matematik 11-sinf 2-qism

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matematika toshkent 2018 umumiy o‘rta ta’lim maktablarining 11-sinflari va o‘rta maxsus, kasb-hunar ta’limi muassasalari uchun darslik o‘zbekiston respublikasi xalq ta’limi vazirligi tomonidan tasdiqlangan 1-nashr algebra va analiz asoslari geometriya ii qism © “zamin nashr” mchj, 2018 © barcha huquqlar himoyalangan algebra va analiz asoslari bo‘limining mualliflari: m.a. mirzaahmedov, sh.n. ismailov, a.q. amanov geometriya bo‘limining muallifi: b.q. xaydarov uo‘k: 51(075.32) kbk: 22.1ya72 m 54 taqrizchilar: r.b. beshimov – mirzo ulug‘bek nomidagi o‘zbekiston milliy universiteti “geometriya va topologiya” kafedrasi mudiri, fizika-matematika fanlari doktori; q.s. jumaniyozov – nizomiy nomidagi tdpu fizika- matematika fakulteti “matematika o‘qitish metodikasi” kafedrasi dotsenti, pedagogika fanlari nomzodi; r.o. ro‘zimov – sergeli tumani 237- umumta’lim maktabi matematika fani o‘qituvchisi; s.b. jumaniyozova – rtm metodisti; i.b. saibova – yashnobod tumani 307- diumning oliy toifali matematika fani o‘qituvchisi. darslikning “algebra va analiz asoslari” bo‘limida ishlatilgan belgilar va ularning talqini:  – masalani yechish (isbotlash) boshlandi ▲ – masalani yechish (isbotlash) tugadi ! …

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iziqlarning kesishish nuqtalarini topamiz: 2) x2+2x–2=x tenglamadan x1=–2, x2=1. 8-rasm. demak, chiziqlar (1; 1), (–2; –2) nuqtalarda kesishadi. ravshanki, (–2; 1) oraliqda y=x funksiya grafigi y=x2+2x–2 funksiya grafigidan yuqorida yotadi (8-rasm). u holda (1) formulada a=–2, b=1, f2(x)=x, f1(x)=x2+2x–2 desak, izlanayotgan yuz (1) ga ko‘ra 4 ( ) 1 1 3 2 2 2 2 2 1 7 10( 2 2) ( 2) ( 2 ) ( ) 4,5. 23 2 6 3 x xs x x x dx x x dx x − − = − + − = − − + = − − + = − − = −∫ ∫ javob: s=4,5 (kv.birlik). ▲ 9-rasm. 2-misol. y x= va 2y x= chiziqlar bilan chegaralangan shakl yuzini hisoblang.  [0;1]x∈ kesmada 2x x≤ (9-rasm). (1) formulada a=0, b=1, 2 1( ) ,f x x= 2 ( )f x x= deymiz. u holda 31 2 32 0 12 1 …

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r, markazi (r; o) nuqtada bo‘lgan doiraning cho- rak qismini ox o‘qi atrofida aylantirishdan (13-rasm) hosil qilinadigan shakl sharning yarmidir. bizning holda mos aylana tenglamasi 2 2 2( )x r y r− + = bo‘ladi, bundan 22 ,y rx x= − [0; ].x r∈ (2) formulaga ko‘ra 3 2 2 3 0 1 2(2 ) ( ) , 02 3 3 r shar rxv rx x dx rx r= π ⋅ − = π ⋅ − = π∫ demak, 34 . 3sharv rπ= p 13-rasm. 14-rasm. 12-rasm. 6 shar segmentining hajmi. 14-rasmda oa=r, o1a=h (segmentning balandligi) bo‘lsin. doira segmentini uning balandligi atrofida aylanishidan shar segmenti hosil bo‘ladi (14-rasm.) shar segmentining hajmini hisoblash shar hajmini topish kabi bo‘ladi, bu holda integrallash [0; h] kesma bo‘yicha bajariladi: 3 2 2 2 3 0 1(2 ) ( ) ( ). 03 3 h segment hxv rx x dx rx rh h= π ⋅ …

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r     = π ⋅ ⋅ − + = π − + = π ⋅ − ⋅ + ⋅ = − π ⋅ ⋅ + π ⋅ ⋅ = π ⋅ − + = π       ∫ ∫ ∫ 3 2 3 2 2 2 2 0 0 0 2 12 ( ) 2 2 2 2 ( 1) . 0 03 2 3 2 3 3 r r r konus r rh h h x x h r rv x x h dx x dx h xdx h h r h r h r r r r     = π ⋅ ⋅ − + = π − + = π ⋅ − ⋅ + ⋅ = − π ⋅ ⋅ + π ⋅ ⋅ = π ⋅ − + = π       ∫ …

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0) va (0;4) nuqtalar orqali o‘tuvchi to‘g‘ri chiziq bilan chegaralangan shakl yuzini toping. 52. f (x)=2x – 2 funksiya grafigi va uning f (0)=1 shartni qanoatlanti ruvchi boshlang‘ich funksiyasi bilan chegaralangan shakl yuzini toping. quyidagi chiziqlar bilan chegaralangan shakl yuzini toping. mos rasm chizing (53–54): 53. 1) 2 ,y x= 21 ;y x= − 2) 1 ,y x = 0,y = 1,x = 4;x = 3) 2 2 ,y x x= − 24 ;y x= − 4) 1 ,y x = 0,y = 1,x = ( 1).x a a= > 54. 1) 2 , 3 xy = 224 ; 3 y x= − 2) 2 ,y x= 22y x= 2y = ; 3) 2 ,y x= 2 , 2 xy = y=2x; 4) 1 ,y x = 2 ,y x= 2 . 2 xy = 55. sin ,y x= x∈[0; p], funksiya grafigining ox o‘qi atrofida aylanishidan hosil bo‘lgan jism …

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matematika toshkent 2018 umumiy o‘rta ta’lim maktablarining 11-sinflari va o‘rta maxsus, kasb-hunar ta’limi muassasalari uchun darslik o‘zbekiston respublikasi xalq ta’limi vazirligi tomonidan tasdiqlangan 1-nashr algebra va analiz asoslari geometriya ii qism © “zamin nashr” mchj, 2018 © barcha huquqlar himoyalangan algebra va analiz asoslari bo‘limining mualliflari: m.a. mirzaahmedov, sh.n. ismailov, a.q. amanov geometriya bo‘limining muallifi: b.q. xaydarov uo‘k: 51(075.32) kbk: 22.1ya72 m 54 taqrizchilar: r.b. beshimov – mirzo ulug‘bek nomidagi o‘zbekiston milliy universiteti “geometriya va topologiya” kafedrasi mudiri, fizika-matematika fanlari doktori; q.s. jumaniyozov – nizomiy nomidagi tdpu fizika- matematika fakulteti “matematika o‘qitish metodikasi” kafedrasi dotsenti, pedagogika ...

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