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To`g`ri javoblar:
1. a 2. d 3. b 4. b 5. c 6. d 7. c 8. b 9. a 10. b 11. c 12. a 13. a 14. c 15. c 16. c 17. a 18. b 19. b 20. d 21. d 22. d 23. a 24. d 25. b 26. b

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Assalomu alaykum hurmatli obunachilar.
Yechimlaringizni kommentariyada qoldiring
@MATEMATIKA
Assalomu alaykum hurmatli obunachilar. Yechimlaringizni kommentariyada qoldiring
1. Dastlab y=f(x) funksiyaning x0=3 nuqtadan o'tadigan urinmasining burchak koeffitsientini topamiz:
Urinma tenglamasi:

y=f'(x0)(x-x0)+f(x0)

Burchak koeffitsient k=tga=f'(x0)=f'(3)=0
Demak, bu urinma x o'qiga parallel.

2. y=f'(x) funksiyaning x1=-2 nuqtadan o'tadigan urinmasining burchak koeffitsientini topamiz:
Bunda ham 1-holdagi kabi y=f'(x) funksiyaning x1=-2 nuqtadagi hosilasini qiymatini topishimiz kerak (ya'ni f''(-2) ni).
Grafikka ko'ra, (-3, -2) da y=f'(x) funksiya o'suvchi, (-2, -1) da esa kamayuvchi. Demak, x1=-2 nuqta y=f'(x) funksiyaning lokal maksimumi. => f''(-2)=0.
Bu urinma ham x o'qiga parallel ekan.

Demak, ikkala urinma parallel.

alfa = parallel to'g'ri chiziqlar orasidagi burchak

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To`g`ri javoblar:
1. a 2. c 3. b 4. c 5. b 6. c 7. b 8. a 9. a 10. b 11. b 12. b 13. d 14. c 15. c 16. c 17. b 18. a 19. a 20. c 21. b 22. b 23. c 24. a 25. b 26. b 27. d 28. a 29. a 30. d 31. d 32. a 33. b 34. a 35. a

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Forwarded from Nurullo Ibroximov
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