ﻤﻭﺍﻀﻴﻊ ﺍﻹﺭﺴﺎل ﺍﻷﻭل ﻴﺘﻀﻤﻥ ﻫﺫﺍ ﺍﻹﺭﺴﺎل ﺍﻟﻤﻭﺍﻀﻴﻊ ﺍﻟﺘﺎﻟﻴﺔ: • ﻋﻤﻭﻤﻴﺎﺕ ﻋﻠﻰ ﺍﻟﺩﻭﺍل• ﺍﻟﺯﻭﺍﻴﺎ ﺍﻟﻤﻭﺠﻬﺔ ﻭﺤﺴﺎﺏ ﺍﻟﻤﺜﻠﺜﺎﺕ • ﺍﻹﺸﺘﻘﺎﻗﻴﺔ • ﺍﻟﻨﻬـﺎﻴﺎﺕ
ﻋﻤﻭﻤﻴﺎﺕ ﻋﻠﻰ ﺍﻟﺩﻭﺍل ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺩﺍﻟﺔ. ﺍﺴﺘﻌﻤﺎل ﺍﻟﺤﺴﺎﺒﻴﺔ ﺍﻟﺒﻴﺎﻨﻴﺔ ﻟﺭﺴﻡ ﻤﻨﺤﻨﻰ ﺩﺍﻟﺔ ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ﺩﺍﻟﺔﺍﻟﺘﻌﺭﻑ ﻋﻠﻰ ﺸﻔﻌﻴﺔ ﺩﺍﻟﺔ)ﺍﺨﺘﺼﺎﺭ ﻤﺠﺎل ﺍﻟﺩﺭﺍﺴﺔ،ﻋﻨﺎﺼﺭ ﺍﻟﺘﻨﺎﻅﺭ( ﺍﻻﻨﺘﻘﺎل ﻤﻥ ﻤﻨﺤﻨﻰ ﺇﻟﻰ ﺁﺨﺭ ﺍﻟﺘﻌﺭﻑ ﻋﻥ ﺍﻟﺩﺍﻟﺔ ﺩﻭﺍل ﻤﺄﻟﻭﻓﺔ fﺇﻥ ﻜﺎﻨﺕ :ﻤﺠﻤﻭﻉ ،ﺠﺩﺍﺀ ،ﻤﺭﻜﺏ ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ ﺘﻌﺎﺭﻴﻑ ﻤﺸﺎﻜل ﻭ ﺘﻤﺎﺭﻴﻥ ﺍﻟﺤﻠﻭل
ﺗﻌﺎرﻳﻒ - Iﺘﺫﻜﻴﺭ ﻋﻠﻰ ﺍﻟﺩﻭﺍل : ﻨﺭﻤﺯ ﻋﺎﺩﺓ ﺒـ D fﻟﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ f ﻭ ﺒـ C fﺇﻟﻰ ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ fﻓﻲ ﻤﻌﻠﻡ ﻤﺨﺘﺎﺭ( ). -1ﺍﻟﺩﺍﻟﺔ – ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ : ﺇﺫﺍ ﻜﺎﻨﺕ fﺩﺍﻟﺔ ﻭ D fﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ .ﻟﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ xﻤﻥ . D fﺍﻟﺩﺍﻟﺔ fﺘﺭﻓﻕ ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻭﺤﻴﺩ ﻨﺭﻤﺯ ﻟﻪ ﺒـ ) . f (xﻨﻘﻭل ﻋﻥ )f (x ﺼﻭﺭﺓ xﺒﺎﻟﺩﺍﻟﺔ . f -2ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﺩﺍﻟﺔ : fﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ . Dﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ fﻓﻲ ﻤﻌﻠﻡ ﻫﻭ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ) M x ; f ( xﺤﻴﺙ xﻋﻨﺼﺭ ﻤﻥ Dﺇﺫﻥ :ﻗﻭﻟﻨﺎ) ( M x ; yﻴﻨﺘﻤﻲ ﺇﻟﻰ ﻤﻨﺤﻨﻰ \" \" fﻴﻌﻨﻲ َ x ∈ D :ﻭ )( ). y = f (x ﻨﻘﻭل ﻋﻥ y = f (x) :ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﻟﻬﺫﺍ ﺍﻟﻤﻨﺤﻨﻰ ﻓﻲ ﺍﻟﻤﻌﻠﻡ ﺍﻟﻤﻌﻁﻰ.
-3ﺍﺘﺠﺎﻩ ﺍﻟﺘﻐﻴﺭ : ﻨﻘﻭل ﻋﻥ ﺍﻟﺩﺍﻟﺔ fﺃﻨﻬﺎ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻤﺠﺎل . Iﺇﺫﺍ ﻜﺎﻥ :ﻤﻥ ﺃﺠل ﻜل x1و x2ﻤﻥ x1 < x2 : Iﻴﺴﺘﻠﺯﻡ ) f (x1 ) < f (x2ﻭ ﺒﺎﻟﻤﺜل ﻨﻌﺭﻑ ﺩﺍﻟﺔ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل : I) f ( x2 x1 < x2ﻴﺴﺘﻠﺯﻡ ) f (x1 ) > f (x2 ) f ( x1) f ( x1 ﺃﻨﻬﺎ ﻨﻘﻭل I ﻤﺠﺎل ﻋﻠﻰ ) f ( x2 ﺃﻭ ﺘﻤﺎﻤﺎ ﻤﺘﻨﺎﻗﺼﺔ ﺩﺍﻟﺔ ﻜﺎﻨﺕ ﺇﺫﺍ - ﺭﺘﻴﺒﺔ ﺘﻤﺎﻤﺎ. ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎO x1 O x1 x2 x2 ﻤﻼﺤﻅﺔ : ﻴﻤﻜﻥ ﺃﻥ ﺘﻜﻭﻥ ﺩﺍﻟﺔ ﻏﻴﺭ ﻤﺘﻨﺎﻗﺼﺔ ﻭ ﻏﻴﺭ ﻤﺘﺯﺍﻴﺩﺓ ﻋﻠﻰ ﻤﺠﺎل. - 4ﺍﻟﺩﻭﺍل ﺍﻟﻤﺭﺠﻌﻴﺔ : ﺘﺫﻜﻴﺭ ﺒﺎﻟﺩﻭﺍل ﺍﻟﻤﺭﺠﻌﻴﺔ ﺍﻟﺘﻲ ﺩﺭﺴﺘﻬﺎ ﻓﻲ ﺍﻟﺴﻨﺔ ﺍﻷﻭﻟﻰ : -ﺍﻟﺩﺍﻟﺔ \" ﻤﺭﺒﻊ \" f : x a x2 : fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ −∞ ; 0ﻭﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ∞] [ ] [0 ; + . x2 < x2 0 ≤ x1 < x2ﻓﺈﻥ : ﺇﺫﺍ ﻜﺎﻥ : 1 2 x2 ≥ x2 ﻓﺈﻥ : x1 < x2 ≤0 ﺇﺫﺍ ﻜﺎﻥ : 1 2 ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ ﻫﻭ ﻗﻁﻊ ﻤﻜﺎﻓﺊ .
y4321-3 -2 -1 0 1 2 3x-1 f :xa 1 -ﺍﻟﺩﺍﻟﺔ \" ﻤﻘﻠﻭﺏ\" : x fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ −∞ ; 0ﻭ ﻋﻠﻰ ∞] [ ] [0 ; +11 ﺇﺫﺍ ﻜﺎﻥ 0 < x1 < x2 :ﻓﺈﻥ :x1 > x2 11 ﻓﺈﻥ : x1 < x2 < 0 ﺇﺫﺍ ﻜﺎﻥ : x1 > x2 ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ ﻫﻭ ﻗﻁﻊ ﺯﺍﺌﺩ :y321-3 -2 -1 0 1 2 3x -1 -2 -3 f :xa -ﺍﻟﺩﺍﻟﺔ \" ﺍﻟﺠﺫﺭ ﺍﻟﺘﺭﺒﻴﻌﻲ \" x :x1 < x2 fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ∞[ [0 ; + ﺇﺫﺍ ﻜﺎﻥ 0 ≤ x1 < x2 :ﻓﺈﻥ :
ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ ﻫﻭ ﺠﺯﺀ ﻤﻥ ﻗﻁﻊ ﻤﻜﺎﻓﺊ :y321 0 1 2 3 4 5 6x -ﺍﻟﺩﺍﻟﺔ \" ﺠﻴﺏ ﺍﻟﺘﻤﺎﻡ \" f : x a cos x :ﺍﻟﺩﺍﻟﺔ fﺩﻭﺭﻴﺔ ﻭﺩﻭﺭﻫﺎ T = 2π :ﺃﻱ cos( x + 2π) = cos x : ﻭ ﻟﺫﻟﻙ ﻴﻜﻔﻰ ﺩﺭﺍﺴﺘﻬﺎ ﻋﻠﻰ ﻤﺠﺎل ﻁﻭﻟﻪ . 2π ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ ﻫﻭ : y ﺩﻭﺭ ﺍﻟﺩﺍﻟﺔ 1-4 -3 -2 -1 0 1 2 3 4 5 6 7 8x -1 -ﺍﻟﺩﺍﻟﺔ \" ﺠﻴﺏ \" f : x a sin x :ﺍﻟﺩﺍﻟﺔ fﺩﻭﺭﻴﺔ ﻭﺩﻭﺭﻫﺎ T = 2π :ﺃﻱ sin( x + 2π ) = sin x : ﻭ ﻟﺫﻟﻙ ﻴﻜﻔﻰ ﺩﺭﺍﺴﺘﻬﺎ ﻋﻠﻰ ﻤﺠﺎل ﻁﻭﻟﻪ . 2π ﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ ﻫﻭ :
y ﺩﻭﺭ ﺍﻟﺩﺍﻟﺔ 1-4 -3 -2 -1 0 1 2 3 4 5 6 7 8x-1 ﺘﻤﺭﻴﻥ ) :ﻜﻴﻔﻴﺔ ﺍﺴﺘﻌﻤﺎل ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﻠﺩﺍﻟﺔ ( fﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﺍﻟﻤﺠﺎل . −3 ; 4ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ) (C fﻤﻌﻁﻰ ﺒﺎﻟﺸﻜل ﺍﻟﻤﻘﺎﺒل[ ]: y 4 3 2 1-3 -2 -1 0 1 2 3 4x -1 -2 -1ﺍﺴﺘﻌﻤل ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺘﺠﻴﺏ ﻋﻠﻰ ﺍﻷﺴﺌﻠﺔ ﺍﻵﺘﻴﺔ : ﺃ( ﺃﻋﻁ ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﺍﻟﻤﺠﺎل ][-3 ; 4 ﺏ( ﻤﺎ ﻫﻲ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ f (x) = 0 .؟ ﺝ( ﻤﺎ ﻫﻲ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ f (x) ≥ 0 ،؟ -2ﻤﺎ ﻫﻭ ﻋﺩﺩ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ f (x) = 3؟ -3ﻟﺘﻜﻥ mﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻜﻴﻔﻲ .ﻨﺎﻗﺵ ﺤﺴﺏ ﻗﻴﻡ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ mﻋﺩﺩ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ f ( x ) = m : ﺤـل : -1ﺃ( ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﻨﺤﻨﻰ \"ﻴﻨﺯل\" ﻋﻠﻰ ﻤﺠﺎل ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ ﻤﺘﻨﺎﻗﺼﺔ .ﻭﺇﺫﺍ ﻜﺎﻥ \"ﻴﺼﻌﺩ\" ﺍﻟﺩﺍﻟﺔ ﻤﺘﺯﺍﻴﺩﺓ ﻭ ﻤﻨﻪ ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ :
x -3 0 3 4f 2,5 4 -2 0ﺏ( ﻟﺩﻴﻨﺎ f ( x ) = 0 :ﻴﻜﺎﻓﺊ x = -2 :ﺃﻭ x = 2ﺃﻭ x = 4 ﺝ( ﻟﺩﻴﻨﺎ f ( x ) ≥ 0 :ﻴﻜﺎﻓﺊ x ∈ [-3 ; -2] U[2 ; 4] : -2ﺍﻟﻤﻌﺎﺩﻟﺔ f ( x ) = 3 :ﺘﻘﺒل ﺤﻠﻴﻥ . -3ﻤﻥ ﻗﺭﺍﺀﺓ ﺍﻟﺒﻴﺎﻥ ﻨﺠﺩ : ﻤﻥ ﺃﺠل m < -2 :ﻻ ﺘﻭﺠﺩ ﺤﻠﻭل . ﻤﻥ ﺃﺠل m = -2 :ﻴﻭﺠﺩ ﺤل ﻭﺍﺤﺩ . ﻤﻥ ﺃﺠل -2 < m < 0 :ﻴﻭﺠﺩ ﺤﻼﻥ . ﻤﻥ ﺃﺠل 0 ≤ m ≤ 2,5 :ﺘﻭﺠﺩ ﺜﻼﺜﺔ ﺤﻠﻭل . ﻤﻥ ﺃﺠل 2,5 < m < 4 :ﻴﻭﺠﺩ ﺤﻼﻥ . ﻤﻥ ﺃﺠل m = 4 :ﻴﻭﺠﺩ ﺤل ﻭﺍﺤﺩ . ﻤﻥ ﺃﺠل m > 4 :ﻻ ﺘﻭﺠﺩ ﺤﻠﻭل . -5ﺸﻔﻌﻴﺔ ﺍﻟﺩﺍﻟﺔ : ﺃ( ﺍﻟﺩﺍﻟﺔ ﺍﻟﺯﻭﺠﻴﺔ : fﺯﻭﺠﻴﺔ ﻴﻌﻨﻲ : -ﺇﺫﺍ ﻜﺎﻥ x ∈ Df :ﻓﺈﻥ - x ∈ Df :ﻭ )f (-x) = f (x -ﺇﺫﺍ ﻜﺎﻨﺕ fﺯﻭﺠﻴﺔ ﻓﺈﻥ ﻤﻨﺤﻨﺎﻫﺎ ﺍﻟﻤﻤﺜل ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﻨﺎﻅﺭ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ. ﺏ( ﺍﻟﺩﺍﻟﺔ ﺍﻟﻔﺭﺩﻴﺔ : fﻓﺭﺩﻴﺔ ﻴﻌﻨﻲ : -ﺇﺫﺍ ﻜﺎﻥ x ∈ Df :ﻓﺈﻥ - x ∈ Df :ﻭ )f (-x) = - f (x -ﺇﺫﺍ ﻜﺎﻨﺕ fﻓﺭﺩﻴﺔ ﻓﺈﻥ ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﻤﺘﻨﺎﻅﺭ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﻤﺒﺩﺃ O
ﻤﻼﺤﻅﺎﺕ :ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ fﺯﻭﺠﻴﺔ ﺃﻭ ﻓﺭﺩﻴﺔ ﻴﻤﻜﻥ ﺘﻘﻠﻴﺹ ﻤﺠﺎل ﺍﻟﺩﺭﺍﺴﺔ ﺇﻟﻰ . ¡ + ∩ D f yy )f(-x)1 f (x f ( x)1 x-x 0 1 x -x x 01 x )f (-xﺘﻨﺎﻅﺭ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ . ﺘﻨﺎﻅﺭ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﻤﺒﺩﺃﺃﻤﺜﻠﺔ :ﺍﻟﺩﺍﻟﺔ ﺠﻴﺏ ﺘﻤﺎﻡ ﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ ﻷﻥ :ﻤﻥ ﺃﺠل ﻜل : x ∈ ℜ cos (-x) = cos x -ﺍﻟﺩﺍﻟﺔ ﺠﻴﺏ ﺩﺍﻟﺔ ﻓﺭﺩﻴﺔ ﻷﻥ :ﻤﻥ ﺃﺠل ﻜل : x ∈ ℜ sin (-x) = - sin x -ﺍﻟﺩﺍﻟﺔ x a x :ﻤﻌﺭﻓﺔ ﻋﻠﻰ ℜﻭ x = -x ﻭﻤﻨﻪ x a x :ﻫﻲ ﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ. x aﻟﻴﺴﺕ ﺯﻭﺠﻴﺔ ﻭ ﻟﻴﺴﺕ ﻓﺭﺩﻴﺔ ﻷﻥ ﻤﺠﻤﻭﻋﺔ -ﺍﻟﺩﺍﻟﺔ x ﺘﻌﺭﻴﻔﻬﺎ [∞D = [0 ; + ﻭ ﺇﺫﺍ ﻜﺎﻥ x > 0 :ﻓﺈﻥ - x < 0 :ﻻ ﺘﻨﺘﻤﻲ ﺇﻟﻰ .D ﺃﻱ ﺸﺭﻁ ﺘﻨﺎﻅﺭ Dﻏﻴﺭ ﻤﺤﻘﻘﺔ . ﺘﻤﺭﻴﻥ : fﻭ gﺩﺍﻟﺘﺎﻥ ﻤﻌﺭﻓﺘﺎﻥ ﻋﻠﻰ ℜﻜﻤﺎ ﻴﻠﻲ :َ f (x) = x2 + xﻭ g (x) = x x2 + 1 -ﺍﺩﺭﺱ ﺸﻔﻌﻴﺔ ﻜل ﻤﻥ fﻭ.g
ﺤـل : fﻤﻌﺭﻓﺔ ﻋﻠﻰ . ℜﻭﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ .x f (-x) = (-x)2 + (-x) = x2 - x ﻭﻤﻨﻪ َ f (-x) ≠ -f (x) :ﻭ )f (-x) ≠ f (x ﻤﺎﻋﺩﺍ ﻤﻥ ﺃﺠل . x = 0 ﺇﺫﻥ :ﺍﻟﺩﺍﻟﺔ fﻟﻴﺴﺕ ﺯﻭﺠﻴﺔ ﻭ ﻟﻴﺴﺕ ﻓﺭﺩﻴﺔ. ﺍﻟﺩﺍﻟﺔ gﻤﻌﺭﻓﺔ ﻋﻠﻰ ℜﻭ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ x )g (-x) = -x (-x)2 + 1 = -x x2 + 1 = -g (x ﻭﻤﻨﻪ ﺍﻟﺩﺍﻟﺔ gﺩﺍﻟﺔ ﻓﺭﺩﻴﺔ . -6ﺍﻟﺩﺍﻟﺔ ﻜﺜﻴﺭﺓ ﺤﺩﻭﺩ : * ﺘﻌﺭﻴﻑ n :ﻋﺩﺩ ﻁﺒﻴﻌﻲ .ﺍﻟﺩﺍﻟﺔ : P : x a an xn + an-1 xn−1 + ....a1 x + a0 ﺘﺴﻤﻰ ﺩﺍﻟﺔ ﻜﺜﻴﺭﺓ ﺍﻟﺤﺩﻭﺩ ﻟﻠﻤﺘﻐﻴﺭ . x ﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ an , an-1 , . . . , a1 , a0 : ﺘﺴﻤﻰ ﻤﻌﺎﻤﻼﺕ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ. ﺇﺫﺍ ﻜﺎﻥ an ≠ 0ﺍﻟﻌﺩﺩ ﺍﻟﻁﺒﻴﻌﻲ nﺘﺴﻤﻲ ﺩﺭﺠﺔ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ. an x1nﻫﻭ ﺍﻟﺤﺩ ﺫﻭ ﺃﻋﻠﻰ ﺩﺭﺠﺔ. ﻤﺜﺎل :ﺩﺍﻟﺔ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻫﻲ ﺩﺍﻟﺔ ﻤﻥ ﺍﻟﺸﻜل : x a a x2 + b x + c ﺤﻴﺙ a , b , cﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ﻭ . a ≠ 0 * ﺘﺴﺎﻭﻱ ﻜﺜﻴﺭﻱ ﺤﺩﻭﺩ :ﻴﺘﺴﺎﻭﻯ ﻜﺜﻴﺭﺍ ﺤﺩﻭﺩ ﺇﺫﺍ ﻜﺎﻨﺕ ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﺩﺭﺠﺔ ﻭ ﻤﻌﺎﻤﻼﺕ ﻭﺤﻴﺩﺍﺕ ﺍﻟﺤﺩ ﻤﻥ ﻨﻔﺱ ﺍﻟﺩﺭﺠﺔ ﻤﺘﺴﺎﻭﻴﺔ . ﻤﺜﺎل P (x) = 5x3 + 2x2 + 7x - 1 : Q (x) = a x3 + b x2 + c x + d ﺇﺫﺍ ﻜﺎﻥ P = Q :ﻓﺈﻥ a = 5 ; b = 2 ; c = 7 ; d = -1 : ﺘﻁﺒﻴﻕ : ﺠﺩ ﻜﺜﻴﺭ ﺤﺩﻭﺩ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ ) Q (xﻴﺤﻘﻕ: ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ : x
)x3 + 2x2 - 6x + 3 = (x - 1) Q (x ﺇﺭﺸﺎﺩ :ﺍﻜﺘﺏ ) Q(xﻋﻠﻰ ﺸﻜﻠﻪ ﺍﻟﻌﺎﻡ . ﺃﻨﺸﺭ ﻭﺒﺴﻁ ﻭﺭﺘﺏ ﺍﻟﺠﺩﺍﺀ ). (x – 1) Q (x rr -7ﺘﻐﻴﻴﺭ ﺍﻟﻤﻌﻠﻡ :ﺍﻟﻤﻌﻠﻡ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ (C f ) . O ; i , jﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ y = f (x) :ﻓﻲ) ( rr ﻫﺫﺍ ﺍﻟﻤﻌﻠﻡ .ﻓﻲ ﺍﻟﻤﻌﻠﻡ) ) ( ( ) ) ( ( ) ( x0 ; y0 ﻓﻲ ﺍﻟﻤﻌﻠﻡ O ; i , j ﻟﺘﻜﻥ Aﻨﻘﻁﺔ ﺇﺤﺩﺍﺜﻴﺎﻫﺎ rrx′ ; y′ Mﻨﻘﻁﺔ ﺇﺤﺩﺍﺜﻴﺎﻫﺎ x ; yﻓﻲ ﺍﻟﻤﻌﻠﻡ O ; i , jﻭ rr( )uuuur rr uuuur . A;i,j rr ﻫﺫﺍ ﻴﻌﻨﻲ uurOMuu=urxi +uuyujur :؛AM = x′ i + y′ j uu ﻭﻟﺩﻴﻨﺎ OM = OA + AM : . x = x′ + x0 ﻭ ﻤﻨﻪ : y = y′ + y0 ( )r r . A ; i , jﻨﺭﻤﺯ ﻟﻬﺎ ﺒـ : ﺘﻐﻴﻴﺭ ﻫﺫﺍ ﺍﻟﻤﻌﻠﻡ ﻴﺅﺩﻱ ﺇﻟﻰ ﻤﻌﺎﺩﻟﺔ ) (C fﻓﻲ ﺍﻟﻤﻌﻠﻡ ﺍﻟﺠﺩﻴﺩ )y′ = g (x′ ( )r -ﺇﺫﺍ ﻜﺎﻨﺕ gﺯﻭﺠﻴﺔ ﻓﺈﻥ ﺍﻟﻤﺤﻭﺭ A ; iﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻰ ) . (C f -ﺇﺫﺍ ﻜﺎﻨﺕ gﻓﺭﺩﻴﺔ ﻓﺈﻥ ﺍﻟﻨﻘﻁﺔ Aﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻰ ) . (C f -8ﻤﺭﻜﺯ ﺍﻟﺘﻨﺎﻅﺭ ﻭ ﻤﺤﻭﺭ ﺍﻟﺘﻨﺎﻅﺭ ( )r r : ﺍﻟﻤﺴﺘﻭﻯ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ . O ; i , j ) (C fﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ f؛ ∆ ﻤﺴﺘﻘﻴﻡ ﻤﻌﺎﺩﻟﺘﻪ ( )x = a : ﻨﺘﻴﺠﺔ : 1
ﻴﻜﻭﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ x = aﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻰ ) (C fﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﻤﻥ ﺃﺠل) ( ﻜل x ∈ D f : a - x ∈ D f , a + x ∈ D f : ﻭ )f (a + x) = f (a – x ﻨﺘﻴﺠﺔ : 2ﺘﻜﻭﻥ ﺍﻟﻨﻘﻁﺔ ) A (a ; bﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻰ ) (C fﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ﻤﻥ ﺃﺠل ﻜل : x ∈ Df : a - x ∈ Df , a + x ∈ Df ﻭ f (a + x) + f (a – x) = 2b ﺘﻁﺒﻴﻕ : 1 fﻫﻲ ﺍﻟﺩﺍﻟﺔ x a 3x2 + 5x - 1 : ﻟﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ .f ﺘﻨﺎﻅﺭ ﻤﺤﻭﺭ ﻫﻭ x = 5 ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺃﻥ ﺒﺭﻫﻥ 6 ﺘﻁﺒﻴﻕ : 2 xa 2x - 1 fﻫﻲ ﺍﻟﺩﺍﻟﺔ x + 1 : ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻨﻘﻁﺔ ) A (-1 ; 2ﻫﻲ ﻤﺭﻜﺯ ﺘﻨﺎﻅﺭ ﻟﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ . f - IIﻋﻤﻠﻴﺎﺕ ﻋﻠﻰ ﺍﻟﺩﻭﺍل : -1ﺘﺴﺎﻭﻱ ﺩﺍﻟﺘﻴﻥ : ﺘﻌﺭﻴﻑ :ﺇﺫﺍ ﺘﺴﺎﻭﺕ ﺩﺍﻟﺘﺎﻥ fﻭ gﻨﻜﺘﺏ f = g : ﻭ ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ﻟﻬﻤﺎ ﻨﻔﺱ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ D ﻭ ﻤﻥ ﺃﺠل ﻜل xﻤﻥ f (x) = g (x) : D -2ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﺠﺒﺭﻴﺔ ﻋﻠﻰ ﺍﻟﺩﻭﺍل : fﻭ gﺩﺍﻟﺘﺎﻥ )x a g (x) ; x a f (x ﻤﻌﺭﻓﺔ ﻤﻥ ﺃﺠل ﺍﻟﺘﻌﺭﻴﻑ ﺍﻟﺘﺭﻤﻴﺯ ﺍﻟﻌﻤﻠﻴﺔ f+g ﺍﻟﻤﺠﻤﻭﻉx ∈ D f ∩ Dg )x a f (x) + g(x ﺍﻟﻔﺭﻕ f-g ﺍﻟﺠﺩﺍﺀx ∈ D f ∩ Dg )x a f (x) − g(x ﺤﺎﺼل ﻗﺴﻤﺔ f.gx ∈ D f ∩ Dg )x a f ( x).g( x fx ∈ D f ∩ Dg xa )f (x g ﻭ g(x) ≠ 0 )g( x
-3ﺘﺭﻜﻴﺏ ﺩﺍﻟﺘﻴﻥ :ﺘﻌﺭﻴﻑ f :ﻭ gﺩﺍﻟﺘﺎﻥ .ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻲ ﻨﺭﻤﺯ ﻟﻬﺎ ﺒـ gofﻫﻲ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺭﻜﺒﺔ ﻭﺍﻟﺘﻲ ﺘﺭﻓﻕ ﺒﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ x ﺍﻟﻌﺩﺩ )) g ( f ( xﺃﻱ :))( gof ) (x) = g ( f (x ))x a f (x) a g ( f (x gof ﻤﺜﺎل : f :x a x-2؛ g:x a x ( gof )(x) = g ( f (x)) = f (x) = x - 2 ﺇﺫﻥ gof : x a x - 2 :ﻤﻌﺭﻓﺔ ﻋﻠﻰ ∞[ [2 ; + ﻤﻼﺤﻅﺔ : ﺍﻟﻜﺘﺎﺒﺔ ) gof ( xﻟﻬﺎ ﻤﻌﻨﻰ ﺇﺫﺍ ﻜﺎﻥ x ∈ D fﻭ f (x) ∈ Dg ﻭﻤﻨﻪ :ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ gofﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ x ﺤﻴﺙ x ∈ D f :ﻭ f (x) ∈ Dgﻭ ﺒﺎﻟﻤﺜل ﺍﻟﺩﺍﻟﺔ fogﻫﻲ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒـ ( )( fog)(x) = f g (x) : ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻤﺔ fog ≠ gof : ﻓﻲ ﺍﻟﻤﺜﺎل ﺍﻟﺴﺎﺒﻕ :( fog)(x) = f ( g (x)) = g(x) - 2 = x - 2 ﺘﻁﺒﻴﻕ := )f (x) = x2 + 1 ; g (x 1 fﻭ gﺩﺍﻟﺘﺎﻥ ﻤﻌﺭﻓﺘﺎﻥ ﻜﻤﺎ ﻴﻠﻲ : x (1ﺍﺤﺴﺏ ) ( fog)(xﺒﻌﺩ ﺘﺤﺩﻴﺩ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ . gof (2ﻨﻔﺱ ﺍﻷﺴﺌﻠﺔ ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﺩﺍﻟﺔ . fog
ﺍﻟﺤـل : (1ﺤﺴﺎﺏ ): ( fog)(x َ f (x) = x2 + 1ﻭ Df = ℜ َﻭ } Dg = ℜ - {0 = )g (x 1 x))x a f (x) a g ( f (xﺤﺘﻰ ﻴﻤﻜﻥ ﺤﺴﺎﺏ ) g f (xﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ ( )f (x) ≠ 0ﻭﻫﺫﺍ ﺼﺤﻴﺢ )ﻷﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ x2 + 1 = 0 :ﻟﻴﺱ ﻟﻬﺎ ﺤﻠﻭل ( ﻭﻤﻨﻪ Dgof = ℜ :ﻟﺤﺴﺎﺏ ) g f (xﻨﻌﻭﺽ xﺒـ ) f (xﻓﻲ ﻋﺒﺎﺭﺓ ) g (xﻓﻨﺠﺩ ( ):= ))g ( f (x 1 = 1 )f (x x2 + 1 (2ﺤﺴﺎﺏ ): ( fog)(x Df } = ℜ - {0 ; = )g (x 1 ﻟﺩﻴﻨﺎ : x Df = ℜ ; f (x) = x 2 + 1 ﻭﻤﻨﻪ ﺍﻟﺩﺍﻟﺔ fogﻤﻌﺭﻓﺔ ﻋﻠﻰ { }ℜ - 0ﻟﺩﻴﻨﺎ x a g (x) a f ( g (x)) :ﻭ ﻤﻨﻪ : 1 1 2 x x f (g ))(x = f = + 1 ))f ( g (x = 1 +1 ﺃﻱ: x2 ﻻﺤﻅ ﺃﻥ gof ≠ fog :
- IIIﺍﺘﺠﺎﻩ ﺍﻟﺘﻐﻴﺭ : - 1ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺠﻤﻭﻉ ﺩﺍﻟﺘﻴﻥ : ﻤﺒﺭﻫﻨﺔ : ﺇﺫﺍ ﻜﺎﻨﺕ fﻭ gﺩﺍﻟﺘﺎﻥ ﻤﺘﺯﺍﻴﺩﺘﺎﻥ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻤﺠﺎل . Iﻓﺈﻥ ﺩﺍﻟﺔ ﺍﻟﻤﺠﻤﻭﻉ ) (f + gﺘﻜﻭﻥ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ . Iﺇﺫﺍ ﻜﺎﻨﺕ fﻭ gﺩﺍﻟﺘﺎﻥ ﻤﺘﻨﺎﻗﺼﺘﺎﻥ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻤﺠﺎل . Iﻓﺈﻥ ﺩﺍﻟﺔ ﺍﻟﻤﺠﻤﻭﻉ ) (f + gﺘﻜﻭﻥ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ . I - 2ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ : λ. f ﻤﺒﺭﻫﻨﺔ : fﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﻤﺠﺎل λ . Iﻋﺩﺩ ﺤﻘﻴﻘﻲ . -ﺇﺫﺍ ﻜﺎﻥ λ > 0ﻓﺈﻥ fﻭ λfﻟﻬﻤﺎ ﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺍﻟﺘﻐﻴﺭ ﻋﻠﻰ ﺍﻟﻤﺠﺎل I -ﺇﺫﺍ ﻜﺎﻥ λ < 0ﻓﺈﻥ fﻭ λfﻟﻬﻤﺎ ﺍﺘﺠﺎﻫﺎﻥ ﻟﻠﺘﻐﻴﺭ ﻤﺘﻌﺎﻜﺴﺎﻥ ﻋﻠﻰ I -ﺇﺫﺍ ﻜﺎﻥ λ = 0ﻓﺈﻥ λfﺜﺎﺒﺘﺔ ﻋﻠﻰ . I ﻤﺜﺎل : ﻟﺘﻜﻥ f : x a x2 + 3 :ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞ [ [I = 0 ; + ﻭﻤﻨﻪ 5. f : x a 5 x2 + 15 : fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ Iﻭ 5. fﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ I -3ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﻤﺭﻜﺏ ﺩﺍﻟﺘﻴﻥ : ﻤﺒﺭﻫﻨﺔ : fﻭ gﺩﺍﻟﺘﺎﻥ ﺭﺘﻴﺒﺘﺎﻥ ﺘﻤﺎﻤﺎ I .ﻤﺠﺎل ﻤﺤﺘﻭﻯ ﻓﻲ . Df Jﻤﺠﺎل ﻤﺤﺘﻭﻯ ﻓﻲ . Dgﺒﺤﻴﺙ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠل ﻜل xﻤﻥ ، Iﻴﻜﻭﻥ ) f ( xﻤﻥ . J (1ﺇﺫﺍ ﻜﺎﻥ ﻟﻠﺩﺍﻟﺘﻴﻥ fﻭ gﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺍﻟﺘﻐﻴﺭ ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ gofﺘﻜﻭﻥ ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل . I (2ﺇﺫﺍ ﻜﺎﻥ ﻟﻠﺩﺍﻟﺘﻴﻥ fﻭ gﺍﺘﺠﺎﻫﺎﻥ ﻟﻠﺘﻐﻴﺭ ﻤﺨﺘﻠﻔﺎﻥ ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ gof ﺘﻜﻭﻥ ﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﺍﻟﻤﺠﺎل . I
ﻤﺜﺎل : hﻫﻲ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒـ h (x) = 1 - x : – 1ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ . h -2ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﺩﺍﻟﺔ hﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ. ﺤـل : ﻨﺭﻤﺯ ﺒـ fﻭ gﻟﻠﺩﺍﻟﺘﻴﻥ ﺍﻟﻤﻌﺭﻓﺘﻴﻥ ﺒـ g (x) = x ; f (x) = 1 - x : ﻭﻤﻨﻪ . h = gof : ﺍﻟﺩﺍﻟﺔ fﻤﻌﺭﻓﺔ ﻋﻠﻰ . ℜﻭﻴﺠﺏ ﺃﻥ ﻴﻜﻭﻥ f (x) ≥ 0ﺤﺘﻰ ﻨﻁﺒﻕ ﺍﻟﺩﺍﻟﺔ g؛ ﺃﻱ : 1 - x ≥ 0ﻭ ﻤﻨﻪ . x ≤ 1 : ﻭﻤﻨﻪ ﻤﺠﻤﻭﻉ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ hﻫﻲ ] ]. Dh = -∞ ; 1 ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ -∞ ; 1ﻭ ∞ [ [ ] ]. f (x) ∈ 0 ; +ﺍﻟﺩﺍﻟﺔ gﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ∞ 0 ; +ﻭﻤﻨﻪ ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ hﻤﺘﻨﺎﻗﺼﺔ ﺘﻤﺎﻤﺎ ﻋل ﺍﻟﻤﺠﺎل[ [ ]. ]-∞ ; 1 ﻤﻼﺤﻅﺔ :ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻤﺔ ﻻ ﻴﻤﻜﻥ ﺍﺴﺘﻨﺘﺎﺝ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ f – g :ﻭ f . g ﻤﻥ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺘﻴﻥ fﻭ. g -4ﺩﺭﺍﺴﺔ ﺘﻐﻴﺭﺍﺕ ﺩﺍﻟﺔ : ﻟﺩﺭﺍﺴﺔ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ fﻋﻠﻰ ﻤﺠﺎل . Iﻴﻤﻜﻥ : (1ﺍﻋﺘﺒﺎﺭﻫﺎ ﻜﻤﺠﻤﻭﻉ ﺩﻭﺍل ﺭﺘﻴﺒﺔ. (2ﺍﻋﺘﺒﺎﺭﻫﺎ ﻜﺠﺩﺍﺀ ﺩﺍﻟﺔ ﺒﻌﺩﺩ ﺤﻘﻴﻘﻲ . (3ﺍﻋﺘﺒﺎﺭﻫﺎ ﻜﻤﺭﻜﺏ ﻋﺩﺩ ﺩﻭﺍل ﻤﺄﻟﻭﻓﺔ . (4ﺍﺴﺘﻌﻤﺎل ﺍﻟﺘﻌﺭﻴﻑ ﻭ ﺩﺭﺍﺴﺔ ﺇﺸﺎﺭﺓ f (b) – f (a) :ﻋﻠﻤﺎ ﺃﻥ a ≤ bﻋﻠﻰ ﺍﻟﻤﺠﺎل .I ﻤﻼﺤﻅﺔ : ﺴﺘﺘﻌﺭﻑ ﻋﻠﻰ ﻁﺭﻴﻘﺔ ﺃﺨﺭﻯ ﻓﻲ ﺩﺭﻭﺱ ﻻﺤﻘﺔ.( )r r - IVﻤﻨﺤﻨﻴﺎﺕ ﻭ ﺘﺤﻭﻴﻼﺕ : fﺩﺍﻟﺔ ﻭ ) (C fﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ . O ; i , j kﻋﺩﺩ ﺤﻘﻴﻘﻲ .ﻜﻴﻑ ﻴﺘﻡ ﺍﺴﺘﻨﺘﺎﺝ ﻤﻨﺤﻨﻴﺎﺕ ﺍﻟﺩﻭﺍل :) x a f (-x) ; x a f (x) ; x a - f (xﻓﻲ ﻨﻔﺱ ﺍﻟﻤﻌﻠﻡ.
• ﺍﻟﺘﻨﺎﻅﺭ : -ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ) x a -f (xﻨﻅﻴﺭ ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ) x a f (xﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل . -ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ) x a f (-xﻨﻅﻴﺭ ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ) x a f (xﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ x aﻤﻥ ﺃﺠل ﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ . - ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ) x a f (xﻴﻨﻁﺒﻕ ﻋﻠﻰ ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ )f (x xﺍﻟﺘﻲ ﺘﺠﻌل f (x) ≥ 0 ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ) x a f (xﻨﻅﻴﺭ ﻤﻨﺤﻨﻰ ﺍﻟﺩﺍﻟﺔ ) x a f (xﻤﻥ ﺃﺠل xﺍﻟﺘﻲ ﺘﺠﻌل f (x) ≤ 0ﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﻤﺤﻭﺭ ﺍﻟﻔﻭﺍﺼل . f )(x = )f (x ; f (x) ≥ 0 ﻷﻥ : )-f (x ; f (x) ≤ 0 * ﺍﻻﻨﺴﺤﺎﺏ : ﻤﺒﺭﻫﻨﺔ : )y= f(x+k -ﺍﻟﻤﻨﺤﻨﻰ ) (C gﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ : r ﻫﻭ ﺼﻭﺭﺓ ﺍﻟﻤﻨﺤﻨﻰ) ( ) (C fﺒﺎﻻﻨﺴﺤﺎﺏ ﺍﻟﺫﻱ ﺸﻌﺎﻋﻪ . -ki -ﺍﻟﻤﻨﺤﻨﻰ ) (Chﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ y = f ( xr) + kﻫﻭ ﺼﻭﺭﺓ ﺍﻟﻤﻨﺤﻨﻰ) ( ) (C fﺒﺎﻻﻨﺴﺤﺎﺏ ﺍﻟﺫﻱ ﺸﻌﺎﻋﻪ . kj yy r r ) (Ch -ki kj) (Cg 1 1 01 x ) (C f 01 x ) (C f
ﺘﻁﺒﻴﻘﺎﺕ : -1ﺘﺭﻜﻴﺏ ﺩﺍﻟﺘﻴﻥ : f = )(x 1 1 : ﺒﺎﻟﺸﻜل ℜ - 1 fﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ 2x - 2 ﺍﻟﻬﺩﻑ :ﻜﺘﺎﺒﺔ fﻜﻤﺭﻜﺏ ﺩﺍﻟﺘﻴﻥ gﻭ (f = goh) hﺜﻡ ﻨﻔﺱ ﺍﻟﻌﻤل ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﺩﺍﻟﺔ fﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ℜ ﺒﺎﻟﺸﻜل f (x) = (x2 + 1)2 : -1ﺇﻴﺠﺎﺩ ﺩﺍﻟﺘﻴﻥ gﻭ hﺒﺤﻴﺙ ﻴﻜﻭﻥ : ﻤﻥ ﺃﺠل ﻜل f (x) = g ( h (x)) ; x ∈ D f ﻟﻬﺫﺍ ﻨﺠﺯﺃ ﺤﺴﺎﺏ )) f (xﻤﻊ ﺍﻷﺨﺫ ﺒﻌﻴﻥ ﺍﻻﻋﺘﺒﺎﺭ ﺃﻭﻟﻭﻴﺎﺕ ﻤﺭﺍﺤل ﻫﺫﺍ ﺍﻟﺤﺴﺎﺏ ( .ﻓﻨﺠﺩ ﻤﺭﺤﻠﺘﻴﻥ : 1 -ﺤﺴﺎﺏ 2 x - 1ﻴﻨﺠﺯ ﺒﺎﻟﺘﺭﺘﻴﺏ ﻋﻠﻰ ﻤﺭﺤﻠﺘﻴﻥ : 1 * ﺤﺴﺎﺏ * X = 2x – 1 :ﺤﺴﺎﺏ . X -ﻹﻴﺠﺎﺩ ﺍﻟﺩﺍﻟﺘﻴﻥ gﻭ hﺒﺤﻴﺙ f = gohﻴﺠﺏ ﺃﻥ ﻨﻌﺭﻑ ﺃﻥ ﺍﻟﺩﺍﻟﺔ h ﻫﻲ ﺍﻟﺘﻲ ﺘﻁﺒﻕ ﻓﻲ ﺍﻷﻭل . -2ﻜﺘﺎﺒﺔ ﺃﺨﺭﻯ ﻟﺩﺍﻟﺔ ﻨﺎﻁﻘﺔ : x2 - x x-2 = ){ }f (x ℜ -ﺒـ : 2 fﻫﻲ ﺩﺍﻟﺔ ﻨﺎﻁﻘﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ rr ﻭ ) (C fﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﻓﻲ ﻤﻌﻠﻡ ( )O ; i , j ﺍﻟﻬﺩﻑ :ﺍﻴﺠﺎﺩ ﺍﻷﻋﺩﺍﺩ ﺍﻟﺤﻘﻴﻘﻴﺔ a , b , cﺒﺤﻴﺙ ﻴﻜﻭﻥ :f (x) = a x + b + c ﻤﻥ ﺃﺠل ﻜل }: x ∈ ℜ -{2 x -2ﺜﻡ ﺩﺭﺍﺴﺔ ﺍﻟﻭﻀﻌﻴﺔ ﺍﻟﻨﺴﺒﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ ) (C fﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻱ ﺍﻟﻤﻌﺎﺩﻟﺔ y = ax + b :) f (xﻤﻌﻁﺎﺓ ax + b +ﺇﻟﻰ ﺤﺎﺼل ﻗﺴﻤﺔ ﻷﻥ c ﺃ( ﻨﺤﻭل ﺍﻟﻜﺘﺎﺒﺔ x-2 ﻋﻠﻰ ﺸﻜل ﺤﺎﺼل ﻗﺴﻤﺔ ﻟﻬﺫﺍ ﻨﻭﺤﺩ ﺍﻟﻤﻘﺎﻤﺎﺕ. -ﻨﻘﻭﻡ ﺒﺘﻭﺤﻴﺩ ﺍﻟﻤﻘﺎﻡ ﻭﻨﺒﺴﻁ ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﺍﻟﻤﻭﺠﻭﺩ ﻓﻲ ﺍﻟﺒﺴﻁ .
-ﺍﻟﻬﺩﻑ ﻫﻭ ﺇﻴﺠﺎﺩ ﺍﻷﻋﺩﺍﺩ a , b , cﺤﻴﺙ ﻤﻥ ﺃﺠل x ≠ 2x2 - x = a x2 + ( b - 2a) x + c - 2b x-2 x-2 ﺏ( ﺍﻟﻌﺒﺎﺭﺘﺎﻥ ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻟﻤﻘﺎﻡ ﻭ ﺤﺘﻰ ﺘﻜﻭﻨﺎﻥ ﻤﺘﺴﺎﻭﻴﺘﺎﻥ ﻴﻜﻔﻲ ﺃﻥ ﻴﻜﻭﻥ ﺍﻟﺒﺴﻁﺎﻥ ﻤﺘﺴﺎﻭﻴﺎﻥ ﺃﻱ :ﻤﻥ ﺃﺠل ﻜل x ≠ 2ﻭx2 - x = ax2 + (b - 2a) x + c - 2b -ﻜﺘﺎﺒﺔ ﺍﻟﺠﻤﻠﺔ ﺍﻟﺘﻲ ﺘﺘﺭﺠﻡ ﺘﺴﺎﻭﻱ ﻜﺜﻴﺭ ﺤﺩﻭﺩ ﻤﻥ ﻨﻔﺱ ﺍﻟﺩﺭﺠﺔ. -ﺘﻌﻴﻴﻥ . a , b , cﺝ( ﻟﻤﻌﺭﻓﺔ ﺇﺫﺍ ﻜﺎﻨﺕ ) (C fﻓﻭﻕ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ y = a x + b ﺘﻘﺎﺭﻥ ﺒﻴﻥ ﺍﻟﻌﺩﺩﻴﻥ ) f (xﻭ a x + bﻟﻬﺫﺍ ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ﺍﻟﻔﺭﻕ ) h (x) = f (x) – (ax + hﻭﻟﺩﻴﻨﺎ ﺸﻜﻼﻥ ﻟـ )f (x -ﻨﺨﺘﺎﺭ ﺍﻟﺸﻜل ﺍﻟﻤﻭﺍﻓﻕ ﻟﺤﺴﺎﺏ ﺍﻟﻔﺭﻕ )h (x -ﻨﺩﺭﺱ ﺇﺸﺎﺭﺓ ) h (xﺜﻡ ﻨﺴﺘﻨﺘﺞ .
ﺘﻤـﺎﺭﻴـﻥ ﻭ ﻤﺸﻜﻼﺕ ﺍﻟﺘﻤﺭﻴﻥ. 1 ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ fﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ : = )f (x 3 (2 = )f (x x+1 (1 x2 )x (x - 1= )f (x 2x - 1 (4 = )f (x x+5 (3 x2 + 1 x2 + xf )(x = x+1 (6 = )f (x x-1 (5 x2 - 4 3 + x2f (x) = x + 2 (8 f (x) = 3 - x (7 = ). f (x 1 (9 x ﺍﻟﺘﻤﺭﻴﻥ. 2 fﻭ gﺩﺍﻟﺘﺎﻥ ﻤﻌﺭﻓﺘﺎﻥ ﻜﻤﺎ ﻴﻠﻲ := )f (x 1 +x-1 ; g (x) = 2x + 3 - 1 x x -1ﻤﺎ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﺍﻟﺔ f + g؟ -2ﺍﺤﺴﺏ )(f + g) (x ﺍﻟﺘﻤﺭﻴﻥ. 3 fﻭ gﺩﺍﻟﺘﺎﻥ ﻤﻌﺭﻓﺘﺎﻥ ﻜﻤﺎ ﻴﻠﻲ : f (x) = 2x + 3 ; = )g (x 1 x -ﺍﺤﺴﺏ ( gof )(x) ; ( fog)(x) : ﺒﻌﺩ ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﺍﻟﺩﻭﺍل :
gof ; fog ; g ; f ﺍﻟﺘﻤﺭﻴﻥ. 4 ﺒﺴﻁ ﻭ ﺭﺘﺏ ﻭ ﺃﻋﻁ ﺩﺭﺠﺔ ﻜﺜﻴﺭﺓ ﺍﻟﺤﺩﻭﺩ :P (x) = x2 + 2x3 - x + x4) Q (x) = x2 - x3 - 6x + (x2 + 2x3 + x4)R (x) = x2 + 2x3 + 4x3 + x4 - (x2 + 2x3 - x ﺍﻟﺘﻤﺭﻴﻥ. 5 (1ﺍﺫﻜﺭ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ Uﻋﻠﻰ ℜﺤﻴﺙ U (x) = x2 : f : x a -5x2 (2ﺍﺴﺘﻨﺘﺞ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﻭﺍل ﺍﻟﺘﺎﻟﻴﺔ : g : x a x2 - 4؛ h : x a 0,5 x 2 + 2 ﺍﻟﺘﻤﺭﻴﻥ. 6 uﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ ﻤﺠﺎل −5 ; 1ﺒﺠﺩﻭل ﺘﻐﻴﺭﺍﺘﻬﺎ ﻜﻤﺎ ﻴﻠﻲ [ ]:x -5 -1 1u0 4 -3ﺝ( h = 0,5.u + 25 ﺸﻜل ﺠﺩﻭل ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﻭﺍل ﺍﻟﺘﺎﻟﻴﺔ : ﺃ( . f = −uﺏ( . g = 2.u ﺍﻟﺘﻤﺭﻴﻥ. 7 ﻟﺘﻜﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻌﺩﺩﻴﺔ fﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ 0 ; 2πﻜﻤﺎ ﻴﻠﻲ[ ]: f (x) = sin x + x ﻋﻠﻰ ﺃﻱ ﻤﺠﺎل ﻴﻤﻜﻥ ﺃﻥ ﻴﻜﻭﻥ -1 : fﻤﺘﺯﺍﻴﺩﺓ ؟ -2ﻤﺘﻨﺎﻗﺼﺔ ؟
