Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 226 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Agar arctga+ arctgb + arctgc= bo’lsa, a+b+c ni toping. A) abc B) 1 C) ab D) ab/c 2 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) o’zaro parallel B) o’zaro perpendikulyar C) aniqlab bo’lmaydi D) ayqash 3 / 30 Tenglamaning ildizlari yig`indisini toping. A) 5 B) 3 C) 6 D) 4 4 / 30 y= funktsiyaning aniqlanish sohasini toping. A) (-∞;2) B) (-∞;2)v(2;∞) C) [2;∞) D) (2;∞) 5 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln|x+4/x-1|+c B) ln(|x+4+|x-1|)+c C) ln|x-1/x+4|+c D) ln(|x+4*|x-1|)+c 6 / 30 Ag ar tgα=2 bo’lsa, u holda ni hisoblang A) 10/3 B) -10/27 C) 10/27 D) -10/3 7 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 10 π B) 12π C) 7 π D) 6 π 8 / 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) 9 / 30 Rasmda ko‘rsatilgan ko‘pyoqlardan qaysi birida 4 ta yoq, 6 ta qirra bor? A) 1, 2 B) 3 C) 2 D) 1, 3 10 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 1, 3 B) 2, 3 C) 1,4 D) 3, 4 11 / 30 Sin9x =4sin3x tenglamani yeching A) πn/3, n€Ζ B) πn, n€Ζ C) π/3+πn, n€Ζ D) π/2+πn, n€Ζ 12 / 30 Tenglamani yeching. (x+2)+(x+4)+(x+6)+…+(x+100)=2800 A) 5 B) 3 C) 4 D) 7 13 / 30 Hisoblang A) 0 B) -1 C) 1 D) 2 14 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 3 B) 5 C) 4 D) 6 15 / 30 sistemadan x+y ning qiymatini toping. A) 12 B) 6 C) -12 D) 35/4 16 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 48√3 B) 135√3/4 C) 67,5 D) 67,5√3 17 / 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) 0 D) 2 18 / 30 Hisoblang. A) 2/34 B) 17/34 C) 15/34 D) 2/17 19 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 18 B) 120 C) 60 D) 47 20 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) juft funksiya B) toq funksiya C) juft ham emas, toq ham emas funksiya D) bunday funksiya mavjud emas 21 / 30 Bekzodda 50 so’m va Sobirda 70 so’m pul bor edi. Anvar Bekzodga o’z pulining 10 foizini bergandan so’ng, Bekzod Sobirga pulining yarmini berdi. So’ng Sobir Anvarga pulining 10 foizini berdi. Anvar o’zidagi pullarini hisoblab, pullari dastlabki holdagi puli bilan teng ekanligini bildi. Anvarda qancha pul bo’lgan? A) 375 B) 127 C) 400 D) 100 22 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 5 B) 6 C) 4 D) 7 23 / 30 tenglamalar sistemasini yeching A) (4;–4) B) (-4;-4) C) (4;4) D) (-4;4) 24 / 30 Ostki asosining yuzi 20π va ustki asosining yuzi 10π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda kesik konus hajmining shar hajmiga nisbatini toping. A) 2√2/3 B) 3√2+2/4 C) 5√3+3/3 D) 3√3+1/5 25 / 30 Quyidagi va to’g’ri chiziqlarning o’zaro holatini aniqlang. A) ayqash to’gri chiziqlar B) o’zaro perpendikulyar C) o’zaro parallel D) o’zaro kesishadi 26 / 30 y= funksiyaning aniqlanish sohasini toping A) (2;∞) B) (-∞;0)v(2;∞) C) (-∞;0])v[2;∞) D) (0,2) 27 / 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) 147 B) 150 C) 231 D) 228 28 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) ln(lg x) B) ln(ln x) C) lg(lg x) D) lg(ln x) 29 / 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 kamayadi B) 2,5% ga ortadi C) 6,25% ga ortadi D) o‘zgarmaydi 30 / 30 integralning qiymatini toping. A) π/2 B) -π/2 C) 0 D) π/4 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz