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. A) 2/17 B) 17/34 C) 15/34 D) 2/34 2 / 30 Tenglamaning ildizlari yig’indisi va ko’paytmasining yig’indisini toping. |x+1|.|x-4|=5 A) 6 B) -3 C) 9 D) -6 3 / 30 tenglamalar sistemasini yeching A) (4;3) B) (4;–3) C) (3;–4) D) (–4;3) 4 / 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 5 / 30 Radiusi 1 ga teng aylana uchta yoyga bo`lingan. Ularga mos markaziy burchaklar 1, 2 va 6 sonlariga proporsional. Yoylardan eng kattasining uzunligini toping. A) 2π/3 B) 3π/4 C) 4π/3 D) 3π/2 6 / 30 Hisoblang A) √3 B) 3√3 C) 2√3 D) 1 7 / 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) 8 B) 6 C) 7 D) 10 8 / 30 Rasmda ko‘rsatilgan ko‘pyoqlardan qaysi birida 4 ta yoq, 6 ta qirra bor? A) 2 B) 1, 3 C) 3 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) cheksiz ko’p B) 2 C) 0 D) 1 10 / 30 bo’lsa, ni x orqali ifodalang. A) 2/x B) 2-x C) x/25 D) 25/x 11 / 30 Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)? A) ²ᴷ⁺⁴v2k+1/k²+1 B) ⁴ᴷ⁺³√-√2k+1 C) (512-1/2⁻⁹)° D) 4k+1/1/8:7-1/56 12 / 30 Tenglamaning nechta ildizi bor? |x+1|=|2x-1| A) 4 B) 2 C) 1 D) 3 13 / 30 sistemadan x+y ning qiymatini toping. A) -12 B) 35/4 C) 6 D) 12 14 / 30 Markazi O nuqtada bo‘lgan aylanaga PA va PB urinmalar o‘tkazilgan bo’lib, A va B nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma PA va PB kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar XQ=YQ bo‘lsa, u holda PXY uchburchak qanday uchburchak bo‘ladi? A) muntazam uchburchak B) ixtiyoriy uchburchak C) teng yonli uchburchak D) to`g`ri burchakli uchburchak 15 / 30 Ag ar tgα=2 bo’lsa, u holda ni hisoblang A) -10/3 B) 10/3 C) 10/27 D) -10/27 16 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusiga teng bo’lgan uchburchak qanday uchburchak deyiladi? A) o’tkir burchakli uchburchak B) teng tomonli uchburchak C) to’gri burchakli uchburchak D) o’tmas burchakli uchburchak 17 / 30 y= funktsiyaning aniqlanish sohasini toping. A) (2;∞) B) (-∞;2)v(2;∞) C) [2;∞) D) (-∞;2) 18 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {1;2} B) {-1;2} C) {-1;3} D) {2;4} 19 / 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) x=y/3=z/2 C) -x-y+z=0 D) 2x+3y+z=0 20 / 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) 108 C) 54 D) 52 21 / 30 A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping. A) 1,5 B) √3/2 C) √5/2 D) 1 22 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) 2π/3 B) 2π C) π/2 D) π 23 / 30 To’rtburchakning uchi M (0, 4) ; N(-4,0) ; P(-3;2) uchlari berilgan. Agar bo’lsa, Q uchining koordinatalarini toping. A) (-7; 1) B) (7; 1) C) (-7;-1) D) (4; -3) 24 / 30 Tenglamani yeching. (x+2)+(x+4)+(x+6)+…+(x+100)=2800 A) 3 B) 7 C) 4 D) 5 25 / 30 A) √6 B) 0 C) 1 D) 2 26 / 30 Tenglamaning ildizlari yig`indisini toping. A) 6 B) 3 C) 4 D) 5 27 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 25 B) 28 C) 20 D) 30 28 / 30 Hisoblang: A) 3 B) 1 C) 2 D) -1 29 / 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 30 / 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) 10 C) 2 yoki 50 D) 50 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz