11-sinf Matematika olimpiada №4 Oktabr 18, 2022Oktabr 18, 2022 da chop etilgan InfoMaster tomonidan 11-sinf Matematika olimpiada №4 ga fikr bildirilmagan. 1 123456789101112131415161718192021222324252627282930 Vaqtingiz tugadi! 11-sinf Matematika olimpiada №4 2022-yil 12-oktyabr: 2022-2023 o'quv yilida tuman bosqichida tushgan savollar! 1 / 30 1. Koordinata boshidan o‘tuvchi, simmetriya o‘qi ordinata o‘qiga parallel, cho‘qqisi y=x va y+x-2=0 to‘g‘ri chiziqlar kesishgan nuqtada bo‘lgan parabolaning tenglamasini toping. A) y=-x²+2x B) y=x²-2x C) y=-x²-2x D) y=-x²+1 2 / 30 2. funksiyaning aniqlanish sohasini toping. A) (2/3;∞) B) (-∞;2/3)∪(2/3;∞) C) (-∞;4/3) D) (-∞;4/3)∪(4/3;∞) 3 / 30 3. y=6sin2x+sin12x funksiyaning hosilasini toping. A) −24sin7xcos5x B) 24sin7xsin5x C) 24cos7xcos5x D) 24sin5xcos7x 4 / 30 4. Agar ?(?) = ?2021 + 3?2020 + 3?2019 − ?2018 + 1 bo‘lsa, ni toping. A) 2 B) 3 C) 1 D) 0 5 / 30 5. АВС uchburchak ∠C=60° burchagining bissektrisasi 5√3 ga teng. Agar АС : ВС = 5 :2 bo‘lsa, ВС ni toping. A) C B) B C) D D) A 6 / 30 6. Soatning minut mili 144° ga burilganda, soatning soat mili qanday burchakka burilada? A) 12° B) 18° C) 9° D) 6° 7 / 30 7. BC = BE = CD ,∠DAE = 20°, ∠BCD = 60° ,∠ADE = ? (chizmaga qarang) A) 40° B) 30° C) 50° D) 20° 8 / 30 8. 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 9 / 30 9. ?(3) ∙ (? − 2) + ?(? − 1) = 3? bo‘lsa, ?(1) ni toping. A) 2 B) 6 C) 4 D) 3 10 / 30 10. 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 11 / 30 11. a2b5 = 620 tenglikni qanoatlantiruvchi nechta (?; ?) butun sonlar juftligi mavjud? A) 14 B) 10 C) 8 D) 12 12 / 30 12. 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) 10 B) 8 C) 6 D) 12 13 / 30 13. 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) 85/6 C) 37/3 D) 35/3 14 / 30 14. (3 − cos²x− 2sinx)(lg²y+ 2lgy+ 4) ≤ 3 bo‘lsa, sin²x+ 20y+ 1 ni toping. A) 5 B) 2 C) 3 D) 4 15 / 30 15. ABC uchburchakda ctgA+ctgB= 3 va AB= 12 bo‘lsa, ABC uchburchak yuzini toping. A) 265 B) 521 C) 216 D) 123 16 / 30 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) 15 D) 16 17 / 30 17. 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) 6 D) 3 18 / 30 18. Tengsizliklar sistemasinining butun yechimlari sonini toping. A) 4 B) 7 C) 6 D) 5 19 / 30 19. To‘g‘ri burchakli uchburchakning bir burchagi 60° ga teng. Bu burchakdan chiqarilgan bissektrisaning uzunligi 2 m. Uchburchakning gipotenuzasi uzunligini toping. A) √5 +1 B) √3 C) √5 D) 2√3 20 / 30 20. formula bilan berilgan sonli ketma-ketlikning dastlabki 20 ta hadining o‘rta arifmetigini toping. A) 23/6 B) 20/3 C) 25/6 D) 22/3 21 / 30 21. Agar ifodaning qiymatini toping. A) 0.8 B) 20 C) 0.05 D) 4 22 / 30 22. Raqamlari yig‘indisi 10 dan kam bo‘lmagan, raqamlari ko‘paytmasi esa 10 dan katta bo‘lmagan uch xonali sonlar nechta? A) 100 B) 106 C) 89 D) 96 23 / 30 23. Soddalashtiring: A) B B) C C) A D) D 24 / 30 24. Tenglamani yeching: A) 10/21 B) 0 C) 29/462 D) ∅ 25 / 30 25. bo’lsa, P(−1) ni toping. A) 100 B) 120 C) 115 D) 105 26 / 30 26. Soddalashtiring: A) -1 B) −сtg²1° C) −сtg13° D) −сtg1° 27 / 30 27. Soddalashtiring: bunda, |?| < 1. A) 4-2a B) 2a+4 C) 2-2a D) -2 28 / 30 28. Uchlari A(2; −3), B(6; −1), C(6; 4) va D(2; 2) nuqtalarda bo‘lgan ABCD to‘rtburchak yuzini toping. A) 15 B) 18 C) 16 D) 20 29 / 30 29. Arifmetik progressiyada ?2 = 12, ?7 − ?4 = 9 bo‘lsa, ?19 ni toping. A) 54 B) 56 C) 63 D) 61 30 / 30 30. un ketma-ketlik quyidagicha berilgan: u1 = 1, un+1 = un + 8n. U holda, u50 −u30 ni toping. A) 6320 B) 5320 C) 6320 D) 6230 0% Author: InfoMaster Matematika olimpiada