Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 154 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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) 30 B) 15 C) 12 D) 25 2 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 3π B) 2π C) 2π/3 D) 4π 3 / 30 sistemadan x+y+z ning qiymatini toping. A) 139/41 B) -139/41 C) 150/41 D) 140/41 4 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) 2,5% ga ortadi B) 6,25% ga ortadi C) 6,25% ga kamayadi D) o‘zgarmaydi 5 / 30 Agar f(x)=sin2x va g(x)=cos2x bo’lsa, u holda f(g(x)) funksiyaning hosilasini toping. A) -4sin2x*cos(cos2x) B) 4sin2x*cos(2cos2x) C) 4sin2x*cos(cos2x) D) -4sin2x*cos(2cos2x) 6 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) (-∞;6) B) (2;∞) C) (-∞;6])v[6;∞) D) 6 7 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) √2e B) 3e C) 1 D) e 8 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) π/2 B) 2π C) π D) 2π/3 9 / 30 Tengsizlik nechta butun yechimga ega? A) 4 B) 1 C) 3 D) cheksiz ko’p 10 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) lg(ln x) B) ln(ln x) C) lg(lg x) D) ln(lg x) 11 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 5 B) 4 C) 6 D) 7 12 / 30 bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya. A) -10 B) -4 C) -14 D) -12 13 / 30 Radiusi 25 bo’lgan doirada 48 ga teng vatar o’tkazilgan. Doira markazidan shu vatargacha masofani toping. A) 9 B) 10 C) 8 D) 7 14 / 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) 1 D) 2 15 / 30 Soddalashtiring. A) 2018 B) a+1 C) 2018a/a+1 D) 2019 16 / 30 Tenglamaning ildizlari yig’indisi va ko’paytmasining yig’indisini toping. |x+1|.|x-4|=5 A) 6 B) -3 C) -6 D) 9 17 / 30 sistemadan x+y ning qiymatini toping. A) 35/4 B) 12 C) 6 D) -12 18 / 30 Quyidagi va to’g’ri chiziqlarning o’zaro holatini aniqlang. A) o’zaro kesishadi B) o’zaro parallel C) o’zaro perpendikulyar D) ayqash to’gri chiziqlar 19 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁵⁰⁵⁰ B) e⁻⁴⁹⁵⁰ C) e⁴⁹⁵⁰ D) e⁻⁵⁰⁵⁰ 20 / 30 Tenglamaning nechta ildizi bor? |x+1|=|2x-1| A) 4 B) 1 C) 3 D) 2 21 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 6 B) 3 C) 4 D) 5 22 / 30 Ifodani soddalashtiring. 2 cos55o.cos40o.sin55o+cos110o.sin40o A) 0 B) 1 C) 0,5 D) 2 23 / 30 Tenglamani yeching. (x+2)+(x+4)+(x+6)+…+(x+100)=2800 A) 7 B) 4 C) 5 D) 3 24 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) b>a>c>d>e B) e>b>a>d>c C) b>c>a>d>e D) a>b>c>d>e 25 / 30 ifodaning qiymatini toping. A) -2 B) -0,5 C) 0 D) 0,5 26 / 30 tenglamalar sistemasini yeching A) (-4;4) B) (4;4) C) (4;–4) D) (-4;-4) 27 / 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) 228 D) 231 28 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (2;∞) B) (-∞;2) C) (-∞;2)v(2;∞) D) (∞;∞) 29 / 30 tenglamalar sistemasini yeching A) (3;–4) B) (–4;3) C) (4;3) D) (4;–3) 30 / 30 sin2x-cos2x=1 tenglama [-π; 2π] oraliqda nechta ildizga ega? A) 9 B) 6 C) 7 D) 10 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
