Uy » Choraklik online testlar » Matematika choraklik » 11-sinf Matematika 4-chorak Matematika choraklik 11-sinf Matematika 4-chorak InfoMaster Aprel 20, 2021 721 Ko'rishlar 83 izohlar SaqlashSaqlanganOlib tashlandi 6 2 OMAD YOR BO'LSIN! Tomonidan yaratilgan InfoMaster 11-sinf Matematika 4-chorak Testni Salomov Sardor tayyorladi. 1 / 25 Moddiy nuqta qonuniyat bilan harakatlanmoqda, eng katta tezlanishga erishiladigan vaqtni toping. A) 4 B) 7 C) 6 D) 5 2 / 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) 7 D) 8 3 / 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 4 / 25 F(x)=0.2sin(5x+12) funksiyasi uchun f(x) ni toping A) f(x)=0.5cos(5x+12) B) f(x)=-cos(5x+12) C) f(x)=cos(5x+12) D) f(x)=-0,5sin(5x+12) 5 / 25 f(x)=3x2-5x+4 va g(x)=4x-5 funksiyalarining urinmalari parallel bo’ladigan nuqtalarini toping. A) 1.5 B) 0.5 C) 0 D) 2 6 / 25 A(-4;3;6) nuqtaning Oxz tekisligiga nisbatan simmrtrik nuqtasini toping. A) (-4;-6;6) B) (-4;-3;6) C) (-4;-3;9) D) (-8;-3;6) 7 / 25 Funksiya hosilasini toping f(x)=(2x+4)(3x+1) A) 16x+14 B) 3x+1 C) 12x+14 D) 12x 8 / 25 Agar f(x)=x² va bo`lsa ,f(g(2)) ni toping A) 0 B) 36 C) 12 D) 16 9 / 25 Koordinatalar boshidan y=x2-4x+3 parabolaning simmetriya o’qigacha bo’lgan masofani toping. A) 4 B) 2 C) 3 D) 1 10 / 25 Agar f(x)=x² va bo`lsa ,f(g(-4)) ni toping A) 12 B) 36 C) -12 D) 0 11 / 25 Funksiyaning kamayish oraliqlarini toping A) (-2;0) va (0;1) B) (-1;0) va (0;1) C) (-3;0) va (0;1) D) (-1;0) va (0;2) 12 / 25 Nargiza 4minutda 260ta so`zni terib, 7ta imloviy xatoga yo`l qo`ydi. Nargizaning matn terish sifatini aniqlang A) 0,5 B) 0,0269 C) 0,01 D) 0,6 13 / 25 funksiya uchun y=2x-1 to`g`ri chiziqqa parallel bo`lgan urinma tenglamasini yozing A) y=3x-2,2 B) y=2x+3 C) y=2x D) y=2x-2,25 14 / 25 Uchlari A(2;-3;4) va B(-2;4;5) nuqtalarda bo’lgan AB kesmasining uzunligini toping A) √66 B) √55 C) √24 D) √19 15 / 25 Murakkab funksiyaning hosilasini toping. y=x2sinx A) y’=2xsinx-x² B) y’=2xsinx+x²cosx C) y’=-2xsinx+x² D) y’=x²cosx-2xsinx 16 / 25 Funksiyaning statsionar nuqtalarini toping A) x=-1, x=3 B) x=0, x=2 C) x=-1, x=2 D) x=3, x=2 17 / 25 Aylanaga o’tkazilgan vatar uni 5:7 nisbatda bo’ladi. Ushbu vatarga tiralgan, aylanaga ichki chizilgan katta burchakni toping A) 105° B) 75° C) 125° D) 55° 18 / 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) 3√3 B) 2√3 C) 3 D) 2 19 / 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) 1 C) 14 D) 23 20 / 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) 4 B) 1 C) 3√2 D) 3 21 / 25 aniq integralni hisoblang A) 6 B) 2 C) 2,5 D) 4 22 / 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) 3/4 B) 4 C) 1 D) 3 23 / 25 y=x2-│2x-4│ funksiya grafigiga x=3 va x= -3 nuqtalarda o’tkazilgan urinmalar kesishish nuqtasi ordinatasini toping A) -6 B) -12 C) -5 D) -9 24 / 25 f(x)=(sinx)cosx bo’lsa, f '(5π/6) ni toping A) (ln2+3)/2)*(√3^√3) B) (ln2+3)/2)*(√2^√3) C) (ln2-3)/2)*(√2^√3) D) (ln2+3)/2)*(√2^√2) 25 / 25 x=1, y=2x va y=2-x chiziqlari bilan chegaralangan sohaning yuzini toping A) ln4 B) ln3 C) log₃e D) log₄e 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