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p f (x) > ϕ(x) ⇐⇒ ½ f (x) > [ϕ(x)] 2k ϕ(x) ≥ 0; ∪ ½ f (x) ≥ 0, ϕ(x) < 0. 2. 2k+1 p f (x) > ϕ(x) ⇐⇒ f (x) > [ϕ(x)] 2k+1 ; 3. 2k p f (x) < ϕ(x) ⇐⇒ f (x) ≥ 0, ϕ(x) ≥ 0, f (x) < [ϕ(x)] 2k 4. 2k+1 p f (x) < ϕ(x) ⇐⇒ f (x) < [ϕ(x)] 2k+1 ; 1. (97-10-34) Tengsizlikning yechimini ko’rsating. (x − 1) p 6 + x − x 2 ≤ 0 A) (−∞; 1] B) [−2; 3] C) [−2; 1] ∪ {3} D) [3; ∞) Yechish: Berilgan tengsizlikni yechish uchun ikki holni qaraymiz. 1) 6 + x − x 2 = 0. Uning ildizlari x 1 = −2, x 2 = 3. Bu sonlar berilgan tengsizlik- ning ham yechimi bo’ladi. 2) ½ 6 + x − x 2 > 0 x − 1 ≤ 0. Birinchi tengsizliklarni −1 ga ko’paytirib, uning chap qismini ko’paytuvchilarga ajratamiz. ½ (x + 2)(x − 3) < 0 x − 1 ≤ 0. Bu holda (−2; 1] yechimni hosil qilamiz. Endi yuqorida topilgan x 1 = −2, x 2 = 3 sonlarni hisobga olib berilgan tengsizlikning [−2; 1] ∪ {3} yechimini hosil qilamiz. Javob: [−2; 1]∪{3} (C). 2. (96-7-34) Tengsizlikning yechimini ko’rsating. (x + 3) p x 2 − x − 2 ≥ 0 A) [−3; ∞) B) [−1; 2] C) [−3; −1] ∪ [2; ∞) D) [2; ∞) 3. (97-3-34) Quyidagilardan qaysi biri (x − 3) p x 2 + x − 2 ≤ 0 tengsizlikning yechimi? A) (−∞; 3] B) (−∞; −2] ∪ [1; 3] C) [−2; 3] D) [−1; 2] ∪ [3; ∞) 4. (97-7-34) Tengsizlikning yechimini ko’rsating. (x − 2) p 3 + 2x − x 2 ≥ 0 A) [2; ∞) B) [−1; 3] C) [3; ∞) D) [2; 3]∪{−1} 5. (00-3-21) Tengsizlikni yeching. √ 3x + 10 > √ 6 − x A) [−1; 6] B) [− 10 3 ; 6] C) (−1; 6] D) [− 10 3 ; −1) ∪ (−1; 6] 6. (01-10-19) Tengsizlikni yeching. √ 3x − 8 > √ 5 − x A) (3, 25; ∞) B) ( 8 3 ; 5) C) (3, 25; 5] D) (3, 25; 5) 7. (01-12-46) Tengsizlikni yeching. p 5x − 2x 2 − 42 > 3 A) {−2} B) {1} C) {2} D) ∅ 8. (02-1-48) Tengsizlikni yeching. √ x + 1 < 4 A) (−∞; 15) B) [0; 15] C) [0; 15) D) [−1; 15) 9. (02-1-68) Tengsizlikni yeching. (x + 3) p 10 − 3x − x 2 ≥ 0 A) [−3; ∞) B) [2; ∞) C) [−3; 2] D) {−5} ∪ [−3; 2] 89 10. (02-10-12) Tengsizlikni yeching. √ 6 + x − x 2 2x + 5 ≥ √ 6 + x − x 2 x + 4 A) [−2; −1]∪{3} B) [−2; 1] C) [1; 3] D) [−2; 3] 11. (98-4-23) Tengsizlik nechta butun yechimga ega? √ x + 2 > x A) 3 B) 2 C) 4 D) 1 Yechish: Berilgan tengsizlikni yechishda 1-qoi- dadan foydalanamiz: ½ x + 2 > x 2 x ≥ 0; ∪ ½ x + 2 ≥ 0, x < 0. 1-sistemani yechamiz. Uning 1-tengsizligi x 2 − x − 2 < 0 ⇐⇒ (x + 1)(x − 2) < 0 kvadrat tengsizlikka teng kuchli bo’lib, uning yechi- mi (−1; 2) dan iborat. Agar sistemaning ikkinchi tengsizligi x ≥ 0 ni hisobga olsak, 1-sistemaning yechimi [0; 2) ekanligini olamiz. 2-sistemani quyidagicha yozib olamiz: ½ x + 2 ≥ 0, x < 0. ⇐⇒ ½ x ≥ −2, x < 0. ⇐⇒ −2 ≤ x < 0. Bu qo’sh tengsizlik 2-sistemaning yechimi [−2; 0) ekanligini bildiradi. Bu yechimlarni birlashtirib, berilgan tengsizlikning yechimi [−2; 2) ekanligini olamiz. Bu yarim intervalda −2, −1, 0, 1 butun sonlar bor. Ularning soni 4 ta. Javob: 4 (C). 12. (98-12-82) Tengsizlikni qanoatlantiruvchi butun sonlar nechta? p 5 − x 2 > x − 1 A) 5 B) 3 C) 4 D) 2 13. (99-2-20) Tengsizlik nechta butun yechimga ega? p x 2 − 6x + 9 < 3 A) 4 B) 6 C) 7 D) 5 14. (00-2-15) Tengsizlikning eng kichik butun musbat yechimini toping? √ x + 5 1 − x < 1 A) 6 B) 3 C) 5 D) 2 15. (00-7-23) Tengsizlikning eng katta butun va eng kichik butun yechimlari ayirmasini toping. p x 2 − 16 < √ 4x + 16 A) 4 B) 5 C) 2 D) 3 16. (01-5-23) Tengsizlikning nechta butun yechimi bor? r 3x − 4 8 − x > 1 A) 4 B) 1 C) 2 D) 3 17. (02-4-26) Tengsizlik nechta butun yechimga ega? √ x − 50 · √ 100 − x > 0 A) 43 B) 54 C) 49 D) 51 Yechish: Arifmetik ildizning manfiymasligidan berilgan tenglama ½ x − 50 > 0 100 − x > 0 ⇐⇒ ½ x > 50 x < 100 sistemaga teng kuchli bo’ladi. Uning yechimi (50; 100) oraliqdan iborat. Bu oraliqda 49 ta bu- tun son bor. Javob: 49 (C). 18. (01-6-26) Tengsizlikning butun yechimlari yig’in- disini toping. √ 2x + 7 6 − 3x ≥ 0 A) −4 B) −3 C) 4 D) −5 19. (02-9-26) Tengsizlikning eng kichik va eng katta butun yechimlari ayirmasini toping. x − 4 √ x − 5 ≤ 0 A) −25 B) −24 C) −27 D) −5 20. (02-9-28) Tengsizlikning butun yechimlari nechta? √ 3 + 2x − x 2 x − 2 ≤ 0 A) 3 B) 4 C) 5 D) 2 21. (02-12-14) Tengsizlikning butun yechimlari nechta? 5 − √ x √ x − 2 > 0 A) 20 B) 19 C) 21 D) 2 22. (02-12-35) Qanday eng kichik butun son √ 12 − x < 2 tengsizlikni qanoatlantiradi? A) 8 B) 9 C) 6 D) 10 23. (03-1-8) Tengsizlikning eng kichik butun yechi- mini toping. r 2 − 3x x + 4 > −2 A) 0 B) −1 C) −2 D) −3 Yechish: Arifmetik ildizning manfiymasligini hi- sobga olsak, 2 − 3x x + 4 ≥ 0 tengsizlikni yechish kifoya. Bu tengsizlikni oraliq- lar usuli yordamida yechib, (−4; 2/3] ni olamiz. Bu yarim intervaldagi eng kichik butun son −3 dir. Javob: −3 (D). 90 24. (03-1-30) Tengsizlikni qanoatlantiruvchi butun son- larning yig’indisini toping. √ x ≥ x − 6 A) 6 B) 15 C) 28 D) 45 25. (03-3-20) Tengsizlikning butun yechimlari nechta? √ x − 4 − √ x − 7 ≥ 1 A) 0 B) 1 C) 2 D) 4 26. (03-3-30) Tengsizlikning butun yechimlari nechta? p 5 − |2x − 1| < 2 A) 2 B) 3 C) 4 D) 6 27. (03-8-37) Tengsizlikning eng kichik natural yechi- mini toping. p x 2 − 3x + 2 ≥ 0 A) 1 B) 2 C) 3 D) 5 28. (03-9-9) Tengsizlikning butun yechimlari nechta? r x 2 − 2 x ≤ 1 A) ∅ B) 1 C) 2 D) 3 29. (03-11-73) Tengsizlikning butun yechimlari nechta? p 8 + 2x − x 2 > 6 − 3x A) 2 B) 3 C) 4 D) 5 8 -bob. Progressiyalar 8.1 Arifmetik progressiya Arifmetik progressiya deb shunday sonlar