مجموعة مواضيع مصححة في الرياضيات

‫ﺘﺼﺤﻴﺢ ﺍﻤﺘﺤﺎﻥ ﺍﻟﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ ﻴﻭﻟﻴﻭﺯ ‪06‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻻﻭل‪:‬‬ ‫‪ (1‬ﺃ‪ -‬ﻟﺩﻴﻨﺎ ‪ AB(0, −3,3) :‬ﻭ )‪ . AC(1, −1, 0‬ﺍﺫﻥ‪0.5 AB ∧ AC(3,3,3) :‬‬ ‫ﺏ‪ -‬ﻟﺩﻴﻨﺎ ‪ AB ∧ AC :‬ﻤﺘﺠﻬﺔ ﻤﻨﻅﻤﻴﺔ ﻋﻠﻰ ﺍﻟﻤﺴﺘﻭﻯ )‪ ( ABC‬ﺍﺫﻥ ‪:‬‬ ‫‪M (x, y, z) ∈ ( ABC) ⇔ AM .( AB ∧ AC) = 0‬‬ ‫‪⇔ 3.(x −1) + 3.( y − 2) + 3.(z + 2) = 0‬‬ ‫‪⇔ 3.(x + y + z −1) = 0‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ﻤﻌﺎﺩﻟﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻟﻠﻤﺴﺘﻭﻯ )‪ ( ABC‬ﻫﻲ ‪0.5 x + y + z −1 = 0 :‬‬ ‫‪ (2‬ﺃ‪ -‬ﻤﺴﺎﻓﺔ ﺍﻟﻤﺭﻜﺯ )‪ Ω(1,1,1‬ﺍﻟﻰ ﺍﻟﻤﺴﺘﻭﻯ )‪ ( ABC‬ﻫﻲ‪d = 1+1+1−1 = 2 :‬‬ ‫‪1² +1² +1² 3‬‬ ‫ﺇﺫﻥ ‪ d = R :‬ﻭﻤﻨﻪ ‪ :‬ﺍﻟﻤﺴﺘﻭﻯ )‪ ( ABC‬ﻤﻤﺎﺱ ﻟﻠﻔﻠﻜﺔ )‪ (S‬ﻓﻲ ﻨﻘﻁﺔ ‪0.5 . H‬‬‫‪⎧x =1+t‬‬‫⎪‬‫⎨‬ ‫‪y‬‬ ‫=‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪t‬‬ ‫ﻫﺫﻩ ﺍﻟﻨﻘﻁﺔ ﻫﻲ ﺘﻘﺎﻁﻊ ﺍﻟﻤﺴﺘﻭﻯ )‪ ( ABC‬ﻤﻊ ﺍﻟﻤﺴﺘﻘﻴﻡ ) ‪ (ΩH‬ﺫﻭ ﺘﻤﺘﻴل ﺒﺭﺍﻤﺘﺭﻱ ‪(t ∈ IR) :‬‬‫‪⎪⎩z = 1+ t‬‬ ‫‪⎧x =1+t‬‬ ‫‪⎪⎪ y = 1+ t‬‬ ‫ﺍﺫﻥ‪0.75 H (1 , 1 , 1) :‬‬ ‫ﻨﺠﺩ ‪t = −2 :‬‬ ‫⎨‬ ‫ﺒﺤل ﺍﻟﻨﻅﻤﺔ ‪:‬‬ ‫‪333‬‬ ‫‪3‬‬ ‫⎪‬ ‫‪z‬‬ ‫=‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪t‬‬ ‫‪⎪⎩x + y + z −1 = 0‬‬ ‫ﺏ‪ -‬ﻟﺘﻜﻥ )‪ M (a, b, c‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ )‪. ( ABC‬‬ ‫‪⎪⎧a + b + c −1 = 0‬‬ ‫‪⎧a + b + c −1 = 0‬‬ ‫‪⎪⎩⎨(a −1)² + (b −1)² + (c −1)²‬‬ ‫⎪‬ ‫ﺃﻱ‪:‬‬ ‫≥‬ ‫‪4‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫≥ )‪⎩⎪⎨d (M , Ω‬‬ ‫‪2‬‬ ‫‪:‬‬ ‫ﻟﺩﻴﻨﺎ ﺇﺫﻥ‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪⎧⎪a + b + c = 1‬‬ ‫ﺃﻱ‪:‬‬ ‫‪⎪⎧a + b + c = 1‬‬ ‫‪⎨⎩⎪a²‬‬ ‫‪4‬‬ ‫‪⎨⎪⎩a²‬‬ ‫‪4‬‬ ‫‪+‬‬ ‫‪b²‬‬ ‫‪+‬‬ ‫‪c²‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫≥‬ ‫‪3‬‬ ‫‪+‬‬ ‫‪b²‬‬ ‫‪+‬‬ ‫‪c²‬‬ ‫‪−‬‬ ‫‪2(a‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪+‬‬ ‫)‪c‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫≥‬ ‫‪3‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‪0.75 a² + b² + c² ≥ 1 :‬‬ ‫‪3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻨﻲ‪:‬‬ ‫‪vn+1‬‬ ‫=‬ ‫‪un+2‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪.un+1‬‬ ‫=‬ ‫‪(2‬‬ ‫‪.un+1‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪.un‬‬ ‫)‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪.un+1‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪25‬‬ ‫‪5‬‬ ‫‪ (1‬ﻟﻜل ‪ n‬ﻤﻥ ‪ IN‬ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫=‬ ‫‪5‬‬ ‫‪.un+1‬‬ ‫‪−‬‬ ‫‪25‬‬ ‫‪.un‬‬ ‫=‬ ‫‪5‬‬ ‫‪.(un+1‬‬ ‫‪−‬‬ ‫‪.u‬‬ ‫)‬ ‫‪5n‬‬ ‫‪vn+1‬‬ ‫=‬ ‫‪1‬‬ ‫ﺍﺫﻥ ‪:‬‬ ‫‪5 .vn‬‬ ‫‪0.5‬‬ ‫‪v0‬‬ ‫=‬ ‫‪u1‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪.u0‬‬ ‫‪=1‬‬ ‫ﺍﻷﻭل‪:‬‬ ‫ﻭﺤﺩﻫﺎ‬ ‫=‪q‬‬ ‫‪1‬‬ ‫) ‪ (vn‬ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‪:‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪0.25‬‬ ‫∈ ‪∀n‬‬ ‫‪IN‬‬ ‫‪: vn‬‬ ‫=‬ ‫‪(1)n‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪5‬‬ ‫‪wn+1‬‬ ‫‪= 5n+1.un+1‬‬ ‫‪= 5n+1.(vn‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪.un‬‬ ‫)‬ ‫‪= 5n+1. 1‬‬ ‫‪+ 5n.un‬‬ ‫‪ (2‬ﺃ‪ -‬ﻟﻜل ‪ n‬ﻤﻥ ‪ IN‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪5‬‬ ‫‪5n‬‬ ‫ﺇﺫﻥ‪ wn+1 = 5 + wn :‬ﻭﺒﺎﻟﺘﺎﻟﻲ‪ (wn ) :‬ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﺃﺴﺎﺴﻬﺎ ‪ r = 5‬ﻭﺤﺩﻫﺎ ﺍﻷﻭل‪0.25 w0 = 0 :‬‬ ‫∈ ‪(∀n‬‬ ‫‪IN‬‬ ‫‪: un‬‬ ‫=‬ ‫‪wn‬‬ ‫)‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪0.25‬‬ ‫ﺏ‪ -‬ﻟﺩﻴﻨﺎ‪(∀n ∈ IN : wn = 5.n) :‬‬ ‫‪5n‬‬‫‪http://arabmaths.ift.fr‬‬

‫‪0.25‬‬ ‫∈ ‪∀n‬‬ ‫‪IN* : un‬‬ ‫=‬ ‫‪5.n‬‬ ‫=‬ ‫‪n‬‬ ‫ﺃﻱ‪:‬‬ ‫‪5n‬‬ ‫‪5n−1‬‬ ‫‪ (3‬ﺃ‪ -‬ﻟﻜل ‪ n‬ﻤﻥ * ‪ IN‬ﻟﺩﻴﻨﺎ ‪ 5.n 0 :‬ﺇﺫﻥ ‪0.25 ∀n ∈ IN*: un+1 0 :‬‬ ‫‪un+1‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪.un‬‬ ‫≤‬ ‫‪0‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪un+1‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪.un‬‬ ‫=‬ ‫‪n +1 −‬‬ ‫‪2.n‬‬ ‫=‬ ‫‪1− n‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫ﻭﻟﻜل ‪ n‬ﻤﻥ * ‪IN‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5n‬‬ ‫‪5n‬‬ ‫‪5n‬‬ ‫‪0.5‬‬ ‫∈ ‪∀n‬‬ ‫‪IN * :‬‬ ‫‪0‬‬ ‫≺‬ ‫‪un+1‬‬ ‫≤‬ ‫‪2‬‬ ‫‪.un‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‪:‬‬ ‫‪5‬‬ ‫‪0‬‬ ‫≺‬ ‫‪u1‬‬ ‫≤‬ ‫(‬ ‫‪2)0‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫ﺏ‪ -‬ﻤﻥ ﺃﺠل ‪ n = 1 :‬ﻟﺩﻴﻨﺎ‪u1 = 1 :‬‬ ‫‪5‬‬ ‫ﻟﻨﻔﺭﺽ ﺃﻥ ﺍﻟﺨﺎﺼﻴﺔ ﻤﺤﻘﻘﺔ ﻤﻥ ﺃﺠل * ‪. n ∈ IN‬‬ ‫‪0‬‬ ‫≺‬ ‫‪un+1‬‬ ‫≤‬ ‫‪( 2 )n‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪0 ≺ un+1‬‬ ‫≤‬ ‫‪2‬‬ ‫‪.un‬‬ ‫≤‬ ‫‪2 .( 2)n−1‬‬ ‫ﻟﺫﻥ‪:‬‬ ‫‪0 ≺ un+1‬‬ ‫≤‬ ‫‪2‬‬ ‫‪.un‬‬ ‫ﺃ(‬ ‫ﻟﺩﻴﻨﺎ ﺤﺴﺏ‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪55‬‬ ‫‪5‬‬ ‫ﺇﺫﻥ ﺍﻟﺨﺎﺼﻴﺔ ﻤﺤﻘﻘﺔ ﺒﺎﻟﻨﺴﺒﺔ ل ‪n +1 :‬‬ ‫‪0.5‬‬ ‫‪∀n‬‬ ‫∈‬ ‫‪IN‬‬ ‫*‬ ‫‪:‬‬ ‫‪0‬‬ ‫≺‬ ‫‪un+1‬‬ ‫≤‬ ‫(‬ ‫‪2‬‬ ‫‪)n‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‪:‬‬ ‫‪5‬‬ ‫) ‪ (un‬ﻤﺘﻘﺎﺭﺒﺔ‬ ‫‪ lim ( 2)n−1 = 0‬ﺇﺫﻥ ﺤﺴﺏ ﻤﺼﺎﺩﻴﻕ ﺍﻟﺘﻘﺎﺭﺏ‪:‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ −1 ≺ 2 ≺ 1‬ﻓﺎﻥ ‪:‬‬ ‫∞‪5n→+‬‬ ‫‪5‬‬ ‫‪0.25‬‬ ‫‪lim‬‬ ‫‪un‬‬ ‫=‬ ‫‪0‬‬ ‫ﻭ‪:‬‬ ‫∞‪n→+‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺜﺎﻟﺙ‪:‬‬ ‫ﺇﺫﺍ ﺴﺤﺒﻨﺎ ﺒﻴﺩﻗﺔ ﺘﺨﻤل ﺍﻟﺭﻗﻡ ‪ 2‬ﻓﺈﻨﻨﺎ ﻨﺴﺤﺏ ﻜﺭﺘﻴﻥ ﻓﻲ ﺁﻥ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﻜﻴﺱ ‪ U2‬ﺇﺫﻥ ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ‪ 0‬ﺃﻭ ﺒﻴﺩﻗﺔ ﺃﻭ‬ ‫ﺒﻴﺩﻗﺘﻴﻥ ﻟﻭﻨﻬﺎ ﺃﺤﻤﺭ‪.‬‬ ‫ﻭ ﺇﺫﺍ ﺴﺤﺒﻨﺎ ﺒﻴﺩﻗﺔ ﺘﺤﻤل ﺍﻟﺭﻗﻡ ‪ 3‬ﻓﺈﻨﻨﺎ ﻨﺴﺤﺏ ‪3‬ﺒﻴﺩﻗﺎﺕ ﻓﻲ ﺁﻥ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﻜﻴﺱ ‪ U2‬ﺇﺫﻥ ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ‪ 0‬ﺃﻭ‬ ‫ﺒﻴﺩﻗﺔ ﺃﻭ ﺒﻴﺩﻗﺘﻴﻥ ﻟﻭﻨﻬﺎ ﺃﺤﻤﺭ‪.‬‬ ‫ﺇﺫﻥ ﺍﻟﻘﻴﻡ ﺍﻟﺘﻲ ﻴﺄﺨﺫﻫﺎ ﺍﻟﻤﺘﻐﻴﺭ ﺍﻟﻌﺸﻭﺍﺌﻲ ‪ X‬ﻫﻲ ‪ 0 :‬ﺃﻭ ‪ 1‬ﺃﻭ ‪ 2‬ﻭﻤﻨﻪ‪0.5 X (Ω) = {0,1, 2} :‬‬‫* ﺍﻟﺤﺩﺙ ‪ [ X = 0] :‬ﻫﻭ ﺍﻟﺤﺩﺙ ‪ \":‬ﻻ ﻨﺤﺼل ﻋﻠﻰ ﺃﻴﺔ ﻜﺭﺓ ﺤﻤﺭﺍﺀ\" ﺃﻱ ﻨﺴﺤﺏ ﺒﻴﺩﻗﺔ ﺘﺤﻤل ﺍﻟﺭﻗﻡ ‪2‬ﻤﻥ ﺍﻟﻜﻴﺱ ‪U1‬‬‫ﻭﻨﺴﺤﺏ ﺒﻴﺩﻗﺘﻴﻥ ﺒﻴﻀﺎﻭﻴﻥ ﻓﻲ ﺃﻥ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﻜﻴﺱ ‪ U2‬ﺃﻭ ﻨﺴﺤﺏ ﺒﻴﺩﻗﺔ ﺘﺤﻤل ﺍﻟﺭﻗﻡ ‪ 3‬ﻤﻥ ﺍﻟﻜﻴﺱ ‪ U1‬ﻭﻨﺴﺤﺏ ‪3‬‬ ‫ﺒﻴﺩﻗﺎﺕ ﺒﻴﻀﺎﺀ ﻓﻲ ﺃﻥ ﻭﺍﺤﺩ ﻤﻥ ﺍﻟﻜﻴﺱ ‪.U2‬‬ ‫‪0.75‬‬ ‫‪p[ X‬‬ ‫= ]‪= 0‬‬ ‫‪3‬‬ ‫‪.‬‬ ‫‪C32‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪C33‬‬ ‫=‬ ‫‪3. 3‬‬ ‫‪+‬‬ ‫‪2. 1‬‬ ‫=‬ ‫‪11‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪5‬‬ ‫‪C52‬‬ ‫‪5‬‬ ‫‪C53‬‬ ‫‪5 10‬‬ ‫‪5 10‬‬ ‫‪50‬‬‫* ﻟﻜﻲ ﻴﺘﺤﻘﻕ ﺍﻟﺤﺩﺙ ‪ [ X = 1] :‬ﻴﺠﺏ ﺴﺤﺏ ﺒﻴﺩﻗﺔ ﺘﺤﻤل ﺍﻟﺭﻗﻡ ‪ 2‬ﻤﻥ ‪ U1‬ﻭﺴﺤﺏ ﺒﻴﺩﻗﺔ ﺤﻤﺭﺍﺀ ﻭﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ ﻤﻥ‬ ‫‪U2‬‬ ‫ﺃﻭ ﺴﺤﺏ ﺒﻴﺩﻗﺔ ﺘﺤﻤل ﺍﻟﺭﻗﻡ ‪ 3‬ﻤﻥ ‪ U1‬ﻭﺴﺤﺏ )ﺒﻴﺩﻗﺔ ﺤﻤﺭﺍﺀ ﻭﺒﻴﺩﻗﺘﻴﻥ ﺒﻴﻀﺎﺀ ( ﻤﻥ ‪ U2‬ﺇﺫﻥ‪:‬‬ ‫‪0.75‬‬ ‫‪p[ X‬‬ ‫= ]‪= 1‬‬ ‫‪3‬‬ ‫‪.‬‬ ‫‪C31.C21‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪C32 .C21‬‬ ‫=‬ ‫‪3. 6‬‬ ‫‪+‬‬ ‫‪2. 6‬‬ ‫=‬ ‫‪30‬‬ ‫=‬ ‫‪3‬‬ ‫‪5‬‬ ‫‪C52‬‬ ‫‪5‬‬ ‫‪C53‬‬ ‫‪5 10‬‬ ‫‪5 10‬‬ ‫‪50‬‬ ‫‪5‬‬‫ﻤﻥ ‪ U2‬ﺃﻭ‬ ‫* ﻟﻜﻲ ﻴﺘﺤﻘﻕ ﺍﻟﺤﺩﺙ ‪ [ X = 2] :‬ﻴﺠﺏ ﺴﺤﺏ ﺒﻴﺩﻗﺔ ﺘﺤﻤل ﺍﻟﺭﻗﻡ ‪ 2‬ﻤﻥ ‪ U1‬ﻭﺴﺤﺏ ﺒﻴﺩﻗﺘﻴﻥ ﺤﻤﺭﺍﻭﻴﻥ‬ ‫ﺴﺤﺏ ﺒﻴﺩﻗﺔ ﺘﺤﻤل ﺍﻟﺭﻗﻡ ‪ 3‬ﻤﻥ ‪ U1‬ﻭﺴﺤﺏ ) ﺒﻴﺩﻗﺘﻴﻥ ﺤﻤﺭﺍﻭﻴﻥ ﻭﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ( ﻤﻥ ‪ U2‬ﺇﺫﻥ‪:‬‬ ‫‪0.75‬‬ ‫‪p[ X‬‬ ‫=‬ ‫= ]‪2‬‬ ‫‪3‬‬ ‫‪.‬‬ ‫‪C22‬‬ ‫‪+‬‬ ‫‪2‬‬ ‫‪.‬‬ ‫‪C2 2 .C31‬‬ ‫=‬ ‫‪3. 1‬‬ ‫‪+‬‬ ‫‪2. 3‬‬ ‫=‬ ‫‪9‬‬ ‫‪5‬‬ ‫‪C52‬‬ ‫‪5‬‬ ‫‪C53‬‬ ‫‪5 10‬‬ ‫‪5 10‬‬ ‫‪50‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ :‬ﺠﺩﻭل ﻗﺎﻨﻭﻥ ﺍﺤﺘﻤﺎل ﺍﻟﻤﺘﻐﻴﺭ ﺍﻟﻌﺸﻭﺍﺌﻲ ‪ X‬ﻫﻭ ‪:‬‬ ‫‪a ∈ X (Ω) 0 1 2‬‬ ‫‪p[ X = a] 11 30‬‬ ‫‪9‬‬ ‫‪50 50 50‬‬

‫‪E( X ) = 0. 11 +1. 30 + 2. 9 = 48 = 24‬‬ ‫‪ (2‬ﺍﻷﻤل ﺍﻟﺭﻴﺎﻀﻲ ﻟﻠﻤﺘﻐﻴﺭ ﺍﻟﻌﺸﻭﺍﺌﻲ ‪ X‬ﻫﻭ ‪:‬‬ ‫‪50 50 50 50 25‬‬ ‫ﺃﻱ‪0.25 E( X ) = 24 :‬‬ ‫‪25‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ ﺍﻟﺭﺍﺒﻊ‪:‬‬ ‫‪ (1‬ﻤﻤﻴﺯ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻭ‪0.25 ∆ = 4 − 4(1+ i) = −4i :‬‬‫ﺍﺫﻥ ‪ . ∆ == 2.(−2i) = 2.(1− i)² = ( 2 .(1− i))² :‬ﺇﺫﻥ ﺍﺤﺩ ﺍﻟﺠﺫﻭﺭ ﺍﻟﻤﺭﺒﻌﺔ ل ∆ ﻫﻭ )‪δ = 2.(1− i‬‬‫)‪( Im(z1‬‬ ‫) ﻷﻥ ‪0 :‬‬ ‫‪z2 = −2 +‬‬ ‫)‪2.(1− i‬‬ ‫ﻭ‬ ‫‪z1 = −2 −‬‬ ‫)‪2.(1− i‬‬ ‫ﻭﺍﻟﺤﻠﻭل ﻫﻲ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪0.25‬‬ ‫‪z2 = −1+‬‬ ‫‪2−‬‬ ‫‪2 .i‬‬ ‫‪ 0.25‬ﻭ‬ ‫‪z1 = −1−‬‬ ‫‪2+‬‬ ‫‪2 .i‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪ (2‬ﺃ‪ -‬ﻟﺩﻴﻨﺎ ‪− 2 + 2 .i = − cos(π ) + i sin(π ) = cos(π − π ) + i sin(π − π ) :‬‬ ‫‪22‬‬ ‫‪44‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪0.5‬‬ ‫‪−‬‬ ‫‪2+‬‬ ‫‪2‬‬ ‫‪.i‬‬ ‫=‬ ‫(‪cos‬‬ ‫‪3π‬‬ ‫)‬ ‫‪+‬‬ ‫‪i‬‬ ‫(‪sin‬‬ ‫‪3π‬‬ ‫)‬ ‫=‬ ‫‪⎡⎣⎢1,‬‬ ‫‪3π‬‬ ‫⎤‬ ‫ﺍﺫﻥ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫⎦⎥‬ ‫ﺏ‪ zM -‬ﻴﺭﻤﺯ ﻟﻠﺤﻕ ﺍﻟﻨﻘﻁﺔ ‪. M‬‬ ‫‪zM1 − zA = −1−‬‬ ‫‪2+‬‬ ‫‪2 .i +1 = −‬‬ ‫‪2+‬‬ ‫‪2‬‬ ‫‪.i‬‬ ‫=‬ ‫‪zB‬‬ ‫‪−‬‬ ‫‪zO‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﺇﺫﻥ ‪0.25 AM1 = OB :‬‬‫‪0.25‬‬ ‫] ‪[M1M2‬‬ ‫‪ A‬ﻫﻲ ﻤﻨﺘﺼﻑ ﺍﻟﻘﻁﻌﺔ‬ ‫ﻓﺎﻥ ‪:‬‬ ‫‪zM1 + zM2‬‬ ‫=‬ ‫‪−2‬‬ ‫=‬ ‫‪−1‬‬ ‫=‬ ‫‪zA‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬‫ﺝ‪ -‬ﺒﻤﺎ ﺃﻥ ‪ AM1 = OB :‬ﻓﺎﻥ ‪ AOBM :‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ ﻭﺒﻤﺎ ﺃﻥ ‪OB = OA = 1 :‬‬ ‫ﻓﺎﻥ‪ AOBM :‬ﻤﻌﻴﻥ‪0.5 .‬‬ ‫‪π‬‬ ‫≡ )‪(e1, OM1) ≡ (e1, OB) + (OB, OM1‬‬ ‫‪3π‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪= 3π‬‬ ‫] ‪+ π [2π‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪4‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪8‬‬ ‫‪0.5‬‬ ‫‪Argz1‬‬ ‫≡‬ ‫‪7π‬‬ ‫] ‪[2π‬‬ ‫‪:‬‬ ‫ﺍﺫﻥ‬ ‫≡ )‪. (e1, OM1‬‬ ‫‪7π‬‬ ‫] ‪[2π‬‬ ‫‪:‬‬ ‫ﺇﺫﻥ‬ ‫‪8‬‬ ‫‪8‬‬ ‫ﻤﺴﺄﻟﺔ‬ ‫‪-I‬‬‫‪ (1‬ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﻴﺯﺓ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ‪ y \"− 2 y '+ y = 0 :‬ﻫﻲ ‪ r² − 2r +1 = 0 :‬ﻭﻫﻲ ﺘﻘﺒل ﺤﻼ ﻤﺯﺩﻭﺠﺎ ‪r = 1:‬‬ ‫‪0.25‬‬‫ﺇﺫﻥ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﻫﻲ ﺍﻟﺩﻭﺍل ﺍﻟﻤﻌﺭﻓﺔ ﺏ‪ y : x (ax + b)ex :‬ﺤﻴﺙ ‪0.5 . (a, b) ∈ IR²‬‬ ‫‪ (2‬ﺃ‪ -‬ﻟﺩﻴﻨﺎ‪ y0 \"(x) = 0) :‬ﻭ ‪ ∀x ∈ IR : ( y0 '(x) = a‬ﺇﺫﻥ‪:‬‬‫‪ y0 ⇔ ∀x ∈ IR : y0 \"(x) − 2 y0 '(x) + y0 (x) = x −1‬ﺤل ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ‬ ‫‪⇔ ∀x ∈ IR : −2a + ax + b = x −1‬‬ ‫⇔‬ ‫‪⎧a =1‬‬ ‫‪⎧a‬‬ ‫=‬ ‫‪1‬‬ ‫⇔‬ ‫‪⎩⎨b − 2a = −1‬‬ ‫‪⎨⎩b‬‬ ‫=‬ ‫‪1‬‬ ‫ﺇﺫﻥ ‪ y0 : x x +1 :‬ﺤل ﺨﺎﺹ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ )‪0.25 . (E‬‬ ‫ﺏ‪ -‬ﺍﻟﺤﻠﻭل ﺍﻟﻌﺎﻤﺔ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﻟﺘﻔﺎﻀﻠﻴﺔ ﻫﻲ ﺍﻟﺩﻭﺍل ‪ y‬ﺍﻟﻤﻌﺭﻓﺔ ﺏ‪:‬‬ ‫‪ y : x (ax + b)ex + x +1‬ﺤﻴﺙ‪0.25 (a, b) ∈ IR² :‬‬ ‫‪∀x‬‬ ‫∈‬ ‫‪IR‬‬ ‫‪:‬‬ ‫‪⎪⎧h(x) = (ax + b)ex‬‬ ‫‪+ x +1‬‬ ‫ﺝ‪ -‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫⎨‬ ‫‪b)ex +1‬‬ ‫‪⎩⎪h '(x) = (ax + a +‬‬

‫⇔ ‪⎧h(0) = 0‬‬ ‫‪b+1= 0‬‬ ‫‪⇔ b = −1‬‬ ‫‪−a‬‬ ‫⇔‬ ‫‪⎧a‬‬ ‫=‬ ‫‪1‬‬ ‫ﺍﺫﻥ‪:‬‬ ‫⇔ ‪⎩⎨h '(0) = 1‬‬ ‫‪a +b+1‬‬ ‫= ‪=1⇔b‬‬ ‫‪⎩⎨b‬‬ ‫=‬ ‫‪−1‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪0.5 ∀x ∈ IR : h(x) = (x −1)ex + x +1 :‬‬ ‫‪ (3‬ﺃ‪ -‬ﻟﺩﻴﻨﺎ ‪∀x ∈ IR : g '(x) = ex + (x −1)ex +1 :‬‬ ‫ﺇﺫﻥ‪0.5 ∀x ∈ IR : g '(x) = xex +1 :‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ ∀x ∈[0, +∞[ : xex +1 0 :‬ﻓﺎﻥ ‪ g‬ﺩﺍﻟﺔ ﺘﺯﺍﻴﺩﻴﺔ ﻗﻁﻌﺎ ﻋﻠﻰ [∞‪0.25 [0, +‬‬ ‫ﺏ‪ -‬ﺒﻤﺎ ﺃﻥ ‪ g‬ﺩﺍﻟﺔ ﺘﺯﺍﻴﺩﻴﺔ ﻗﻁﻌﺎ ﻋﻠﻰ [∞‪ [0, +‬ﻓﺎﻥ ‪ g(0) :‬ﻫﻲ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺩﻨﻭﻴﺔ ﻟﻠﺩﺍﻟﺔ ‪ g‬ﻋﻠﻰ [∞‪[0, +‬‬ ‫ﺇﺫﻥ‪∀x ∈[0, +∞[ : g(x) ≥ g(0) :‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ g(0) = 0 :‬ﻓﺎﻥ‪0.25 . ∀x ∈[0, +∞[ : g(x) ≥ 0 :‬‬ ‫‪ (1 - II‬ﻟﻜل ‪ x‬ﻤﻥ * ‪ IR‬ﻟﺩﻴﻨﺎ * ‪ −x ∈ IR‬ﻭ ‪:‬‬ ‫‪−x x‬‬ ‫‪f‬‬ ‫)‪(−x‬‬ ‫=‬ ‫‪− xe− x‬‬ ‫=‬ ‫‪ex‬‬ ‫=‬ ‫‪−‬‬ ‫‪ex‬‬ ‫=‬ ‫‪−‬‬ ‫‪x‬‬ ‫‪.‬‬ ‫‪e2x‬‬ ‫‪−‬‬ ‫‪xex‬‬ ‫‪(e−x −1)²‬‬ ‫‪ex‬‬ ‫‪(ex −1)²‬‬ ‫‪(ex −1)²‬‬ ‫‪1‬‬ ‫‪(1‬‬ ‫‪−‬‬ ‫‪e‬‬ ‫‪x‬‬ ‫‪ex‬‬ ‫‪e‬‬ ‫(‬ ‫‪−1)²‬‬ ‫‪x‬‬ ‫‪)²‬‬ ‫ﺍﺫﻥ‪ ∀x∈IR*: f (−x) = −f (x) :‬ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ f‬ﺩﺍﻟﺔ ﻓﺭﺩﻴﺔ ‪0.5‬‬ ‫‪05‬‬ ‫‪lim‬‬ ‫= )‪f (x‬‬ ‫‪lim‬‬ ‫‪xex‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x².ex‬‬ ‫=‬ ‫‪lim‬‬ ‫‪ex‬‬ ‫‪.‬‬ ‫‪e‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫‪1)²‬‬ ‫=‬ ‫∞‪+‬‬ ‫ﺃ‪ -‬ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪(2‬‬ ‫‪(ex −1)²‬‬ ‫‪x.(ex −1)²‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪x→0+‬‬ ‫‪x→0+‬‬ ‫‪x→0+‬‬ ‫‪x→0+‬‬ ‫(‬ ‫‪x‬‬ ‫ﺍﺫﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻭ ﻤﻌﺎﺩﻟﺔ ‪ x = 0‬ﻤﻘﺎﺭﺏ ﻟﻠﻤﻨﺤﻨﻰ ‪0.25‬‬ ‫ﻭ ∞‪lim ex = +‬‬ ‫ﻷﻥ ‪lim ex −1 = 1 :‬‬ ‫‪xx→0+‬‬ ‫‪xx→0+‬‬ ‫‪0.25‬‬ ‫‪lim‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪.‬‬ ‫‪(1‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫‪)²‬‬ ‫=‬ ‫‪0‬‬ ‫ﺏ‪-‬‬ ‫‪ex.(1− e−x )²‬‬ ‫‪ex‬‬ ‫‪e−‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫ﻷﻥ ‪ lim ex = +∞ :‬ﻭ ‪ . lim (1− e−x )² = 1‬ﺍﺫﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻭ ﻤﻌﺎﺩﻟﺔ ‪ y = 0‬ﻤﻘﺎﺭﺏ ﻟﻠﻤﻨﺤﻨﻰ ‪0.25‬‬ ‫∞‪x→+‬‬ ‫∞‪xx→+‬‬ ‫‪ (3‬ﺃ‪ -‬ﻟﺩﻴﻨﺎ‪:‬‬‫∈ ‪∀x‬‬ ‫‪IR* :‬‬ ‫‪f‬‬ ‫)‪' x‬‬ ‫=‬ ‫‪(ex‬‬ ‫‪+‬‬ ‫‪xex )(ex −1)² − 2xe2x (ex‬‬ ‫)‪−1‬‬ ‫=‬ ‫‪(ex‬‬ ‫‪−1)[(ex‬‬ ‫‪−1)(ex‬‬ ‫‪+‬‬ ‫‪xex ) −‬‬ ‫‪2 xe2 x‬‬ ‫‪(ex −1)4‬‬ ‫=‬ ‫‪e2x‬‬ ‫‪+ xe2x − ex − xex‬‬ ‫‪− 2xe2x‬‬ ‫‪= ex. ex‬‬ ‫‪− xex − x −1‬‬ ‫‪(ex −1)3‬‬ ‫‪(ex −1)3‬‬ ‫‪0.75‬‬ ‫∈ ‪∀x‬‬ ‫‪IR* :‬‬ ‫‪f‬‬ ‫)‪'(x‬‬ ‫=‬ ‫‪−‬‬ ‫‪ex.‬‬ ‫)‪.g(x‬‬ ‫ﺇﺫﻥ‪:‬‬ ‫‪(ex −1)3‬‬ ‫ﺏ‪ -‬ﻋﻠﻰ ﺍﻟﻤﺠﺎل ‪ ]0, +∞[ :‬ﻟﺩﻴﻨﺎ‪ ex −1 0 :‬ﻭ ‪ ex 0‬ﻭ ‪ g(x) 0‬ﺇﺫﻥ‪ f '(x) ≺ 0 :‬ﻭﺒﺎﻟﺘﺎﻟﻲ‪ :‬ﺠﺩﻭل‬ ‫ﺘﻐﻴﺭﺍﺕ ﻫﻭ ﻜﺎﻟﺘﺎﻟﻲ‪0.5 :‬‬ ‫‪x0‬‬ ‫∞‪+‬‬‫)‪f’(x‬‬ ‫‪-‬‬ ‫∞‪+‬‬‫)‪f(x‬‬ ‫‪0‬‬ ‫‪ (4‬ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ‪0.5 : f‬‬

Cf 3 −dt −== (1∫ ∫3 1 − 1)dt [ln(t 1) ln t]3 : ‫ ﻟﺩﻴﻨﺎ‬-‫( ﺃ‬52 t(t −1) 2 t −1 t 2 = ⎡⎢⎣ln( t −t=1+)−⎤⎥⎦ 3 −ln(2) =ln(1 ) ln 2 ln 3 ln 2 = 2ln 2 − ln 3 0.5 2 32 x = ln t :‫ ﺇﺫﻥ‬t = ex : ‫ ﻨﻀﻊ‬-‫ﺏ‬ f (x) = t.ln t ‫ ﻭ‬dx = 1 dt ‫ﻭ‬ (t −1)² t ⎧x = 2=⇒ t ln 2 : ‫ﻭﻟﺩﻴﻨﺎ‬ ⎨ ln 3 ⎩ x = 3 ⇒= t ln 3 t.ln t .1 dt = 3 ln t dt : ‫ﺇﺫﻥ‬ f (x)dx = ∫ ∫ ∫0.5 ln2 (t −1)² t 2 (t −1)² ⎨⎪⎧⎪⎩⎪⎪uv('(tt))==t1−t−11 ‫ﻭ‬ ⎧u(t) = ln t ⎪ ⎨⎪⎩v 1 :‫ ﻨﻀﻊ‬-‫( ﺃ‬6 '(t) = −1)² (t ∫ ∫3ln t dt = ⎡ − ln t ⎤3 − 3 −1 dt : ‫ﺇﺫﻥ‬ (t −1)² ⎢⎣ t −1 ⎥⎦ 2 2 t(t −1) 2 =+−− ln 3 + ln 2 2 ln 2 ln 3 2 ∫0.5 3 ln t dt = 3ln 2 − 3 ln 3 : ‫ﺇﺫﻥ‬ 2 (t −1)² 2 ln 3A = ∫ f (x)dx : ‫ ﺍﻟﻤﺴﺎﺤﺔ ﺍﻟﻤﻁﻠﻭﺒﺔ ﻫﻲ‬:‫[ ﻓﺎﻥ‬2,3] ‫ ﻤﻭﺠﺒﺔ ﻭﻤﺘﺼﻠﺔ ﻋﻠﻰ ﺍﻟﻤﺠﺎل‬f ‫ ﺒﻤﺎ ﺃﻥ ﺍﻟﺩﺍﻟﺔ‬-‫ﺏ‬ln 2 0.5 A = 0.45 u.a ‫ﺃﻱ‬.‫ ﺒﻭﺤﺩﺓ ﺍﻟﻤﺴﺎﺤﺔ‬A = 3ln 2 − 3 ln 3 : ‫ﺇﺫﻥ ﺍﻟﻤﺴﺎﺤﺔ ﻫﻲ‬ 2

‫ا‬ ‫ت‬ ‫ا آا‬ ‫ا‬ ‫ا دة‪ :‬ا ت‬‫‪1‬‬ ‫ا )ة( ‪ :‬ا م ا‬ ‫‪2‬‬‫‪C :NS22‬‬ ‫ا نا ا آ ر‬ ‫) ا ورة ا د ‪( 2007‬‬‫ةا ز‪3 :‬‬ ‫اع‬‫‪7:‬‬ ‫ا‬ ‫‪ +‬ا م ا را‬ ‫‪+‬ا ما‬ ‫لا ا ا (‬ ‫)‬ ‫ا ا ول ‪ 3 ) :‬ن(‬ ‫د ه‪:‬‬ ‫) ‪ (o, i , j, k‬ا )‪ (S‬ا‬ ‫ا ءا بإ‬‫‪ x2 + y2 + z2 - 2x - 4y - 6z + 8 = 0‬و ا ى )‪ (P‬ا ي د ه ‪. x – y + 2z + 1 = 0 :‬‬ ‫‪1‬‬ ‫وي ‪. 6‬‬ ‫‪ (1‬ان آ ا )‪ (S‬ه ا )‪ Ω(1, 2,3‬أن‬ ‫‪0,75‬‬ ‫‪0,5‬‬ ‫أن ا ى )‪ (P‬س )‪. (S‬‬ ‫‪(2‬‬ ‫‪0,75‬‬ ‫)∆( ا ر ‪ Ω‬و ا دي )‪. (P‬‬ ‫‪ (3‬أ‪ -‬د را‬ ‫س )‪ (P‬و)‪. (S‬‬ ‫ب‪ -‬د ث إ ا ت ‪ω‬‬ ‫ا ا ‪ 3 ) :‬ن(‬ ‫ا ا ي ا د ا ي ‪(3 – 2i )2‬‬ ‫‪ (1 0,5‬أ‪ -‬أآ‬ ‫‪ ℂ‬ا د ‪z2 – 2(4 + i )z + 10 + 20i = 0 :‬‬ ‫ا اد ا‬ ‫ب‪-‬‬ ‫‪(2 1‬‬‫) ‪ (O,u,v‬ا ‪ A‬و ‪ B‬و‪ C‬ا أ‬ ‫ا ىا يا بإ‬ ‫ا ا ه ‪ a = 1 + 3i :‬و‪ b = 7 – i‬و ‪. c = 5 + 9i‬‬ ‫و ا او ‪.‬‬ ‫أ‪ -‬أن ‪c − a = i :‬‬ ‫‪0,5‬‬ ‫‪1‬‬ ‫‪b−a‬‬ ‫ب‪ -‬ا أن ا ‪ ABC‬وي ا‬ ‫ا ا ‪( 2,5 ) :‬‬ ‫‪. ℝ −{−1} x‬‬ ‫ان ‪x2 = x −1+ 1 :‬‬ ‫‪(1‬‬ ‫‪0,5‬‬ ‫‪(2‬‬ ‫‪1‬‬ ‫‪x+1 x+1‬‬ ‫‪(3‬‬ ‫‪1‬‬ ‫أن ‪. ∫2 x2 dx = ln 3 :‬‬ ‫‪0 x +1‬‬ ‫اء ‪ ،‬أن ‪∫2 x ln(x +1)dx = 3 ln 3 :‬‬ ‫ل‬ ‫‪02‬‬

‫ا‬ ‫ا نا ا آ ر‬ ‫ات‬ ‫ا دة ‪:‬‬ ‫‪2‬‬ ‫) ا ورة ا د ‪(2007‬‬ ‫‪2‬‬‫‪C :NS22‬‬ ‫ا )ة( ا م ا ا ‪ +‬ا م ا ع‬ ‫‪ +‬ا م ا را‬ ‫ا‬ ‫ا ا ا ‪( 2,5 ):‬‬ ‫ت ا اد ‪ 0‬و‪ 0‬و‪ 0‬و‪ -1‬و ‪ 1‬و‪ 1‬و ‪1‬‬ ‫يآ‬ ‫ا ت (‪.‬‬ ‫)ا‬ ‫ا و ن وا ث ت ا ‪.‬‬ ‫ا ا‪:‬‬ ‫ت ا ا \"‪.‬‬ ‫ا د‪ 0‬ا‬ ‫اث ا ‪:‬‬ ‫ا‬ ‫\"‪.‬‬ ‫أ ادا‬ ‫أ‬ ‫‪\":A‬‬ ‫‪\":B‬‬ ‫ا م\"‪.‬‬ ‫ا تا‬ ‫ثت‬ ‫‪\":C‬‬ ‫ع ا اد ا‬ ‫أن ا ل ا ث ‪ C‬ه ‪2‬‬ ‫‪ 2,5‬ا ا ل آ ا ‪ A‬و‪C‬‬ ‫‪7‬‬ ‫‪. g(x) = e−x + x −1 :‬‬ ‫‪ℝ‬‬ ‫‪ 9 ) :‬ن(‬ ‫‪ (I‬ا ا ا د ‪ g‬ا‬‫]‪. ]−∞,0‬‬ ‫[∞‪[0, +‬و‬ ‫‪ (1‬ا )‪ ℝ x g ‘(x‬ا أن ‪ g‬ا‬ ‫‪0,75‬‬ ‫‪0,5‬‬‫‪ (2‬أن ‪ ) ℝ x g(x) ≥ 0‬أن ‪ ( g(0) = 0‬ا أن ‪. ℝ x e−x + x ≥ 1‬‬ ‫‪0,5‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪:‬‬ ‫ا ‪x‬ا‬ ‫‪ (II‬ا ا ا د ‪f‬‬ ‫‪0,25‬‬ ‫‪+ e−x‬‬ ‫)‪. (o, i , j‬‬ ‫و )‪ (C‬ا ا ا ‪f‬‬ ‫‪1,5‬‬ ‫‪0,75‬‬ ‫ا ا ‪ f‬ه ‪ ) ℝ‬ا ل ا ال ‪. ( (2(I‬‬ ‫‪ (1‬أن‬ ‫‪0,5‬‬ ‫‪ℝ* x‬‬ ‫أن ‪f (x) = 1 :‬‬ ‫‪ (2‬أ –‬ ‫‪0,5‬‬ ‫‪1‬‬ ‫‪0,75‬‬ ‫‪1+‬‬ ‫‪xex‬‬ ‫‪0,25‬‬ ‫ب – أن ‪ lim f (x) = 0 :‬و ‪ lim f (x) = 1‬أول ه ه ا ‪.‬‬ ‫‪1‬‬ ‫∞‪x→+‬‬ ‫∞‪x→−‬‬ ‫‪0,5‬‬ ‫‪.ℝ x‬‬ ‫‪(1+ x)e−x‬‬ ‫أن ‪:‬‬ ‫‪(3‬أ–‬ ‫‪0,5‬‬ ‫‪x + e−x 2‬‬ ‫‪0,75‬‬ ‫== )‪( )f '(x‬‬ ‫ول ات ا ا ‪. f‬‬ ‫ب‪ -‬ادرس إ رة )‪f ‘(x‬‬ ‫)‪ (C‬ا ‪ O‬أ ا ‪.‬‬ ‫‪ (4‬أ‪ -‬اآ د ا س‬‫أن ‪ ℝ x x − f (x) = xg(x) :‬ادرس إ رة )‪. ℝ x − f (x‬‬ ‫ب‪-‬‬ ‫‪g(x) +1‬‬ ‫)‪ (C‬و ا )∆( ا ي د ه ‪. y = x :‬‬ ‫ج‪ -‬ا ا ا‬ ‫‪.( 1 ≃ −0, 6‬‬ ‫)∆( و)‪ (C‬ا )‪) (o, i , j‬‬ ‫‪ (5‬أ‬‫‪.ℕ n‬‬ ‫‪1− e‬‬ ‫ا د )‪ (Un‬ا‬ ‫ا‬ ‫‪(III‬‬ ‫‪(1‬‬ ‫‪ U0 = 1‬و ) ‪Un+1 = f ( Un‬‬ ‫أن ‪ℕ n 0 ≤ Un ≤ 1‬‬ ‫‪(2‬‬ ‫ل ا ال ‪(4(II‬ب(‪.‬‬ ‫‪ (3‬ا‬ ‫)ا‬ ‫أن ا )‪(Un‬‬ ‫د‪.‬‬ ‫أن )‪ (Un‬ر‬

‫اﻟﺮﻳﺎﺿﻴﺎت‬ ‫اﻟﻤﺎدة ‪:‬‬ ‫ﺗﺼﺡﻴﺢ اﻻﻣﺘﺡﺎن اﻟﻮﻃﻨﻲ‬ ‫اﻟﻤﻮﺣﺪ ﻟﻠﺒﻜﺎﻟﻮرﻳﺎ‬‫‪ 3‬ﺳﺎﻋﺎت‬ ‫‪7‬‬ ‫ﻣﺪة اﻻﻧﺠﺎز ‪:‬‬ ‫اﻟﺪورة اﻟﻌﺎدﻳﺔ ‪2007‬‬ ‫اﻟﻤﻌﺎﻣﻞ ‪:‬‬ ‫ﺷﻌﺒﺔ اﻟﻌﻠﻮم اﻟﺘﺠﺮﻳﺒﻴﺔ‬ ‫اﻟﺘﻤﺮﻳﻦ اﻷول ‪:‬‬ ‫‪( )JJG JJG JJG‬‬ ‫ﻧﻌﺘﺒﺮ ﻓﻲ اﻟﻔﻀﺎء ‪ E‬اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ‪: O ,i , j , k‬‬ ‫اﻟﻔﻠﻜﺔ ‪(S ) : x 2 + y 2 + z 2 − 2x − 4y − 6z + 8 = 0‬‬ ‫و اﻟﻤﺴﺘﻮى ‪( )P : x − y + 2z +1= 0‬‬ ‫‪ .1‬ﺑﻤﺎ أن ‪:‬‬‫‪x 2 + y 2 + z 2 − 2x − 4y − 6z + 8 = 0 ⇔ x 2 − 2x 2+1−1+ y − 4y + 4 − 4 + z 2 − 6z + 9 − 9 + 8 = 0‬‬ ‫⇔‬ ‫= ‪(x −1)2 + ( y − 2)2 + (z − 3)2‬‬ ‫‪2‬‬ ‫‪6‬‬ ‫ﻓﺈن ‪ (S ) :‬ﻓﻠﻜﺔ ﻣﺮآﺰهﺎ ‪ Ω 1,2,3‬وﺷﻌﺎﻋﻬﺎ ‪( ). R = 6‬‬ ‫ﻣﻤﺎس ﻟﻠﻔﻠﻜﺔ ) ‪( )( ) ( ) ( ). (S‬‬ ‫‪P‬‬ ‫‪ . d Ω, P‬إذن‬ ‫‪1− 2 +‬‬ ‫‪2×3‬‬ ‫‪+1‬‬ ‫‪6‬‬ ‫=‬ ‫‪ .2‬ﻟﺪﻳﻨﺎ ‪6 = R :‬‬ ‫=‬ ‫=‬ ‫‪12 + (−1)2 + 22 6‬‬‫‪JJG‬‬ ‫هﻮ اﻟﻤﺴﺘﻘﻴﻢ اﻟﻤﺎر ﻣﻦ اﻟﻨﻘﻄﺔ ‪ Ω‬واﻟﻌﻤﻮدي ﻋﻠﻰ اﻟﻤﺴﺘﻮى ‪P‬‬‫ﻣﺘﺠﻬﺔ) ( ) ( ) (‬‫‪n‬‬ ‫‪ .3‬أ‪ -‬ﻟﺪﻳﻨﺎ‬ ‫‪1, −1, 2‬‬ ‫وﻟﺪﻳﻨﺎ‬ ‫∆‬‫ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ اﻟﻤﺴﺘﻮى ‪ P‬ﻓﻬﻲ ﻣﻮﺟﻬﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ ∆ ‪ ،‬وﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ ﺗﻤﺜﻴﻼ ﺑﺎراﻣﺘﺮﻳﺎ ﻟﻠﻤﺴﺘﻘﻴﻢ ∆) ( ) ( ) (‬ ‫‪⎧x =1+t‬‬ ‫⎪‬ ‫⎨‬ ‫‪y‬‬ ‫‪= 2−t‬‬ ‫\∈ ‪/ t‬‬ ‫آﻤﺎ ﻳﻠﻲ ‪:‬‬ ‫‪⎩⎪z = 3+ 2t‬‬ ‫ب‪ -‬ﻟﺘﻜﻦ ) ‪ ω (x , y ,z‬ﻧﻘﻄﺔ ﺗﻤﺎس آﻞ ﻣﻦ ‪ P‬و ) ‪ . (S‬ﻟﺪﻳﻨﺎ ‪ ω ∈ P :‬و ) ‪ . ω ∈(S‬إذن ‪( ) ( ):‬‬ ‫‪⎧x =1+t‬‬ ‫⎪‬ ‫و ‪ ، x − y + 2z +1= 0‬وﻣﻨﻩ ﻓﺈن ‪:‬‬ ‫⎨‬ ‫‪y‬‬ ‫‪= 2−t‬‬ ‫\∈ ‪/ t‬‬ ‫‪⎪⎩z = 3+ 2t‬‬ ‫‪1+t − (2 −t ) + 2(3+ 2t ) +1= 0 ⇔ 6t + 6 = 0‬‬ ‫‪⇔ t = −1‬‬ ‫‪⎧x = 0‬‬ ‫⎪‬ ‫‪( ). ω 0,3,1‬‬ ‫‪:‬‬ ‫ﻓﺈن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫‪،‬‬ ‫⎨‬ ‫‪y‬‬ ‫‪=3‬‬ ‫وﻋﻠﻴﻩ ﻓﺈن ‪:‬‬ ‫‪⎪⎩z =1‬‬ ‫اﻟﺘﻤﺮﻳﻦ اﻟﺜﺎﻧﻲ ‪:‬‬ ‫‪ .1‬أ‪ -‬ﻟﺪﻳﻨﺎ ‪( ). 3− 2i 2 = 9 −12i − 4 = 5 −12i :‬‬ ‫ب‪ -‬ﻧﻌﺘﺒﺮ ﻓﻲ اﻟﻤﺠﻤﻮﻋﺔ ^ اﻟﻤﻌﺎدﻟﺔ ‪( ) ( ). E : z 2 − 2 4 + i z +10 + 20i = 0 :‬‬ ‫اﻟﻤﻤﻴﺰ اﻟﻤﺨﺘﺼﺮ اﻟﻤﻌﺎدﻟﺔ ) ‪ (E‬هﻮ ‪:‬‬‫‪( )∆′ = b′2 −ac = −(4 + i ) 2 −1×(10 + 20i ) =16 + 8i −1−10 − 20i = 5−12i = (3− 2i )2‬‬

‫‪z2‬‬ ‫=‬ ‫‪4+i‬‬ ‫‪− 3+ 2i‬‬ ‫‪=1+ 3i‬‬ ‫و‬ ‫= ‪z1‬‬ ‫‪4+i‬‬ ‫‪+ 3− 2i‬‬ ‫‪=7−i‬‬ ‫إذن ﻟﻠﻤﻌﺎدﻟﺔ ) ‪ (E‬ﺣﻠﻴﻦ هﻤﺎ ‪:‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ﻣﺠﻤﻮﻋﺔ ﺣﻠﻮل اﻟﻤﻌﺎدﻟﺔ ) ‪ (E‬هﻲ ‪. S = {1+ 3i , 7 − i } :‬‬ ‫‪JJG JJG‬‬ ‫اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ وﻣﺒﺎﺷﺮ ‪O ,u ,v‬‬‫‪( ) ( )B‬‬ ‫‪P‬‬ ‫‪ .2‬ﻓﻲ اﻟﻤﺴﺘﻮى اﻟﻌﻘﺪي‬‫‪A‬و‬ ‫‪ ،‬ﻧﻌﺘﺒﺮ اﻟﻨﻘﻁ‬ ‫و ‪ C‬اﻟﺘﻲ أﻟﺡﺎﻗﻬﺎ ﻋﻠﻰ اﻟﺘﻮاﻟﻲ هﻲ ‪ a =1+ 3i :‬و ‪ b = 7 − i‬و ‪. c = 5 + 9i‬‬ ‫‪.‬‬ ‫‪c −a‬‬ ‫=‬ ‫) ‪(5+9i ) −(1+ 3i‬‬ ‫=‬ ‫‪5+ 9i‬‬ ‫‪−1− 3i‬‬ ‫=‬ ‫‪4 + 6i‬‬ ‫=‬ ‫‪i‬‬ ‫‪(6 − 4i‬‬ ‫)‬ ‫=‬ ‫‪i‬‬ ‫أ‪ -‬ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪b −a‬‬ ‫) ‪(7 −i ) −(1+3i‬‬ ‫‪7−i‬‬ ‫‪−1− 3i‬‬ ‫‪6 − 4i‬‬ ‫‪6− 4i‬‬ ‫‪.‬‬ ‫‪AB =AC‬‬ ‫و ﻣﻨﻩ ﻓﺈن ‪:‬‬ ‫‪AC‬‬ ‫=‬ ‫‪c −a‬‬ ‫‪=i‬‬ ‫‪=1‬‬ ‫‪ .‬إذن ‪:‬‬ ‫‪c‬‬ ‫‪−a‬‬ ‫=‬ ‫‪i‬‬ ‫ب‪ -‬ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪AB‬‬ ‫‪b −a‬‬ ‫‪b‬‬ ‫‪−a‬‬ ‫‪( )JJJJG JJJJJG‬‬ ‫وﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪AB,AC‬‬ ‫≡‬ ‫‪arg‬‬ ‫⎛‬ ‫‪c‬‬ ‫‪−‬‬ ‫‪a‬‬ ‫⎞‬ ‫‪⎡⎣2π‬‬ ‫⎤‬ ‫⎝⎜‬ ‫‪b‬‬ ‫‪−‬‬ ‫‪a‬‬ ‫⎠⎟‬ ‫⎦‬ ‫≡‬ ‫) ‪arg(i‬‬ ‫⎦⎤ ‪⎣⎡2π‬‬ ‫⎦⎤ ‪⎣⎡2π‬‬ ‫‪( )JJJJG JJJJJG‬‬ ‫‪π‬‬ ‫‪2‬‬ ‫≡ ‪AB,AC‬‬ ‫‪ A BC‬ﻣﺜﻠﺚ ﻣﺘﺴﺎوي اﻟﺴﺎﻗﻴﻦ وﻗﺎﺋﻢ اﻟﺰاوﻳﺔ ﻓﻲ ‪. A‬‬ ‫ﻓﺈن‬ ‫وﺑﺎﻟﺘﺎﻟﻲ‬ ‫‪.‬‬ ‫‪i‬‬ ‫=‬ ‫‪⎡⎢1,‬‬ ‫‪π‬‬ ‫⎤‬ ‫ﻷن‬ ‫‪2‬‬ ‫⎥‬ ‫⎣‬ ‫⎦‬ ‫اﻟﺘﻤﺮﻳﻦ اﻟﺜﺎﻟﺚ ‪:‬‬ ‫‪ .1‬ﻟﻴﻜﻦ ‪ ، x ∈\ − −1‬ﻟﺪﻳﻨﺎ ‪{ }:‬‬ ‫‪x2‬‬ ‫=‬ ‫‪x‬‬ ‫‪2 −1+1‬‬ ‫=‬ ‫‪x 2 −1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫=‬ ‫‪(x‬‬ ‫‪−1)(x‬‬ ‫)‪+1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫=‬ ‫‪x‬‬ ‫‪−1+‬‬ ‫‪1‬‬‫‪x +1‬‬ ‫‪x +1‬‬ ‫‪x +1‬‬ ‫‪+1‬‬ ‫‪+1‬‬ ‫‪+1‬‬ ‫‪x‬‬ ‫‪x +1‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪xx+21dx‬‬ ‫‪2‬‬ ‫⎛‬ ‫‪1‬‬ ‫⎞‬ ‫‪⎡x 2‬‬ ‫‪⎤2‬‬ ‫‪ .2‬ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪∫ ∫.‬‬‫‪0‬‬ ‫=‬ ‫‪0‬‬ ‫⎜‬ ‫‪x‬‬ ‫‪−1+‬‬ ‫‪+‬‬ ‫⎟‬ ‫‪dx‬‬ ‫=‬ ‫⎢‬ ‫‪−x‬‬ ‫‪+ ln‬‬ ‫‪x‬‬ ‫=‬ ‫‪ln 3‬‬ ‫⎝‬ ‫‪x‬‬ ‫‪1‬‬ ‫⎠‬ ‫⎣⎢‬ ‫‪2‬‬ ‫⎥‪+1‬‬ ‫‪⎦⎥0‬‬ ‫‪.‬‬ ‫‪u (x ) = x 2‬‬ ‫‪ .‬إذن ‪:‬‬ ‫‪ .3‬ﻧﻀﻊ ‪u ′(x ) = x :‬‬ ‫‪2‬‬ ‫‪v ′(x ) = (x +1)′ = 1‬‬ ‫)‪v (x ) = ln (x +1‬‬ ‫‪x +1 x +1‬‬‫ﻟﺪﻳﻨﺎ ‪ u :‬و ‪ v‬ﻣﺘﺼﻠﺘﻴﻦ وﻗﺎﺑﻠﺘﻴﻦ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤‪ ⎡⎣0, 2‬وﻟﺪﻳﻨﺎ‪ u ′ :‬و ‪ v ′‬ﻣﺘﺼﻠﺘﻴﻦ ﻋﻠﻰ اﻟﻤﺠﺎل ⎡⎣‪. ⎦⎤0, 2‬‬ ‫ﺣﺴﺐ ﺗﻘﻨﻴﺔ اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء ‪ ،‬ﻟﺪﻳﻨﺎ‪:‬‬ ‫‪2‬‬ ‫‪ln (x‬‬ ‫‪+1)dx‬‬ ‫=‬ ‫⎡‬ ‫‪x2‬‬ ‫‪ln (x‬‬ ‫‪∫+‬‬ ‫‪1)⎥⎤2‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪xx+21dx‬‬ ‫= ‪= 2ln 3− 1 ln 3‬‬ ‫‪3 ln 3‬‬ ‫⎢‬ ‫‪2‬‬ ‫‪⎥⎦0‬‬ ‫‪2‬‬ ‫‪0‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪∫0 x‬‬ ‫⎢⎣‬ ‫اﻟﺘﻤﺮﻳﻦ اﻟﺮاﺑﻊ ‪:‬‬ ‫ﻳﺡﺘﻮي آﻴﺲ ﻋﻠﻰ ﺳﺒﻊ ﺑﻴﺪﻗﺎت ) ﻻ ﻳﻤﻜﻦ اﻟﺘﻤﻴﻴﺰ ﺑﻴﻨﻬﺎ ﺑﺎﻟﻠﻤﺲ ( ﺗﺡﻤﻞ اﻷﻋﺪاد ‪:‬‬ ‫ﻧﺴﺡﺐ ﻋﺸﻮاﺋﻴﺎ وﻓﻲ ﺁن واﺣﺪ ﺛﻼث ﺑﻴﺪﻗﺎت ﻣﻦ اﻟﻜﻴﺲ ‪ ،‬ﻧﻌﺘﺒﺮ اﻷﺣﺪاث اﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪ : A‬ﻻ ﺗﻮﺟﺪ أﻳﺔ ﺑﻴﺪﻗﺔ ﺗﺡﻤﻞ اﻟﻌﺪد ‪ 0‬ﻣﻦ ﺑﻴﻦ اﻟﺒﻴﺪﻗﺎت اﻟﺜﻼﺛﺔ اﻟﻤﺴﺡﻮﺑﺔ ‬ ‫‪ : B‬ﺳﺡﺐ ﺛﻼث ﺑﻴﺪﻗﺎت ﺗﺡﻤﻞ أﻋﺪادا ﻣﺨﺘﻠﻔﺔ ﻣﺜﻨﻰ ﻣﺜﻨﻰ ‬ ‫‪ :C‬ﻣﺠﻤﻮع اﻷﻋﺪاد اﻟﻤﺴﺠﻠﺔ ﻋﻠﻰ اﻟﺒﻴﺪﻗﺎت اﻟﺜﻼﺛﺔ اﻟﻤﺴﺡﻮﺑﺔ ﻣﻨﻌﺪم ‬

‫اﺣﺘﻤﺎﻻت اﻷﺣﺪاث ‪ A‬و ‪ B‬و ‪ C‬هﻲ ‪:‬‬ ‫‪p‬‬ ‫‪(A‬‬ ‫)‬ ‫=‬ ‫‪Card‬‬ ‫) ‪(A‬‬ ‫=‬ ‫‪C‬‬ ‫‪3‬‬ ‫=‬ ‫‪4‬‬ ‫‪Card‬‬ ‫)‪(Ω‬‬ ‫‪C‬‬ ‫‪4‬‬ ‫‪35‬‬ ‫‪3‬‬ ‫‪7‬‬ ‫‪p (B‬‬ ‫)‬ ‫=‬ ‫‪Card‬‬ ‫) ‪(B‬‬ ‫=‬ ‫‪C‬‬ ‫‪1‬‬ ‫‪×C11‬‬ ‫‪×C‬‬ ‫‪1‬‬ ‫=‬ ‫‪9‬‬ ‫‪Card‬‬ ‫)‪(Ω‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪35‬‬ ‫‪C‬‬ ‫‪3‬‬ ‫‪7‬‬ ‫‪p‬‬ ‫‪(C‬‬ ‫)‬ ‫=‬ ‫‪Card‬‬ ‫‪( ) ( )( )C‬‬‫=‬‫‪C‬‬‫‪3‬‬ ‫‪+‬‬ ‫‪C‬‬ ‫‪1‬‬ ‫‪×C‬‬ ‫‪1‬‬ ‫‪×C‬‬ ‫‪1‬‬ ‫= ‪= 10‬‬ ‫‪2‬‬ ‫‪Card‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪7‬‬ ‫‪Ω‬‬ ‫‪35‬‬ ‫‪C‬‬ ‫‪3‬‬ ‫‪7‬‬ ‫ﻣﺴﺄﻟﺔ ‪:‬‬ ‫‪ .I‬ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ ‪ g‬اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ ‪. ∀x ∈\ : g (x ) = e −x + x −1 :‬‬ ‫‪ .1‬ﻟﻴﻜﻦ \∈ ‪ ، x‬ﻟﺪﻳﻨﺎ ‪. g ′(x ) = (e −x + x −1)′ = −e −x +1 :‬‬ ‫‪g ′(x ) = 0 ⇔ −e −x +1 = 0‬‬ ‫‪⇔ e −x =1‬‬ ‫‪⇔ −x = 0‬‬ ‫‪⇔ x =0‬‬‫‪x ≤ 0 ⇒ −x ≥ 0‬‬ ‫‪x ≥ 0 ⇒ −x ≤ 0‬‬ ‫‪⇒ e −x ≥1‬‬ ‫و‬ ‫‪⇒ e −x ≤1‬‬ ‫‪⇒ −e −x +1≤ 0‬‬ ‫‪⇒ −e −x +1≥ 0‬‬ ‫‪⇒ g′(x )≤ 0‬‬ ‫‪⇒ g′(x )≥ 0‬‬ ‫إذن ‪ g :‬ﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡∞‪ ⎡⎣0, +‬و ﺗﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤‪. ⎤⎦−∞,0‬‬‫‪ .2‬ﻟﺪﻳﻨﺎ ‪ . g 0 = e 0 + 0 −1 =1−1 = 0 :‬ﻟﻴﻜﻦ \∈ ‪ ، x‬ﻟﺪﻳﻨﺎ ‪ g‬ﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎡⎣∞‪ ⎡⎣0,+‬و ﺗﻨﺎﻗﺼﻴﺔ) (‬ ‫ﻋﻠﻰ اﻟﻤﺠﺎل ⎤⎦‪ . ⎦⎤−∞,0‬إذن ‪:‬‬ ‫)‪x ≤ 0 ⇒ g (x ) ≥ g (0‬‬ ‫)‪ x ≥ 0 ⇒ g (x ) ≥ g (0‬و‬ ‫‪⇒ g (x )≥ 0‬‬ ‫‪⇒ g (x )≥0‬‬ ‫وﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ أن ‪ . ∀x ∈\ : g (x ) ≥ 0 :‬أي ‪.∀x ∈\ : e −x + x −1≥ 0 :‬‬ ‫‪∀x ∈\ : e −x + x ≥1‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ‪:‬‬‫‪( ) ( )JJG JJG‬‬‫‪.f‬‬ ‫‪x‬‬ ‫=‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪ f‬اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ ﻟﻠﻤﺘﻐﻴﺮ اﻟﺡﻘﻴﻘﻲ ‪ x‬اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ ‪:‬‬ ‫‪ .II‬ﻟﺘﻜﻦ‬ ‫‪+e −x‬‬‫وﻟﻴﻜﻦ ) ‪ (C‬اﻟﻤﻨﺡﻨﻰ اﻟﻤﻤﺜﻞ ﻟﻠﺪاﻟﺔ ‪ f‬ﻓﻲ اﻟﻤﺴﺘﻮى اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ‪. O ,i , j‬‬‫‪ .1‬ﻟﻴﻜﻦ \∈ ‪ ، x‬ﻟﺪﻳﻨﺎ ‪ ، x ∈Df ⇔ x +e −x ≠ 0 :‬وﺑﻤﺎ أن ‪ ، ∀x ∈\ : e −x + x ≥1‬ﻓﺈن ‪:‬‬ ‫\ = ‪Df‬‬ ‫‪ . ∀x ∈\ : e −x + x ≠ 0‬إذن ‪:‬‬‫‪( ). 1‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫=‬ ‫‪1‬‬ ‫=‬ ‫‪xe‬‬ ‫=‬ ‫‪e −x‬‬ ‫‪x‬‬ ‫‪+1‬‬ ‫=‬ ‫‪+e −x‬‬ ‫‪=f‬‬ ‫) ‪(x‬‬ ‫‪ .2‬أ‪ -‬ﻟﻴﻜﻦ *\ ∈ ‪ ، x‬ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪ex‬‬ ‫‪xex +1‬‬ ‫‪xe x‬‬ ‫‪+1‬‬ ‫‪xe x‬‬ ‫‪x‬‬‫‪1+‬‬ ‫‪x‬‬ ‫‪xe x‬‬

‫‪.‬‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫∞‪= −‬‬ ‫و‬ ‫‪lim xe x‬‬ ‫‪= 0−‬‬ ‫ﻷن ‪:‬‬ ‫‪، lim f‬‬ ‫‪(x‬‬ ‫)‬ ‫=‬ ‫‪lim‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫ب‪-‬‬ ‫‪ex‬‬ ‫∞‪x →−‬‬ ‫‪x →−∞ 1+‬‬ ‫‪1‬‬ ‫∞‪x →−‬‬ ‫∞‪x →−‬‬ ‫‪xe x‬‬ ‫‪.‬‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫‪=0‬‬ ‫و‬ ‫‪lim xe x‬‬ ‫∞‪= +‬‬ ‫ﻷن ‪:‬‬ ‫‪، lim f‬‬ ‫(‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫‪lim‬‬ ‫‪1‬‬ ‫‪=1‬‬ ‫‪ex‬‬ ‫∞‪x →+‬‬ ‫‪1‬‬ ‫∞‪x →+‬‬ ‫∞‪x →+‬‬ ‫∞‪x →+‬‬ ‫‪1+‬‬ ‫‪xe x‬‬‫ﻧﺴﺘﻨﺘﺞ ﻣﻤﺎ ﺳﺒﻖ أن اﻟﻤﻨﺡﻨﻰ ) ‪ (C‬ﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ أﻓﻘﻴﺎ ‪ ،‬ﺑﺠﻮار ∞‪ ، −‬ﻣﻌﺎدﻟﺘﻩ ‪ y = 0‬؛ وﻳﻘﺒﻞ ﻣﻘﺎرﺑﺎ أﻓﻘﻴﺎ ‪،‬‬ ‫ﺑﺠﻮار ∞‪ ، +‬ﻣﻌﺎدﻟﺘﻩ ‪. y =1‬‬ ‫‪ .3‬أ‪ -‬ﻟﻴﻜﻦ \ ∈ ‪ ، x‬ﻟﺪﻳﻨﺎ ‪:‬‬ ‫= ) ‪f ′(x‬‬ ‫‪⎛ x ⎞′‬‬ ‫⎠⎟ ‪⎝⎜ x +e −x‬‬ ‫‪( ) ( )x ′ x +e −x − x x +e −x ′‬‬ ‫= ‪( )..........‬‬ ‫‪x +e −x 2‬‬ ‫= ‪..........‬‬ ‫‪( )x +e −x − x 1−e −x‬‬ ‫‪( )x +e −x 2‬‬ ‫= ‪..........‬‬ ‫‪x +e −x − x + xe −x‬‬ ‫‪( )x +e −x 2‬‬ ‫= ) ‪f ′(x‬‬ ‫‪(x +1)e −x‬‬ ‫‪(x +e −x )2‬‬‫ب‪ -‬إﺷﺎرة ‪ f ′ x‬ﻋﻠﻰ \ هﻲ إﺷﺎرة ‪ ، x +1‬وﻣﻨﻩ ﻧﺴﺘﻨﺘﺞ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ ‪ f‬ﻋﻠﻰ \ آﻤﺎ ﻳﻠﻲ‪( ) ( ):‬‬ ‫‪f‬‬ ‫)‪(−1‬‬ ‫=‬ ‫‪−1‬‬ ‫=‬ ‫‪1‬‬ ‫‪−1+e‬‬ ‫‪1−e‬‬‫‪ .4‬أ‪ -‬ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس )∆( ﻟﻠﻤﻨﺡﻨﻰ ) ‪ (C‬ﻓﻲ اﻟﻨﻘﻄﺔ ‪ O‬هﻲ ‪( )( ) ( ). y = f ′ 0 x − 0 + f 0 :‬‬ ‫‪. (∆) : y = x‬‬ ‫أي ‪:‬‬‫ب‪ -‬ﻟﻴﻜﻦ \∈ ‪ ، x‬ﻟﺪﻳﻨﺎ ‪ ، g (x ) = e −x + x −1 :‬إذن ‪ ، g (x ) +1 = e −x + x :‬وﻣﻨﻩ ﻓﺈن ‪:‬‬‫‪( )x − f‬‬‫‪(x‬‬ ‫)‬ ‫=‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪x‬‬ ‫‪=x‬‬ ‫‪⎜⎛1−‬‬ ‫‪1‬‬ ‫⎞‬ ‫=‬ ‫‪x‬‬ ‫‪x +e −x −1‬‬ ‫=‬ ‫) ‪xg (x‬‬ ‫‪+e −x‬‬ ‫‪+e −x‬‬ ‫⎟‬ ‫‪x +e −x‬‬ ‫‪g (x )+1‬‬ ‫‪x‬‬ ‫⎝‬ ‫‪x‬‬ ‫⎠‬

‫إذن ‪ :‬إﺷﺎرة ‪ f (x ) − x‬ﻋﻠﻰ \ هﻲ إﺷﺎرة ‪. x‬‬ ‫ﺟـ‪ -‬ﺣﺴﺐ اﻟﺴﺆال اﻟﺴﺎﺑﻖ ‪ ،‬ﻟﺪﻳﻨﺎ ‪:‬‬‫‪ (C ) 9‬ﻳﻮﺟﺪ ﺗﺡﺖ اﻟﻤﺴﺘﻘﻴﻢ )∆( ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡∞‪. ⎣⎡0, +‬‬‫‪( ).‬‬‫⎦⎤‪⎤⎦ −∞, 0‬‬ ‫ﻋﻠﻰ اﻟﻤﺠﺎل‬ ‫)∆(‬ ‫ﻓﻮق اﻟﻤﺴﺘﻘﻴﻢ‬ ‫ﻳﻮﺟﺪ‬ ‫)‪(C‬‬ ‫‪9‬‬ ‫‪JJG JJG‬‬ ‫‪ .5‬إﻧﺸﺎء اﻟﻤﻨﺡﻨﻰ ) ‪ (C‬واﻟﻤﺴﺘﻘﻴﻢ )∆( ﻓﻲ اﻟﻤﻌﻠﻢ ‪: O ,i , j‬‬ ‫‪ .III‬ﻟﺘﻜﻦ `∈‪ un n‬اﻟﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدﻳﺔ اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ ‪( ):‬‬‫⎪⎧‬ ‫‪n‬‬ ‫‪u0‬‬ ‫‪=1‬‬ ‫‪n‬‬ ‫)‬ ‫;‬ ‫`∈ ‪n‬‬ ‫= ‪+1‬‬‫‪⎩⎪⎨u‬‬ ‫‪f (u‬‬ ‫‪ .1‬ﻣﻦ أﺟﻞ ‪ ، n = 0‬ﻟﺪﻳﻨﺎ ‪ ، u0 =1‬إذن ‪. 0 ≤ u0 ≤1 :‬‬‫ﻟﻴﻜﻦ `∈ ‪ ، n‬ﻧﻔﺘﺮض أن ‪ 0 ≤un ≤1‬وﻧﺒﻴﻦ أن ‪. 0 ≤un+1 ≤1‬‬‫ﻟﺪﻳﻨﺎ ‪ 0 ≤un ≤1‬و ‪ f‬ﺗﺰاﻳﺪﻳﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤‪ ، ⎣⎡0,1‬إذن )‪ f (0) ≤ f (un ) ≤ f (1‬وﺑﻤﺎ أن ) ‪ (C‬ﻳﻮﺟﺪ ﺗﺡﺖ‬‫اﻟﻤﺴﺘﻘﻴﻢ )∆( ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤‪ ، ⎡⎣0,1‬ﻓﺈن ‪. 0 ≤un+1 ≤ f (1) ≤1‬‬ ‫‪∀n ∈` : 0 ≤un ≤1‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ‪:‬‬‫‪ .2‬ﺣﺴﺐ اﻟﺴﺆال ‪.4.II‬ب ‪ ،‬ﻟﺪﻳﻨﺎ ‪ .∀x ∈ ⎡⎣0,1⎤⎦ : x − f (x ) ≥ 0 :‬إذن ‪.∀x ∈ ⎣⎡0,1⎤⎦ : f (x ) ≤ x :‬‬ ‫وﺑﻤﺎ أن ‪ ، ∀n ∈` : 0 ≤un ≤1 :‬ﻓﺈن ‪،∀n ∈` : f (un ) ≤un :‬‬‫‪∀n ∈` : un+1 ≤un‬‬ ‫وﻣﻨﻩ ﻓﺈن ‪:‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن `∈‪ un n‬ﻣﺘﺘﺎﻟﻴﺔ ﺗﻨﺎﻗﺼﻴﺔ ‪( ).‬‬‫‪ .3‬ﺑﻤﺎ أن `∈‪ un n‬ﻣﺘﺘﺎﻟﻴﺔ ﺗﻨﺎﻗﺼﻴﺔ وﻣﺼﻐﻮرة ﺑﺎﻟﻌﺪد ‪ 0‬ﻓﺈﻧﻬﺎ ﻣﺘﻘﺎرﺑﺔ ‪ .‬ﻟﺘﻜﻦ ‪ l‬ﻧﻬﺎﻳﺘﻬﺎ‪( ).‬‬

‫ﻟﺪﻳﻨﺎ ‪:‬‬ ‫‪ f 9‬داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎤⎦‪. ⎡⎣0,1‬‬ ‫‪ . ∀x ∈⎡⎣0,1⎦⎤ : 0 ≤ f (x ) ≤ x ≤1 9‬إذن ‪( ). f ⎣⎡0,1⎦⎤ ⊂ ⎣⎡0,1⎦⎤ :‬‬ ‫‪. ∀n ∈` : un ∈⎣⎡0,1⎤⎦ 9‬‬ ‫‪ un n∈` 9‬ﻣﺘﻘﺎرﺑﺔ ﻧﻬﺎﻳﺘﻬﺎ ‪( ). l‬‬ ‫إذن ‪ f l = l :‬و ⎦⎤‪( ). l ∈ ⎡⎣0,1‬‬‫‪.‬‬ ‫‪l‬‬ ‫‪=0‬‬ ‫إذن ‪:‬‬ ‫‪.‬‬ ‫‪⎧x −f‬‬ ‫‪(x )>0‬‬ ‫⇔‬ ‫‪x >0‬‬ ‫‪⎪⎪⎨x − f‬‬ ‫‪(x )<0‬‬ ‫⇔‬ ‫ﺣﺴﺐ اﻟﺴﺆال ‪.4.II‬ب ‪ ،‬ﻟﺪﻳﻨﺎ ‪x < 0 :‬‬ ‫‪⎪⎪⎩x − f (x ) = 0 ⇔ x = 0‬‬ ‫‪.‬‬ ‫‪lim u‬‬ ‫‪n‬‬ ‫=‬ ‫‪0‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ‪:‬‬ ‫∞‪n →+‬‬

‫ا‬ ‫ت‬ ‫ا آا‬ ‫‪1‬‬ ‫ا دة‪ :‬ا ت‬ ‫‪2‬‬ ‫ا )ة( ‪ :‬ا م ا‬‫‪C :RS22‬‬ ‫ةا‬ ‫ا آر‬ ‫ا نا‬ ‫ز‪3 :‬‬ ‫ا‬ ‫راآ ‪(2007‬‬ ‫) ا ورة ا‬ ‫‪7:‬‬ ‫ع‬ ‫ا‬ ‫‪ +‬ا م ا را‬ ‫‪+‬ا ما‬ ‫ا‬ ‫لا ا ا (‬ ‫)‬ ‫ا ا ول ) ‪ 3,5‬ن (‬ ‫ﻨﻌﺘﺒﺭ ﻓﻲ ﺍﻝﻔﻀﺎﺀ ﺍﻝﻤﻨﺴﻭﺏ ﺍﻝﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﻤﻨﻅﻡ ) ‪ (o, i , j, k‬ﺍﻝﻨﻘﻁ )‪ A(2, 0, −1‬ﻭ )‪B(2, 4, 2‬‬ ‫‪1‬‬ ‫ﻭ )‪ C(3,3, 3‬ﻭ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﺍﻝﺘﻲ ﻤﻌﺎﺩﻝﺘﻬﺎ ﺍﻝﺩﻴﻜﺎﺭﺘﻴﺔ ﻫﻲ ‪x2 + y2 + z2 − 4x − 4 y − 8z + 20 = 0 :‬‬ ‫‪0,75‬‬ ‫‪ ( 1‬ﺒﻴﻥ ﺍﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻫﻲ ﺍﻝﻨﻘﻁﺔ )‪ Ω(2, 2, 4‬ﻭﺃﻥ ﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ ‪2‬‬ ‫‪1‬‬ ‫‪0,25‬‬ ‫‪ ( 2‬ﻝﻴﻜﻥ )‪ (P‬ﺍﻝﻤﺴﺘﻭﻯ ﺍﻝﻤﺎﺭ ﻤﻥ ﺍﻝﻨﻘﻁﺔ ‪ A‬ﻭ ﺍﻝﻌﻤﻭﺩﻱ ﻋﻠﻰ ﺍﻝﻤﺴﺘﻘﻴﻡ )‪. (BC‬‬ ‫‪0,5‬‬ ‫ﺒﻴﻥ ﺃﻥ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ )‪ (P‬ﻫﻲ ‪x − y + z −1 = 0 :‬‬ ‫‪ ( 3‬ﺃ – ﺒﻴﻥ ﺃﻥ ﺍﻝﻤﺴﺘﻭﻯ )‪ (P‬ﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻭﻓﻕ ﺩﺍﺌﺭﺓ )‪ (Γ‬ﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ ‪. 1‬‬ ‫ﺏ – ﺤﺩﺩ ﺘﻤﺜﻴﻼ ﺒﺎﺭﺍﻤﺘﺭﻴﺎ ﻝﻠﻤﺴﺘﻘﻴﻡ )∆( ﺍﻝﻤﺎﺭ ﻤﻥ ‪ Ω‬ﻭ ﺍﻝﻌﻤﻭﺩﻱ ﻋﻠﻰ )‪. (P‬‬ ‫ﺝ‪ -‬ﺤﺩﺩ ﻤﺜﻠﻭﺙ ﺍﺤﺩﺍﺜﻴﺎﺕ ﺍﻝﻨﻘﻁﺔ ‪ω‬ﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ )‪. (Γ‬‬ ‫ا ا ) ‪ 2,5‬ن (‬ ‫ﻴﺤﺘﻭﻱ ﻜﻴﺱ ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩﻗﺎﺕ ﺒﻴﻀﺎﺀ ﻭ ﺃﺭﺒﻊ ﺒﻴﺩﻗﺎﺕ ﺴﻭﺩﺍﺀ ) ﻻ ﻴﻤﻜﻥ ﺍﻝﺘﻤﻴﻴﺯ ﺒﻴﻥ ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺒﺎﻝﻠﻤﺱ(‪.‬‬ ‫‪0,75‬‬ ‫ﻨﺴﺤﺏ ﻋﺸﻭﺍﺌﻴﺎ ﻭﻓﻲ ﺁﻥ ﻭﺍﺤﺩ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﺍﻝﻜﻴﺱ ‪.‬‬ ‫‪0,75‬‬ ‫‪ ( 1‬ﻤﺎ ﻫﻭ ﺍﺤﺘﻤﺎل ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺘﻴﻥ ﺒﺎﻝﻀﺒﻁ ﻝﻭﻨﻬﻤﺄﺒﻴﺽ ؟‬ ‫‪1‬‬ ‫‪ ( 2‬ﻤﺎ ﻫﻭ ﺍﺤﺘﻤﺎل ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩﻗﺎﺕ ﻤﻥ ﻨﻔﺱ ﺍﻝﻠﻭﻥ ؟‬ ‫‪1‬‬ ‫‪ ( 3‬ﻤﺎ ﻫﻭ ﺍﺤﺘﻤﺎل ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ ﻋﻠﻰ ﺍﻷﻗل ؟‬ ‫‪0,5‬‬ ‫‪0,5‬‬ ‫ا ا )‪3‬ن(‬ ‫‪1‬‬ ‫‪.ℕ‬‬ ‫ﻤﻥ‬ ‫‪n‬‬ ‫ﻝﻜل‬ ‫‪un+1‬‬ ‫=‬ ‫‪1‬‬ ‫‪(un‬‬ ‫‪−‬‬ ‫‪4n‬‬ ‫)‪−1‬‬ ‫ﻭ‬ ‫‪u0‬‬ ‫‪=2‬‬ ‫ﺍﻝﻤﺘﺘﺎﻝﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﺒﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﻝﺘﻜﻥ ) ‪(un‬‬ ‫‪5‬‬ ‫ﻨﻀﻊ ‪ vn = un + n −1‬ﻝﻜل ‪ n‬ﻤﻥ ‪. ℕ‬‬ ‫‪.1‬‬ ‫ﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ‬ ‫) ‪(vn‬‬ ‫‪ ( 1‬ﺒﻴﻥ ﺃﻥ‬ ‫‪5‬‬ ‫‪ ( 2‬ﺃ – ﺍﺤﺴﺏ ‪ vn‬ﺒﺩﻻﻝﺔ ‪. n‬‬ ‫‪.‬‬ ‫‪lim‬‬ ‫‪un‬‬ ‫ﺜﻡ ﺍﺤﺴﺏ‬ ‫‪n‬‬ ‫ﺒﺩﻻﻝﺔ‬ ‫‪un‬‬ ‫ﺏ‪ -‬ﺍﺴﺘﻨﺘﺞ‬ ‫∞‪x→+‬‬ ‫‪ ( 3‬ﻨﻀﻊ ‪ Tn = v0 + v1 + .............. + vn‬ﻭ ‪ Sn = u0 + u1 + ............. + un‬ﺤﻴﺙ ‪ n‬ﻋﻨﺼﺭ ﻤﻥ ‪. ℕ‬‬ ‫‪.ℕ‬‬ ‫‪ n‬ﻤﻥ‬ ‫ﻝﻜل‬ ‫‪Sn‬‬ ‫=‬ ‫‪Tn‬‬ ‫‪−‬‬ ‫‪(n‬‬ ‫‪+ 1)(n‬‬ ‫‪−‬‬ ‫)‪2‬‬ ‫ﻭ ﺃﻥ‬ ‫‪Tn‬‬ ‫=‬ ‫‪1‬‬ ‫‪‬‬ ‫‪5‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪‬‬ ‫ﺒﻴﻥ ﺃﻥ ‪:‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪5n‬‬ ‫‪‬‬‫‪Prof : MISSOURI mohamed‬‬