ﺍﻟﺘﻤﺭﻴﻥ. 8 uﺩﺍﻟﺔ ﺤﻴﺙ u (x) = 2x – 8 : (1ﺍﺩﺭﺱ ﺇﺸﺎﺭﺘﻬﺎ ؟ ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ ؟ (2 (3 1 ﺍﺩﺭﺱ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ u : ﺍﻟﺘﻤﺭﻴﻥ. 9= )f(x x+5 ﻟﺘﻜﻥ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻌﺩﺩﻴﺔ fﺍﻟﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ: -2x 4 1 B f (x) = - 2 + -2x + 4 (1ﺍﻜﺘﺏ ) f (xﻋﻠﻰ ﺍﻟﺸﻜل : ﻤﻌﻴﻨﺎ ﺍﻟﻌﺩﺩ . B (2ﺍﺩﺭﺱ ﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ fﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞] [2 ; += )U (x 1 ﺍﻟﺘﻤﺭﻴﻥ. 10 x ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺘﻴﻥ U :ﻭ Vﺤﻴﺙ ; V (x) = 3x + 5 : -ﺍﻜﺘﺏ :)( VoU )(x) ; ( UoV )(x) ; ( UoU )(x) ; ( VoV )(x ﺍﻟﺘﻤﺭﻴﻥ. 11 ﺃﺼﺤﻴﺢ ﺃﻡ ﺨﻁﺄ : (1ﺍﻟﺩﺍﻟﺘﺎﻥ َ x a x2 + 1 :ﻭ x a x2 - 1 ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺍﻟﺘﻐﻴﺭ ﻋﻠﻰ . ℜ (2ﺠﺩﺍﺀ ﺩﺍﻟﺘﻴﻥ ﺘﺂﻟﻔﻴﺘﻴﻥ ﻫﻭ ﺩﺍﻟﺔ ﺘﺂﻟﻔﻴﺔ . x aﻤﺘﺯﺍﻴﺩﺓ ﻋﻠﻰ ∞] [0 ; + 1 (3ﺍﻟﺩﺍﻟﺔ x 2
ﻴﺴﺘﻨﺘﺞ ﻤﻥ ﺍﻟﻤﻨﺤﻨﻰ =y 1 (4ﺍﻟﻤﻨﺤﻨﻰ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ x+1 r 1 . j ﺒﺎﻻﻨﺴﺤﺎﺏ ﺍﻟﺫﻱ ﺸﻌﺎﻋﻪ =y x ﺫﻱ ﺍﻟﻤﻌﺎﺩﻟﺔ 4+ 7 4x - 5 ﻴﻤﻜﻥ ﻜﺘﺎﺒﺘﻪ ﻋﻠﻰ ﺍﻟﺸﻜل x - 3 : (5ﺍﻟﻜﺴﺭ x - 3 (6ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﺘﺎﻤﺎ ﻋﻠﻰ Iﻓﺈﻥ : ﺍﻟﺩﺍﻟﺔ f 2ﻤﺘﺯﺍﻴﺩﺓ ﺘﻤﺎﻤﺎ ﻋﻠﻰ .I (7ﻟﺘﻜﻥ َ f(x) = x :ﻭ g(x) = x4 + x2 + 1 ﻭ h(x) = x4 + x2 + 1ﺇﺫﻥ h = gof :( )x4+ 6x3+ 7x2 - 6x + 1 ﺍﻟﺘﻤﺭﻴﻥ. 12 ﺍﺤﺴﺏ x2+ p x + q 2 :ﺜﻡ ﺘﺤﻘﻕ ﺃﻥ ( ): ﻫﻭ ﻤﺭﺒﻊ ﻜﺜﻴﺭ ﺤﺩﻭﺩ ﻤﻥ ﺍﻟﺩﺭﺠﺔ ﺍﻟﺜﺎﻨﻴﺔ. y ) (C ﺍﻟﺘﻤﺭﻴﻥ. 13 4 3 ﺘﻐﻴﻴﺭ ﺍﻟﻤﻌﻠﻡ : 2 1 ﺍﻟﻤﻨﺤﻨﻰ ) (Cﻓﻲ ﺍﻟﺸﻜل ﺍﻟﻤﻘﺎﺒل-2 -1 0 ﻓﻲ) (y = x2 - 2x - 1 ﻤﻌﺎﺩﻟﺘﻪ : -1 r r -2 ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭ ﻤﺘﺠﺎﻨﺱ O ; i , j 1 2 3 4x -1ﻤﻥ ﺍﻟﺸﻜل ﻴﻤﻜﻥ ﺘﺨﻤﻴﻥ ﻤﺤﻭﺭ ﺍﻟﺘﻨﺎﻅﺭ ﻟﻤﺤﻭﺭ ﺍﻟﺘﺭﺍﺘﻴﺏ . ﻤﺎﻫﻲ ﻤﻌﺎﺩﻟﺘﻪ ؟ ( )r r -2ﺇﺨﺘﺭ ﻨﻘﻁﺔ Aﻤﻥ ﻫﺫﺍ ﺍﻟﻤﺤﻭﺭ ﻜﻤﺒﺩﺃ ﻟﻤﻌﻠﻡ ﺠﺩﻴﺩ . A ; i , j
ﻭﺍﻜﺘﺏ ﻤﻌﺎﺩﻻﺕ ﺘﻐﻴﻴﺭ ﺍﻟﻤﻌﻠﻡ . -3ﺍﺴﺘﻌﻤل ﻫﺫﻩ rﺍﻟﻤﻌrﺎﺩﻻﺕ ﻟﺘﻌﻴﻴﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺍﻟﺠﺩﻴﺩﺓ ﻟﻠﻤﻨﺤﻨﻰ ) (Cﻓﻲ) ( ﺍﻟﻤﻌﻠﻡ . A ; i , j -ﺃﺜﺒﺕ ﺃﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﻤﺨﻤﻥ ﻫﻡ ﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻰ ) . (C= )g (x 1 َﻭ ﺍﻟﺘﻤﺭﻴﻥ. 14 1+x ﻟﺘﻜﻥ f (x) = 1 - x + x2 - x3 : ﺃ( ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺤﺎﺴﺒﺔ ﺃﻭ ﻤﺠﺩﻭل ﺃﻋﻁ ﺠﺩﻭل ﻗﻴﻡ fﻭ gﻤﻥ ﺃﺠل x ﻴﺘﻐﻴﺭ ﻤﻥ 0,1ﺇﻟﻰ . 0,01ﻭ ﻗﺎﺭﻥ ﺍﻟﻨﺘﺎﺌﺞ . ﺏ( -ﺍﺤﺴﺏ )d (x) = g (x) – f (x -ﺍﺤﺴﺏ ) d (xﻤﻥ ﺃﺠل . x = 0,01 ، -ﺍﺸﺭﺡ ﻜﻴﻑ ﺘﺒﻴﻥ ﺍﻟﺩﺍﻟﺔ dﺃﻥ ) f (xﻭ ) g (xﻟﻬﻤﺎ ﻗﻴﻡ ﻤﺘﻘﺎﺭﺒﺔ ﺠﺩﺍ ﻤﻥ ﺃﺠل xﻗﺭﻴﺏ ﻤﻥ . 0 ﺍﻟﺘﻤﺭﻴﻥ. 15 ﺍﺴﺘﻌﻤﺎل ﻤﺠﺩﻭل Exelﺍﻟﺸﺎﺸﺔ ﺃﺩﻨﺎﻩ ﺘﻌﻁﻲ ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺩﺍﻟﺔ . fﺍﻟﻨﺘﻴﺠﺔ ﻤﻌﻁﺎﺓ ﻓﻲ ﺍﻟﺨﻠﻴﺔ B1 == -1ﻓﻲ ﺃﻱ ﺨﻠﻴﺔ ﻴﻜﻭﻥ ﺍﻟﻤﺒﺭﻤﺞ ﻗﺩ ﺘﻭﻗﻊ ﺤﺠﺯ ﻗﻴﻤﺔ x؟ -2ﺃ( ﻤﺎﻫﻤﺎ ﺍﻟﺩﺍﻟﺘﺎﻥ hﻭ gﺍﻟﻠﺘﺎﻥ ﺍﺴﺘﻌﻤﻠﻬﻤﺎ ﺍﻟﻤﺒﺭﻤﺞ ﻟﺘﺸﻜﻴل ﺠﺩﻭﻟﻪ . ﺏ( ﺍﻜﺘﺏ fﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺩﺍﻟﺘﻴﻥ . ﺝ( ﺃﻋﻁ ﻋﺒﺎﺭﺍﺕ h (x) :ﻭ ) g (xﻭ )f (x
ﺍﻟﺘﻤﺭﻴﻥ. 16 = )g (x x2 + 1 ﻟﺘﻜﻥ gﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻟـ : x2 - 1 ﺃ( ﻤﺎ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ ؟ ﺏ( ﺍﺩﺭﺱ ﺇﺸﺎﺭﺓ x 2 - 1 :ﺤﺴﺏ ﻗﻴﻡ . x g(x) = 1 + 2 ﺝ( ﺒﻴﻥ ﺃﻥ : x2 - 1ﺩ( ﻤﺎ ﻫﻲ ﻭﻀﻌﻴﺔ ) (C gﺍﻟﻤﻨﺤﻨﻰ ﺍﻟﻤﻤﺜل ﻟﻠﺩﺍﻟﺔ gﺒﺎﻟﻨﺴﺒﺔ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻱ ﺍﻟﻤﻌﺎﺩﻟﺔ y = 1؟ ﺍﻟﺘﻤﺭﻴﻥ. 17 ﻟﺘﻜﻥ fﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﻌﺭﻓﺔ ﺒـ f (x) = 2x3 - x 2 + 1 : ﺃﻋﻁ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ :)f ( 2x ; f x ; )f ( -x 2
ﺍﻟﺤـﻠـــﻭل . 1ﺍﻟﺘﻤﺭﻴﻥDf = ℜ - {0 ; 1} (2 ؛Df = ℜ - {0} (1Df = ℜ - {-1 ; 0} (4 ؛ Df = ℜ (3Df = ℜ (6 ؛Df =ℜ - {-2 ; 2} (5Df = ]-∞ ; 3] (8 ؛Df = [-2 ; +∞[ (7 . Df = ]0 ; +∞[ (9 . 2ﺍﻟﺘﻤﺭﻴﻥDf +g = ℜ - {0} ﻭﻤﻨﻪDf = Dg =ℜ - {0} (1 { }ℜ- 0 ﻤﻥx ( ﻤﻥ ﺃﺠل ﻜلf + g) (x) = 3x + 2 (2 . 3ﺍﻟﺘﻤﺭﻴﻥ( gof ) (x) = 1 3 ; ( fog) (x) = 2 +3 2x + xDfog = ℜ - {0} ; Dg = ℜ - {0} ; Df = ℜ Dgof =ℜ - − 3 2 p (x) = x4 + 2x3 + x2 − x . 4ﺍﻟﺘﻤﺭﻴﻥq (x) = x4 + x3 + 2x2 - 6x : ﻟﺩﻴﻨﺎ R (x) = x4 - x
ﺩﺭﺠﺔ ﻜﺜﻴﺭﺍﺕ ﺍﻟﺤﺩﻭﺩ ﻫﺫﻩ ﻫﻲ .4 : ﺍﻟﺘﻤﺭﻴﻥ. 5 (1ﺍﻟﺩﺍﻟﺔ ) uﺍﻟﺩﺍﻟﺔ \"ﻤﺭﺒﻊ\"( ﻤﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ ﺍﻟﻤﺠﺎل −∞ ; 0ﻭ ﻤﺘﺯﺍﻴﺩﺓ] ] ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞[ [. 0 ; + (2ﺍﻟﺩﺍﻟﺔ fﻤﺘﺯﺍﻴﺩﺓ ﻋﻠﻰ ﺍﻟﻤﺠﺎل −∞ ; 0ﻭ ﻤﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ] ] ﺍﻟﻤﺠﺎل [∞[0 ; +ﺍﻟﺩﺍﻟﺘﺎﻥ gﻭ hﻤﺘﻨﺎﻗﺼﺘﺎﻥ ﻋﻠﻰ −∞ ; 0ﻭﻤﺘﺯﺍﻴﺩﺘﺎﻥ ﻋﻠﻰ ∞[ [ ] ]0 ; +x -5 -1 ﺍﻟﺘﻤﺭﻴﻥ. 6 1u0 4 -3 -4-u 3 8 0 232u 0 -6h 26،5 25 ﺍﻟﺘﻤﺭﻴﻥ. 7 ﺍﻟﺩﺍﻟﺔ x a sin xﻤﺘﺯﺍﻴﺩﺓ ﻋﻠﻰ ﻜل ﻤﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ : 3π ; 2π ﻭ ; 0 π 2 2 ﻭﻜﺫﻟﻙ ﺍﻟﺩﺍﻟﺔ x a x :ﻤﺘﺯﺍﻴﺩﺓ ﻋﻠﻰ ﻜل ﻤﻥ ﻫﺫﻴﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ . ﻭﻤﻨﻪ ﺍﻟﺩﺍﻟﺔ fﻤﺠﻤﻭﻉ ﺩﺍﻟﺘﻴﻥ ﻤﺘﺯﺍﻴﺩﺘﻴﻥ ﻓﻬﻲ ﻤﺘﺯﺍﻴﺩﺓ ﻋﻠﻰ ﻜل ﻤﻥ ﻫﺫﻴﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ .ﻋﻠﻰ ﺍﻟﻤﺠﺎﻻﺕ ﺍﻷﺨﺭﻯ ﻻ ﻴﻤﻜﻥ ﺍﻻﺴﺘﻨﺘﺎﺝ .
ﺍﻟﺘﻤﺭﻴﻥ. 8 u (x) = 0 (1ﻤﻥ ﺃﺠل x = 4 u (x) > 0ﺇﺫﺍ ﻜﺎﻥ x > 4 u (x) < 0ﺇﺫﺍ ﻜﺎﻥ x < 4 (2ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ﻫﻲ Du = ¡ : x aﺘﻜﻭﻥ ﻤﻌﺭﻓﺔ ﻋﻠﻰ -∞ ; 4ﻭ ∞] [ ] [4 ; + 1 ( 3ﺍﻟﺩﺍﻟﺔ 2x - 8 ﻓﻬﻲ ﻤﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ -∞ ; 4ﻭ ﻤﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ ∞] [ ] [. 4 ; + ﺍﻟﺘﻤﺭﻴﻥ. 9 x+5 + 1 = 7 ﺃﻱ : )f(x - - 1 : ﻟﻨﺤﺴﺏ (1−2x + 4 2 -2x + 4 2 f ( x ) = − 1 + 7 ﺇﺫﻥ : 2 -2x+4 (2ﺍﻟﺩﺍﻟﺔ fﻤﺘﻨﺎﻗﺼﺔ ﻋﻠﻰ ﺍﻟﻤﺠﺎل ∞] [2 ; + ﺍﻟﺘﻤﺭﻴﻥ. 10])(VoU ) ( x) = V [U (x = 3 +5 x)(UoV )( x = U [V ])(x = 3 5 3x +ﻟﺩﻴﻨﺎ (VoV ) ( x) = 9 x + 20 ; (UoU ) ( x) = x : ﺍﻟﺘﻤﺭﻴﻥ. 11 (1ﺼﺤﻴﺢ ( 2 ،ﺨﻁﺄ ( 3 ،ﺨﻁﺄ ( 4 ،ﺨﻁﺄ (5ﺼﺤﻴﺢ ( 6 ،ﺨﻁﺄ ( 7 ،ﺨﻁﺄ .
ﺍﻟﺘﻤﺭﻴﻥ. 12 ﻟﻨﺘﺤﻘﻕ ﺃﻨﻪ ﻴﻭﺠﺩ pﻭ qﺒﺤﻴﺙ ﻴﻜﻭﻥ ﻤﻥ ﺃﺠل ﻋﺩﺩ ﺤﻘﻴﻘﻲ : x( x2 + Px + q)2 = x4 + 6x2 + 7x2 - 6x + 1 ﺒﻌﺩ ﺍﻟﻨﺸﺭ ﻭ ﺍﻟﻤﻁﺎﺒﻘﺔ ﻨﺠﺩ ﺍﻟﺠﻤﻠﺔ : 2p = 6 p + 2q = 7P=3 ; q = -1 ﻭ ﻤﻨﻪ ﺍﻟﺤﻼﻥ ﻫﻤﺎ : 2pq = -6 q2 = 1 ﺇﺫﻥ :ﻜﺜﻴﺭ ﺍﻟﺤﺩﻭﺩ ﺍﻟﻤﻌﻁﻰ ﻫﻭ ﻤﺭﺒﻊ x2 + 3 x - 1 : ﺍﻟﺘﻤﺭﻴﻥ. 13 x = 1 (1 x = x′ + 1 A (1 ; - 2) (2 ﻭﻤﻨﻪ : y = y′ − 2 (3ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻭ ﺒﻌﺩ ﺍﻻﺨﺘﺼﺎﺭ ﻨﺠﺩ : )y′ = x′2 = g( x′ gﺩﺍﻟﺔ ﺯﻭﺠﻴﺔ .ﻭﻤﻨﻪ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ x = 1 :ﻫﻭ ﻤﺤﻭﺭ ﺘﻨﺎﻅﺭ ﻟﻠﻤﻨﺤﻨﻰ ) . (C ﺍﻟﺘﻤﺭﻴﻥ. 14ﺃ( ﺒﺎﺴﺘﻌﻤﺎل ﻤﺠﺩﻭل ﺇﻜﺴﺎل \"Exel \" :ﻨﺤﺼل ﻋﻠﻰ ﺠﺩﻭل ﺍﻟﻘﻴﻡ ﺍﻟﺘﺎﻟﻲ :
= )d (x x4 ﺏ( ﺒﻌﺩ ﺍﻟﺤﺴﺎﺏ ﻨﺠﺩ : 1+x ﻨﻼﺤﻅ ﺃﻥ ) d (xﻤﻭﺠﺏ ﻤﻥ ﺃﺠل xﻗﺭﻴﺏ ﻤﻥ . Oﻤﻥ ﺠﻬﺔ ﺃﺨﺭﻯ ﺇﺫﺍ ﻜﺎﻥ x = 10-2ﻓﺈﻥ ) d (xﻗﺭﻴﺏ ﻤﻥ 10−8 ﻭﻤﻨﻪ ﺍﻟﻔﺎﺭﻕ ﺒﻴﻥ ) g (xﻭ ) f (xﺼﻐﻴﺭ ﺠﺩﺍ .ﻓﻲ ﺍﻟﺤﺎﻟﺔ ﺍﻟﻌﺎﻤﺔ ﺇﺫﺍﻜﺎﻥ ) x = 10-nﻤﻊ nﻁﺒﻴﻌﻲ ( n ≥ 1 .ﻓﺈﻥ ) d (xﻴﻜﻭﻥ ﻗﺭﻴﺏﻤﻥ x = 10-4nﺒﺠﻭﺍﺭ . x = 0ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻨﺎﻅﺭﻴﺔ \" gﻗﺭﻴﺒﺔ \" ﻤﻥ ﺍﻟﺩﺍﻟﺔ ﻜﺜﻴﺭﺓ ﺍﻟﺤﺩﻭﺩ fﻤﻊ )g (x) > f (x ﺍﻟﺘﻤﺭﻴﻥ. 15 -1ﻨﺘﻴﺠﺔ ﺍﻟﺨﻠﻴﺔ B1ﺘﺴﺘﻌﻤل ﻤﺤﺘﻭﻯ A4ﻭ ﻤﺤﺘﻭﻯ A4ﻨﺘﺤﺼل ﻋﻠﻴﻪ ﻤﻥ A2ﺇﺫﻥ :ﻗﻴﻤﺔ xﺘﺤﺠﺯ ﻓﻲ . A2h :x a -2ﺃ( ﺍﻟﺩﺍﻟﺔ hﺍﻟﺘﻲ ﺘﻤﺭ ﻤﻥ A2ﺇﻟﻰ A4ﻫﻲ x 2 + 1 : ) A2ﻴﻠﻌﺏ ﺩﻭﺭ (xg :x a ﺍﻟﺩﺍﻟﺔ gﺍﻟﺘﻲ ﺘﻤﺭ ﻤﻥ A4ﺇﻟﻰ B2ﻫﻲ x : ﺏ( f = goh
ﺝ( h (x) = x2 + 1 ; g (x) = x f (x) = x2 + 1 ﺍﻟﺘﻤﺭﻴﻥ. 16 ﺃ( ﻴﻤﻜﻥ ﺤﺴﺎﺏ ) g (xﺒﺸﺭﻁ ﺃﻥ ﻴﻜﻭﻥ : x2 - 1 ≠ 0ﺃﻱ x ≠ 1 :ﻭ x ≠ -1ﻤﺠﻤﻭﻋﺔ ﺍﻟﺘﻌﺭﻴﻑ ﻫﻲ D = ]-∞ ; -1[ U ]-1 ; 1[ U ]1 ; + ∞[ : ﺏ( )x2 - 1 = (x - 1) (x + 1 ﺒﺎﺴﺘﻌﻤﺎل ﺠﺩﻭل ﺍﻹﺸﺎﺭﺓ ﻟﻠﺠﺩﺍﺀ (x – 1) (x + 1) :ﻨﺠﺩ : x ∞− -1 ∞1 +x-1 - - + +x+1 - - +x2 - 1 + + ﻭﻤﻨﻪ x2 - 1 < 0 :ﻤﻥ ﺃﺠل ] [x ∈ -1 ; 1 x2 - 1 > 0ﻤﻥ ﺃﺠل [∞x ∈ ]-∞ ; -1[ U ]1 ; + = )g (x x2 - 1 + 2 ﺝ( ﻴﻤﻜﻥ ﻜﺘﺎﺒﺔ : x2 - 1 = )g (x x2 - 1 + 1 ﺃﻱ : x2 - 1 x2 - 1 g )(x = 1 + 2 1 x2 - ) ﻴﻤﻜﻥ ﻜﺫﻟﻙ ﺘﺤﻭﻴل ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﻤﻌﻁﺎﺓ ﻭ ﺇﻴﺠﺎﺩ )( g (x ﺤﺴﺏ ﺇﺸﺎﺭﺓ x 2 - 1 :ﻴﻜﻭﻥ ﺍﻟﻤﻨﺤﻨﻰ ﻓﻭﻕ ﺃﻭ ﺘﺤﺕ ﺍﻟﻤﺴﺘﻘﻴﻡ )(d ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ y = 1ﺇﺫﺍ ﻜﺎﻥ x ∈ -1 ; 1 :ﻓﺈﻥ ] [g (x) < 1
ﻭﻤﻨﻪ C :ﺘﻘﻊ ﺘﺤﺕ ) .(dﺇﺫﺍ ﻜﺎﻥ ∞] [ ] [ ( )x ∈ -∞ ; -1 U 1 ; + ﻓﺈﻥ g (x) > 1ﻭﻤﻨﻪ Cﺘﻘﻊ ﻓﻭﻕ )( ). (d ﺍﻟﺘﻤﺭﻴﻥ. 17 ﻟﺤﺴﺎﺏ ) f (2xﻴﻜﻔﻲ ﺘﻌﻭﻴﺽ xﺒـ 2xﻓﻲ ﻋﺒﺎﺭﺓ ) f (xﻓﻨﺠﺩ :f (2x) = 2 (2x)3 - (2x)2 + 1 = 16 x3 - 4x2 + 1 ﻭ ﻫﻲ ﺩﺍﻟﺔ ﺠﺩﻴﺩﺓ gﺤﻴﺙ )g (x) = f (2xf x = 2 x 3 - x 2 ﻭ ﺒﺎﻟﻤﺜل + 1 : 2 2 2 f x = x3 - x2 +1 2 4 4ﻭ ﻜﺫﻟﻙ f (-x) = 2 (-x)3 - (-x)2 + 1 : f (-x) = -2x3 - x2 + 1
ﺍﻟﺯﻭﺍﻴـﺎ ﺍﻟﻤﻭﺠﻬﺔ ﻭ ﺤﺴﺎﺏ ﺍﻟﻤﺜﻠﺜﺎﺕ ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ ﺤﺴﺎﺏ ﺃﻗﻴﺎﺱ ﺯﻭﺍﻴﺎ ﻤﻭﺠﻬﺔ ﺤل ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻤﺜﻠﺜﻴﺔ ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ﺍﻟﻤﺜﻠﺜﻴﺔ ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ ﺘﻌﺎﺭﻴﻑ ﻤﺸﺎﻜل ﻭ ﺘﻤﺎﺭﻴﻥ ﺍﻟﺤﻠﻭل
ﺗﻌﺎرﻳﻒ .Iﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ﻟﺸﻌﺎﻋﻴﻥ : َﻭ ﺸﻌﺎﻋﺎﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﻭﺠﻪ (C) .ﺍﻟﺩﺍﺌﺭﺓvr ur ﺍﻟﻤﺜﻠﺜﻴﺔ ﺍﻟﺘﻲ ﻤﺭﻜﺯﻫﺎ . O Y X)N(s B OXﻨﺼﻑ ﻤﺴﺘﻘﻴﻡ ﻟﻪ ﻨﻔﺱ) [ )M(t ﻤﻨﺤﻰ ﻭ ﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺍﻟﺸﻌﺎﻉ ur ﻭﻴﻘﻁﻊ ﺍﻟﺩﺍﺌﺭﺓ ) (Cﻓﻲ ﺍﻟﻨﻘﻁﺔ . M OYﻨﺼﻑ ﻤﺴﺘﻘﻴﻡ ﻟﻪ ﻨﻔﺱ[ )A' OA'vr r vr(C) B ur ﻤﻨﺤﻨﻰ ﻭﻨﻔﺱ ﺍﺘﺠﺎﻩ ﺍﻟﺸﻌﺎﻉ ﻫﻲ ﺒﺎﻟﺘﻌﺭﻴﻑ ﺍﻟﺯﺍﻭﻴﺔv َﻭur .N ﻓvrﻲ,ﺍﻟﻨﻘurﻁ(ﺔ ﺍﻟﺩﺍﺌﺭﺓ )(C ﻭﻴﻘﻁﻊ ﻟﻠﺸﻌﺎﻋﻴﻥ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ )ﺍﻟﻤﻭﺠﻬﺔ ) ( OM , ONﻟﻨﺼﻔﻲ ﺍﻟﻤﺴﺘﻘﻴﻤﻴﻥ َ OXﻭ [ ) [ ). OY .IIﺃﻗﻴﺎﺱ ﺍﻟﺯﻭﺍﻴﺎ ﺍﻟﻤﻭﺠﻬﺔ ﻟﺸﻌﺎﻋﻴﻥ :ﺃﻗﻴﺎﺱ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ) (ur , vrﺍﻟﻤﻌﺭﻓﺔ ﺴﺎﺒﻘﺎ ﻫﻲ ﺍﻷﻋﺩﺍﺩ s – tﺤﻴﺙ sﻭ tﻫﻤﺎ ﺍﻟﻌﺩﺩﺍﻥ ﺍﻟﻤﺭﻓﻘﺎﻥ ﺒﺎﻟﻨﻘﻁﺘﻴﻥ Nﻭ Mﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻋﻨﺩ ﻟﻑ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﻌﺩﺩﻱ ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ). (Cﻟ-ﻜلﺘﻌﻴﺯﻴﺍﻭﻥﻴ ﺃﺔﻗﻴﻤﺎﻭﺱﺠﻬﺯﺔﺍﻭﻴ)ﺔvrﻤ,ﻭﺠﻬurﺔ( :ﻤﺎﻻ ﻨﻬﺎﻴﺔ ﻤﻥ ﺍﻷﻗﻴﺎﺱ .ﺇﺫﺍ ﻜﺎﻥ αﻫﻭ ﺃﺤﺩ ﻫﺫﻩ ﺍﻷﻗﻴﺎﺱ ﻓﺈﻥ ﺃﻗﻴﺎﺱ ﻫﺫﻩ ﺍﻟﺯﺍﻭﻴﺔ ﺘﻌﻁﻰ ﺒﺎﻟﻌﺒﺎﺭﺓ θ = α + 2kπ :ﺤﻴﺙ. k ∈ Z : ﺃﻤﺜﻠﺔ : ( )uuur uuur (1ﺃﻗﻴﺎﺱ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ : OA , OB = π + 2kπ ; k∈Z 2
uuuuruuur (2ﺃﻗﻴﺎﺱ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ : OB,( )OB′ = π u-uur32π + 2kπ ; k∈Z uuuur 2 ( )OB′ , OB = -π + 2kπ ﺇﺫﻥ : ﻓﺈﻥ : N 3π ، M π (3ﺇﺫﺍ ﻜﺎﻨﺕ 4 6 ( )uuuur uuur = 3π - π + 2kπ ; ∈k Z OM , ON 4 6 ( )uuuur uuur = 7π + 2kπ ﺇﺫﻥ : 12 OM , ON 9π −π (4ﻫل ﺍﻟﻌﺩﺩﺍﻥ 5ﻭ 5ﻫﻤﺎ ﻗﻴﺴﺎﻥ ﻟﻨﻔﺱ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ؟ 9π - -π = 9π + π = 10π = 2π ﺍﻟﺤل : 5 5 5 5 9π = −π + 2π ﻭﻋﻠﻴﻪ : 5 5 9π −π ﺇﺫﻥ ﺍﻟﻌﺩﺩﺍﻥ 5ﻭ 5ﻫﻤﺎ ﻗﻴﺴﺎﻥ ﻟﻨﻔﺱ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ . 5π 3π (5ﻫل ﺍﻟﻌﺩﺩﺍﻥ 2ﻭ 3ﻫﻤﺎ ﻗﻴﺴﺎﻥ ﻟﻨﻔﺱ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ؟ 5π − π = 10π − π = 9π = 3π ﺍﻟﺤل : 3 6 6 6 2 5π π ﻭﻋﻠﻴﻪ 6 :ﻭ 3ﻫﻤﺎ ﻗﻴﺴﺎﻥ ﻟﺯﺍﻭﻴﺘﻴﻥ ﻤﻭﺠﻬﺘﻴﻥ ﻤﺨﺘﻠﻔﺘﻴﻥ . -ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻟﺯﺍﻭﻴﺔ ﻤﻭﺠﻬﺔ :
ﻤﻥ ﺒﻴﻥ ﺃﻗﻴﺎﺱ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ﻟﺸﻌﺎﻋﻴﻥ ﻴﻭﺠﺩ ﻗﻴﺱ ﺭﺌﻴﺴﻲ ﻭﺤﻴﺩ θﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺠﺎل ]. ]−π ; π θﺘﺴﻤﻰ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻟﻬﺫﻩ ﺍﻟﻘﻭﺱ ﺍﻟﻤﻭﺠﻬﺔ . ( )uuur uuur ﺃﻤﺜﻠﺔ : OA , OBﻓﻲ π 2 (1ﻫﻲ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻟﻠﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ( )uuur uuur ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ . OB , OA −π 2 (2ﻫﻭ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻟﻠﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ (3ﻋﻴﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻟﻠﺯﺍﻭﻴﺔ ﺍﻟﻤﻭﺠﻬﺔ ﺍﻟﺘﻲ ﺃﺤﺩ ﺃﻗﻴﺎﺴﻬﺎ . 2007π Rd 6 ﺍﻟﺤل :2007π = (6 × 334 + 3) π = 6 × 334π + 3π 6 6 6 6 = π + 334 π 2 π ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻟﻬﺫﻩ ﺍﻟﺯﺍﻭﻴﺔ ﻫﻭ . 2 1954π 5 Rad (4ﻨﻔﺱ ﺍﻟﺴﺅﺍل ﺍﻟﺴﺎﺒﻕ ﻤﻥ ﺃﺠل :1954π = (5 × 390 + )4 π = π + 390π ﺍﻟﺤل : 5 5 5 π ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ . 5(ur )vr (vr ) wr . )ﻏﻴﺭ wrﻤ,ﻌﺩﻭurﻤ(ﺔ ﺃﺸﻌﺔ ﻜﻴﻔﻴﺔ ﻭ ﺜﻼﺜﺔ ur : wr-ﺨﻭ,ﺍ vrﺹ Z (1 = + 2kπ , , + , ∈; k
vr - vr (2ﺍﻟ)ﺯﺍrﻭuﻴﺘﺎ,ﻥvr()ur=, -vr()vrﻭ(vr , (urur) , r u ﺍﻟﻫ))ﺯﺍﻤﺎvrrﻭvﻴﺘﺯ,,ﺎﺍﻥﻭﻴrrﺘuuﺎ)((ﻥ-vrﻤﺘ,وﻌﺎπﻜ)urﺴ(ﺘ=vrﺎ-ﻥو)vr(vr(,-u,urru)r.,ﻤﺘ-ﻘﺎ(+ﻴﺴﺘﺯﺎﺍπﻭﻥﻴﺘﺎ=ﻥ)ﻤﺘurﻜﺎ,ﻤﻠﺘﺎvrﻥ.(- (3 (4 (5ﻤﺠﻤﻭﻉ ﺯﻭﺍﻴﺎ ﺍﻟﻤﺜﻠﺙ uuur uuur uuur uuur uuur uuur : (AB , AC) + (BC , BA) + (CA , CB) = π + 2kπ ; k ∈ Z C (6ﺸﺭﻭﻁ ﻭﻗﻭﻉ ﺜﻼﺙ ﻨﻘﻁ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ : A , B , Cﺜﻼﺙ ﻨﻘﺎﻁ ﻤﺘﻤﺎﻴﺯﺓ ﻤﺜﻨﻰ ﻤﺜﻨﻰC AA BB ﺘﻜﻭﻥ ﺍﻟﻨﻘﻁ A , B , Cﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ :A (CuuCAur , uuur =Bπ + 2kπ uuur uuur )CB (CA , CB) = 2kπ ; k ∈ Zأو BI (7ﻤﻨﺼﻑ ﺯﺍﻭﻴﺔ : OIﻤﻨﺼﻑ ﺍﻟﺯﺍﻭﻴﺔ [ )AOB ( ) ( )A uuur uur ﻴﻌﻨﻲ ﺃﻥ : uur uuur OB , OI = OI , OA + 2kπ ; k ∈ Z O (8ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﺤﻴﻁﻴﺔ ﻭ ﺍﻟﺯﺍﻭﻴﺔ ﺍﻟﻤﺭﻜﺯﻴﺔ ﺍﻟﻠﺘﺎﻥ ﻴﺤﺼﺭﺍﻥ ﻨﻔﺱ ﺍﻟﻘﻭﺱ : M B , Aﻨﻘﻁﺘﺎﻥ ﻤﻥ ﺍﻟﺩﺍﺌﺭﺓ ) (Cﻤﺭﻜﺯﻫﺎ O O ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ Mﻤﻥ )(C) (C A ﻟﺩﻴﻨﺎ : uuuur uuur ( ) ( )uuur uuur OA , OB = 2 MA , MB + 2kπ B .IIIﺤﺴﺎﺏ ﺍﻟﻤﺜﻠﺜﺎﺕ : ( )r r ) (Cﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ﺍﻟﻤﺭﻓﻘﺔ ﺒﺎﻟﻤﻌﻠﻡ ﺍﻟﻤﺘﻌﺎﻤﺩ ﺍﻟﻤﺘﺠﺎﻨﺱ . O ; i , j
: ( ﺒﺤﻴﺙC) ﺘﻭﺠﺩ ﻨﻘﻁﺔ ﻭﺤﻴﺩﺓ ﻤﻥ، rθ ﺎuﺴﻬuﻴuﻗur(ur , vr) ﻤﻥ ﺃﺠل ﻜل ﺯﺍﻭﻴﺔ ﻤﻭﺠﻬﺔ ( i , OM )= θr + 2kπ ; k∈Z r BM ( )O;i , j M (ur , vr) ﻓﻲ ﺍﻟﻤﻌﻠﻡ ﻓﺎﺼﻠﺔ - sin θ cos r r θ :ﻫﻲ = cos θ ( )O ; i , j ﻓﻲ ﺍﻟﻤﻌﻠﻡM ﺘﺭﺘﻴﺏ- O cos θ A uuuursin (ur , vrr) = sin θr :ﻫﻲ OM = cos θ i + sinθ j :ﺇﺫﻥ : k ﻭ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺼﺤﻴﺢθ ﻨﺘﺎﺌﺞ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ cos2 θ + sin2 θ = 1 (1 -1 ≤ cos θ ≤ 1 ﻭ-1 ≤ sin θ ≤ 1 (2 cos(θ + k.2π) = cos θ َﻭsin(θ + k.2π) = sin θ (3 : ﻭ ﺍﻷﻋﺩﺍﺩ ﺍﻟﻤﺭﻓﻘﺔ ﺒﻪx ﺠﻴﺏ ﻭ ﺠﻴﺏ ﺘﻤﺎﻡ ﺍﻟﻌﺩﺩ : ﻟﺩﻴﻨﺎx ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ cos (- x ) = cos x ; sin(- x ) = - sin x (1 cos( π - x ) = -cos x ; sin( π - x ) = sin x (2 cos( π + x ) = - cos x ; sin( π + x ) = - sin x (3 cos π - x = sin x ; sin π - x = cos x (4 2 2 cos π + x = - sin x ; sin π + x = cos x (5 2 2 : ﺘﻁﺒﻴﻘﺎﺕcos -π ; sin 2π ; cos 5π ; cos 5π : ﺍﺤﺴﺏ ﻤﺎ ﻴﻠﻲ 3 3 4 6 : ﺍﻟﺤل
* cos 5π = cos π - π = - cos π =- 3 6 6 6 2* cos 5π = cos π + π = - cos π =- 2 4 4 4 2* sin 2π = sin π + π = cos π = 3 3 2 6 6 2* cos -π = cos π = 1 3 3 2 : ﻤﻌﺎﺩﻻﺕ ﻭ ﻤﺘﺭﺍﺠﺤﺎﺕ ﻤﺜﻠﺜﻴﺔ.IV : ﺍﻟﻤﻌﺎﺩﻻﺕ- : ﺘﻜﺎﻓﺊcos α = cos β : ( ﺍﻟﻤﻌﺎﺩﻟﺔ1 α = β + 2kπ أو k∈Z α = -β + 2kπ : ﺘﻜﺎﻓﺊsin α = sin β : ( ﺍﻟﻤﻌﺎﺩﻟﺔ2 α = β + 2kπ أو k∈Z α = π -β + 2kπ α = β + kπ ; k ∈ Z : ﺘﻜﺎﻓﺊtan α = tan β : ( ﺍﻟﻤﻌﺎﺩﻟﺔ3 : ﻜل ﻤﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔℜ ﺃﻤﺜﻠﺔ ﺤل ﻓﻲ 2cos x + 1 = 0 (1 sin 2x - π = sin x + π (2 3 6 tan 3x = tan x + π (3 3
: ﺍﻟﺤل cos x = - 1 : ﺘﻜﺎﻓﺊ2 cos x + 1 = 0 : ( ﺍﻟﻤﻌﺎﺩﻟﺔ1 2 x = 2π + 2kπ 2π = 3 3 x k∈Z :ﻭﻋﻠﻴﻪ cos x = cos : ﺃﻱ -2π 3 + 2kπ : ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ S = 2π + 2kπ ; -2π + 2kπ ; k ∈ Z 3 3 ﺘﻜﺎﻓﺊ sin 2x - π = sin x + π : ( ﺍﻟﻤﻌﺎﺩﻟﺔ2 3 6 2x - π =x+ π + 2k π 3 6 k ∈Z2x π π - 3 = π - x + 6 + 2k π x = π + 2kπ x = π + 2kπ 2 2 : ﺃﻱ : ﻭ ﻤﻨﻪ 7π 2kπ 3 x 7π x = 18 + 3 = 6 + 2kπ : ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ S = π + 2k π ; 7π + 2k π ; k ∈ Z 2 18 3 : ( ﺘﻜﻭﻥ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻤﻌﺭﻓﺔ ﺇﺫﺍ ﻜﺎﻥ3
َﻭ cos 3 x ≠ 0 cos x + π ≠ 0 3 3x ≠ π +kπ ; x + π ≠ π +kπ ; k∈Z ﺃﻱ : 2 3 2 ≠x π + kπ ﻭ x ≠ π +kπ ﻭﻋﻠﻴﻪ : 6 3 6 kπ k ≠x π + 3 ﻭ x ≠ π + 3 )(3π ﺇﺫﻥ : 6 6 x ≠ π + kπ ; k∈Z ﻭﻤﻨﻪ : 6 3 tanﺘﻜﺎﻓﺊ : 3 x = tan x + π ﺍﻟﻤﻌﺎﺩﻟﺔ : 3 . x = π + k′ π ﺃﻱ = 3x x+ π + k′ π 6 2 3 k′π kπ π + 2 ≠ π + 3 ﻭﻴﻜﻭﻥ ﺤﻼ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺇﺫﺍ ﻜﺎﻥ : 6 6 ﺃﻱ 3k′ ≠ 2k :ﺃﻱ ﻴﻜﻭﻥ k′ﻋﺩﺩ ﺼﺤﻴﺢ ﻓﺭﺩﻱ . -ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﺍﻟﻤﺜﻠﺜﻴﺔ : ﻤﺜﺎل: 1 ﺤل ﻓﻲ 0 ; 2πﺍﻟﻤﺘﺭﺍﺠﺤﺨﺔ [ ]2 cos x - 1 < 0 : ﺍﻟﺤل : π cos x < 1 ﻟﺩﻴﻨﺎ : 3 2 cos 5π = cos π = 1 ﻭﻨﻌﻠﻡ ﺃﻥ: 3 3 2 1 2 5π 3
cos x < 1 ﻭﻋﻠﻴﻪ ﻴﻜﻭﻥ 2 π ; 5π ﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ xﻓﻲ ﺍﻟﻤﺠﺎل 3 3 . S = π ; 5π ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻫﻲ : 3 3 ﻤﺜﺎل: 2 [ ]2 π 3 sin 2x - 6 < 2 ﺤل ﻓﻲ 0 ; πﺍﻟﻤﺘﺭﺍﺠﺤﺔ : ﺍﻟﺤل : ﻟﺩﻴﻨﺎ 0 ≤ x ≤ π :ﻭ ﻋﻠﻴﻪ 0 ≤ 2x ≤ 2π : -π ≤ 2x - π ≤ 2π - π ﻭﻤﻨﻪ : 6 6 6 -π 11π 6 ≤y ≤ 6 ﻨﺠﺩ : 2x - π =y ﻭ ﺒﻭﻀﻊ : 62π π ﻭﻤﻨﻪ ﻨﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ :33 3 −π ≤ y ≤ 11π < sin yﺤﻴﺙ : 3 2 6 6 2 -π sin π = sin 2π = 3 ﻨﻌﻠﻡ ﺃﻥ: 6 3 3 2 11π < sin yﻋﻨﺩﻤﺎ ﻴﻜﻭﻥ: 3 6 ﺇﺫﻥ ﺘﻜﻭﻥ 2 2π ≤ y ≤ 11π -π ≤ y ≤ π 3 ﺃﻭ 6 6 32π 11π -π3 ≤ 2x - π ≤ 6 ﺃﻭ 6 ≤ 2x - π ≤ π ﺇﺫﻥ : 6 6 3 5π 6 ≤ 2x ≤ 2π ﺃﻭ ≤0 2x ≤ π ﻭ ﻤﻨﻪ : 2
