Tajribi Math Sx (110)

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‫اﻻﻣﺘﺤــﺎن اﻟﺘﺠﺮﻳﺒـﻲ‬ ‫ﻟﻠﺴﻨﺔ اﻟﺜﺎﻧﻴﺔ ﻣﻦ ﺳﻠﻚ اﻟﺒﺎآﺎﻟﻮرﻳــﺎ‬ ‫‪2006 -2005‬‬ ‫اﻟﺜﺎﻧﻮﻳﺔ ‪ :‬اﻟﺤﺴﻦ اﻟﺜﺎﻧﻲ اﻟﺘﺄهﻴﻠﻴﺔ‬

‫اﻟﺸﻌﺒﺔ ‪ :‬اﻟﻌﻠـــــﻮم ﺗﺠﺮﻳـﺒﻴـــــــﺔ‬

‫اﻟﻤــﺎدة ‪ :‬اﻟﺮﻳﺎﺿﻴـــــــــﺎت‬ ‫اﻟﻤﺴﺘﻮى ‪ :‬اﻟﺜــﺎﻧﻴــﺔ ﺛﺎﻧـﻮي‬

‫ﺗﻤﺮﻳﻦ ‪1‬‬ ‫‪.I‬‬

‫‪1‬‬ ‫⎞‪⎛ 1‬‬ ‫‪g ( x ) = L n ⎜1 + ⎟ −‬‬ ‫‪⎝ x ⎠ x +1‬‬

‫ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ ‪ g‬اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ ‪:‬‬

‫‪ .1‬ﺑﻴﻦ أن ‪Dg = ]−∞, −1[ ∪ ]0, +∞[ :‬‬ ‫و ) ‪lim g ( x‬‬

‫‪ .2‬أﺣﺴﺐ ‪lim g ( x ) :‬‬

‫∞‪x→+‬‬

‫‪ .3‬ﺑﻴﻦ أن ‪lim g ( x ) = +∞ :‬‬

‫∞‪x→−‬‬

‫‪x →−1−‬‬

‫‪ .4‬أﻋﻂ ﺟﺪول ﺗﻐﻴﺮات اﻟﺪاﻟﺔ ‪ g‬واﺳﺘﻨﺘﺞ أن ‪0 :‬‬ ‫‪.II‬‬

‫و ) ‪lim g ( x‬‬

‫‪x → 0+‬‬

‫)‪: g ( x‬‬

‫‪x≠0‬‬

‫ﻧﻌﺘﺒﺮ اﻟﺪاﻟﺔ اﻟﻌﺪدﻳﺔ ‪ f‬اﻟﻤﻌﺮﻓﺔ ﺑﻤﺎ ﻳﻠﻲ ‪:‬‬

‫;‬

‫‪ .1‬ﺑﻴﻦ أن ‪D f = ]−∞, −1[ ∪ [ 0, +∞[ :‬‬ ‫‪ .2‬أﺣﺴﺐ ‪lim f ( x ) :‬‬

‫و ) ‪lim f ( x‬‬

‫‪x →−1−‬‬

‫∞‪x →+‬‬

‫‪1‬‬ ‫) ﺿﻊ‬ ‫‪x‬‬

‫) ‪( ∀x ∈ D‬‬ ‫‪g‬‬

‫⎧‬ ‫⎞‪⎛ 1‬‬ ‫⎟ ‪⎪ f ( x ) = xL n ⎜ 1 +‬‬ ‫⎠‪⎝ x‬‬ ‫⎨‬ ‫‪⎪ f ( 0) = 0‬‬ ‫⎩‬

‫= ‪(X‬‬

‫‪ .3‬أدرس اﺗﺼﺎل ‪ f‬ﻋﻠﻰ ﻳﻤﻴﻦ ‪. 0‬‬ ‫‪ .4‬أدرس ﻗﺎﺑﻠﻴﺔ اﺷﺘﻘﺎق ‪ f‬ﻋﻠﻰ ﻳﻤﻴﻦ ‪ 0‬ﺛﻢ أﻋﻂ ﺗﺄوﻳﻼ هﻨﺪﺳﻴﺎ‪.‬‬ ‫‪ .5‬أدرس ﺗﻐﻴﺮات ‪. f‬‬ ‫‪.6‬‬

‫أرﺳﻢ )‬

‫‪ ( C f‬ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ ‪ f‬ﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ) ‪. ( O, i , j‬‬ ‫‪n‬‬

‫‪v‬‬ ‫⎞‪⎛ 1‬‬ ‫‪ .III‬ﻧﻀﻊ ﻟﻜﻞ ‪ n‬ﻣﻦ * ‪ un = ⎜ 1 + ⎟ :‬و ) ‪ vn = Ln ( un‬و ‪. wn = n‬‬ ‫‪n‬‬ ‫⎠‪⎝ n‬‬ ‫‪ .1‬ﺗﺤﻘﻖ أن ‪ vn = f ( n ) :‬وأن ) ‪ ( vn‬ﻣﺘﺘﺎﻟﻴﺔ ﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ و ‪ 0 ≺ vn ≺ 1‬ﻟﻜﻞ ‪ n‬ﻣﻦ‬ ‫‪ .2‬أﺣﺴﺐ ‪lim vn :‬‬

‫∞‪n →+‬‬

‫‪ .3‬أﺣﺴﺐ ‪:‬‬

‫*‬

‫‪.‬‬

‫ﺛﻢ اﺳﻨﺘﺞ ‪. lim un‬‬ ‫∞‪n →+‬‬

‫‪ S n = w1 + w2 + w3 + ................. + wn‬ﺑﺪﻻﻟﺔ ‪ n‬واﺣﺴﺐ ‪. lim S n‬‬ ‫∞‪n →+‬‬

‫ﺗﻤﺮﻳﻦ ‪2‬‬ ‫ﻓﻲ اﻟﻔﻀﺎء اﻟﻤﻨﺴﻮب إﻟﻰ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻣﺒﺎﺷﺮ ‪O, i , j , k‬‬

‫)‬

‫واﻟﻤﺴﺘﻮى ذو اﻟﻤﻌﺎدﻟﺔ ‪:‬‬

‫(‬

‫ﻧﻌﺘﺒﺮ اﻟﻨﻘﻄﺔ )‪A (1,1,1‬‬

‫‪( P) : x − 2 y + 2z + 8 = 0‬‬

‫‪ .1‬ﺗﺤﻘﻖ أن ‪A ∉ ( P ) :‬‬ ‫‪ .2‬ﺣﺪد إﺣﺪاﺛﻴﺎت اﻟﻨﻘﻄﺔ ‪ H‬اﻟﻤﺴﻘﻂ اﻟﻌﻤﻮدي ﻟــ ‪ A‬ﻋﻠﻰ ) ‪. ( P‬‬ ‫‪ .3‬ﻟﺘﻜﻦ ) ‪ ( S‬اﻟﻔﻠﻜﺔ ذات اﻟﻤﺮآﺰ ‪ A‬وﺗﻘﻄﻊ اﻟﻤﺴﺘﻮى ) ‪ ( P‬وﻓﻖ اﻟﺪاﺋﺮة ذات اﻟﻤﺮآﺰ ‪ H‬واﻟﺸﻌﺎع ‪r = 4‬‬

‫‪ .a‬ﺣﺪد ‪ R‬ﺷﻌﺎع اﻟﻔﻠﻜﺔ ) ‪. ( S‬‬ ‫‪ .b‬ﺣﺪد ﻣﻌﺎدﻟﺔ دﻳﻜﺎرﺗﻴﺔ ﻟﻠﻔﻠﻜﺔ ) ‪. ( S‬‬

‫اﻻﻣﺘﺤــﺎن اﻟﺘﺠﺮﻳﺒـﻲ‬ ‫ﻟﻠﺴﻨﺔ اﻟﺜﺎﻧﻴﺔ ﻣﻦ ﺳﻠﻚ اﻟﺒﺎآﺎﻟﻮرﻳــﺎ‬ ‫‪2006 -2005‬‬ ‫اﻟﺜﺎﻧﻮﻳﺔ ‪ :‬اﻟﺤﺴﻦ اﻟﺜﺎﻧﻲ اﻟﺘﺄهﻴﻠﻴﺔ‬

‫اﻟﺸﻌﺒﺔ ‪ :‬اﻟﻌﻠـــــﻮم ﺗﺠﺮﻳـﺒﻴـــــــﺔ‬

‫اﻟﻤــﺎدة ‪ :‬اﻟﺮﻳﺎﺿﻴـــــــــﺎت‬ ‫اﻟﻤﺴﺘﻮى ‪ :‬اﻟﺜــﺎﻧﻴــﺔ ﺛﺎﻧـﻮي‬

‫ﺗﻤﺮﻳﻦ ‪3‬‬ ‫‪ .1‬أﺣﺴﺐ اﻟﺘﻜﺎﻣﻼت اﻟﺘﺎﻟﻴﺔ ‪:‬‬ ‫‪π‬‬

‫‪dx‬‬

‫‪x2‬‬

‫) ‪(1 − 2 x‬‬

‫‪3 2‬‬

‫‪ .2‬ﺑﻴﻦ أن ‪:‬‬ ‫‪ .3‬أﺣﺴﺐ اﻟﺘﻜﺎﻣﻞ ‪:‬‬

‫‪2‬‬

‫∫= ‪I‬‬

‫و‬

‫‪dx‬‬

‫‪1‬‬

‫‪x2 − 2 x‬‬ ‫‪3x 2 − x3‬‬

‫‪2‬‬

‫∫= ‪J‬‬

‫‪2‬‬

‫‪cos x‬‬ ‫∫=‪K‬‬ ‫‪dx‬‬ ‫‪cos 2 x − 2‬‬ ‫‪0‬‬

‫و‬

‫‪1‬‬

‫‪e −2 x‬‬ ‫‪ex‬‬ ‫‪−x‬‬ ‫=‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪+‬‬ ‫‪e‬‬ ‫‪1 + e− x‬‬ ‫‪1 + ex‬‬ ‫‪Ln 2‬‬ ‫‪e −2 x‬‬ ‫∫ =‪L‬‬ ‫‪dx‬‬ ‫‪−x‬‬ ‫‪+‬‬ ‫‪e‬‬ ‫‪1‬‬ ‫‪0‬‬

‫‪ .4‬ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺟﺰاء أﺣﺴﺐ اﻟﺘﻜﺎﻣﻞ ‪:‬‬

‫‪Ln (1 + e − x ) dx‬‬

‫‪−x‬‬

‫‪Ln 2‬‬

‫‪∫e‬‬

‫=‪R‬‬

‫‪0‬‬

‫ﺗﻤﺮﻳﻦ ‪4‬‬ ‫ﻧﻌﺘﺒﺮ اﻟﺤﺪودﻳﺔ ‪:‬‬

‫‪P ( z ) = z + ( 3 − 2i ) z + (1 − 4i ) z − 1 − 2i‬‬ ‫‪2‬‬

‫‪3‬‬

‫‪ .1‬أﺣﺴﺐ )‪P ( −1‬‬

‫) ∈‪(z‬‬

‫‪ .2‬أوﺟﺪ اﻟﻌﺪدﻳﻦ اﻟﻌﻘﺪﻳﻴﻦ ‪ a‬و ‪ b‬ﺑﺤﻴﺚ ‪P ( z ) = ( z + 1) ( z 2 + az + b ) :‬‬

‫‪ .3‬ﺣﻞ ﻓﻲ‬

‫اﻟﻤﻌﺎدﻟﺔ ‪P ( z ) = 0 :‬‬

‫‪ .4‬ﻧﻀﻊ ‪ A ( −1) :‬و ) ‪ B ( i‬و ) ‪. C ( −2 + i‬‬ ‫‪ .a‬أﺛﺒﺖ أن ‪ ABC‬ﻣﺜﻠﺜﺎ ﻗﺎﺋﻢ اﻟﺰاوﻳﺔ وﻣﺘﺴﺎوي اﻟﺴﺎﻗﻴﻦ‪.‬‬

‫‪ .b‬اﺳﺘﻨﺘﺞ ﻗﻴﺎﺳﺎ ﻟﻠﺰاوﻳﺔ )‬

‫(‬

‫‪. BA, BC‬‬

‫) ﺿﻊ ‪.( t = sin x‬‬

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