Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 141 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 7 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) lg(ln x) B) lg(lg x) C) ln(ln x) D) ln(lg x) 2 / 30 bo’lsa, ni x orqali ifodalang. A) 2-x B) x/25 C) 25/x D) 2/x 3 / 30 Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping. A) 113° B) 112,5° C) 113,5° D) 100° 4 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 67,5√3 B) 135√3/4 C) 67,5 D) 48√3 5 / 30 Agar bank qo’yilgan pulga 40% yillik bersa, qo’yilgan 4500 so’m pul bir yildan so’ng qancha bo’ladi? A) 6100 B) 6300 C) 6200 D) 6000 6 / 30 Tenglamaning nechta ildizi bor? |x+1|=|2x-1| A) 2 B) 1 C) 3 D) 4 7 / 30 Tengsizlikni yeching. A) 0 B) to'g'ri javob yo'q C) (0;+∞) D) (-1/³ √2 ;0) 8 / 30 ifodaning qiymatini toping. A) -2 B) 0 C) 0,5 D) -0,5 9 / 30 x2-5|x|-6=0 tenglama ildizini toping A) ±;±6 B) -1;-6 C) -1;6 D) ±6 10 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -1 B) -0,5 C) -0,25 D) -0,75 11 / 30 Tengsizlik nechta butun yechimga ega? A) cheksiz ko’p B) 1 C) 3 D) 4 12 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 4 B) 3 C) 5 D) 6 13 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 120 B) 60 C) 47 D) 18 14 / 30 tenglamalar sistemasini yeching A) (3;–4) B) (–4;3) C) (4;–3) D) (4;3) 15 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 7 π B) 10 π C) 6 π D) 12π 16 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {1;2} B) {-1;2} C) {2;4} D) {-1;3} 17 / 30 sonning oxirgi raqamini toping. A) 4 B) 8 C) 2 D) 6 18 / 30 Tenglamani yeching. (x+2)+(x+4)+(x+6)+…+(x+100)=2800 A) 4 B) 5 C) 3 D) 7 19 / 30 AA1A2A3A4A5A6 muntazam oltiburchakli piramidaning hajmi ga va balandligi 2 ga teng bo‘lsa, u holda AA2A6 kesim yuzini toping. A) √5 B) 3 C) √15 D) 2 20 / 30 tenglamani yeching. A) 2019 B) 2018 C) 2017 D) 0 21 / 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) 36 B) 50 C) 54 D) 40 22 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 108 B) 48√3 C) 54 D) 52√3 23 / 30 Uchburchakning balandligi 12 ga teng bo’lib, u asosni 5:16 nidbatda bo’ladi. Agar asosning uzunligi 21 ga teng bo’lsa, uchburchakning perimetrini toping A) 108 B) 48 C) 54 D) 52 24 / 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(2cos2x) D) -4sin2x*cos(cos2x) 25 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 3 B) 2 C) 1 D) 5 26 / 30 a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega A) a≠5 B) a ning bunday qiymati yo’q C) a≠12 D) a=12 27 / 30 x(t)=t2+6t+5 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakat boshlangandan necha sekund o’tgach boshlang’ich nuqtaga nisbatan 77 metr masofaga siljiydi? A) 6 B) 7 C) 8 D) 10 28 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 30 B) 25 C) 20 D) 28 29 / 30 funktsiyaning aniqlanish sohasini toping. A) (-∞;1)v(2;∞) B) [1;2] C) (1;2) D) (-∞;1]v[2;∞) 30 / 30 Ostki asosining yuzi 32π va ustki asosining yuzi 18π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda sharning sirtini toping. A) 96π B) 72π C) 100π D) 56π 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz