Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 225 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 ABC muntazam uchburchak ichidan ixtiyoriy P nuqta olinib, undan BC, CA va AB tomonlarga mos ravishda PD, PE va PF perpendikulyarlar tushirilgan bo’lsa,ni toping. A) 0,5 B) 1 C) 1/√2 D) 1/√3 2 / 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) (-7; 1) D) (4; -3) 3 / 30 Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping. A) 113,5° B) 113° C) 112,5° D) 100° 4 / 30 tenglamalar sistemasini yeching A) (-4;4) B) (4;–4) C) (-4;-4) D) (4;4) 5 / 30 ifodaning qiymatini toping. A) 0 B) 0,5 C) -0,5 D) -2 6 / 30 tenglamalar sistemasini yeching A) (4;–3) B) (3;–4) C) (–4;3) D) (4;3) 7 / 30 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakatning 200- metrida qanday tezlikka (m/s) erishadi? A) 42 B) 38 C) 36 D) 32 8 / 30 Hisoblang A) √3 B) √3/2 C) 1/2 D) 2 9 / 30 A) 3 B) 5 C) 1 D) 2 10 / 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) 100π C) 56π D) 72π 11 / 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) 6,25% ga kamayadi C) 2,5% ga ortadi D) o‘zgarmaydi 12 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) π/2 B) 2π C) 2π/3 D) π 13 / 30 tenglamani yeching. A) 2018 B) 2019 C) 2017 D) 0 14 / 30 x2-5|x|-6=0 tenglama ildizini toping A) -1;6 B) ±;±6 C) -1;-6 D) ±6 15 / 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) 8√2π/5 B) 64√2π/3 C) (5√3+3)π/3 D) 3√2π/4 16 / 30 A) √6 B) 2 C) 1 D) 0 17 / 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(1+2cosα/4sin²) B) πa²/4cos²α C) πa²(1-2sinα/4sin²) D) πa²/4sin²α 18 / 30 Bog’bon uch kun davomida o’nta daraxt ko’chati o’tqazishi lozim. Agar bog’bon bir kunda eng kamida bitta ko’chat o’tqazadigan bo’lsa, u shu ishni kunlar bo’yicha necha xil usul bilan taqsimlashi mumkin? A) 30 B) 32 C) 25 D) 36 19 / 30 y=cos(2sinx) funksiyaning qiymatlar sohasini toping. A) [cos2;1] B) [-1;1] C) [0;1] D) [0;cos2] 20 / 30 AA1A2A3A4A5A6 muntazam oltiburchakli piramidaning hajmi ga va balandligi 2 ga teng bo‘lsa, u holda AA2A6 kesim yuzini toping. A) 2 B) √5 C) 3 D) √15 21 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) lg(ln x) B) lg(lg x) C) ln(lg x) D) ln(ln x) 22 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) bunday funksiya mavjud emas B) juft funksiya C) toq funksiya D) juft ham emas, toq ham emas funksiya 23 / 30 Bankda qo`yilgan pul bir yildan kegin foydasi bilan 2600 so`m bo`ldi; Agar bank yillik 30% foyda to`lasa, boshida qancha pul qo`yilgan bo`ladi ? A) 2200 B) 1900 C) 2000 D) 2100 24 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) (-∞;6])v[6;∞) B) (2;∞) C) 6 D) (-∞;6) 25 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 5 ga, bir katetining gipotenuzadagi proyeksiyasi 1,6 ga teng. Ikkinchi katetning kvadratini toping. A) 18 B) 16 C) 17 D) 14 26 / 30 ABCD to’gri to’rtburchak ichidan olingan O nuqtadan A, B, C, D uchlarigacha bo’lgan masofalar mos ravishda 1; 2; 1,5; 1,2 ga teng bo’ladigan barcha to’g’ri to’rtburchaklar sonini toping A) 2 B) 0 C) cheksiz ko’p D) 1 27 / 30 sonning oxirgi raqamini toping. A) 6 B) 8 C) 2 D) 4 28 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) o’zaro perpendikulyar B) o’zaro parallel C) ayqash D) aniqlab bo’lmaydi 29 / 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) 2 yoki 50 D) 10 yoki 50 30 / 30 A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping. A) 1 B) √5/2 C) √3/2 D) 1,5 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
