Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 126 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 3 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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 B) 67,5√3 C) 135√3/4 D) 48√3 2 / 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/√3 D) 1/√2 3 / 30 sonning oxirgi raqamini toping. A) 4 B) 8 C) 6 D) 2 4 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) o’zaro perpendikulyar B) aniqlab bo’lmaydi C) o’zaro parallel D) ayqash 5 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 5 ga, bir katetining gipotenuzadagi proyeksiyasi 1,6 ga teng. Ikkinchi katetning kvadratini toping. A) 18 B) 17 C) 14 D) 16 6 / 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) 24 B) 12 C) 14 D) 10 7 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 1 B) 3 C) 2 D) 5 8 / 30 x2-5|x|-6=0 tenglama ildizini toping A) ±6 B) ±;±6 C) -1;-6 D) -1;6 9 / 30 sistemada xy ning qiymatini toping. A) 60 B) 75 C) 64 D) 80 10 / 30 Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping A) 16 B) 20 C) 32 D) 24 11 / 30 Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping. A) 112,5° B) 100° C) 113° D) 113,5° 12 / 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+2cosα/4sin²) D) πa²(1-2sinα/4sin²) 13 / 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) 4 B) 5 C) 10 D) 6 14 / 30 y= funksiyaning aniqlanish sohasini toping A) (2;∞) B) (-∞;0)v(2;∞) C) (-∞;0])v[2;∞) D) (0,2) 15 / 30 Hisoblang A) √3/2 B) √3 C) 1/2 D) 2 16 / 30 Tengsizlikni yeching. A) (-1/³ √2 ;0) B) to'g'ri javob yo'q C) (0;+∞) D) 0 17 / 30 Agar sinx+cosx=a bo’lsa, ning qiymatini toping. A) -1/2 B) 1/2 C) 2/5 D) 1/3 18 / 30 Hisoblang A) 1 B) 3√3 C) √3 D) 2√3 19 / 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 20 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) 2,5% ga ortadi B) 6,25% ga ortadi C) o‘zgarmaydi D) 6,25% ga kamayadi 21 / 30 integralning qiymatini toping. A) 0 B) -π/2 C) π/2 D) π/4 22 / 30 Soddalashtiring. (0<m<7) A) 7 B) 7-2m C) 2m-7 D) m 23 / 30 Tenglamaning ildizlari yig`indisini toping. A) 3 B) 4 C) 5 D) 6 24 / 30 Bir vaqtning o’zida 9,13, . . . ,405 va 15,21, . . . ,255 ketma–ketliklarning hadlari bo’lgan sonlarning eng kattasi va eng kichigining ayirmasini toping A) 228 B) 150 C) 231 D) 147 25 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {-1;2} B) {-1;3} C) {1;2} D) {2;4} 26 / 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) 4 B) 3√3 C) 2√3 D) 2√5 27 / 30 Tenglamaning nechta ildizi bor? |x+1|=|2x-1| A) 1 B) 2 C) 3 D) 4 28 / 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) 3√3+1/5 C) 3√2+2/4 D) 2√2/3 29 / 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) 1 D) √2 30 / 30 Hisoblang: A) -1 B) 3 C) 1 D) 2 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
