Uy » Choraklik online testlar » Matematika choraklik » 11-sinf Matematika 4-chorak Matematika choraklik 11-sinf Matematika 4-chorak InfoMaster Aprel 20, 2021 283 Ko'rishlar 83 izohlar SaqlashSaqlanganOlib tashlandi 4 0 OMAD YOR BO'LSIN! 11-sinf Matematika 4-chorak Testni Salomov Sardor tayyorladi. 1 / 25 f(x)=(sinx)cosx bo’lsa, f '(5π/6) ni toping A) (ln2+3)/2)*(√3^√3) B) (ln2+3)/2)*(√2^√2) C) (ln2-3)/2)*(√2^√3) D) (ln2+3)/2)*(√2^√3) 2 / 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;+∞) 3 / 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) 4 / 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) 17,64 dm; 4,2 dm C) 17,64 dm²; 4,9 dm D) 17,68 dm²; 4,2 dm 5 / 25 To’g’riburchakli parallelepiped asosining tomonlari 7:24 nisbatda, diagonal kesimining yuzi 50dm2 ga teng. Yon sirtining yuzini toping. A) 124 dm² B) 135 dm² C) 224 dm² D) 124 dm 6 / 25 f(x)=sin2x funksiyasining barcha boshlang’ich funksiyalarini toping. A) F(x)=0,5cos2x+C B) F(x)=-2cos2x+C C) F(x)=2sin4x+c D) F(x)=-0,5cos2x+C 7 / 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^(-π) D) e^(-0.5π) 8 / 25 Moddiy nuqta qonuniyat bilan harakatlanmoqda, eng katta tezlanishga erishiladigan vaqtni toping. A) 6 B) 4 C) 5 D) 7 9 / 25 To’g’riburchakli parallelepiped bir uchidan chiquvchi uchta qirralari uzunliklari 4,6 va 9 ga teng. Unga tengdosh kub qirrasining uzunligini toping. A) 7 B) 6 C) 6 D) 8 10 / 25 sonini taqriban hisoblang. A) 4,08 B) 3,08 C) 5,08 D) 5,01 11 / 25 Trapetsiyaning kichik asosi 6 ga va yon tomoni 5 va 7 ga teng. Uning kata asosi ham butun son bo’lsa, perimetri eng ko’pi bilan nechaga teng bo’lishi mumkin A) 3√2 B) 4 C) 3 D) 1 12 / 25 Aylanaga o’tkazilgan vatar uni 5:7 nisbatda bo’ladi. Ushbu vatarga tiralgan, aylanaga ichki chizilgan katta burchakni toping A) 105° B) 55° C) 75° D) 125° 13 / 25 Silindr shaklidagi idishga 6 sm3 suv solindi. Idishga detal to’liq cho’ktirilganda, suv sathi 1,5 marta ko’tarildi. Detal hajmini aniqlang A) 3 B) 14 C) 23 D) 1 14 / 25 Funksiya hosilasini toping: f(x)=5 A) 0 B) x C) 5 D) 10 15 / 25 Uchlari A(2;-3;4) va B(-2;4;5) nuqtalarda bo’lgan AB kesmasining uzunligini toping A) √24 B) √55 C) √19 D) √66 16 / 25 Funksiya hosilasini toping: A) 12x²-12 B) 21x²-10x C) 12x+12 D) 14x+12 17 / 25 funksiyaning [-4;2] oraliqdagi eng katta qiymatini toping A) 16 B) 14 C) 18 D) 17 18 / 25 A(1;-2;2), B(1;4;0), C(-4;1;1) va D(-5;-5;3) nuqtalar berilgan. AC va BD vektorlar orasidagi burchakni toping A) 15° B) 52° C) 90° D) 80° 19 / 25 Agar f(x)=x² va bo`lsa ,f(g(2)) ni toping A) 16 B) 0 C) 36 D) 12 20 / 25 A(2;0;-3) va B(3;4;0) nuqtalar orasidagi masofani toping A) √26 B) √24 C) 16 D) 25 21 / 25 Funksiyaning statsionar nuqtalarini toping A) x=0, x=2 B) x=-1, x=2 C) x=3, x=2 D) x=-1, x=3 22 / 25 Uchlari A(2;3;4) va B(-4;-5;-6) nuqtalarda bo’lgan AB kesmasining o’rtasining koordinatalarini toping. A) (-1;-3;-1) B) (-1;-1;-1) C) (-2;-4;-1) D) (-1;-1;-4) 23 / 25 funksiya grafigiga x=4 abssissali nuqtada o`tkazilgan urinma tenglamasini tuzing A) y=35x+30 B) y=35x C) y=20x+20 D) y=10x 24 / 25 f(x)=2x+3 funksiyasi uchun A(1;5) nuqtadan o’tuvchi boshlang’ich funksiyasini toping. A) F(x)=x²-3x-1 B) F(x)=x²+3x+1 C) F(x)=x²+3x-1 D) F(x)=x²-3x+1 25 / 25 Funksiyaning kamayish oraliqlarini toping A) (-1;0) va (0;1) B) (-3;0) va (0;1) C) (-1;0) va (0;2) D) (-2;0) va (0;1) 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