π5 ≤x ≤ π ﺃﻭ 0≤x ≤ π : ﻭ ﻋﻠﻴﻪ 12 4 S= 0 ; π U 5π ; π : ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ 4 12 : ﺩﺴﺎﺘﻴﺭ ﺍﻟﺘﺤﻭﻴل.V : ﻟﺩﻴﻨﺎb ﻭa ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩﻴﻥ ﺤﻘﻴﻘﻴﻴﻥ cos (a + b) = cos a .cos b - sin a .sin b (1 cos (a - b) = cos a .cos b + sin a .sin b (2 sin (a + b) = sin a .cos b + cos a .sin b (3 sin (a - b) = sin a .cos b - cos a .sin b (4 sin 5π ; Cos π ﺍﺤﺴﺏ: ﺘﻁﺒﻴﻘﺎﺕ 12 12 : ﺍﻟﺤل : ﻭ ﻤﻨﻪ π = π - π : * ﻟﺩﻴﻨﺎ 12 3 4cos π = cos π - π 12 3 4 = cos π .cos π + sin π .sin π 3 4 3 4 = 1 . 2 + 3 . 2 2 2 2 2 . cos π = 2+ 6 12 4 : ﺇﺫﻥ
: ﻭ ﻤﻨﻪ 5π = π + π : * ﻟﺩﻴﻨﺎ 12 4 6sin 5π = sin π + π 12 4 6 = sin π . cos π + sin π . cos π 4 6 6 4 = 2 . 3 + 1 . 2 2 2 2 2 . sin 5π = 2+ 6 : ﺇﺫﻥ 12 4 : ( ﻨﺠﺩ2) ( ﻭ1) ﺒﺠﻤﻊ ﺍﻟﺩﺴﺘﻭﺭﻴﻥ cos(a + b) + cos (a - b) = 2cos a .cos b (5 : ( ﻨﺠﺩ2) ( ﻭ1) ﺒﻁﺭﺡ ﺍﻟﺩﺴﺘﻭﺭﻴﻥ cos(a + b) -cos(a - b) = -2 sin a .sin b (6 : ( ﻨﺠﺩ4) ( ﻭ3) ﺒﺠﻤﻊ ﺍﻟﺩﺴﺘﻭﺭﻴﻥ sin(a + b) + sin(a - b) = 2 sin a .cos b (7 : ( ﻨﺠﺩ3) ( ﻭ4) ﺒﻁﺭﺡ sin(a + b) - sin(a - b) = 2cos a .sin b (8 a= x+y ; b= x-y : ؛ ﻨﺠﺩa – b = y ﻭa + b = x ﺒﻭﻀﻊ 2 2 : ﻭﻋﻠﻴﻪ cos x + cos y = 2 cos x + y . cos x - y (9 2 2 cos x - cos y = -2 sin x + y . sin x - y (10 2 2 sin x + sin y = 2 sin x + y . cos x - y (11 2 2
sin x- sin y = 2 cos x + y . sin x - y (12 2 2 cos (2a) = cos2 a - sin2 a (13 : b = a ﻭ ﺒﻭﻀﻊ : ( ﻨﺠﺩ1) ﻤﻥ ﺍﻟﺩﺴﺘﻭﺭ cos2 a + sin2 a = 1 : ﻟﻜﻥsin2 a = 1 - cos2 a ﺃﻭcos2 a = 1 - sin2 a : ﻭ ﻤﻨﻪ : ( ﻨﺠﺩ13) ﺒﺎﻟﺘﻌﻭﻴﺽ ﻓﻲ ﺍﻟﺩﺴﺘﻭﺭcos (2a) = 1 - 2 sin2 a ﺃﻭcos (2a) = 2cos2 a - 1 sin(2a) = 2 sin a . cos a (14 : ( ﻨﺠﺩ3) ﻤﻥ ﺍﻟﺩﺴﺘﻭﺭ : ﺘﻁﺒﻴﻘﺎﺕcos x + cos x - π = 2 : ﺍﻟﻤﻌﺎﺩﻟﺔℜ ( ﺤل ﻓﻲ1 2 2 : ﺍﻟﺤل π x + x - π x - x + π 2 2 2cos x + cos x - = 2 cos . cos 2 2 = 2 cos x - π .cos π 4 4 = 2× 2 cos x - π 2 4 = 2 cos x - π 4 2 cos x- π = 2 : ﻭﻤﻨﻪ ﺍﻟﻤﻌﺎﺩﻟﺔ ﺘﻜﺎﻓﺊ 4 2 cos x - π = cos π : ﺃﻱ cos x - π = 1 : ﺃﻱ 4 3 4 2
x = 7π + 2kπ x - π = π + 2kπ 12 4 3x k∈Z : ﺃﻱ :ﻭﻋﻠﻴﻪ -π π -π = 12 + 2kπ x - 4 = 3 + 2kπ : ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ S = 7π + 2kπ , −π + 2kπ ; k ∈ Z 12 12 cos π ( ﺍﺤﺴﺏ2 12 cos 2 × π =2 cos2 π -1 : ﻟﺩﻴﻨﺎ: ﺍﻟﺤل 12 12 : ﻭﻋﻠﻴﻪ cos π = 2 cos 2 π -1 : ﺃﻱ 6 12 π 1 + 3 cos π +1 π 12 2 2 6 12 cos 2 = : ﺃﻱ = cos2 2cos2 π = 8 +4 3 :ﺃﻱ cos 2 π = 2+ 3 :ﻭﻤﻨﻪ 12 16 12 4 2 2+2. 2 . 6 + 62 16 ( ) ( )cos2 π = : ﺇﺫﻥ 12 2+ 6 2 16 : ﺃﻱ ( )0 π π 2 π = < 12 < 2 : ﻭ ﺒﻤﺎ ﺃﻥ cos 12 . cos π = 2+ 6 : ﻭﻤﻨﻪ cos π >0 : ﻓﺈﻥ 12 4 12 a cos x + b sin x = c : ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻤﻥ ﺍﻟﺸﻜل- (*) a cos x + b sin x : ( ﺘﺤﻭﻴل ﺍﻟﻌﺒﺎﺭﺓ1
a cos a + b sin x = a2 + b2 cos (x - θ) : ﻟﺩﻴﻨﺎ cos θ = a : ﺤﻴﺙ a2 + b2 sin θ = b a2 + b2 a2 + b2 cos (x - θ) = c : ( ﺘﺼﺒﺢ ﺍﻟﻤﻌﺎﺩﻟﺔ2 . ﻭﺘﺤل ﺒﻁﺭﻴﻘﺔ ﻋﺎﺩﻴﺔ : ﻤﺜﺎل 3 cos x + sin x = 1 : ﺍﻟﻤﻌﺎﺩﻟﺔℜ ﺤل ﻓﻲ . ﺜﻡ ﻤﺜل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ : ﺍﻟﺤل 3 cos x + sin x : ( ﺘﺤﻭﻴل ﺍﻟﻌﺒﺎﺭﺓ1( )3 cos x + sin x = 3 2 + (1)2 cos(x - θ) θ = π : ﺃﻱ cos θ = 3 : ﺤﻴﺙ 6 θ = 2 1 sin 2 3 cos x + sin x = 2cos x - π : ﻭﻤﻨﻪ 6 2 cos x - π = 1 : ( ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ2 6 cos x - π = cos π : ﻭﻤﻨﻪ cos x - π = 1 : ﺃﻱ 6 3 6 2
x = π + 2k π x- π = π + 2kπ 2 6 3x k ∈Z ﺃﻱ : ﺇﺫﻥ : -π π -π = 6 + 2k π x- 6 = 3 + 2kπ (3ﺘﻤﺜﻴل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ : x = -π ﺃﻭ x = π ﻤﻥ ﺃﺠل : k = 0 6 2 ﻭﻋﻠﻴﻪ ﺼﻭﺭ ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ﻫﻤﺎ ﺍﻟﻨﻘﻁﺘﺎﻥ : M1 M0 -π ; M1 π 6 2 M0
ﺘﻤـﺎﺭﻴـﻥ ﻭ ﻤﺸﻜﻼﺕ . 1ﺍﻟﺘﻤﺭﻴﻥ [ ]: ﺤﻴﺙBC ﻨﻘﻁﺔ ﻤﻥI . ﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻭ ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥABCuur uuur uuur uuurAI , AB = α + 2kπ ; AB , AC π( ) ( )C = 2 + 2kπ α ∈ ℜ ; k ∈ Z : ﺤﻴﺙ uur uu: uﺔrﺯﻭﺍﻴﺎ ﺍﻟﺘﺎﻟﻴuﺍﻟuﺱur ﺃﻗﻴﺎuﺏuﺴur ﺍﺤ- AI , AC ; AB , BC ( ) ( )I ( )uur uuurAα B AI , BC A . 2ﺍﻟﺘﻤﺭﻴﻥED . ﻤﺜﻠﺙ ﻤﺘﻘﺎﻴﺱ ﺍﻷﻀﻼﻉABC ﻤﺜﻠﺜﺎﻥACD ﻭABE ﻗﺎﺌﻤﺎﻥ ﻭ ﻤﺘﻘﺎﻴﺴﺎ ﺍﻟﺴﺎﻗﻴﻥB uuur uuurC uuur uuur : ﻋﻴﻥ ﺃﻗﻴﺎﺱ ﺍﻟﺯﻭﺍﻴﺎ ﺍﻵﺘﻴﺔ- uuur uuur( ) ( ) ( )AB , AC ; DC , DA ; EB , EAuuur uuur uuur uuur uuur uuur (1( ) ( ) ( )CB , CD ; AE , AB ; BC , BE uuur uuur uuur uuur uuur uuur uuur uuur( ) ( ) ( ) ( )EA , ED ; EA , CB ; EC , BA ; EC , DB (2 . 3ﺍﻟﺘﻤﺭﻴﻥ AB = 2 Cm : ﺇﻟﻴﻙ ﺍﻟﺸﻜل ﺍﻟﻤﻘﺎﺒل ﺤﻴﺙ uuur uuur AB , AC 5π( )C = 6 + 2kπ ; k∈ZD AB
uuur uuur uuur uuur -1ﺍﺤﺴﺏ ﻜل ﻤﻥ uuur uuur :( ) ( ) ( )AC , CB ; CA , CB ; BC , BA -2ﺍﺤﺴﺏ . BC , DB , DA , DC cos π ; sin π -ﺍﺴﺘﻨﺘﺞ 12 12 ﺍﻟﺘﻤﺭﻴﻥ. 4 ABCﻤﺜﻠﺙ ﻤﺘﻘﺎﻴﺱ ﺍﻷﻀﻼﻉ . ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘـﻁ Mﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ( )uuuur uuuur : MB , MC = 0 + k (2π) ; k ∈ Z (1 uuuur uuuur MC , MB u=uurπ2 uuuur uuuur ( )uuur )+ k (2π ; k∈Z (2 ( ) ( )AB , AM = AM , AC + k (2π) ; k ∈ Z (3 ﺍﻟﺘﻤﺭﻴﻥ. 5ﻤﺜل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ﺍﻟﻨﻘﻁ ﺍﻟﻤﺭﻓﻘﺔ ﺒﺎﻷﻋﺩﺍﺩ ﺍﻵﺘﻴﺔ ﻋﻨﺩ ﻟﻑ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﻌﺩﺩﻱ ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ .- 5π ; -13π ; 3π ; 15π ; 25π ; 21π 3 2 2 6 4 ﺍﻟﺘﻤﺭﻴﻥ. 6 -ﻋﻴﻥ ﺍﻷﻗﻴﺎﺱ ﺍﻟﺭﺌﻴﺴﻴﺔ ﻟﻠﺯﻭﺍﻴﺎ ﺍﻟﻤﻭﺠﻬﺔ ﺍﻟﺘﻲ ﺇﺤﺩﻯ ﺃﻗﻴﺎﺱ ﻜل ﻤﻨﻬﺎ :344π ; ; 99π ; 120π ; -13π 2007π ; 177π ; 1830π 7 5 4 3 ﺍﻟﺘﻤﺭﻴﻥ. 7 ﻓπﻴﻤ0ﺎ ﻴ5ﻠ-ﻲ: ﻟﻠﻌﺩﺩ α tan αﻤﻥ ﺃﺠل ﻗﻴﻤﺔ ﺍﺤﺴﺏ ; cos α ; sin α 3 . α = (3 ؛ =α 1427π (2 ؛ α = 2007π (1 3 6
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