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−2x − 2x 2 e −2x = 2xe −2x (1 − x). 1-xossaga ko’ra, f 0 (x) = 2xe −2x (1 − x) > 0 teng- sizlikni yechamiz. Barcha x ∈ R lar uchun 2e −2x > 0 ekanligini hisobga olsak, f 0 (x) > 0 tengsizlik x(1 − x) > 0 tengsizlikka teng kuchli bo’ladi. Bu tengsizlikni oraliqlar usuli bilan yechib (0; 1) ni olamiz. [0; 1] kesmada ham funksiyaning o’sish xossasi saqlanadi. Javob: [0; 1] (D). 15. (00-1-44) Qaysi oraliqda f (x) = ln(4x−x 2 ) funksiya kamayadi? A) (0; 2) B) (−∞; 0) C) (0; 4) D) (2; 4) 16. (00-7-37) Ushbu f (x) = 2 3 x 3 − 4x 2 + 3 funksiya kamayadigan oraliqdagi barcha butun qiymatlar yig’indisini toping. A) 9 B) 8 C) 10 D) 7 17. (01-3-13) Funksiyaning kamayish oralig’ini top- ing. y = x 2 2 − 12 ln(x − 4) A) [6; ∞) B) (4; ∞) C) (2; 4) D) (4; 6] 18. (01-11-37) Qaysi oraliqda f (x) = 1 5 x 5 − 4x 2 funk- siya kamayadi? A) [0; 2] B) (0; 2] C) [0; 2) D) (0; 2) 19. (02-1-65) y = 2x 3 +3x 2 −2 funksiyaning kamayish oraliqlarini aniqlang. A) (0; 8) B) (−∞; −1] C) [−1; ∞) D) [−1; 0] 20. (02-5-43) y = 1 4 x 4 − 5 3 x 3 + 3x 2 + 10 funksiyaning kamayish oraliqlarini aniqlang. A) (2; 3) B) (−∞; 0] ∪ [2; 3] C) (−∞; 3) D) (−∞; 0) ∪ (3; +∞) 21. (02-9-31) f (x) = −2x 3 + 15x 2 + 12 funksiya o’sa- digan kesmaning uzunligini aniqlang. A) 5 B) 4 C) 6 D) 4,5 22. (02-12-55) Funksiya qaysi oraliqda kamayadi? f (x) = 1 5 x 5 − 4x 2 A) (−2; 0] B) [0; 2] C) [−2; 0) D) (0; 3) Funksiyaning ekstremumlari 176 23. (96-7-29) f (x) = 3x − x 3 funksiyaning maksimu- mini toping. A) −1 B) 2 C) −2 D) 4 Yechish: Ferma teoremasiga ko’ra, funksiya mak- simumga erishadigan nuqtalarda uning hosilasi nolga aylanadi. Shu maqsadda f 0 (x) = 3 − 3x 2 = 3(1 − x 2 ) = 3(1 − x)(1 + x) = 0 tenglamani yechamiz. Bu tenglamaning yechimi x 1 = −1 va x 2 = 1 lardir. Oson tekshirish mumkinki (−∞; −1) oraliqda f 0 (x) > 0 va (−1; 1) oraliqda f 0 (x) < 0. 4-xossaga ko’ra, berilgan funk- siya x = −1 nuqtada maksimumga erishadi. De- mak, f (−1) = −2 funksiyaning maksimumi bo’- ladi. Javob: −2 (C). 24. (97-3-29) Ushbu g(x) = 12x − x 3 funksiyaning minimumini toping. A) −32 B) −16 C) 0 D) 16 25. (97-10-29) Ushbu y = −4x 3 + 12x funksiyaning minimumini toping. A) 0 B) −8 C) −16 D) 8 26. (97-11-15) Ushbu y = x 2 − 8x + 7 funksiyaning qiymatlari sohasini toping. A) (2; ∞) B) [−9; ∞) C) [9; ∞) D) [−4; ∞) 27. (98-1-29) Ushbu f (x) = x 3 +2, 5x 2 −2x funksiya- ning maksimum nuqtasidagi qiymatini hisoblang. A) −8 B) 6 C) 10,5 D) −12 28. (98-9-8) t ning qanday qiymatida −t 2 + 14t − 31 uchhad eng katta qiymatga erishadi? A) 6 B) 5 C) 8 D) 7 29. (99-3-16) Ushbu x 2 − ax + a − 1 = 0 tenglamaning ildizlari x 1 va x 2 bo’lsin. a ning qanday qiymatida x 2 1 +x 2 2 yig’indi eng kichik qiy- matga ega bo’ladi? A) 1 B) 2 C) 1,5 D) 2,5 30. (99-4-21) Agar 2x + y = 6 bo’lsa, xy ning eng katta qiymati nechaga teng bo’ladi? A) 2,5 B) 4,5 C) 3 D) −2, 5 31. (99-9-48) Ushbu y = −x 2 + 6x − 12 funksiyaning qiymatlari sohasini toping. A) (−3; ∞) B) [−3; ∞) C) (−∞; −3) D) (−∞; −3] 32. (99-4-25) Ushbu f (x) = √ 2 − x − x 2 funksiya- ning eng katta qiymatini toping. A) √ 2 B) 1,5 C) 3 D) 2 √ 2 Yechish: y = √ t funksiya [0; ∞) o’suvchi bo’lganligi uchun, berilgan funksiyaning eng katta qiymati ildiz ostidagi g(x) = 2 − x − x 2 funksiyaning eng katta qiymatida erishadi. Ferma teoremasiga ko’ra, g funksiya maksimumga erishadigan nuq- talar g 0 (x) = −1 − 2x = 0 tenglama ildizlari ichida bo’ladi. Bu tenglama esa yagona x 0 = −0, 5 ildizga ega. Shuning uchun f (x 0 ) = = p 2 − (−0, 5) − (0, 5) 2 = p 2, 5 − 0, 25 = 1, 5 f ning eng katta qiymati bo’ladi. Javob: 1, 5 (B). 33. (99-3-28) Funksiyaning qiymatlar sohasini toping. y = p 3x 2 − 4x + 5 A) [0; ∞) B) [ √ 3; ∞) C) [ r 3 2 ; ∞) D) [ r 11 3 ; ∞) 34. (03-2-7) y = √ x 2 + 2x + 4 funksiyaning qiymat- lar sohasini ko’rsating. A) [0; ∞) B) [2; ∞) C) (0; ∞) D) [ √ 3; ∞) 35. (99-8-37) Ushbu y = √ 3 − x 2 − 2x funksiyaning eng katta qiymatini toping. A) −2 B) 4 C) 2 D) 3 36. (00-7-35) Ushbu f (x) = 3 1+x +3 1−x funksiyaning eng kichik qiymatini toping. A) 9 B) 4 C) 8 D) 6 37. (99-10-42) Ushbu y = √ x 2 − 2x + 10 funksiya- ning qiymatlar sohasini toping. A) [3; ∞) B) (3; ∞) C) [5; ∞) D) [2; ∞) 38. (01-1-34) Ushbu f (x) = 3x 5 − 5x 3 − 3 funksiyan- ing ekstremum nuqtalaridagi qiymatlari yig’indisini hisoblang. A) −9 B) −6 C) −8 D) −4 Yechish: Dastlab berilgan funksiyaning ekstre- mum nuqtalari topamiz. Ferma teoremasiga ko’ra, funksiyaning ekstremum nuqtalari f 0 (x) = 15x 4 − 15x 2 = 15x 2 (x − 1)(x + 1) = 0 tenglama ildizlari ichida bo’ladi. Bu tenglama x 0 = −1, x 1 = 0, x 2 = 1 ildizlarga ega. Argu- ment x x 0 = −1 yoki x 2 = 1 nuqtadan o’tganda funksiya hosilasi ishorasini o’zgartiradi. 3 va 4- xossalarga ko’ra, funksiya bu nuqtalarda ekstre- mumga erishadi. Argument x x 1 = 0 nuqtadan o’tganda funksiya hosilasi ishorasini o’zgartirmay- di. 5-xossaga ko’ra, bu nuqta funksiya uchun eks- tremum emas. Masalani oxiriga yetkazish uchun f (x 0 ) + f (x 2 ) yig’indini hisoblashimiz kerak. f (−1)+f (1) = 3(−1) 5 −5(−1) 3 −3+3−5−3 = −6. Javob: −6 (B). 39. (01-2-60) x(x + 1)(x + 2)(x + 3) ko’paytmaning eng kichik qiymatini toping. A) 3 B) 2 C) 1 D) −1 40. (01-12-38) y = −x 2 + bx + c funksiya x = −1 nuqtada 5 ga teng eng katta qiymatni qabul qilsa, y(1) ni toping. A) −1 B) 0 C) 1 D) 1,5 177 41. (02-4-6) y = x 2 + 4x + 11 funksiyaning eng kichik qiymatini toping. A) 4 B) 11 C) 11 4 D) 7 42. (02-2-4) a ning qanday qiymatida (a − 7) 2 + (a − 8) 2 + (a − 12) 2 ifoda eng kichik qiymatga ega bo’ladi? A) 9 B) 10 C) 8 D) 11 43. (02-11-52) f (x) = 0, 9x 5 −4, 5x 3 + 4 funksiyaning minimum nuqtasini toping. A) −1 B) 1 C) √ 2 D) √ 3 44. (03-5-30) f (x) = 9 x + 5 · 3 −2x funksiyaning qiy- matlar to’plamini ko’rsating. A) [2 √ 5; ∞) B) (0; ∞) C) [5; ∞) D) [6; ∞) Yechish: Ikkala g(x) = 9 x va ϕ(x) = 5 · 3 −2x funksiyalarning qiymatlari to’plami (0; ∞) dan iborat. Shuning uchun yig’indi f (x) = 9 x + 5 · 3 −2x funksiyaning qiymatlar to’plami [m; ∞), bu yerda m berilgan funksiyaning eng kichik qiy- mati. Berilgan funksiyani f (x) = 9 x + 5 · 9 −x shaklda yozib, uning kritik nuqtalari topamiz. Shu maqsadda f 0 (x) = 9 x ·ln 9−5·9 −x ·ln 9 = ln 9(9 x −5·9 −x ) = 0 tenglamani yechamiz. Bu ko’rsatkichli tenglama bo’lib, uning ildizi x 0 = log 9 √ 5 dir. Funsiya hosilasi yagona x 0 nuqtada nolga aylanyapti, shun- ing uchun u minimum nuqta bo’ladi. Uni hisoblay- miz: f (x 0 ) = 9 log 9 √ 5 +5·9 − log 9 √ 5 = √ 5+5· 1 √ 5 = 2 √ 5. Berilgan funksiyaning qiymatlar to’plami [2 √ 5; ∞) ekan. Javob: [2 √ 5; ∞) (A). 45. (03-1-57) Qaysi sonni o’zining kvadrati bilan yig’indisi eng kichik bo’ladi? A) −1 B) −0, 4 C) −0, 8 D) −0, 5 46. (03-3-52) Agar m va M sonlar y = x+ 1 x funksiya- ning mos ravishda minimum va maksimum nuq- talaridagi qiymatlari bo’lsa, m−2M ning qiyma- tini toping. A) −6 B) 6 C) −4 D) 4 47. (03-7-81) y = −x 4 +2x 2 +5 funksiyaning qiymat- lar to’plamini toping. A) (−∞; 6] B) (−∞; 6) C) [5; 6] D) (−∞; 5] 48. (03-9-46) f (x) = 0, 6x 5 − 2x 3 − 1 funksiyaning maksimum va minimum nuqtalaridagi qiymatlari yig’indisini toping. A) −3 B) −2 C) −1 D) 1 49. (03-11-7) m va n natural sonlar. 6 x = 1 m + 1 n va m + n = 18 bo’lsa, x ning eng katta qiymatini toping. A) 27 B) 24 C) 18 D) 30 Eng katta va eng kichik qiymatlar Berilgan f (x) funksiyaning [a; b] kesmadagi eng katta va eng kichik qiymatlarini topish uchun, dastlab uning shu oraliqqa tegishli kritik nuqta- lari topiladi, ya’ni f 0 (x) = 0 tenglamaning [a; b] oraliqqa tegishli ildizlari topilib, so’ngra beril- gan funksiyaning bu ildizlardagi qiymatlari va kesmaning chetki x = a, x = b nuqtalardagi qiymatlari hisoblanib, ular o’zaro taqqoslanadi. Bu qiymatlardan eng kattasi funksiyaning [a; b] kesmadagi eng katta qiymati, eng kichigi esa funk- siyaning eng kichik qiymati bo’ladi. 50. (97-9-90) Ushbu f (x) = 3x 2 −4x−4 funksiyaning [0; 3] kesmadagi eng katta qiymatini toping. A) 10 B) 20 C) 11 D) 16 Yechish: Yuqorida keltirilgan qoidaga ko’ra f 0 (x) = 6x − 4 = 0 tenglamani yechamiz. Uning yechimi x = 2 3 ∈ [0; 3]. Endi berilgan funksiyanung x = 0, x = 3, x = 2 3 nuqtalardagi qiymatlarini hisoblaymiz: f (0) = −4, f (3) = 11, f ( 2 3 ) = − 16 3 . Sonlardan eng kattasi 11 bo’lganligi uchun funksi- yaning [0; 3] kesmadagi eng katta qiymati 11 ga teng bo’ladi. Javob: 11 (C). 51. (97-4-30) Ushbu f (x) = x 2 − 3x + 1, 25 funksiya- ning [−1; 1] oraliqdagi eng katta qiymatini to- ping. A) 0 B) −0, 75 C) 5,25 D) 6,25 52. (98-5-27) Ushbu y = x 2 − 2x + 5 funksiyaning [0; 1] kesmadagi eng katta qiymatini toping. A) 5 B) 4 C) −2 D) 0 53. (98-9-38) Ushbu f (x) = x 3 + 2x − 5 funksiyaning [−1; 1] kesmadagi eng katta va eng kichik qiymat- lari ayirmani toping. A) −6 B) 6 C) −5 D) 5 54. (98-10-72) y = 2x 3 +3x 2 −12x funksiyaning [0; 2] kesmadagi eng kichik qiymatini toping. A) 0 B) −2 C) −5 D) −7 55. (98-11-33) y = 0, 25x 4 − x 3 3 − x 2 funksiyaning [−2, 5; ∞) oraliqdagi eng kichik qiymatini aniqlang. A) − 3 8 B) 3 8 C) 8 3 D) − 8 3 56. (99-2-42) y = 3x 4 − 4x 3 funksiyaning [0; 2] kes- madagi eng kichik qiymatini toping. A) 0 B) −16 C) −1 D) 1 178 57. (99-3-53) Ushbu y = x 3 − 3x 2 + 1 funksiyaning [−1; 4] kesmadagi eng katta va eng kichik qiymat- lari ayirmasini toping. A) 20 B) 14 C) 15 D) 18 58. (99-7-28) Ushbu y = x 2 − 2x − 1 funksiyaning [−1; 1] kesmadagi eng katta qiymatini toping. A) 4 B) 2 C) 0 D) 6 59. (00-1-15) Agar m > 0, n > 0 va m + n = 16 bo’lsa, mn ning eng katta qiymatini toping. A) 62 B) 72 C) 64 D) 60 Yechish: Masala shartlaridan m = 16 − n va mn = (16 − n)n, n ∈ (0; 16) kelib chiqadi. Agar f (n) = (16 − n)n = 16n − n 2 deb belgilab olsak, bu funksiyaning (0; 16) intervaldagi eng katta qiy- matini topish talab qilinadi. f 0 (n) = 16 − 2n = 0 tenglama yagona n = 8 yechimga ega. f (0) = f (16) = 0, f (8) = 8 · 8 = 64. Demak, mn ning eng katta qiymati 64 ekan. Javob: 64 (C). 60. (00-3-66) Ushbu f (x) = 3x − x 3 funksiyaning [−2; 3] kesmadagi eng katta va eng kichik qiymat- lari ayirmasini toping. A) 20 B) 18 C) 16 D) 12 61. (00-3-67) Bir tomondan imorat bilan chegaralan- gan, qolgan tomonlari uzunligi 120 m panjaradan iborat to’gri to’rtburchak shaklidagi yer maydo- nining eng katta yuzini toping. A) 1600 B) 1500 C) 1800 D) 2000 62. (00-4-52) Ushbu y = 4x 2 + 1 x funksiyaning [0, 25; 1] kesmadagi eng katta qiymatini toping. A) 3 B) 4,25 C) 4,5 D) 5 63. (00-10-28) y = 12x − x 3 funksiyaning [−1; 3] kesmadagi eng katta va eng kichik qiymatlari ayirmasini toping. A) 27 B) 15 C) 5 D) 32 64. (01-3-18) Ushbu f (x) = x 2 (x − 6) funksiyaning [−1; 3] dagi eng katta va eng kichik qiymatlarini aniqlang. A) 2; −4 B) 0; −32 C) 6; −21 D) 0; −27 65. (01-7-49) Ushbu y = 4x 2 + 1 x funksiyaning [ 1 4 ; 1] kesmadagi eng katta va eng kichik qiymatlari yig’indisini toping. A) 7 1 4 B) 9 1 4 C) 10 1 4 D) 8 66. (01-9-49) Ushbu y = 1 3 x 3 + x 2 − 8x funksiyaning [1; 3] kesmadagi eng katta va eng kichik qiymat- larining ko’paytmasini toping. A) 48 B) −37 C) 50 D) 56 67. (01-11-39) Ushbu y = 1 3 x 3 − 4x funksiyaning [0; 2] kesmadagi eng katta va eng kichik qiymat- larining ayirmasini toping. A) 5 1 3 B) 15 2 3 C) 10 2 3 D) 15 1 5 14.3 Hosilaning geometrik va mexanik ma’nosi. Urinma va tezlik y = f (x) funksiya grafigiga (x 0 ; y 0 ) nuqtada o’tkazilgan urinmaning burchak koeffitsiyenti k, bu urinmaning Ox o’qining musbat yo’nalishi bilan tashkil qilgan burchagi α bo’lsin (14.2-chizma). U holda quyidagilar o’rinli. 1. k = f 0 (x 0 ) = tgα. 2. y = f (x) funksiya grafigining (x 0 ; y 0 ) nuq- tasidan o’tuvchi urinma tenglamasi y = y 0 + f 0 (x 0 )(x − x 0 ). (1) 3. y = f (x) va y = g(x) funksiyalar grafiklari- ga abssissasi x 0 bo’lgan nuqtada o’tkazilgan urinmalarning parallellik sharti: f 0 (x 0 ) = g 0 (x 0 ). S(t) qonuniyat bo’yicha harakatlanayotgan mod- diy nuqtaning tezligi ϑ(t), tezlanishi esa a(t) bo’l- sin. U holda quyidagilar o’rinli. 4. ϑ(t) = S 0 (t). 5. a(t) = ϑ 0 (t). 1. (96-13-23) y = x 1 − x funksiya grafigiga abssis- sasi x 0 = 3 bo’lgan nuqtada o’tkazilgan urinma- ning Ox o’qi bilan tashkil qilgan burchagi α bo’lsa, tg2α ni toping. A) 7 15 B) 2 5 C) 8 15 D) 3 5 Yechish: 1-xossaga ko’ra tgα = f 0 (x 0 ) ekani ma’lum. Endi funksiyaning hosilasini topamiz: y 0 = 1 − x + x (1 − x) 2 = 1 (1 − x) 2 So’ngra x ning o’rniga x = 3 ni qo’yib tgα = 1 (1 − 3) 2 = 1 4 ni topamiz. Ikkilangan burchak formulasidan tg2α = 2tgα 1 − tg 2 α = 2 · 1 4 1 − 1 16 = 8 15 ni hosil qilamiz. Javob: 8 15 (C). 179 2. (96-6-46) Ushbu y = x 2 − 3x + 2 parabolaga abs- sissasi x 0 = 2 bo’lgan nuqtada o’tkazilgan urin- maning burchak koeffitsiyenti nimaga teng. A) 1 B) 2 C) −3 D) 3 3. (97-2-46) y = ln x + x 2 funksiyaning grafigiga x 0 = 1 2 nuqtada o’tkazilgan urinmaning burchak koeffitsiyentini toping. A) 3 B) 6 C) 4 D)6,5 4. (97-8-46) Ushbu y = 1 3 x 3 − ln x funksiyaning grafi- giga x 0 = 2 nuqtada o’tkazilgan urinmaning bur- chak koeffitsiyentini toping. A) 4 B) 3 C) 2 D) 3,5 5. (97-4-29) y = 3x 2 + 2x funksiya grafigiga ab- ssissasi x 0 = −3 nuqtada o’tkazilgan urinma Ox o’qining musbat yo’nalishi bilan qanday burchak hosil qiladi? A) arctg3 B) π − arctg16 C) π − arctg3 D) −arctg16 6. (98-9-39) Ushbu f (x) = √ 3 3 · x 3 − 1 funksiyaning grafigiga Download 1.09 Mb. Do'stlaringiz bilan baham: |
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