M u n d a r I j a
Download 1.09 Mb. Pdf ko'rish
|
abiturshtabalgebra
≤ arccos x ≤
π 2 tengsizlikni olamiz. Kosinus funksiyaning [0; π 2 ] da kamayuvchi ekanligidan cos π 3 ≥ x ≥ cos π 2 ⇐⇒ 1 2 ≥ x ≥ 0 ni olamiz. Demak, berilgan tengsizlikning yechimi [0; 1 2 ] kesmadan iborat. Bu kesmaning o’rtasi 0, 25 dir. Javob: 0, 25 (C). 7. (01-5-18) Tenglama nechta ildizga ega? x · arctgx = 1 A) 2 B) 1 C) 0 D) 3 8. (01-5-19) Tenglama nechta ildizga ega? cos(10arctgx) = 1 A) 5 B) cheksiz ko’p C) 1 D) 3 9. (01-9-14) Tenglama ildizlari ko’paytmasini toping. 4arctg(x 2 − 3x + 3) − π = 0 A) 2 B) 3 C) −3 D) 1 10. (01-12-27) Tengsizlikni yeching. lg(arcsin x) > −1 A) (0; π 2 ] B) [sin 0, 1; 1] C) (sin 0, 1; 1) D) (sin 0, 1; 1] 11. (00-10-25) Tenglamaning nechta ildizi bor? arctg|x| = π 2 A) 2 B) 1 C) 0 D) cheksiz ko’p 169 12. (02-4-37) Tengsizlikni qanoatlantiruvchi x ning eng katta butun qiymatini toping. arctgx < 0 A) −2 B) −1 C) 0 D) 1 13. (99-8-35) Ushbu y = arcsin x + π 2 funksiyaning qiy- matlar to’plamini toping. A) [0; π] B) [− π 2 ; π 2 ] C) [ π 2 − 1; π 2 + 1] D) [0; π 2 ] Yechish: 1-xossaga ko’ra y = arcsin x funksiyan- ing qiymatlar to’plami [− π 2 ; π 2 ] kesmadan iborat. Shuning uchun y = arcsin x+ π 2 funksiyaning qiy- matlar to’plami [0; π] kesmadan iborat bo’ladi. Javob: [0; π] (A). 14. (98-6-49) Ushbu x = arccos 0, 9; y = arccos(−0, 7); z = arccos(−0, 2) sonlarni o’sib borish tartibida yozing. A) y < z < x B) x < y < z C) y < x < z D) x < z < y 15. (07-156-36) cos(2 arccos 4 5 ) ning qiymatini top- ing. A) 7 25 B) 24 25 C) − 24 25 D) − 7 25 16. (07-158-36) cos(2 arccos 4 9 ) ning qiymatini top- ing. A) 49 81 B) 8 9 C) − 49 81 D) − 8 9 17. (99-3-30) Funksiyaning aniqlanish sohasini toping. y = arcsin x − 3 2 − lg(4 − x) A) [1; 4] B) [1; 5] C) (1; 4) D) [1; 4) Yechish: y = arcsin x funksiyaning aniqlanish sohasi [−1; 1] kesmadan, y = lg x funksiyaning aniqlanish sohasi (0; ∞) dan iborat. Bularga ko’ra berilgan funksiyaning aniqlanish sohasi ( −1 ≤ x − 3 2 ≤ 1 4 − x > 0 sistema yechimidan iborat bo’ladi. Bu sisteman- ing yechimlari [1; 4) kesmadan iborat. Javob: [1; 4) (D). 18. (99-8-73) y = arcsin x 3 8 funksiyaning aniqlanish sohasini toping. A) [−2; 2] B) [−1; 1] C) (−2; 2) D) [1; 2] 19. (99-10-41) Funksiyaning aniqlanish sohasini to- ping. y = √ x + 0, 2 arccos x A) (−0, 2; 1) B) (−0, 2; 1] C) [−0, 2; 1] D) [−0, 2; 1) 20. (03-6-62) Funksiyaning aniqlanish sohasini toping. y = arcsin 2 2 + sin x A) −π + 2πk ≤ x ≤ π + 2πk, k ∈ Z B) x ≤ π + 2πk, k ∈ Z C) x > 2πk, k ∈ Z D) 2πk ≤ x ≤ π + 2πk, k ∈ Z 21. (03-6-67) y = arccos |x − 2| funksiyaning aniqla- nish sohasini toping. A) 1 ≤ x ≤ 3 B) x > 1 C) x < 3 D) 2 ≤ x ≤ 3 22. (03-7-58) Funksiyaning aniqlanish sohasini toping. y = √ x 2 − 5x + 6 lg(x + 5) 2 + 1 arccos(x + 3) A) (−4; −2] B) (−∞; 2) ∪ [3; ∞) C) (−∞; −3) ∪ (−3; 2] D) (−4; −2) 23. (02-4-35) y = (x − 10)arctgx funksiya grafigining Ox o’qi bilan kesishish nuqtasi abssissasining eng kichik qiymatini toping. A) −2 B) −1 C) 0 D) 1 Yechish: Ox o’qidagi nuqtalarning ordinatasi, ya’ni y = 0 bo’ladi. Shunday qilib, (x−10)arctgx = 0 tenglama yechimlarini topamiz. Bu tenglaman- ing yechimlari x−10 = 0 va arctgx = 0 tenglama yechimlaridan iborat. Demak, berilgan tenglama x 1 = 10 va x 2 = 0 yechimlarga ega. Ularning kichigi x = 0. Javob: 0 (C). 24. (02-7-5) y = arcsin(3x − 7) funksiyaning aniqla- nish sohasiga tegishli x ning butun qiymatlari nechta? A) 2 B) 3 C) 1 D) 4 25. (02-11-48) y = arccos(log 3 x − 1) funksiyaning aniqlanish sohasiga tegishli butun sonlar nechta? A) 12 B) 9 C) 8 D) 7 14 -bob. Hosila va integral Funksiyaning uzluksizligi, hosilasi va integrali funksiya limiti tushunchasi bilan uzviy bog’liq. Shuning uchun funksiya limiti tushunchasi beramiz. Bizga f : [a; b] → R funksiya va x 0 ∈ [a; b] berilgan bo’lsin. Agar ixtiy- oriy ε > 0 uchun shunday δ > 0 mavjud bo’lib, 0 < |x − x 0 | < δ shartni qanoatlantiruvchi batcha x ∈ [a; b] larda |f (x) − A| < ε tengsizligi bajarilsa, f funksiya x → x 0 da A limitga ega deyiladi va quyidagicha yozi- ladi: lim x→x 0 f (x) = A. Xususan, agar lim x→x 0 f (x) = f (x 0 ) bo’lsa, f funksiya x 0 nuqtada uzluksiz deyiladi. Agar funksiya aniqlan- ish sohasining barcha nuqtalarida uzluksiz bo’lsa, u uzluksiz funksiya deyiladi. Uzluksiz funksiyaga misol- lar: y = ax + b − chiziqli funksiya, y = ax 2 + bx + 170 c − kvadratik funksiya, ixtiyoriy ko’phad y = a 0 + a 1 x + a 2 x 2 + · · · + a n x n , ko’rsatkichli funksiya y = a x , logarifmik funksiya y = log a x va trigonometrik funksiyalar y = sin x, y = cos x, y = tgx, y = ctgx funksiyalar o’zlarining aniqlanish sohasida uzluksizdir. Ishora funksiyasi y = signx = −1 agarx < 0 0 agarx = 0 1 agarx > 0 yagona x = 0 nuqtada uzilishga ega, qolgan barcha nuqtalarda uzluksiz. x ning butun qismi y = [x] va x ning kasr qismi y = {x} lar barcha butun nuqtalarda uzilishga ega. Endi hosila ta’rifini beramiz. x va x+∆x lar [a; b] kesmaga qarashli bo’lsin. Agar f (x + ∆x) − f (x) ∆x nisbat ∆x → 0 da chekli limitga ega bo’lsa, u holda f funksiya x nuqtada hosilaga ega deyiladi va bu quyidagi- cha yoziladi: f 0 (x) = lim ∆x→0 f (x + ∆x) − f (x) ∆x . Matematika, mexanika va fizikada hosilaning ko’pgina tadbiqlari uchraydi. Endi elementar funksiyalar uchun hosilalar jadvalini va hosilani hisoblash qoidalarini kelti- ramiz. 14.1 Elementar funksiyalarning hosilasi Hosilalar jadvali 1. c 0 = 0, c = const. 2. (x α ) 0 = α x α−1 , x 0 = 1. 3. ( √ x) 0 = 1 2 √ x . 4. ( 1 x ) 0 = − 1 x 2 . 5. (a x ) 0 = a x ln a, (e x ) 0 = e x . 6. (log a x) 0 = 1 x ln a , (ln x) 0 = 1 x . 7. (sin x) 0 = cos x, (cos x) 0 = − sin x. 8. (tgx) 0 = 1 cos 2 x , (ctgx) 0 = − 1 sin 2 x . Yig’indi va ayirmaning hosilasi 9. (u(x) ± v(x)) 0 = u 0 (x) ± v 0 (x) 10. (C · u(x)) 0 = C · u 0 (x), C- o’zgarmas son. 1. Agar f (x) = 2x + 3 bo’lsa, f 0 (x) ni hisoblang. A) 1 B) 2 C) 0 D) 5 Yechish: 9 va 10-qoidalarga ko’ra f 0 (x) = (2x + 3) 0 = 2 x 0 + 3 0 . Endi 1 va 2-formulalardan foy- dalanamiz: f 0 (x) = 2 · 1 + 0 = 2. Javob: 2 (B). 2. Agar f (x) = 2x 2 +3x+7 bo’lsa, f 0 (x) ni hisoblang. A) 2x + 3 B) 4x + 3 C) 4x D) 3 3. Agar f (x) = x 2 + 2 √ x bo’lsa, f 0 (x) ni hisoblang. A) 2x + 2 √ x B) 2x + √ x C) x D) 2x + x −1/2 4. Agar f (x) = − 2 x bo’lsa, f 0 (x) ni hisoblang. A) 2x −2 B) −2 x 2 C) 4 x 2 D) x −1/2 5. Agar f (x) = 2 x + 7 bo’lsa, f 0 (x) ni hisoblang. A) 2 x · ln 2 B) 2 x ln 2 C) 2 x−1 D) −2 x · ln 2 6. Agar f (x) = ln x + 5 bo’lsa, f 0 (x) ni hisoblang. A) 1 x B) 1 x · ln 2 C) − 1 x D) ln 2 x 7. Agar f (x) = x − tgx bo’lsa, f 0 (x) ni hisoblang. A) −tg 2 x B) 1 + 1 cos 2 x C) 1 − 1 sin 2 x D) 1 − ctgx 8. (96-7-28) Agar f (x) = 5 sin x + 3 cos x bo’lsa, f 0 ( π 4 ) ni hisoblang. A) − √ 2 B) √ 2 C) −2 √ 3 D) 4 √ 2 Yechish: 7 va 10-qoidalarga ko’ra f 0 (x) = 5(sin x) 0 + 3(cos x) 0 = 5 cos x − 3 sin x. Endi x = π 4 deb f 0 ( π 4 ) ni hisoblaymiz: f 0 ( π 4 ) = 5 cos π 4 − 3 sin π 4 = √ 2 2 (5 − 3) = √ 2. Javob: √ 2 (B). 9. (97-6-19) Agar g(x) = ctgx + 12x 3 π 2 + π bo’lsa, g 0 ( π 6 ) ni hisoblang. A) −1 B) −3 C) 5 D) 3 10. (97-7-28) Agar f (x) = 2 sin x − 4 √ 3 cos x bo’lsa, f 0 ( π 3 ) ni hisoblang. A) 7 B) −5 C) 2 + 4 √ 3 D) 2 √ 3 − 2 11. (97-9-34) Ushbu y = 1 3 6 x − 6 funksiyaning x = 1 nuqtadagi hosilasini toping. A) ln 12 B) ln 36 C) ln 6 D) ln 6 e 12. (97-12-55) Agar f (x) = 1 3 x 3 − 16x bo’lsa, f 0 (4) ni toping. A) 1 B) 2 C) 3 D) 0 171 13. (98-1-28) Agar f (x) = e x + 5x bo’lsa, f 0 (ln 3) ni hisoblang. A) 8 B) 5 C) e 3 + 5 D) e 3 14. (98-5-26) Ushbu y = sin 2 x + cos 2 x funksiyaning hosilasini toping. A) 2 sin 2x B) 0 C) 4 sin x D) sin 4x 15. (99-7-27) Ushbu y = tgx · ctgx funksiyaning hosilasini toping. A) 1 B) 2 C) − 1 sin 2 x · cos 2 x D) 0 16. (96-1-30) Agar f (x) = x 3 − 3x − 4 bo’lsa, f 0 (x) x − 5 ≥ 0 tengsizlikning eng kichik butun yechimini toping. A) 1 B) −1 C) −5 D) 0 Yechish: 2-qoidaga ko’ra f 0 (x) = 3x 2 − 3. Endi 3x 2 − 3 x − 5 ≥ 0 ⇐⇒ 3(x − 1)(x + 1) x − 5 ≥ 0 tengsizlikni yechamiz. Bu tengsizlik oraliqlar usuli bilan oson yechiladi (14.1-chizma) va uning yechimi [−1; 1]∪(5; ∞) to’plamdan iborat. Bu to’plamdagi eng kichik butun son −1 dir. Javob: −1 (B). 17. (96-10-32) Agar f (x) = x 3 − 12x + 7 bo’lsa, f 0 (x) x − 4 ≤ 0 tengsizlikning eng katta butun yechimini toping. A) 2 B) −4 C) 3 D) −2 18. (98-5-25) x ning qanday qiymatlarida f (x) = sin x va g(x) = 5x + 3 funksiyalar uchun f 0 (x) < g 0 (x) tengsizlik bajariladi? A) (−∞; 5) B) (2πn; π 2 + 2πn), n ∈ Z C) (−∞; ∞) D) (0; ∞) 19. (00-6-27) Agar f (x) = −4x 3 −11x 2 −8x+7 bo’lsa, f 0 (x) ≥ 0 tengsizlikning nechta butun yechimi bor? A) 4 B) 3 C) 2 D) 1 20. (97-5-34) Ushbu y = 2 x − 1 funksiyaning x = 1 nuqtadagi hosilasini toping. A) 1 B) ln 2 C) ln 4 e D) ln 4 21. (01-6-42) Agar f (x) = x 2/3 + 85 1 3 ln x bo’lsa, f 0 (8) ning qiymatini toping. A) 10 B) 12 C) 9 D) 11 22. (02-2-29) Agar f (x) = 3x + 3 x bo’lsa, f 0 (x) < 0 tengsizlikni yeching. A) (−1; 0) ∪ (0; 1) B) (−∞; −1) C) (1; ∞) D) (0; 1) 23. (02-2-32) Nechta nuqtada f (x) = x 3 funksiya va uning hosilasi qiymatlari teng bo’ladi? A) 2 B) 1 C) ∅ D) 3 Yechish: 2-qoidaga ko’ra f 0 (x) = 3x 2 bo’ladi. Masala shartiga ko’ra f (x) = f 0 (x) ⇐⇒ x 3 = 3x 2 ⇐⇒ x 2 (x − 3) = 0 tenglamani yechamiz. Bu tenglamaning ildizlari x 1 = 0 va x 2 = 3 lardir. Javob: 2 (A). 24. (03-2-9) Agar f (x) = x 3 + x − √ 2, g(x) = 3x 2 + x + √ 2 bo’lsa, f 0 (x) > g 0 (x) tengsizlikning eng kichik natural yechimini toping. A) 3 B) 2 C) 6 D) 5 25. (03-8-48) f (x) = x 4 + x 3 − 13, 5x 2 + 2003 bo’lsa, f 0 (x) ≤ 0 tengsizlikning eng kichik natural yechi- mini toping. A) 1 B) 2 C) 3 D) 4 26. (03-12-20) Agar f (x) = ln x bo’lsa, f 0 (x) ≤ x tengsizlikni yeching. A) [−1; 0) ∪ [1; ∞) B) (−1; 0) ∪ [1; ∞) C) (−∞; −1] ∪ [1; ∞) D) [1; ∞) 27. (03-12-73) Agar f (x) = x 3 + 5x 2 + 4x + 2 bo’lsa, f 0 (x) = f (1) tenglamaning eng kichik ildizini toping. A) −6 B) − 1 3 C) −2 D) −4 14.1.1 Murakkab funksiyaning hosilasi Ko’paytma va bo’linmaning hosilasi 1. (u(x) · v(x)) 0 = u 0 (x)v(x) + u(x)v 0 (x). 2. µ u(x) v(x) ¶ 0 = u 0 (x)v(x) − u(x)v 0 (x) v 2 (x) , v(x) 6= 0. Murakkab funksiyalarning hosilalari 3. (f (g(x))) 0 = f 0 (g(x))g 0 (x) umumiy qoida. 4. (u α (x)) 0 = α u α−1 (x) u 0 (x). 5. (e u(x) ) 0 = e u(x) u 0 (x), (a u(x) ) 0 = a u(x) ln a u 0 (x). 6. (log a u(x)) 0 = u 0 (x) u(x) ln a , (ln u(x)) 0 = u 0 (x) u(x) . 7. (sin u(x)) 0 = cos u(x) u 0 (x). 172 8. (cos u(x)) 0 = − sin u(x) u 0 (x). 9. (tgu(x)) 0 = u 0 (x) cos 2 u(x) . 10. (ctgu(x)) 0 = − u 0 (x) sin 2 u(x) . 1. (96-9-79) Agar f (x) = 3x · 2 x bo’lsa, f 0 (0) ni toping. A) −3 B) 3 C) 1 D) −1 Yechish: 1-qoidaga ko’ra f 0 (x) = (3x) 0 · 2 x + 3x · (2 x ) 0 = 3 · 2 x + 3x · 2 x · ln 2 bo’ladi. Endi f 0 (0) ni hisoblaymiz: f 0 (0) = 3 · 2 0 + 3 · 0 · 2 0 · ln 2 = 3 + 0 = 3. Javob: 3 (B). 2. (96-10-30) Agar f (x) = 2x · 3 x bo’lsa, f 0 (0) ni toping. A) −1 B) 2 C) −2 D) 3 3. (98-12-39) Ushbu y = e x · x 2 funksiyaning hosi- lasini toping. A) e x (x 2 + 2x) B) e x (x 2 + 2) C) e x (2x + 1) D) e x (x 2 + x) 4. (00-5-46) y = (x − 3)(x 2 + 3x + 9) funksiyaning Download 1.09 Mb. Do'stlaringiz bilan baham: |
Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling
ma'muriyatiga murojaat qiling