Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 192 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 4 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 x2-5|x|-6=0 tenglama ildizini toping A) ±6 B) -1;6 C) -1;-6 D) ±;±6 2 / 30 sin2x-cos2x=1 tenglama [-π; 2π] oraliqda nechta ildizga ega? A) 6 B) 7 C) 10 D) 9 3 / 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) [2;4]v{2}v[3;∞) D) [1;6] 4 / 30 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakatning 200- metrida qanday tezlikka (m/s) erishadi? A) 38 B) 32 C) 42 D) 36 5 / 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) a>b>c>d>e D) b>a>c>d>e 6 / 30 Tomoni 12 ga teng bo`lgan teng tomonli uchburchakga ichki chizilgan aylana uzunligini toping. A) 4√3π B) 12π C) 2√3π D) 3π 7 / 30 Konsert zalining birinchi qatorida 40 ta o’rindiq bor. Har bir keyingi qatordagi o’rindiqlar soni oldingi qatordan 4 ga ko’p. Agar konsert zalida jami 40 ta qator bo’lsa, u holda shu zaldagi barcha o’rindiqlar sonini toping. A) 4760 B) 4716 C) 4680 D) 4720 8 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 8 B) 12 C) 4 D) 6 9 / 30 y=cos(2sinx) funksiyaning qiymatlar sohasini toping. A) [0;1] B) [0;cos2] C) [-1;1] D) [cos2;1] 10 / 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) 24 B) 14 C) 12 D) 10 11 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) x-1/2=y-2/3=z-3/4 B) x=y/3=z/2 C) 2x+3y+z=0 D) -x-y+z=0 12 / 30 Agar sinx+cosx=a bo’lsa, ning qiymatini toping. A) -1/2 B) 2/5 C) 1/3 D) 1/2 13 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 5 B) 2 C) 1 D) 3 14 / 30 Soddalashtiring. A) a+1 B) 2019 C) 2018 D) 2018a/a+1 15 / 30 Hisoblang A) 3√3 B) √3 C) 1 D) 2√3 16 / 30 Tengsizlik nechta butun yechimga ega? A) 1 B) 4 C) 3 D) cheksiz ko’p 17 / 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 18 / 30 A) 2 B) 1 C) 0 D) √6 19 / 30 Ag ar tgα=2 bo’lsa, u holda ni hisoblang A) 10/27 B) 10/3 C) -10/3 D) -10/27 20 / 30 Hisoblang. 11+192+1993+19994+199995+1999996+19999997+199999998+1999999999 A) 222222222222 B) 22222222220 C) 222220175 D) 2222222175 21 / 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) 231 C) 150 D) 228 22 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁻⁴⁹⁵⁰ B) e⁴⁹⁵⁰ C) e⁵⁰⁵⁰ D) e⁻⁵⁰⁵⁰ 23 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 2π/3 B) 4π C) 2π D) 3π 24 / 30 Tenglamaning ildizlari yig’indisi va ko’paytmasining yig’indisini toping. |x+1|.|x-4|=5 A) -3 B) 6 C) 9 D) -6 25 / 30 Muntazam uchburchakli piramidaning yon qirrasi asos tekisligi bilan 45o li burchak tashkil etgan bo‘lsa, u holda piramidaning yon sirti yuzining uning asosi yuziga nisbatini toping. A) 2√5 B) 2√3 C) 4 D) 3√3 26 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 4 B) 5 C) 6 D) 7 27 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 1,4 B) 3, 4 C) 2, 3 D) 1, 3 28 / 30 Hisoblang A) 1 B) 0 C) -1 D) 2 29 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 18 B) 47 C) 120 D) 60 30 / 30 x(t)=t2+7t-6 qonuniyat bo’yicha harakatlanayotgan moddiy nuqtaning tezligi harakat boshlangandan necha sekund o’tgach 87 m/s ga teng bo’ladi? A) 40 B) 54 C) 50 D) 36 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz