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abiturshtabalgebra
x = 3 nuqtadagi hosilasini toping.
A) 0 B) 3 C) 27 D) −27 5. (96-9-22) Ushbu f (x) = x 2 x 2 − 1 funksiya uchun f 0 (−2) ni hisoblang. A) 4 9 B) − 4 9 C) 3 4 D) − 3 4 Yechish: 2-qoidaga ko’ra f 0 (x) = (x 2 ) 0 · (x 2 − 1) − x 2 (x 2 − 1) 0 (x 2 − 1) 2 . Endi hosilalarni hisoblaymiz va soddalashtiramiz: f 0 (x) = −2x (x 2 − 1) 2 =⇒ f 0 (−2) = −2(−2) ((−2) 2 − 1) 2 = 4 9 . Javob: 4 9 (A). 6. (96-3-81) Agar f (x) = x 1 − x bo’lsa, f 0 (2) ni toping. A) −1 B) −2 C) 2 D) 1 7. (96-13-22) F 0 (1) ni toping. F (x) = x 2 x 2 + 1 A) − 1 2 B) 1 2 C) 2 3 D) − 2 3 8. (98-10-69) f 0 (1) ni toping. f (x) = √ x + 1 √ x A) 1 2 B) 2 C) − 1 2 D) 1 9. (02-1-27) f 0 (1) ni toping. f (x) = 8x √ x + 2 x A) 2 B) 1 C) 0 D) 3 10. (97-5-33) y = e sin 2 x funksiya hosilasini toping. A) e sin 2 x B) e sin 2 x · sin 2x C) 2e sin 2 x · sin x D) sin 2 x · e sin 2 x−1 Yechish: Murakkab funksiyaning hosilasini hisob- lash qoidasi 5-ga ko’ra y 0 = e sin 2 x · 2 sin x · cos x ekanini topamiz. 2 sin x · cos x = sin 2x formulaga ko’ra y 0 = e sin 2 x · sin 2x bo’ladi. Javob: e sin 2 x · sin 2x (B). 11. (96-3-34) f (x) = e sin 2x funksiya hosilasini to- ping. A) sin 2x · e sin 2x−1 B) 2 cos 2x · e sin 2x C) 2 cos 2x · e cos 2x D) cos 2x · e sin 2x 12. (96-12-36) f (x) = e cos 2x funksiyaning hosilasini toping. A) 2 sin 2x · e cos 2x B) cos 2x · e cos 2x−1 C) −2 sin 2x · e −2 sin 2x D) −2 sin 2x · e cos 2x 13. (97-6-48) g(x) = 1 3 ctg3x funksiyaning x = π 18 dagi hosilasini hisoblang. A) −2 B) 4 3 C) 4 D) −4 14. (00-6-26) Agar f (x) = 3x 2 · e sin x − 8 bo’lsa, f 0 (π) ning qiymatini toping. A) 3π(2 + π) B) 3π 2 (3 − π) C) 2π(3 + π) D) 3π(2 − π) 15. (99-4-33) Agar f (x) = (x − 2) 2 · (x + 4) bo’lsa, f 0 (x) ≤ 0 tengsizlikni eching. A) [−4; 2] B) [2; 4] C) [−2; 2] D) [−3; 2] 16. (96-1-28) Agar f (x) = x · 2 x+1 bo’lsa, f 0 (0) ni toping. A) −2 B) −1 C) 1 D) 2 17. (96-3-29) y = cos(x 3 − 5) funksiyaning hosilasini toping. A) −3x 2 sin(x 3 − 5) B) 3x 2 sin(x 3 − 5) C) − sin(3x 2 − 5) D) sin(3x 2 − 5) 173 18. (96-3-33) f (x) = ln(x 2 − 3 sin x) funksiyaning hosilasini toping. A) 3 x 2 − 3 sin x B) 2x + 3 cos x x 2 − 3 sin x C) 2x − 3 cos x x 2 − 3 sin x D) 2x x 2 − 3 sin x Yechish: Murakkab funksiyaning hosilasini hisob- lash qoidasi 6-ga ko’ra f 0 (x) = (x 2 − 3 sin x) 0 x 2 − 3 sin x = 2x − 3 cos x x 2 − 3 sin x ekanini topamiz. Javob: (C). 