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## corrigé bac 2008.pdf

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‫⎤‬
‫⎡‬
‫\ = ⎣⎡ ∞‪ J = f ⎤⎦ 0, +∞ ⎡⎣ = ⎥ lim f ( x ) , lim f ( x ) ⎢ = ⎤⎦ −∞, +‬ﻧﺤﻮ اﻟﻤﺠﺎل ⎣⎡ ∞‪ ، I = ⎤⎦ 0, +‬وﺑﻤﺎ أن ‪ ، 0 ∈ J‬ﻓﺈن‬
‫‪0‬‬
‫∞‪x →+‬‬
‫→ ‪⎥ xx‬‬
‫⎢‬
‫‪⎦ &gt;0‬‬
‫⎣‬
‫اﻟﻤﻌﺎدﻟﺔ ‪ f ( x ) = 0‬ﺕﻘﺒﻞ ﺡﻼ وﺡﻴﺪا ‪ α‬ﻓﻲ اﻟﻤﺠﺎل ⎣⎡ ∞‪. I = ⎤⎦ 0, +‬‬

‫)‬

‫(‬

‫‪2‬‬
‫‪⎛1⎞ 1‬‬
‫‪⎛1⎞ 1‬‬
‫‪1‬‬
‫‪1−e‬‬
‫= ‪ f ⎜ ⎟ = − 1‬و ‪ ) f ⎜ ⎟ = − ( ln 2 ) &gt; 0‬ﻷﻧﻪ ﺡﺴﺐ اﻟﻤﻌﻄﻴﺎت‬
‫وﺑﻤﺎ أن ‪&lt; 0 :‬‬
‫‪e‬‬
‫‪2‬‬
‫‪⎝ 2⎠ 2‬‬
‫‪⎝e ⎠ e‬‬

‫&lt;‬

‫‪2‬‬

‫‪1‬‬
‫‪1‬‬
‫ﻓﺈﻧﻪ ﺡﺴﺐ ﻣﺒﺮهﻨﺔ اﻟﻘﻴﻢ اﻟﻮﺳﻴﻄﻴﺔ ‪ ،‬ﻟﺪیﻨﺎ ‪&lt; α &lt; :‬‬
‫‪e‬‬
‫‪2‬‬
‫‪ .5‬إﻧﺸﺎء اﻟﻤﻨﺤﻨﻰ ) ‪: (C‬‬
‫‪ I (e ,e −1) . α ≈ 0,4948664145‬ﻧﻘﻄﺔ اﻧﻌﻄﺎف ﻟﻠﻤﻨﺤﻨﻰ ) ‪(C‬‬

‫)‬

‫(‬

‫‪.( ln 2‬‬

‫‪.‬‬

‫‪ .6‬أ‪ -‬ﻟﺪیﻨﺎ ‪∈ ⎤⎦ 0, +∞ ⎡⎣ : H ′ ( x ) = ( x ln x − x )′ = x ′ ln x + xln ′x − 1 = ln x :‬‬
‫هﻲ داﻟﺔ أﺻﻠﻴﺔ ﻟﻠﺪاﻟﺔ ‪ ln : x 6 ln x‬ﻋﻠﻰ اﻟﻤﺠﺎل ⎣⎡ ∞‪ ، ⎤⎦ 0, +‬وﻟﺪیﻨﺎ ‪:‬‬

‫‪ . ∀x‬إذن ‪:‬‬

‫‪≈ 2,7‬‬

‫‪.‬‬

‫‪.e‬‬

‫‪H : x 6 x ln x − x‬‬

‫‪∫1 ln ( x ) dx = ⎡⎣H ( x )⎤⎦1 = H (e ) − H (1) = 0 − ( −1) = 1‬‬
‫‪e‬‬

‫‪e‬‬

‫ب‪ -‬ﺑﺎﺳﺘﻌﻤﺎل اﻟﻤﻜﺎﻣﻠﺔ ﺑﺎﻷﺝﺰاء‪ ،‬ﻟﺪیﻨﺎ ‪:‬‬
‫‪e‬‬

‫‪dx = ∫ H ′ ( x ) ln ( x ) dx = ⎡⎣ H ( x ) ln ( x )⎤⎦ − ∫ H ( x ) ln ′ ( x ) dx‬‬
‫‪1‬‬
‫‪1‬‬
‫‪1‬‬
‫‪e x ln x − x‬‬
‫∫ ‪= H (e ) ln (e ) − H (1) ln (1) −‬‬
‫‪dx‬‬
‫‪1‬‬
‫‪x‬‬
‫‪e‬‬
‫‪e‬‬
‫‪= − ∫ ( ln ( x ) − 1) dx = − ∫ ln ( x ) dx + (e − 1) = e − 2‬‬
‫‪e‬‬

‫‪e‬‬

‫‪1‬‬

‫‪4 ‬‬
‫‪ ‬‬

‫اﻷﺳﺘﺎذ ‪ :‬ﻣﺤﻤﺪ اﻟﺤﻴــــــــــــــــــــﺎن‬

‫‪2‬‬

‫) ‪∫1 ln ( x‬‬
‫‪e‬‬

‫‪1‬‬

‫اﻟﺜﺎﻧﻴﺔ ﺑﻜﺎﻟﻮریﺎ ﺷﻌﺒﺔ اﻟﻌﻠﻮم اﻟﺘﺠﺮیﺒﻴﺔ ﺑﻤﺴﺎﻟﻜﻬﺎ وﺷﻌﺒﺔ اﻟﻌﻠﻮم واﻟﺘﻜﻨﻮﻟﻮﺝﻴﺎت ﺑﻤﺴﻠﻜﻴﻬﺎ‬