ketma-ketligi- ga aytiladiki, unda ikkinchi hadidan boshlab har bir hadi o’zidan oldingi hadga shu ketma-ketlik uchun o’zgar- mas bo’lgan biror d sonni qo’shish natijasida hosil bo’ladi. Masalan, 1) 1, 2, 3, 4, . . . ; 2) 10, 12, 14, 16, . . . ketma- ketliklar arifmetik progressiya tashkil qiladi. Chunki har bir son, ikkinchisidan boshlab, mos ravishda 1 va 2 sonlarini oldingisiga qo’shish natijasida hosil bo’ladi. Arifmetik progressiyani tashkil qiluvchi sonlar un- ing hadlari deyiladi va umumiy ko’rinishda a 1 , a 2 , a 3 , . . . , a n−1 , a n , . . . (1) yoziladi. Arifmetik progressiyaning keyingi hadini hosil qilish uchun oldingi hadiga qo’shiladigan d son arifmetik progressiya ayirmasi deyiladi. Agar d > 0 bo’lsa, pro- gressiya o’suvchi, d < 0 bo’lsa, progressiya kamayuvchi deyiladi. Agar d = 0 bo’lsa, arifmetik progressiyaning barcha hadlari o’zaro teng bo’ladi. d = 0 hol odatda qaralmaydi. Arifmetik progressiyaning n − hadi a n quyidagi for- mula yordamoda topiladi: a n = a 1 + (n − 1)d. Arifmetik progressiya hadlarining xossalari. 1-xossa. Arifmetik progressiyaning ikkinchi hadi- dan boshlab istalgan hadi o’ziga qo’shni bo’lgan ikki hadning o’rta arifmetik qiymatiga teng, ya’ni a n = a n−1 + a n+1 2 . 2-xossa. Chekli arifmetik progressiyada boshidan va oxiridan teng uzoqlikda to’rgan hadlar yig’indisi chetki hadlar yig’indisiga teng, ya’ni a 1 + a n = a 2 + a n−1 = a 3 + a n−2 = · · · = a k + a n−k+1 . 3-xossa. Arifmetik progressiyaning dastlabki n ta yig’indisi, ya’ni a 1 +a 2 +a 3 +· · ·+a n−1 +a n ni S n bilan belgilaymiz. Arifmetik progressiyaning dastlabki n ta yig’indisi S n = a 1 + a 2 + · · · + a n−1 + a n chetki hadlar yig’indisining yarmi bilan hadlar soni ko’paytmasiga teng, ya’ni S n = a 1 + a n 2 n. Arifmetik progressiya xossalarini jamlab, ularni quyidagi tartibda keltiramiz. 1. a n = a 1 + (n − 1)d; a n = a n−1 + d. 2. a n − a m = (n − m)d, n > m. 3. a n = a n−1 + a n+1 2 = a n−k + a n+k 2 , k < n. 4. a k + a m = a p + a q , k + m = p + q. 5. S n = a 1 + a n 2 n, S n = 2a 1 + d(n − 1) 2 n. 6. S n − S n−1 = a n . 1. Arifmetik progressiyada a 1 = 5, d = 2 bo’lsa, a 7 ni toping. A) 12 B) 18 C) 17 D) 10 Yechish: 1-xossadan a 7 = a 1 + 6d ni olamiz. a 1 va d lar o’rniga berilganlarni qo’yib, a 7 = 5 + 6 · 2 = 5 + 12 = 17. Javob: 17 (C) . 