Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 210 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 7 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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) 6 B) 10 C) 7 D) 8 2 / 30 Ag ar tgα=2 bo’lsa, u holda ni hisoblang A) 10/3 B) -10/27 C) 10/27 D) -10/3 3 / 30 Tenglamaning nechta ildizi bor? |x+1|=|2x-1| A) 2 B) 1 C) 3 D) 4 4 / 30 Hisoblang: A) 3 B) 1 C) -1 D) 2 5 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 25 B) 30 C) 28 D) 20 6 / 30 funktsiyaning aniqlanish sohasini toping. A) (1;2) B) [1;2] C) (-∞;1]v[2;∞) D) (-∞;1)v(2;∞) 7 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 2 B) 5 C) 1 D) 3 8 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 10 π B) 6 π C) 12π D) 7 π 9 / 30 Soddalashtiring. A) 2018a/a+1 B) 2019 C) 2018 D) a+1 10 / 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) (5√3+3)π/3 B) 64√2π/3 C) 3√2π/4 D) 8√2π/5 11 / 30 y= funktsiyaning aniqlanish sohasini toping. A) (2;∞) B) (-∞;2) C) [2;∞) D) (-∞;2)v(2;∞) 12 / 30 ABC muntazam uchburchak ichidan ixtiyoriy P nuqta olinib, undan BC, CA va AB tomonlarga mos ravishda PD, PE va PF perpendikulyarlar tushirilgan bo’lsa,ni toping. A) 1/√3 B) 1 C) 0,5 D) 1/√2 13 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 47 B) 60 C) 120 D) 18 14 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) π B) 2π C) 2π/3 D) π/2 15 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) o’zaro parallel B) ayqash C) o’zaro perpendikulyar D) aniqlab bo’lmaydi 16 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 4 B) 12 C) 8 D) 6 17 / 30 Hisoblang A) 1 B) 0 C) 2 D) -1 18 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 54 B) 52√3 C) 108 D) 48√3 19 / 30 Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping. A) 112,5° B) 113° C) 113,5° D) 100° 20 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 2 B) 3,5 C) 4 D) 3 21 / 30 Tengsizlik nechta butun yechimga ega? A) 4 B) cheksiz ko’p C) 3 D) 1 22 / 30 A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping. A) 1 B) 1,5 C) √5/2 D) √3/2 23 / 30 Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)? A) (512-1/2⁻⁹)° B) 4k+1/1/8:7-1/56 C) ²ᴷ⁺⁴v2k+1/k²+1 D) ⁴ᴷ⁺³√-√2k+1 24 / 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) 52 D) 54 25 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) -x-y+z=0 B) x-1/2=y-2/3=z-3/4 C) x=y/3=z/2 D) 2x+3y+z=0 26 / 30 a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega A) a≠12 B) a ning bunday qiymati yo’q C) a≠5 D) a=12 27 / 30 |x2-5x-14|+20≥5|x+2|+4|x-7| tengsizlikni yeching. A) (-∞;-6]v{2}v[12;∞) B) [-2;4]v{6} C) [1;6] D) [2;4]v{2}v[3;∞) 28 / 30 Yuzasi 10 ga teng bo’lgan kvadratning ketma-ket ikki uchidan o’tuvchi aylana chizilgan. Uchinchi uchidan aylanaga urunma o’tkazilgan. Urunma tomondan ikki marta katta bo’lsa, aylana radiusini toping. A) 6 B) 4 C) 10 D) 5 29 / 30 y= funksiyaning aniqlanish sohasini toping A) (-∞;0])v[2;∞) B) (2;∞) C) (0,2) D) (-∞;0)v(2;∞) 30 / 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(1+2cosα/4sin²) B) πa²/4cos²α C) πa²/4sin²α D) πa²(1-2sinα/4sin²) 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz