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0 = 1 nuqtada o’tkazilgan urinmaning Ox o’qi bilan tashkil qilgan burchagini toping. A) 60 0 B) 30 0 C) 45 0 D) 120 0 7. (98-11-76) Qaysi nuqtada y = 1 + e x−1 funksiya- ning grafigiga o’tkazilgan urinma Ox o’qi bilan 45 0 li burchak hosil qiladi? A) x = 1 B) x = 0 C) x = −1 D) x = 2 8. (99-2-41) Abssissasi x 0 = 3 bo’lgan nuqtadan f (x) = √ 3 ln x funksiya grafigiga o’tkazilgan ur- inma Oy o’qi bilan qanday burchak tashkil qi- ladi? A) arctg3 B) 60 0 C) 30 0 D) arctg2 9. (99-10-44) Ushbu y = √ 3·x 2 −3 √ 3·x+4 funksiya- ning grafigiga x 0 = 2 nuqtada o’tkazilgan urinma Oy o’qi bilan qanday burchak tashkil qiladi? A) 120 0 B) 60 0 C) 30 0 D) 150 0 10. (98-10-73) y = 2x 3 +3x 2 −6x funksiyaning grafigiga o’tkazilgan urinma x ning qanday qiymatlarida y = 6x + 1 to’g’ri chiziqqa parallel bo’ladi? A) −2 va 3 B) 1 va 3 C) −2 va 1 D) 2 va −1 Yechish: Parallellik sharti 3-xossaga ko’ra 6 = f 0 (x 0 ) = 6x 2 + 6x − 6 tenglik bajarilishi kerak. Bu tenglik bajarilishi uchun x = −2 yoki x = 1 bo’lishi kerak. Javob: −2 va 1 (C). 11. (98-11-37) y = x 2 − 2x + 1 funksiya grafigining qanday nuqtasidan o’tkazilgan urinma y = −4(x+ 1) to’g’ri chiziqqa parallel bo’ladi? A) (−1; 1 4 ) B) (−1; 4) C) (1; 1 4 ) D) (1; 4) 12. (98-11-77) Agar f (x) funksiyaning grafigiga x 0 = 2 nuqtada o’tkazilgan urinmaning tenglamasi 2x− 3y = 6 bo’lsa, f 0 (2) qanchaga teng bo’ladi? A) 2 3 B) 3 2 C) 2 D) 3 13. (99-3-54) y = x 2 + 1 egri chiziqqa o’tkazilgan urinma y = 2x + 3 to’g’ri chiziqqa parallel. Uri- nish nuqtasining ordinatasini toping. A) 0 B) 2 C) 4 D) 1 2 14. (00-9-41) y = (2x+1) 2 egri chiziqqa (x 0 ; y 0 ) nuq- tadan o’tkazilgan urinma y = 2x + 1 2 to’g’ri chi- ziqqa parallel. Shu nuqtadan koordinata boshi- gacha bi’lgan masofani toping. A) √ 2 2 B) √ 2 4 C) √ 2 8 D) 1 15. (00-10-32) Qaysi nuqtada y = x 2 +2x+8 funksiya- ning grafigiga o’tkazilgan urinma y + 2x − 8 = 0 to’g’ri ciziqqa parallel bo’ladi? A) (−2; 8) B) (2; 8) C) (−2; −8) D) (2; −8) 16. (01-3-25) Ushbu y = x 2 + ln(x − 1) funksiyaning grafigiga x = 2 nuqtada o’tkazilgan urinmaning burchak koeffitsiyenti toping. A) 12 B) 5 C) 3 D)1 Yechish: Agar abssissasi x = 2 nuqtada o’tkazil- gan urinmaning burchak koeffitsiyentini k desak, u holda 1-xossaga ko’ra k = f 0 (2) = (2x + 1 x − 1 )| x=2 = 2 · 2 + 1 2 − 1 = 5 tenglik o’rinli bo’ladi. Javob: 5 (B). 17. (02-6-53) f (x) = 0, 5x 2 + x − 1, 5 funksiya grafigining abssissasi 2 ga teng bo’lgan nuqtasiga o’tkazilgan urinmaning burchak koeffitsiyenti to- ping. A) 1 B) 2 C) 3 D)4 18. (02-12-54) y = −5x + 3 to’g’ri chiziq, f (x) = x 2 − x funksiyaning grafigiga o’tkazilgan urin- maga parallel. Urinish nuqtasining koordinata- larini toping. A) (−2; 6) B) (1; 0) C) (2; 4) D) (0; 0) 19. (03-2-8) Qaysi to’g’ri chiziq y = 4 − x 2 funksiya grafigiga x 0 = 2 nuqtada o’tkazilgan urinmaga parallel bo’ladi? A) y = 4 − 4x B) y = 2x + 8 C) y = x + 8 D) y = 4x + 8 20. (03-3-51) y = sin x 2 (x ∈ (0; π)) funksiyaning grafigiga (x 0 , y 0 ) nuqtada o’tkazilgan urinmaning burchak koeffitsiyenti √ 3 4 ga teng. x 0 · y 0 ni hisoblang. A) 2 3 B) 1 6 C) 2π 3 D) π 6 21. (03-11-19) (x + 3) 2 + (y − 5) 2 = 45 aylananing A(0; 11) nuqtasiga o’tkazilgan urinmaning bur- chak koeffitsiyentini toping. A) − 1 2 B) −2 C) 1 2 D) 2 180 Urinma tenglamasi 22. (99-4-31) y = e 2−x · cos πx 2 funksiyaga abssissasi x 0 = 2 bo’lgan nuqtada o’tkazilgan urinmaning tenglamasini ko’rsating. A) y = x − 1 B) y = 1 − x C) y = 2x − 1 D) y = x − 3 Yechish: Masalani yechishda 2-xossadan foydala- namiz. Shu maqsadda y 0 va f 0 (x 0 ) larni hisoblay- miz: y 0 = e 2−2 · cos π2 2 = e 0 · cos π = −1, f 0 (2) = (−e 2−x ·cos πx 2 − π 2 e 2−x ·sin πx 2 )| x=2 = 1. Endi topilganlarni (1) formulaga qo’yamiz: y = −1 + 1 · (x − 2) = x − 3. Javob: y = x − 3 (D). 23. (96-1-29) Ushbu f (x) = 2x 2 −1 funksiya grafigiga abssissasi x 0 = 0 bo’lgan nuqtada o’tkazilgan ur- inma tenglamasini ko’rsating. A) y = −1 B) y = 2 C) y = 2x + 1 D) y = 1 24. (96-9-80) y = −2x 2 − 1 funksiya grafigiga abssis- sasi x 0 = 0 bo’lgan nuqtada o’tkazilgan urinma tenglamasini ko’rsating. A) y = 1 B) y = −2x C) y = x − 1 D) y = −1 25. (96-10-31) y = 1 − 2x 2 funksiya grafigiga abssis- sasi x 0 = 0 nuqtada o’tkazilgan urinma tengla- masini ko’rsating. A) y = 1 B) y = −1 C) y = −x D) y = 1 − 4x 26. (99-9-52) Abssissasi x 0 = 0 nuqtadan y = x 3 funksiya grafigiga o’tkazilgan urinmaning tengla- masini ko’rsating. A) y = x B) y = −0, 5x C) y = 0 D) y = 0, 5x 27. (00-5-48) y = 4 − x 2 parabolaga abssissasi x 0 = 1 nuqtada urinma o’tkazilgan. Bu urinmaning Oy o’qi bilan kesishadigan nuqtasining koordi- natasini toping. A) (0; 5) B) (0; 1) C) (0; −5) D) (0; −1) Yechish: Dastlab urinma tenglamasini topamiz. Shu maqsadda y 0 va f 0 (x 0 ) larni hisoblaymiz: y 0 = 4 − 1 2 = 3, f 0 (1) = −2x| x=1 = −2. Bu qiymatlarni (1) formulaga qo’yib, y = 3 − 2(x−1) urinma tenglamasini olamiz. Oy o’qining tenglamasi x = 0. Urinma tenglamasida x = 0 deb y = 5 ni olamiz. Javob: (0; 5) (A). 