19. (96-12-34) Funksiyaning hosilasini toping. f (x) = ln(x 2 + 3 sin x) A) 3 x 2 + 3 sin x B) 2x + 3 sin x x 2 + 3 sin x C) 2x + 3 cos x x 2 + 3 sin x D) 2x − 3 cos x x 2 + 3 sin x 20. (96-6-56) Agar f (x) = ln sin x bo’lsa, f 0 ( π 6 ) ni toping. A) − √ 3 B) √ 3 3 C) √ 3 D) − √ 3 3 21. (97-3-28) Agar f (x) = 2 √ 3 cos 4x−2 cos x bo’lsa, f 0 ( π 6 ) ni hisoblang. A) −11 B) 13 C) √ 3 + 1 D) √ 3 − 2 22. (97-6-62) Agar f (x) = ln sin x bo’lsa, f 0 ( π 4 ) ni hisoblang. A) −1 B) 3 C) − √ 3 D) 1 23. (97-8-57) Agar f (x) = (x 2 + 1) 2 bo’lsa, f 0 ( 1 2 ) ni toping. A) 2,5 B) −1 2 5 C) −1 4 5 D) 2 5 24. (97-10-28) Agar f (x) = 3 cos 2x − sin 2x bo’lsa, f 0 ( π 8 ) ni hisoblang. A) −4 √ 2 B) √ 2 C) 2 √ 2 D) 4 √ 3 25. (97-12-62) Agar f (x) = 0, 5tg2x bo’lsa, f 0 ( π 6 ) ni toping. A) 4 3 B) − 1 4 C) 4 D) 2 26. (98-7-39) Ushbu y = − 1 7 sin(7x − 5) funksiyaning hosilasini toping. A) − 1 7 · cos(7x − 5) B) −7 cos(7x − 5) C) cos(7x − 5) D) − cos(7x − 5) 27. (98-7-40) Ushbu y = log 5 2x funksiyaning hosi- lasini toping. A) 1 x ln 2 B) 1 x ln 5 C) 2 x ln 5 D) 2 x ln 2 28. (98-8-28) Agar f (x) = 3x − 2e −x bo’lsa, f 0 (ln 2) ni hisoblang. A) 1 B) 2 C) 5 D) 4 29. (02-1-64) Agar f (x) = x · sin 2x bo’lsa, f 0 (π) + f (π) + 2 ni hisoblang. A) 2π B) 2 C) 2 + 2π D) 2 − 2π 30. (99-1-24) Funksiyaning hosilasini hisoblang. y = 2 − cos 2x A) 2 sin 2x B) sin 2x C) 4 cos 2x D) − sin 2x 31. (99-1-25) y 1 = cos 2 3x, y 2 = − sin 2 3x va y 3 = 2 sin 6x funksiyalardan qaysilarining hosi- lalari teng? A) y 1 ; y 2 B) y 1 ; y 3 C) y 2 ; y 3 D) y 1 ; y 2 ; y 3 32. (99-10-43) Agar f (x) = sin 2 3x bo’lsa, f 0 ( π 12 ) ni hisoblang. A) −3 B) 3 C) 2 D) −2 33. (00-2-27) Agar f (x) = 5 sin(2x + 2 x ) bo’lsa, f 0 (1) ni toping. A) 5 B) 0 C) 2,5 D) − 1 5 34. (00-3-62) Ushbu f (x) = sin 2x+ln cos 2x funksiya uchun f 0 ( π 6 ) ni toping. A) 1 2 (1 − √ 3) B) 1 − 2 √ 3 C) − 3 2 D) 3 2 35. (00-8-67) Ushbu f (x) = sin ³ 1 x − 1 ´ funksiyaning hosilasini toping. A) 1 x cos( 1 x − 1) B) − 1 x cos( 1 x − 1) C) 1 x cos( 1 x + 1) D) − 1 x 2 cos( 1 x − 1) 36. (01-8-26) Agar f (x) = e 1−2x · cos 2x bo’lsa, f 0 (0) ning qiymatini toping. A) −2e B) 0 C) e D) 2e 37. (01-10-49) Agar f (x) = sin 4 3x, ϕ(x) = 6 sin 6x bo’lsa, f 0 (x) = ϕ(x) tenglik o’rinli bo’ladigan x ning barcha qiymatlarini toping. A) πn 3 , n ∈ Z B) πn 6 , n ∈ Z C) πn 4 , n ∈ Z D) π 3 + πn 4 , n ∈ Z 38. (02-1-66) f (x) = sin 2 2x funksiya berilgan. f 0 (x) 2 cos 2x ni toping. A) sin 2x B) cos 2x C) − sin 2x D) 2 sin 2x 39. (02-2-28) Agar f (x) = √ sin 2x bolsa, f 0 ( π 4 ) ni toping. A) 0 B) 1 C) 1 2 D) √ 2 2 174 Yechish: Berilgan funksiyani f (x) = (sin 2x) 1/2 shaklda yozib, unga murakkab funksiyalarning hosilalarini hisoblashdagi 6-qoidani qo’llab f 0 (x) = 1 2 2 cos 2x √ sin 2x = cos 2x √ sin 2x ekanini topamiz. Endi hosilaning π 4 nuqtadagi qiymatini hisoblaymiz: f 0 ( π 4 ) = cos(2 · π 4 ) r sin(2 · π 4 ) = cos π 2 r sin π 2 = 0 1 = 0. Javob: 0 (A). 40. (02-3-46) Agar f (x) = √ tgx bolsa, f 0 ( π 4 ) ni to- ping. A) 1 B) 1 2 C) 1 4 D) 3 4 41. (02-9-32) Agar f (x) = cos(x + π 2 ), tg( α 2 ) = 1 2 bolsa, f 0 (α) ni hisoblang. A) −0, 6 B) 3 5 C) 0,8 D) − 1 3 42. (03-1-50) y = sin 4 2x, y 0 =? A) 2 sin 2 2x sin 4x B) 4 sin 2 4x sin 2x C) 4 sin 2x sin 2 4x D) 4 sin 2 2x sin 4x 43. (03-2-10) Agar f (x) = e 1−x · sin πx 2 bolsa, f 0 (1) ning qiymatini toping. A) 1 B) 2 C) − √ 2 D) −1 44. (03-6-21) f (x) = |x 2 − 14x + 45|. f 0 (6)−? A) 0 B) 5 C) 2 D) 7 14.2 Funksiyani hosila yordamida tek- shirish. Maksimum va minimum Funksiyalarni tekshirishda ularning o’sish yoki kamay- ish oraliqlarini topish muhim ahamiyatga ega. Bizga [a; b] kesmada aniqlangan va uning har nuqtasida hosi- lasi mavjud bo’lgan f : [a; b] → R funksiya berilgan bo’lsin. Aniqlanish sohasining har nuqtasida hosilasi mavjud bo’lgan funksiyalar, differensiallanuvchi funksi- yalar deyiladi. Agar biror δ > 0 va barcha x ∈ (x 0 −δ; x 0 +δ)∩[a; b] lar uchun f (x) ≤ f (x 0 ) (f (x) ≥ f (x 0 )) tengsizligi ba- jarilsa, x = x 0 nuqta f : [a; b] → R funksiyaning mak- simum nuqtasi (minimum nuqtasi) deyiladi. Funksiya- ning minimum va maksimum nuqtalari shu funksiya- ning ekstrimum nuqtalari, funksiyaning bu nuqtalardagi qiymatlari esa funksiyaning ekstrimumlari deyiladi. Funksiyaning hosilasi nolga aylanadigan yoki hosi- lasi mavjud bo’lmagan nuqtalari funksiyaning kritik nuqtalari deyiladi. Ekstrimumning zaruriy sharti haqida Ferma teore- masini keltiramiz. Agar differensiallanuvchi y = f (x) funksiya x = x 0 nuqtada minimumga yoki maksimumga erishsa, u holda f 0 (x 0 ) = 0 bo’ladi. Demak, funksiyan- ing ekstrimum nuqtalarini, uning hosilasi nolga aylanadi- gan nuqtalari ichidan izlashimiz kerak ekan. 1. Agar [a; b] kesmada differensiallanuvchi y = f (x) funksiya uchun f 0 (x) > 0, x ∈ (a 1 ; b 1 ) ⊂ [a; b] bo’lsa, u holda f funksiya (a 1 ; b 1 ) oraliq- da o’suvchi bo’ladi. 2. Agar [a; b] kesmada differensiallanuvchi y = f (x) funksiya uchun f 0 (x) < 0, x ∈ (a 1 ; b 1 ) ⊂ [a; b] bo’lsa, u holda f funksiya (a 1 ; b 1 ) oraliq- da kamayuvchi bo’ladi. 3. Agar x 0 ∈ (a; b) nuqtaning shunday δ > 0 atrofi mavjud bo’lib, barcha x ∈ (x 0 − δ; x 0 ) larda f 0 (x) < 0 va barcha x ∈ (x 0 ; x 0 + δ) larda f 0 (x) > 0 bo’lsa, u holda x = x 0 nuqta f funksiya uchun minimum nuqta bo’ladi. 4. Agar x 0 ∈ (a; b) nuqtaning shunday δ > 0 atrofi mavjud bo’lib, barcha x ∈ (x 0 − δ; x 0 ) larda f 0 (x) > 0 va barcha x ∈ (x 0 ; x 0 + δ) larda f 0 (x) < 0 bo’lsa, u holda x = x 0 nuqta f funksiya uchun maxsimum nuqta bo’ladi. 5. Agar f 0 (x 0 ) = 0 bo’lib, biror δ > 0 va barcha x ∈ (x 0 − δ; x 0 ) ∪ (x 0 ; x 0 + δ) larda f 0 (x) > 0 ( yoki f 0 (x) < 0 ) bo’lsa, u holda x = x 0 nuqta f funksiya uchun bo’rilish nuqtasi bo’ladi. Bo’rilish nuqta funksiya uchun ekstrimum nuqta bo’la olmaydi. 1. (98-6-18) Ushbu y = x 2 2 − ln x funksiyaning o’sish oraliqlarini toping. A) [−1; 0) ∪ [1; ∞) B) [1; ∞) C) [−1; ∞) D) (−∞; −1) ∪ [1; ∞) Yechish: 1-xossaga ko’ra, agar f (x) funksiya uchun f 0 (x) > 0, x ∈ (a; b) bo’lsa, u holda f (x) funksiya (a; b) oraliqda o’suvchi bo’ladi. Berilgan funksiyaning hosilasini topamiz. f 0 (x) = x − 1 x = x 2 − 1 x = (x − 1)(x + 1) x f 0 (x) > 0 tengsizlik oraliqlar usuli bilan oson yechiladi. Uning yechimi (−1; 0) ∪ (1; ∞). Beril- gan funksiyaning aniqlanish sohasi x > 0 bo’lgani uchun (−1; 0) oraliqni chiqarib tashlaymiz. Bun- dan tashqari x = 1 nuqta funksiyaning aniqlanish sohasiga tegishli bo’lgani uchun uni ham funksiyan- ing o’sish oralig’iga qo’shib qo’yamiz. Javob: [1; ∞) (B). 175 2. (97-9-25) Ushbu y = x 2 − 2 funksiyaning kamay- ish oralig’ini ko’rsating. A) (−∞; −2) B) (−∞; 2) C) (2; ∞) D) (−∞; 0] 3. (97-11-20) Ushbu y = 2x 3 +3x 2 −12x+7 funksiya- ning kamayish oralig’ini aniqlang. A) (−∞; −2] ∪ [1; ∞) B) [−2; 1] C) [−1; 2] D) [−2; ∞) 4. (96-11-21) Ushbu f (x) = x 2 +2x+4 funksiyaning o’sish oralig’ini toping. A) (−∞; −1) B) [−1; ∞) C) (1; ∞; ) D) (0; ∞) 5. (96-12-21) f (x) = x 2 − 2x + 3 funksiyaning o’sish oralig’ini toping. A) (0; ∞) B) (−∞; 1) C) [1; ∞) D) (−∞; −1) 6. (01-2-35) Ushbu y = x + 1 x − 1 funksiyaning ka- mayish oraliqlarini toping. A) [0; 1) ∪ (1; 2] B) (0; 2) C) (0; 1) D) (1; 2) 7. (96-3-20) Ushbu f (x) = −x 2 +2x−1 funksiyaning o’sish oralig’ini toping. A) (1; ∞) B) (0; ∞) C) (−∞; −1) D) (−∞; 1] 8. (96-6-44) a ning qanday qiymatlarida f (x) = ax+ sin x funksiya o’zining aniqlanish sohasida o’sadi. Shunday a larning barchasini toping. A) |a| > 1 B) 0 < a < 1 C) a ≥ 1 D) a = 0 Yechish: Berilgan funksiyaning aniqlanish so- hasi D(f ) = R. 1-xossaga ko’ra, f 0 (x) = a + cos x > 0, x ∈ R bo’lishi kerak. Bu tengsizlik a > 1 shartda bajariladi. a = 1 da bu funksiya o’suvchidir. Javob: a ≥ 1 (C). 9. (96-10-14) Quyidagi funksiyalardan qaysi biri (0; ∞) oraliqda kamayuvchi bo’ladi? A) y = x + 8 B) y = 3 − x C) y = − 4 x D) y = 2x 2 10. (96-1-14) Quyidagi funksiyalardan qaysi biri (−∞; 0) oraliqda o’suvchi bo’ladi? A) y = 3x + 2 B) y = 3 x C) y = 6 − 3x D) y = x 2 11. (96-9-64) Quyidagi funksiyalardan qaysi biri (−∞; 0) oraliqda o’suvchi bo’ladi? A) y = 0, 5 − 2x B) y = 5 x C) y = 2 + 3x D) y = 2 √ −x 12. (97-8-44) k ning qanday qiymatlarida f (x) = sin x− kx funksiya o’zining aniqlanish sohasida o’sadi? A) (−∞; 1) B) (1; ∞) C) (−1; 0) D) (−∞; −1] 13. (02-11-54) Quyidagi funksiyalardan qaysi biri o’zi- ning aniqlanish sohasida o’suvchi bo’ladi? A) y = sin x B) y = ln x x C) y = 1 x 2 + 1 D) y = 2x 7 − 8 14. (01-1-35) Ushbu y = x 2 e −2x funksiyaning o’sish oraliqlarini toping. A) (−∞; −1) B) [−1; 1] C) (−∞; −1) ∪ [0; 1] D) [0; 1] Yechish: Ko’paytmaning hosilasini hisoblash qoi- dasidan foydalanib, berilgan funksiyaning hosi- lasini topamiz: f 0 (x) = 2 Download 1.09 Mb. Do'stlaringiz bilan baham: |
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