+2t+4 qonuniyat bilan harakatlanayotgan jismning t=2 sekunddagi tezligini toping. A. 70 B. 60 C. 74 D


Download 35.87 Kb.
Sana01.05.2023
Hajmi35.87 Kb.
#1419138

1.Moddiy nuqta S(t)=5t3+3t2+2t+4 qonuniyat bilan harakatlanayotgan jismning t=2 sekunddagi tezligini toping. A. 70 B. 60 C. 74 D. 84
2. Limitni hisoblang: . A.3 B. -4 C. -3 D. 4
3. Hosilani hisoblang:
A.f(x)=12x2-10x B. f(x)=-12x2+10x C. f(x)=21x2-10x D. f(x)=-21x2+10x
4. f(x)=sin2x funksiyasining barcha boshlang’ich funksiyalarini toping.
A.F(x)=2sin4x+c B. F(x)=-2cos2x+C C. F(x)=-0,5cos2x+C D. F(x)=0,5cos2x+C
5. f(x)=excosx funksiyasi uchun f ’(π/2) ni aniqlang. A. e-0.5π B. e0,5π C. eπ D. e
6. . f(x)=2x+3 funksiyasi uchun A(1;5) nuqtadan o’tuvchi boshlang’ich funksiyasini toping.
A.F(x)=x2+3x+1 B. F(x)=x2-3x-1 C. F(x)=x2-3x+1 D. F(x)=x2+3x-1
7. . f(x)=(3x+2)2 funksiyasining barcha boshlang’ich funksiyalarini toping.
A.F(x)=6x3+3x2-4 B. F(x)=6x3+6x+4 C. F(x)=6x3+3x2+4 D. F(x)=6x3+6x2+4
8. . aniq integralni hisoblang. A.4 B. 2.5 C. 2 D. 6
9. . To’g’ri to’rtburchak shaklidagi yer maydoninig atrofini o’rashmoqchi. 480 m panjara yordamida eng ko’pi bilan necha kvadrat metr yer maydonni o’rash mumkin?
A.14400 B. 25600 C. 19600 D. 8100
10. Murakkab funksiyaning hosilasini toping. y=x2sinx
A.y’=2xsinx-x2 B. y’=-2xsinx+x2 C. y’=x2cosx-2xsinx D. y’=2xsinx+x2cosx
11. . Ushbu y = 6 x -6 funksiyaning x = 1 nuqtadagi hosilasini toping. A) ln12 B) ln36 C) ln 6 D) ln E) 6
12. Ushbu f(x) = ln(x2 – 3 sinx ) funksiyaning hosilasini toping.
A) B) C) D) E)
13. . f(x) = bo`lsa , f ' ( ) = ? A) 1 B) C) D) E)
14. Ushbu f(x) = 1 - funksiyaning boshlang`ich funksiyasining umumiy ko`rinishini toping.
A) x + ctgx + c B) x - tgx + c C) x - tg3x + c D) tg3x + c E) x - ctg3x + c
15. . y = (x-3)(x2 + 3x + 9) funksiyaning x = 3 nuqtadagi hosilasini aniqlang.
A) 0 B) 3 C) 27 D) -27 E) 9
Download 35.87 Kb.

Do'stlaringiz bilan baham:




Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling