Barcha fanlardan online testlar

Matematika olimpiada Tuman 2022-2023 o'quv yili

11-sinf Matematika olimpiada №4

InfoMaster

Avgust 19, 2024

1022 Ko'rishlar 2 izohlar

SaqlashSaqlanganOlib tashlandi 1

Tomonidan yaratilgan InfoMaster

11-sinf Matematika olimpiada №4

2022-yil 12-oktyabr:

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

1 / 30

1. Sonlarni kamayish tartibida joylashtiring:

A) a>b>c

B) c>b>a

C) b>c>a

D) c>a>b

2 / 30

2. Agar f(g(x))= − x+13 bo`lsa, f(x) va g(x) to`g`ri chiziqlar orasidagi burchakni toping .

A) 0

B) π/2

C) π/3

D) π/4

3 / 30

3. a+b=−1 a+c=6 va b+c=1 bo`lsa, a² (3b+3c+2a)+b² (3a+3c+2b)+c² (3a+3b+2c) ning son qiymatini toping.

A) 100

B) 45

C) 27

D) 216

4 / 30

4. Hisoblang: (1 − tg5°)(1 − tg20°)2 + 2(2tg20° + tg5°)

A) 1

B) 2

C) 4

D) 3

5 / 30

5. Tenglama nechta yechimga ega? Bunda, x ∈ (1; 47)

A) 16

B) 15

C) 14

D) 12

6 / 30

6. y = x² − 2x + 3 parabolaga y = 3 to‘g‘ri chiziqqa nisbatan simmetrik bo‘lgan parabola tenglamasini tuzing.

A) y =-x² + 2x + 3

B) y =-x² + 2x - 3

C) y = -x² − 2x + 3

D) y = x² − 2x - 3

7 / 30

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

A) 18

B) 20

C) 16

D) 15

8 / 30

8. Soddalashtiring:

A) D

B) B

C) A

D) C

9 / 30

9. ?(3) ∙ (? − 2) + ?(? − 1) = 3? bo‘lsa, ?(1) ni toping.

A) 4

B) 2

C) 3

D) 6

10 / 30

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

A) 2-2a

B) 2a+4

C) 4-2a

D) -2

11 / 30

11. formula bilan berilgan sonli ketma-ketlikning dastlabki 20 ta hadining o‘rta arifmetigini toping.

A) 22/3

B) 25/6

C) 20/3

D) 23/6

12 / 30

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

A) 106

B) 96

C) 89

D) 100

13 / 30

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

A) 18°

B) 9°

C) 6°

D) 12°

14 / 30

14. 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) 91

B) 41

C) 81

D) 40

15 / 30

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

A) 56

B) 63

C) 54

D) 61

16 / 30

16. 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) 6

B) 12

C) 10

D) 8

17 / 30

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

A) 10

B) 12

C) 14

D) 8

18 / 30

18. (3 − cos²x− 2sinx)(lg²y+ 2lgy+ 4) ≤ 3 bo‘lsa, sin²x+ 20y+ 1 ni toping.

A) 5

B) 3

C) 2

D) 4

19 / 30

19. 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) 35/3

B) 85/6

C) 83/6

D) 37/3

20 / 30

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

A) 11

B) 13

C) 12

D) 10

21 / 30

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

A) 19

B) 15

C) 17

D) 16

22 / 30

22. ? soniga teskari bo‘lgan son ? ning 9% ini tashkil qiladi. ? ni toping. (?u yerda, ? > 0)

A) 16/3

B) 13/3

C) 8/3

D) 10/3

23 / 30

23. Tengsizliklar sistemasinining butun yechimlari sonini toping.

A) 5

B) 6

C) 7

D) 4

24 / 30

24. To‘g‘ri burchakli uchburchakning bir burchagi 60° ga teng. Bu burchakdan chiqarilgan bissektrisaning uzunligi 2 m. Uchburchakning gipotenuzasi uzunligini toping.

A) 2√3

B) √3

C) √5 +1

D) √5

25 / 30

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

A) 5320

B) 6230

C) 6320

D) 6320

26 / 30

26. bo’lsa, P(−1) ni toping.

A) 120

B) 100

C) 115

D) 105

27 / 30

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.