Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 156 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 1 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Tengsizlikni yeching. A) to'g'ri javob yo'q B) (0;+∞) C) 0 D) (-1/³ √2 ;0) 2 / 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) 67,5 C) 135√3/4 D) 48√3 3 / 30 Hisoblang. 11+192+1993+19994+199995+1999996+19999997+199999998+1999999999 A) 2222222175 B) 222220175 C) 222222222222 D) 22222222220 4 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 3 B) 1 C) 2 D) 5 5 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) (2;∞) B) 6 C) (-∞;6])v[6;∞) D) (-∞;6) 6 / 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 7 / 30 Rasmda ko‘rsatilgan ko‘pyoqlardan qaysi birida 4 ta yoq, 6 ta qirra bor? A) 3 B) 1, 2 C) 2 D) 1, 3 8 / 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 9 / 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) 3√2π/4 B) (5√3+3)π/3 C) 8√2π/5 D) 64√2π/3 10 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 4 B) 6 C) 5 D) 3 11 / 30 (-3;4) nuqtaga absissa, ordinata o’qlariga va koordinata boshiga nisbatan simmetrik bo’lgan nuqtalarni tutashtirishdan hosil bo’lgan uchburchakning eng kata tomonini toping. A) 12 B) 10 C) 14 D) 24 12 / 30 A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping. A) 1,5 B) √5/2 C) 1 D) √3/2 13 / 30 Agar geometrik progressiyaning ketma–ket dastlabki uchta hadining yig’indisi 62 ga, ularning o’nli logarifmlari yig’indisi 3 ga teng bo’lsa, shu geometrik progressiyaning birinchi hadini toping. A) 10 B) 50 C) 10 yoki 50 D) 2 yoki 50 14 / 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) bunday to’gri to’rtburchak mavjud emas B) √37 C) √7 D) 2 15 / 30 A) 1 B) 2 C) √6 D) 0 16 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (2;∞) B) (∞;∞) C) (-∞;2) D) (-∞;2)v(2;∞) 17 / 30 Ifodani soddalashtiring. 2 cos55o.cos40o.sin55o+cos110o.sin40o A) 0 B) 0,5 C) 1 D) 2 18 / 30 Tomoni 12 ga teng bo`lgan teng tomonli uchburchakga ichki chizilgan aylana uzunligini toping. A) 2√3π B) 3π C) 12π D) 4√3π 19 / 30 Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping. A) 113,5° B) 100° C) 112,5° D) 113° 20 / 30 Markazi O nuqtada bo‘lgan aylanaga PA va PB urinmalar o‘tkazilgan bo’lib, A va B nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma PA va PB kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar XQ=YQ bo‘lsa, u holda PXY uchburchak qanday uchburchak bo‘ladi? A) ixtiyoriy uchburchak B) muntazam uchburchak C) to`g`ri burchakli uchburchak D) teng yonli uchburchak 21 / 30 ifodaning qiymatini toping. A) 0,5 B) -2 C) 0 D) -0,5 22 / 30 bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya. A) -4 B) -14 C) -12 D) -10 23 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 12 B) 10 C) √46 D) 11 24 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 8 B) 4 C) 12 D) 6 25 / 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) πa²/4cos²α B) πa²/4sin²α C) πa²(1-2sinα/4sin²) D) πa²(1+2cosα/4sin²) 26 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) √3 B) √2 C) 2 D) 1 27 / 30 y=cos(2sinx) funksiyaning qiymatlar sohasini toping. A) [-1;1] B) [0;cos2] C) [cos2;1] D) [0;1] 28 / 30 Hisoblang A) 2√3 B) 1 C) √3 D) 3√3 29 / 30 To’rtburchakning uchi M (0, 4) ; N(-4,0) ; P(-3;2) uchlari berilgan. Agar bo’lsa, Q uchining koordinatalarini toping. A) (7; 1) B) (-7; 1) C) (4; -3) D) (-7;-1) 30 / 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) 12 D) 25 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