28. (00-6-28) y = 3 ln x−0, 5x funksiya grafigiga abs- sissasi x 0 = 3 nuqtada o’tkazilgan urinmaning tenglamasini tuzing. A) y = 0, 5x − 1, 5 B) y = 3x − ln 3 C) y = x − 3 ln 3 D) y = 0, 5x + 3 ln 3 − 3 29. (00-10-58) Ushbu f (x) = cos 2x funksiyaga ( π 4 ; f ( π 4 )) nuqtadan o’tkazilgan urinma tenglamasini ko’rsating. A) y = π 2 − 2x B) y = π − 3x C) y = π 2 + 3x D) y = π − 2x 30. (01-4-35) f (x) = x 3 funksiya grafigining A(−1; −1) nuqtasiga o’tkazilgan urinma tenglamasini ko’rsating. A) y = 3x − 2 B) y = 3x + 2 C) y = x + 2 D) y = x − 2 31. (02-1-67) y = x − 3x 2 funksiyaning grafigiga x 0 = 2 nuqtada o’tkazilgan urinmaning tenglamasini yozing. A) y = 1 − 6x B) y = −11x + 12 C) y = 3x + 1 D) y = x − 3 32. (03-6-69) y = x 2 − 2x parabolaga uning biror nuqtasidan o’tkazilgan urinmaning burchak ko- effisenti 4 ga teng. Shu urinmaning tenglamasini toping. A) y = 4x − 4 B) y = 4x + 9 C) y = 4x + 4 D) y = 4x − 9 Hosilaning mexanik ma’nosi 33. (99-2-39) Moddiy nuqta S(t) = 3t 3 −3t 2 +12t(m) qonuniyat bo’yicha harakatlanyapti. Uning tez- lanishi 0 ga teng bo’lgan paytda tezligi necha m/min bo’ladi? A) 8 B) 7 C) 9 D) 11 Yechish: 4 va 5-xossalarga ko’ra, nuqtaning tez- ligi uchun ϑ(t) = S 0 (t), tezlanishi uchun esa, a(t) = ϑ 0 (t) formulalar o’rinlidir. Demak, ϑ(t) = 9t 2 − 6t + 12, a(t) = ϑ 0 (t) = 18t − 6. Tezlanish 0 ga tengligidan 18t − 6 = 0 ⇐⇒ t = 6 18 = 1 3 ekanini topamiz. Uni tezlikning ifodasiga qo’yib ϑ( 1 3 ) = 9 · ³ 1 3 ´ 2 − 6 · 1 3 + 12 = 1 − 2 + 12 = 11 ni topamiz. Javob: 11 (D). 34. (96-3-83) To’g’ri chiziq bo’ylab x(t) = −t 3 +6t 2 + 15t qonuniyat bo’yicha harakatlanayotgan mod- diy nuqta harakat boshlangandan necha sekund o’tgach to’xtaydi. A) 1 B) 2 C) 3 D) 5 35. (96-9-14) To’g’ri chiziq bo’ylab x(t) = − 1 3 t 3 + 3 2 t 2 + 4t qonun bo’yicha harakatlanayotgan moddiy nuqta harakat boshlangandan necha sekund o’tgach to’x- taydi. A) 5 B) 3 C) 2 D) 4 36. (98-9-40) Moddiy nuqta S(t) = e t + cos t + 5t qo- nuniyat bo’yicha harakatlanyapti. Shu nuqtaning t = 0 dagi tezligini toping. A) 5 B) 8 C) 4 D) 6 181 Yechish: 4-xossaga ko’ra, nuqtaning tezligi uchun ϑ(t) = S 0 (t) = e t −sin t+5 formula o’rinlidir. De- mak, ϑ(0) = e 0 −sin 0+5 = 1−0+5 = 6. Javob: 6 (D). 37. (98-12-107) Moddiy nuqta S(t) = − 1 6 t 3 + 3t 2 − 5 qonuniyat bo’yicha harakatlanyapti. Uning te- zlanishi nolga teng bo’lganda, tezligi qanchaga teng bo’ladi? A) 24 B) 18 C) 12 D) 6 38. (99-3-57) Ikki moddiy nuqta S 1 (t) = 2, 5t 2 −6t+1 va S 2 (t) = 0, 5t 2 + 2t − 3 qonuniyat bo’yicha harakatlanyapti. Qaysi vaqtda birinchi nuqta- ning tezligi ikkinchisinikidan uch marta ko’p bo’lishi mumkin? A) 2 B) 3 C) 4 D) 6 39. (99-9-51) Moddiy nuqta to’g’ri chiziq bo’ylab S(t) = 6t 2 − 2t 3 + 5 qonuniyat bo’yicha harakat- lanyapti. Uning tezlanishi 0 ga teng bo’lgandagi oniy tezligi nimaga teng. A) 8 B) 6 C) 7 D) 9 40. (02-3-50) S(t) = t √ t qonuniyat bilan harakat- lanayotgan moddiy nuqtaning t = 2 sekunddagi tezlanishini hisoblang. A) 3 8 √ 2 B) 3 4 √ 2 C) 3 16 √ 2 D) 3 √ 2 41. (02-11-51) To’g’ri chiziq bo’ylab S(t) = 3t + 2 t + 3 qo- nuniyat bo’yicha harakatlanayotgan moddiy nuq- taning t = 2 sekunddagi tezligini (m/sek) aniqlang. A) 0,2 B) 0,25 C) 0,28 D) 0,32 42. (03-4-44) Ikki moddiy nuqta S 1 (t) = 2t 3 − 5t 2 − 3t(m) va S 2 (t) = 2t 3 −3t 2 −11t+7(m) qonuniyat- lar bo’yicha harakatlanyapti. Bu ikki nuqtaning tezliklari teng bo’lgan paytda birinchi nuqtaning tezlanishini (m/s 2 ) toping. A) 10 B) 8 C) 14 D) 9 14.4 Boshlang’ich funksiya va integral Biror intervalda ikki f va F funksiyalar berilgan bo’lib, ular F 0 (x) = f (x) (1) munosabat bilan bog’langan bo’lsin. Bobning boshida ta’kidlanganidek, f funksiya F funksiyaning hosilasi deyiladi. Funksiya hosilasi mavzusida F funksiyani bil- gan holda f funksiyani topish usullarini ko’rib chiqdik. Endi teskari masalani o’rganamiz, ya’ni f funksiya ma’- lum bo’lsa, hosilasi f ga teng bo’lgan F funksiyani top- ish usullari bilan tanishamiz. Agar F funksiya biror intervalda differensiallanuvchi bo’lib, (1) tenglik bajar- ilsa, F funksiya shu intervalda f funksiya uchun bosh- lang’ich funksiya deyiladi. Ko’p hollarda biror amalga teskari amal kiritilganda, u yagona ravishda aniqlan- maydi. Shunga o’xshash hol berilgan funksiyaga bosh- lang’ich funksiyani topishda ham sodir bo’ladi. Agar F funksiya f funksiya uchun boshlang’ich funksiya bo’lsa, u holda ixtiyoriy C o’zgarmas uchun F (x) + C ham f funksiya uchun boshlang’ich funksiya bo’ladi. Berilgan f funksiya uchun boshlang’ich funksiyani topish jara- yoni f funksiyani integrallash deyiladi. Agar F funksiya f funksiya uchun boshlang’ich funk- siya bo’lsa, u holda ixtiyoriy boshqa boshlang’ich funksi- ya F (x)+C ko’rinishga ega bo’ladi va u aniqmas integ- ral degan maxsus nomga ega bo’lib, quyidagicha yozi- ladi: Z f (x)dx = F (x) + C. Endi elementar funksiyalar uchun aniqmas integrallar jadvalini keltiramiz. 1. Z x α dx = x α+1 α + 1 + C, (α 6= −1). 2. Z x −1 dx = ln |x| + C. 3. R sin xdx = − cos x + C; 4. R cos xdx = sin x + C. 5. Z 1 sin 2 x dx = −ctgx + C. 6. Z 1 cos 2 x dx = tgx + C. 7. Z a x dx = a x ln a + C. 8. R e x dx = e x + C; 9. R tgxdx = − ln | cos x| + C. 10. R ctgxdx = ln | sin x| + C. Boshlang’ich funksiyani hisoblash qoidalari 11. R f 0 (x)dx = f (x) + C. 12. R (f (x) + g(x))dx = R f (x)dx + R g(x)dx. 13. R Cf (x)dx = C R f (x)dx. 14. Agar R f (x)dx = F (x) + C bo’lsa, Z f (ax + b)dx = 1 a F (ax + b) + C. 1. Ushbu f (x) = 2x − 1 funksiyaning boshlang’ich funksiyasining umumiy ko’rinishini toping. A) 2x 2 − x + C B) x 2 − x + C C) x 2 − 1 + C D) x 2 + x + C Yechish: Berilgan funksiyani f (x) = 2x 1 − x 0 shaklda yozib, 1 va 12-13 formulalardan foydalanib Z f (x)dx = 2 · x 1+1 2 − x 1 + C = x 2 − x + C ni olamiz. Javob: x 2 − x + C (B). 2. Ushbu f (x) = 3x 2 − sin x funksiyaning aniqmas integralini toping. A) x 3 + cos x + C B) x 3 − cos x + C C) 3x 3 − cos x + C D) x 3 + tgx + C 182 3. Ushbu f (x) = e x +cos x funksiyaning boshlang’ich funksiyasining umumiy ko’rinishini toping. A) e x − cos x + C B) e x + sin x + C C) e x − sin x + C D) e x + cos x + C 4. Ushbu f (x) = 1 + cos −2 x funksiyaning bosh- lang’ich funksiyasini toping. A) x + 2cos −3 x + C B) x − tgx + C C) x − ctgx + C D) x + tgx + C 5. (99-8-40) Ushbu f (x) = 3 4 √ x funksiyaning bosh- lang’ich funksiyasini toping. A) 3 2 √ x + C B) 3 √ x + C C) 4 3 √ x + C D) − 3 2 √ x + C 6. Ushbu f (x) = x −1 −tgx funksiyaning boshlang’ich funksiyasining umumiy ko’rinishini toping. A) ln |x cos x| + C B) ln |x| + ctgx + C C) ln | cos x x | + C D) ln |x| − ctgx + C 7. Ushbu f (x) = 3(x − 1)(x + 3) funksiyaning bosh- lang’ich funksiyasini toping. A) x 3 +3x 2 −9x+C B) x 3 −3x 2 −9x+C C) x 3 +3x 2 +9x+C D) x 3 +3x 2 −3x+C Yechish: Qavsni ochib berilgan funksiyani f (x) = 3x 2 +6x−9 shaklda yozib, 1 va 12-13 formulalar- dan foydalanib Z f (x)dx = x 3 + 3x 2 − 9x + C ni olamiz. Javob: x 3 + 3x 2 − 9x + C (A). 8. Ushbu f (x) = (3x + 1) 2 funksiyaning aniqmas integralini toping. A) x 3 + 3x 2 + x + C B) x 3 − 3x 2 − x + C C) x 3 − 3x 2 + x + C D) x 3 − 3x 2 + x + C 9. Ushbu f (x) = (e 0,5x +e −0,5x ) 2 funksiyaning aniq- mas integralini toping. A) e x + e −x + x + C B) e x − e −x + x + C C) e x +e −x +2x+C D) e x −e −x +2x+C 10. Ushbu f (x) = √ x(x+1) funksiyaning boshlang’ich funksiyasini toping. A) 2 5 x 2 √ x+ 2 3 x √ x+C B) 2 5 x 5/2 − 2 3 x 3/2 +C C) 5 2 Download 1.09 Mb. Do'stlaringiz bilan baham: |
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