Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 221 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 8 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 integralning qiymatini toping. A) 0 B) π/2 C) -π/2 D) π/4 2 / 30 Anvar tub son o’yladi va o’ylagan sonini 5 ga ko’paytirib, 8 ni ayirgan edi, yana tub son hosil bo’ldi. Anvar qanday son o’ylagan? A) 374389 B) aniqlab bo’lmaydi C) 2 D) 412967 3 / 30 Agar geometrik progressiyaning ketma–ket dastlabki uchta hadining yig’indisi 62 ga, ularning o’nli logarifmlari yig’indisi 3 ga teng bo’lsa, shu geometrik progressiyaning birinchi hadini toping. A) 10 B) 10 yoki 50 C) 2 yoki 50 D) 50 4 / 30 tenglamani yeching. A) 2017 B) 2019 C) 2018 D) 0 5 / 30 bo’lsa, ni x orqali ifodalang. A) 2/x B) 25/x C) 2-x D) x/25 6 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusiga teng bo’lgan uchburchak qanday uchburchak deyiladi? A) o’tkir burchakli uchburchak B) to’gri burchakli uchburchak C) teng tomonli uchburchak D) o’tmas burchakli uchburchak 7 / 30 Hisoblang A) 2 B) √3/2 C) 1/2 D) √3 8 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 6 B) 8 C) 4 D) 12 9 / 30 sistemadan x+y ning qiymatini toping. A) 6 B) 35/4 C) -12 D) 12 10 / 30 Soddalashtiring. A) 2019 B) 2018a/a+1 C) 2018 D) a+1 11 / 30 Ostki asosining yuzi 16π ga va ustki asosining yuzi 4π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda sharning hajmini toping. A) 64√2π/3 B) 8√2π/5 C) 3√2π/4 D) (5√3+3)π/3 12 / 30 Teng yonli uchburchakning tomonlari 5, 5 va 6 ga teng. Bu uchburchakning bissektiritsalari va medianalari kesishgan nuqtalar A) 1/6 B) 1 C) 1,2 D) 1/2 13 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 1,4 B) 2, 3 C) 3, 4 D) 1, 3 14 / 30 Bog’bon uch kun davomida o’nta daraxt ko’chati o’tqazishi lozim. Agar bog’bon bir kunda eng kamida bitta ko’chat o’tqazadigan bo’lsa, u shu ishni kunlar bo’yicha necha xil usul bilan taqsimlashi mumkin? A) 32 B) 25 C) 30 D) 36 15 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 24 B) 32 C) 20 D) 16 16 / 30 Hisoblang. A) 15/34 B) 2/34 C) 17/34 D) 2/17 17 / 30 To’g’ri to’rtburchakning perimetri 50 ga teng. Bir tomoni boshqa tomonidan 5 ga ko’p. To’g’ri to’rtburchakning yuzini toping. A) 60 B) 150 C) 225 D) 50 18 / 30 Radiusi 25 bo’lgan doirada 48 ga teng vatar o’tkazilgan. Doira markazidan shu vatargacha masofani toping. A) 8 B) 7 C) 9 D) 10 19 / 30 y= funksiyaning aniqlanish sohasini toping A) (2;∞) B) (-∞;0])v[2;∞) C) (-∞;0)v(2;∞) D) (0,2) 20 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) 1; 10 B) (-∞;1]v[10;∞) C) (-∞;1] D) [10;∞) 21 / 30 Ostki asosining yuzi 32π va ustki asosining yuzi 18π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda sharning sirtini toping. A) 100π B) 96π C) 72π D) 56π 22 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) 1 B) e C) √2e D) 3e 23 / 30 AA1A2A3A4A5A6 muntazam oltiburchakli piramidaning hajmi ga va balandligi 2 ga teng bo‘lsa, u holda AA2A6 kesim yuzini toping. A) √15 B) 3 C) √5 D) 2 24 / 30 Tomoni 25 ga diagonallaridan biri 4 ga teng bo’lgan rombning yuzini toping. A) 8√5 B) 24√5 C) 16 D) 32 25 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,25 B) -0,75 C) -1 D) -0,5 26 / 30 funktsiyaning aniqlanish sohasini toping. A) (-∞;1)v(2;∞) B) (-∞;1]v[2;∞) C) (1;2) D) [1;2] 27 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) 1 B) 2 C) √2 D) √3 28 / 30 A) 2 B) 0 C) √6 D) 1 29 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 5 B) 6 C) 3 D) 4 30 / 30 Tengsizlikni yeching. A) (-4;2)v(2;3) B) (2;4) C) (-3;2)v(2;4) D) (-3;2) 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
