Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 123 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 2 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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 2 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (-∞;2)v(2;∞) B) (2;∞) C) (∞;∞) D) (-∞;2) 3 / 30 Hisoblang A) 2√3 B) 1 C) √3 D) 3√3 4 / 30 A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping. A) √5/2 B) 1,5 C) √3/2 D) 1 5 / 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) 6 / 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) 25 B) 30 C) 12 D) 15 7 / 30 sin2x-cos2x=1 tenglama [-π; 2π] oraliqda nechta ildizga ega? A) 10 B) 9 C) 7 D) 6 8 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 4 B) 5 C) 6 D) 8 9 / 30 Hisoblang. 11+192+1993+19994+199995+1999996+19999997+199999998+1999999999 A) 222220175 B) 22222222220 C) 2222222175 D) 222222222222 10 / 30 Tengsizlikni yeching. A) (0;+∞) B) 0 C) (-1/³ √2 ;0) D) to'g'ri javob yo'q 11 / 30 tenglamalar sistemasini yeching A) (4;–4) B) (-4;4) C) (4;4) D) (-4;-4) 12 / 30 Tomoni 12 ga teng bo`lgan teng tomonli uchburchakga ichki chizilgan aylana uzunligini toping. A) 12π B) 4√3π C) 2√3π D) 3π 13 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 3π B) 4π C) 2π/3 D) 2π 14 / 30 sonning oxirgi raqamini toping. A) 6 B) 2 C) 4 D) 8 15 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁴⁹⁵⁰ B) e⁻⁵⁰⁵⁰ C) e⁻⁴⁹⁵⁰ D) e⁵⁰⁵⁰ 16 / 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) 108 B) 48 C) 54 D) 52 17 / 30 Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping. A) 113° B) 112,5° C) 113,5° D) 100° 18 / 30 Soddalashtiring. A) a+1 B) 2019 C) 2018 D) 2018a/a+1 19 / 30 Hisoblang: A) -1 B) 2 C) 1 D) 3 20 / 30 Rasmda ko‘rsatilgan ko‘pyoqlardan qaysi birida 4 ta yoq, 6 ta qirra bor? A) 1, 2 B) 1, 3 C) 2 D) 3 21 / 30 x2-5|x|-6=0 tenglama ildizini toping A) ±6 B) ±;±6 C) -1;-6 D) -1;6 22 / 30 Agar f(x)=ax3-5x2+b va bo’lsa, a ni toping. A) 2 B) 1 C) 3 D) 0 23 / 30 AA1A2A3A4A5A6 muntazam oltiburchakli piramidaning hajmi ga va balandligi 2 ga teng bo‘lsa, u holda AA2A6 kesim yuzini toping. A) √15 B) 2 C) 3 D) √5 24 / 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 yoki 50 B) 50 C) 10 D) 2 yoki 50 25 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) 2,5% ga ortadi B) 6,25% ga ortadi C) 6,25% ga kamayadi D) o‘zgarmaydi 26 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 6 π B) 12π C) 7 π D) 10 π 27 / 30 ifodaning qiymatini toping. A) -0,5 B) 0 C) -2 D) 0,5 28 / 30 sistemadan x+y ning qiymatini toping. A) 12 B) -12 C) 35/4 D) 6 29 / 30 Sin9x =4sin3x tenglamani yeching A) πn, n€Ζ B) πn/3, n€Ζ C) π/3+πn, n€Ζ D) π/2+πn, n€Ζ 30 / 30 Tomoni 25 ga diagonallaridan biri 4 ga teng bo’lgan rombning yuzini toping. A) 32 B) 8√5 C) 24√5 D) 16 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
