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 Bir vaqtning o’zida 9,13, . . . ,405 va 15,21, . . . ,255 ketma–ketliklarning hadlari bo’lgan sonlarning eng kattasi va eng kichigining ayirmasini toping A) 228 B) 147 C) 150 D) 231 2 / 30 Agar f(x)=ax3-5x2+b va bo’lsa, a ni toping. A) 2 B) 3 C) 0 D) 1 3 / 30 Soddalashtiring. A) a+1 B) 2019 C) 2018 D) 2018a/a+1 4 / 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 5 / 30 Tomoni 25 ga diagonallaridan biri 4 ga teng bo’lgan rombning yuzini toping. A) 8√5 B) 24√5 C) 16 D) 32 6 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) (-∞;1]v[10;∞) B) (-∞;1] C) [10;∞) D) 1; 10 7 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (2;∞) B) (-∞;2) C) (-∞;2)v(2;∞) D) (∞;∞) 8 / 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) 52 C) 108 D) 54 9 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {-1;3} B) {2;4} C) {1;2} D) {-1;2} 10 / 30 Soddalashtiring. (0<m<7) A) m B) 7-2m C) 2m-7 D) 7 11 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) (-∞;6) B) 6 C) (2;∞) D) (-∞;6])v[6;∞) 12 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) 6,25% ga ortadi B) 2,5% ga ortadi C) o‘zgarmaydi D) 6,25% ga kamayadi 13 / 30 sistemada xy ning qiymatini toping. A) 80 B) 64 C) 60 D) 75 14 / 30 Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)? A) ⁴ᴷ⁺³√-√2k+1 B) ²ᴷ⁺⁴v2k+1/k²+1 C) (512-1/2⁻⁹)° D) 4k+1/1/8:7-1/56 15 / 30 (-3;4) nuqtaga absissa, ordinata o’qlariga va koordinata boshiga nisbatan simmetrik bo’lgan nuqtalarni tutashtirishdan hosil bo’lgan uchburchakning eng kata tomonini toping. A) 10 B) 24 C) 12 D) 14 16 / 30 O’qishni bilmaydigan bola alifbening A,A,A, N,N, S- 6 ta harflarini ixtiyoriy ravishda terib chiqadi. Bunda ANANAS so’zining hosil bo’lish ehtimolini toping. A) 1/60 B) 6/720 C) 5/24 D) 5/720 17 / 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) πa²/4cos²α B) πa²(1+2cosα/4sin²) C) πa²/4sin²α D) πa²(1-2sinα/4sin²) 18 / 30 Tenglamaning nechta ildizi bor? |x+1|=|2x-1| A) 3 B) 1 C) 4 D) 2 19 / 30 Teng yonli uchburchakning tomonlari 5, 5 va 6 ga teng. Bu uchburchakning bissektiritsalari va medianalari kesishgan nuqtalar A) 1/2 B) 1 C) 1/6 D) 1,2 20 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 3 B) 4 C) 3,5 D) 2 21 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -1 B) -0,75 C) -0,5 D) -0,25 22 / 30 Hisoblang: A) 2 B) 1 C) -1 D) 3 23 / 30 bo’lsa, ni x orqali ifodalang. A) 2/x B) 25/x C) x/25 D) 2-x 24 / 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 25 / 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) cheksiz ko’p C) 2 D) 0 26 / 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 27 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 10 B) 12 C) 11 D) √46 28 / 30 integralning qiymatini toping. A) 0 B) -π/2 C) π/2 D) π/4 29 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) e B) 1 C) √2e D) 3e 30 / 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) 3√2π/4 B) 64√2π/3 C) 8√2π/5 D) (5√3+3)π/3 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz