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−1 ≤ a ≤ 1 2nπ+arcsin a ≤ x ≤ − arcsin a+(2n+1)π, n ∈ Z. 2. sin x ≤ a, −1 ≤ a ≤ 1 (2n − 1)π − arcsin a ≤ x ≤ arcsin a + 2nπ, n ∈ Z. 3. cos x ≥ a, −1 ≤ a ≤ 1 2nπ − arccos a ≤ x ≤ arccos a + 2nπ, n ∈ Z. 4. cos x ≤ a, −1 ≤ a ≤ 1 2nπ + arccos a ≤ x ≤ 2(n + 1)π − arccos a, n ∈ Z. 5. tgx ≥ b, arctgb + nπ ≤ x < π 2 + nπ, n ∈ Z. 6. tgx ≤ b, − π 2 + πn < x ≤ arctgb + nπ, n ∈ Z. 7. ctgx ≥ b, nπ < x ≤ arcctgb + nπ, n ∈ Z. 165 8. ctgx ≤ b, arcctgb + nπ ≤ x < π + nπ, n ∈ Z. 9. sin x ≥ a, a > 1 bo’lsa x ∈ ∅. 10. sin x ≤ a, a ≥ 1 bo’lsa x ∈ (−∞; ∞). 11. cos x ≥ a, a > 1 bo’lsa x ∈ ∅. 12. cos x ≤ a, a ≥ 1 bo’lsa x ∈ (−∞; ∞). Agar sin x ≤ a, cos x ≤ a, tgx ≤ a, ctgx ≤ a teng- sizliklarda ≤ belgisi < belgi bilan almashsa, u holda yechimlarda ham ≤ belgisi o’rniga < belgisi qo’yiladi. 1. (97-6-47) Ushbu y = √ 2 sin x − 1 funksiyaning aniqlanish sohasini toping. A) ³ − π 6 + 2πn; π 6 + 2πn ´ , n ∈ Z B) [ π 6 + 2πn; 5π 6 + 2πn], n ∈ Z C) ³ π 6 + 2πn; 5π 6 + 2πn ´ , n ∈ Z D) [− π 6 + 2πn; π 6 + 2πn], n ∈ Z Yechish: y = √ 2 sin x − 1 funksiya 2 sin x − 1 ≥ 0 bo’lganda aniqlangan. Bu tengsizlikni sin x ≥ 1 2 ko’rinishda yozamiz. 1-formulaga ko’ra uning javobi 2πn + π 6 ≤ x ≤ 5π 6 + 2πn, n ∈ Z. Javob: (B). 2. (96-9-51) Ushbu sin 2 x − 5 2 sin x + 1 < 0 tengsiz- lik x (x ∈ [0; 2π]) ning qanday qiymatlarida o’rinli? A) [0; π 6 ] ∪ [ 5π 6 ; 2π] B) ( π 6 ; 5π 6 ) C) (0; π 3 ) ∪ ( 2π 3 ; 2π] D) [0; π 3 ) ∪ ( 2π 3 ; 2π] 3. (99-1-43) Tengsizlikni yeching. 2 sin x ≥ √ 2 A) π 4 + 2πn ≤ x ≤ 3π 4 + 2πn, n ∈ Z B) − 5π 4 + 2πn ≤ x ≤ π 4 + 2πn, n ∈ Z C) π 4 + 2πn ≤ x ≤ 3π 4 + 2πn, n ∈ Z D) π 4 + πn ≤ x ≤ 3π 4 + πn, n ∈ Z 4. (96-9-105) Tengsizlikni yeching. 2 sin 2x ≥ ctg π 4 A) [ π 6 + 2πn; 5π 6 + 2πn], n ∈ Z B) ( π 12 + πn; 5π 12 + πn), n ∈ Z C) [ π 12 + πn; 5π 12 + πn], n ∈ Z D) [ π 12 + 2πn; 5π 12 + 2πn], n ∈ Z 5. (97-9-101) Tengsizlikni yeching. sin x · cos x > √ 2 4 A) π 8 + 2πk < x < 3π 8 + 2πk, k ∈ Z B) π 4 + πk < x < 3π 4 + πk, k ∈ Z C) π 8 + πk < x < 3π 8 + πk, k ∈ Z D) π 8 + πk ≤ x ≤ 3π 8 + πk, k ∈ Z 6. (98-5-51) Tengsizlikni yeching. sin 5x · cos 4x + cos 5x · sin 4x > 1 2 A) π 6 + 2πn < x < 5π 6 + 2πn, n ∈ Z B) π 54 + 2πn < x < 5π 54 + 2πn, n ∈ Z C) π 36 + 2πn 9 < x < 5π 36 + 2πn 9 , n ∈ Z D) π 54 + 2πn 9 < x < 5π 54 + 2πn 9 , n ∈ Z 7. (98-8-60) Tengsizlikni yeching. 