11-sinf Matematika 4-chorak Aprel 17, 2021Aprel 20, 2021 da chop etilgan InfoMaster tomonidan 11-sinf Matematika 4-chorak ga fikr bildirilmagan. 63 OMAD YOR BO'LSIN! 11-sinf Matematika 4-chorak Testni Salomov Sardor tayyorladi. 1 / 25 funksiyasining kamayish oralig’ini toping. A) (-∞;-1) B) (-1/3;+∞) C) [-1;-1/3] D) (-1;1/3) 2 / 25 funksiya grafigiga x=1 abssisssali nuqtadao`tkazilgan urinma tenglamasini yozing A) y=x B) y=-x C) y=1/2 D) y= 2x 3 / 25 Uchlari A(4;0;1), B(5;-2;1), C(4;8;5) nuqtalarda bo’lgan uchburchakning AL bissektrisasi uzunligini toping A) 4 B) (4√2)/5 C) (4√2)3 D) (2√5)/3 4 / 25 f(x)=3x2-5x+4 va g(x)=4x-5 funksiyalarining urinmalari parallel bo’ladigan nuqtalarini toping. A) 0.5 B) 1.5 C) 0 D) 2 5 / 25 Funksiyaning statsionar nuqtalarini toping A) x=-1, x=3 B) x=3, x=2 C) x=0, x=2 D) x=-1, x=2 6 / 25 A(2;0;-3) va B(3;4;0) nuqtalar orasidagi masofani toping A) √26 B) √24 C) 25 D) 16 7 / 25 A) 14 B) 18 C) 20 D) 13 8 / 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) 80° D) 90° 9 / 25 Funksiya hosilasini toping f(x)=(2x+4)(3x+1) A) 12x B) 16x+14 C) 3x+1 D) 12x+14 10 / 25 To’g’riburchakli parallelepiped asosining tomonlari 7:24 nisbatda, diagonal kesimining yuzi 50dm2 ga teng. Yon sirtining yuzini toping. A) 135 dm² B) 124 dm² C) 124 dm D) 224 dm² 11 / 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) 12 / 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) 13 / 25 Funksiyaning aniqlanish sohasini toping : y=logx-2(x2+7x-8) A) (8;+∞) B) (2;3)U(3;+∞) C) (-∞;-1)U(8;+∞) D) (-∞;-8)U(2;3)U(3;+∞) 14 / 25 Agar f(x)=6+5tg22x bo’lsa, f '(π) ni toping A) 1 B) 3 C) 0 D) 5 15 / 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√2 B) 1 C) 2√3 D) 4√2 16 / 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) √3/2 C) (8√3)/2 D) (81√3)/2 17 / 25 x=1, y=2x va y=2-x chiziqlari bilan chegaralangan sohaning yuzini toping A) log₄e B) ln3 C) ln4 D) log₃e 18 / 25 Massasi 735 gr va konsentratsiyasi 16% bo’lgan yodning spirtdagi eritmasiga qancha toza spirt qo’shilsa, uning konsentratsiyasi 10% bo’ladi A) 441 B) 252 C) 256 D) 420 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) 1 B) 23 C) 14 D) 3 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) 3 C) 1 D) 3√2 21 / 25 Uchlari soni 14 ta, qirralari soni esa 21 ta bo’lgan ko’pyoqning yoqlari sonini toping. A) 8 B) 9 C) 4 D) 16 22 / 25 (x2+23x+23)(x2+x+23)=23x2 tenglama haqiyqiy ildizlari yig’indisini toping A) 24 B) -24 C) 23 D) -23 23 / 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 24 / 25 chiziqlar bilan chegaralangan shaklning yuzini toping A) 1 B) 2 C) 0,5 D) √2 25 / 25 cos150*cos450 *cos750 ni hisoblang A) √8/2 B) 2√2 C) √2 D) √2/2 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 Author: InfoMaster Matematika choraklik Teglar:11-sinf, matematika