Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 198 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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) 228 B) 150 C) 147 D) 231 2 / 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 ning bunday qiymati yo’q D) a≠12 3 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 52√3 B) 54 C) 108 D) 48√3 4 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) 2 B) 4 C) √2 D) 2√2 5 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 4 B) 5 C) 6 D) 3 6 / 30 ABCD to’gri to’rtburchak ichidan olingan O nuqtadan A, B, C, D uchlarigacha bo’lgan masofalar mos ravishda 3, 4, 5, 6 ga teng bo’lsa, u holda AB tomon uzunligini toping. A) √7 B) bunday to’gri to’rtburchak mavjud emas C) 2 D) √37 7 / 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) 8√2π/5 B) (5√3+3)π/3 C) 3√2π/4 D) 64√2π/3 8 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusidan katta bo’lgan uchburchaklar sonini toping. A) 2019 B) cheksiz ko’p C) 0 D) 1 9 / 30 Ushbu (y6+y3+1)(y3+1)(y3-1)-y6+y3+1 ifodani soddalashtish natijasida ko’phad hosil qilindi. Uning nechta hadi bor? A) 4 B) 2 C) 3 D) 1 10 / 30 Agar arctga+ arctgb + arctgc= bo’lsa, a+b+c ni toping. A) ab/c B) abc C) 1 D) ab 11 / 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) 5 C) 10 D) 4 12 / 30 Radiusi 25 bo’lgan doirada 48 ga teng vatar o’tkazilgan. Doira markazidan shu vatargacha masofani toping. A) 8 B) 7 C) 9 D) 10 13 / 30 funktsiyaning aniqlanish sohasini toping. A) [1;2] B) (-∞;1]v[2;∞) C) (1;2) D) (-∞;1)v(2;∞) 14 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁻⁴⁹⁵⁰ B) e⁴⁹⁵⁰ C) e⁵⁰⁵⁰ D) e⁻⁵⁰⁵⁰ 15 / 30 tenglamalar sistemasini yeching A) (3;–4) B) (4;–3) C) (4;3) D) (–4;3) 16 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 20 B) 25 C) 28 D) 30 17 / 30 sistemadan x+y+z ning qiymatini toping. A) 139/41 B) 140/41 C) -139/41 D) 150/41 18 / 30 Bekzodda 50 so’m va Sobirda 70 so’m pul bor edi. Anvar Bekzodga o’z pulining 10 foizini bergandan so’ng, Bekzod Sobirga pulining yarmini berdi. So’ng Sobir Anvarga pulining 10 foizini berdi. Anvar o’zidagi pullarini hisoblab, pullari dastlabki holdagi puli bilan teng ekanligini bildi. Anvarda qancha pul bo’lgan? A) 127 B) 100 C) 375 D) 400 19 / 30 Tomoni 12 ga teng bo`lgan teng tomonli uchburchakga ichki chizilgan aylana uzunligini toping. A) 2√3π B) 12π C) 3π D) 4√3π 20 / 30 Agar f(x)=sin2x va g(x)=cos2x bo’lsa, u holda f(g(x)) funksiyaning hosilasini toping. A) -4sin2x*cos(2cos2x) B) 4sin2x*cos(cos2x) C) 4sin2x*cos(2cos2x) D) -4sin2x*cos(cos2x) 21 / 30 A) √6 B) 1 C) 0 D) 2 22 / 30 Agar sinx+cosx=a bo’lsa, ning qiymatini toping. A) -1/2 B) 1/2 C) 1/3 D) 2/5 23 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) lg(lg x) B) ln(ln x) C) lg(ln x) D) ln(lg x) 24 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusiga teng bo’lgan uchburchak qanday uchburchak deyiladi? A) to’gri burchakli uchburchak B) teng tomonli uchburchak C) o’tmas burchakli uchburchak D) o’tkir burchakli uchburchak 25 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 4π B) 3π C) 2π D) 2π/3 26 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,75 B) -0,5 C) -0,25 D) -1 27 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 20 B) 32 C) 24 D) 16 28 / 30 A) 2 B) 3 C) 5 D) 1 29 / 30 Hisoblang. 11+192+1993+19994+199995+1999996+19999997+199999998+1999999999 A) 22222222220 B) 2222222175 C) 222220175 D) 222222222222 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) b>a>c>d>e C) a>b>c>d>e D) e>b>a>d>c 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
