Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 188 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 3 Vaqtingiz tugadi! Created by InfoMaster Matematika abituriyentlar uchun №1 1 / 30 x2-5|x|-6=0 tenglama ildizini toping A) ±6 B) -1;-6 C) ±;±6 D) -1;6 2 / 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) 2 D) 1 3 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 6 B) 8 C) 12 D) 4 4 / 30 Quyidagi va to’g’ri chiziqlarning o’zaro holatini aniqlang. A) o’zaro perpendikulyar B) o’zaro parallel C) o’zaro kesishadi D) ayqash to’gri chiziqlar 5 / 30 |x2-5x-14|+20≥5|x+2|+4|x-7| tengsizlikni yeching. A) [2;4]v{2}v[3;∞) B) [1;6] C) (-∞;-6]v{2}v[12;∞) D) [-2;4]v{6} 6 / 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) πa²/4sin²α C) πa²(1-2sinα/4sin²) D) πa²/4cos²α 7 / 30 bo’lsa, ni x orqali ifodalang. A) 2/x B) x/25 C) 25/x D) 2-x 8 / 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) 3√2+2/4 B) 2√2/3 C) 3√3+1/5 D) 5√3+3/3 9 / 30 Agar bank qo`yilgan pulga 40% yillik foyda bersa, qo`yilgan 5000 so`m pul bir yildan keyin qancha bo`ladi ? A) 7000 B) 6200 C) 6900 D) 7200 10 / 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) 10 B) 7 C) 6 D) 8 11 / 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) 67,5√3 C) 135√3/4 D) 67,5 12 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) x=y/3=z/2 B) x-1/2=y-2/3=z-3/4 C) 2x+3y+z=0 D) -x-y+z=0 13 / 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 14 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁵⁰⁵⁰ B) e⁻⁴⁹⁵⁰ C) e⁴⁹⁵⁰ D) e⁻⁵⁰⁵⁰ 15 / 30 Hisoblang. 11+192+1993+19994+199995+1999996+19999997+199999998+1999999999 A) 2222222175 B) 222222222222 C) 222220175 D) 22222222220 16 / 30 Hisoblang A) 3√3 B) 2√3 C) 1 D) √3 17 / 30 tenglamani yeching. A) 2018 B) 0 C) 2019 D) 2017 18 / 30 Markazi nuqtada bo‘lgan aylanaga va urinmalar o‘tkazilgan bo’lib, va nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma va kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar uchburchakning perimetri 48 va aylana radiusi 7 ga teng bo‘lsa, u holda kesma uzunligini toping A) 12 B) 25 C) 30 D) 15 19 / 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) 50 B) 225 C) 60 D) 150 20 / 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} 21 / 30 a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega A) a≠12 B) a=12 C) a ning bunday qiymati yo’q D) a≠5 22 / 30 Tenglamaning ildizlari yig`indisini toping. A) 5 B) 3 C) 6 D) 4 23 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 5 ga, bir katetining gipotenuzadagi proyeksiyasi 1,6 ga teng. Ikkinchi katetning kvadratini toping. A) 14 B) 17 C) 16 D) 18 24 / 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 25 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) 2√2 B) 2 C) √2 D) 4 26 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 4 B) 8 C) 6 D) 5 27 / 30 ifodaning qiymatini toping. A) -0,5 B) 0 C) -2 D) 0,5 28 / 30 sonning oxirgi raqamini toping. A) 4 B) 8 C) 2 D) 6 29 / 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) 6,25% ga ortadi C) 2,5% ga ortadi D) 6,25% ga kamayadi 30 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) b>c>a>d>e B) e>b>a>d>c C) b>a>c>d>e D) a>b>c>d>e 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