‫ا‬ ‫ا نا ا آ ر‬ ‫ات‬ ‫ا دة ‪:‬‬ ‫) ا ورةا راآ ‪(2007‬‬ ‫‪2‬‬ ‫اع‬ ‫ا )ة( ا م ا ا ‪ +‬ا م‬ ‫‪2‬‬ ‫‪ +‬ا م ا را‬ ‫ا‬‫‪C :RS22‬‬‫‪z2 − ( 2 + 2)z + 2 +‬‬ ‫ﺍﻝﺘﻤﺭﻴﻥ ﺍﻝﺭﺍﺒﻊ )‪ 3‬ﻥ (‬ ‫‪0,25‬‬ ‫‪ (1‬ﺘﺤﻘﻕ ﻤﻥ ﺃﻥ ‪. ( 2 + 2i)2 = −2 + 4 2i :‬‬ ‫‪0,75‬‬ ‫‪ (2‬ﺤل ﻓﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻝﻌﻘﺩﻴﺔ ‪ ℂ‬ﺍﻝﻤﻌﺎﺩﻝﺔ ‪2 − 2i = 0 :‬‬ ‫‪ (3‬ﻨﻌﺘﺒﺭ ﺍﻝﻌﺩﺩﻴﻥ ﺍﻝﻌﻘﺩﻴﻴﻥ ‪ z1 = 1− i‬ﻭ ‪. z2 = 1+ 2 + i‬‬ ‫‪0,5‬‬ ‫‪1‬‬ ‫ﺃ – ﺤﺩﺩ ﺍﻝﺸﻜل ﺍﻝﻤﺜﻠﺜﻲ ﻝﻠﻌﺩﺩ ﺍﻝﻌﻘﺩﻱ ‪. z1‬‬ ‫‪0,5‬‬ ‫ﺏ – ﺒﻴﻥ ﺃﻥ ‪ z2 ) z1.z2 = 2z2 :‬ﻫﻭ ﻤﺭﺍﻓﻕ ﺍﻝﻌﺩﺩ ‪. ( z2‬‬ ‫ﺍﺴﺘﻨﺘﺞ ﺃﻥ ‪arg(z1) + 2 arg(z2 ) ≡ 0[2π ] :‬‬ ‫ﺝ – ﺤﺩﺩ ﻋﻤﺩﺓ ﻝﻠﻌﺩﺩ ‪. z2‬‬ ‫ﻤﺴﺄﻝﺔ ) ‪ 8‬ﻥ(‬‫ﻝﺘﻜﻥ ‪ g‬ﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞‪ ]0, +‬ﺒﻤﺎ ﻴﻠﻲ ‪. g(x) = x − 1 − 2 ln x :‬‬ ‫‪(I‬‬ ‫‪x‬‬‫‪ g‬ﻋﻠﻰ [∞‪. ]0, +‬‬ ‫ﻤﻥ [∞‪ ]0, +‬ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﻤﻨﺤﻰ ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ‬ ‫‪x‬‬ ‫ﻝﻜل‬ ‫‪g‬‬ ‫)‪'(x‬‬ ‫=‬ ‫‪(x‬‬ ‫‪− 1) 2‬‬ ‫ﺒﻴﻥ ﺃﻥ‬ ‫‪(1‬‬ ‫‪1‬‬ ‫‪x2‬‬ ‫‪0,5‬‬‫‪ (2‬ﺒﻴﻥ ﺃﻥ ‪ g(x) ≤ 0‬ﻝﻜل ‪ x‬ﻤﻥ ]‪ ]0,1‬ﻭ ﺃﻥ ‪ g(x) ≥ 0‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪ )[1, +‬ﻻﺤﻅ ﺃﻥ ‪. ( g(1) = 0‬‬ ‫‪0,75‬‬ ‫‪0,25‬‬‫‪ (II‬ﻨﻌﺘﺒﺭ ﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ‪ f‬ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞‪ ]0, +‬ﺒﻤﺎ ﻴﻠﻲ ‪. f (x) = x + 1 − (ln x)2 − 2 :‬‬ ‫‪0,5‬‬ ‫‪x‬‬ ‫‪10,5‬‬ ‫ﻝﻴﻜﻥ )‪ (C‬ﺍﻝﻤﻨﺤﻨﻰ ﺍﻝﻤﻤﺜل ﻝﻠﺩﺍﻝﺔ ‪ f‬ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﻤﻨﻅﻡ ) ‪. (o, i , j‬‬ ‫‪1,5‬‬ ‫= ‪ ( t‬ﺜﻡ ﺍﺤﺴﺏ )‪. lim f (x‬‬ ‫‪x‬‬ ‫ﻴﻤﻜﻥ ﻭﻀﻊ‬ ‫)‬ ‫‪(ln x)2‬‬ ‫ﺃ – ﺒﻴﻥ ﺃﻥ‬ ‫‪(1‬‬ ‫‪1‬‬ ‫‪lim‬‬ ‫∞‪x→+‬‬ ‫∞‪xx→+‬‬ ‫‪0,5‬‬ ‫‪0,75‬‬ ‫ﺏ – ﺘﺤﻘﻕ ﻤﻥ ﺃﻥ ‪ f (1) = f (x) :‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪. ]0, +‬‬ ‫‪0,75‬‬ ‫‪x‬‬ ‫ﺝ – ﺍﺤﺴﺏ )‪ ) lim f (x‬ﻴﻤﻜﻥ ﻭﻀﻊ ‪ ( t = 1‬ﺜﻡ ﺃﻭل ﺍﻝﻨﺘﻴﺠﺔ ﻫﻨﺩﺴﻴﺎ ‪.‬‬ ‫‪x x→0‬‬ ‫‪x≻0‬‬‫ﺩ – ﺒﻴﻥ ﺃﻥ )‪ (C‬ﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ ﻫﻲ ‪. y = x :‬‬‫‪ (2‬ﺒﻴﻥ ﺃﻥ ‪ f '(x) = g(x) :‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪ ، ]0, +‬ﺜﻡ ﻀﻊ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ ‪. f‬‬ ‫‪x‬‬ ‫‪ (3‬ﺃﻨﺸﺊ ﺍﻝﻤﻨﺤﻨﻰ )‪ (C‬ﻓﻲ ﺍﻝﻤﻌﻠﻡ ) ‪. (o, i , j‬‬ ‫‪ (4‬ﺃ ‪ -‬ﺒﻴﻥ ﺃﻥ ﺍﻝﺩﺍﻝﺔ ‪ G : x ln x − x‬ﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ ‪ g : x → ln x‬ﻋﻠﻰ [∞‪. ]0, +‬‬ ‫ﺏ‪ -‬ﺒﺎﺴﺘﻌﻤﺎل ﻤﻜﺎﻤﻠﺔ ﺒﺎﻷﺠﺯﺍﺀ ‪ ،‬ﺒﻴﻥ ﺃﻥ ‪∫. e (ln x)2dx = e − 2 :‬‬ ‫‪1‬‬ ‫ﺝ – ﺤﺩﺩ ﻤﺴﺎﺤﺔ ﺤﻴﺯ ﺍﻝﻤﺴﺘﻭﻯ ﺍﻝﻤﺤﺼﻭﺭ )‪ (C‬ﻭ ﻤﺤﻭﺭ ﺍﻷﻓﺎﺼﻴل ﻭ ﺍﻝﻤﺴﺘﻘﻴﻤﻴﻥ‬ ‫ﺍﻝﻠﺫﻴﻥ ﻤﻌﺎﺩﻝﺘﺎﻫﻤﺎ ‪ x = 1 :‬ﻭ ‪. x = e‬‬

‫ﺍﻝﺸﻌﺏ‪ :‬ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ ﺍﻷﺼﻴﻠﺔ‬ ‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ‬ ‫ﺍﻝﻌﻠﻭﻡ ﺍﻝﺘﺠﺭﻴﺒﻴﺔ‬ ‫ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬ ‫ﺍﻝﻌﻠﻭﻡ ﺍﻝﺯﺭﺍﻋﻴﺔ‬ ‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬ ‫ا ول ‪:‬‬ ‫ا‬‫ﻓﻲ ﺍﻝﻔﻀﺎﺀ ﺍﻝﻤﻨﺴﻭﺏ ﺇﻝﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﻤﻨﻅﻡ ) ‪ (o, i , j, k‬ﻝﺩﻴﻨﺎ ﺍﻝﻨﻘﻁ )‪ A(2, 0, −1‬ﻭ )‪B(2, 4, 2‬‬‫ﻭ )‪ C(3,3, 3‬ﻭ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﺍﻝﺘﻲ ﻤﻌﺎﺩﻝﺘﻬﺎ ﺍﻝﺩﻴﻜﺎﺭﺘﻴﺔ ﻫﻲ ‪x2 + y2 + z2 − 4x − 4y − 8z + 20 = 0 :‬‬ ‫‪1‬‬ ‫‪0,75‬‬ ‫‪ ( 1‬ﻨﺒﻴﻥ ﺍﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻫﻲ ﺍﻝﻨﻘﻁﺔ )‪ Ω(2, 2, 4‬ﺃﻥ ﺸﻌﺎﻋﻬﺎ ﻴﺴﺎﻭﻱ ‪2‬‬ ‫‪1‬‬‫‪∀M (x, y, z) ∈ (S) ⇔ x2 + y2 + z2 − 4x − 4 y − 8z + 20 = 0‬‬ ‫‪0,25‬‬‫‪⇔ (x2 − 4x) + ( y2 − 4 y) + (z2 − 8z) + 20 = 0‬‬ ‫‪0,5‬‬‫‪⇔ (x2 − 4x + 4) − 4 + ( y2 − 4 y + 4) − 4 + (z2 − 8z + 16) − 16 + 20 = 0‬‬ ‫‪⇔ (x − 2)2 + ( y − 2)2 + (z − 4)2 = 22‬‬ ‫ﺇﺫﻥ ﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻫﻲ ﺍﻝﻨﻘﻁﺔ )‪ Ω(2, 2, 4‬ﻭ ﺸﻌﺎﻋﻬﺎ ‪.R= 2‬‬ ‫‪ ( 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 ‪.1‬‬‫ﻝﺩﻴﻨﺎ ﻤﻌﺎﺩﻝﺔ ﺩﻴﻜﺎﺭﺘﻴﺔ ﻝﻠﻤﺴﺘﻭﻯ )‪ (P‬ﻫﻲ ‪ x − y + z −1 = 0 :‬ﻭﻤﺭﻜﺯ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻫﻲ ﺍﻝﻨﻘﻁﺔ )‪Ω (2, 2, 4‬‬ ‫ﻭﻝﺩﻴﻨﺎ ‪R = 2‬‬ ‫ﻝﺩﻴﻨﺎ ‪d (Ω, (P)) = 2 − 2 + 4 −1 = 3 = 3‬‬ ‫‪12 + (−1)2 +12 3‬‬‫ﺒﻤﺎ ﺃﻥ ‪ d (Ω, (P)) ≺ R‬ﺇﺫﻥ ﺍﻝﻤﺴﺘﻭﻯ )‪ (P‬ﻴﻘﻁﻊ ﺍﻝﻔﻠﻜﺔ )‪ (S‬ﻭﻓﻕ ﺩﺍﺌﺭﺓ )‪ (Γ‬ﺸﻌﺎﻋﻬﺎ ‪ r‬ﺤﻴﺙ ‪:‬‬ ‫‪r = R2 − d 2 = 22 − 32 = 4 − 3 = 1‬‬ ‫ﺏ‪ -‬ﻨﺤﺩﺩ ﺘﻤﺜﻴﻼ ﺒﺎﺭﺍ ﻤﺘﺭﻴﺎ ﻝﻠﻤﺴﺘﻘﻴﻡ )∆( ﺍﻝﻤﺎﺭ ﻤﻥ ‪ Ω‬ﻭ ﺍﻝﻌﻤﻭﺩﻱ ﻋﻠﻰ )‪. (P‬‬‫ﻝﺩﻴﻨﺎ 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 ‫ﺝ‪ -‬ﻨﺤﺩﺩ ﻤﺜﻠﻭﺙ ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻝﻨﻘﻁﺔ ‪ ω‬ﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ )‪. (Γ‬‬ ‫‪ ω‬ﻤﺭﻜﺯ ﺍﻝﺩﺍﺌﺭﺓ )‪ (Γ‬ﻫﻲ ﺘﻘﺎﻁﻊ )∆( ﻭ )‪. (P‬‬‫)∆( ∈ ‪ ω‬و )‪{ω} = (∆) ∩ (P) ⇔ ω ∈ (P‬‬ ‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪t‬‬ ‫‪‬‬‫‪⇔ (1) :‬‬ ‫و‪x − y + z −1 = 0‬‬ ‫‪(2) : y=2-t‬‬ ‫‪z = 4 + t‬‬ ‫‪(2 + t) − (2 − t)(2 + t) − (2 − t) + (4 + t) −1 = 0‬‬ ‫ض )‪(1) (2‬‬ ‫‪:‬‬ ‫‪t = −1‬‬

‫ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬ ‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ‬ ‫ﺇﺫﻥ )‪.ω(1,3,3‬‬ ‫ﻨﻌﻭﺽ ﻗﻴﻤﺔ ‪ t = -1‬ﻓﻲ )‪ (2‬ﻨﺤﺼل ﻋﻠﻰ‬ ‫‪x‬‬ ‫=‬ ‫‪2‬‬ ‫‪+‬‬ ‫)‪(−1‬‬ ‫=‬ ‫‪1‬‬ ‫‪ y‬‬ ‫=‬ ‫‪2‬‬ ‫‪−‬‬ ‫)‪(−1‬‬ ‫=‬ ‫‪3‬‬ ‫‪z = 4 + (−1) = 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 ‫‪ (1‬ﺍﻝﺤﺩﺙ ‪ \" A‬ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺘﻴﻥ ﺒﺎﻝﻀﺒﻁ ﻝﻭﻨﻬﻤﺎ ﺃﺒﻴﺽ \"ﺃﻱ ) ‪( B, B, N‬‬ ‫‪0,75‬‬ ‫‪0,75‬‬ ‫ﺇﺫﻥ ‪p( A) = card ( A) = 12‬‬ ‫ﻝﺩﻴﻨﺎ ‪card ( A) = C32 ⋅C41 = 12‬‬ ‫‪card (Ω) 35‬‬ ‫‪1‬‬ ‫‪ (2‬ﺍﻝﺤﺩﺙ ‪ \" B‬ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺜﻼﺙ ﺒﻴﺩ ﻗﺎﺕ ﻤﻥ ﻨﻔﺱ ﺍﻝﻠﻭﻥ \"‪.‬ﺍﻱ ) ‪ ( N , N, N‬أو )‪( B, B, B‬‬ ‫‪1‬‬ ‫ﺇﺫﻥ ‪P(B) = card (B) = 5 = 1‬‬ ‫‪card (B) = C33 + C43 = 1 + 4 = 5‬‬ ‫‪0,5‬‬ ‫‪card (Ω) 35 7‬‬ ‫‪0,5‬‬ ‫‪ (3‬ﺍﻝﺤﺩﺙ‪ \" C‬ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ ﻋﻠﻰ ﺍﻷﻗل \"‬ ‫ﺍﻝﺤﺩﺙ ﺍﻝﻤﻀﺎﺩ ‪\" C‬ﻋﺩﻡ ﺍﻝﺤﺼﻭل ﻋﻠﻰ ﺃﻴﺔ ﺒﻴﺩﻗﺔ ﺒﻴﻀﺎﺀ \" ﻴﻌﻨﻲ )ﺍﻝﺒﻴﺩ ﻗﺎﺕ ﺍﻝﺜﻼﺙ ﺍﻝﻤﺴﺤﻭﺒﺔ ﺴﻭﺩﺍﺀ(‬ ‫ﺇﺫﻥ ‪:‬‬ ‫)‪P(C) = card (C‬‬ ‫‪=4‬‬ ‫‪:‬‬ ‫ﺇﺫﻥ‬ ‫‪card (C) = C43‬‬ ‫‪=4‬‬ ‫ﻝﺩﻴﻨﺎ‬ ‫)‪card (Ω‬‬ ‫‪35‬‬ ‫)‪p(C) = 1 − p(C‬‬ ‫‪=1− 4‬‬ ‫‪35‬‬ ‫‪= 31‬‬ ‫‪35‬‬ ‫ا‪:‬‬ ‫ا‬ ‫‪.ℕ‬‬ ‫ﻤﻥ‬ ‫‪n‬‬ ‫ﻝﻜل‬ ‫‪un+1‬‬ ‫=‬ ‫‪1‬‬ ‫‪(un‬‬ ‫‪−‬‬ ‫‪4n‬‬ ‫)‪−1‬‬ ‫ﻭ‬ ‫‪u0‬‬ ‫‪=2‬‬ ‫ﺍﻝﻤﺘﺘﺎﻝﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﺒﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﻝﺘﻜﻥ ) ‪(un‬‬ ‫‪5‬‬ ‫ﻨﻀﻊ ‪ vn = un + n −1‬ﻝﻜل ‪ n‬ﻤﻥ ‪. ℕ‬‬ ‫‪.1‬‬ ‫ﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ‬ ‫) ‪(vn‬‬ ‫‪ ( 1‬ﻨﺒﻴﻥ ﺃﻥ‬ ‫‪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 ‫=‪q‬‬ ‫‪1‬‬ ‫ﺃﺴﺎﺴﻬﺎ‬ ‫ﻫﻨﺩﺴﻴﺔ‬ ‫ﻤﺘﺘﺎﻝﻴﺔ‬ ‫) ‪(vn‬‬ ‫ﺍﺩﻥ‬ ‫‪5‬‬ ‫‪ ( 2‬ﺃ – ﻨﺤﺴﺏ ‪ vn‬ﺒﺩﻻﻝﺔ ‪. n‬‬‫‪vn‬‬ ‫=‬ ‫‪‬‬ ‫‪1‬‬ ‫‪n‬‬ ‫ﺃﻱ‬ ‫‪vn = v0 ⋅ qn‬‬ ‫‪ v0 = u0 + 0 −1 = 2 −1 = 1‬ﺇﺫﻥ‬ ‫ﻭﺤﺩﻫﺎ ﺍﻷﻭل‬ ‫‪q=1‬‬ ‫ﻤﺘﺘﺎﻝﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺃﺴﺎﺴﻬﺎ‬ ‫) ‪(vn‬‬ ‫ﻝﺩﻴﻨﺎ‬ ‫‪‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪5‬‬

2007‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‬ . lim un ‫ﺤﺴﺎﺏ‬ ‫ﺜﻡ‬ n ‫ﺒﺩﻻﻝﺔ‬ un ‫ﺍﺴﺘﻨﺘﺎﺝ‬ -‫ﺏ‬ x→+∞ un =  1 n − n +1 ‫ﻓﺎﻥ‬ ‫ﻭﻤﻨﻪ‬ . un = vn − n +1 ‫ﺇﺫﻥ‬ vn = un + n −1 ‫ﻝﺩﻴﻨﺎ‬ 0,5  5 lim un = −∞ ‫ ﺇﺫﻥ‬lim (−n +1) = −∞ ‫ﻝﺩﻴﻨﺎ‬ ‫ﻭ‬ lim  1 n =0 ‫ ﺇﺫﻥ‬−1 ≺ 1 ≺ 1 ‫ﻝﺩﻴﻨﺎ‬ x→+∞  5  5x→+∞ x→+∞ Sn = u0 + u1 + ............. + un ‫ ﻭ‬Tn = v0 + v1 + .............. + vn (3 Tn = 1  5 − 1  : ‫ﺃﻥ‬ ‫ﻨﺒﻴﻥ‬ 1 4  5n  0,25 Tn = v0 + v1 + .............. + vn = v0 ⋅ 1 − qn+1 1− q 1 −  1 n+1  5  = 1 ⋅ 1− 1 5 = 5  −  1 n+1  4 1  5   = 1  − 5  1 n+1  4  5  5   = 1  −5⋅ 1  4  5 5n+1  = 1  − 1  4  5 5n  Sn = Tn − (n +1)(n − 2) ‫ﻨﺒﻥ ﺃﻥ‬ 2 : ‫ ﺍﺫﻥ‬un = vn − n +1 ‫ﻝﺩﻴﻨﺎ‬Sn = u0 + u1 + ............. + un= (v0 − (−1)) + (v1 − 0) + (v2 −1) + ................ + (vn − (n −1) )= (v0 + v1 + .............. + vn ) − ((−1) + 0 +1 + 2 + ............. + (n −1) )= Tn − ( (−1) + (n − 1) ) ( n + 1) 2= Tn − ( n + 1) ( n − 2 ) 2 : ‫اا‬ ‫ا‬ ( 2 + 2i)2 = −2 + 4 2i : ‫( ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ‬1( 2 + 2i)2 = 2 2 ⋅ 2i + (2i)2 2 +2 = 2 + 4 2i − 4 = −2 + 4 2i

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬ ‫‪ (2‬ﻨﺤل ﻓﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻝﻌﻘﺩﻴﺔ ‪ ℂ‬ﺍﻝﻤﻌﺎﺩﻝﺔ ‪z2 − ( 2 + 2)z + 2 + 2 − 2i = 0 :‬‬ ‫‪0,75‬‬ ‫ﻝﺩﻴﻨﺎ ﻤﻤﻴﺯ ﺍﻝﻤﻌﺎﺩﻝﺔ ﻫﻭ‬ ‫‪( ) ( )∆ = −‬‬ ‫‪2‬‬ ‫‪−‬‬ ‫‪2−‬‬ ‫‪2+2‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪2+‬‬ ‫‪2i‬‬ ‫‪= 2 + 4 2 + 4 − 8 − 4 2 + 4 2i‬‬ ‫‪= −2 + 4 2i‬‬ ‫‪= ( 2 + 2i)2‬‬‫‪( ) ( )2 + 2 + 2 + 2i‬‬ ‫‪2 + 2 − 2 + 2i‬‬‫ا د ه ‪ z1 = 2 = 1− i :‬و ‪z2 = 2 = 2 +1+ 2i‬‬ ‫إذن‬ ‫‪(3‬ﻝﺩﻴﻨﺎ ﺍﻝﻌﺩﺩﻴﻥ ﺍﻝﻌﻘﺩﻴﻴﻥ ‪ z1 = 1− i‬ﻭ ‪. z2 = 1+ 2 + i‬‬ ‫ﺃ – ﻨﺤﺩﺩ ﺍﻝﺸﻜل ﺍﻝﻤﺜﻠﺜﻲ ﻝﻠﻌﺩﺩ ﺍﻝﻌﻘﺩﻱ ‪. z1‬‬ ‫‪ z1 = 1− i = 2‬إذن ‪:‬‬‫= ‪z1 = 1− i‬‬ ‫‪‬‬ ‫‪2−‬‬ ‫‪2‬‬ ‫‪i‬‬ ‫‪‬‬ ‫=‬ ‫‪2‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪sin‬‬ ‫‪π‬‬ ‫‪‬‬ ‫=‬ ‫‪2‬‬ ‫‪‬‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪−‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪sin‬‬ ‫‪‬‬ ‫‪−‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪0,5‬‬ ‫‪2 ‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪ cos‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪1‬‬ ‫ﺏ – ﻨﺒﻴﻥ ﺃﻥ ‪ z2 ) z1.z2 = 2z2 :‬ﻫﻭ ﻤﺭﺍﻓﻕ ﺍﻝﻌﺩﺩ ‪. ( z2‬‬ ‫‪0,5‬‬ ‫ﻝﺩﻴﻨﺎ ‪:‬‬ ‫‪1‬‬‫‪( )z1 ⋅ z2 = (1− i) 1+ 2 + i‬‬‫‪=1+ 2 +i −i −i 2 +1‬‬‫‪=2+ 2 −i 2‬‬‫‪( )= 2 2 +1− i‬‬ ‫‪= 2z2‬‬ ‫ا‬ ‫ج ‪arg(z1) + 2 arg(z2 ) ≡ 0[2π ] :‬‬‫‪ z1.z2 = 2z2‬ﺍﺫﻥ ‪ arg ( z1.z2 ) ≡ arg 2z2 [2π ] :‬ﺃﻱ ] ‪( ) ( )arg(z1) + arg(z2) ≡ arg 2 + arg ( z2 )[2π‬‬‫و أن ] ‪ arg ( z2 ) ≡ − arg ( z2 )[2π‬و ] ‪ arg 2 ≡ 0[2π‬ن ] ‪( )arg(z1) + 2 arg(z2 ) ≡ 0[2π‬‬‫‪arg‬‬ ‫(‬ ‫‪z2‬‬ ‫)‬ ‫≡‬ ‫‪−‬‬ ‫‪π‬‬ ‫[‬ ‫‪2π‬‬ ‫]‬ ‫ﺇﺫﻥ‬ ‫‪arg‬‬ ‫(‬ ‫‪z1‬‬ ‫)‬ ‫≡‬ ‫‪−‬‬ ‫‪π‬‬ ‫[‬ ‫‪2π‬‬ ‫]‬ ‫ﻭ‬ ‫ﺝ – ﻨﺤﺩﺩ ﻋﻤﺩﺓ ﻝﻠﻌﺩﺩ ‪z2‬‬ ‫‪8‬‬ ‫‪4‬‬ ‫ﻝﺩﻴﻨﺎ ] ‪arg(z1) + 2 arg(z2 ) ≡ 0[2π‬‬ ‫‪:‬‬ ‫‪ (I‬ﻝﺩﻴﻨﺎ ‪ g‬ﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞‪ ]0, +‬ﺒﻤﺎ ﻴﻠﻲ ‪. g(x) = x − 1 − 2 ln x :‬‬ ‫‪x‬‬‫[∞‪. ]0, +‬‬ ‫‪ g‬ﻋﻠﻰ‬ ‫ﻤﻥ [∞‪ ]0, +‬ﺜﻡ ﻨﺴﺘﻨﺘﺞ ﻤﻨﺤﻰ ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ‬ ‫ﻝﻜل ‪x‬‬ ‫‪g‬‬ ‫)‪'( x‬‬ ‫=‬ ‫‪(x‬‬ ‫‪− 1) 2‬‬ ‫‪ (1‬ﻨﺒﻴﻥ ﺃﻥ‬ ‫‪x2‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬ ‫‪∀x ∈ ]0, +∞[ :‬‬ ‫)‪g '(x‬‬ ‫‪=1+‬‬ ‫‪1‬‬ ‫‪− 2 ln‬‬ ‫‪x‬‬ ‫‪x2‬‬ ‫‪=1+‬‬ ‫‪1‬‬ ‫⋅‪− 2‬‬ ‫‪1‬‬ ‫‪x2‬‬ ‫‪x‬‬ ‫=‬ ‫‪x2‬‬ ‫‪+1−‬‬ ‫‪2x‬‬ ‫‪x2‬‬ ‫=‬ ‫(‬ ‫‪x‬‬ ‫‪− 1)2‬‬ ‫‪x2‬‬ ‫‪ ( x −1)2 ≥ 0‬ﻭ ‪. x2 ≻ 0‬‬ ‫ﻝﻜل ‪ x‬ﻤﻥ [∞‪]0, +‬‬ ‫ا ل [∞‪]0, +‬‬ ‫نا ا ‪ g‬ا‬ ‫إذن ‪ ∀x ∈ ]0, +∞[ : g '( x) ≥ 0‬و‬ ‫ا ل ]‪ ]0,1‬إذن‬ ‫ا ا ‪ g‬ﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞‪ ]0, +‬و‬ ‫‪(2‬‬ ‫)‪∀x ∈ ]0,1] ⇒ 0 ≺ x ≤ 1⇒ g(x) ≤ g(1‬‬ ‫أن ‪ g(1) = 0‬ن ‪ x g(x) ≤ 0‬ا ل ]‪]0,1‬‬ ‫‪0,5‬‬ ‫‪0,75‬‬ ‫ا ا ‪ g‬ﺘﺯﺍﻴﺩﻴﺔ ﻋﻠﻰ ﺍﻝﻤﺠﺎل [∞‪ [1, +‬ﺇﺫﻥ )‪∀x ∈[1, +∞[ ⇒ 1 ≤ x ⇒ g(1) ≤ g(x‬‬ ‫‪0,25‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ g(1) = 0‬ن ‪ x g(x) ≥ 0‬ا ل [∞‪[1, +‬‬ ‫‪ (II‬ﺍﻝﺩﺍﻝﺔ ﺍﻝﻌﺩﺩﻴﺔ ‪ f‬ﺍﻝﻤﻌﺭﻓﺔ ﻋﻠﻰ [∞‪ ]0, +‬ﺒﻤﺎ ﻴﻠﻲ ‪f (x) = x + 1 − (ln x)2 − 2 :‬‬ ‫‪x‬‬ ‫= ‪ ( t‬ﺜﻡ ﻨﺤﺴﺏ )‪lim f (x‬‬ ‫‪x‬‬ ‫) ﻴﻤﻜﻥ ﻭﻀﻊ‬ ‫‪(ln x)2‬‬ ‫ﺃ – ﻨﺒﻴﻥ ﺃﻥ‬ ‫‪(1‬‬ ‫‪lim‬‬ ‫∞‪x→+‬‬ ‫∞‪xx→+‬‬ ‫ﻨﻀﻊ ‪ t = x‬ﺇﺫﻥ ‪ x = t2‬ﻋﻨﺩﻤﺎ ∞‪ x → +‬ﻓﺎﻥ ∞‪t → +‬‬ ‫= ‪( )(ln x)2‬‬ ‫‪ln t2‬‬ ‫‪2‬‬ ‫‪(2ln t )2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪x‬‬ ‫‪t2‬‬ ‫‪‬‬ ‫‪‬‬ ‫=‬ ‫‪t2‬‬ ‫=‬ ‫‪4‬‬ ‫‪ln‬‬ ‫‪t‬‬ ‫ﻝﺩﻴﻨﺎ‬ ‫‪t‬‬ ‫‪(ln x)2‬‬ ‫=‬ ‫‪lim‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪ln‬‬ ‫‪t‬‬ ‫‪2‬‬ ‫‪=0‬‬ ‫‪ lim ln t‬إذن‬ ‫‪=0‬‬ ‫‪‬‬ ‫‪t‬‬ ‫‪‬‬ ‫∞‪tt →+‬‬ ‫‪lim‬‬ ‫∞‪t →+‬‬ ‫∞‪xx→+‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪− (ln‬‬ ‫‪x)2‬‬ ‫‪−2‬‬ ‫=‬ ‫‪‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪(ln x)2‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪x 1+‬‬ ‫‪x2‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪lim‬‬ ‫‪f‬‬ ‫∞‪( x) = +‬‬ ‫ﺇﺫﻥ‬ ‫‪lim 1‬‬ ‫و‪=0‬‬ ‫‪(ln x)2‬‬ ‫‪=0‬‬ ‫و‬ ‫‪lim‬‬ ‫‪2 =0‬‬ ‫ﻝﺩﻴﻨﺎ‬ ‫‪xx→+∞ 2‬‬ ‫∞‪x→+‬‬ ‫‪lim‬‬ ‫∞‪xx→+‬‬ ‫∞‪xx→+‬‬ ‫ب – ﻨﺘﺤﻘﻕ ﻤﻥ ﺃﻥ ‪ f (1) = f (x) :‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪. ]0, +‬‬ ‫‪x‬‬ ‫ﻝﻜل ‪ x‬ﻤﻥ [∞‪ ]0, +‬ﻝﺩﻴﻨﺎ‬‫‪f‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪‬‬ ‫‪ln‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪ 2‬‬ ‫‪−2‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪ ‬‬ ‫‪x‬‬ ‫‪= 1 + x − (− ln x)2 − 2‬‬ ‫‪x‬‬ ‫‪= 1 + x − (ln x)2 − 2‬‬ ‫‪x‬‬ ‫)‪= f (x‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬ ‫ج – ﻨﺤﺴﺏ )‪lim f (x‬‬ ‫‪0,5‬‬ ‫‪0,5‬‬ ‫‪x→0‬‬ ‫‪1,5‬‬ ‫‪x≻0‬‬‫‪lim‬‬ ‫‪f‬‬ ‫)‪(x‬‬ ‫=‬ ‫‪lim‬‬ ‫‪f‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫=‬ ‫‪lim‬‬ ‫‪f‬‬ ‫)‪(t‬‬ ‫=‬ ‫∞‪+‬‬ ‫∞‪ t → +‬ﻭ ﻤﻨﻪ ﻓﺎﻥ‬ ‫‪ x → 0+‬ﻓﺎﻥ‬ ‫ﺇﺫﻥ ﻋﻨﺩﻤﺎ‬ ‫‪t=1‬‬ ‫ﻨﻀﻊ‬ ‫‪‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪x‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫∞‪t →+‬‬‫‪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 :‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪(ln x)2‬‬ ‫‪2‬‬ ‫‪‬‬ ‫)‪(x‬‬ ‫‪1‬‬ ‫‪x2‬‬ ‫‪x‬‬ ‫‪‬‬ ‫ﻭ‪=1‬‬ ‫ﻝﺩﻴﻨﺎ ∞‪lim f ( x) = +‬‬ ‫‪lim‬‬ ‫‪f‬‬ ‫=‬ ‫‪lim‬‬ ‫‪+‬‬ ‫‪−‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬ ‫∞‪x→+‬‬‫∞‪ lim f ( x) − x = lim 1 − (ln x)2 − 2 = −‬ﺇﺫﻥ )‪ (C‬ﻴﻘﺒل ﻓﺭﻋﺎ ﺸﻠﺠﻤﻴﺎ ﺍﺘﺠﺎﻫﻪ ﺍﻝﻤﻘﺎﺭﺏ ﻫﻭ ﺍﻝﻤﺴﺘﻘﻴﻡ ﺍﻝﺫﻱ ﻤﻌﺎﺩﻝﺘﻪ‬ ‫∞‪x→+‬‬ ‫∞‪xx→+‬‬ ‫ﻫﻲ ‪y = x :‬‬ ‫‪ (2‬ﺒﻴﻥ ﺃﻥ ‪ f '(x) = g(x) :‬ﻝﻜل ‪ x‬ﻤﻥ [∞‪]0, +‬‬ ‫‪x‬‬ ‫‪f‬‬ ‫)‪'(x‬‬ ‫‪=1−‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪2 ( ln‬‬ ‫‪x ) ( ln‬‬ ‫')‪x‬‬ ‫‪x2‬‬ ‫‪=1−‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪2 ( ln‬‬ ‫⋅)‪x‬‬ ‫‪1‬‬ ‫‪x2‬‬ ‫‪x‬‬ ‫=‬ ‫‪1‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪ln‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪‬‬ ‫)‪= g(x‬‬ ‫‪x‬‬ ‫ﺇﺸﺎﺭﺓ )‪ f '( x‬ﻫﻲ ﺇﺸﺎﺭﺓ )‪g ( x‬‬ ‫ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻝﺩﺍﻝﺔ ‪f‬‬‫‪x0‬‬ ‫‪1‬‬ ‫∞‪+‬‬‫‪f '(x) - φ‬‬ ‫‪+‬‬‫∞‪f ( x) +‬‬ ‫∞‪+‬‬ ‫‪0‬‬ ‫‪ (3‬ا‬ ‫‪1‬‬

‫ﺘﺼﺤﻴﺢ ﻤﻭﻀﻭﻉ ﺍﻻﻤﺘﺤﺎﻥ ﺍﻝﻭﻁﻨﻲ ﺍﻝﻤﻭﺤﺩ ﻝﻠﺒﺎﻜﺎﻝﻭﺭﻴﺎ ﻤﺎﺩﺓ ﺍﻝﺭﻴﺎﻀﻴﺎﺕ ﺍﻝﺩﻭﺭﺓ ﺍﻻﺴﺘﺩﺭﺍﻜﻴﺔ‪2007‬‬ ‫‪0,5‬‬ ‫‪0,75‬‬ ‫‪ ( 4‬ﺃ ‪ -‬ﻨﺒﻴﻥ ﺃﻥ ﺍﻝﺩﺍﻝﺔ ‪ G : x ln x − x‬ﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ ‪ g : x → ln x‬ﻋﻠﻰ [∞‪]0, +‬‬ ‫‪0,75‬‬ ‫ﻝﺩﻴﻨﺎ‬ ‫‪∀x ∈ ]0, +∞[ : G '( x) = x 'ln x + x (ln x) '−1‬‬ ‫‪= ln x + x ⋅ 1 −1‬‬ ‫‪x‬‬ ‫‪= ln x +1−1‬‬ ‫‪= ln x‬‬ ‫ﺇﺫﻥ ﺍﻝﺩﺍﻝﺔ ‪ G‬ﺩﺍﻝﺔ ﺃﺼﻠﻴﺔ ﻝﻠﺩﺍﻝﺔ ‪.g‬‬ ‫‪∫e‬‬ ‫‪1‬‬ ‫‪x)2dx‬‬ ‫=‬ ‫‪e−2‬‬ ‫‪:‬‬ ‫ﺃﻥ‬ ‫ﻨﺒﻴﻥ‬ ‫‪،‬‬ ‫ﺒﺎﻷﺠﺯﺍﺀ‬ ‫ﻤﻜﺎﻤﻠﺔ‬ ‫ﺒﺎﺴﺘﻌﻤﺎل‬ ‫ﺏ‪-‬‬ ‫‪(ln‬‬ ‫‪‬‬ ‫('‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫‪2‬‬ ‫‪ln‬‬ ‫‪x‬‬ ‫‪u‬‬ ‫(‬ ‫‪x‬‬ ‫)‬ ‫=‬ ‫(‬ ‫‪ln‬‬ ‫‪x‬‬ ‫‪)2‬‬ ‫‪u‬‬ ‫‪‬‬ ‫‪x‬‬ ‫ﺇﺫﻥ‬ ‫ﻨﻀﻊ‬ ‫‪v ( x) = x‬‬ ‫‪v '( x) = 1‬‬ ‫ﺇﺫﻥ‬ ‫‪2‬‬ ‫‪e‬‬ ‫‪e 2 ln x ⋅ x dx‬‬ ‫‪1‬‬ ‫‪1x‬‬ ‫‪x) dx‬‬‫‪∫ ∫e‬‬ ‫‪1‬‬‫‪(ln‬‬ ‫=‬ ‫‪x (ln‬‬ ‫‪x)2‬‬ ‫‪−‬‬ ‫‪x (ln‬‬ ‫‪e‬‬ ‫‪e‬‬ ‫=∫‬ ‫‪x)2‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪2‬‬ ‫‪x‬‬ ‫‪dx‬‬ ‫‪ln‬‬ ‫‪1‬‬ ‫=‬ ‫‪‬‬ ‫‪x‬‬ ‫(‬ ‫‪ln‬‬ ‫‪x)2‬‬ ‫‪e‬‬ ‫‪−‬‬ ‫‪2[x ln‬‬ ‫‪x‬‬ ‫‪−‬‬ ‫‪]x e‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫))‪( )= e(ln e)2 −1(ln1)2 − 2((e ln e − e) − (1ln1−1‬‬ ‫‪=e−2‬‬‫ﺝ – 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 f ( x) dx‬‬ ‫‪e‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪1 ‬‬ ‫‪x‬‬ ‫‪2 ‬‬ ‫‪1‬‬ ‫= ‪∫ ∫A‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪−‬‬ ‫‪(ln‬‬ ‫‪x)2‬‬ ‫‪−‬‬ ‫‪dx‬‬ ‫=∫ ∫‬ ‫‪e‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪‬‬ ‫‪dx‬‬ ‫‪−‬‬ ‫‪e (ln x)2‬‬ ‫‪dx‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2 ‬‬ ‫‪1‬‬ ‫=‬ ‫‪ x2‬‬ ‫‪+ ln‬‬ ‫‪x‬‬ ‫‪e‬‬ ‫)‪− (e − 2‬‬ ‫‪‬‬ ‫‪− 2x‬‬ ‫‪ 2 1‬‬ ‫=‬ ‫‪ e2‬‬ ‫‪+‬‬ ‫‪ln‬‬ ‫‪e‬‬ ‫‪−‬‬ ‫‪2e‬‬ ‫‪‬‬ ‫‪−‬‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪+ ln1 − 2  − e +‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪= e2 + 1 − 2e − 1 + 2 − e + 2‬‬ ‫‪22‬‬ ‫‪= e2 − 3e + 9‬‬ ‫‪22‬‬ ‫ﺇﺫﻥ ‪ A = e2 − 3 e + 9‬ﺒﻭﺤﺩﺓ ﻗﻴﺎﺱ ﺍﻝﻤﺴﺎﺤﺔ‬ ‫‪22‬‬





‫‪))a[†„J)r„J)ÉK[T†¤J)\[nT‬‬ ‫‪)KŠ‚„Kj†R)’RcXT)‡„t)Kc„K‚R)’KV„J‬‬ ‫))‪2008‬‬ ‫‪)Kc„K‚R„„)’aKu„J)‘ca„J‬‬ ‫‪))KŠ‚„j†R)UKX„‚T„J)‡„u„J)’Rul‬‬ ‫‪))C)…¦J)Éhhhhhhhhhhhhhhhhhhhhc†T„J‬‬ ‫‪( )JG JJG JJG‬‬‫ﻧﻌﺘﺒﺮ ﻓﻲ اﻟﻔﻀﺎء اﻟﻤﻨﺴﻮب ﻹﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ وﻣﺒﺎﺷﺮ ‪   O ,i , j , k‬اﻟﻨﻘﻄﺘﻴﻦ )‪ A (0, −1,1‬و )‪B (1, −1,0‬‬ ‫‪. x 2 + y 2 + z 2 − 2x − 4z + 2 = 0‬‬ ‫واﻟﻔﻠﻜﺔ ) ‪   (S‬اﻟﺘﻲ ﻣﻌﺎدﻟﺘﻬﺎ ‪:‬‬ ‫‪ .1‬ﻟﺪیﻨﺎ ‪. x 2 + y 2 + z 2 − 2x − 4z + 2 = 0 ⇔ (x −1)2 + y 2 + (z − 2)2 = 32 :‬‬‫إذن ) ‪ (S‬ﻓﻠﻜﺔ ﻣﺮآﺰهﺎ )‪ Ω (1,0, 2‬وﺷﻌﺎﻋﻬﺎ ‪ . R = 3‬وﻟﺪیﻨﺎ ‪ ، 02 + (−1)2 +12 − 2× 0 − 4×1+ 2 = 0 :‬إذن ) ‪. A ∈(S‬‬‫‪JJJG JJJJG‬‬ ‫‪−1‬‬ ‫‪−1‬‬ ‫‪JG‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪JG‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪JJG‬‬ ‫‪JG‬‬ ‫‪JG‬‬ ‫‪JJG‬‬ ‫⎜⎜⎛ ‪OJJJAJG‬‬ ‫⎞‪0‬‬ ‫‪JJJJJG‬‬ ‫⎞ ‪⎛1‬‬ ‫‪i‬‬ ‫‪0‬‬ ‫‪j‬‬ ‫‪−1‬‬ ‫‪−1‬‬ ‫‪k‬‬ ‫‪i‬‬ ‫‪j‬‬ ‫‪k‬‬ ‫⎟⎟‪−1‬‬ ‫⎜‬ ‫⎟⎟‪−1‬‬‫‪OA ∧OB‬‬ ‫=‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪+‬‬ ‫=‬ ‫‪+‬‬ ‫‪+‬‬ ‫‪ ،‬وﻣﻨﻪ ﻓﺈن ‪:‬‬ ‫و‬ ‫‪OB‬‬ ‫⎜‬ ‫‪ .2‬ﻟﺪیﻨﺎ ‪:‬‬ ‫‪01‬‬ ‫‪⎝⎜1JJJ⎟⎠JG‬‬ ‫⎠⎟ ‪⎝⎜ 0‬‬ ‫‪.‬‬ ‫‪JJJJG‬‬ ‫)‪(1,1,1‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ‪:‬‬ ‫‪∧OB‬‬ ‫‪OA‬‬ ‫‪JJJJG‬‬ ‫)‪∧OJJJBJG(1,1,1‬‬ ‫ﺕﻜﺘﺐ ﻋﻠﻰ ﺷﻜﻞ‬ ‫ﻣﺘﺠﻬﺔ ﻣﻨﻈﻤﻴﺔ ﻋﻠﻰ اﻟﻤﺴﺘﻮى ) ‪ . (OA B‬إذن ﻣﻌﺎدﻟﺔ اﻟﻤﺴﺘﻮى ) ‪(OA B‬‬ ‫‪OA‬‬ ‫ﻟﺪیﻨﺎ ‪:‬‬ ‫‪.3‬‬ ‫‪ ، x + y + z +d = 0‬وﺑﻤﺎ أن ) ‪ ، O ∈(OAB‬ﻓﺈن ‪ x + y + z = 0‬هﻲ ﻣﻌﺎدﻟﺔ دیﻜﺎرﺕﻴﺔ ﻟﻠﻤﺴﺘﻮى ) ‪. (OAB‬‬ ‫‪( ).d Ω,(OAB ) = 1+ 0 + 2 = 3 = 3 = R‬‬ ‫ﻟﻨﺤﺴﺐ ﻣﺴﺎﻗﺔ اﻟﻨﻘﻄﺔ ‪ A‬ﻋﻦ اﻟﻤﺴﺘﻮى ) ‪: (OA B‬‬ ‫‪12 +12 +12 3‬‬ ‫وﻋﻠﻴﻪ ﻓﺈن اﻟﻤﺴﺘﻮى ) ‪ (OAB‬ﻣﻤﺎس ﻟﻠﻔﻠﻜﺔ ) ‪ (S‬ﻓﻲ اﻟﻨﻘﻄﺔ ‪ A‬ﻋﻠﻰ اﻋﺘﺒﺎر أن ) ‪ A ∈(S‬و ) ‪. A ∈(OAB‬‬ ‫‪))C)hhhhhhhhhhhhhhhhhhhhKV„J)Éc†T„J‬‬‫‪ .1‬ﻧﻌﺘﺒﺮ ﻓﻲ اﻟﻤﺠﻤﻮﻋﺔ ^ اﻟﻤﻌﺎدﻟﺔ ‪ . z 2 − 6z + 34 = 0 :‬ﻣﻤﻴﺰ هﺬﻩ اﻟﻤﻌﺎدﻟﺔ هﻮ ‪. ∆ = (−3)2 −1× 34 = 9 − 34 = −25 = (5i )2 :‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ﻟﻠﻤﻌﺎدﻟﺔ اﻟﺴﺎﺑﻘﺔ ﺡﻠﻴﻦ ﻋﻘﺪیﻴﻦ ﻣﺘﺮاﻓﻘﻴﻦ هﻤﺎ ‪:‬‬‫‪z‬‬ ‫=‬ ‫‪−b‬‬ ‫‪′‬‬ ‫‪−i‬‬ ‫‪−∆′ = − (−3) − 5i = 3− 5i‬‬ ‫و‬ ‫‪z‬‬ ‫=‬ ‫‪−b‬‬ ‫‪′‬‬ ‫‪+i‬‬ ‫‪−∆′ = −(−3) + 5i = 3+ 5i‬‬ ‫‪a‬‬ ‫‪a‬‬ ‫‪2‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ﻣﺠﻤﻮﻋﺔ ﺡﻠﻮل اﻟﻤﻌﺎدﻟﺔ هﻲ ‪{ }.S = 3 − 5i , 3 + 5i :‬‬ ‫‪( )JJG JJG‬‬‫اﻟﺘﻮاﻟﻲ‬ ‫ﻋﻠﻰ‬ ‫أﻟﺤﺎﻗﻬﺎ‬ ‫اﻟﺘﻲ‬ ‫‪C‬‬ ‫و‬ ‫‪B‬‬ ‫و‬ ‫‪A‬‬ ‫اﻟﻨﻘﻂ‬ ‫ﻧﻌﺘﺒﺮ‬ ‫‪،‬‬ ‫‪O ,e1,e2‬‬ ‫‪ .2‬ﻓﻲ اﻟﻤﺴﺘﻮى اﻟﻌﻘﺪي اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ وﻣﺒﺎﺷﺮ‬‫‪JG‬‬‫‪ a = 3+ 5i‬و ‪ b = 3− 5i‬و ‪ . c = 7 + 3i‬ﻟﺘﻜﻦ اﻟﻨﻘﻄﺔ )‪ M ′(z ′‬ﺻﻮرة اﻟﻨﻘﻄﺔ ) ‪ M (z‬ﺑﺎﻻزاﺡﺔ ‪ T‬ذات اﻟﻤﺘﺠﻬﺔ ‪ u‬اﻟﺘﻲ‬ ‫‪( )JJJJJJG JG‬‬ ‫‪JG‬‬ ‫ﻟﺤﻘﻬﺎ ‪. 4 − 2i‬‬ ‫‪M ′ =T (M ) ⇔ MM ′ = u ⇔ z ′ = z + aff u ⇔ z ′ = z + 4 − 2i‬‬ ‫أ‪ -‬ﻟﺪیﻨﺎ ‪:‬‬ ‫وﺑﻤﺎ أن ‪ ، a + 4 − 2i = 3 + 5i + 4 − 2i = 7 + 3i = c :‬ﻓﺈن‪ C =T (A ) :‬أي ‪ C‬هﻲ ﺻﻮرة ‪ A‬ﺑﺎﻻزاﺡﺔ ‪. T‬‬ ‫‪( )JJJG JJJG‬‬ ‫ب‪ -‬ﻟﺪیﻨﺎ ‪. b −c = 3 − 5i − 7 − 3i = −4 − 8i = 2i (−4 + 2i ) = 2i :‬‬ ‫‪CA ,CB‬‬ ‫‪a −c 3 + 5i − 7 − 3i −4 + 2i −4 + 2i‬‬ ‫≡‬ ‫‪arg‬‬ ‫⎛‬ ‫‪b‬‬ ‫‪−c‬‬ ‫⎞‬ ‫⎦⎤ ‪⎡⎣2π‬‬ ‫‪b −c‬‬ ‫‪⎣⎢⎡2,‬‬ ‫‪π‬‬ ‫⎤‬ ‫⎜⎝‬ ‫‪a‬‬ ‫‪−c‬‬ ‫⎟⎠‬ ‫‪a −c‬‬ ‫‪2‬‬ ‫⎦⎥‬ ‫‪:‬‬ ‫إذن‬ ‫‪.‬‬ ‫‪= 2i‬‬ ‫=‬ ‫‪:‬‬ ‫ﺝـ‪ -‬ﻟﺪیﻨﺎ‬ ‫‪( )JJJG JJJG‬‬ ‫‪π‬‬ ‫‪2‬‬ ‫⎤⎦ ‪⎡⎣2π‬‬ ‫≡ ‪CA ,CB‬‬ ‫وﻣﻨﻪ ﻓﺈن ‪ ABC‬ﻣﺜﻠﺚ ﻗﺎﺋﻢ اﻟﺰاویﺔ ﻓﻲ ‪ C‬وﻟﺪیﻨﺎ ‪ . CB = b −c = 2 :‬إذن ‪. BC = 2A C :‬‬ ‫‪CA a −c‬‬

‫‪)) )))C)Whhhhhhhhhhhhhhhhhhhh„KV„J)Éc†T„J‬‬ ‫یﺤﺘﻮي ﺻﻨﺪوق ﻋﻠﻰ ﺳﺖ آﺮات ﺡﻤﺮاء وﺙﻼث آﺮات ﺧﻀﺮاء ) ﻻ یﻤﻜﻦ اﻟﺘﻤﻴﻴﺰ ﺑﻴﻨﻬﺎ ﺑﺎﻟﻠﻤﺲ (‬‫‪.‬‬ ‫‪C‬‬ ‫‪p‬‬ ‫‪:‬‬ ‫‪ .1‬ﻧﺴﺤﺐ ﻋﺸﻮاﺋﻴﺎ وﻓﻲ ‪‬א‪ ) ‬اﻟﺘﺮﺕﻴﺐ ﻏﻴﺮ ﻣﻬﻢ ( ﺙﻼث آﺮات ﻣﻦ اﻟﺼﻨﺪوق‪ .‬ﺕﺜﺒﻴﺖ اﻟﺼﻨﻒ ‪ :‬א‪‬‬ ‫‪n‬‬ ‫‪.‬‬ ‫‪C‬‬ ‫‪2‬‬ ‫‪×C‬‬ ‫‪1‬‬ ‫= ‪= 15×3‬‬ ‫‪15‬‬ ‫أ‪ -‬اﺡﺘﻤﺎل اﻟﺤﺼﻮل ﻋﻠﻰ آﺮﺕﻴﻦ ﺡﻤﺮاویﻦ وآﺮة ﺧﻀﺮاء ‪ RRV‬هﻮ ‪:‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪84‬‬ ‫‪28‬‬ ‫‪C‬‬ ‫‪3‬‬ ‫‪9‬‬ ‫اﺡﺘﻤﺎل اﻟﺤﺼﻮل ﻋﻠﻰ آﺮة ﺧﻀﺮاء واﺡﺪة ﻋﻠﻰ اﻷﻗﻞ ‪ RRV‬أو ‪ RV V‬أو ‪ V V V‬هﻮ ‪:‬‬ ‫ب‪ -‬ﻃﺮیﻘﺔ ‪: 1‬‬ ‫‪.‬‬ ‫‪C‬‬ ‫‪62C‬‬ ‫‪1‬‬ ‫‪+C‬‬ ‫‪61C‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪C‬‬ ‫‪3‬‬ ‫= ‪= 15×3+ 6×3+1‬‬ ‫‪16‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪84‬‬ ‫‪21‬‬ ‫‪C‬‬ ‫‪3‬‬ ‫‪9‬‬ ‫ﻧﻀﻊ اﻟﺤﺪث ‪ : A‬اﻟﺤﺼﻮل ﻋﻠﻰ آﺮة ﺧﻀﺮاء واﺡﺪة ﻋﻠﻰ اﻷﻗﻞ  ‪.‬‬ ‫ﻃﺮیﻘﺔ ‪: 2‬‬ ‫اﻟﺤﺪث اﻟﻤﻀﺎد ﻟﻠﺤﺪث ‪ A‬هﻮ ‪ :A :‬اﻟﺤﺼﻮل ﻋﻠﻰ ﺙﻼث آﺮات ﺡﻤﺮاء ‪.  - RRR -‬‬ ‫‪( ). p (A ) =1− p‬‬ ‫‪A‬‬ ‫=‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪C‬‬ ‫‪3‬‬ ‫‪=1−‬‬ ‫‪20‬‬ ‫=‬ ‫‪64‬‬ ‫=‬ ‫‪16‬‬ ‫ﻟﺪیﻨﺎ ‪:‬‬ ‫‪C‬‬ ‫‪6‬‬ ‫‪84‬‬ ‫‪84‬‬ ‫‪21‬‬ ‫‪3‬‬ ‫‪9‬‬ ‫‪ .2‬ﻧﺴﺤﺐ ﻋﺸﻮاﺋﻴﺎ ‪ ) ‬اﻟﺘﺮﺕﻴﺐ ﻣﻬﻢ واﻟﺘﻜﺮار ﻏﻴﺮ وارد ( ﺙﻼث آﺮات ﻣﻦ اﻟﺼﻨﺪوق‪.‬‬ ‫‪.‬‬ ‫‪A‬‬ ‫‪p‬‬ ‫‪:‬‬ ‫ﺕﺜﺒﻴﺖ اﻟﺼﻨﻒ ‪ :‬א‪‬א‪‬‬ ‫‪n‬‬ ‫‪.‬‬ ‫‪A63‬‬ ‫‪= 120‬‬ ‫=‬ ‫‪5‬‬ ‫اﺡﺘﻤﺎل اﻟﺤﺼﻮل ﻋﻠﻰ ﺙﻼث آﺮات ﺡﻤﺮاء هﻮ ‪:‬‬ ‫‪A93‬‬ ‫‪504‬‬ ‫‪21‬‬ ‫‪))C)vhhhhhhhhhhhhhhhhhhhhRJc„J)Éc†T„J‬‬ ‫‪))C)…¦J)•iX„J‬‬ ‫ﻟﺘﻜﻦ ‪ g‬اﻟﺪاﻟﺔ اﻟﻌﺪدیﺔ اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎡⎣∞‪ ⎤⎦0, +‬ﺑﻤﺎ یﻠﻲ ‪. g (x ) = x − 2 ln x :‬‬ ‫‪.‬‬ ‫‪g ′(x‬‬ ‫=)‬ ‫‪(x‬‬ ‫‪− 2ln x‬‬ ‫‪)′‬‬ ‫‪=1−‬‬ ‫‪2‬‬ ‫=‬ ‫‪x −2‬‬ ‫ﻟﻴﻜﻦ ⎣⎡∞‪ ، x ∈ ⎦⎤0, +‬ﻟﺪیﻨﺎ ‪:‬‬ ‫‪ .1‬أ‪-‬‬ ‫‪x‬‬ ‫‪x‬‬ ‫⎡⎣∞‪ . ∀x ∈⎦⎤0, +‬إذن إﺷﺎرة ) ‪ g ′(x‬ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡∞‪ ⎤⎦0, +‬هﻲ إﺷﺎرة ‪. x − 2‬‬ ‫‪:‬‬ ‫ب‪ -‬ﻧﻌﻠﻢ أن ‪g ′(x ) = x − 2 :‬‬ ‫‪x‬‬ ‫وﻟﺪیﻨﺎ ‪ x ∈⎤⎦0, 2⎤⎦ ⇒ x ≤ 2 ⇒ x − 2 ≤ 0 :‬و ‪ . x ∈ ⎣⎡2, +∞⎣⎡ ⇒ x ≥ 2 ⇒ x − 2 ≥ 0‬إذن ‪:‬‬ ‫‪ g‬ﺕﻨﺎﻗﺼﻴﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎤⎦‪ ⎤⎦0, 2‬وﺕﺰایﺪیﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡∞‪ . ⎣⎡2, +‬ﺧﻼﺻﺔ ‪:‬‬

