Uy » Choraklik online testlar » Matematika choraklik » 11-sinf Matematika 4-chorak Matematika choraklik 11-sinf Matematika 4-chorak InfoMaster Aprel 20, 2021 697 Ko'rishlar 83 izohlar SaqlashSaqlanganOlib tashlandi 6 0 OMAD YOR BO'LSIN! Tomonidan yaratilgan InfoMaster 11-sinf Matematika 4-chorak Testni Salomov Sardor tayyorladi. 1 / 25 Nargiza 4minutda 260ta so`zni terib, 7ta imloviy xatoga yo`l qo`ydi. Nargizaning matn terish sifatini aniqlang A) 0,0269 B) 0,6 C) 0,5 D) 0,01 2 / 25 Koordinatalar boshidan y=x2-4x+3 parabolaning simmetriya o’qigacha bo’lgan masofani toping. A) 1 B) 4 C) 3 D) 2 3 / 25 funksiyaning [-4;2] oraliqdagi eng katta qiymatini toping A) 18 B) 16 C) 17 D) 14 4 / 25 funksiya grafigiga x=4 abssissali nuqtada o`tkazilgan urinma tenglamasini tuzing A) y=20x+20 B) y=35x+30 C) y=35x D) y=10x 5 / 25 y=-3x3+2x2-4x+5 funksiyaga x0=-1 nuqtada o’tkazilgan urinma tenglamasining burchak koeffitsiyentini toping. A) 17 B) -17 C) -13 D) 15 6 / 25 Murakkab funksiyaning hosilasini toping. y=x2sinx A) y’=-2xsinx+x² B) y’=2xsinx+x²cosx C) y’=x²cosx-2xsinx D) y’=2xsinx-x² 7 / 25 Moddiy nuqtaning berilgan t vaqtdagi tezligini hisoblang: , t=5 A) 80 B) 90 C) 75 D) 70 8 / 25 Funksiyaning statsionar nuqtalarini toping A) x=-1, x=2 B) x=3, x=2 C) x=-1, x=3 D) x=0, x=2 9 / 25 Funksiyaning kamayish oraliqlarini toping A) (-1;0) va (0;2) B) (-1;0) va (0;1) C) (-3;0) va (0;1) D) (-2;0) va (0;1) 10 / 25 funksiya hosilasining x0=-2 nuqtadagi qiymatini hisoblang. A) 38/961 B) 38/31 C) -38/961 D) -38/31 11 / 25 Muntazam uchburchak tomoni 3 ga teng. Uchburchak tomonlari o’rtalari tutashtirilib, muntazam uchburchaklar hosil qilindi. Ichma-ich joylashgan uchburchaklar yuzlari yig’indisini toping A) 2√3 B) 3 C) 3√3 D) 2 12 / 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) 1 B) 2 C) √2 D) √22 13 / 25 Uchlari soni 14 ta, qirralari soni esa 21 ta bo’lgan ko’pyoqning yoqlari sonini toping. A) 16 B) 4 C) 9 D) 8 14 / 25 Ixtiyoriy kvadratga tashqi va ichki chizilgan aylanalar yuzlari nisbati a ga teng. Ixtiyoriy aylanaga ichki va tashqi chizilgan kvadratlar yuzlari nisbati b ga teng bolsa, a/b ni toping A) 4 B) 1/2 C) 335 D) 1 15 / 25 Ushbu funksiyasining boshlang’ich funksiyasini toping A) ln(|x+2|*|x+1|) B) lnIx-1I C) (x+2)(x+1) D) lnI(x-2)(x-3)I 16 / 25 aniq integralni hisoblang A) 2 B) 6 C) 4 D) 2,5 17 / 25 Havo shariga t[0;10] minut oralig’ida V(t)=t3+13t2+t+20 (m3) havo purkamoqda. t=10 minutdagi havo hajmini toping. A) 2300 B) 1023 C) 2330 D) 2025 18 / 25 F(x)=0.2sin(5x+12) funksiyasi uchun f(x) ni toping A) f(x)=cos(5x+12) B) f(x)=-cos(5x+12) C) f(x)=-0,5sin(5x+12) D) f(x)=0.5cos(5x+12) 19 / 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^(-0.5π) B) e^(-π) C) e^(0.5π) D) e^(π) 20 / 25 Tekislikdagi 1 m masofada yotgan nuqtadan ikkita teng og’ma o’tkazilgan. Agar og’malar perpendikulyar va tekislikka o’tkazilgan perpendikulyar bilan 600 ga teng burchaklar tashkil etsa, og’malarning asoslari orasidagi masofani toping A) 2√3 B) 2√2 C) 4√2 D) 1 21 / 25 Funksiyaning aniqlanish sohasini toping : y=logx-2(x2+7x-8) A) (-∞;-1)U(8;+∞) B) (-∞;-8)U(2;3)U(3;+∞) C) (8;+∞) D) (2;3)U(3;+∞) 22 / 25 funksiyasining kamayish oralig’ini toping A) ∅ B) (-√5;√5) C) (-∞;0] D) (0;+∞) 23 / 25 x=1, y=2x va y=2-x chiziqlari bilan chegaralangan sohaning yuzini toping A) log₃e B) ln3 C) log₄e D) ln4 24 / 25 2cosx+sinx=-2 tenglamaning [-π;π] kesmada nechta ildizi bor A) 1 B) 2 C) 3 D) Ø 25 / 25 Hisoblang: cos2330+cos2450-sin2600+cos2570 A) 1,5 B) 0,75 C) 0,5 D) 2,25 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 Author: InfoMaster Foydali bo'lsa mamnunmiz