دروس مادة الرياضيات للفصل الاول شعب علمية سنة ثانية ثانوي

‫‪. α = -177π (6‬‬ ‫؛‬ ‫=‪α‬‬ ‫‪1962π‬‬ ‫؛ ‪(5‬‬ ‫‪α = 410π (4‬‬ ‫‪4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫‪ -‬ﻋﻴﻥ ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ﻨﻘﻁﺔ ‪ M‬ﻤﺭﻓﻘﺔ ﺒﺎﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ‪x‬‬ ‫‪.‬‬ ‫‪x‬‬ ‫∈‬ ‫‪‬‬ ‫‪0‬‬ ‫;‬ ‫‪π‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪‬‬ ‫‪6 ‬‬ ‫‪ -‬ﻋﻴﻥ ﺍﻟﻨﻘﻁ ‪ F , I , L , T , S‬ﺍﻟﻤﻌﻴﻨﺔ ﺒﺎﻷﻋﺩﺍﺩ ‪:‬‬ ‫; ‪π+x ; π-x‬‬ ‫‪π‬‬ ‫‪+x‬‬ ‫;‬ ‫‪π‬‬ ‫‪-x‬‬ ‫;‬ ‫‪-x‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪ -‬ﺍﺤﺴﺏ ﺇﺤﺩﺍﺜﻴﻲ ﻜل ﻤﻥ ﻫﺫﻩ ﺍﻟﻨﻘﻁ ﺒﺩﻻﻟﺔ ‪ x‬ﻭ ﺍﺫﻜﺭ ﺇﺸﺎﺭﺓ ﺇﺤﺩﺍﺜﻴﻲ ﻜل ﻤﻥ ﻫﺫﻩ ﺍﻟﻨﻘﻁ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬ ‫ﺒﺴﻁ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬‫‪cos‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪11π‬‬ ‫‪‬‬ ‫;‬ ‫‪sin‬‬ ‫)‪( x + 4π‬‬ ‫;‬ ‫)‪cos ( x + 25π‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫)‪cos ( x + 2007π‬‬ ‫;‬ ‫‪sin‬‬ ‫‪ 15π‬‬ ‫‪-‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪ 2‬‬ ‫‪‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬ ‫ﺃﻜﺘﺏ ﺒﺩﻻﻟﺔ ‪ tan x‬ﺍﻷﻋﺩﺍﺩ ﺍﻵﺘﻴﺔ ‪:‬‬‫)‪tan( x + 20π‬‬ ‫;‬ ‫‪tan‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪13π‬‬ ‫‪‬‬ ‫;‬ ‫)‪tan( x + 13π‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪tan‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪-‬‬ ‫‪x‬‬ ‫‪‬‬ ‫;‬ ‫‪tan‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪x‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬‫ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺤﺎﺴﺒﺔ ‪ :‬ﺍﺤﺴﺏ ﺍﻟﻘﻴﻡ ﺍﻟﻤﺩﻭﺭﺓ ﺇﻟﻰ ‪ 10- 5‬ﻟﻸﻋﺩﺍﺩ ﺍﻵﺘﻴﺔ ‪:‬‬ ‫‪cos‬‬ ‫‪π‬‬ ‫;‬ ‫‪sin‬‬ ‫‪π‬‬ ‫;‬ ‫‪cos‬‬ ‫‪π‬‬ ‫;‬ ‫‪sin‬‬ ‫‪π‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪7‬‬ ‫‪15‬‬ ‫‪2π‬‬ ‫‪13π‬‬‫‪cos‬‬ ‫‪9‬‬ ‫;‬ ‫‪sin‬‬ ‫‪π‬‬ ‫;‬ ‫‪cos‬‬ ‫‪5‬‬ ‫‪10‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬‫‪cos x = 0,3‬‬ ‫َﻭ‬ ‫‪-π‬‬ ‫‪<x‬‬ ‫‪<0‬‬ ‫ﺍﺤﺴﺏ ‪ sin x‬ﻭ ‪ tan x‬ﻋﻠﻤﺎ ﺃﻥ ‪:‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 13‬‬‫‪x ∈ 0‬‬ ‫;‬ ‫‪π‬‬ ‫َﻭ‬ ‫‪sin‬‬ ‫‪x‬‬ ‫=‬ ‫‪3‬‬ ‫ﺍﺤﺴﺏ ‪ cos x‬ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ‪:‬‬ ‫‪2 ‬‬ ‫‪5‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬ ‫ﺍﺤﺴﺏ ‪ cos x‬ﻭ ‪ tan x‬ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ‪:‬‬ ‫‪sin x = -0,6‬‬ ‫َﻭ‬ ‫‪3π‬‬ ‫‪<x‬‬ ‫‪< 2π‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 15‬‬

‫‪x‬‬ ‫∈‬ ‫‪π‬‬ ‫‪; π‬‬ ‫َﻭ‬ ‫= ‪cos x‬‬ ‫‪6- 2‬‬ ‫‪ 2‬‬ ‫ﻨﻀﻊ ‪4 :‬‬ ‫‪ -‬ﻤﺎ ﻫﻲ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﻘﻴﻘﻴﺔ ﻟـ ‪ sin x‬؟‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 16‬‬ ‫ﺤل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 ; 2π‬ﻜل ﻤﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪[ ]:‬‬‫)‪1‬‬ ‫‪cos‬‬ ‫‪x‬‬ ‫=‬ ‫‪1‬‬ ‫;‬ ‫= ‪2) cos x‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪3) sin x‬‬ ‫=‬ ‫‪-2‬‬ ‫;‬ ‫= ‪4) sin x‬‬ ‫‪-1‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 17‬‬ ‫ﺤل ﻓﻲ ‪ -π , π‬ﻜل ﻤﻥ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ﺜﻡ ﻤﺜل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ‪] ].‬‬‫)‪1‬‬ ‫) ‪sin (2 x‬‬ ‫=‬ ‫‪-‬‬ ‫‪3‬‬ ‫;‬ ‫)‪2‬‬ ‫)‪cos(2 x‬‬ ‫=‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪2‬‬‫‪3) cos x‬‬ ‫=‬ ‫‪-3‬‬ ‫;‬ ‫‪4) sin x‬‬ ‫=‬ ‫‪-1‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 18‬‬ ‫ﺤل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , π‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪[ ]:‬‬ ‫‪4cos2 x - 2 (1 + 2 ) cos x + 2 = 0‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 19‬‬ ‫ﺤل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , 2π‬ﻜل ﻤﻥ ﺍﻟﻤﺘﺭﺍﺠﺤﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﺜﻡ ﻤﺜل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ‪[ ]:‬‬

‫‪1 ) cos x ≤ 0‬‬ ‫;‬ ‫≤ ‪2) cos x‬‬ ‫‪2‬‬ ‫;‬ ‫≥ ‪3) sin x‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 20‬‬ ‫ﺍﻜﺘﺏ ﻜل ﻤﻥ ﺍﻟﻌﺒﺎﺭﺘﻴﻥ ﺍﻟﺘﺎﻟﻴﺘﻴﻥ ﺒﺩﻻﻟﺔ ‪َ cos x‬ﻭ ‪: sin x‬‬ ‫)‪sin (3x) ; cos (4x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 21‬‬ ‫ﺒﺭﻫﻥ ﻋﻠﻰ ﺼﺤﺔ ﺍﻟﻤﺴﺎﻭﻴﺎﺕ ﺍﻵﺘﻴﺔ ‪:‬‬‫‪1) cos4 x - sin4 x = cos2 x‬‬‫‪2) 2cos4 x + 2 sin4 x + sin2 (2x) = 2‬‬‫‪3) cos 2 x‬‬ ‫×‬ ‫‪sin 2 x‬‬ ‫=‬ ‫‪1-‬‬ ‫‪cos 4x‬‬ ‫‪8‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 22‬‬‫= ‪. cos x + sin x‬‬ ‫‪2‬‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫‪ (1‬ﺒﺭﻫﻥ ﺃﻥ ‪:‬‬ ‫‪‬‬ ‫‪4 ‬‬ ‫= ‪. cos x + sin x‬‬ ‫‪2‬‬ ‫‪ (2‬ﺤل ﻓﻲ ‪ ℜ‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬ ‫‪2‬‬ ‫‪ (3‬ﻤﺜل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ ‪.‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 23‬‬ ‫‪ -1‬ﺒﺴﻁ ﺍﻟﻌﺒﺎﺭﺓ ‪p (x) = sin 2x + sin x + sin 3 x :‬‬ ‫‪ -2‬ﺤل ﻓﻲ ‪ ℜ‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪. p (x) = 0 :‬‬

‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 24‬‬ ‫‪ (1‬ﺍﻜﺘﺏ ﺍﻟﻌﺒﺎﺭﺓ ‪p( x ) = cos x - 3 sin x :‬‬ ‫ﻋﻠﻰ ﺍﻟﺸﻜل ‪p( x ) = a cos ( x - θ ) :‬‬ ‫‪ (2‬ﺤل ﻓﻲ ‪ ℜ‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪. p (x) = 0 :‬‬ ‫‪ (3‬ﺤل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , π‬ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪[ ]p ( x ) ≥ 0 :‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 25‬‬‫‪sin x -‬‬ ‫> ‪3 cos x‬‬ ‫ﺤل ﻓﻲ ‪ ℜ‬ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪2 :‬‬ ‫‪3‬‬ ‫=‬ ‫‪tan‬‬ ‫‪π‬‬ ‫ﻴﻤﻜﻥ ﻭﻀﻊ ‪:‬‬ ‫‪3‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 26‬‬ ‫ﺤل ﻓﻲ ‪ ℜ‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪cos x + 3 sin x = 1 :‬‬ ‫ﻴﻤﻜﻥ ﺇﺩﺨﺎل ﻋﺩﺩ ‪ α‬ﺤﻴﺙ ‪. tan α = 3 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 27‬‬ ‫ﺤل ﻓﻲ ‪ ℜ‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪:‬‬‫‪cos2‬‬ ‫‪‬‬ ‫‪2x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫‪‬‬ ‫=‬ ‫‪sin2‬‬ ‫‪‬‬ ‫‪-x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪‬‬ ‫ﺜﻡ ﻤﺜل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ‪.‬‬

‫ﺍﻟﺤﻠـــــــﻭل‬ . 1‫ﺍﻟﺘﻤﺭﻴﻥ‬uur uuur uur uuur uuur uuur : ‫ ﺤﺴﺎﺏ ﺃﻗﻴﺎﺱ ﺍﻟﺯﻭﺍﻴﺎ‬-( ) ( ) ( )1) AI , AC = AI , AB + AB , AC + 2kπ ; k ∈Z ( )uur uuur = α + π + 2kπ ; k∈Z : ‫ﻭ ﻤﻨﻪ‬ AI , AC 2uuur uuur uuur uuur( ) ( )2) AB , BC = π + BA , BC + 2kπ ; k ∈ Z ( )uuur uuur = π - π + 2kπ , k∈Z : ‫ﻭ ﻤﻨﻪ‬ AB , BC 4 ( )uuur uuur = 3π + 2kπ ; k∈Z : ‫ﺃﻱ‬ 4 AB , BC uur uuur uur uuur uuur uuur( ) ( ) ( )3) AI , BC = AI , AB + AB , BC + 2kπ ; k ∈Z( )uur uuur =α + 3π + 2kπ ; k∈Z : ‫ﻭ ﻤﻨﻪ‬ AI , BC 4 . 2‫ﺍﻟﺘﻤﺭﻴﻥ‬uuur uuur ( k ∈ Z : ‫ ) ﻓﻲ ﻜل ﻤﻤﺎ ﻴﻠﻲ ﻨﺄﺨﺫ‬: ‫ﺤﺴﺎﺏ ﺃﻗﻴﺎﺱ ﺍﻟﺯﻭﺍﻴﺎ‬AB , AC( )1) • = π + 2kπ A 3 uuur uuur -π DC , DA 2( )• = + 2Ekπ D BC

uuur uuur uuur uuur uuur uuur( ) ( ) ( )• CB , CD = CB , CA + CA , CD uuur uuur =uuuπ4r + 2kπ uuur uuur( )• AE , AB uuur uuur uuur( ) ( ) ( )• BC , BE = BC , BA + BA , BE= π + π + 2kπ = 7π + 2kπ 3 4 12 uuur uuur uuur uuur uuur uuur( ) ( ) ( )2) • ED , EA + DA , DE + AE , AD = π +2kπ( ) ( )uuur uuur: ‫ﺴﺎﻗﻴﻥ ﻓﺈﻥ‬u‫ﺍﻟ‬u‫ﻱ‬ur‫ﻤﺘﺴﺎﻭ‬uuAurED ‫ﺒﻤﺎ ﺃﻥ ﺍﻟﻤﺜﻠﺙ‬ ED , EA = DA , DE uuur uuur uuur uuur uuur uuur :u‫ﺩ‬u‫ﺠ‬u‫ﻨ‬r‫ﺸﻜل‬u‫ﺍﻟ‬u‫ﻥ‬u‫ﻤ‬r‫ﻭ‬( ) ( ) ( ) ( )AE , AD = AE , AB + AB , AC + AC , AD= π + π + π = 10π = 5π 4 3 4 12 6 ( )uuur uuur 5π + 6 =π : ‫ﻭ ﻤﻨﻪ‬ 2 × ED , EA uuur uuur 5π EA , ED 6 ( )- 2× + = π : ‫ﻭﻋﻠﻴﻪ‬ ( )uuur uuur =π - 5π = π : ‫ﺃﻱ‬ 6 6 - 2 × EA , ED uuur uuur -π EA , ED 12( ). = + 2kπ : ‫ﺇﺫﻥ‬ ( ) ( ) ( )uuur uuur uuur uuur uuur uuur• EA , CB = EA , EB + EB , CB + 2kπ uuur uuur -π BE , BC( )= 2 + + 2kπ

= -π + - 7π + 2kπ 2 12 - 13π = 1uu2ur + 2kπuuur uuur uuur uuur uuur( ) ( ) ( )• EC , BA = EC , EB + EB , BA + 2kπ uuur uuur -π BE , BA ( )=4 + π + + 2kπ = -π +π + -π + 2kπ 4 4 = uπ2uur+ 2kπ uuuruuur uuur uuur uuur( ) ( ) ( )• EC , BA = EC , EB + EB , BA + 2kπ uuur uuur -π BE , BA ( )=4 +π + + 2kπ = -π +π + -π + 2kπ 4 4uuur uuur uuur =uπu2ur+ 2kπ uuur uuur( ) ( ) ( )• EC , DB = EC , BA + BA , DB + 2kπ uuur uuur π BA , BD ( )= 2 +π + + 2kπ = 3π + -π + 2kπ = 4π + 2kπ 2 6 3 . 3‫ﺍﻟﺘﻤﺭﻴﻥ‬ ( k ∈ Z ‫ ) ﻨﻌﺘﺒﺭ ﻓﻴﻤﺎ ﻴﻠﻲ‬: ‫( ﺍﻟﺤﺴﺎﺏ‬1 : ‫ ﻟﺩﻴﻨﺎ‬ABC ‫ﻓﻲ ﺍﻟﻤﺜﻠﺙ‬

uuur uuur uuur uuur uuur uuur( ) ( ) ( )• AB , AC + BC , BA + CA , CB = π uuur uuur ( )56πuu+ur2× BC , BA =π : ‫ﻤﻨﻪ‬ ‫ﻭ‬ uuur uuur uuur ( ) ( )BC , BA = CA , CB : ‫ﻷﻥ‬ uuur uuur CA , CB ( ) ( )uuur uuur = = π + 2kπ : ‫ﻭﻤﻨﻪ‬ 12 BC , BA uuur uuur ( ) ( )uuur uuur• AC , CB = π + CA , CB + 2kπ =π + π + 2kπ 12 13π = 12 + 2kπ : ‫( ﺍﻟﺤﺴﺎﺏ‬2 : ‫ ﻟﺩﻴﻨﺎ‬DAC ‫ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻘﺎﺌﻡ‬ uuur uuur DC AC , AD π ( )sin AC = 6 + 2kπ DAC = ‫َﻭ‬ *. DC = 1 :‫ﺃﻱ‬ DC = 2 × 1 sin π = DC :‫ﻭﻤﻨﻪ‬ 2 :‫ﺇﺫﻥ‬ 6 2 DA DA cos π = 2 : ‫ﻭﻤﻨﻪ‬ cos DAC = AC * 6. DA = 3 : ‫ﺃﻱ‬ DA = 2cos π = 2× 3 : ‫ﻭﻋﻠﻴﻪ‬ 6 2 . DB = 3 + 2 : ‫ ﻭﻋﻠﻴﻪ‬DB = DA + AB * BC2 = DB2 + DC2 : ‫ ﻟﺩﻴﻨﺎ‬DCB ‫ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻘﺎﺌﻡ‬( )BC2 = 8 + 4 3 ‫ ﺃﻱ‬BC2 = 3 + 2 2 + (1)2 : ‫ﻭﻤﻨﻪ‬ . BC = 2 2 + 3 : ‫ ﺃﻱ‬BC = 8 + 4 3 : ‫ﺇﺫﻥ‬

: cos π ; sin π ‫ ﺍﺴﺘﻨﺘﺎﺝ‬- 12 12 DC sin DBC = BC : ‫ ﻟﺩﻴﻨﺎ‬DBC ‫ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺍﻟﻘﺎﺌﻡ‬ sin π = 1= 1 : ‫ﺇﺫﻥ‬ 12 8 + 4 3 8 + 2 12 =1 ( ) ( )2 2 + 2 2 . 6 + 6 2 = 1 =1 ( 2 + )6 2 2+ 6 =( 2+ 2- 6 6) 6) ( 2 - = 2- 6 = 2- 6 = 6- 2 2-6 −4 4cos π = cos DBC = DB = 3 +2 12 BC 8+4 3 = 3 +2 = ( 3 + 2) ( 2- 6) 2+ 6 ( 2 + 6) ( 2- 6) cos π = 6 - 18 + 2 2 - 2 6 : ‫ﻭ ﻤﻨﻪ‬ 12 2-6 = 6 -3 2 +2 2 -2 6 -4 =- 6- 2= 6+ 2 -4 4

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫‪( )uuuur uuuur‬‬ ‫ﺘﻌﻴﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪: M‬‬ ‫‪MB , MC = 0 + k (2π) ; k ∈ Z ( 1‬‬ ‫ﺘﻌﻨﻲ ﺃﻥ ‪ M‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪ (BC‬ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﻘﻁﻌﺔ ‪[ ]BC‬‬ ‫‪( )uuuur uuuur‬‬ ‫ﻭﻟﺘﻜﻥ ‪( ). γ1‬‬ ‫‪MC , MB‬‬ ‫=‬ ‫‪π‬‬ ‫)‪+ k (2π‬‬ ‫;‬ ‫‪k∈Z‬‬ ‫‪(2‬‬ ‫‪2‬‬ ‫ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ‪ M‬ﻨﻘﻁﺔ ﻤﻥ ﻗﻭﺱ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻘﻁﺭ ‪ BC‬ﺒﺎﺴﺘﺜﻨﺎﺀ] [‬‫‪( ) ( )uuur uuuur uuuur uuur‬‬ ‫‪َ B‬ﻭ ‪ C‬ﻭﻟﺘﻜﻥ ‪( ). γ 2‬‬‫‪AB , AM = AM , AC + k (2π) ; k ∈ Z ( 3‬‬‫‪( ) ( )uuur uuur‬‬‫ﻫﺫﺍ ﻴﻌﻨﻲ ﺃﻥ ‪ M‬ﻨﻘﻁﺔ ﻤﻥ ﻤﻨﺘﺼﻑ ﺍﻟﺯﺍﻭﻴﺔ ‪ AB , AC‬ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﻨﻘﻁﺔ ‪ A‬ﻭﻟﺘﻜﻥ ‪. γ 3‬‬‫) ‪(γ1‬‬ ‫) ‪(γ 3‬‬ ‫) ‪C (γ 2‬‬ ‫) ‪A B (γ1‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬‫‪• -5π = -π - 4π .‬‬ ‫‪M‬‬ ‫‪5‬‬ ‫(‬ ‫‪25π‬‬ ‫)‬ ‫‪6‬‬‫•‬ ‫‪-13π‬‬ ‫=‬ ‫‪-π‬‬ ‫)‪- 4π .M1(-5π‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪3π‬‬ ‫‪3π‬‬‫•‬ ‫‪2‬‬ ‫=‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪0‬‬ ‫×‬ ‫‪π‬‬ ‫‪.‬‬ ‫‪M6‬‬ ‫(‬ ‫‪21π‬‬ ‫)‬ ‫‪M‬‬ ‫(‬ ‫‪-13π‬‬ ‫)‬ ‫‪4‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪15π‬‬ ‫‪3π‬‬ ‫‪3π‬‬‫•‬ ‫‪2‬‬ ‫=‬ ‫‪π‬‬ ‫= ‪+ 7π‬‬ ‫‪2‬‬ ‫‪+ 6π .‬‬ ‫‪M‬‬ ‫‪3‬‬ ‫(‬ ‫‪2‬‬ ‫)‬ ‫‪2‬‬ ‫‪25π‬‬ ‫‪M‬‬ ‫‪4‬‬ ‫(‬ ‫‪15π‬‬ ‫)‬ ‫‪6‬‬ ‫‪2‬‬ ‫‪π‬‬‫•‬ ‫=‬ ‫‪6‬‬ ‫‪+ 4π‬‬ ‫‪.‬‬‫•‬ ‫‪21π‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 5π‬‬ ‫=‬ ‫‪5π‬‬ ‫‪+ 4π .‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫ﺘﻌﻴﻴﻥ ﺍﻷﻗﻴﺎﺱ ﺍﻟﺭﺌﻴﺴﻴﺔ ‪:‬‬ ‫‪1830π‬‬ ‫=‬ ‫‪(3 × 610) π‬‬ ‫‪= 0 + 610π‬‬ ‫‪ ( 1‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ ‪. 0 :‬‬ ‫‪177π‬‬ ‫=‬ ‫‪(4 × 44 + 1) π‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 44π‬‬ ‫‪ ( 2‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪π‬‬ ‫ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ ‪. 4 :‬‬

‫‪2007π‬‬ ‫=‬ ‫‪(5 × 401+2) π‬‬ ‫=‬ ‫‪2π‬‬ ‫‪+ 401π‬‬ ‫‪ ( 3‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪5‬‬ ‫‪2π‬‬ ‫‪-3π‬‬ ‫=‬ ‫‪5‬‬ ‫‪-π‬‬ ‫‪+ 402π‬‬ ‫=‬ ‫‪5‬‬ ‫‪+ 402π‬‬ ‫‪- 3π‬‬ ‫ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ ‪. 5 :‬‬ ‫‪ ( 4‬ﻟﺩﻴﻨﺎ ‪ -13π = π - 14π :‬ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ ‪. π :‬‬ ‫‪ ( 5‬ﻟﺩﻴﻨﺎ ‪ 120π = 0 + 120π :‬ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ ‪. 0 :‬‬ ‫‪ ( 6‬ﻟﺩﻴﻨﺎ ‪99π = π + 98π :‬‬ ‫ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ ‪. π :‬‬‫‪344π‬‬ ‫=‬ ‫‪(7 × 49 + 1) π‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 49π‬‬ ‫‪ ( 7‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪7‬‬ ‫‪-‬‬ ‫‪6π‬‬ ‫=‬ ‫‪π‬‬ ‫‪-‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪50π‬‬ ‫=‬ ‫‪7‬‬ ‫‪+ 50π‬‬ ‫‪7‬‬ ‫‪- 6π‬‬ ‫ﺇﺫﻥ ﺍﻟﻘﻴﺱ ﺍﻟﺭﺌﻴﺴﻲ ﻫﻭ ‪. 7 :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬ ‫‪α‬‬ ‫=‬ ‫‪2007π‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 334π‬‬ ‫‪ ( 1‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪6‬‬ ‫‪2‬‬‫= ‪sinα‬‬ ‫‪sin‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪‬‬ ‫=‬ ‫‪1‬‬ ‫ﻭ‬ ‫‪cos‬‬ ‫‪α‬‬ ‫=‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪π‬‬ ‫‪‬‬ ‫=‬ ‫‪0‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬‫‪α‬‬ ‫=‬ ‫‪1427π‬‬ ‫=‬ ‫‪2π‬‬ ‫‪+‬‬ ‫‪475π‬‬ ‫‪ ( 2‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪2π‬‬ ‫‪-π‬‬‫=‬ ‫‪3‬‬ ‫‪-π‬‬ ‫‪+ 476π‬‬ ‫=‬ ‫‪3‬‬ ‫‪+ 476π‬‬ ‫‪cos‬‬ ‫‪α‬‬ ‫=‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪-π‬‬ ‫‪‬‬ ‫=‬ ‫‪cos ‬‬ ‫‪π‬‬ ‫‪‬‬ ‫=‬ ‫‪1‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪2‬‬

sin α = sin  -π  = - sin  π = - 3 3   3  2 -50π = -2π - 16π : ‫ ( ﻟﺩﻴﻨﺎ‬3 3 3 : ‫ﻭ ﻤﻨﻪ‬cos α = cos  -50π  = cos  -2π  = cos 2π  3   3  3 = cos  π - π = - cos π = -1  3  3 2sin α = sin  -2π  =- sin 2π =- sin  π - π  3  3  3  =- sin π = -3 3 2 α = 410 π = 0 + 410 π : ‫ ( ﻟﺩﻴﻨﺎ‬4 cos α = cos 0 = 1 ; sin α = sin 0 = 0 : ‫ﻭ ﻤﻨﻪ‬ 1962 π = 981 π = π + 490 π : ‫ ( ﻟﺩﻴﻨﺎ‬5 2 2 2cos α = cos π =0 ; sin α = sin π =1 : ‫ﻭ ﻤﻨﻪ‬ 2 2 α = -177 π = π - 188 π : ‫ ( ﻟﺩﻴﻨﺎ‬6cos α = cos π = - 1 ; sin α = sin π = 0 : ‫ﻭ ﻤﻨﻪ‬ L( π +x) T( π -x ) . 8‫ﺍﻟﺘﻤﺭﻴﻥ‬ 2 2 ‫ ( ﺘﻌﻴﻴﻥ ﺍﻟﻨﻘﻁ‬1I(π- x) M(x)F(π + x) S(-x) : ‫ ( ﺘﻌﻴﻴﻥ ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﻨﻘﻁ‬2

