Uy » Choraklik online testlar » Matematika choraklik » 11-sinf Matematika 4-chorak Matematika choraklik 11-sinf Matematika 4-chorak InfoMaster Aprel 20, 2021 292 Ko'rishlar 83 izohlar SaqlashSaqlanganOlib tashlandi 4 4 OMAD YOR BO'LSIN! 11-sinf Matematika 4-chorak Testni Salomov Sardor tayyorladi. 1 / 25 To’g’riburchakli parallelepiped bir uchidan chiquvchi uchta qirralari uzunliklari 4,6 va 9 ga teng. Unga tengdosh kub qirrasining uzunligini toping. A) 6 B) 6 C) 8 D) 7 2 / 25 Kub to’la sirtining yuzi 105,84 sm2 bo’lsa, uning har bir yog’i yuzini va qirrasining uzunligini toping. A) 17,64 dm²; 4,2 dm B) 17,68 dm²; 4,2 dm C) 17,64 dm²; 4,9 dm D) 17,64 dm; 4,2 dm 3 / 25 f(x)=excosx funksiyasi uchun f ’(π/2) ni aniqlang. A. e-0.5π B. e0,5π C. eπ D. e-π A) e^(-π) B) e^(π) C) e^(0.5π) D) e^(-0.5π) 4 / 25 F(x)=0.2sin(5x+12) funksiyasi uchun f(x) ni toping A) f(x)=-0,5sin(5x+12) B) f(x)=-cos(5x+12) C) f(x)=cos(5x+12) D) f(x)=0.5cos(5x+12) 5 / 25 Moddiy nuqta qonuniyat bilan harakatlanmoqda, eng katta tezlanishga erishiladigan vaqtni toping. A) 5 B) 4 C) 6 D) 7 6 / 25 Moddiy nuqta S(t)=5t3+3t2+2t+4 qonuniyat bilan harakatlanayotgan jismning t=2 sekunddagi tezligini toping. A) 74 B) 84 C) 60 D) 70 7 / 25 Moddiy nuqtaning berilgan t vaqtdagi tezligini hisoblang: , t=5 A) 90 B) 75 C) 80 D) 70 8 / 25 Koordinatalar boshidan y=x2-4x+3 parabolaning simmetriya o’qigacha bo’lgan masofani toping. A) 2 B) 3 C) 4 D) 1 9 / 25 funksiya uchun y=2x-1 to`g`ri chiziqqa parallel bo`lgan urinma tenglamasini yozing A) y=2x+3 B) y=2x C) y=2x-2,25 D) y=3x-2,2 10 / 25 Funksiya hosilasini toping: A) 12x²-12 B) 14x+12 C) 21x²-10x D) 12x+12 11 / 25 funksiyaning [-4;1] kesmadagi eng kata qiymatini toping. A) -13 B) -4 C) 131 D) 13 12 / 25 Funksiya hosilasini toping: f(x)=5 A) x B) 0 C) 10 D) 5 13 / 25 y=-3x3+2x2-4x+5 funksiyaga x0=-1 nuqtada o’tkazilgan urinma tenglamasining burchak koeffitsiyentini toping. A) -13 B) -17 C) 15 D) 17 14 / 25 y>0 bo’lsin. To’g’riburchakning uchlari to’g’ri burchakli koordinatalar sistemasida qyuidagicha berilgan: A(0;0), B(0;y), C(-9;y) va D(-11;0). To’rtburchak diagonallarining o’tralari orasidagi masofani toping A) 2 B) 1 C) √2 D) √22 15 / 25 Ushbu funksiyasining boshlang’ich funksiyasini toping A) lnIx-1I B) lnI(x-2)(x-3)I C) ln(|x+2|*|x+1|) D) (x+2)(x+1) 16 / 25 sin3x+cos3x≥0 tengsizligini yeching A) [-π/9+n π/3;2 π/9+n π/3] B) (-π/2+n π;- π/2+n π) C) [-π/9+2πn/3; 2π/9+2πn/3] D) ø 17 / 25 Aylanaga o’tkazilgan vatar uni 5:7 nisbatda bo’ladi. Ushbu vatarga tiralgan, aylanaga ichki chizilgan katta burchakni toping A) 55° B) 105° C) 75° D) 125° 18 / 25 chiziqlar bilan chegaralangan shaklning yuzini toping A) 1 B) 0,5 C) √2 D) 2 19 / 25 y=x2-│2x-4│ funksiya grafigiga x=3 va x= -3 nuqtalarda o’tkazilgan urinmalar kesishish nuqtasi ordinatasini toping A) -6 B) -5 C) -12 D) -9 20 / 25 Funksiyaning aniqlanish sohasini toping : y=logx-2(x2+7x-8) A) (-∞;-8)U(2;3)U(3;+∞) B) (2;3)U(3;+∞) C) (-∞;-1)U(8;+∞) D) (8;+∞) 21 / 25 Agar f(x)=6+5tg22x bo’lsa, f '(π) ni toping A) 5 B) 0 C) 1 D) 3 22 / 25 x=1, y=2x va y=2-x chiziqlari bilan chegaralangan sohaning yuzini toping A) ln4 B) ln3 C) log₄e D) log₃e 23 / 25 Massasi 735 gr va konsentratsiyasi 16% bo’lgan yodning spirtdagi eritmasiga qancha toza spirt qo’shilsa, uning konsentratsiyasi 10% bo’ladi A) 420 B) 441 C) 256 D) 252 24 / 25 B,C,D,E nuqtalar aylanadagi, A esa aylanadan tashqaridagi nuqtalar. D nuqta AE kesmada, B esa AC kesmada yotadi. Agar AE=12 va AC=16 bo’lsa, BE vatar uzunligining CD vatar uzunligiga nisbatini toping A) 4 B) 3/4 C) 1 D) 3 25 / 25 Bir burchagi 600 bo’lgan to’g’ri burchakli uchburchakka tomoni 6 ga teng romb shunday ichki chizilganki, 600 li burchak ular uchun umumiy, rombning barcha uchlari uchburchakning tomonlarida yotadi. Uchburchak yuzini toping A) √3/2 B) (1√3)/2 C) (8√3)/2 D) (81√3)/2 0% Testni qayta ishga tushiring Baholash mezoni 86%-100% 5 baho 71%-85% 4 baho 56%-70% 3 baho 55% va kamiga 2 baho Fikr-mulohaza yuboring Tomonidan Wordpress Quiz plugin Author: InfoMaster Foydali bo'lsa mamnunmiz