Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 148 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Hisoblang A) 2√3 B) √3 C) 3√3 D) 1 2 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 60 B) 120 C) 47 D) 18 3 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 3π B) 4π C) 2π D) 2π/3 4 / 30 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakatning 200- metrida qanday tezlikka (m/s) erishadi? A) 32 B) 38 C) 36 D) 42 5 / 30 Tenglamaning ildizlari yig`indisini toping. A) 5 B) 3 C) 4 D) 6 6 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) 6 B) (-∞;6) C) (2;∞) D) (-∞;6])v[6;∞) 7 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) e B) 3e C) 1 D) √2e 8 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {2;4} B) {1;2} C) {-1;2} D) {-1;3} 9 / 30 Agar geometrik progressiyaning ketma–ket dastlabki uchta hadining yig’indisi 62 ga, ularning o’nli logarifmlari yig’indisi 3 ga teng bo’lsa, shu geometrik progressiyaning birinchi hadini toping. A) 2 yoki 50 B) 10 C) 10 yoki 50 D) 50 10 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 6 B) 4 C) 5 D) 8 11 / 30 200 kishidan iborat turistlar guruxida 140 kishi ingliz tilini, 90 kishi nemis tilini va 46 kishi ikkala tilni biladi. Ikkala tilni xam bilmaydigan turistlar necha foizni tashkil qiladi. A) 12 B) 4 C) 8 D) 16 12 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 6 B) 5 C) 3 D) 4 13 / 30 Tenglamani yeching. (x+2)+(x+4)+(x+6)+…+(x+100)=2800 A) 5 B) 4 C) 3 D) 7 14 / 30 Tenglamaning ildizlari yig’indisi va ko’paytmasining yig’indisini toping. |x+1|.|x-4|=5 A) -6 B) -3 C) 6 D) 9 15 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 25 B) 28 C) 30 D) 20 16 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 2, 3 B) 1, 3 C) 1,4 D) 3, 4 17 / 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) 150 B) 147 C) 231 D) 228 18 / 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) 4π/3 B) 2π/3 C) 3π/4 D) 3π/2 19 / 30 Soddalashtiring. (0<m<7) A) m B) 7 C) 7-2m D) 2m-7 20 / 30 tenglamalar sistemasini yeching A) (4;–4) B) (-4;-4) C) (-4;4) D) (4;4) 21 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln(|x+4+|x-1|)+c B) ln|x-1/x+4|+c C) ln|x+4/x-1|+c D) ln(|x+4*|x-1|)+c 22 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) -x-y+z=0 B) x=y/3=z/2 C) 2x+3y+z=0 D) x-1/2=y-2/3=z-3/4 23 / 30 Anvar tub son o’yladi va o’ylagan sonini 5 ga ko’paytirib, 8 ni ayirgan edi, yana tub son hosil bo’ldi. Anvar qanday son o’ylagan? A) 412967 B) aniqlab bo’lmaydi C) 374389 D) 2 24 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 10 B) 12 C) √46 D) 11 25 / 30 Sin9x =4sin3x tenglamani yeching A) π/2+πn, n€Ζ B) π/3+πn, n€Ζ C) πn, n€Ζ D) πn/3, n€Ζ 26 / 30 A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping. A) 1,5 B) 1 C) √5/2 D) √3/2 27 / 30 sistemada xy ning qiymatini toping. A) 75 B) 80 C) 64 D) 60 28 / 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√3+1/5 B) 2√2/3 C) 5√3+3/3 D) 3√2+2/4 29 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 5 ga, bir katetining gipotenuzadagi proyeksiyasi 1,6 ga teng. Ikkinchi katetning kvadratini toping. A) 16 B) 14 C) 18 D) 17 30 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) 2 B) 1 C) √3 D) √2 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
