Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 194 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Hisoblang. 11+192+1993+19994+199995+1999996+19999997+199999998+1999999999 A) 2222222175 B) 222220175 C) 222222222222 D) 22222222220 2 / 30 Agar sinx+cosx=a bo’lsa, ning qiymatini toping. A) 2/5 B) 1/3 C) 1/2 D) -1/2 3 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) 2π B) π/2 C) 2π/3 D) π 4 / 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) aniqlab bo’lmaydi B) 412967 C) 2 D) 374389 5 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 32 B) 20 C) 24 D) 16 6 / 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) 54 B) 48 C) 108 D) 52 7 / 30 Quyidagi va to’g’ri chiziqlarning o’zaro holatini aniqlang. A) o’zaro parallel B) ayqash to’gri chiziqlar C) o’zaro perpendikulyar D) o’zaro kesishadi 8 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {1;2} B) {-1;3} C) {2;4} D) {-1;2} 9 / 30 ABCD to’gri to’rtburchak ichidan olingan O nuqtadan A, B, C, D uchlarigacha bo’lgan masofalar mos ravishda 1; 2; 1,5; 1,2 ga teng bo’ladigan barcha to’g’ri to’rtburchaklar sonini toping A) 1 B) 0 C) 2 D) cheksiz ko’p 10 / 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° 11 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 67,5√3 B) 48√3 C) 135√3/4 D) 67,5 12 / 30 tenglamani yeching. A) 0 B) 2017 C) 2018 D) 2019 13 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) ayqash B) o’zaro parallel C) aniqlab bo’lmaydi D) o’zaro perpendikulyar 14 / 30 Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)? A) 4k+1/1/8:7-1/56 B) ²ᴷ⁺⁴v2k+1/k²+1 C) (512-1/2⁻⁹)° D) ⁴ᴷ⁺³√-√2k+1 15 / 30 Soddalashtiring. (0<m<7) A) 2m-7 B) 7 C) 7-2m D) m 16 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁴⁹⁵⁰ B) e⁻⁴⁹⁵⁰ C) e⁵⁰⁵⁰ D) e⁻⁵⁰⁵⁰ 17 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 7 B) 6 C) 4 D) 5 18 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) (-∞;1]v[10;∞) B) (-∞;1] C) [10;∞) D) 1; 10 19 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 8 B) 12 C) 4 D) 6 20 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) bunday funksiya mavjud emas B) juft ham emas, toq ham emas funksiya C) toq funksiya D) juft funksiya 21 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 4 B) 2 C) 3,5 D) 3 22 / 30 integralning qiymatini toping. A) -π/2 B) π/4 C) π/2 D) 0 23 / 30 Tengsizlikni yeching. A) (-3;2)v(2;4) B) (2;4) C) (-4;2)v(2;3) D) (-3;2) 24 / 30 Agar arctga+ arctgb + arctgc= bo’lsa, a+b+c ni toping. A) ab/c B) 1 C) abc D) ab 25 / 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) 72π B) 96π C) 56π D) 100π 26 / 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 27 / 30 Hisoblang A) √3 B) 2 C) √3/2 D) 1/2 28 / 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) πa²/4sin²α B) πa²/4cos²α C) πa²(1-2sinα/4sin²) D) πa²(1+2cosα/4sin²) 29 / 30 Agar f(x)=sin2x va g(x)=cos2x bo’lsa, u holda f(g(x)) funksiyaning hosilasini toping. A) 4sin2x*cos(cos2x) B) -4sin2x*cos(2cos2x) C) 4sin2x*cos(2cos2x) D) -4sin2x*cos(cos2x) 30 / 30 x(t)=t2+6t+5 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakat boshlangandan necha sekund o’tgach boshlang’ich nuqtaga nisbatan 77 metr masofaga siljiydi? A) 10 B) 7 C) 6 D) 8 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
