Math bac cours 6

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‫اﻟﺪاﻟﺔ اﻷﺳﻴﺔ‬ ‫اﻟﺜﺎﻧﻴﺔ ﺳﻠﻚ ﺑﻜﺎﻟﻮرﻳﺎ ﻋﻠﻮم ﺗﺠﺮﻳﺒﻴﺔ‬ ‫‪ -І‬اﻟﺪاﻟﺔ اﻷﺳﻴﺔ اﻟﻨﻴﺒﺮﻳﺔ‬ ‫‪ -1‬ﺗﻌﺎرﻳﻒ و ﺧﺎﺻﻴﺎت أوﻟﻴﺔ‬ ‫ﻧﻌﻠﻢ أن داﻟﺔ ‪ ln‬ﺗﻘﺎﺑﻞ ﻣﻦ [∞‪ ]0; +‬ﻧﺤﻮ‬

‫و ﺑﺎﻟﺘﺎﻟﻲ ﺗﻘﺒﻞ داﻟﺔ ﻋﻜﺴﻴﺔ ﻣﻦ‬

‫ﻧﺤﻮ‬

‫[∞‪]0; +‬‬

‫أ‪ -‬ﺗﻌﺮﻳﻒ‬ ‫اﻟﺪاﻟﺔ اﻟﻌﻜﺴﻴﺔ ﻟﺪاﻟﺔ اﻟﻠﻮﻏﺎرﻳﺘﻢ اﻟﻨﻴﺒﻴﺮي ﺗﺴﻤﻰ اﻟﺪاﻟﺔ اﻷﺳﻴﺔ اﻟﻨﻴﺒﻴﺮﻳﺔ ﻧﺮﻣﺰ ﻟﻬﺎ )ﻣﺆﻗﺘﺎ( ﺑﺎﻟﺮﻣﺰ ‪exp‬‬

‫‪exp ( x ) = y ⇔ ln y = x‬‬ ‫ب‪ -‬ﺧﺎﺻﻴﺎت أوﻟﻴﺔ‬

‫‪exp ( 0 ) = 1‬‬

‫*‬ ‫*‬ ‫*‬

‫[∞‪∀y ∈ ]0; +‬‬

‫‪exp (1) = e‬‬

‫) ‪exp ( x‬‬

‫∈ ‪∀x‬‬

‫‪ln ( exp ( x ) ) = x‬‬

‫∈ ‪∀x‬‬

‫‪0‬‬

‫* ‪exp ( ln ( x ) ) = x‬‬

‫∈ ‪∀x‬‬

‫[∞‪∀x ∈ ]0; +‬‬

‫* اﻟﺪاﻟﺔ ‪ exp‬ﺗﺰاﻳﺪﻳﺔ ﻗﻄﻌﺎ ﻋﻠﻰ‬

‫*‬

‫‪exp ( a ) = exp ( b ) ⇔ a = b‬‬

‫‪2‬‬

‫∈ ) ‪∀ ( a; b‬‬

‫*‬

‫) ‪exp ( a‬‬

‫‪2‬‬

‫∈ ) ‪∀ ( a; b‬‬

‫‪b‬‬

‫‪exp ( b ) ⇔ a‬‬

‫‪ -2‬اﻟﺘﻤﺜﻴﻞ اﻟﻤﺒﻴﺎﻧﻲ ﻟﺪاﻟﺔ ‪exp‬‬ ‫ﻓﻲ ﻣﻌﻠﻢ ﻣﺘﻌﺎﻣﺪ ﻣﻤﻨﻈﻢ ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ ‪ ln‬و ﻣﻨﺤﻨﻰ اﻟﺪاﻟﺔ ‪ exp‬ﻣﺘﻤﺎﺛﻼن ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﻨﺼﻒ اﻷول‬

‫‪ -3‬ﺧﺎﺻﻴﺔ أﺳﺎﺳﻴﺔ‬

‫) ‪exp ( a + b ) = exp ( a ) × exp (b‬‬

‫اﻟﺒﺮهﺎن‬

‫‪2‬‬

‫∈ ) ‪∀ ( a; b‬‬

‫‪ln(exp ( a ) × exp ( b )) = ln exp ( a ) + ln exp ( b ) = a + b‬‬ ‫‪ln exp ( a + b ) = a + b‬‬ ‫) ‪ln ( exp ( a ) × exp ( b ) ) = ln exp ( a + b‬‬ ‫) ‪exp ( a + b ) = exp ( a ) × exp ( b‬‬

‫‪1‬‬


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