11-sinf Matematika 4-chorak Aprel 17, 2021Aprel 20, 2021 da chop etilgan InfoMaster tomonidan 11-sinf Matematika 4-chorak ga fikr bildirilmagan. 62 OMAD YOR BO'LSIN! 11-sinf Matematika 4-chorak Testni Salomov Sardor tayyorladi. 1 / 25 Moddiy nuqta qonuniyat bilan harakatlanmoqda, eng katta tezlanishga erishiladigan vaqtni toping. A) 5 B) 6 C) 7 D) 4 2 / 25 f(x)=sin2x funksiyasining barcha boshlang’ich funksiyalarini toping. A) F(x)=-2cos2x+C B) F(x)=2sin4x+c C) F(x)=-0,5cos2x+C D) F(x)=0,5cos2x+C 3 / 25 Tandirdan olingan nonning temperaturasi 20 minut ichida 1000 dan 600 gacha pasayadi. Tashqi muhit temperaturasi 250. Nonning temperaturasi qancha vaqtda 300 gacha pasayadi? A) 41 B) 61 C) 71 D) 51 4 / 25 Qavariq ko’pburchakning har bir uchidan uchtadan qirra chiqadi. Agar bu ko’pyoqning qirralari soni 12 ta bo’lsa, uning nechta uchi va yog’i mavjud? A) 8;10 B) 4;18 C) 4;11 D) 4;10 5 / 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 6 / 25 Moddiy nuqta S(t)=5t3+3t2+2t+4 qonuniyat bilan harakatlanayotgan jismning t=2 sekunddagi tezligini toping. A) 70 B) 84 C) 74 D) 60 7 / 25 A(2;0;-3) va B(3;4;0) nuqtalar orasidagi masofani toping A) √26 B) 25 C) 16 D) √24 8 / 25 Karim 3 minutda 213ta so`zni terib , 6ta imloviy xatoga yo`l qo`ydi. Karimning matn terish sifatini toping. A) 0,623 B) 0,3265 C) 0,0321 D) 0,0282 9 / 25 y=-3x3+2x2-4x+5 funksiyaga x0=-1 nuqtada o’tkazilgan urinma tenglamasining burchak koeffitsiyentini toping. A) 15 B) -13 C) 17 D) -17 10 / 25 funksiya grafigiga x=1 abssisssali nuqtadao`tkazilgan urinma tenglamasini yozing A) y=x B) y= 2x C) y=1/2 D) y=-x 11 / 25 tgx funksiya hosilasini toping A) sinx B) cosx C) 1/cos²x D) 1/sinx 12 / 25 Funksiyaning statsionar nuqtalarini toping A) x=-1, x=2 B) x=0, x=2 C) x=3, x=2 D) x=-1, x=3 13 / 25 funksiya hosilasining x0=-2 nuqtadagi qiymatini hisoblang. A) -38/961 B) 38/31 C) -38/31 D) 38/961 14 / 25 Funksiya hosilasini toping: f(x)=5 A) x B) 5 C) 10 D) 0 15 / 25 cos150*cos450 *cos750 ni hisoblang A) √2/2 B) 2√2 C) √8/2 D) √2 16 / 25 tengsizlikning yechimi bo’lgan eng kichik natural sonni toping A) 4 B) 3 C) 1 D) 5 17 / 25 chiziqlar bilan chegaralangan shaklning yuzini toping A) 2 B) 1 C) √2 D) 0,5 18 / 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) √22 B) 1 C) 2 D) √2 19 / 25 sin3x+cos3x≥0 tengsizligini yeching A) [-π/9+n π/3;2 π/9+n π/3] B) [-π/9+2πn/3; 2π/9+2πn/3] C) (-π/2+n π;- π/2+n π) D) ø 20 / 25 aniq integralni hisoblang A) 2 B) 2,5 C) 6 D) 4 21 / 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) (1√3)/2 B) (8√3)/2 C) (81√3)/2 D) √3/2 22 / 25 Agar f(x)=6+5tg22x bo’lsa, f '(π) ni toping A) 3 B) 0 C) 1 D) 5 23 / 25 x=1, y=2x va y=2-x chiziqlari bilan chegaralangan sohaning yuzini toping A) ln3 B) ln4 C) log₄e D) log₃e 24 / 25 2cosx+sinx=-2 tenglamaning [-π;π] kesmada nechta ildizi bor A) 2 B) 1 C) 3 D) Ø 25 / 25 integralni hisoblang A) 5ln2e B) 5 C) 5ln3e³ D) 5ln4e 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 Matematika choraklik Teglar:11-sinf, matematika