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Matematika olimpiada Tuman 2022-2023 o'quv yili

11-sinf Matematika olimpiada №4

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Avgust 19, 2024

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11-sinf Matematika olimpiada №4

2022-yil 12-oktyabr:

2022-2023 o'quv yilida tuman bosqichida tushgan savollar!

1 / 30

1. tengsizlik yechimlari joylashgan interval uzunligini toping.

A) 0,75

B) 0,5

C) 0,125

D) 0,25

2 / 30

2. y=6sin2x+sin12x funksiyaning hosilasini toping.

A) 24sin5xcos7x

B) 24sin7xsin5x

C) −24sin7xcos5x

D) 24cos7xcos5x

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3. Perimetri 4 ga o`tkir burchagi 30^oga va shu burchak qarshisidagi tomoni √3 ga teng bo`lgan uchburchakka ichki chizlgan aylana radiusini toping.

A) 4√3+7

B) (√3-1)/2

C) 7-4√3

D) (1+√3)/2

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4. Aylanaga АВС uchburchak ichki chizilgan. Uchburchakning СС1medianasining davomi aylanani ? nuqtada kesib o‘tadi. ni toping.

A) √5

B) 2

C) 4

D) 1

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5. To‘g‘ri burchakli uchburchakka ichki va tashqi chizilgan aylana radiuslari nisbati 2 : 5 bo‘lsa, uchburchakning katta o‘tkir burchagining kosinusini toping.

A) 0.6

B) 0.2

C) 0.5

D) 0.8

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6. Hisoblang:

A) D

B) C

C) B

D) A

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7. Raqamlari yig‘indisi 10 dan kam bo‘lmagan, raqamlari ko‘paytmasi esa 10 dan katta bo‘lmagan uch xonali sonlar nechta?

A) 106

B) 89

C) 100

D) 96

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8. Agar bo‘lsa sin²α ni toping.

A) C

B) B

C) D

D) A

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9. ABCD - kvadrat, AC-diagonal. Agar AE=BE+CE ( a + b = c ) bo‘lsa, ∠AEB burchakni toping(chizmaga qarang!).

A) 60

B) 45

C) 55

D) 50

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10. Uchlari A(2; −3), B(6; −1), C(6; 4) va D(2; 2) nuqtalarda bo‘lgan ABCD to‘rtburchak yuzini toping.

A) 16

B) 15

C) 18

D) 20

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11. O‘tkir burchakli uchburchakning ikki tomon uzunliklari ayirmasi 8, bu tomonlarning uchinchi tomondagi proyeksiyalari mos ravishda 8 va 20 ga teng. Berilgan uchburchakka tashqi chizilgan aylana radiusini toping.

A) 83/6

B) 37/3

C) 35/3

D) 85/6

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12. Muntazam uchburchak 36 ta yuzi 1 ga teng bo‘lgan kichkina muntazam uchburchaklardan iborat (chizmaga qarang!). ABC uchburchak yuzini toping.

A) 13

B) 12

C) 11

D) 10

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13. Arifmetik progressiyada ?₂ = 12, ?₇ − ?₄ = 9 bo‘lsa, ?₁₉ni toping.

A) 63

B) 56

C) 61

D) 54

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14. a²b⁵ = 6²⁰ tenglikni qanoatlantiruvchi nechta (?; ?) butun sonlar juftligi mavjud?

A) 14

B) 8

C) 12

D) 10

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15. ?(3) ∙ (? − 2) + ?(? − 1) = 3? bo‘lsa, ?(1) ni toping.

A) 3

B) 4

C) 2

D) 6

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16. ∠CAD = 2∠ACD, AC =13a, BD = 5a va AB = CD bo’lsa, 20⋅cos ∠ACD ni toping.(chizmaga qarang!)

A) 17

B) 19

C) 16

D) 15

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17. Agar ifodaning qiymatini toping.

A) 4

B) 0.8

C) 20

D) 0.05

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18. Soddalashtiring: bunda, |?| < 1.

A) 2-2a

B) -2

C) 4-2a

D) 2a+4

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19. Nargizada 2 ta olma va 3 ta nok bor. U 5 kun ketma-ket har kuni singliga bittadan meva beradi. Bu ishni necha usul bilan amalga oshirish mumkin?

A) 12

B) 8

C) 10

D) 6

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20. bo’lsa, P(−1) ni toping.

A) 120

B) 115

C) 100

D) 105

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21. Soddalashtiring:

A) −сtg²1°

B) −сtg13°

C) −сtg1°

D) -1

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22. u_nketma-ketlik quyidagicha berilgan: u₁ = 1, u_n+1 = u_n + 8n. U holda, u₅₀ −u₃₀ ni toping.

A) 6230

B) 5320

C) 6320

D) 6320

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23. BC = BE = CD ,∠DAE = 20°, ∠BCD = 60° ,∠ADE = ? (chizmaga qarang)

A) 30°

B) 20°

C) 40°

D) 50°

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24. Tenglamani yeching:

A) 0

B) 29/462

C) ∅

D) 10/21

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25. Soatning minut mili 144° ga burilganda, soatning soat mili qanday burchakka burilada?

A) 6°

B) 12°

C) 9°

D) 18°

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26. Soddalashtiring:

A) C

B) B

C) D

D) A

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27. Doskaga ?₁, ?₂, ?₃, . . . , ?₂₀₀ sonlari yozilgan. Ma’lumki, ?₁ = 3, ?₂ = 9. Agar ixtiyoriy n natural son uchun ?_n+2 = ?_n+1 − ?_n tenglik o‘rinli bo‘lsa, ?₂₀₀ ni toping.

A) 3

B) 6

C) 9

D) 8

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28. (3 − cos²x− 2sinx)(lg²y+ 2lgy+ 4) ≤ 3 bo‘lsa, sin²x+ 20y+ 1 ni toping.

A) 2

B) 4

C) 5

D) 3

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29. ABC uchburchakda ctgA+ctgB= 3 va AB= 12 bo‘lsa, ABC uchburchak yuzini toping.

A) 123

B) 521

C) 265

D) 216

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30. Natural son roppa-rosa 2 ta tub bo‘luvchiga, bu sonning kvadrati esa 45 ta turli natural bo‘luvchiga ega. Berilgan sonning kubi eng ko‘pi bilan nechta natural bo‘luvchiga ega bo‘lishi mumkin?

A) 40

B) 91

C) 81

D) 41