1 − 2 sin 4x < cos 2 4x A) (πk; π 2 + πk), k ∈ Z B) (− π 2 + 2πk; π 2 + 2πk), k ∈ Z C) ( πk 2 ; π 4 + πk 2 ), k ∈ Z D) (− π 4 + 2πk; π 4 + 2πk), k ∈ Z Yechish: Agar cos 2 4x = 1−sin 2 4x ayniayatdan foydalansak, berilgan tengsizlikni 1−2 sin 4x < 1−sin 2 4x ⇐⇒ sin 4x(sin 4x−2) < 0 ko’rinishda yozishimiz mumkin. Barcha x ∈ R lar uchun sin 4x−2 < 0 tengsizligi o’rinli, shuning uchun berilgan tengsizlik sin 4x > 0 tengsizlikka teng kuchli. 1-formulaga ko’ra uning yechimi 2πn < 4x < π(2n + 1), n ∈ Z. Bu tengsizlikning barcha qismlarini 4 ga bo’lib πn 2 < x < π 4 + πn 2 , n ∈ Z. Javob: (C). 8. (98-1-60) Tengsizlikni yeching. 1 − 2 cos 2x > sin 2 2x A) ( π 2 + πk; π + πk), k ∈ Z B) ( π 3 + 2πk; 2π 3 + 2πk), k ∈ Z C) ( π 4 + πk; 3π 4 + πk), k ∈ Z D) (− π 2 + πk; π 2 + πk), k ∈ Z 9. (98-12-59) Tengsizlikni yeching. sin 2 3x − cos 2 3x ≤ − √ 3 2 166 A) h − π 36 + πn 3 ; π 36 + πn 3 i , n ∈ Z B) ³ − π 36 + πn 3 ; π 36 + πn 3 ´ , n ∈ Z C) h − π 6 + 2πn; π 6 + 2πn i , n ∈ Z D) ³ − π 6 + 2πn; π 6 + 2πn i , n ∈ Z 10. (99-3-38) Tengsizlikni yeching. 4 cos 2 x − 3 ≥ 0 A) [− π 3 + 2πk; π 3 + 2πk], k ∈ Z B) [− π 3 + πk; π 3 + πk], k ∈ Z C) [− π 6 + πk; π 6 + πk], k ∈ Z D) [− π 6 + 2πk; π 6 + 2πk], k ∈ Z 11. (99-7-49) Tengsizlikni yeching. cos 5x · cos 4x + sin 5x · sin 4x < √ 3 2 A) π 3 + 2πn < x < 5π 3 + 2πn, n ∈ Z B) π 6 + 2πn < x < 11π 6 + 2πn, n ∈ Z C) π 3 + πn < x < 5π 3 + πn, n ∈ Z D) π 6 + πn < x < 11π 6 + πn, n ∈ Z 12. (96-12-111) x (x ∈ [0; 2π]) ning qanday qiymat- larida tengsizlik to’g’ri? cos 2 x − 5 2 cos x + 1 > 0 A) [0; π 3 ] ∪ ( 5π 3 ; 2π] B) ( π 3 ; π 2 ] ∪ [ 3π 2 ; 5π 3 ) C) ( π 3 ; 5π 3 ) D) ( π 3 ; π 2 ] Yechish: Berilgan tengsizlikda cos x = y belgi- lash olib, uni quyidagicha yozib olamiz y 2 − 2, 5y + 1 > 0 ⇐⇒ (y − 0, 5)(y − 2) > 0. Yana eski belgilashga qaytib berilgan tengsizlikka teng kuchli bo’lgan (cos x − 0, 5)(cos x − 2) > 0 tengsizlikka ega bo’lamiz. Barcha x ∈ R larda cos x − 2 < 0 tengsizligi o’rinli bo’lgani uchun, berilgan tengsizlik cos x − 0, 5 < 0 ⇐⇒ cos x < 1 2 tengsizlikka teng kuchli. 4-formulaga ko’ra uning yechimi 2πn + π 3 < x < 2π(n + 1) − π 3 , n ∈ Z ko’rinishda bo’ladi. Bu yechimning [0; 2π] kesma- dagi qismini olish uchun n = 0 deymiz, u holda ( π 3 ; 5π 3 ) oraliq [0; 2π] kesmaning qismi bo’ladi. Agar n 6= 0 bo’lsa, u holda (2πn + π 3 ; 2π(n + 1) − π 3 ) interval [0; 2π] kesma bilan umumiy qismga ega emas. Javob: ( π 3 ; 5π 3 ) (C). 13. (96-13-26) Ushbu cos 2 x − 5 2 cos x + 1 ≤ 0 tengsizlik x (x ∈ [0; 2π]) ning qanday qiymatla- rida o’rinli? A) [0; π 3 ] ∪ [ 5π 3 ; 2π] B) [0; π 3 ] C) [ 5π 3 ; 2π] D) [ π 3 ; π 2 ] ∪ [ 3π 2 ; 5π 3 ] 14. (97-4-41) Tengsizlikni yeching. cos 2 x < √ 2 2 + sin 2 x A) π 8 + 2πn < x < 7π 8 + 2πn, n ∈ Z B) π 8 + πn < x < 7π 8 + πn, n ∈ Z C) − π 8 + 2πn < x < π 8 + 2πn, n ∈ Z D) π 4 + 2πn < x < 7π 4 + 2πn, n ∈ Z 15. (98-6-55) Ushbu cos 2x ≤ − 1 2 tengsizlikning [0; π] kesmadagi yechimini toping. A) [ π 3 ; 2π 3 ] B) [0; 2π 3 ] C) [− 2π 3 ; 4π 3 ] D) [ 4π 3 ; 2π] 16. (00-3-55) Quyidagi tengsizlik −1 − 2 √ 3 cos x > 0 [−π; π] kesmada nechta butun yechimga ega? A) 4 B) 3 C) 6 D) 2 17. (00-6-56) Tengsizlikni yeching. cos x < sin x A) ( π 4 + πk; 3π 4 + πk), k ∈ Z B) ( π 4 + πk; 5π 4 + πk), k ∈ Z C) ( π 4 + 2πk; 3π 4 + 2πk), k ∈ Z D) ( π 4 + 2πk; 5π 4 + 2πk), k ∈ Z 18. (96-1-59) Tengsizlikni yeching. tg ³ x + π 4 ´ ≥ 1 A) h − π 4 + πk; π 2 + πk i , k ∈ Z B) [πk; π 2 + πk) k ∈ Z 167 C) h π 4 + 2πk; π 2 + 2πk i , k ∈ Z D) h πk; π 4 + πk ´ , k ∈ Z 19. (96-12-91) x (x ∈ [0; 2π]) ning qaysi qiymatlarida funksiya aniqlangan? y = q 1 − log 1 2 cos x A) [ π 3 ; π 2 ) B) ( 3π 2 ; 5π 3 ] C) [0; π 3 ] D) [ 5π 3 ; 2π] ∪ [0; π 3 ] 20. (96-13-34) Ushbu y = q 1 + log 1 2 sin x funksiya- ning aniqlanish sohasiga tegishli bo’lgan x ning [0; 2π] kesmadagi barcha qiymatlarini aniqlang. A) h π 6 ; 5π 6 i B) ³ 0; π 6 i ∪ h 5π 6 ; π ´ C) ³ 0; π 6 i D) (0; π) 21. (01-4-3) Ushbu y = arccos(2 sin x) funksiyaning aniqlanish sohasiga tegishli bo’lgan x ning [−π; π] kesmadagi barcha qiymatlarini aniqlang. A) [− π 6 ; π 6 ] B) [− π 4 ; π 4 ] C) [− π 3 ; π 3 ] D) [−π; − 5π 6 ] ∪ [− π 6 π 6 ] ∪ [ 5π 6 ; π] 22. (01-10-39) Tengsizlikni yeching. sin 2x < cos 2x A) (− 3π 8 + 2πn; π 8 + 2πn), n ∈ Z B) (− 3π 4 + 2πn; π 4 + 2πn), n ∈ Z C) (− π 8 + πn; π 8 + πn), n ∈ Z D) (− 3π 8 + πn; π 8 + πn), n ∈ Z 23. (01-11-22) Tengsizlikning [0; 2π] kesmadagi eng katta va eng kichik yechimlari yig’indisini hisoblang. 2 1 2 ≤ 2 sin x ≤ 2 √ 3 2 A) 2π 3 B) π C) 4π 5 D) π 2 24. (02-1-62) Tengsizlikni yeching. cos(sin x) < 0 A) ( π 2 + 2πn; 3π 2 + 2πn), n ∈ Z B) ( π 2 + πn; 3π 2 + πn), n ∈ Z C) (0; 3π 2 + 2πn), n ∈ Z D) yechimga ega emas Yechish: Ma’lumki, ixtiyoriy x ∈ R da sin x ∈ [−1; 1] bo’ladi. Kosinus funksiya esa (− π 2 ; π 2 ) da musbat qiymatlar qabul qiladi. Demak, ixtiyoriy t = sin x ∈ (− π 2 ; π 2 ) da ham cos(sin x) = cos t > 0 bo’ladi. Bu yerdan berilgan tengsizlik yechimga ega emasligi kelib chiqadi. Javob: yechimga ega emas. (D). 25. (02-6-45) Tengsizlikni yeching. sin x > √ 3 · cos x A) ( π 3 + 2πn; 4π 3 + 2πn), n ∈ Z B) ( π 6 + πn; 2π 3 + πn), n ∈ Z C) ( π 6 + 2πn; 7π 6 + 2πn), n ∈ Z D) ( π 4 + πn; 3π 4 + πn), n ∈ Z 26. (02-8-19) Funksiyaning aniqlanish sohasini toping. y = p log 3 sin x A) π 2 + 2πn, n ∈ Z B) π 2 + πn, n ∈ Z C) (0; 1) D) (0; π) 27. (02-10-62) Tengsizlikni yeching. r cos 2 x − cos x + 1 4 ≥ 1 2 A) [ π 2 + 2πn; 3π 2 + 2πn] ∪ {2πn}, n ∈ Z B) [− π 2 + 2πn; π 2 + 2πn] ∪ {2πn}, n ∈ Z C) (− π 2 + 2πn; π + 2πn] ∪ {2πn}, n ∈ Z D) [ 2π 3 + πn; 7π 6 + πn], n ∈ Z 28. (03-2-31) Tengsizlikni yeching. cos(π sin x) > 0 A) ³ πk; π 3 + πk ´ , k ∈ Z B) ³ − π 6 + πk; π 6 + πk ´ , k ∈ Z C) ³ − π 3 + 2πk; π 3 + 2πk ´ , k ∈ Z D) ³ πk; π 6 + πk ´ , k ∈ Z 13.6 Aralash tipdagi masalalar 1. y = arcsin x funksiyaning aniqlanish sohasi [−1; 1], qiymatlar sohasi esa - [− π 2 ; π 2 ]. y = arcsin x funksiya [−1; 1] da o’suvchi. 