11-sinf Matematika olimpiada №4 Oktabr 18, 2022Oktabr 18, 2022 da chop etilgan InfoMaster tomonidan 11-sinf Matematika olimpiada №4 ga 4 fikr bildirilgan 0 123456789101112131415161718192021222324252627282930 Vaqtingiz tugadi! 11-sinf Matematika olimpiada №4 2022-yil 12-oktyabr: 2022-2023 o'quv yilida tuman bosqichida tushgan savollar! 1 / 30 1. tenglama nechta ildizga ega? A) 3 B) 4 C) 2 D) 1 2 / 30 2. Bitta tekislikda yotuvchi n ta turli tog`ri chiziq o`tkazilgan. Ularning ixtiyoriy ikkitasi parallel emas va ixtiyoriy uchtasi bitta nuqtadan o`tmaydi ,ya`ni bitta nuqtada kesishmaydi. Bu to`g`ri chiziqlar tekislikni 1276 qismga ajratadi. n ni aniqlang? A) 49 B) 51 C) 52 D) 50 3 / 30 3. ni hisoblang. A) 13 B) 12 C) -13 D) -12 4 / 30 4. Aylanaga АВС uchburchak ichki chizilgan. Uchburchakning СС1medianasining davomi aylanani ? nuqtada kesib o‘tadi. ni toping. A) 2 B) 1 C) √5 D) 4 5 / 30 5. Ikki xonali sonni qandaydir natural songa bo‘lganda bo‘linmada 3 va qoldiqda 8 hosil bo‘ladi. Agar bo‘linuvchining raqamlari o‘rni almashtirilsa va bo‘luvchi son o‘zgarishsiz qoldirilsa, u holda bo‘linmada 2 va qoldiqda 5 hosil bo‘ladi. Bo‘linuvchning dastlabki qiymatini toping. A) 73 B) 65 C) 55 D) 53 6 / 30 6. Soddalashtiring: A) −сtg13° B) -1 C) −сtg1° D) −сtg²1° 7 / 30 7. Soatning minut mili 144° ga burilganda, soatning soat mili qanday burchakka burilada? A) 6° B) 18° C) 12° D) 9° 8 / 30 8. 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 9 / 30 9. To‘g‘ri burchakli uchburchakning bir burchagi 60° ga teng. Bu burchakdan chiqarilgan bissektrisaning uzunligi 2 m. Uchburchakning gipotenuzasi uzunligini toping. A) √3 B) √5 C) 2√3 D) √5 +1 10 / 30 10. un ketma-ketlik quyidagicha berilgan: u1 = 1, un+1 = un + 8n. U holda, u50 −u30 ni toping. A) 6230 B) 6320 C) 5320 D) 6320 11 / 30 11. Agar ifodaning qiymatini toping. A) 4 B) 20 C) 0.05 D) 0.8 12 / 30 12. bo’lsa, P(−1) ni toping. A) 105 B) 115 C) 100 D) 120 13 / 30 13. Soddalashtiring: bunda, |?| < 1. A) 2a+4 B) 4-2a C) -2 D) 2-2a 14 / 30 14. BC = BE = CD ,∠DAE = 20°, ∠BCD = 60° ,∠ADE = ? (chizmaga qarang) A) 50° B) 40° C) 20° D) 30° 15 / 30 15. ABC uchburchakda ctgA+ctgB= 3 va AB= 12 bo‘lsa, ABC uchburchak yuzini toping. A) 521 B) 265 C) 216 D) 123 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) 8 B) 6 C) 10 D) 12 17 / 30 17. 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) 96 D) 100 18 / 30 18. Tenglamani yeching: A) 29/462 B) ∅ C) 0 D) 10/21 19 / 30 19. 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) 8 B) 9 C) 3 D) 6 20 / 30 20. ABCD - kvadrat, AC-diagonal. Agar AE=BE+CE ( a + b = c ) bo‘lsa, ∠AEB burchakni toping(chizmaga qarang!). A) 45 B) 60 C) 50 D) 55 21 / 30 21. Agar bo‘lsa sin²α ni toping. A) B B) C C) D D) A 22 / 30 22. a2b5 = 620 tenglikni qanoatlantiruvchi nechta (?; ?) butun sonlar juftligi mavjud? A) 12 B) 14 C) 8 D) 10 23 / 30 23. Tengsizliklar sistemasinining butun yechimlari sonini toping. A) 7 B) 6 C) 5 D) 4 24 / 30 24. ∠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 25 / 30 25. Hisoblang: A) C B) D C) B D) A 26 / 30 26. (3 − cos²x− 2sinx)(lg²y+ 2lgy+ 4) ≤ 3 bo‘lsa, sin²x+ 20y+ 1 ni toping. A) 5 B) 3 C) 4 D) 2 27 / 30 27. ? 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 28 / 30 28. ? = ?² − 2? + 3 parabolaga ? = 3 to‘g‘ri chiziqqa nisbatan simmetrik bo‘lgan parabola tenglamasini tuzing. A) ? =-?² + 2? - 3 B) ? = -?² − 2? + 3 C) ? = ?² − 2? - 3 D) ? =-?² + 2? + 3 29 / 30 29. formula bilan berilgan sonli ketma-ketlikning dastlabki 20 ta hadining o‘rta arifmetigini toping. A) 25/6 B) 23/6 C) 20/3 D) 22/3 30 / 30 30. Soddalashtiring: A) C B) B C) D D) A 0% Author: InfoMaster Matematika olimpiada
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