Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 160 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 2 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Ifodani soddalashtiring. 2 cos55o.cos40o.sin55o+cos110o.sin40o A) 2 B) 0,5 C) 1 D) 0 2 / 30 x2-5|x|-6=0 tenglama ildizini toping A) -1;6 B) ±6 C) -1;-6 D) ±;±6 3 / 30 Hisoblang A) 2 B) 0 C) 1 D) -1 4 / 30 Hisoblang: A) 1 B) 2 C) 3 D) -1 5 / 30 funktsiyaning aniqlanish sohasini toping. A) (-∞;1]v[2;∞) B) (-∞;1)v(2;∞) C) (1;2) D) [1;2] 6 / 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) aniqlab bo’lmaydi B) 374389 C) 2 D) 412967 7 / 30 bo’lsa, ni x orqali ifodalang. A) x/25 B) 2/x C) 2-x D) 25/x 8 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) b>a>c>d>e B) b>c>a>d>e C) e>b>a>d>c D) a>b>c>d>e 9 / 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) cheksiz ko’p B) 1 C) 0 D) 2 10 / 30 Hisoblang A) 2√3 B) √3 C) 3√3 D) 1 11 / 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) 2√2/3 B) 5√3+3/3 C) 3√3+1/5 D) 3√2+2/4 12 / 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) 1 B) 1/√2 C) 0,5 D) 1/√3 13 / 30 Rombning yuzi 96 ga, dioganallaridan biri 16 ga teng. Romb tomonini toping. A) 12 B) 11 C) 9 D) 10 14 / 30 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakatning 200- metrida qanday tezlikka (m/s) erishadi? A) 32 B) 42 C) 36 D) 38 15 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) juft ham emas, toq ham emas funksiya B) bunday funksiya mavjud emas C) toq funksiya D) juft funksiya 16 / 30 Hisoblang. A) 2/34 B) 17/34 C) 2/17 D) 15/34 17 / 30 Agar arctga+ arctgb + arctgc= bo’lsa, a+b+c ni toping. A) ab B) ab/c C) 1 D) abc 18 / 30 tenglamalar sistemasini yeching A) (4;–3) B) (3;–4) C) (4;3) D) (–4;3) 19 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) √2e B) 1 C) e D) 3e 20 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln(|x+4+|x-1|)+c B) ln(|x+4*|x-1|)+c C) ln|x+4/x-1|+c D) ln|x-1/x+4|+c 21 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) 2 B) 1 C) √3 D) √2 22 / 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) 2100 B) 2000 C) 2200 D) 1900 23 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 20 B) 32 C) 16 D) 24 24 / 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(1+2cosα/4sin²) C) πa²(1-2sinα/4sin²) D) πa²/4cos²α 25 / 30 Soddalashtiring. A) 2018 B) 2018a/a+1 C) 2019 D) a+1 26 / 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 yoki 50 B) 50 C) 2 yoki 50 D) 10 27 / 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) 30 B) 25 C) 15 D) 12 28 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,5 B) -0,75 C) -1 D) -0,25 29 / 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 30 / 30 Qaysi javobda faqat juft funksiyalar ko’rsatilgan? A) 2, 3 B) 3, 4 C) 1,4 D) 1, 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