2. y = arccos x funksiyaning aniqlanish sohasi [−1; 1], qiymatlar sohasi esa - [0; π]. y = arccos x funksiya [−1; 1] da kamayuvchi. 3. y = arctgx funksiyaning aniqlanish sohasi (−∞; ∞), qiymatlar sohasi esa - (− π 2 ; π 2 ). y = arctgx funksiya (−∞; ∞) da o’suvchi. 4. y = arcctgx funksiyaning aniqlanish sohasi (−∞; ∞), qiymatlar sohasi esa - (0; π). y = arcctgx funksiya (−∞; ∞) da kamayu- vchi. 168 5. y = arcsin x va y = arctgx - toq funksiyalar, y = arccos x va y = arcctgx funksiyalar esa juft ham emas, toq ham emas va ular uchun arccos(−x) = π − arccos x, arcctg(−x) = π − arcctgx tengliklar o’rinli. 6. arcsin x + arccos x = π 2 , x ∈ [−1; 1]. 7. arcsin a > arcsin b ⇔ a > b b ≥ −1 a ≤ 1. 8. arccos a > arccos b ⇔ a < b a ≥ −1 b ≤ 1. 9. arctga > arctgb ⇔ a > b. 10. arcctga > arcctgb ⇔ a < b. 1. (98-6-51) Tengsizlikni yeching. arcsin x < arcsin(1 − x) A) [0; 1 2 ) B) [−1; 1] C) (−∞; 1 2 ] D) [0; 2] Yechish: y = arcsin x, −1 ≤ x ≤ 1 funksiya o’suvchi ekani ma’lum. U holda berilgan tengsi- zlik quyidagi x < 1 − x −1 ≤ x ≤ 1 −1 ≤ 1 − x ≤ 1 sistemaga ekvivalent bo’ladi. Uni yechamiz. 2x < 1 −1 ≤ x ≤ 1 0 ≤ x ≤ 2 Demak, 0 ≤ x < 1 2 Javob: [0; 1 2 ) (A). 2. (98-6-53) Tenglamaning eng kichik musbat ildizini toping. arcsin(2 sin x) = π 2 A) 1 3 B) 5π 6 C) 1 2 D) π 6 3. (98-11-30) Tenglamaning yechimi nechta? arctg|x| = − π 6 A) 1 B) 0 C) 2 D) cheksiz ko’p 4. (98-11-74) Tengsizlikni yeching. arccos x > arccos x 2 A) (0; 1) B) [−1; 0) C)[−1; 1] D) (−∞; 0) ∪ (1; ∞) 5. (00-1-33) Tenglamaning ildizlari yig’indisini toping. 2(arccos x) 2 + π 2 = 3π arccos x A) √ 2 2 B) −1 C) 1 D) − √ 2 2 6. (01-4-4) Ushbu arccos 2 x − 5π 6 · arccos x + π 2 6 ≤ 0 tengsizlik o’rinli bo’ladigan kesmaning o’rtasini toping. A) 0,5 B) 0,4 C) 0,25 D) π 4 Yechish: Berilgan tengsizlikning chap qismini ko’paytuvchilarga ajratamiz (arccos x − π 3 )(arccos x − π 2 ) ≤ 0. Bu tengsizlikka oraliqlar usulini qo’llab π 3 Download 1.09 Mb. Do'stlaringiz bilan baham: |
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