺘﻴﺔ ﻫﻲ ‪x 2 + y 2 + z 2 − 4 x − 4 y − 8 z + 20 = 0 :‬‬
‫‪1‬‬

‫‪ ( 1‬ﻨﺒﻴﻥ ﺍﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻫﻲ ﺍﻝﻨﻘﻁﺔ ) ‪ Ω ( 2, 2, 4‬ﺃﻥ ﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ ‪2‬‬

‫‪∀M ( x, y, z ) ∈ ( S ) ⇔ x 2 + y 2 + z 2 − 4 x − 4 y − 8 z + 20 = 0‬‬
‫‪⇔ ( x 2 − 4 x) + ( y 2 − 4 y ) + ( z 2 − 8 z ) + 20 = 0‬‬
‫‪⇔ ( x 2 − 4 x + 4) − 4 + ( y 2 − 4 y + 4) − 4 + ( z 2 − 8 z + 16) − 16 + 20 = 0‬‬
‫‪⇔ ( x − 2) 2 + ( y − 2) 2 + ( z − 4) 2 = 2 2‬‬
‫ﺇﺫﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻫﻲ ﺍﻝﻨﻘﻁﺔ ) ‪ Ω ( 2, 2, 4‬ﻭ ﺸﻌﺎﻋﻬﺎ ‪.R= 2‬‬

‫‪0,75‬‬

‫‪1‬‬

‫‪ ( 2‬ﻨﺒﻴﻥ ﺃﻥ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ )‪ ( P‬ﻫﻲ ‪x − y + z − 1 = 0 :‬‬
‫ ‬
‫ﻤﻌﺎﺩﻝﺔ ﺍﻝﻤﺴﺘﻭﻯ )‪ ( P‬ﺘﻜﺘﺏ ﻋﻠﻰ ﺍﻝﺸﻜل ‪ ax + by + cz + d = 0‬ﺤﻴﺙ )‪ n (a, b, c‬ﻤﺘﺠﻬﺔ ﻤﻨﻅﻤﻴﺔ ﻋﻠﻴﻪ‪.‬‬
‫ ‬
‫ﻝﺩﻴﻨﺎ )‪ B(2, 4, 2‬ﻭ )‪ C (3,3, 3‬ﺇﺫﻥ )‪BC (1, −1,1‬‬
‫ ‬
‫ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻭﻯ )‪ ( P‬ﻋﻤﻭﺩﻱ ﻋﻠﻰ ﺍﻝﻤﺴﺘﻘﻴﻡ )‪ (BC‬ﺇﺫﻥ ﺍﻝﻤﺘﺠﻬﺔ )‪ BC (1, −1,1‬ﻤﻨﻅﻤﻴﺔ ﻋﻠﻰ )‪( P‬‬
‫ﻭﻤﻨﻪ ﻓﺎﻥ ﻤﻌﺎﺩﻝﺔ )‪ ( P‬ﻫﻲ ‪x − y + z + d = 0‬‬
‫ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻭﻯ )‪ ( P‬ﻴﻤﺭ ﻤﻥ ﺍﻝﻨﻘﻁﺔ )‪ A(2, 0, −1‬ﺇﺫﻥ ‪ 2 − 0 + ( −1) + d = 0‬ﺃﻱ ‪d = −1‬‬
‫ﺇﺫﻥ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ‪ ( P‬ﻫﻲ ‪. x − y + z − 1 = 0 :‬‬
‫‪ ( 3‬ﺃ – ﻨﺒﻴﻥ ﺃﻥ ﺍﻝﻤﺴﺘﻭﻯ ) ‪ ( P‬ﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ ) ‪ ( S‬ﻭﻓﻕ ﺩﺍﺌﺭﺓ )‪ (Γ‬ﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ ‪.1‬‬

‫ﻝﺩﻴﻨﺎ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ‪ ( P‬ﻫﻲ ‪ x − y + z − 1 = 0 :‬ﻭﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻫﻲ ﺍﻝﻨﻘﻁﺔ ) ‪Ω ( 2, 2, 4‬‬
‫‪3‬‬
‫ﻝﺩﻴﻨﺎ ‪= 3‬‬
‫‪3‬‬

‫=‬

‫‪2 − 2 + 4 −1‬‬

‫= )) ‪d (Ω, ( P‬‬

‫ﻭﻝﺩﻴﻨﺎ‬

‫‪R=2‬‬

‫‪1 + (−1) + 1‬‬
‫ﺒﻤﺎ ﺃﻥ ‪ d (Ω, ( P )) ≺ R‬ﺇﺫﻥ ﺍﻝﻤﺴﺘﻭﻯ ) ‪ ( P‬ﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ ) ‪ ( S‬ﻭﻓﻕ ﺩﺍﺌﺭﺓ )‪ (Γ‬ﺸﻌﺎﻋﻬﺎ ‪ r‬ﺤﻴﺙ ‪:‬‬

