Uy » Online olimpiada » Matematika olimpiada » 11-sinf Matematika olimpiada №4 Matematika olimpiadaTuman 2022-2023 o'quv yili 11-sinf Matematika olimpiada №4 InfoMaster Avgust 19, 2024 1057 Ko'rishlar 2 izohlar SaqlashSaqlanganOlib tashlandi 1 1 Savollar 123456789101112131415161718192021222324252627282930 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster 11-sinf Matematika olimpiada №4 2022-yil 12-oktyabr: 2022-2023 o'quv yilida tuman bosqichida tushgan savollar! 1 / 30 1. ko‘phadning barcha koeffitsientlari va ozod hadining yig‘indisini toping. A) -238 B) 125 C) 245 D) -247 2 / 30 2. 3cos2x-3 sin2x=0 trigonometrik tenglamani yeching. A) π/6+πk; k∈Z B) π/12+2πk; k∈Z C) π/12+πk; k∈Z D) π/12+πk/2; k∈Z) 3 / 30 3. Bir savdogar 1 mеtrini 1200 tiyindan olgan matosini yuvib, kuritgandan so`ng 1800 tiyindan sotyapti. Mato yuvilib kuritilganidan so`ng 20 foizi qisqardi. Bularga ko`ra savdogar nеcha foiz foyda ko`radi? A) 10 B) 20 C) 15 D) 18 4 / 30 4. Tenglamalar sistemasi nechta yechimga ega? A) 10 B) 6 C) 8 D) 12 5 / 30 5. Quyidagi sonlar orasidan musbat sonni aniqlang. A) B B) A C) C D) D 6 / 30 6. 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) 81 C) 91 D) 41 7 / 30 7. Agar ifodaning qiymatini toping. A) 0.8 B) 20 C) 4 D) 0.05 8 / 30 8. bo’lsa, P(−1) ni toping. A) 100 B) 115 C) 105 D) 120 9 / 30 9. 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 10 / 30 10. 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) 6 B) 9 C) 3 D) 8 11 / 30 11. Soddalashtiring: A) −сtg²1° B) -1 C) −сtg1° D) −сtg13° 12 / 30 12. To‘g‘ri burchakli uchburchakning bir burchagi 60° ga teng. Bu burchakdan chiqarilgan bissektrisaning uzunligi 2 m. Uchburchakning gipotenuzasi uzunligini toping. A) √5 B) 2√3 C) √3 D) √5 +1 13 / 30 13. Soddalashtiring: A) A B) C C) B D) D 14 / 30 14. 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 15 / 30 15. ? soniga teskari bo‘lgan son ? ning 9% ini tashkil qiladi. ? ni toping. (?u yerda, ? > 0) A) 8/3 B) 16/3 C) 13/3 D) 10/3 16 / 30 16. 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 17 / 30 17. 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) 85/6 D) 35/3 18 / 30 18. Tengsizliklar sistemasinining butun yechimlari sonini toping. A) 7 B) 6 C) 5 D) 4 19 / 30 19. 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) 18 C) 15 D) 20 20 / 30 20. ?(3) ∙ (? − 2) + ?(? − 1) = 3? bo‘lsa, ?(1) ni toping. A) 2 B) 4 C) 6 D) 3 21 / 30 21. Arifmetik progressiyada ?2 = 12, ?7 − ?4 = 9 bo‘lsa, ?19 ni toping. A) 63 B) 61 C) 54 D) 56 22 / 30 22. Hisoblang: A) C B) B C) A D) D 23 / 30 23. ABCD - kvadrat, AC-diagonal. Agar AE=BE+CE ( a + b = c ) bo‘lsa, ∠AEB burchakni toping(chizmaga qarang!). A) 60 B) 50 C) 45 D) 55 24 / 30 24. 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) 8 C) 12 D) 10 25 / 30 25. (3 − cos²x− 2sinx)(lg²y+ 2lgy+ 4) ≤ 3 bo‘lsa, sin²x+ 20y+ 1 ni toping. A) 5 B) 4 C) 2 D) 3 26 / 30 26. Soddalashtiring: bunda, |?| < 1. A) 2-2a B) 2a+4 C) -2 D) 4-2a 27 / 30 27. Tenglamani yeching: A) 29/462 B) 10/21 C) 0 D) ∅ 28 / 30 28. a2b5 = 620 tenglikni qanoatlantiruvchi nechta (?; ?) butun sonlar juftligi mavjud? A) 12 B) 8 C) 14 D) 10 29 / 30 29. un ketma-ketlik quyidagicha berilgan: u1 = 1, un+1 = un + 8n. U holda, u50 −u30 ni toping. A) 5320 B) 6320 C) 6320 D) 6230 30 / 30 30. ∠CAD = 2∠ACD, AC =13a, BD = 5a va AB = CD bo’lsa, 20⋅cos ∠ACD ni toping.(chizmaga qarang!) A) 17 B) 15 C) 19 D) 16 0% Author: InfoMaster Foydali bo'lsa mamnunmiz