11-sinf Matematika olimpiada №4 Oktabr 18, 2022Avgust 19, 2024 da chop etilgan InfoMaster tomonidan 11-sinf Matematika olimpiada №4 ga 2 fikr bildirilgan 0 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. tenglama nechta ildizga ega? A) 4 B) 3 C) 1 D) 2 2 / 30 2. Tengsizlikni yeching. A) (1/2;∞) B) (4/3;3) C) (5/3;2) D) (2/3;2) 3 / 30 3. f(x)=x³/3-x²-35x+2 funksiya uchun f `(x)=0 bo`lsa x ni toping. A) −7 va −5 B) −7 va 5 C) 5 va 7 D) −5 va 7 4 / 30 4. Soddalashtiring: A) I B) II C) IV D) III 5 / 30 5. Tenglama nechta yechimga ega? Bunda, x ∈ (1; 47) A) 16 B) 15 C) 12 D) 14 6 / 30 6. Arifmetik progressiyada ?2 = 12, ?7 − ?4 = 9 bo‘lsa, ?19 ni toping. A) 63 B) 54 C) 56 D) 61 7 / 30 7. ABC uchburchakda ctgA+ctgB= 3 va AB= 12 bo‘lsa, ABC uchburchak yuzini toping. A) 265 B) 123 C) 521 D) 216 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) 35/3 B) 37/3 C) 85/6 D) 83/6 9 / 30 9. 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 10 / 30 10. 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 11 / 30 11. Tenglamani yeching: A) ∅ B) 29/462 C) 0 D) 10/21 12 / 30 12. Soatning minut mili 144° ga burilganda, soatning soat mili qanday burchakka burilada? A) 9° B) 18° C) 6° D) 12° 13 / 30 13. a2b5 = 620 tenglikni qanoatlantiruvchi nechta (?; ?) butun sonlar juftligi mavjud? A) 8 B) 14 C) 10 D) 12 14 / 30 14. BC = BE = CD ,∠DAE = 20°, ∠BCD = 60° ,∠ADE = ? (chizmaga qarang) A) 30° B) 20° C) 40° D) 50° 15 / 30 15. ABCD - kvadrat, AC-diagonal. Agar AE=BE+CE ( a + b = c ) bo‘lsa, ∠AEB burchakni toping(chizmaga qarang!). A) 60 B) 45 C) 50 D) 55 16 / 30 16. un ketma-ketlik quyidagicha berilgan: u1 = 1, un+1 = un + 8n. U holda, u50 −u30 ni toping. A) 6320 B) 5320 C) 6230 D) 6320 17 / 30 17. ∠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 18 / 30 18. Soddalashtiring: A) B B) C C) A D) D 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) √3 B) √5 C) 2√3 D) √5 +1 20 / 30 20. Hisoblang: A) D B) C C) B D) A 21 / 30 21. 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 22 / 30 22. ? soniga teskari bo‘lgan son ? ning 9% ini tashkil qiladi. ? ni toping. (?u yerda, ? > 0) A) 8/3 B) 10/3 C) 16/3 D) 13/3 23 / 30 23. 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) 20 C) 15 D) 18 24 / 30 24. Agar bo‘lsa sin²α ni toping. A) D B) B C) C D) A 25 / 30 25. Soddalashtiring: bunda, |?| < 1. A) -2 B) 2a+4 C) 4-2a D) 2-2a 26 / 30 26. (3 − cos²x− 2sinx)(lg²y+ 2lgy+ 4) ≤ 3 bo‘lsa, sin²x+ 20y+ 1 ni toping. A) 4 B) 3 C) 2 D) 5 27 / 30 27. 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) 12 D) 10 28 / 30 28. 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 29 / 30 29. Soddalashtiring: A) −сtg1° B) −сtg13° C) -1 D) −сtg²1° 30 / 30 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) 41 C) 91 D) 81 0% tomonidan Wordpress Quiz plugin Matematika olimpiada, Tuman 2022-2023 o'quv yili