‫‪0,25‬‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪r = R 2 − d 2 = 22 − 32 = 4 − 3 = 1‬‬
‫ﺏ‪ -‬ﻨﺤﺩﺩ ﺘﻤﺜﻴﻼ ﺒﺎﺭﺍ ﻤﺘﺭﻴﺎ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( ﺍﻝﻤﺎﺭ ﻤﻥ ‪ Ω‬ﻭ ﺍﻝﻌﻤﻭﺩﻱ ﻋﻠﻰ ) ‪. ( P‬‬
‫ ‬
‫ﻝﺩﻴﻨﺎ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ ) ‪ ( P‬ﻫﻲ ‪ x − y + z − 1 = 0 :‬ﺇﺫﻥ )‪ n (1, −1,1‬ﻤﺘﺠﻬﺔ ﻤﻨﻅﻤﻴﺔ ﻋﻠﻴﻪ‪.‬‬
‫ ‬
‫ﻝﺩﻴﻨﺎ ﺍﻝﻤﺴﺘﻘﻴﻡ ) ∆( ﻋﻤﻭﺩﻱ ﻋﻠﻰ ) ‪ ( P‬ﺇﺫﻥ )‪ n (1, −1,1‬ﻤﻭﺠﻬﺔ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( ‪.‬‬
‫ ‬
‫ﺇﺫﻥ ﺍﻝﺘﻤﺜﻴل ﺍﻝﺒﺎراﻤﺘﺭﻱ ﻝﻠﻤﺴﺘﻘﻴﻡ ) ∆( ﺍﻝﻤﺎﺭ ﻤﻥ ﺍﻝﻨﻘﻁﺔ ) ‪ Ω ( 2, 2, 4‬ﻭ ﺍﻝﻤﻭﺠﻪ ﺒﺎﻝﻤﺘﺠﻬﺔ )‪ n (1, −1,1‬ﻫﻭ‪:‬‬
‫‪x = 2 + t‬‬
‫‪‬‬
‫‪y = 2 −t‬‬
‫‪z = 4 + t‬‬
‫‪‬‬

‫‪0,5‬‬

‫ﺝ‪ -‬ﻨﺤﺩﺩ ﻤﺜﻠﻭﺙ ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻝﻨﻘﻁﺔ ‪ ω‬ﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ )‪. (Γ‬‬
‫‪ ω‬ﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ )‪ (Γ‬ﻫﻲ ﺘﻘﺎﻁﻊ ) ∆( ﻭ ) ‪. ( P‬‬


ض )‪ (1) (2‬‬

‫)∆( ∈ ‪ ω‬و )‪{ω} = (∆) ∩ ( P) ⇔ ω ∈ ( P‬‬

‫‪x = 2 + t‬‬
‫‪‬‬
‫‪ (2) :  y=2-t‬و ‪⇔ (1) : x − y + z − 1 = 0‬‬
‫‪z = 4 + t‬‬
‫‪‬‬
‫‪(2 + t ) − (2 − t )(2 + t ) − (2 − t ) + (4 + t ) − 1 = 0‬‬
‫‪:‬‬
‫ ‪t = −1‬‬

‫‪1‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬
‫ﻨﻌﻭﺽ ﻗﻴﻤﺔ ‪ t = -1‬ﻓﻲ )‪ (2‬ﻨﺤﺼل ﻋﻠﻰ‬
‫‪ x = 2 + (−1) = 1‬‬
‫‪‬‬
‫‪ y = 2 − (−1) = 3‬‬
‫‪ z = 4 + (−1) = 3‬‬
‫‪‬‬

‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬

‫ﺇﺫﻥ )‪. ω (1, 3,3‬‬

‫ا
ا ‪:‬‬
‫ﻴﺤﺘﻭﻱ ﻜﻴﺱ ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﺒﻴﻀﺎﺀ ﻭ ﺃﺭﺒﻊ ﺒﻴﺩ ﻗﺎﺕ ﺴﻭﺩﺍﺀ ) ﻻ ﻴﻤﻜﻥ ﺍﻝﺘﻤﻴﻴﺯ ﺒﻴﻥ ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺒﺎﻝﻠﻤﺱ(‪.‬‬
‫ﻨﺴﺤﺏ ﻋﺸﻭﺍﺌﻴﺎ ﻭﻓﻲ ﺁﻥ ﻭﺍﺤﺩ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﺍﻝﻜﻴﺱ ‪.‬‬
‫ﻝﺩﻴﻨﺎ ‪card (Ω) = C73 = 35‬‬
‫‪0,75‬‬

‫‪ (1‬ﺍﻝﺤﺩﺙ ‪ " A‬ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺘﻴﻥ ﺒﺎﻝﻀﺒﻁ ﻝﻭﻨﻬﻤﺎ ﺃﺒﻴﺽ "ﺃﻱ‬

‫) ‪( B, B, N‬‬

‫‪card ( A) 12‬‬
‫ﺇﺫﻥ‬
‫ﻝﺩﻴﻨﺎ ‪card ( A) = C32 ⋅ C41 = 12‬‬
‫=‬
‫‪card (Ω) 35‬‬
‫‪ (2‬ﺍﻝﺤﺩﺙ ‪ " B‬ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﻨﻔﺱ ﺍﻝﻠﻭﻥ "‪.‬ﺍﻱ ) ‪ ( N , N , N‬أو‬
‫= )‪p ( A‬‬

‫‪0,75‬‬

‫‪1‬‬

‫) ‪( B, B, B‬‬

‫‪card ( B ) 5 1‬‬
‫=‬
‫‪ card ( B ) = C33 + C43 = 1 + 4 = 5‬ﺇﺫﻥ =‬
‫‪card (Ω) 35 7‬‬
‫‪ (3‬ﺍﻝﺤﺩﺙ‪ " C‬ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ ﻋﻠﻰ ﺍﻷﻗل "‬
‫ﺍﻝﺤﺩﺙ ﺍﻝﻤﻀﺎﺩ ‪" C‬ﻋﺩﻡ ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺃﻴﺔ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ " ﻴﻌﻨﻲ )ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺍﻝﺜﻼﺙ ﺍﻝﻤﺴﺤﻭﺒﺔ ﺴﻭﺩﺍﺀ(‬
‫‪card (C ) 4‬‬
‫=‬
‫= ) ‪ P(C‬ﺇﺫﻥ ‪:‬‬
‫ﻝﺩﻴﻨﺎ ‪ card (C ) = C43 = 4‬ﺇﺫﻥ ‪:‬‬
‫‪card (Ω) 35‬‬
‫= ) ‪P( B‬‬

‫) ‪p (C ) = 1 − p (C‬‬
‫‪4‬‬
‫‪= 1−‬‬
‫‪35‬‬
‫‪31‬‬
‫=‬
‫‪35‬‬
‫ا
ا ‪:‬‬

‫‪1‬‬

‫‪1‬‬
‫ﻝﺘﻜﻥ ) ‪ (un‬ﺍﻝﻤﺘﺘﺎﻝﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﺒﻤﺎ ﻴﻠﻲ ‪ u0 = 2 :‬ﻭ )‪ un +1 = (un − 4n − 1‬ﻝﻜل ‪ n‬ﻤﻥ ‪. ℕ‬‬
‫‪5‬‬
‫ﻨﻀﻊ ‪ vn = un + n − 1‬ﻝﻜل ‪ n‬ﻤﻥ ‪. ℕ‬‬
‫‪1‬‬
‫‪ ( 1‬ﻨﺒﻴﻥ ﺃﻥ ) ‪ (vn‬ﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪.‬‬
‫‪5‬‬
‫‪∀n ∈ ℕ: vn +1 = un +1 + (n + 1) − 1‬‬
‫‪1‬‬
‫‪= (un − 4n − 1) + n‬‬
‫‪5‬‬
‫‪1‬‬
‫)‪= (un − 4n − 1 + 5n‬‬
‫‪5‬‬
‫‪1‬‬
‫)‪= (un + n − 1‬‬
‫‪5‬‬
‫‪1‬‬
‫‪= vn‬‬
‫‪5‬‬

‫‪0,5‬‬
‫‪1‬‬
‫ﺍﺩﻥ ) ‪ (vn‬ﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ‬
‫‪5‬‬
‫‪ ( 2‬ﺃ – ﻨﺤﺴﺏ ‪ vn‬ﺒﺩﻻﻝﺔ ‪. n‬‬

‫‪0,5‬‬

‫=‪q‬‬

‫‪n‬‬

‫‪1‬‬
‫‪1‬‬
‫ﻝﺩﻴﻨﺎ ) ‪ (vn‬ﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ = ‪ q‬ﻭﺤﺩﻫﺎ ﺍﻷﻭل ‪ v0 = u0 + 0 − 1 = 2 − 1 = 1‬ﺇﺫﻥ ‪ vn = v0 ⋅ q n‬ﺃﻱ ‪vn =  ‬‬
‫‪5‬‬
‫‪5‬‬

‫‪2‬‬

2007‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬
. lim un ‫ ﺜﻡ ﺤﺴﺎﺏ‬n ‫ ﺒﺩﻻﻝﺔ‬un ‫ ﺍﺴﺘﻨﺘﺎﺝ‬-‫ﺏ‬
x →+∞

n

1
un =   − n + 1 ‫ ﻭﻤﻨﻪ ﻓﺎﻥ‬. un = vn − n + 1 ‫ ﺇﺫﻥ‬vn = un + n − 1 ‫ﻝﺩﻴﻨﺎ‬
5

0,5

n

1
1
lim un = −∞ ‫ ﺇﺫﻥ‬lim ( − n + 1) = −∞ ‫ ﻭ ﻝﺩﻴﻨﺎ‬lim   = 0 ‫ ﺇﺫﻥ‬−1 ≺ ≺ 1 ‫ﻝﺩﻴﻨﺎ‬
x →+∞
x →+∞ 5
x →+∞
5
 
S n = u0 + u1 + ............. + un ‫ ﻭ‬Tn = v0 + v1 + .............. + vn (3
1
1
Tn =  5 − n  : ‫ﻨﺒﻴﻥ ﺃﻥ‬
4
5 
Tn = v0 + v1 + .............. + vn
= v0 ⋅

1

1 − q n +1
1− q
n +1

1
1−  
5
= 1⋅  
1
1−
5
n +1
5 1 
= 1 −   
4   5  
n +1
1
1 
= 5 − 5  
4 
 5  
1
1 
=  5 − 5 ⋅ n +1 
4
5 
1
1 
= 5 − n 
4
5 
(n + 1)(n − 2)
‫ﻨﺒﻥ ﺃﻥ‬
2
: ‫ ﺍﺫﻥ‬un = vn − n + 1 ‫ﻝﺩﻴﻨﺎ‬

S n = Tn −
S n = u0 + u1 + ............. + un

= ( v0 − (−1) ) + ( v1 − 0 ) + ( v2 − 1) + ................ + ( vn − (n − 1) )

= ( v0 + v1 + .............. + vn ) − ( (−1) + 0 + 1 + 2 + ............. + (n − 1) )
= Tn −
= Tn −

( (−1) + (n − 1) )( n + 1)

2
( n + 1)( n − 2 )
2

: ‫ا
ا ا‬
( 2 + 2i ) = −2 + 4 2i : ‫( ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ‬1
2

( 2 + 2i )2 = 2 + 2 2 ⋅ 2i + ( 2i )
2

2

= 2 + 4 2i − 4
= −2 + 4 2i

3

0,25

‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬
‫‪0,75‬‬

‫‪ (2‬ﻨﺤل ﻓﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻝﻌﻘﺩﻴﺔ ‪ ℂ‬ﺍﻝﻤﻌﺎﺩﻝﺔ ‪z 2 − ( 2 + 2) z + 2 + 2 − 2i = 0 :‬‬
‫ﻝﺩﻴﻨﺎ ﻤﻤﻴﺯ ﺍﻝﻤﻌﺎﺩﻝﺔ ﻫﻭ‬

‫)‬

‫(‬

‫‪2‬‬

‫)‬

‫‪2 + 2  − 4 2 + 2 − 2i‬‬
‫‪‬‬

‫(‬

‫‪∆ = −‬‬
‫‪‬‬

‫‪= 2 + 4 2 + 4 − 8 − 4 2 + 4 2i‬‬
‫‪= −2 + 4 2i‬‬
‫‪= ( 2 + 2i ) 2‬‬

‫إذن ا د ه ‪) = 1 − i :‬‬

‫‪2 + 2i‬‬

‫(‬

‫‪2 +2−‬‬

‫=)‬

‫= ‪ z1‬و ‪2 + 1 + 2i‬‬

‫‪2‬‬
‫‪(3‬ﻝﺩﻴﻨﺎ ﺍﻝﻌﺩﺩﻴﻥ ﺍﻝﻌﻘﺩﻴﻴﻥ ‪ z1 = 1 − i‬ﻭ ‪. z2 = 1 + 2 + i‬‬

‫‪2 + 2i‬‬

‫(‬

‫‪2 +2+‬‬

‫‪2‬‬

‫= ‪z2‬‬

‫ﺃ – ﻨﺤﺩﺩ ﺍﻝﺸﻜل ﺍﻝﻤﺜﻠﺜﻲ ﻝﻠﻌﺩﺩ ﺍﻝﻌﻘﺩﻱ ‪. z1‬‬
‫ ‪ z1 = 1 − i = 2 #$%‬إذن ‪:‬‬
‫‪0,5‬‬

‫‪ 2‬‬
‫‪‬‬
‫‪2 ‬‬
‫‪π‬‬
‫‪π‬‬
‫‪‬‬
‫‪ π‬‬
‫‪ π ‬‬
‫‪−‬‬
‫‪z1 = 1 − i = 2 ‬‬
‫‪i  = 2  cos + i sin  = 2  cos  −  + i sin  −  ‬‬
‫‪2 ‬‬
‫‪4‬‬
‫‪4‬‬
‫‪‬‬
‫‪ 4‬‬
‫‪ 4 ‬‬
‫‪‬‬
‫‪ 2‬‬
‫ﺏ – ﻨﺒﻴﻥ ﺃﻥ ‪ z2 ) z1.z2 = 2 z2 :‬ﻫﻭ ﻤﺭﺍﻓﻕ ﺍﻝﻌﺩﺩ ‪. ( z2‬‬
‫ﻝﺩﻴﻨﺎ ‪:‬‬

‫)‬

‫‪1‬‬

‫(‬

‫‪z1 ⋅ z2 = (1 − i ) 1 + 2 + i‬‬

‫‪= 1+ 2 + i − i − i 2 +1‬‬

‫)‬
‫)‬

‫ ‪ z1.z2 = 2 z2 #$%‬ﺍﺫﻥ ‪2 z2 [ 2π ] :‬‬

‫و* أن ] ‪ arg ( z2 ) ≡ − arg ( z2 ) [ 2π‬و‬

‫‪ arg ( z1 .z2 ) ≡ arg‬ﺃﻱ‬

‫] ‪( 2 ) ≡ 0 [2π‬‬

‫ﺝ – ﻨﺤﺩﺩ ﻋﻤﺩﺓ ﻝﻠﻌﺩﺩ ‪z2‬‬
‫‪0,5‬‬

‫ﻝﺩﻴﻨﺎ ] ‪arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π‬‬

‫‪2 +1− i‬‬

‫(‬

‫‪= 2‬‬

‫‪= 2 z2‬‬

‫ا('‪ '#‬ج ‪arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π ] :‬‬

‫(‬

‫‪= 2+ 2 −i 2‬‬

‫ﻭ‬

‫] ‪[ 2π‬‬

‫‪π‬‬
‫‪4‬‬

‫‪arg‬‬

‫] ‪( 2 ) + arg ( z ) [2π‬‬
‫‪2‬‬

‫‪arg( z1 ) + arg( z2 ) ≡ arg‬‬

‫ ن ] ‪arg( z1 ) + 2 arg( z2 ) ≡ 0 [ 2π‬‬

‫‪ arg ( z1 ) ≡ −‬ﺇﺫﻥ‬

‫] ‪[ 2π‬‬

‫‪π‬‬
‫‪8‬‬

‫‪arg ( z2 ) ≡ −‬‬

‫ ‪:‬‬
‫‪ (I‬ﻝﺩﻴﻨﺎ ‪ g‬ﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞‪ ]0, +‬ﺒﻤﺎ ﻴﻠﻲ ‪:‬‬
‫‪( x − 1) 2‬‬
‫‪ (1‬ﻨﺒﻴﻥ ﺃﻥ‬
‫‪x2‬‬

‫‪1‬‬
‫‪− 2 ln x‬‬
‫‪x‬‬

‫‪. g ( x) = x −‬‬

‫= )‪ g '( x‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪ ]0, +‬ﺜﻡ ﻨﺴﺘﻨﺘﺞ ﻤﻨﺤﻰ ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ ‪ g‬ﻋﻠﻰ [∞‪. ]0, +‬‬

‫‪1‬‬

‫‪4‬‬

‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬
‫‪1‬‬
‫‪g '( x) = 1 + 2 − 2 ln x‬‬
‫‪x‬‬
‫‪1‬‬
‫‪1‬‬
‫⋅‪= 1+ 2 − 2‬‬
‫‪x‬‬
‫‪x‬‬
‫‪2‬‬
‫‪x + 1 − 2x‬‬
‫=‬
‫‪x2‬‬
‫‪2‬‬

‫)‪( x − 1‬‬
‫‪x2‬‬

‫‪∀x ∈ ]0, +∞[ :‬‬

‫=‬

‫ ‪ #$%‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪]0, +‬‬
‫إذن ‪ ∀x ∈ ]0, +∞[ : g ' ( x ) ≥ 0‬و* ' ن ا ‪%‬ا ‪-. g‬ا‪ $%$‬ا ‪ ,‬ل [∞‪]0, +‬‬
‫‪ #$% (2‬ا ‪%‬ا ‪ g‬ﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞‪ ]0, +‬و ‪ /
0‬ا ‪ ,‬ل ]‪ ]0,1‬إذن‬
‫)‪∀x ∈ ]0,1] ⇒ 0 ≺ x ≤ 1 ⇒ g ( x) ≤ g (1‬‬
‫* أن ‪ g (1) = 0‬ن ‪ 23 x 1 g ( x) ≤ 0‬ا ‪ ,‬ل ]‪]0,1‬‬
‫ ‪ #$%‬ا ‪%‬ا ‪ g‬ﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞‪ [1, +‬ﺇﺫﻥ )‪∀x ∈ [1, +∞[ ⇒ 1 ≤ x ⇒ g (1) ≤ g ( x‬‬
‫ﺒﻤﺎ ﺃﻥ ‪ g (1) = 0‬ن ‪ 23 x 1 g ( x) ≥ 0‬ا ‪ ,‬ل [∞‪[1, +‬‬
‫‪ ( x − 1) ≥ 0‬ﻭ ‪. x 2 ≻ 0‬‬
‫‪2‬‬

‫‪0,5‬‬

‫‪1‬‬
‫‪ (II‬ﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ‪ f‬ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞‪ ]0, +‬ﺒﻤﺎ ﻴﻠﻲ ‪− (ln x) 2 − 2 :‬‬
‫‪x‬‬
‫‪(ln x) 2‬‬
‫‪ ) lim‬ﻴﻤﻜﻥ ﻭﻀﻊ ‪ ( t = x‬ﺜﻡ ﻨﺤﺴﺏ )‪lim f ( x‬‬
‫‪ (1‬ﺃ – ﻨﺒﻴﻥ ﺃﻥ‬
‫∞‪x →+‬‬
‫∞‪x →+‬‬
‫‪x‬‬
‫ﻨﻀﻊ ‪ t = x‬ﺇﺫﻥ ‪ x = t 2‬ﻋﻨﺩﻤﺎ ∞‪ x → +‬ﻓﺎﻥ ∞‪t → +‬‬
‫‪f ( x) = x +‬‬

‫‪0,75‬‬

‫‪2‬‬

‫ﻝﺩﻴﻨﺎ‬
‫‪2‬‬

‫‪0,25‬‬

‫‪ ln t ‬‬
‫‪= 4‬‬
‫‪‬‬
‫‪ t ‬‬
‫‪2‬‬

‫‪2‬‬

‫) ‪( ln t ) = ( 2ln t‬‬
‫=‬

‫) ‪( ln x‬‬

‫‪2 2‬‬

‫‪t2‬‬

‫‪t2‬‬

‫‪2‬‬

‫) ‪( ln x‬‬
‫‪x‬‬

‫‪ln t‬‬
‫‪ ln t ‬‬
‫ ‪= 0 #$%‬‬
‫‪= lim 4 ‬‬
‫‪ tlim‬إذن ‪ = 0‬‬
‫∞‪x →+‬‬
‫‪t‬‬
‫∞‪→+‬‬
‫∞‪→+‬‬
‫‪x‬‬
‫‪t‬‬
‫‪ t ‬‬
‫‪2‬‬
‫‪‬‬
‫‪1‬‬
‫‪1 ( ln x ) 2 ‬‬
‫‪2‬‬
‫‪− ‬‬
‫‪f ( x ) = x + − ( ln x ) − 2 = x 1 + 2 −‬‬
‫‪ x‬‬
‫‪x‬‬
‫‪x‬‬
‫‪x ‬‬
‫‪‬‬
‫‪2‬‬
‫) ‪( ln x‬‬
‫‪1‬‬
‫‪2‬‬
‫‪ lim‬و ‪ lim 2 = 0‬ﺇﺫﻥ ∞‪lim f ( x ) = +‬‬
‫ﻝﺩﻴﻨﺎ ‪ lim = 0‬و ‪= 0‬‬
‫∞‪x →+‬‬
‫‪x →+∞ x‬‬
‫∞‪x →+‬‬
‫‪x‬‬
‫∞‪→+‬‬
‫‪x‬‬
‫‪x‬‬
‫‪1‬‬
‫ب – ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ ‪ f ( ) = f ( x) :‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪. ]0, +‬‬
‫‪x‬‬
‫ﻝﻜل ‪ x‬ﻤﻥ [∞‪ ]0, +‬ﻝﺩﻴﻨﺎ‬

‫‪lim‬‬

‫‪2‬‬

‫‪ 1  1 1   1 ‬‬
‫‪f   = + −  ln    − 2‬‬
‫‪ x  x 1   x ‬‬
‫‪x‬‬
‫‪1‬‬
‫‪2‬‬
‫‪= + x − ( − ln x ) − 2‬‬
‫‪x‬‬
‫‪1‬‬
‫‪2‬‬
‫‪= + x − ( ln x ) − 2‬‬
‫‪x‬‬
‫)‪= f ( x‬‬

‫‪5‬‬

‫‪0,5‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬
‫ج – ﻨﺤﺴﺏ )‪lim f ( x‬‬

‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬

‫‪x →0‬‬
‫‪x≻0‬‬

‫‪1‬‬
‫‪1‬‬
‫ﻨﻀﻊ = ‪ t‬ﺇﺫﻥ ﻋﻨﺩﻤﺎ ‪ x → 0+‬ﻓﺎﻥ ∞‪ t → +‬ﻭ ﻤﻨﻪ ﻓﺎﻥ ∞‪lim f ( x) = lim f   = lim f (t ) = +‬‬
‫‪x →0‬‬
‫‪x →0‬‬
‫‪t‬‬
‫∞‪→+‬‬
‫‪x‬‬
‫‪x‬‬
‫‪x ≻0‬‬
‫‪x≻0‬‬
‫ﺇﺫﻥ ﺍﻝﻤﻨﺤﻨﻰ ) ‪ (C‬ﻴﻘﺒل ﻤﻘﺎﺭﺒﺎ ﺭﺃﺴﻲ ﻤﻌﺎﺩﻝﺘﻪ ‪x = 0‬‬
‫ﺩ‪ – -‬ﻨﺒﻴﻥ ﺃﻥ ) ‪ (C‬ﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ ﻫﻲ ‪y = x :‬‬

‫‪0,5‬‬

‫ﻝﺩﻴﻨﺎ ∞‪lim f ( x ) = +‬‬

‫∞‪x →+‬‬

‫‪2‬‬
‫‪‬‬
‫‪1 ( ln x ) 2 ‬‬
‫ﻭ‪−  =1‬‬
‫‪lim f ( x ) = lim 1 + 2 −‬‬
‫∞‪x →+‬‬
‫‪x →+∞ ‬‬
‫‪x‬‬
‫‪x‬‬
‫‪x ‬‬
‫‪‬‬

‫‪1‬‬
‫∞‪− (ln x) 2 − 2 = −‬‬
‫∞‪x →+‬‬
‫‪x →+∞ x‬‬
‫ﻫﻲ ‪y = x :‬‬
‫)‪g ( x‬‬
‫‪ (2‬ﺒﻴﻥ ﺃﻥ ‪:‬‬
‫= )‪ f '( x‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪]0, +‬‬
‫‪x‬‬

‫‪ lim f ( x ) − x = lim‬ﺇﺫﻥ ) ‪ (C‬ﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ‬

‫‪1‬‬
‫' ) ‪− 2 ( ln x )( ln x‬‬
‫‪x2‬‬
‫‪1‬‬
‫‪1‬‬
‫⋅ ) ‪= 1 − 2 − 2 ( ln x‬‬
‫‪x‬‬
‫‪x‬‬
‫‪1‬‬
‫‪1‬‬
‫‪‬‬
‫‪=  x − − 2 ln x ‬‬
‫‪x‬‬
‫‪x‬‬
‫‪‬‬
‫)‪g ( x‬‬
‫=‬
‫‪x‬‬

‫‪1,5‬‬

‫‪f '( x) = 1−‬‬

‫ﺇﺸﺎﺭﺓ ) ‪ f ' ( x‬ﻫﻲ ﺇﺸﺎﺭﺓ ) ‪g ( x‬‬
‫ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ ‪f‬‬

‫∞‪+‬‬

‫‪0‬‬

‫‪1‬‬

‫‪φ‬‬

‫‪+‬‬
‫∞‪+‬‬

‫‪0‬‬
‫‪ (3‬ا ‪ # #‬‬
‫‪1‬‬

‫‪6‬‬

‫‪x‬‬
‫)‪f '( x‬‬

‫‬‫∞‪+‬‬

‫)‪f ( x‬‬

‫‪0,5‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬
‫‪ ( 4‬ﺃ ‪ -‬ﻨﺒﻴﻥ ﺃﻥ ﺍﻝﺩﺍﻝﺔ ‪ G : x ln x − x‬ﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ ‪ g : x → ln x‬ﻋﻠﻰ [∞‪]0, +‬‬
‫ﻝﺩﻴﻨﺎ‬
‫‪∀x ∈ ]0, +∞[ : G ' ( x ) = x 'ln x + x ( ln x ) '− 1‬‬
‫‪1‬‬
‫‪= ln x + x ⋅ − 1‬‬
‫‪x‬‬
‫‪= ln x + 1 − 1‬‬
‫‪= ln x‬‬

‫ﺇﺫﻥ ﺍﻝﺩﺍﻝﺔ ‪ G‬ﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ ‪.g‬‬

‫ﺏ‪ -‬ﺒﺎﺴﺘﻌﻤﺎل ﻤﻜﺎﻤﻠﺔ ﺒﺎﻷﺠﺯﺍﺀ ‪ ،‬ﻨﺒﻴﻥ ﺃﻥ ‪(ln x) 2 dx = e − 2 :‬‬
‫‪0,75‬‬

‫‪u ( x ) = ( ln x )2‬‬
‫ﻨﻀﻊ‬
‫‪‬‬
‫‪v ' ( x ) = 1‬‬

‫ﺇﺫﻥ‬

‫‪e‬‬

‫∫‬

‫‪1‬‬

‫‪ln x‬‬
‫‪‬‬
‫‪u ' ( x ) = 2‬‬
‫‪x‬‬
‫‪‬‬
‫‪v ( x ) = x‬‬
‫‪‬‬

‫ﺇﺫﻥ‬
‫‪ln x‬‬
‫‪2‬‬
‫‪⋅ x dx‬‬
‫‪x‬‬

‫‪e‬‬

‫‪e‬‬

‫‪ −‬‬
‫‪1 ∫1‬‬

‫‪2‬‬

‫‪2‬‬

‫) ‪∫ ( ln x ) dx =  x ( ln x‬‬
‫‪e‬‬

‫‪1‬‬

‫‪e‬‬

‫‪2‬‬
‫‪=  x ( ln x )  − 2 ∫ ln x dx‬‬
‫‪‬‬
‫‪1‬‬
‫‪1‬‬
‫‪e‬‬

‫‪e‬‬
‫‪2‬‬
‫‪=  x ( ln x )  − 2 [ x ln x − x ]1‬‬
‫‪‬‬
‫‪1‬‬
‫‪e‬‬

‫) )‪) − 2 (( e ln e − e ) − (1ln1 − 1‬‬
‫‪0,75‬‬

‫‪2‬‬

‫(‬

‫)‪= e ( ln e ) − 1( ln1‬‬
‫‪2‬‬

‫‪=e−2‬‬
‫ﺝ – ﻤﺴﺎﺤﺔ ﺤﻴﺯ ﺍﻝﻤﺴﺘﻭﻯ ﺍﻝﻤﺤﺼﻭﺭ ) ‪ (C‬ﻭ ﻤﺤﻭﺭ ﺍﻷﻓﺎﺼﻴل ﻭ ﺍﻝﻤﺴﺘﻘﻴﻤﻴﻥ ﺍﻝﻠﺫﻴﻥ ﻤﻌﺎﺩﻝﺘﺎﻫﻤﺎ ‪ x = 1 :‬ﻭ ‪x = e‬‬

‫ﻝﺩﻴﻨﺎ‬

‫‪ f‬ﺩﺍﻝﺔ ﻤﻭﺠﺒﺔ ﻭ ﻤﺘﺼﻠﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل ]‪ [1, e‬ﺇﺫﻥ ﺍﻝﻤﺴﺎﺤﺔ ﺍﻝﻤﻁﻠﻭﺒﺔ ﻫﻲ‬

‫‪e‬‬
‫‪e‬‬
‫‪1‬‬
‫‪‬‬
‫‪A = ∫ f ( x ) dx = ∫  x + − (ln x) 2 − 2  dx‬‬
‫‪1‬‬
‫‪1‬‬
‫‪x‬‬
‫‪‬‬
‫‪‬‬
‫‪e‬‬
‫‪e‬‬
‫‪1‬‬
‫‪2‬‬
‫‪‬‬
‫‪= ∫  x + − 2  dx − ∫ ( ln x ) dx‬‬
‫‪1‬‬
‫‪1‬‬
‫‪x‬‬
‫‪‬‬
‫‪‬‬
‫‪e‬‬

‫‪ x2‬‬
‫‪‬‬
‫) ‪=  + ln x − 2 x  − ( e − 2‬‬
‫‪2‬‬
‫‪1‬‬
‫‪ e2‬‬
‫‪ 1‬‬
‫‪‬‬
‫‪=  + ln e − 2e  −  + ln1 − 2  − e + 2‬‬
‫‪‬‬
‫‪ 2‬‬
‫‪ 2‬‬
‫‪1‬‬
‫‪e2‬‬
‫=‬
‫‪+ 1 − 2e − + 2 − e + 2‬‬
‫‪2‬‬
‫‪2‬‬
‫‪2‬‬
‫‪9‬‬
‫‪e‬‬
‫=‬
‫‪− 3e +‬‬
‫‪2‬‬
‫‪2‬‬
‫‪9‬‬
‫‪e2‬‬
‫= ‪ A‬ﺒﻭﺤﺩﺓ ﻗﻴﺎﺱ ﺍﻝﻤﺴﺎﺤﺔ‬
‫ﺇﺫﻥ ‪− 3 e +‬‬
‫‪2‬‬
‫‪2‬‬

‫‪7‬‬


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