Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 224 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 10 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Hisoblang A) √3 B) 1 C) 2√3 D) 3√3 2 / 30 Soddalashtiring. A) 2019 B) 2018a/a+1 C) 2018 D) a+1 3 / 30 Tengsizlikni yeching. A) (-3;2)v(2;4) B) (2;4) C) (-3;2) D) (-4;2)v(2;3) 4 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 16 B) 32 C) 20 D) 24 5 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) √3 B) 1 C) √2 D) 2 6 / 30 x2-5|x|-6=0 tenglama ildizini toping A) ±6 B) -1;-6 C) -1;6 D) ±;±6 7 / 30 Tengsizlik nechta butun yechimga ega? A) 4 B) 3 C) cheksiz ko’p D) 1 8 / 30 y= funksiyaning aniqlanish sohasini toping A) (2;∞) B) (-∞;0)v(2;∞) C) (-∞;0])v[2;∞) D) (0,2) 9 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) 2√2 B) √2 C) 4 D) 2 10 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 28 B) 25 C) 30 D) 20 11 / 30 |x2-5x-14|+20≥5|x+2|+4|x-7| tengsizlikni yeching. A) [-2;4]v{6} B) (-∞;-6]v{2}v[12;∞) C) [2;4]v{2}v[3;∞) D) [1;6] 12 / 30 Sin9x =4sin3x tenglamani yeching A) πn, n€Ζ B) π/3+πn, n€Ζ C) πn/3, n€Ζ D) π/2+πn, n€Ζ 13 / 30 Ushbu (y6+y3+1)(y3+1)(y3-1)-y6+y3+1 ifodani soddalashtish natijasida ko’phad hosil qilindi. Uning nechta hadi bor? A) 1 B) 3 C) 4 D) 2 14 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 18 B) 60 C) 120 D) 47 15 / 30 A) 0 B) 2 C) √6 D) 1 16 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 4 B) 2 C) 3,5 D) 3 17 / 30 Bankda qo`yilgan pul bir yildan kegin foydasi bilan 2600 so`m bo`ldi; Agar bank yillik 30% foyda to`lasa, boshida qancha pul qo`yilgan bo`ladi ? A) 1900 B) 2000 C) 2100 D) 2200 18 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) (-∞;1] B) 1; 10 C) (-∞;1]v[10;∞) D) [10;∞) 19 / 30 bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya. A) -4 B) -12 C) -14 D) -10 20 / 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(1-2sinα/4sin²) C) πa²/4cos²α D) πa²(1+2cosα/4sin²) 21 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) √46 B) 11 C) 10 D) 12 22 / 30 a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega A) a≠12 B) a≠5 C) a=12 D) a ning bunday qiymati yo’q 23 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 48√3 B) 52√3 C) 108 D) 54 24 / 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√3 D) 67,5 25 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 2, 3 B) 3, 4 C) 1, 3 D) 1,4 26 / 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/2 B) 3π/4 C) 4π/3 D) 2π/3 27 / 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) to`g`ri burchakli uchburchak B) muntazam uchburchak C) teng yonli uchburchak D) ixtiyoriy uchburchak 28 / 30 To’rtburchakli muntazam piramidaning yon qirrasidagi ikki yoqli burchak 120 ga teng. Diagonal kesimining yuzasi S ga teng bo’lsa, uning yon sirtini toping. A) 3S B) 2S C) 4S D) 0,5S 29 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) o’zaro perpendikulyar B) ayqash C) aniqlab bo’lmaydi D) o’zaro parallel 30 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,25 B) -0,5 C) -1 D) -0,75 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz