Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 158 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 1 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 6 B) 5 C) 4 D) 3 2 / 30 “Tutgan balig‘ining og‘irligi qancha?” degan savolga baliqchi: “Baliqning dumi 3 kg, boshi uning dumi hamda tanasi yarmining og‘irligiga teng, tanasi esa boshi va dumining og‘irligiga teng”, deb javob berdi. Baliqning og‘irligini (kg) toping. A) 3 B) 6 C) 18 D) 12 3 / 30 Tengsizlik nechta butun yechimga ega? A) 1 B) 3 C) 4 D) cheksiz ko’p 4 / 30 AA1A2A3A4A5A6 muntazam oltiburchakli piramidaning hajmi ga va balandligi 2 ga teng bo‘lsa, u holda AA2A6 kesim yuzini toping. A) √5 B) √15 C) 3 D) 2 5 / 30 y= funksiyaning aniqlanish sohasini toping A) (0,2) B) (-∞;0])v[2;∞) C) (-∞;0)v(2;∞) D) (2;∞) 6 / 30 Hisoblang A) √3/2 B) 1/2 C) 2 D) √3 7 / 30 ABCD kvadrat ichidan olingan O nuqtadan A, B, C uchlarigacha bo’lgan masofalar mos ravishda 3, 4, 5 ga teng bo’lsa, u holda OD kesma uzunligini toping. A) 3 B) 6 C) √37 D) √32 8 / 30 Radiusi 1 ga teng aylana uchta yoyga bo`lingan. Ularga mos markaziy burchaklar 1, 2 va 6 sonlariga proporsional. Yoylardan eng kattasining uzunligini toping. A) 2π/3 B) 3π/2 C) 4π/3 D) 3π/4 9 / 30 Agar f(x)=sin2x va g(x)=cos2x bo’lsa, u holda f(g(x)) funksiyaning hosilasini toping. A) -4sin2x*cos(2cos2x) B) 4sin2x*cos(2cos2x) C) -4sin2x*cos(cos2x) D) 4sin2x*cos(cos2x) 10 / 30 Tenglamani yeching. |x2-11x+10|=x2-11x+10 A) [10;∞) B) 1; 10 C) (-∞;1] D) (-∞;1]v[10;∞) 11 / 30 sin2x-cos2x=1 tenglama [-π; 2π] oraliqda nechta ildizga ega? A) 9 B) 6 C) 7 D) 10 12 / 30 Tomoni 12 ga teng bo`lgan teng tomonli uchburchakga ichki chizilgan aylana uzunligini toping. A) 3π B) 12π C) 4√3π D) 2√3π 13 / 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 14 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) 2√2 B) 2 C) √2 D) 4 15 / 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) 10 D) 4 16 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 8 B) 12 C) 4 D) 6 17 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 48√3 B) 135√3/4 C) 67,5 D) 67,5√3 18 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) (-∞;6) B) (-∞;6])v[6;∞) C) 6 D) (2;∞) 19 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 6 π B) 10 π C) 7 π D) 12π 20 / 30 O’qishni bilmaydigan bola alifbening A,A,A, N,N, S- 6 ta harflarini ixtiyoriy ravishda terib chiqadi. Bunda ANANAS so’zining hosil bo’lish ehtimolini toping. A) 6/720 B) 5/24 C) 1/60 D) 5/720 21 / 30 sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping. A) {1;2} B) {-1;2} C) {-1;3} D) {2;4} 22 / 30 y= funktsiyaning aniqlanish sohasini toping. A) (2;∞) B) (-∞;2) C) [2;∞) D) (-∞;2)v(2;∞) 23 / 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) 0 C) 1 D) 2 24 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,25 B) -0,5 C) -0,75 D) -1 25 / 30 sistemadan x+y ning qiymatini toping. A) 35/4 B) -12 C) 12 D) 6 26 / 30 m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi? A) 3 B) 2 C) 5 D) 1 27 / 30 Hisoblang: A) 1 B) -1 C) 2 D) 3 28 / 30 Rombning yuzi 96 ga, dioganallaridan biri 16 ga teng. Romb tomonini toping. A) 12 B) 9 C) 11 D) 10 29 / 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) 7 B) 6 C) 8 D) 10 30 / 30 ifodaning qiymatini toping. A) -0,5 B) 0,5 C) 0 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