Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 131 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 4 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {-1;3} B) {1;2} C) {-1;2} D) {2;4} 2 / 30 bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya. A) -12 B) -4 C) -10 D) -14 3 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 5 B) 6 C) 7 D) 4 4 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 2, 3 B) 1,4 C) 3, 4 D) 1, 3 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 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) 3√3 B) 2√5 C) 2√3 D) 4 7 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) lg(ln x) B) ln(lg x) C) ln(ln x) D) lg(lg x) 8 / 30 Hisoblang A) 0 B) 2 C) 1 D) -1 9 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) o’zaro parallel B) ayqash C) o’zaro perpendikulyar D) aniqlab bo’lmaydi 10 / 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) 60 C) 225 D) 150 11 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) π/2 B) π C) 2π D) 2π/3 12 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) 6 B) (-∞;6) C) (-∞;6])v[6;∞) D) (2;∞) 13 / 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) 374389 B) aniqlab bo’lmaydi C) 412967 D) 2 14 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 6 B) 4 C) 3 D) 5 15 / 30 Tengsizlikni yeching. A) (-1/³ √2 ;0) B) to'g'ri javob yo'q C) 0 D) (0;+∞) 16 / 30 Quyidagi va to’g’ri chiziqlarning o’zaro holatini aniqlang. A) o’zaro parallel B) o’zaro perpendikulyar C) o’zaro kesishadi D) ayqash to’gri chiziqlar 17 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (-∞;2)v(2;∞) B) (∞;∞) C) (-∞;2) D) (2;∞) 18 / 30 Rombning yuzi 96 ga, dioganallaridan biri 16 ga teng. Romb tomonini toping. A) 11 B) 9 C) 10 D) 12 19 / 30 bo’lsa, ni x orqali ifodalang. A) 25/x B) 2-x C) x/25 D) 2/x 20 / 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) 15 B) 30 C) 25 D) 12 21 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) 6,25% ga ortadi B) 2,5% ga ortadi C) o‘zgarmaydi D) 6,25% ga kamayadi 22 / 30 Agar sinx+cosx=a bo’lsa, ning qiymatini toping. A) 1/3 B) -1/2 C) 1/2 D) 2/5 23 / 30 Tengsizlikni yeching. A) (2;4) B) (-3;2) C) (-4;2)v(2;3) D) (-3;2)v(2;4) 24 / 30 tenglamalar sistemasini yeching A) (-4;4) B) (-4;-4) C) (4;4) D) (4;–4) 25 / 30 Ifodani soddalashtiring. 2 cos55o.cos40o.sin55o+cos110o.sin40o A) 2 B) 0,5 C) 0 D) 1 26 / 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) 4 C) 5 D) 10 27 / 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) 36 C) 54 D) 50 28 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 48√3 B) 52√3 C) 108 D) 54 29 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) (-∞;1]v[10;∞) B) (-∞;1] C) [10;∞) D) 1; 10 30 / 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) 1 D) 3 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