‫‪ .2‬ﺑﻤﺎ أن ‪ ، e > 2 ⇒1> ln 2 ⇒1− ln 2 > 0 :‬ﻓﺈن ‪. g (2) = 2(1− ln 2) > 0 :‬‬ ‫وﻟﺪیﻨﺎ ‪ g (2) = 2 1− ln 2 :‬ﻗﻴﻤﺔ دﻧﻮیﺔ ﻣﻄﻠﻘﺔ ﻟﻠﺪاﻟﺔ ‪ g‬ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡∞‪ ⎤⎦0, +‬ﻋﻨﺪ اﻟﻌﺪد ‪ . 2‬وﻣﻨﻪ ﻓﺈ ن‪( ):‬‬ ‫‪∀x ∈⎤⎦0,+∞⎣⎡ : g (x ) ≥ g (2) > 0‬‬ ‫‪))C)KV„J)•iX„J‬‬ ‫ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدیﺔ ‪ f‬اﻟﻤﻌﺮﻓﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎡⎣∞‪ ⎤⎦0, +‬ﺑﻤﺎ یﻠﻲ ‪. f (x ) = x − (ln x )2 :‬‬ ‫‪ .1‬ﻟﺪیﻨﺎ ‪ ، lim f (x ) = lim x − (ln x )2 = −∞ :‬ﻷن ‪. lim ln x = −∞ :‬‬ ‫‪x →0‬‬ ‫‪x →0‬‬ ‫‪x →0‬‬ ‫‪x >0 x >0 x >0‬‬ ‫اﻟﻤﻨﺤﻨﻰ ) ‪ (C‬یﻘﺒﻞ ﻣﻘﺎرﺑﺎ ﻋﻤﻮدیﺎ ﻣﻌﺎدﻟﺘﻪ ‪. x = 0‬‬ ‫‪ .2‬أ‪ -‬ﻧﻀﻊ ‪ . t = x :‬إذن ‪ . x → +∞ :‬وﺡﻴﺚ أن ‪ ، lim lnt = 0‬ﻓﺈن ‪:‬‬ ‫‪t →+∞ t‬‬ ‫∞‪t →+‬‬ ‫‪( ) ( )lim‬‬‫‪ln x‬‬‫‪2‬‬ ‫‪⎞2‬‬ ‫⎛‬ ‫‪ln‬‬ ‫‪t‬‬ ‫‪2‬‬ ‫‪⎞2‬‬ ‫‪⎞2‬‬ ‫‪x‬‬ ‫⎟‬ ‫⎜‬ ‫‪t‬‬ ‫⎟‬ ‫⎟‬ ‫∞‪x →+‬‬ ‫=‬ ‫⎛‬ ‫‪ln x‬‬ ‫⎠‬ ‫=‬ ‫⎝⎜⎜‬ ‫⎟⎟⎠‬ ‫=‬ ‫⎛‬ ‫×‪2‬‬ ‫‪lnt‬‬ ‫⎠‬ ‫=‬ ‫‪lim‬‬ ‫⎜‬ ‫‪x‬‬ ‫‪t‬‬ ‫‪lim‬‬ ‫‪lim‬‬ ‫⎜‬ ‫‪t‬‬ ‫‪0‬‬ ‫⎝‬ ‫⎝‬ ‫∞‪x →+‬‬ ‫∞‪→+‬‬ ‫∞‪t →+‬‬‫‪( ). lim‬‬‫‪ln x‬‬ ‫‪2‬‬ ‫‪(x‬‬ ‫)‬ ‫=‬ ‫‪lim‬‬ ‫‪x‬‬ ‫‪− (ln x‬‬ ‫‪)2‬‬ ‫‪= lim x‬‬ ‫⎛‬ ‫‪−‬‬ ‫(‬ ‫‪ln‬‬ ‫‪x‬‬ ‫‪)2‬‬ ‫⎞‬ ‫∞‪+‬‬ ‫ب‪ -‬ﻟﺪیﻨﺎ ‪:‬‬ ‫∞‪x →+‬‬ ‫‪x‬‬ ‫∞‪x →+‬‬ ‫‪ ، lim f‬ﻷن ‪= 0 :‬‬ ‫∞‪x →+‬‬ ‫‪⎜1‬‬ ‫=⎟‬ ‫∞‪x →+‬‬ ‫‪⎜⎝ x‬‬ ‫⎠⎟‬ ‫‪.‬‬ ‫‪lim‬‬ ‫‪f‬‬ ‫= ) ‪(x‬‬ ‫‪lim‬‬ ‫⎛‬ ‫‪(ln‬‬ ‫‪x‬‬ ‫‪)2‬‬ ‫⎞‬ ‫=‬ ‫‪1‬‬ ‫وﻟﺪیﻨﺎ ‪:‬‬ ‫⎟‬ ‫‪⎜1−‬‬ ‫‪x →+∞ x‬‬ ‫⎝⎜ ∞‪x →+‬‬ ‫⎠⎟ ‪x‬‬‫‪ ، lim f‬وﺡﺴﺐ اﻟﺴﺆال اﻟﺴﺎﺑﻖ ‪ ،‬ﻓﺈن اﻟﻤﻨﺤﻨﻰ‬ ‫‪(x‬‬ ‫‪)−x‬‬ ‫‪= lim x‬‬ ‫‪− (ln x‬‬ ‫‪)2 − x‬‬ ‫=‬ ‫‪lim −‬‬ ‫‪(ln‬‬ ‫‪x‬‬ ‫‪)2‬‬ ‫=‬ ‫∞‪−‬‬ ‫ﻟﺪیﻨﺎ ‪:‬‬ ‫ﺝـ‪-‬‬ ‫∞‪x →+‬‬ ‫∞‪x →+‬‬ ‫∞‪x →+‬‬ ‫) ‪ (C‬یﻘﺒﻞ ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ ﺑﺠﻮار ∞‪ +‬اﺕﺠﺎهﻪ اﻟﻤﺴﺘﻘﻴﻢ )∆( اﻟﺬي ﻣﻌﺎدﻟﺘﻪ ‪. y = x :‬‬‫د‪ -‬ﻟﺪیﻨﺎ ‪ . ∀x ∈⎦⎤0, +∞⎣⎡ : f (x ) − x = − (ln x )2 ≤ 0 :‬إذن اﻟﻤﻨﺤﻨﻰ ) ‪ (C‬یﻮﺝﺪ ﺕﺤﺖ اﻟﻤﺴﺘﻘﻴﻢ )∆( ‪.‬‬‫) ‪( ).f ′(x ) = x − (ln x )2 ′ =1− 2ln′(x )ln x =1− 2ln x = x − 2ln x = g (x‬‬ ‫‪ .3‬أ‪ -‬ﻟﻴﻜﻦ ⎣⎡∞‪ ، x ∈⎦⎤0, +‬ﻟﺪیﻨﺎ ‪:‬‬ ‫‪xx‬‬ ‫‪x‬‬‫وﺡﺴﺐ إﺷﺎرة ) ‪ g (x‬ﻓﻲ اﻟﺠﺰء اﻷول ‪ ،‬ﻟﺪیﻨﺎ ‪ .∀x ∈⎤⎦0, +∞⎡⎣ : f ′(x ) > 0 :‬إذن ‪ f‬ﺕﺰایﺪیﺔ ﻋﻠﻰ ⎡⎣∞‪. ⎦⎤0, +‬‬ ‫ب‪ -‬ﺝﺪول ﺕﻐﻴﺮات اﻟﺪاﻟﺔ ‪ : f‬‬ ‫ﺝـ‪ -‬ﻣﻌﺎدﻟﺔ اﻟﻤﻤﺎس ﻟﻠﻤﻨﺤﻨﻰ ) ‪ (C‬ﻓﻲ اﻟﻨﻘﻄﺔ اﻟﺘﻲ أﻓﺼﻮﻟﻬﺎ ‪ 1‬هﻲ ‪. y = f ′(1)(x −1) + f (1) ⇔ y = x :‬‬‫‪ .4‬ﻟﺪیﻨﺎ ‪ f :‬ﻣﺘﺼﻠﺔ وﺕﺰایﺪیﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل ⎡⎣∞‪ . ⎤⎦0, +‬إذن‪ f :‬ﺕﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ ‪ f −1‬ﻣﻌﺮﻓﺔ ﻣﻦ اﻟﻤﺠﺎل ‪ J‬ﺡﻴﺚ ‪:‬‬

‫⎡ ⎤) (‬ ‫(‬ ‫⎢)‬ ‫\‬‫‪ J = f‬ﻧﺤﻮ اﻟﻤﺠﺎل ⎡⎣∞‪ ، I = ⎦⎤0, +‬وﺑﻤﺎ أن ‪ ، 0∈J‬ﻓﺈن‬ ‫⎣⎡∞‪⎦⎤0, +‬‬ ‫‪= ⎥lim f‬‬ ‫‪(x ), lim f‬‬ ‫‪x‬‬ ‫=‬ ‫‪⎦⎤−∞,‬‬ ‫⎣⎡∞‪+‬‬ ‫=‬ ‫⎦⎥‬ ‫‪x‬‬ ‫‪→0‬‬ ‫∞‪x →+‬‬ ‫⎢⎣‬ ‫‪x‬‬ ‫‪>0‬‬ ‫اﻟﻤﻌﺎدﻟﺔ ‪ f (x ) = 0‬ﺕﻘﺒﻞ ﺡﻼ وﺡﻴﺪا ‪ α‬ﻓﻲ اﻟﻤﺠﺎل ⎣⎡∞‪. I = ⎦⎤0, +‬‬ ‫(‪.‬‬ ‫‪(ln‬‬ ‫‪2)2‬‬ ‫<‬ ‫‪1‬‬ ‫‪ ) f‬ﻷﻧﻪ ﺡﺴﺐ اﻟﻤﻌﻄﻴﺎت‬ ‫⎛‬ ‫‪1‬‬ ‫⎞‬ ‫=‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪(ln‬‬ ‫‪2)2‬‬ ‫>‬ ‫‪0‬‬ ‫و‬ ‫‪f‬‬ ‫‪⎛1‬‬ ‫⎞‬ ‫=‬ ‫‪1‬‬ ‫=‪−1‬‬ ‫‪1−e‬‬ ‫<‬ ‫‪0‬‬ ‫‪:‬‬ ‫أن‬ ‫وﺑﻤﺎ‬ ‫‪2‬‬ ‫⎜‬ ‫‪2‬‬ ‫⎟‬ ‫‪2‬‬ ‫⎜‬ ‫⎟‬ ‫‪e‬‬ ‫‪e‬‬ ‫⎝‬ ‫⎠‬ ‫⎝‬ ‫‪e‬‬ ‫⎠‬ ‫ﻓﺈﻧﻪ ﺡﺴﺐ ﻣﺒﺮهﻨﺔ اﻟﻘﻴﻢ اﻟﻮﺳﻴﻄﻴﺔ ‪ ،‬ﻟﺪیﻨﺎ ‪. 1 < α < 1 :‬‬ ‫‪e2‬‬‫‪ I (e,e −1) .α ≈ 0,4948664145‬ﻧﻘﻄﺔ اﻧﻌﻄﺎف ﻟﻠﻤﻨﺤﻨﻰ ) ‪.e ≈ 2,7 . (C‬‬ ‫‪ .5‬إﻧﺸﺎء اﻟﻤﻨﺤﻨﻰ ) ‪: (C‬‬‫‪ .6‬أ‪ -‬ﻟﺪیﻨﺎ ‪ .∀x ∈⎤⎦0, +∞⎡⎣ : H ′(x ) = (x ln x − x )′ = x ′ln x + xln′x −1 = ln x :‬إذن ‪H : x 6 x ln x − x :‬‬ ‫هﻲ داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ ln : x 6 ln x‬ﻋﻠﻰ اﻟﻤﺠﺎل ⎡⎣∞‪ ، ⎤⎦0, +‬وﻟﺪیﻨﺎ ‪:‬‬ ‫‪e‬‬ ‫‪ln‬‬ ‫(‬ ‫‪x‬‬ ‫‪)dx‬‬ ‫‪= ⎣⎡H‬‬ ‫‪(x‬‬ ‫‪)⎤⎦1e‬‬ ‫‪=H‬‬ ‫‪(e ) − H‬‬ ‫= )‪(1) = 0 − (−1‬‬ ‫‪1‬‬ ‫‪∫1‬‬ ‫ب‪ -‬ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺝﺰاء‪ ،‬ﻟﺪیﻨﺎ ‪:‬‬ ‫‪e‬‬ ‫‪ln‬‬ ‫(‬ ‫‪x‬‬ ‫‪)2 dx‬‬ ‫=∫‬ ‫‪e‬‬ ‫‪′(x‬‬ ‫‪)ln (x‬‬ ‫‪)dx‬‬ ‫‪= ⎣⎡H‬‬ ‫‪(x‬‬ ‫‪)ln (x‬‬ ‫‪e‬‬ ‫‪∫−‬‬ ‫‪e‬‬ ‫‪(x‬‬ ‫‪) ln ′ ( x‬‬ ‫‪)dx‬‬‫‪∫1‬‬ ‫‪H‬‬ ‫‪)⎦⎤1‬‬ ‫‪H‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫=‬ ‫‪H‬‬ ‫‪(e‬‬ ‫‪)ln (e‬‬ ‫‪)−‬‬ ‫‪H‬‬ ‫‪(1)ln (1) −‬‬ ‫‪e‬‬ ‫‪x‬‬ ‫‪ln x‬‬ ‫‪−x‬‬ ‫‪dx‬‬ ‫‪x‬‬ ‫‪∫1‬‬ ‫=‬ ‫‪∫− e‬‬ ‫(‬ ‫‪ln‬‬ ‫(‬ ‫‪x‬‬ ‫‪) −1)dx‬‬ ‫=‬ ‫‪∫− e‬‬ ‫‪ln‬‬ ‫(‬ ‫‪x‬‬ ‫‪)dx‬‬ ‫‪+ (e‬‬ ‫= )‪−1‬‬ ‫‪e‬‬ ‫‪−2‬‬ ‫‪1‬‬ ‫‪1‬‬

‫‪ -‬ﺡﺴﺐ اﻟﺴﺆال أﻋﻼﻩ ‪-‬‬‫ﺝـ‪ -‬ﻣﺴﺎﺡﺔ اﻟﺤﻴﺰ اﻟﻤﺴﺘﻮي اﻟﻤﺤﺼﻮر ﺑﻴﻦ اﻟﻤﻨﺤﻨﻰ ) ‪ (C‬واﻟﻤﺴﺘﻘﻴﻢ ∆ واﻟﻤﺴﺘﻘﻴﻤﻴﻦ اﻟﻤﻌﺮﻓﻴﻦ ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ ‪ x =1‬و ‪ x = e‬هﻲ ‪( ):‬‬‫‪A =e‬‬ ‫‪f‬‬ ‫‪(x )−x‬‬ ‫‪dx‬‬ ‫=‬ ‫‪e‬‬ ‫(‬ ‫‪x‬‬ ‫‪−f‬‬ ‫‪(x ))dx‬‬ ‫=‬ ‫‪e‬‬ ‫‪(ln‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪= e −2‬‬ ‫)‪≈ 0,7(u.a.‬‬ ‫‪∫1‬‬ ‫‪∫1‬‬ ‫‪∫1‬‬ ‫‪) dx‬‬ ‫‪))C)W„KV„J)•iX„J‬‬ ‫⎧⎪‬ ‫‪n‬‬ ‫‪u0‬‬ ‫=‬ ‫‪2‬‬ ‫‪n‬‬ ‫)‬ ‫;‬ ‫`∈ ‪n‬‬ ‫ﻧﻌﺘﺒﺮ اﻟﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدیﺔ `∈‪ un n‬اﻟﻤﻌﺮﻓﺔ آﻤﺎ یﻠﻲ ‪( ):‬‬ ‫= ‪+1‬‬ ‫‪f‬‬ ‫‪⎪⎩⎨u‬‬ ‫‪(u‬‬ ‫‪ .1‬ﻟﻨﺒﻴﻦ ﺑﺎﻟﺘﺮﺝﻊ أن ‪. ∀n ∈` : 1≤un ≤ 2 :‬‬ ‫‪ 9‬ﻣﻦ أﺝﻞ ‪ ، n = 0‬ﻟﺪیﻨﺎ ‪ ، u0 = 2 :‬إذن ‪  . 1≤ u0 ≤ 2 :‬‬ ‫‪ 9‬ﻟﻴﻜﻦ `∈ ‪  . n‬‬ ‫ﻧﻔﺘﺮض أن ‪  .1 ≤ un ≤ 2 :‬‬ ‫ﻟﻨﺒﻴﻦ أن ‪  : 1 ≤ un+1 ≤ 2 :‬‬‫ﻧﻌﻠﻢ أن ‪ f‬ﺕﺰایﺪیﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡∞‪ . ⎤⎦0, +‬إذن ‪1≤un ≤ 2 ⇒ f (1) ≤ f (un ) ≤ f (2) ⇒1≤un+1 ≤ 2 :‬‬‫ﻷن ‪. f (2) − 2 = −(ln 2)2 ≤ 0 ⇒ f (2) ≤ 2 :‬‬ ‫‪ 9‬وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ‪. ∀n ∈` : 1≤un ≤ 2 :‬‬‫‪ .2‬ﻟﻴﻜﻦ `∈ ‪ . n‬ﻟﺪیﻨﺎ ‪ . un+1 −un = f (un ) −un = − ln (un ) 2 ≤ 0 :‬إذن ‪ (un )n∈` :‬ﻣﺘﺘﺎﻟﻴﺔ ﺕﻨﺎﻗﺼﻴﺔ‪( ).‬‬ ‫‪ .3‬ﺑﻤﺎ أن `∈‪ (un )n‬ﻣﺘﺘﺎﻟﻴﺔ ﺕﻨﺎﻗﺼﻴﺔ وﻣﺼﻐﻮرة ﺑﺎﻟﻌﺪد ‪ ، 1‬ﻓﺈﻧﻬﺎ ﻣﺘﻘﺎرﺑﺔ‪ .‬وﻟﺪیﻨﺎ ‪:‬‬ ‫‪ f‬داﻟﺔ ﻣﺘﺼﻠﺔ ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤‪  . ⎣⎡1, 2‬‬ ‫‪9‬‬ ‫‪9‬‬‫‪ ، f‬ﻷن ‪( )  . f (2) ≤ 2 :‬‬ ‫‪( ) ( )⎡⎣1,2⎤⎦ = ⎣⎡f‬‬ ‫‪2‬‬ ‫⎤‬ ‫⊂‬ ‫⎦⎤‪⎡⎣1, 2‬‬ ‫‪ f‬داﻟﺔ ﻣﺘﺼﻠﺔ وﺕﺰایﺪیﺔ ﻗﻄﻌﺎ ﻋﻠﻰ اﻟﻤﺠﺎل ⎦⎤‪ . ⎣⎡1, 2‬إذن ‪:‬‬ ‫‪1 ,f‬‬ ‫⎦‬ ‫‪  . u0 = 2∈ ⎡⎣1, 2⎤⎦ 9‬‬ ‫‪ (un )n∈` 9‬ﻣﺘﺘﺎﻟﻴﺔ ﻣﺘﻘﺎرﺑﺔ ﻧﻬﺎیﺘﻬﺎ ‪  . l‬‬ ‫ﺡﺴﺐ ﻣﺼﺎدیﻖ اﻟﺘﻘﺎرب ‪ ،‬ﻟﺪیﻨﺎ ‪ f l = l :‬و ⎤⎦‪( ). l ∈ ⎣⎡1, 2‬‬‫وﻟﺪیﻨﺎ ‪ . f l = l ⇔ l − ln (l ) 2 = l ⇔ ln (l ) = 0 ⇔ l =1 :‬وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈن ‪( ). nl→im+∞un =1 ( ):‬‬ ‫‪))“hhhhhhhhhhhhhhhhhhŠTJ‬‬‫ﻣﺴﺎﺡﺔ اﻟﺤﻴﺰ اﻟﻤﺴﺘﻮي اﻟﻤﺤﺼﻮر ﺑﻴﻦ اﻟﻤﻨﺤﻨﻰ ) ‪ (C‬واﻟﻤﺴﺘﻘﻴﻢ ∆ واﻟﻤﺴﺘﻘﻴﻤﻴﻦ اﻟﻤﻌﺮﻓﻴﻦ ﺑﺎﻟﻤﻌﺎدﻟﺘﻴﻦ‪ x =1‬و ‪ x = e‬ﺑﺎﺳﺘﻌﻤﺎل ‪( )Maple 7‬‬‫;‪> f:=x->x-(ln(x))^2‬‬ ‫‪f := x → x − ln( x )2‬‬‫;))‪> A:=Int(abs('f'(x)-x),x=1..exp(1))=int(abs(f(x)-x),x=1..exp(1‬‬ ‫‪e‬‬ ‫‪A := ⎮⌠⌡ −f( x ) + x dx = e − 2‬‬ ‫‪1‬‬‫;)‪> A:=evalf(rhs(A),20‬‬ ‫‪A := .7182818284590452354‬‬

‫ﺕﻤﺜﻴﻞ اﻟﺤﺪود اﻟﺴﺘﺔ اﻷوﻟﻰ ﻟﻠﻤﺘﺘﺎﻟﻴﺔ اﻟﻌﺪدیﺔ `∈‪ (un )n‬ﻋﻠﻰ ﻣﺤﻮر اﻷﻓﺎﺻﻴﻞ ﺑﺎﺳﺘﻌﻤﺎل ‪: Archimède II‬‬ ‫‪ ‬‬