2. Arifmetik progressiyada a 1 = 3, d = 4 bo’lsa, a 9 ni toping. A) 36 B) 35 C) 39 D) 34 3. Arifmetik progressiyada a 2 = 5, a 3 = 8 bo’lsa, shu progressiyaning ayirmasini toping. A) 2 B) 3 C) 5 D) 1, 6 4. Agar arifmetik progressiyada a 5 = 16, d = 5 bo’lsa, a 3 ni toping. A) 36 B) 35 C) 39 D) 34 5. Agar arifmetik progressiyada a 3 = 5, a 9 = 25 bo’lsa, a 6 ni toping. A) 16 B) 15 C) 19 D) 14 6. Agar arifmetik progressiyada a 1 + a 9 = 20 bo’lsa, a 7 + a 3 ni toping. A) 16 B) 15 C) 20 D) 25 91 7. Agar arifmetik progressiyada a 9 − a 1 = 32 bo’lsa, d ni toping. A) 6 B) 5 C) 2 D) 4 8. (96-9-78) Arifmetik progressiyada a 4 − a 2 = 4 va a 7 = 14. Shu progressiyaning beshinchi hadini toping. A) 12 B) 8 C) 7 D) 10 Yechish: 2-xossadan hamda birinchi shartdan 2d = 4 ni olamiz. Yana 2-xossadan foydalanamiz: a 7 − a 5 = 2d ⇐⇒ a 5 = a 7 − 2d = 14 − 4 = 10. Javob: a 5 = 10 (D) . 9. (96-1-27) Arifmetik progressiyada a 2 = 12 va a 5 = 3. Shu progressiyaning o’ninchi hadini to- ping. A) −6 B) 0 C) −12 D) −30 10. (98-12-36) Arifmetik progressiya uchun quyidagi formulalardan qaysilari to’g’ri? 1) a 1 − 2a 2 + a 3 = 0 2) a 1 = a 3 − a 2 3) n = a n − a 1 + d d A) 1; 3 B) 1 C) 2 D) 1; 2 11. (99-1-22) Arifmetik progressiyada a 20 = 0 va a 21 = −41 bo’lsa, a 1 ni toping. A) 779 B) −779 C) 41 D) −41 12. (99-9-26) Arifmetik progressiyada a 2 − a 1 = 6 bo’lsa, a 8 − a 6 ni toping. A) 10 B) 12 C) 9 D) 18 13. (00-5-32) Arifmetik progressiyada a 2 = 9 va a 26 = 105 bo’lsa, shu progressiya birinchi hadi va ayir- masining o’rta proporsional qiymatini toping. A) 20 B) 4, 5 C) 2 √ 5 D) 9 14. (00-10-22) 4; 9; 14; ... arifmetik progressiyaning sakkizinchi hadi to’rtinchi hadidan nechtaga or- tiq? A) 16 B) 18 C) 20 D) 22 15. (02-4-16) Arifmetik progressiyada a 1 = 3 va d = 2 bo’lsa, a 1 − a 2 + a 3 − a 4 + ... + a 25 − a 26 + a 27 ning qiymatini hisoblang. A) 31 B) 30 C) 29 D) 28 16. (02-9-18) − 1 4 ; − 1 5 , ... arifmetik progressiyan- ing nechta hadi manfiy? A) 10 B) 6 C) 5 D) 7 Yechish: Demak, a 1 = − 1 4 , a 2 = − 1 5 . U holda progressiyaning ayirmasi d = − 1 5 + 1 4 = 1 20 . 1-qoi- dadan a n = − 1 4 + (n − 1) 1 20 ekanligi kelib chiqadi. Masala shartiga ko’ra a n = − 1 4 + (n − 1) 1 20 < 0 tengsizlikning yechimi bo’lgan eng katta natural sonni topishimiz kerak. Bu chiziqli tengsizlikn- ing yechimi n < 6 dir. Demak, arifmetik pro- gressiyaning 5 ta hadi manfiy ekan. Javob: 5 (C). 17. (02-11-38) Arifmetik progressiyaning to’rtinchi hadi va o’n birinchi hadlari mos ravishda 2 va 30 ga teng. Shu progressiyaning uchinchi va o’ninchi hadlari yig’indisini toping. A) 16 B) 18 C) 24 D) 28 18. (03-2-67) Kinoteatrning birinchi qatorida 21 ta o’rin bor. Har bir keyingi qatorda o’rinlar soni oldingi qatordagidan 2 tadan ko’p. 40 - qatorda nechta o’rin bor? A) 42 B) 80 C) 99 D) 100 19. (03-3-36) Arifmetik progressiyada a 2 + a 5 − a 3 = 10 va a 1 + a 6 = 17 bo’lsa, uning o’ninchi hadini toping. A) 24 B) 26 C) 28 D) 29 3-4 xossalariga oid misollar 20. (97-4-27) Arifmetik progressiyaning dastlabki 6 ta hadlari 7, a 2 , a 3 , a 4 , a 5 va 22 bo’lsa, a 2 + a 3 + a 4 + a 5 ni hisoblang. A) 65 B) 60 C) 82 D) 58 Yechish. Shartga ko’ra a 1 = 7, a 6 = 22 ekan. Arifmetik progressiyaning 4-xossasiga ko’ra a 2 + a 5 = a 3 + a 4 = a 1 + a 6 = 7 + 22 = 29. U holda a 2 + a 3 + a 4 + a 5 = 29 + 29 = 58. Javob: 58 (D). 21. (97-12-36) Ikkinchi, to’rtinchi va oltinchi hadlar- ining yig’indisi −18 ga teng arifmetik progres- siyaning to’rtinchi hadini toping. A) 6 B) −5 C) −6 D) −4 22. (98-3-20) Birinchi hadi 1 ga, o’n birinchi hadi 13 ga teng bo’lgan arifmetik progressiyaning oltinchi hadini toping. A) 4 B) 5 C) 6 D) 7 23. (98-10-67) Ikkinchi hadi 5 ga sakkizinchi hadi 15 ga teng bo’lgan arifmetik progressiyaning besh- inchi hadini toping. A) 7, 5 B) 12, 5 C) 10 D) 8, 5 Yechish. Shartga ko’ra a 2 = 5, a 8 = 15. Arif- metik progressiyaning 3-xossasiga ko’ra a 2 + a 8 2 = a 5 ⇐⇒ 5 + 15 2 = 10 = a 5 . Javob: 10 (C). 24. (02-1-40) Uchta sonning o’rta arifmetigi 2,6 ga, birinchi son esa 2,4 ga teng. Agar keyingi har bir son avvalgisidan ayni bir songa farq qilsa, keyingi sondan oldingisining ayirmasini toping. A) 1 3 B) 0, 1 C) 1 4 D) 0, 2 25. (02-5-29) Arifmetrik progressiyaning birinchi va to’rtinchi hadi yig’indisi 26 ga teng, ikkinchi hadi esa beshinchi hadidan 6 ga ko’p. Shu progres- siyaning uchinchi va beshinchi hadi yig’indisini toping. A) 20 B) 21 C) 22 D) 23 92 26. (08-103-27) Arifmetrik progressiyaning to’rtinchi va o’n birinchi hadlari mos ravishda 15 va 43 ga teng. Shu progressiyaning uchinchi va o’ninchi hadi yig’indisini toping. A) 68 B) 60 C) 50 D) 24 Dastlabki n ta hadi yig’indisiga oid misol- lar 27. (96-3-27) Arifmetik progressiya uchinchi va to’q- qizinchi hadlarining yig’indisi 8 ga teng. Shu pro- gressiyaning dastlabki 11 ta hadi yig’indisini to- ping. A) 22 B) 33 C) 44 D) 55 Yechish. Shartga ko’ra a 3 + a 9 = 8. 4-xossaga ko’ra a 1 + a 11 = a 3 + a 9 = 8. 5-xossaga ko’ra S 11 = a 1 + a 11 2 · 11 = 8 2 · 11 = 44. Javob: 44 (C). 28. (96-11-28) Arifmetik progressiyada a 3 + a 5 = 12. S 7 ni toping. A) 18 B) 36 C) 42 D) 48 29. (96-12-28) Arifmetik progressiyada a 4 + a 6 = 10. S 9 ni toping. A) 25 B) 30 C) 35 D) 45 30. (98-10-18) Arifmetik progressiyada a 2 + a 19 = 40. Shu progressiyaning dastlabki 20 ta hadlari yig’indisini toping. A) 300 B) 360 C) 400 D) 420 31. (96-6-36) Ikkinchi va o’n to’qqizinchi hadlarining yig’indisi 12 ga teng bo’lgan arifmetik progres- siyaning dastlabki yigirmata hadining yig’indisini toping. A) 110 B) 120 C) 130 D) 115 32. (96-10-29) Arifmetik progressiyada Download 1.09 Mb. 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