Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 164 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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(1-2sinα/4sin²) C) πa²/4sin²α D) πa²/4cos²α 2 / 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) √2 D) √3 3 / 30 Soddalashtiring. A) 2019 B) 2018 C) 2018a/a+1 D) a+1 4 / 30 sistemada xy ning qiymatini toping. A) 80 B) 64 C) 75 D) 60 5 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) (-∞;1]v[10;∞) B) 1; 10 C) [10;∞) D) (-∞;1] 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) 374389 B) 412967 C) aniqlab bo’lmaydi D) 2 7 / 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 kamayadi B) 2,5% ga ortadi C) o‘zgarmaydi D) 6,25% ga ortadi 8 / 30 Agar f(x)=ax3-5x2+b va bo’lsa, a ni toping. A) 3 B) 1 C) 0 D) 2 9 / 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) 8 B) 6 C) 7 D) 10 10 / 30 integralning qiymatini toping. A) 0 B) π/4 C) -π/2 D) π/2 11 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) -x-y+z=0 B) 2x+3y+z=0 C) x-1/2=y-2/3=z-3/4 D) x=y/3=z/2 12 / 30 Konsert zalining birinchi qatorida 40 ta o’rindiq bor. Har bir keyingi qatordagi o’rindiqlar soni oldingi qatordan 4 ga ko’p. Agar konsert zalida jami 40 ta qator bo’lsa, u holda shu zaldagi barcha o’rindiqlar sonini toping. A) 4680 B) 4720 C) 4760 D) 4716 13 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 12 B) √46 C) 10 D) 11 14 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 10 π B) 7 π C) 12π D) 6 π 15 / 30 Agar arctga+ arctgb + arctgc= bo’lsa, a+b+c ni toping. A) ab/c B) 1 C) ab D) abc 16 / 30 y= funksiyaning aniqlanish sohasini toping A) (2;∞) B) (0,2) C) (-∞;0)v(2;∞) D) (-∞;0])v[2;∞) 17 / 30 Hisoblang: A) 1 B) -1 C) 3 D) 2 18 / 30 Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping A) 28 B) 30 C) 20 D) 25 19 / 30 y= funktsiyaning aniqlanish sohasini toping. A) (-∞;2) B) (-∞;2)v(2;∞) C) [2;∞) D) (2;∞) 20 / 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) 36 D) 25 21 / 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) 52 D) 54 22 / 30 Yuzasi 10 ga teng bo’lgan kvadratning ketma-ket ikki uchidan o’tuvchi aylana chizilgan. Uchinchi uchidan aylanaga urunma o’tkazilgan. Urunma tomondan ikki marta katta bo’lsa, aylana radiusini toping. A) 6 B) 5 C) 4 D) 10 23 / 30 Sin9x =4sin3x tenglamani yeching A) π/3+πn, n€Ζ B) π/2+πn, n€Ζ C) πn/3, n€Ζ D) πn, n€Ζ 24 / 30 Hisoblang A) 1/2 B) √3 C) √3/2 D) 2 25 / 30 sistemadan x+y ning qiymatini toping. A) 35/4 B) 12 C) 6 D) -12 26 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 6 B) 5 C) 4 D) 7 27 / 30 Soddalashtiring. (0<m<7) A) 7 B) 7-2m C) m D) 2m-7 28 / 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(cos2x) C) -4sin2x*cos(2cos2x) D) 4sin2x*cos(2cos2x) 29 / 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 30 / 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) 50 B) 10 yoki 50 C) 2 yoki 50 D) 10 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz