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 Tenglamaning ildizlari yig’indisi va ko’paytmasining yig’indisini toping. |x+1|.|x-4|=5 A) 9 B) -6 C) -3 D) 6 2 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) (-∞;6])v[6;∞) B) (2;∞) C) 6 D) (-∞;6) 3 / 30 integralning qiymatini toping. A) -π/2 B) π/2 C) 0 D) π/4 4 / 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) 5 / 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) 7 B) 10 C) 6 D) 8 6 / 30 ABCD kvadrat ichidan olingan O nuqtadan A, B, C uchlarigacha bo’lgan masofalar mos ravishda 3, 4, 5 ga teng bo’lsa, u holda OD kesma uzunligini toping. A) 6 B) √37 C) √32 D) 3 7 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln|x-1/x+4|+c B) ln|x+4/x-1|+c C) ln(|x+4*|x-1|)+c D) ln(|x+4+|x-1|)+c 8 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁴⁹⁵⁰ B) e⁻⁵⁰⁵⁰ C) e⁵⁰⁵⁰ D) e⁻⁴⁹⁵⁰ 9 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 2, 3 B) 1,4 C) 3, 4 D) 1, 3 10 / 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) 5√3+3/3 B) 2√2/3 C) 3√2+2/4 D) 3√3+1/5 11 / 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 uchburchakka ichki chizilgan aylana markazi bo‘lsa, u holda burchakni toping. A) 72° B) 30° C) 60° D) 90° 12 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 48√3 B) 67,5 C) 135√3/4 D) 67,5√3 13 / 30 tenglamalar sistemasini yeching A) (4;–4) B) (-4;-4) C) (-4;4) D) (4;4) 14 / 30 bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya. A) -4 B) -14 C) -10 D) -12 15 / 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) 72π B) 100π C) 56π D) 96π 16 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 11 B) 12 C) √46 D) 10 17 / 30 Hisoblang A) √3/2 B) 2 C) 1/2 D) √3 18 / 30 Soddalashtiring. (0<m<7) A) m B) 2m-7 C) 7 D) 7-2m 19 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 8 B) 6 C) 5 D) 4 20 / 30 Ifodani soddalashtiring. 2 cos55o.cos40o.sin55o+cos110o.sin40o A) 1 B) 2 C) 0 D) 0,5 21 / 30 funktsiyaning aniqlanish sohasini toping. A) (-∞;1]v[2;∞) B) (-∞;1)v(2;∞) C) (1;2) D) [1;2] 22 / 30 Sin9x =4sin3x tenglamani yeching A) πn, n€Ζ B) πn/3, n€Ζ C) π/2+πn, n€Ζ D) π/3+πn, n€Ζ 23 / 30 y=cos(2sinx) funksiyaning qiymatlar sohasini toping. A) [cos2;1] B) [-1;1] C) [0;1] D) [0;cos2] 24 / 30 Tomoni 25 ga diagonallaridan biri 4 ga teng bo’lgan rombning yuzini toping. A) 24√5 B) 16 C) 32 D) 8√5 25 / 30 Tomoni 12 ga teng bo`lgan teng tomonli uchburchakga ichki chizilgan aylana uzunligini toping. A) 12π B) 3π C) 4√3π D) 2√3π 26 / 30 ifodaning qiymatini toping. A) -0,5 B) 0 C) -2 D) 0,5 27 / 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) 2 B) √37 C) bunday to’gri to’rtburchak mavjud emas D) √7 28 / 30 “Tutgan balig‘ining og‘irligi qancha?” degan savolga baliqchi: “Baliqning dumi 3 kg, boshi uning dumi hamda tanasi yarmining og‘irligiga teng, tanasi esa boshi va dumining og‘irligiga teng”, deb javob berdi. Baliqning og‘irligini (kg) toping. A) 6 B) 18 C) 3 D) 12 29 / 30 Tengsizlik nechta butun yechimga ega? A) cheksiz ko’p B) 1 C) 4 D) 3 30 / 30 O’q kesimining diagonallari o’zaro perpendikulyar bo’lgan kesik konus yasovchisi va asos tekisligi orasidagi burchak ga teng. Agar o’q kesimining diagonali ga teng bo’lsa, kesik konus asosining yuzini toping. A) πa²/4sin²α B) πa²/4cos²α C) πa²(1-2sinα/4sin²) D) πa²(1+2cosα/4sin²) 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