M( cos x , sin x) ; S(cos (-x) , sin (-x)) : ‫ﻟﺩﻴﻨﺎ‬ T cos ( π -x) , sin( π -x)  ; L cos ( π + x) , sin( π + x) 2 2  2 2I (cos(π- x) , sin(π- x)) ; F(cos(π + x) , sin(π + x)) : ‫ﻭﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﻤﺫﻜﻭﺭﺓ ﻓﻲ ﺍﻟﺩﺭﺱ ﻨﺠﺩ‬ M( cos x , sin x) ; S ( cos x , - sin x) T ( sin x , cos x) ; L (- sin x , cos x) I (-cos x , sin x) ; F (-cos x , - sin x) . ‫ ﻤﻭﺠﺒﺔ ﻭﺘﺭﺘﻴﺒﻬﺎ ﺴﺎﻟﺒﺔ‬S ‫* ﻓﺎﺼﻠﺔ‬ : ‫ﺇﺸﺎﺭﺓ ﺍﻹﺤﺩﺍﺜﻴﺎﺕ‬ . ‫ ﺴﺎﻟﺒﺔ ﻭﺘﺭﺘﻴﺒﻬﺎ ﻤﻭﺠﺒﺔ‬L ‫* ﻓﺎﺼﻠﺔ‬ . ‫ ﻤﻭﺠﺒﻴﻥ‬M ‫* ﺇﺤﺩﺍﺜﻴﻲ‬ . ‫ ﺴﺎﻟﺒﺔ ﻭﺘﺭﺘﻴﺒﻬﺎ ﻤﻭﺠﺒﺔ‬I ‫* ﻓﺎﺼﻠﺔ‬ . ‫ ﻤﻭﺠﺒﻴﻥ‬T ‫* ﺇﺤﺩﺍﺜﻴﻲ‬ . ‫ ﺴﺎﻟﺒﻴﻥ‬F ‫* ﺇﺤﺩﺍﺜﻴﻲ‬ . 9‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ﺘﺒﺴﻴﻁ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺎﻟﻴﺔ‬1) cos (x +25π) = cos (x+ π +24π) = cos(x + π)= -cos x2) sin (x + 4π) = sin x3) cos  x+ 11π  = cos  x + (2 × 5 + 1) π  2   2  = cos  x + π + 5π   2  = cos  x + π +π + 4π   2  cos  x+ 11π  = cos  x + π + π + 4π  : ‫ﻭ ﻤﻨﻪ‬  2   2 

= cos  x + π + π  = - cos  x + π  = sin x  2   2  4) sin  15π - x  = sin  (2 × 7 + 1) π - x   2   2  = sin  π +7π - x  = sin  π - x +π + 6π  2   2  = sin  π +( π - x)  = - sin  π - x  = - cos x  2   2 5) cos (x + 2007π) = cos (x + π + 2006π) = cos (x + π ) = - cos x . 10‫ﺍﻟﺘﻤﺭﻴﻥ‬• tan (x + 13π) = tan x• tan  x - 13π  = tan  x - (2×6 + 1)π  2  2  = tan x - π - 6π  = tan  x - π 2   2  = tan - π - x   =- tan  π - x  2   2 • tan ( x + 20 π) = tan x = - cot x tan  π  sin π + x  cos x 2  cos  2  - sin x• + x = π = =- cot x  2  + x 

 π  sin π - x  cos x  2  cos  2  sin x• tan - x = π = = cot x  2  - x  . 11‫ﺍﻟﺘﻤﺭﻴﻥ‬ : 10-5 ‫ﺍﻟﻘﻴﻡ ﺍﻟﻤﺩﻭﺭﺓ ﺇﻟﻰ‬ cos π ≈ 0,99995 ; sin π ≈ 0,00987 5 5 cos π ≈ 0,99998 ; sin π ≈ 0,00329 7 15 cos 2π ≈ 0,99994 ; sin π ≈ 0,00493 9 10 cos 13π ≈ 0,99178 . 5 . 12‫ﺍﻟﺘﻤﺭﻴﻥ‬ : sin x ‫* ﺤﺴﺎﺏ‬ sin2 x = 1 - cos2 x : ‫ ﻭ ﻤﻨﻪ‬sin2 x + cos2 x = 1 : ‫ﻟﺩﻴﻨﺎ‬ sin2 x = 0,91 : ‫ ﺃﻱ‬sin2 x = 1- (0,3)2 : ‫ﺇﺫﻥ‬ sin x = - 0,91 ‫ ﺃﻭ‬sin x = 0,91 : ‫ﺇﺫﻥ‬ sin x = - 0,91 : ‫ﻭﻋﻠﻴﻪ‬ -π <x <0 : ‫ﻟﻜﻥ‬ 2 sin x ≈ - 0,95 : ‫ ﺇﺫﻥ‬sin x < 0 : ‫ﻷﻥ‬ : tan x ‫* ﺤﺴﺎﺏ‬ tan x ≈- 0, 95 ≈- 3,18 : ‫ﻭﻋﻠﻴﻪ‬ tan x = sin x : ‫ﻟﺩﻴﻨﺎ‬ 0, 3 cis x . 13‫ﺍﻟﺘﻤﺭﻴﻥ‬

cos2 x = 1 -  3 2 : ‫ ﻭ ﻋﻠﻴﻪ‬cos2 x = 1 - sin2 x : ‫ﻟﺩﻴﻨﺎ‬  5  cos x = -4 ‫ﺃﻭ‬ cos x = 4 : ‫ﻭﻋﻠﻴﻪ‬ cos2 x = 16 : ‫ﺇﺫﻥ‬ 5 5 25 . cos x = 4 : ‫ﻭﻤﻨﻪ‬ cos x ≥0 : ‫ﺃﻱ‬ x ∈ 0 , π : ‫ﻟﻜﻥ‬ 5 2  . 14‫ﺍﻟﺘﻤﺭﻴﻥ‬ cos2 x = 1 - (-0,6)2 : ‫ ﻭﻋﻠﻴﻪ‬cos2 x = 1 - sin2 x : ‫ ﻟﺩﻴﻨﺎ‬- cos x = - 0,8 ‫ ﺃﻭ‬cos x = 0,8 : ‫ ﻭﻋﻠﻴﻪ‬cos2 x = 0,64 : ‫ﺇﺫﻥ‬ cos x = 0,8 : ‫ ﺇﺫﻥ‬cos x ≥0 : ‫ﻭﻋﻠﻴﻪ‬ x ∈  3π , 2π ‫ﻟﻜﻥ‬  2 tan x = -3 : ‫ﺇﺫﻥ‬ tan x = sin x = - 0,6 = -6 : ‫ ﻟﺩﻴﻨﺎ‬- 4 cos x 0,8 8 . 15‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ ﻭﻋﻠﻴﻪ‬sin2 x = 1 - cos2 x : ‫ﻟﺩﻴﻨﺎ‬sin2 x = 1 -  6- 2 2 = 1 - 6-2 6 . 2 +2  4  16   = 16 - 8 +2 6. 2= 8+2 6 . 2 16 4 2 6+ 2 ( ) ( ) ( )2 2 6 +2 6 . 2 + 2sin2 x = 4 = 4

‫‪sin x = -‬‬ ‫‪6+‬‬ ‫‪2‬‬ ‫أو‬ ‫= ‪sin x‬‬ ‫‪6+‬‬ ‫‪2‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪sin x > 0‬‬ ‫ﻓﺈﻥ ‪:‬‬ ‫‪x‬‬ ‫∈‬ ‫‪‬‬ ‫‪π‬‬ ‫‪,‬‬ ‫‪π‬‬ ‫‪‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫= ‪sin x‬‬ ‫‪6+‬‬ ‫‪2‬‬ ‫‪4‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 16‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻻﺕ ﻓﻲ ‪[ ]: 0 , 2π‬‬‫‪cos‬‬ ‫‪x‬‬ ‫=‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪π‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪cos‬‬ ‫‪x‬‬ ‫=‬ ‫‪1‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪‬‬ ‫‪3 ‬‬ ‫‪2‬‬ ‫‪x‬‬ ‫=‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫‪‬‬ ‫=‬ ‫‪3‬‬ ‫‪x‬‬ ‫;‬ ‫‪k∈Z‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪-π‬‬ ‫‪3‬‬ ‫‪+‬‬ ‫‪2kπ‬‬‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫أو‬ ‫‪x‬‬ ‫=‬ ‫‪π‬‬ ‫ﻴﻭﻀﻊ ‪ k = 0 :‬ﻨﺠﺩ ‪:‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪7π‬‬ ‫‪5π‬‬‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪3‬‬ ‫أو‬ ‫‪x‬‬ ‫=‬ ‫‪3‬‬ ‫ﻴﻭﻀﻊ ‪ k = 1 :‬ﻨﺠﺩ ‪:‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , 2π‬ﻫﻲ ‪[ ]:‬‬ ‫‪S‬‬ ‫=‬ ‫‪π‬‬ ‫‪,‬‬ ‫‪5π ‬‬ ‫‪.‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪‬‬‫‪cos x‬‬ ‫‪= cos‬‬ ‫‪π‬‬ ‫ﻭ ﻫﻲ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫= ‪cos x‬‬ ‫‪3‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪6‬‬ ‫‪2‬‬

‫‪x‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫‪‬‬ ‫=‬ ‫‪6‬‬ ‫‪x‬‬ ‫;‬ ‫‪k∈Z‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪-π‬‬ ‫‪6‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫ﺃﻭ‬ ‫‪x‬‬ ‫=‬ ‫‪π‬‬ ‫ﻤﻥ ﺃﺠل ‪ k = 0 :‬ﻨﺠﺩ ‪:‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪11π‬‬ ‫‪x‬‬ ‫=‬ ‫‪6‬‬ ‫ﺃﻭ‬ ‫‪) x‬ﻤﺭﻓﻭﺽ(‬ ‫=‬ ‫‪π‬‬ ‫‪+2π‬‬ ‫ﻤﻥ ﺃﺠل ‪ k = 1 :‬ﻨﺠﺩ ‪:‬‬ ‫‪6‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , 2π‬ﻫﻲ ‪[ ]:‬‬ ‫‪S‬‬ ‫=‬ ‫‪π‬‬ ‫‪,‬‬ ‫‪11π ‬‬ ‫‪.‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪‬‬ ‫‪sin x = sin‬‬ ‫‪ -π ‬‬ ‫‪sin‬‬ ‫‪x‬‬ ‫=‬ ‫‪-2‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪ 4  :‬‬ ‫‪2‬‬‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫‪+ 2kπ‬‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫‪+ 2kπ‬‬‫‪‬‬ ‫‪4‬‬ ‫‪ ‬ﺃﻱ ‪:‬‬ ‫‪4‬‬‫‪‬‬ ‫‪; k∈Z‬‬ ‫ﻭ ﻋﻠﻴﻪ‪:‬‬ ‫‪5π‬‬ ‫‪x‬‬ ‫‪π‬‬‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪4‬‬ ‫‪+ 2kπ‬‬ ‫‪=π‬‬ ‫‪+‬‬ ‫‪4‬‬ ‫‪+ 2kπ‬‬‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪5π‬‬ ‫ﺃﻭ‬ ‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫ﻟﻤﺎ ‪ k = 0 :‬ﻨﺠﺩ ‪:‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪5π‬‬ ‫‪7π‬‬ ‫‪) x‬ﻤﺭﻓﻭﺽ(‬ ‫=‬ ‫‪4‬‬ ‫‪+ 2π‬‬ ‫ﺃﻭ‬ ‫‪x‬‬ ‫=‬ ‫‪4‬‬ ‫ﻟﻤﺎ ‪ k = 1 :‬ﻨﺠﺩ ‪:‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , 2π‬ﻫﻲ ‪[ ]:‬‬ ‫‪S‬‬ ‫=‬ ‫‪ 5π‬‬ ‫‪,‬‬ ‫‪7π ‬‬ ‫‪.‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪sin x = sin‬‬ ‫‪ −π ‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪sin‬‬ ‫‪x‬‬ ‫=‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪ (4‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪ 6 ‬‬ ‫‪2‬‬

‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫‪+ 2kπ‬‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬‫‪‬‬ ‫‪6‬‬ ‫‪ ‬ﺃﻱ ‪:‬‬ ‫‪6‬‬‫‪‬‬ ‫‪; k∈Z‬‬ ‫ﻭ ﻋﻠﻴﻪ‪:‬‬ ‫‪7π‬‬ ‫‪x‬‬ ‫‪π‬‬‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪6‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫=‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪6‬‬ ‫‪+‬‬ ‫‪2kπ‬‬‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪7π‬‬ ‫ﺃﻭ‬ ‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫ﻟﻤﺎ ‪ k = 0 :‬ﻨﺠﺩ ‪:‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪7π‬‬ ‫‪11π‬‬ ‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪6‬‬ ‫‪+ 2π‬‬ ‫ﺃﻭ‬ ‫‪x‬‬ ‫=‬ ‫‪6‬‬ ‫ﻟﻤﺎ ‪ k = 0 :‬ﻨﺠﺩ ‪:‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , 2π‬ﻫﻲ ‪[ ]:‬‬ ‫‪S‬‬ ‫=‬ ‫‪ 7π‬‬ ‫‪,‬‬ ‫‪11π ‬‬ ‫‪.‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 17‬‬ ‫ﺤل ﺍﻟﻤﻌﺎﺩﻻﺕ ﻓﻲ ﺍﻟﻤﺠﺎل ‪] ]: -π , π‬‬‫ﻭﻋﻠﻴﻪ‪:‬‬ ‫‪sin 2x = sin‬‬ ‫‪ -π ‬‬ ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪sin‬‬ ‫‪2x‬‬ ‫=‬ ‫‪-3‬‬ ‫‪(1‬‬ ‫‪ 3 ‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪−π‬‬ ‫‪+ kπ‬‬ ‫‪2x‬‬ ‫=‬ ‫‪−π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪‬‬ ‫‪; k∈Z‬‬ ‫‪2π‬‬ ‫‪2x‬‬ ‫‪π‬‬ ‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪3‬‬ ‫‪+ kπ‬‬ ‫‪=π‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪+ 2kπ‬‬ ‫‪‬‬ ‫‪x‬‬ ‫=‬ ‫‪2π‬‬ ‫ﺃﻭ‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫ﻟﻤﺎ ‪ k = 0 :‬ﻨﺠﺩ ‪:‬‬ ‫‪3‬‬ ‫‪6‬‬ ‫‪2π‬‬ ‫‪5π‬‬ ‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪3‬‬ ‫‪+π‬‬ ‫ﺃﻭ‬ ‫‪x‬‬ ‫=‬ ‫‪6‬‬ ‫ﻟﻤﺎ ‪ k = 1 :‬ﻨﺠﺩ ‪:‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ -π , π‬ﻫﻲ ‪] ]:‬‬

S =  -π , 2π , 5π  .  3   6 6  cos 2x = cos π : ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ‬ cos(2x) = 1 : ‫( ﻟﺩﻴﻨﺎ‬2 3 2 x = π + kπ 2x = π + 2kπ 6  = 3 : ‫ﻭﻋﻠﻴﻪ‬ 2x ; k∈Z : ‫ﺇﺫﻥ‬ π π x = - 6 + kπ - 3 + 2kπ x = -π ‫ﺃﻭ‬ x = π : ‫ ﻨﺠﺩ‬k = 0 : ‫ﻟﻤﺎ‬ 6 6 5π x = 6 ‫)ﻤﺭﻓﻭﺽ( ﺃﻭ‬ x = π +π : ‫ ﻨﺠﺩ‬k = 1 : ‫ﻟﻤﺎ‬ 6 ] ]: ‫ ﻫﻲ‬-π , π ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻓﻲ ﺍﻟﻤﺠﺎل‬ S =  -π , π , 5π  .  6   6 6 : ‫ﻭﻋﻠﻴﻪ‬ cos x = cos 5π : ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ‬ cos x = -3 : ‫( ﻟﺩﻴﻨﺎ‬3 6 2 x = 5π + 2kπ  6 : ‫ﻭ ﻤﻨﻪ‬ x ; k∈Z  = - 5π 6 + 2kπ x = - 5π ‫ﺃﻭ‬ x = 5π : ‫ ﻨﺠﺩ‬k = 0 : ‫ﻟﻤﺎ‬ 6 6 . S =  -5π , 5π   6 6  sin x = sin  -π  : ‫ﻭﻫﻲ ﺘﻜﺎﻓﺊ‬ sin x = -1 : ‫( ﻟﺩﻴﻨﺎ‬4  6  2

‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫‪‬‬ ‫=‬ ‫‪6‬‬ ‫‪x‬‬ ‫;‬ ‫‪k∈Z‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫‪π‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪6‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫)ﻤﺭﻓﻭﺽ(‬ ‫‪x‬‬ ‫=‬ ‫‪7π‬‬ ‫ﺃﻭ‬ ‫‪x‬‬ ‫=‬ ‫‪-π‬‬ ‫ﻟﻤﺎ ‪ k = 0 :‬ﻨﺠﺩ ‪:‬‬ ‫‪6‬‬ ‫‪6‬‬ ‫‪-‬‬ ‫‪5π‬‬ ‫‪-π‬‬‫‪x‬‬ ‫=‬ ‫‪6‬‬ ‫ﺃﻭ‬ ‫‪) x‬ﻤﺭﻓﻭﺽ(‬ ‫=‬ ‫‪6‬‬ ‫‪+ 2π‬‬ ‫ﻟﻤﺎ ‪ k = 1 :‬ﻨﺠﺩ ‪:‬‬ ‫‪S‬‬ ‫=‬ ‫‪‬‬ ‫‪π‬‬ ‫‪,‬‬ ‫‪-5π ‬‬ ‫‪.‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 18‬‬ ‫ﺤل ﻓﻲ ﺍﻟﻤﺠﺎل ‪ 0 , π‬ﺍﻟﻤﻌﺎﺩﻟﺔ ‪[ ]:‬‬ ‫‪4cos2 x - 2 (1 + 2 ) cos x + 2 = 0‬‬ ‫ﺒﻭﻀﻊ ‪ cos x = y‬ﻨﺠﺩ ‪4 y2 - 2 ( 1 + 2 ) y + 2 = 0 :‬‬ ‫‪( ) ( )2‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪∆′ = 1 + 2 - 4 2‬‬ ‫‪( ) ( )2 2‬‬ ‫‪=1-2 2 + 2 = 1- 2‬‬ ‫ﻭﻋﻠﻴﻪ ‪∆′ = 1 - 2 :‬‬ ‫ﺇﺫﻥ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺤﻠﻴﻥ ﻫﻤﺎ ‪:‬‬‫‪y1 = 1 +‬‬ ‫‪2-‬‬ ‫‪2 +1‬‬ ‫;‬ ‫‪y2 = 1 +‬‬ ‫‪2+‬‬ ‫‪2 -1‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪y1‬‬ ‫=‬ ‫‪1‬‬ ‫;‬ ‫= ‪y2‬‬ ‫‪2‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬

cos x = 1 ‫ﺃﻭ‬ cos x = 2 : ‫ﻭﻋﻠﻴﻪ‬ 2 2 1 cos x = cos π : ‫ﺘﻜﺎﻓﺊ‬ cos x = 2 * 3 x = π + 2kπ  = 3 x ; k∈Z : ‫ﻭ ﻤﻨﻪ‬ -π 3 + 2kπ cos x = cos π : ‫ﺘﻜﺎﻓﺊ‬ cos x = 2 * 4 2 x = π + 2kπ  = 4 x ; k∈Z : ‫ﻭﻤﻨﻪ‬ -π 4 + 2kπ x = π ‫ﺃﻭ‬ x = π : ‫ ﻨﺠﺩ‬k = 0 ‫ﻤﻥ ﺃﺠل‬ 4 3 [ ]Sπ π =  3 , 4  : 0 , π ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻓﻲ ﺍﻟﻤﺠﺎل‬π . 19‫ﺍﻟﺘﻤﺭﻴﻥ‬2 [ ]: ‫ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ‬0 , 2π ‫ﺤل ﻓﻲ ﺍﻟﻤﺠﺎل‬ π ≤ x ≤ 3π : ‫ﻤﻌﻨﺎﻩ‬ cos x ≤ 0 (1 2 2 3π S = π , 3π  : ‫ﻭﻋﻠﻴﻪ‬ 2  2 2 

π : cos x ≤ 2 4 2 (2 cos π = cos 7π = 2 : ‫ﻟﺩﻴﻨﺎ‬ 4 4 2 7π x ∈  π , 7π  : ‫ﻭﻋﻠﻴﻪ‬ 4  4 4  S = π , 7π   4 4  sin π = sin 2π = 3 : ‫ ﻟﺩﻴﻨﺎ‬، sin x ≥ 3 3 3 2 2 (32π π x∈ π , 2π  : ‫ﻭﻋﻠﻴﻪ‬3 3  3 3  S= π , 2π   3 3  . 20‫ﺍﻟﺘﻤﺭﻴﻥ‬1) sin( 3x ) = sin( 2x + x ) : cos x ‫ َﻭ‬sin x ‫ﺍﻟﻜﺘﺎﺒﺔ ﺒﺩﻻﻟﺔ‬ = sin( 2x ).cos x + cos( 2x ).sin x = ( 2 sin x.cos x) .cos x + (1 − 2 sin2 x) .sin x = 2 sin x .cos2 x + sin x − 2 sin3 x2) cos( 4x ) = cos( 2 × 2x ) = 1 − 2.sin2( 2x ) = 1 − 2 × ( 2.sin x.cos x)2 = 1 − 8.sin2 x .cos2 x

. 21‫ﺍﻟﺘﻤﺭﻴﻥ‬1) cos4 x - sin4 x = ( cos2 x)2 - ( sin2 x)2 = ( cos2 x - sin2 x) ( cos2 x + sin2 x) = ( cos 2x) . (1) = cos 2x2) P(x) = 2cos4 x + 2 sin4 x + sin2 2 x = 2 cos4 x + 2 sin4 x + ( 2 sin x .cos x)2 = 2 ( cos2 x)2 + 2 sin2 x . cos2 x + ( sin2 x)2  = 2 ( cos2 x + sin2 x)2  = 2 × 1 = 23) ( cos 2 x) × ( sin2 x) = 1 + cos 2x × 1 - cos 2x 2 2 = 1 - ( cos 2x)2 4 1 - 1 + cos 4x  2 - 1- cos 4x  2  8 = = 4 1 - cos 4x = 8 . 22‫ﺍﻟﺘﻤﺭﻴﻥ‬ : ‫ ( ﺍﻹﺜﺒﺎﺕ‬1 2 cos  x+ π  = 2× cos x .cos π + sin x . sin π   4  4 4 

= 2  2 cos x + 2 sin x   2 2    = cos x + sin x = p(x) p(x) = 2 : ‫( ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬2 2 2 cos  x+ π  = 2 : ‫ ﺘﻜﺎﻓﺊ‬p(x) = 2  4  2 2 cos  x + π  = cos π : ‫ﺃﻱ‬ cos  x + π = 1 : ‫ﻭﻋﻠﻴﻪ‬  4  3  4  2 x + π = π + 2kπ  4 3 x + ; k∈Z : ‫ﻭﻤﻨﻪ‬ π −π 4 = 3 + 2kπ  x = π + 2kπ  12 : ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ‬  : ‫ﻭﻋﻠﻴﻪ‬ -7π  x = 12 + 2kπ  S= π + 2kπ , -7π + 2kπ ; k ∈ Z 12 12  π 12 : ‫ﺘﻤﺜﻴل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ‬ x = -7π ‫أو‬ x = π :k=0 • 12 12-7π12 . 23‫)*( ﺍﻟﺘﻤﺭﻴﻥ‬ : p (x) ‫( ﺘﺒﺴﻴﻁ‬1 p(x) = sin 2x + sin x + sin 3x

= sin 2x + 2 sin  x + 3x  cos  x - 3x   2   2  = sin 2x + 2 sin 2x . cos(-x) p(x) = sin 2x (1 + 2cos x ) : ‫ﻭﻤﻨﻪ‬ : p (x) = 0 : ‫( ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬2 1 + 2cos x = 0 ‫ ﺃﻭ‬sin 2x = 0 : ‫ ﺘﻜﺎﻓﺊ‬p (x) = 0 cos x = -1 ‫ﺃﻭ‬ sin 2x = 0 : ‫ﺃﻱ‬ 2 kπ x = 2 : ‫ ﺃﻱ‬2x = kπ ; k ∈ Z : ‫ ﺘﻜﺎﻓﺊ‬sin 2x = 0 * cos x = cos 2π : ‫ﺘﻜﺎﻓﺊ‬ cos x = -1 3 2* x = 2π + 2kπ  = 3 x ; k∈Z : ‫ﻭﻋﻠﻴﻪ‬ −2π 3 + 2kπ : ‫ﻭﻤﻨﻪ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ‬S=  kπ , 2π + 2kπ , −2π + 2kπ ; k ∈ Z  3 3   2 . 24‫)*( ﺍﻟﺘﻤﺭﻴﻥ‬cos x - a cos (x - θ) : ‫ ﻋﻠﻰ ﺍﻟﺸﻜل‬p (x) ‫ﻜﺘﺎﺒﺔ‬ ( )3 sin x = (1)2 + - 3 2 cos (x - θ) = 2 cos (x - θ)

‫‪-π‬‬ ‫‪cos‬‬ ‫‪θ‬‬ ‫=‬ ‫‪1‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪θ‬‬ ‫=‬ ‫ﺃﻱ‪:‬‬ ‫‪‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪-3‬‬ ‫‪‬‬ ‫‪sin‬‬ ‫‪θ‬‬ ‫=‬ ‫‪2‬‬ ‫‪‬‬‫‪cos x -‬‬ ‫‪3‬‬ ‫‪sin‬‬ ‫‪x‬‬ ‫‪=2‬‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪ (2‬ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ ‪p (x) = 0 :‬‬‫‪x+‬‬ ‫‪π‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ kπ‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫‪=0‬‬ ‫ﻭ ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪3 ‬‬ ‫‪x‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ kπ‬‬ ‫ﻭ ﻤﻨﻪ ‪; k ∈ Z :‬‬ ‫‪6‬‬‫‪.‬‬ ‫‪S‬‬ ‫=‬ ‫‪‬‬ ‫‪π‬‬ ‫‪+ kπ‬‬ ‫‪; k ∈ Z‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﻌﺎﺩﻟﺔ ﻫﻲ ‪:‬‬ ‫‪‬‬ ‫‪6‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ (3‬ﺤل ﻓﻲ ‪ 0 , π‬ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪[ ]p (x) ≥ 0 :‬‬ ‫‪cos‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫‪‬‬ ‫≥‬ ‫‪0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬‫‪π‬‬ ‫≤‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫≤‬ ‫‪4π‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪0 ≤ x ≤ π :‬‬‫‪3‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪4π‬‬ ‫‪π‬‬ ‫≤‬ ‫‪y‬‬ ‫≤‬ ‫‪3‬‬ ‫ﻭﺘﺼﺒﺢ ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ‪ cos y ≥ 0 :‬ﻤﻊ‬ ‫‪3‬‬ ‫‪π‬‬ ‫‪≤y‬‬ ‫≤‬ ‫‪π‬‬ ‫ﻭﻤﻨﻪ ‪ cos y ≥ 0 :‬ﺘﻜﺎﻓﺊ ‪:‬‬ ‫‪3‬‬ ‫‪2‬‬‫‪0‬‬ ‫‪≤x‬‬ ‫≤‬ ‫‪π‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪π‬‬ ‫≤‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪π‬‬ ‫≤‬ ‫‪π‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪6‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪[ ]S‬‬‫‪=0 ,‬‬ ‫‪π‬‬ ‫‪6 ‬‬ ‫‪ 0 , π‬ﻫﻲ‪:‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺤﻠﻭل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ ﻓﻲ ﺍﻟﻤﺠﺎل‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 25‬‬

sin x - 3 cos x > 2 : ‫ﺤل ﺍﻟﻤﺘﺭﺍﺠﺤﺔ‬sin x - tan π cos x > 2 : ‫ﻨﺠﺩ‬ 3 = tan π : ‫ﺒﻭﻀﻊ‬ 3 3 sin x - sin π . cos x > 2 : ‫ﺇﺫﻥ‬ cos 3 π 3 cos π .sin x - sin π . cos x 3 3 > 2 : ‫ﻭﻋﻠﻴﻪ‬ π cos 3 sin x .cos π - cos x .sin π > 2 .cos π : ‫ﻭﻋﻠﻴﻪ‬ 3 3 3 sin x - π > 2 3  2 : ‫ﻭﻋﻠﻴﻪ‬ sin y > 2 x - π =y : ‫ﻨﻀﻊ‬ 2 : ‫ﻭﻋﻠﻴﻪ‬ 3 sin π = sin 3π = 2 : ‫ﻭﻟﺩﻴﻨﺎ‬ 4 4 2π + 2kπ <y < 3π + 2kπ : ‫ ﻤﻌﻨﺎﻩ‬sin y > 24 4 2 : ‫ﺇﺫﻥ‬ 3π π + 2kπ <x - π < 4 + 2kπ : ‫ ﻭﻋﻠﻴﻪ‬k ∈ Z : ‫ﺤﻴﺙ‬ 4 3 3π π + π + 2kπ <x < π + 4 + 2kπ : ‫ﺇﺫﻥ‬ 3 4 3 7π 13π 12 + 2kπ < x < 12 + 2kπ : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬

: ‫ﻭ ﻤﻨﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﺤﻠﻭل ﻫﻲ ﺍﺘﺤﺎﺩ ﻜل ﺍﻟﻤﺠﺎﻻﺕ ﻤﻥ ﺍﻟﺸﻜل‬Z‫ ﻴﻤﺴﺢ ﻜل ﺍﻷﻋﺩﺍﺩ ﺍﻟﺼﺤﻴﺤﺔ‬k ‫ﻟﻤﺎ‬  7π + 2kπ ; 13π + 2kπ  12 12 . 26‫)*( ﺍﻟﺘﻤﺭﻴﻥ‬ cos x + 3sin x = 1 ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬cos x + tan α .sin x = 1 : ‫ ﻨﺠﺩ‬tan α = 3 ‫ﺒﻭﻀﻊ‬ cos x + sin α . sin x = 1 : ‫ﺇﺫﻥ‬ cos α cos α .cos x + sinα . sin x cos α =1 : ‫ﻭﻋﻠﻴﻪ‬cos (x - α) = cos α : ‫ﻭﺒﺎﻟﺘﺎﻟﻲ‬ cos (x - α) = 1 : ‫ﺇﺫﻥ‬ cos α x -α = α + 2kπ ; k∈Z : ‫ﻭﻋﻠﻴﻪ‬ x -α = -α + 2kπ x = 2α + 2kπ ; k∈Z : ‫ﻭﻤﻨﻪ‬ x = 2kπ S = {2α + 2kπ , 2kπ ; k ∈ Z} α = 1,25 rd : ‫ﺤﻴﺙ‬ . 27‫ﺍﻟﺘﻤﺭﻴﻥ‬ cos2  2x + π = sin2  -x + π  : ‫ﺤل ﺍﻟﻤﻌﺎﺩﻟﺔ‬  2   4 : ‫ﻭﻋﻠﻴﻪ‬ cos2  2x + π  - sin2  - x + π  = 0 : ‫ﻭﺘﻜﺎﻓﺊ‬  2   4 

cos 2x+ π  - sin -x+ π  cos 2x+ π  + sin - x+ π  = 0 2 4  2 4  cos  2x+ π  - sin -x+ π  = 0 . . . (1) : ‫ﻭﻋﻠﻴﻪ ﺇﻤﺎ‬  2  4  cos  2x+ π  + sin -x+ π  = 0 . . . (2) : ‫ﻭ ﺇﻤﺎ‬ 2  4  cos  2x+ π  = sin - x+ π  : ‫( ﺘﻜﺎﻓﺊ‬1)  2  4  cos  2x + π  = cos  π +x - π : ‫ﻭﻋﻠﻴﻪ‬  2   2 4  : ‫ﻭﻤﻨﻪ‬ cos  2x + π = cos  π + x  : ‫ﺃﻱ‬  2   4  x = -π + 2kπ 2x + π = π +x + 2kπ 4  2 4 + 2kπ : ‫ﺃﻱ‬ 2x ; k∈Z -π 2kπ π π x = 4 + 3 + 2 = - 4 -x cos  2x+ π  = - sin -x+ π  ‫( ﺘﻜﺎﻓﺊ‬2) 2  4  cos  2x+ π  = sin x - π  : ‫ﻭﻋﻠﻴﻪ‬  2  4  cos  2x + π = cos  π - x + π : ‫ﺇﺫﻥ‬  2  2 4  : ‫ﻭﻤﻨﻪ‬ cos  2x + π = cos  3π - x  : ‫ﺃﻱ‬  2   4 

 3x = π + 2kπ 2x + π = 3π - x + 2kπ  4 : ‫ ﺃﻱ‬ 2 = 4 + 2kπ  ; k∈Z -5π 2x π 3π  x = 4 + 2kπ + 2 - 4 +x   x = π + 2kπ  12 3: ‫ﻫﻲ‬ ‫ﺍﻟﺤﻠﻭل‬ ‫ﻤﺠﻤﻭﻋﺔ‬ ‫ﺇﺫﻥ‬  : ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬ -5π  x = 4 + 2kπ  S =  -π + 2kπ ; -π + 2kπ ; π + 2kπ ; -5π + 2kπ ; k ∈ Z  4 3 12 3 4   4 : ‫ﺘﻤﺜﻴل ﺍﻟﺤﻠﻭل ﻋﻠﻰ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺜﻠﺜﻴﺔ‬ x = -5π , x = π , x = -π :k=0 4 12 4 x = 7π , x = 3π , x = 5π : k = 1 3π 5π 4 4 1213π -5π 4 12 x = 17π , x = 13π : k = 2 44 12 12 π 1213π -π12 12 -π 17π 4 12

‫ﺍﻻﺸﺘﻘـﺎﻗﻴﺔ‬ ‫ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‬ ‫ﺍﻟﺘﻌ ُﺭﻑ ﻋﻠﻰ ﺍﺸﺘﻘﺎﻗﻴﺔ ﺩﺍﻟﺔ ﻋﻨﺩ ﻗﻴﻤﺔ‬ ‫ﺤﺴﺎﺏ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ‬ ‫ﺩﺭﺍﺴﺔ ﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺩﺍﻟﺔ ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ‬ ‫ﺤل ﻤﺴﺎﺌل ﺍ ٍﻻﺴﺘﻤﺜﺎل ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﻤﺸﺘﻘﺎﺕ‬‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫ﺘﻌﺎﺭﻴﻑ‬ ‫ﻤﺸﺎﻜل ﻭ ﺘﻤﺎﺭﻴﻥ‬ ‫ﺍﻟﺤﻠﻭل‬

‫ﺗﻌﺎرﻳﻒ‬ ‫‪ -I‬ﺍﻟﻌﺩﺩ ﺍﻟﻤﺸﺘﻕ ‪:‬‬ ‫‪ -1‬ﺍﻟﻨﻬﺎﻴﺔ ﻋﻨﺩ ﻋﺩﺩ ‪: a‬‬‫= )‪f (x‬‬ ‫‪x2 - 1‬‬ ‫ﻤﺜﺎل ‪ : 1‬ﻟﺘﻜﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﺒﺎﺭﺓ‪:‬‬ ‫‪x-1‬‬ ‫ﻭﺍﻟﺩﺍﻟﺔ ‪ g‬ﺍﻟﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﺒﺎﺭﺓ ‪. g (x) = x + 1 :‬‬‫• ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺒﻴﺎﻨﻴﺔ ﺃﻨﺸﺊ ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ‪ C f‬ﻟﻠﺩﺍﻟﺔ ‪( ). f‬‬ ‫ﻫل ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﻌﺭﻓﺔ ﻋﻨﺩ ‪. 1‬‬ ‫ﻤﺎﺫﺍ ﺘﻼﺤﻅ ﻋﻨﺩﻤﺎ ﻴﻘﺘﺭﺏ ‪ x‬ﻤﻥ ‪. 1‬‬‫• ﺃﻨﺸﺊ ﺘﻤﺜﻴﻼ ﺒﻴﺎﻨﻴﺎ ‪ Cg‬ﻟﻠﺩﺍﻟﺔ ‪ g‬ﻓﻲ ﻤﻌﻠﻡ ﺁﺨﺭ‪ .‬ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ )‪( )g(1‬‬‫• ﺒﻴﻥ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ x‬ﻤﻥ ‪ ℜ − 1‬ﻓﺈﻥ ‪{ }:‬‬ ‫)‪ f (x) = g (x‬ﻤﺎﺫﺍ ﺘﺴﺘﻨﺘﺞ ؟‬ ‫ﺍﻟﺤل‪:‬‬ ‫• ﺇﻨﺸﺎﺀ ‪ : C f‬ﺒﺎﺴﺘﻌﻤﺎل ﺁﻟﺔ ﺒﻴﺎﻨﻴﺔ ﻭ ﺒﻌﺩ) (‬ ‫ﻨﻜﺘﺏ ‪:‬‬ ‫ﺍﻟﻀﻐﻁ ﻋﻠﻰ ﺍﻟﺯﺭ‬‫ﻨﺭﺴﻡ ﺍﻟﻤﻨﺤﻨﻰ ﻜﻤﺎ ﻴﻅﻬﺭ ﻋﻠﻰ ﺸﺎﺸﺔ ﺍﻵﻟﺔ ‪.‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺯﺭ‬ ‫‪. lim f ( x) = 2‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻏﻴﺭ ﻤﻌﺭﻓﺔ ﻋﻨﺩ‪.1‬‬ ‫‪x→1‬‬ ‫ﻨﻼﺤﻅ ﺃﻨﻪ ﻋﻨﺩﻤﺎ ﻴﻘﺘﺭﺏ‬ ‫‪ x‬ﻤﻥ ‪ 1‬ﻓﺈﻥ )‪f (x‬‬ ‫ﺘﻘﺘﺭﺏ ﻤﻥ ‪. 2‬‬ ‫ﺃﻱ ﺃﻨﻪ ﻟﻤﺎ ﻴﺘﻨﺎﻫﻰ ‪x‬‬ ‫ﻨﺤﻭ‪ 1‬ﻓﺈﻥ )‪f (x‬‬ ‫ﺘﺘﻨﺎﻫﻰ ﻨﺤﻭ ‪ 2‬ﻭﻨﻜﺘﺏ ‪:‬‬

