Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 141 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 7 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 x2-5|x|-6=0 tenglama ildizini toping A) -1;6 B) ±;±6 C) -1;-6 D) ±6 2 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) x=y/3=z/2 B) 2x+3y+z=0 C) x-1/2=y-2/3=z-3/4 D) -x-y+z=0 3 / 30 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakatning 200- metrida qanday tezlikka (m/s) erishadi? A) 36 B) 42 C) 32 D) 38 4 / 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) 5/720 C) 6/720 D) 5/24 5 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 7 π B) 6 π C) 10 π D) 12π 6 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,5 B) -0,25 C) -1 D) -0,75 7 / 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) 2 B) 1 C) cheksiz ko’p D) 0 8 / 30 sistemadan x+y ning qiymatini toping. A) 6 B) -12 C) 12 D) 35/4 9 / 30 y= funktsiyaning aniqlanish sohasini toping. A) [2;∞) B) (2;∞) C) (-∞;2)v(2;∞) D) (-∞;2) 10 / 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) 52 C) 108 D) 48 11 / 30 Tengsizlik nechta butun yechimga ega? A) 4 B) 1 C) 3 D) cheksiz ko’p 12 / 30 Soddalashtiring. A) a+1 B) 2018 C) 2018a/a+1 D) 2019 13 / 30 Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)? A) ⁴ᴷ⁺³√-√2k+1 B) ²ᴷ⁺⁴v2k+1/k²+1 C) 4k+1/1/8:7-1/56 D) (512-1/2⁻⁹)° 14 / 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) 3π/4 B) 3π/2 C) 4π/3 D) 2π/3 15 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusidan katta bo’lgan uchburchaklar sonini toping. A) 1 B) cheksiz ko’p C) 2019 D) 0 16 / 30 To’g’ri to’rtburchakning perimetri 50 ga teng. Bir tomoni boshqa tomonidan 5 ga ko’p. To’g’ri to’rtburchakning yuzini toping. A) 225 B) 150 C) 60 D) 50 17 / 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 18 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 2 B) 1 C) 5 D) 3 19 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) o‘zgarmaydi B) 2,5% ga ortadi C) 6,25% ga kamayadi D) 6,25% ga ortadi 20 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln(|x+4+|x-1|)+c B) ln|x+4/x-1|+c C) ln(|x+4*|x-1|)+c D) ln|x-1/x+4|+c 21 / 30 n ning qanday qiymatida ushbu 81 .82 .83 .….8n=51222 tenglik o’rinli bo’ladi? A) 12 B) 11 C) 14 D) 10 22 / 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(cos2x) C) 4sin2x*cos(2cos2x) D) -4sin2x*cos(2cos2x) 23 / 30 sistemadan x+y+z ning qiymatini toping. A) 140/41 B) 139/41 C) -139/41 D) 150/41 24 / 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) 14 B) 24 C) 10 D) 12 25 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 6 B) 8 C) 5 D) 4 26 / 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) 7 B) 8 C) 6 D) 10 27 / 30 A) 3 B) 2 C) 1 D) 5 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(1+2cosα/4sin²) C) πa²(1-2sinα/4sin²) D) πa²/4cos²α 29 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) 2 B) √3 C) 1 D) √2 30 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 3 B) 5 C) 6 D) 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
