Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 125 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 3 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Ushbu (y6+y3+1)(y3+1)(y3-1)-y6+y3+1 ifodani soddalashtish natijasida ko’phad hosil qilindi. Uning nechta hadi bor? A) 3 B) 2 C) 4 D) 1 2 / 30 Uchburchakning balandligi 12 ga teng bo’lib, u asosni 5:16 nidbatda bo’ladi. Agar asosning uzunligi 21 ga teng bo’lsa, uchburchakning perimetrini toping A) 48 B) 54 C) 108 D) 52 3 / 30 Ag ar tgα=2 bo’lsa, u holda ni hisoblang A) 10/27 B) -10/27 C) -10/3 D) 10/3 4 / 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) 50 C) 225 D) 150 5 / 30 Markazi nuqtada bo‘lgan aylanaga va urinmalar o‘tkazilgan bo’lib, va nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma va kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar uchburchakning perimetri 48 va aylana radiusi 7 ga teng bo‘lsa, u holda kesma uzunligini toping A) 12 B) 25 C) 15 D) 30 6 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 60 B) 18 C) 47 D) 120 7 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) b>a>c>d>e B) b>c>a>d>e C) a>b>c>d>e D) e>b>a>d>c 8 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) √2 B) 4 C) 2 D) 2√2 9 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) √3 B) √2 C) 2 D) 1 10 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) ayqash B) o’zaro perpendikulyar C) aniqlab bo’lmaydi D) o’zaro parallel 11 / 30 sistemada xy ning qiymatini toping. A) 60 B) 80 C) 75 D) 64 12 / 30 Agar f(x)=sin2x va g(x)=cos2x bo’lsa, u holda f(g(x)) funksiyaning hosilasini toping. A) -4sin2x*cos(2cos2x) B) -4sin2x*cos(cos2x) C) 4sin2x*cos(cos2x) D) 4sin2x*cos(2cos2x) 13 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) ln(ln x) B) ln(lg x) C) lg(lg x) D) lg(ln x) 14 / 30 tenglamalar sistemasini yeching A) (–4;3) B) (4;–3) C) (3;–4) D) (4;3) 15 / 30 funktsiyaning aniqlanish sohasini toping. A) (1;2) B) (-∞;1)v(2;∞) C) (-∞;1]v[2;∞) D) [1;2] 16 / 30 O’q kesimining diagonallari o’zaro perpendikulyar bo’lgan kesik konus yasovchisi va asos tekisligi orasidagi burchak ga teng. Agar o’q kesimining diagonali ga teng bo’lsa, kesik konus asosining yuzini toping. A) πa²(1-2sinα/4sin²) B) πa²(1+2cosα/4sin²) C) πa²/4sin²α D) πa²/4cos²α 17 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) x-1/2=y-2/3=z-3/4 B) 2x+3y+z=0 C) x=y/3=z/2 D) -x-y+z=0 18 / 30 Hisoblang A) 1 B) √3 C) 2√3 D) 3√3 19 / 30 Hisoblang. A) 2/17 B) 17/34 C) 2/34 D) 15/34 20 / 30 sistemadan x+y+z ning qiymatini toping. A) 139/41 B) 150/41 C) 140/41 D) -139/41 21 / 30 AA1A2A3A4A5A6 muntazam oltiburchakli piramidaning hajmi ga va balandligi 2 ga teng bo‘lsa, u holda AA2A6 kesim yuzini toping. A) √5 B) 2 C) √15 D) 3 22 / 30 Soddalashtiring. A) 2018a/a+1 B) 2019 C) a+1 D) 2018 23 / 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) 412967 B) 2 C) 374389 D) aniqlab bo’lmaydi 24 / 30 Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping. A) 113° B) 100° C) 113,5° D) 112,5° 25 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) 1 B) e C) 3e D) √2e 26 / 30 Sin9x =4sin3x tenglamani yeching A) πn/3, n€Ζ B) π/3+πn, n€Ζ C) π/2+πn, n€Ζ D) πn, n€Ζ 27 / 30 Markazi nuqtada bo‘lgan aylanaga va urinmalar o‘tkazilgan bo’lib, va nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma va kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar uchburchakka ichki chizilgan aylana markazi bo‘lsa, u holda burchakni toping. A) 30° B) 72° C) 60° D) 90° 28 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,75 B) -0,25 C) -0,5 D) -1 29 / 30 a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega A) a=12 B) a ning bunday qiymati yo’q C) a≠12 D) a≠5 30 / 30 y= funktsiyaning aniqlanish sohasini toping. A) [2;∞) B) (-∞;2) C) (-∞;2)v(2;∞) D) (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