‫‪ -‬ﺇﻨﺸﺎﺀ ‪( ): Cg‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﻨﻔﺱ ﺍﻵﻟﺔ ﻨﺠﺩ ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪. g (1) = 2 :‬‬ ‫= )‪f (x‬‬ ‫‪x2 − 1‬‬ ‫ﻟﺩﻴﻨﺎ‬ ‫•‬ ‫‪x−1‬‬ ‫= )‪f (x‬‬ ‫)‪( x − 1) ( x+1‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬ ‫•‬ ‫‪x−1‬‬ ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ‪ f (x) = x + 1 :‬ﻤﻥ ﺃﺠل ‪x ≠ 1‬‬ ‫ﺍﻟﻨﺘﻴﺠﺔ ‪:‬‬ ‫= )‪lim f (x‬‬ ‫‪lim‬‬ ‫‪x2 − 1‬‬ ‫=‬ ‫‪lim‬‬ ‫)‪(x + 1) = 2 = g (1‬‬ ‫‪x−1‬‬ ‫‪x→1‬‬ ‫‪x→1‬‬ ‫‪x→1‬‬ ‫ﻤﺜﺎل ‪: 2‬‬‫‪lim‬‬ ‫‪x2 − 4x‬‬ ‫=‬ ‫‪lim‬‬ ‫)‪x( x − 4‬‬ ‫=‬ ‫)‪lim ( x − 4‬‬ ‫‪=-4‬‬ ‫‪x‬‬ ‫‪x‬‬‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪x→0‬‬ ‫‪ -2‬ﺍﻟﻌﺩﺩ ﺍﻟﻤﺸﺘﻕ ‪:‬‬ ‫‪ h‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﺤﻴﺙ ‪ x0 + h‬ﻋﻨﺼﺭ ﻤﻥ ﻤﺠﺎل ‪. I‬‬ ‫ﻫﻲ ﻨﺴﺒﺔ ﺘﺯﺍﻴﺩ‬ ‫‪f ( x0‬‬ ‫) ‪+ h) - f (x0‬‬ ‫ﻓﺈﻥ ﺍﻟﻨﺴﺒﺔ ‪:‬‬ ‫‪h≠o‬‬ ‫ﺇﺫﺍ ﻜﺎﻥ‬ ‫‪h‬‬ ‫ﺍﻟﺩﺍﻟﺔ ﺒﻴﻥ ‪ x0‬ﻭ ‪. x0 + h‬‬ ‫) ‪f ( x ) - f (x0‬‬ ‫ﻓﺈﻥ ﺍﻟﻨﺴﺒﺔ ﺘﻜﺘﺏ ‪:‬‬ ‫‪x‬‬ ‫‪= x0‬‬ ‫‪+h‬‬ ‫ﺒﻭﻀﻊ‬ ‫‪x - x0‬‬ ‫ﻨﻘﻭل ﺃﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ x0‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﺍﻗﺘﺭﺒﺕ ﺍﻟﻨﺴﺒﺔ‬

‫‪ l‬ﻋﻨﺩﻤﺎ ﻴﻘﺘﺭﺏ ‪ h‬ﻤﻥ ﺍﻟﻌﺩﺩ ‪0‬‬ ‫ﻤﻥ ﻋﺩﺩ ﺜﺎﺒﺕ‬ ‫) ‪f ( x0 + h) - f (x0‬‬ ‫‪h‬‬ ‫‪lim‬‬ ‫‪f ( x0‬‬ ‫‪+‬‬ ‫)‪h‬‬ ‫‪-‬‬ ‫‪f‬‬ ‫) ‪(x0‬‬ ‫=‬ ‫‪l‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪lim‬‬ ‫) ‪f ( x ) - f (x0‬‬ ‫‪=l‬‬ ‫ﻭﺒﻌﺒﺎﺭﺓ ﺃﺨﺭﻯ ‪:‬‬ ‫‪x - x0‬‬ ‫‪x→ x0‬‬ ‫ﻨﺭﻤﺯ ﻟﻠﻌﺩﺩ ‪ l‬ﺒﺎﻟﺭﻤﺯ ) ‪ f ′( x0‬ﻭﻴﺴﻤﻰ ﺍﻟﻌﺩﺩ ﺍﻟﻤﺸﺘﻕ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ‬ ‫ﺍﻟﻌﺩﺩ ‪. x0‬‬ ‫ﻤﺜﺎل ‪: 1‬‬ ‫ﺍﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ ‪ f ( x) = x2 :‬ﻋﻨﺩ ﺍﻟﻌﺩﺩ ‪4‬‬ ‫ﺤـل ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪f (4) = 16 ; Df = ℜ :‬‬‫)‪lim f ( x) - f (4) = lim x2 -16 = lim ( x - 4) ( x + 4‬‬‫‪x→4 x - 4‬‬ ‫‪x→4 x - 4 x→4‬‬ ‫‪x-4‬‬ ‫‪= lim ( x + 4) = 8‬‬ ‫‪x→4‬‬ ‫ﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 4‬ﺤﻴﺙ ‪. f ′(4) = 8 :‬‬ ‫ﻤﺜﺎل ‪: 2‬‬ ‫ﺍﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ ‪f(x) = - 4x + 5 :‬‬ ‫ﻋﻨﺩ ﺍﻟﻌﺩﺩ ‪. -3‬‬ ‫ﺤـل‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪f (−3) = 17 ; Df = ℜ :‬‬‫‪lim f ( x) - f (-3) = lim -4x + 5 - 17‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬‫‪x→−3‬‬ ‫‪x+3‬‬ ‫‪x→−3‬‬ ‫‪x+3‬‬ ‫=‬ ‫‪lim‬‬ ‫‪−4x -‬‬ ‫‪12‬‬ ‫=‬ ‫‪lim‬‬ ‫)‪-4( x + 3‬‬ ‫=‬ ‫‪-4‬‬ ‫‪x+‬‬ ‫‪3‬‬ ‫‪x+3‬‬ ‫‪x→−3‬‬ ‫‪x→−3‬‬ ‫ﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ -3‬ﺤﻴﺙ ‪. f ′(−3) = -4‬‬

‫‪ -‬ﺍﻟﺘﻔﺴﻴﺭ ﺍﻟﻬﻨﺩﺴﻲ ﻟﻠﻌﺩﺩ ﺍﻟﻤﺸﺘﻕ ‪:‬‬ ‫‪rr‬‬ ‫‪ C f‬ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ‪ f‬ﻓﻲ ﻤﻌﻠﻡ )‪( ). (O , i , j‬‬ ‫‪ A‬ﻭ‪ M‬ﻨﻘﻁﺘﺎﻥ ﻤﻥ ‪ C f‬ﺤﻴﺙ ‪( ) ( )A x0 , f ( x0 ) :‬‬ ‫ﻭ ))‪. M ( x0 + h , f(x0 + h‬‬ ‫‪M‬‬ ‫ﻤﻥ ﺃﺠل ‪ : h ≠ 0‬ﻤﻌﺎﻤل ﺘﻭﺠﻴﻪ ﺍﻟﻤﺴﺘﻘﻴﻡ‬ ‫‪A‬‬ ‫) ‪f ( x0 + h) - f ( x0‬‬ ‫)‪ (AM‬ﻫﻭ ‪:‬‬‫‪O‬‬ ‫‪h‬‬ ‫ﺇﺫﺍ ﻜﺎﻨﺕ ‪ f‬ﻗﺎﺒﻠﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ x0‬ﻓﺈﻥ ‪:‬‬ ‫‪lim‬‬ ‫‪f ( x0‬‬ ‫‪+‬‬ ‫) ‪h) - f (x0‬‬ ‫) ‪= f (′x0‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫ﺒﻴﺎﻨﻴﺎ ﻋﻨﺩﻤﺎ ﻴﺅﻭل ‪ h‬ﺇﻟﻰ ‪ 0‬ﻓﺈﻥ ‪ M‬ﺘﻘﺘﺭﺏ ﻤﻥ ‪ A‬ﻭﻋﻠﻴﻪ ﺍﻟﻤﺴﺘﻘﻴﻡ‬ ‫)‪ (AM‬ﻴﺼﺒﺢ ﻤﻤﺎﺴﺎ ﻟﻠﻤﻨﺤﻨﻰ ‪ C f‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪( ). A‬‬ ‫‪ -‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻋﻨﺩ ‪y = f ′(x0 ) × x + b : A‬‬ ‫ﻭﺒﻤﺎ ﺃﻥ ‪ A‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﻤﺎﺱ ﻓﺈﻥ ‪ y0 = f (x0 ) :‬ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪ y0 = f ′(x0 ) × x0 + b‬ﻭﻋﻠﻴﻪ ‪b = y0 - f ′(x0 ) × x0 :‬‬ ‫ﺇﺫﻥ ‪y = f ′(x0 ) . x + y0 - f ′(x0 ) . x0 :‬‬ ‫ﻭﻋﻠﻴﻪ ‪ y - y0 = f ′(x0 ) .(x - x0 ) :‬ﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻟﻠﻤﻨﺤﻨﻰ‬ ‫‪ C f‬ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ ) ‪ A (x0 ; y0‬ﺤﻴﺙ ‪( )y0 = f (x0 ) :‬‬ ‫ﻤﺜﺎل ‪ f :‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﺒﺎﻟﻌﺒﺎﺭﺓ‪f ( x) = x2 - 2 x :‬‬ ‫ﺍﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ ‪. 4‬‬ ‫‪-‬‬ ‫‪-‬‬‫ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ∆ ﻟﻠﻤﻨﺤﻨﻰ ‪ C f‬ﺍﻟﻤﻤﺜل ﻟﺘﻐﻴﺭﺍﺕ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ) ( ) (‬ ‫ﺍﻟﻔﺎﺼﻠﺔ ‪. 4‬‬ ‫ﺍﻟﺤل ‪:‬‬‫‪ -‬ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪f (4) = (4)2 - 2(4) = 8 ; Df = ℜ : 4‬‬

‫‪lim‬‬ ‫) ‪f(x) - f( 4‬‬ ‫=‬ ‫‪lim‬‬ ‫‪x2- 2x -‬‬ ‫‪8‬‬ ‫=‬ ‫‪lim‬‬ ‫(‬ ‫‪x‬‬ ‫()‪- 4‬‬ ‫‪x+‬‬ ‫)‪2‬‬ ‫‪x-4‬‬ ‫‪x-4‬‬ ‫‪x-‬‬ ‫‪4‬‬‫‪x→4‬‬ ‫‪x→4‬‬ ‫‪x→4‬‬ ‫‪= lim (x + 2) = 6‬‬ ‫‪x→4‬‬ ‫ﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ 4‬ﺤﻴﺙ ‪f ′(4) = 6 :‬‬ ‫‪ -‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻟﻠﻤﻨﺤﻨﻰ ) ‪ (C f‬ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ ‪4‬‬ ‫ﻟﺩﻴﻨﺎ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ∆ ﻫﻲ ‪( )y - y0 = f ′(x0 ) .(x - x0 ) :‬‬ ‫ﻭ ﻤﻨﻪ ‪y - 8 = 6( x - 4) :‬‬ ‫ٍﺇﺫﻥ ‪. (∆ ) : y = 6 x - 16 :‬‬ ‫‪ - 3‬ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ‪:‬‬ ‫ﺘﻌﺭﻴﻑ ‪ f :‬ﺩﺍﻟﺔ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻨﺩ ﻜل ﻋﺩﺩ ‪ x‬ﻤﻥ ﻤﺠﺎل ‪ I‬ﻤﻥ‬ ‫ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻔﻬﺎ ‪. D f‬‬ ‫ﺍﻟﺩﺍﻟﺔ ﺍﻟﺘﻲ ﺘﺭﻓﻕ ﺒﻜل ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ x‬ﻤﻥ ‪ I‬ﺍﻟﻌﺩﺩ ﺍﻟﻤﺸﺘﻕ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ‬ ‫‪ x‬ﺘﺴﻤﻰ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻭﻴﺭﻤﺯ ﻟﻬﺎ ﺒﺎﻟﺭﻤﺯ ‪ f ′‬ﺤﻴﺙ ‪f ′: x a f ′(x) :‬‬ ‫ﻤﺜﺎل ‪:‬‬ ‫ﺍﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻟﻠﺩﺍﻟﺔ ‪ f : x a x 2‬ﻋﻨﺩ ﻋﺩﺩ ‪ x0‬ﻤﻥ ‪. ℜ‬‬ ‫ﺍﺴﺘﻨﺘﺞ ﺩﺍﻟﺘﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ‪.‬‬ ‫ﺤـل ‪:‬‬‫‪lim‬‬ ‫‪f ( x0 + h) - f ( x0 ) = lim‬‬ ‫‪( x0‬‬ ‫‪+ h)2‬‬ ‫‪-‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪0‬‬‫‪h→0 h‬‬ ‫‪h→0 h‬‬ ‫‪= lim‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+ 2x0h‬‬ ‫‪+ h2‬‬ ‫‪-‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪= lim‬‬ ‫‪2x0h + h2‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪h→0 h‬‬ ‫‪h→0 h‬‬ ‫‪h(2x0 +‬‬ ‫‪h) = lim‬‬ ‫‪h‬‬ ‫‪h→0‬‬ ‫‪( )= lim‬‬ ‫‪h→ 0‬‬ ‫‪2x0 + h = 2x0‬‬ ‫ﻭﻋﻠﻴﻪ ‪ f‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ x0‬ﺤﻴﺙ ‪f ′( x0 ) = 2x0 :‬‬ ‫ﻭﻤﻨﻪ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻫﻲ ﺍﻟﺩﺍﻟﺔ ' ‪ f‬ﺤﻴﺙ ‪:‬‬ ‫‪f ′: x a f ′(x) = 2 x‬‬

: ‫ ﻤﺸﺘﻘﺎﺕ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺄﻟﻭﻓﺔ‬- 4 f : x a a : ‫ ﺤﻴﺙ‬f ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬1 lim f (x) - f (x0 ) = lim a-a = 0 x - x0 x - x0 x→ x0 x→ x0 . f′ :x a 0 : ‫ﻭﻋﻠﻴﻪ‬ : ‫ ﺤﻴﺙ‬f ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬2 Df = ℜ* ; f : x a 1 x 1- 1 x0 - xlim f (x) - f (x0 ) = lim x x0 = lim x x0 x - x0 x - x0 x→ x0 x - x0x→ x0 x→ x0 = lim - (x - x0 ) × 1 = lim - 1 = −1 x→ x0 x x0 - x0 x x0 x02 x x→ x0 . f′ :x a -1 x2 : ‫ﻭ ﻋﻠﻴﻪ‬ Df = ℜ ‫ ؛‬n ∈ ℜ ‫ ﻤﻊ‬f : x a xn : ‫ ﺤﻴﺙ‬f ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬3 lim f (x) - f (x0 ) = lim xn - x0n x→ x0 x - x0 x→ x0 x - x0= lim (x - x0 ) (xn - 1 + x0 xn - 2 + x02 xn -3 + ... + x n−1 ) 0x→ x0 x - x0 = lim (xn - 1 + x0 xn - 2 + x02 xn -3 + ... + x n−1 ) x→ x0 0 = x0n - 1 + x0n - 1 + ... + xn - 1 ( ‫ ﻤﺭﺓ‬n) 0 = n . x0n - 1 . f ′ : x a n.xn - 1 : ‫ﻭﻋﻠﻴﻪ‬

f :x a x : ‫ ﺤﻴﺙ‬f ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬4 ] [ [ [0 ; + ∞ ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬f ‫ ﺍﻟﺩﺍﻟﺔ‬، D f = 0 ; + ∞lim f (x) - f (x0 ) = lim x - x0 : ‫ﻟﺩﻴﻨﺎ‬x→ x0 x - x0 x→ x0 x - x0lim f (x) - f (x0 ) = lim ( x - x0 ) ( x + x0 )x→ x0 x - x0 x→x0 ( x - x0 ) ( x + x0 )lim f (x) - f (x0 ) = lim x - x0 : ‫ﻭ ﻤﻨﻪ‬x→ x0 x - x0 x→x0 ( x - x0 )( x + x0 ) = lim 1 =1 x→ x0 x + x0 2 x0 . f′ :x a 1 : ‫ﻭﻋﻠﻴﻪ‬ 2x f : x a Sin x : ‫ ﺤﻴﺙ‬f ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬5lim f (x) - f (x0 ) = lim Sin x - Sin x0x→ x0 x - x0 x→ x0 x - x0 2Sin  x - x0  Cos  x + x0   2   2  = lim x→ x0 x - x0 2Sin  x - x0   2  = lim × Cos  x + x0  x→ x0  2  x -x0 = cos x0 . f ' : x a Cos x : ‫ﻭ ﻋﻠﻴﻪ‬

f : x a Cos x : ‫ ﺤﻴﺙ‬f ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬6 lim f (x) - f (x0 ) = lim Cos x - Cos x0 x→ x0 x - x0 x→ x0 x - x0 -2Sin  x - x0  Sin  x + x0   2   2 lim f (x) - f (x0 ) = lim x - x0 x→ x0x→ x0 x - x0 Sin  x - x0   2  = lim - Sin  x + x0  . = - Sin x0 x→ x0  2  x - x0 2 . f ′ : x a - Sin x : ‫ﻭ ﻋﻠﻴﻪ‬ : ‫ ﺤﺴﺎﺏ ﻤﺸﺘﻘﺎﺕ ﺍﻟﺩﻭﺍل‬- 5 . I ‫ ﺩﺍﻟﺘﺎﻥ ﺘﻘﺒﻼﻥ ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻤﺠﺎل‬g ‫ ﻭ‬f : f + g ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬1lim (f+g)(x) - (f+g)(x0 ) = lim f(x)+g(x) - f(x0 )- g(x0 ) x - x0 x→ x0 x - x0x→ x0 = lim f(x) - f(x0 ) + lim g (x) - g (x0 ) x→ x0 x - x0 x→ x0 x - x0 = f ′(x0 ) + g′ (x0 ) : ‫ ﺤﻴﺙ‬I ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬f + g ‫ﻭﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ‬ . ( f + g)′ = f ′+ g′ : f . g ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬2

lim (f.g) (x) - (f.g) (x0 ) = lim f(x).g (x) - f (x0 ).g (x0 ) x - x0 x - x0x→ x0 x→ x0= lim f (x).g (x) - f (x) . g (x0 ) + f (x) . g (x0 ) - f (x0 ).g (x0 ) x→x0 x - x0= lim f(x)  g(x) - g(x0 ) + g (x0 )  f(x)- f(x0 ) x - x0 x - x0 x→ x0 = f (x0 ) . g′ (x0 ) + g (x0 ) . f ′(x0 ) : ‫ ﺤﻴﺙ‬I ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬f . g ‫ﻭ ﻋﻠﻴﻪ ﺍﻟﺩﺍﻟﺔ‬ . ( f . g)′ = f ′. g + f . g′ 1 I ‫ ﻻ ﺘﻨﻌﺩﻡ ﻋﻠﻰ‬g ‫ ﺤﻴﺙ‬: g ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬3  1 (x) -  1 (x0 ) 1 - 1     g (x) g (x0 )lim  g   g  = limx→ x0 x - x0 x→ x0 x - x0 = lim -( g (x) - g (x0 )) × 1 x → x0 x - x0 g (x) . g (x0 ) = lim -1 . ( g (x) - g (x0 )) x → x0 g (x) . g (x0 ) x - x0 = -1 × g′ (x0 ) = - g′ (x0 )  g (x0 ) 2  g (x0 ) 2 1 : ‫ ﺤﻴﺙ‬I ‫ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ‬g ‫ﻭﺒﺎﻟﺘﺎﻟﻲ ﺍﻟﺩﺍﻟﺔ‬ .  1 ′ = -g′   g2  g 

f : g ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬4  f ′ =  f × 1 ′ =f ′. 1 + f .  1 ′ : ‫ﻟﺩﻴﻨﺎ‬  g   g  g  g   f ′ = f′ +f  - g′  : ‫ﺃﻱ‬  g  g      g2  .  f ′ = f ′.g - f .g′ : ‫ﻭ ﻋﻠﻴﻪ‬  g  g2 g : x a f (a x + b ) : ‫( ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬5lim g (x) - g (x0 ) = lim f (a x + b) - f (a x0 + b) x - x0 x → x0 x - x0x → x0 ‫ ﻨﺠﺩ‬u0 = a x0 + b ، u = a x + b : ‫ﺒﻭﻀﻊ‬lim f (u) - f (u0 ) = lim f (u) - f (u0 ) × u - u0 x - x0 u - u0 x - x0x → x0 x → x0 = lim f (u) - f (u0 ) × a x + b - a x0 - b x → x0 u - u0 x - x0 = lim f (u) - f (u0 ) × a (x - x0 ) x → x0 u - u0 x - x0 = lim a × f (u) - f (u0 ) u - u0 x → x0 = a . f ′(u) = a . f ′(a x + b) g : x a f ( a x + b) : ‫ﺇﺫﻥ ﻤﺸﺘﻘﺔ ﺍﻟﺩﺍﻟﺔ‬ . g ' : x a a . f ′( a x + b) : ‫ﻫﻲ ﺍﻟﺩﺍﻟﺔ‬

‫ﻤﺜﺎل ‪ :‬ﻋﻴﻥ ﻤﺸﺘﻕ ﻜل ﻤﻥ ﺍﻟﺩﻭﺍل ‪:‬‬‫‪f :x a‬‬ ‫‪1‬‬ ‫;‬ ‫‪g : x a ( x + 2)3‬‬ ‫‪x+1‬‬‫‪h:x a x-2‬‬ ‫ﺤـل ‪:‬‬ ‫‪Df‬‬ ‫}‪= ℜ - {- 1‬‬ ‫‪.‬‬ ‫= )‪f (x‬‬ ‫‪x‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫•‬ ‫‪+‬‬ ‫‪-1‬‬ ‫‪f′‬‬ ‫= )‪(x‬‬ ‫‪(x‬‬ ‫‪+ 1)2‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫• ‪ g (x ) = ( x + 2)3‬؛ ‪Dg = ℜ‬‬‫ﻭ ﻤﻨﻪ ‪g′ (x) = 3 × 1 (x + 2)2 = 3 (x + 2)2 :‬‬ ‫• ‪Dh = [2 ; +∞[ , h (x) = x - 2‬‬ ‫ﺍﻟﺩﺍﻟﺔ ‪ h‬ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ∞ ‪ 2 ; +‬ﺤﻴﺙ ‪] ]:‬‬ ‫‪h′ (x) = 1‬‬ ‫‪2 x-2‬‬ ‫‪ - 6‬ﺍﻟﺭﺒﻁ ﺒﻴﻥ ﺇﺸﺎﺭﺓ ﺍﻟﻤﺸﺘﻕ ﻭ ﺍﺘﺠﺎﻩ ﺍﻟﺘﻐﻴﺭ ‪:‬‬ ‫ﻟﺘﻜﻥ ‪ f‬ﺩﺍﻟﺔ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻤﺠﺎل ‪. I‬‬ ‫‪ -‬ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻤﻌﺩﻭﻤﺔ ﻋﻠﻰ ‪ I‬ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺜﺎﺒﺘﺔ ﻋﻠﻰ ‪. I‬‬ ‫‪ -‬ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﻤﻭﺠﺒﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪ I‬ﻓﺈﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﺘﺯﺍﻴﺩﺓ‬ ‫ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪. I‬‬ ‫‪ -‬ﺇﺫﺍ ﻜﺎﻨﺕ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺸﺘﻘﺔ ﺴﺎﻟﺒﺔ ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪ I‬ﻓﺎﻥ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻤﺘﻨﺎﻗﺼﺔ‬ ‫ﺘﻤﺎﻤﺎ ﻋﻠﻰ ‪. I‬‬ ‫‪ - 7‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺩﻴﺔ ﻟﺩﺍﻟﺔ ‪:‬‬ ‫‪ f‬ﺩﺍﻟﺔ ﺘﻘﺒل ﺍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ﻤﺠﺎل ‪ x0 . I‬ﻋﻨﺼﺭ ﻤﻥ ‪. I‬‬ ‫ﺇﺫﺍ ﺍﻨﻌﺩﻤﺕ ﺍﻟﺩﺍﻟﺔ ‪ f ′‬ﻋﻨﺩ ‪ x0‬ﻤﻐﻴﺭﺓ ﺇﺸﺎﺭﺘﻬﺎ ﻓﺈﻥ ) ‪ f (x0‬ﻗﻴﻤﺔ‬

‫ﺤﺩﻴﺔ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻠﻰ ‪. I‬‬ ‫‪x‬‬ ‫‪x0‬‬ ‫‪0‬‬ ‫‪-‬‬‫)‪f '(x‬‬ ‫‪+‬‬‫)‪f (x‬‬ ‫)‪f (x0‬‬ ‫ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ) ‪ f (x0‬ﻗﻴﻤﺔ ﺤﺩﻴﺔ ﻋﻅﻤﻰ ﻟﻠﺩﺍﻟﺔ ‪. f‬‬ ‫‪x‬‬ ‫‪x0‬‬ ‫‪0‬‬ ‫‪+‬‬‫)‪f '(x‬‬ ‫‪-‬‬‫)‪f (x‬‬ ‫)‪f (x0‬‬ ‫ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ) ‪ f (x0‬ﻗﻴﻤﺔ ﺤﺩﻴﺔ ﺼﻐﺭﻯ ﻟﻠﺩﺍﻟﺔ ‪. f‬‬

‫ﺘﻤـﺎﺭﻴـﻥ ﻭ ﻤﺸﻜﻼﺕ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬ ‫‪ f‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻜﻤﺎ ﻴﻠﻲ ‪. f (x) = 4 x2 + x - 5 :‬‬ ‫‪ -‬ﺃﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍﺸﺘﻘﺎﻕ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ ‪. 3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬ ‫‪ -‬ﺃﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍ ٍﻻﺸﺘﻘﺎﻕ ﻋﻨﺩ ‪ x0‬ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪f (x) = x4 ; x0 = 1 (1‬‬ ‫‪f (x) = x3 + 1 ; x0 = - 1 (2‬‬ ‫= )‪f (x‬‬ ‫‪2‬‬ ‫‪; x0 = 3 (3‬‬ ‫‪x‬‬‫)‪(Cf‬‬ ‫‪y‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫‪6‬‬ ‫‪5‬‬ ‫‪A‬‬ ‫ﻨﻌﺘﺒﺭ ‪ C f‬ﺍﻟﺘﻤﺜﻴل ﺍﻟﺒﻴﺎﻨﻲ ﻟﺩﺍﻟﺔ ‪( )f‬‬ ‫‪4‬‬ ‫‪3‬‬ ‫‪2x‬‬ ‫ﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ ℜ‬ﻭﻟﻴﻜﻥ ∆ ﻤﻤﺎﺴﺎ) (‬ ‫‪2‬‬ ‫‪1‬‬ ‫ﻟﻠﻤﻨﺤﻨﻰ ‪ C f‬ﻜﻤﺎ ﻫﻭ ﻤﺒﻴﻥ) (‬‫‪-2 -1 0‬‬ ‫‪1‬‬ ‫ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﻤﻘﺎﺒل ‪:‬‬ ‫‪-1‬‬ ‫‪ -‬ﺍﺤﺴﺏ ﻤﻥ ﺨﻼل ﺍﻟﺸﻜل )‪. f ′(2‬‬ ‫‪ -‬ﺜﻡ ٍﺍﺴﺘﻨﺘﺞ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ∆) (‬ ‫‪-2‬‬ ‫‪-3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫)∆( ‪-4‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ f‬ﺤﻴﺙ ‪f (x) = - x + 4 :‬‬ ‫‪B-5‬‬ ‫‪ (1‬ﺍﺩﺭﺱ ﻗﺎﺒﻠﻴﺔ ﺍ ٍﻻﺸﺘﻘﺎﻕ ﻟﻠﺩﺍﻟﺔ ‪ f‬ﻋﻨﺩ ﺍﻟﻌﺩﺩ ‪. -4‬‬ ‫‪-6‬‬

‫‪ (2‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻟﻠﻤﻨﺤﻨﻰ ‪ C f‬ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ ‪( ). -4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬‫‪ (1‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻟﻠﻤﻨﺤﻨﻰ ‪ C f‬ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪( ):‬‬‫‪ y = - x2 + 5‬ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ ‪.1‬‬‫‪ (2‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻟﻠﻤﻨﺤﻨﻰ ‪ C g‬ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪( ):‬‬ ‫ﻋﻨﺩ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ ﺍﻟﻔﺎﺼﻠﺔ ‪. -5‬‬ ‫‪y‬‬ ‫=‬ ‫‪-‬‬ ‫‪5‬‬ ‫‪x‬‬ ‫‪ (3‬ﻋﻴﻥ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻟﻤﻤﺎﺴﻴﻥ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬‫ﺃﺜﺒﺕ ﺃﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪ y = -x + 2 :‬ﻫﻭ ﻤﻤﺎﺱ ﻟﻠﻤﻨﺤﻨﻰ ‪ Cf‬ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪( ):‬‬ ‫ﻓﻲ ﺍﻟﻨﻘﻁﺔ ﺍﻟﺘﻲ ﻓﺎﺼﻠﺘﻬﺎ ‪. 1‬‬ ‫=‪y‬‬ ‫‪1‬‬ ‫‪x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬‫ﻨﻌﺘﺒﺭ ﺍﻟﻌﺒﺎﺭﺓ ﺍﻟﺠﺒﺭﻴﺔ ‪p (x) = -4 x3 + 6 x2 - 2 :‬‬ ‫‪ (1‬ﺍﺤﺴﺏ )‪ p (1‬ﺜﻡ ٍﺍﺴﺘﻨﺘﺞ ﺘﺤﻠﻴﻼ ﻟـ )‪. p (x‬‬ ‫‪ (2‬ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺘﻴﻥ ‪ f‬ﻭ ‪ g‬ﺍﻟﻤﻌﺭﻓﺘﻴﻥ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬‫= )‪g (x‬‬ ‫‪2‬‬ ‫َﻭ‬ ‫‪f ( x) = −4x2 + 6x‬‬ ‫‪x‬‬‫‪-‬ﻋﻴﻥ ﻨﻘﻁ ﺘﻘﺎﻁﻊ ‪َ Cf‬ﻭ ‪ Cg‬ﺍﻟﻤﻨﺤﻨﻴﻴﻥ ﺍﻟﻤﻤﺜﻠﻴﻥ ﻟﻠﺩﺍﻟﺘﻴﻥ ‪َ f‬ﻭ‪( ) ( ).g‬‬‫‪ (3‬ﺒﻴﻥ ﺃﻥ ‪َ Cf‬ﻭ ‪ Cg‬ﻴﻘﺒﻼﻥ ﻤﻤﺎﺴﺎ ﻤﺸﺘﺭﻜﺎ ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪ B‬ﺫﺍﺕ) ( ) (‬ ‫ﺍﻟﻔﺎﺼﻠﺔ ‪. 1‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫‪[ [f (x) = x3‬‬ ‫‪rr‬‬ ‫ﺒﺎﻟﻌﺒﺎﺭﺓ ‪:‬‬ ‫∞‪0 ; +‬‬ ‫‪ f‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﻋﻠﻰ‬ ‫‪ Cf‬ﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﻓﻲ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ‪( ) ( ). O ; i , j‬‬ ‫‪ A‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﻨﺤﻨﻰ ‪ Cf‬ﻓﺎﺼﻠﺘﻬﺎ ‪( ). x0‬‬ ‫‪ (1‬ﺃﻨﺸﺊ ‪ Cf‬ﺒﺂﻟﺔ ﺒﻴﺎﻨﻴﺔ ‪ uHr.‬ﺍ‪u‬ﻟﻤ‪u‬ﺴﻘﻁ ﺍﻟﻌﻤﻭ‪r‬ﺩ‪u‬ﻱ‪ u‬ﻟﻠﻨﻘﻁﺔ ‪ A‬ﻋﻠﻰ ﻤﺤﻭﺭ) (‬ ‫ﺍﻟﺘﺭﺍﺘﻴﺏ ‪ I .‬ﻨﻘﻁﺔ ﺤﻴﺙ ‪. HI = 3 HO :‬‬ ‫‪ (2‬ﺃﻨﺸﺊ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪. (AI‬‬ ‫‪ (3‬ﻋﻴﻥ ﺒﺩﻻﻟﺔ ‪ٍ x0‬ﺍﺤﺩﺍﺜﻴﻲ ﺍﻟﻨﻘﻁ ‪. I , H , A‬‬ ‫‪ (4‬ﺒﻴﻥ ﺃﻥ )‪ (AI‬ﻫﻭ ﻤﻤﺎﺱ ‪ Cf‬ﻓﻲ ‪( ). A‬‬ ‫‪ (5‬ﺍﺴﺘﻌﻤل ﻫﺫﻩ ﺍﻟﻁﺭﻴﻘﺔ ﻹﻨﺸﺎﺀ ﺍﻟﻤﻤﺎﺱ ﻟﻠﻤﻨﺤﻨﻰ ‪ Cf‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ﺫﺍﺕ) (‬ ‫ﺍﻟﻔﺎﺼﻠﺔ ‪. 1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬ ‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺘﻌﺭﻴﻑ ﻭ ﻤﺠﻤﻭﻋﺔ ﻗﺎﺒﻠﻴﺔ ﺍﻻﺸﺘﻘﺎﻕ ﻟﻠﺩﻭﺍل ﺍﻟﻤﻌﺭﻓﺔ‬ ‫ﻜﻤﺎ ﻴﻠﻲ ﻭﺍﺤﺴﺏ ﺩﻭﺍﻟﻬﺎ ﺍﻟﻤﺸﺘﻘﺔ ‪.‬‬‫‪ f : x a -x2+ 5x - 3 (1‬؛ ‪f : x a ( x - 2) ( x + 5) (2‬‬ ‫‪f:x a‬‬ ‫‪1‬‬ ‫‪(4‬‬ ‫؛‬ ‫‪f : x a 4 ( x - 5)2 (3‬‬ ‫‪5+x‬‬‫‪f :x a‬‬ ‫‪x+3-‬‬ ‫‪2‬‬ ‫‪ f : x a‬؛ ‪(6‬‬ ‫‪-x2 - x + 4‬‬ ‫‪(5‬‬ ‫‪x+3‬‬ ‫‪− 4x2 + 5x -1‬‬‫‪f:x a‬‬ ‫‪2x - 3‬‬ ‫‪(8‬‬ ‫‪ f:x a‬؛‬ ‫‪2x - 5 (7‬‬ ‫‪x-4‬‬‫‪ f : x a Sin x - Cos x (9‬؛ ‪f : x a Sin x .Cos x (10‬‬‫‪f:xa‬‬ ‫‪Cos x - 1‬‬ ‫‪(12‬‬ ‫؛‬ ‫‪f:xa‬‬ ‫‪3‬‬ ‫‪Cos‬‬ ‫‪‬‬ ‫‪2x‬‬ ‫‪-‬‬ ‫‪π‬‬ ‫‪(11‬‬ ‫‪Cos x - 2‬‬ ‫‪‬‬ ‫‪2 ‬‬

‫‪f:xa‬‬ ‫‪Sin x‬‬ ‫‪(13‬‬ ‫‪1 + 2 Sin x‬‬ ‫‪f : x a Sin x + 3 Cos x (14‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬‫‪ f‬ﺩﺍﻟﺔ ﻤﻌﺭﻓﺔ ﺒﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ‪ Cf‬ﻋﻠﻰ ﺍﻟﻤﺠﺎل ‪ -3 ; 3‬ﻭﻫﻲ ﻗﺎﺒﻠﺔ ﻟﻼﺸﺘﻘﺎﻕ ﻋﻠﻰ) ( ] [‬ ‫]‪. [-3 ; 3‬‬ ‫‪y‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪1‬‬‫‪-4 -3 -2 -1 0 1 2 3 4 x‬‬‫‪-1‬‬ ‫‪ (1‬ﺍﺤﺴﺏ ﻜل ﻤﻥ ‪. f (3-)2; f (-3) ; f (-1) ; f (0) ; f (2) :‬‬ ‫‪ (2‬ﺍﺤﺴﺏ ‪. f ′(-1 ) ; f ′( 2 ) :‬‬‫‪ (3‬ﻨﻌﺘﺒﺭ ﺍﻟﺩﺍﻟﺔ ‪ g‬ﺍﻟﻤﻌﺭﻓﺔ ﻋﻠﻰ ‪ −3 ; 0 ∪ 0 ; 3‬ﺒﺎﻟﻌﺒﺎﺭﺓ ‪[ [ ] ]:‬‬ ‫‪.‬‬ ‫)‪g(x‬‬ ‫=‬ ‫‪‬‬ ‫‪1‬‬ ‫‪‬‬ ‫)‪(x‬‬ ‫‪‬‬ ‫‪f‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬‫‪ -‬ﺍﺤﺴﺏ ﻜل ﻤﻥ ‪g (-3) ، g (-1) ، g (2) ، g (3) :‬‬ ‫‪ -‬ﺍﺤﺴﺏ )‪. g′ (2‬‬‫‪ -‬ﺍﺩﺭﺱ ٍﺍﺘﺠﺎﻩ ﺘﻐﻴﺭ ﺍﻟﺩﺍﻟﺔ ‪ . g‬ﻭﺍﻜﺘﺏ ﺠﺩﻭل ﺍﻟﺘﻐﻴﺭﺍﺕ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬‫‪ f‬ﺩﺍﻟﺔ ﺘﻘﺒل ﺍ ٍﻻﺸﺘﻘﺎﻕ ﻋﻠﻰ ‪ -5 ; 7‬ﻭﻤﻌﺭﻓﺔ ﺒﺘﻤﺜﻴﻠﻬﺎ ﺍﻟﺒﻴﺎﻨﻲ ﺍﻵﺘﻲ ‪[ ]:‬‬ ‫‪y‬‬ ‫‪4‬‬ ‫‪3‬‬