Uy » Choraklik online testlar » Matematika choraklik » 11-sinf Matematika 2-chorak Matematika choraklik 11-sinf Matematika 2-chorak InfoMaster Aprel 20, 2021 379 Ko'rishlar 16 izohlar SaqlashSaqlanganOlib tashlandi 1 0 OMAD YOR BO'LSIN! Tomonidan yaratilgan InfoMaster 11-sinf Matematika 2-chorak Testni Mamadoliyev Diyorbek tayyorladi. 1 / 25 f(x)=sin2x funksiyasining barcha boshlang’ich funksiyalarini toping. A) F(x)=0,5cos2x+C B) F(x)=-0,5cos2x+C C) F(x)=-2cos2x+C D) F(x)=2sin4x+c 2 / 25 Kub to’la sirtining yuzi 105,84 sm2 bo’lsa, uning har bir yog’i yuzini va qirrasining uzunligini toping. A) 17,64 dm; 4,2 dm B) 17,64 dm²; 4,2 dm C) 17,68 dm²; 4,2 dm D) 17,64 dm²; 4,9 dm 3 / 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) 4 / 25 f(x)=(3x+2)2 funksiyasining barcha boshlang’ich funksiyalarini toping. A) F(x)=6x³+6x+4 B) F(x)=6x³+6x²+4 C) F(x)=6x³+3x²-4 D) F(x)=6x³+3x²+4 5 / 25 To’g’riburchakli parallelepiped bir uchidan chiquvchi uchta qirralari uzunliklari 4,6 va 9 ga teng. Unga tengdosh kub qirrasining uzunligini toping. A) 8 B) 6 C) 6 D) 7 6 / 25 To’g’ri to’rtburchak shaklidagi yer maydoninig atrofini o’rashmoqchi. 480 m panjara yordamida eng ko’pi bilan necha kvadrat metr yer maydonni o’rash mumkin? A) 8100 B) 14400 C) 19600 D) 25600 7 / 25 To’g’riburchakli parallelepiped asosining tomonlari 7:24 nisbatda, diagonal kesimining yuzi 50dm2 ga teng. Yon sirtining yuzini toping. A) 124 dm B) 224 dm² C) 124 dm² D) 135 dm² 8 / 25 funksiya hosilasining x0=-2 nuqtadagi qiymatini hisoblang. A) -38/31 B) -38/961 C) 38/961 D) 38/31 9 / 25 Karim uyida 14 bet matn terib, 8ta imloviy xatoga yo’l qo’ydi. Agar 1 betda o’rtacha 380 ta so’z bo’lsa, Karim 100ta so’z terganda o’rtacha qancha xato qiladi? A) 1,5 B) 0,15 C) 0,25 D) 0,5 10 / 25 tgx funksiya hosilasini toping A) sinx B) cosx C) 1/sinx D) 1/cos²x 11 / 25 funksiya grafigiga x=4 abssissali nuqtada o`tkazilgan urinma tenglamasini tuzing A) y=35x B) y=20x+20 C) y=10x D) y=35x+30 12 / 25 A(-4;3;6) nuqtaning Oxz tekisligiga nisbatan simmrtrik nuqtasini toping. A) (-4;-3;9) B) (-8;-3;6) C) (-4;-6;6) D) (-4;-3;6) 13 / 25 f(x)=3x2-5x+4 va g(x)=4x-5 funksiyalarining urinmalari parallel bo’ladigan nuqtalarini toping. A) 0 B) 1.5 C) 0.5 D) 2 14 / 25 Koordinatalar boshidan y=x2-4x+3 parabolaning simmetriya o’qigacha bo’lgan masofani toping. A) 1 B) 3 C) 2 D) 4 15 / 25 Karim 3 minutda 213ta so`zni terib , 6ta imloviy xatoga yo`l qo`ydi. Karimning matn terish sifatini toping. A) 0,0321 B) 0,3265 C) 0,623 D) 0,0282 16 / 25 Murakkab funksiyaning hosilasini toping. y=x2sinx A) y’=x²cosx-2xsinx B) y’=2xsinx-x² C) y’=2xsinx+x²cosx D) y’=-2xsinx+x² 17 / 25 Funksiya hosilasini toping: A) 12x²-12 B) 21x²-10x C) 12x+12 D) 14x+12 18 / 25 Uchlari A(4;0;1), B(5;-2;1), C(4;8;5) nuqtalarda bo’lgan uchburchakning AL bissektrisasi uzunligini toping A) (2√5)/3 B) 4 C) (4√2)/5 D) (4√2)3 19 / 25 Funksiya hosilasini toping: f(x)=5 A) 0 B) x C) 10 D) 5 20 / 25 Uchlari A(2;-3;4) va B(-2;4;5) nuqtalarda bo’lgan AB kesmasining uzunligini toping A) √66 B) √55 C) √19 D) √24 21 / 25 Agar A(1;2;3), B(3;7;6) bo’lsa, A̅B̅ vector koordinatalarini toping A) (4;9;9) B) (2;5;3) C) (-2;-5;-3) D) (4;9;9) 22 / 25 Funksiya hosilasini toping f(x)=(2x+4)(3x+1) A) 12x B) 12x+14 C) 16x+14 D) 3x+1 23 / 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 24 / 25 Moddiy nuqta S(t)=5t3+3t2+2t+4 qonuniyat bilan harakatlanayotgan jismning t=2 sekunddagi tezligini toping. A) 70 B) 84 C) 60 D) 74 25 / 25 y=-3x3+2x2-4x+5 funksiyaga x0=-1 nuqtada o’tkazilgan urinma tenglamasining burchak koeffitsiyentini toping. A) -17 B) 17 C) 15 D) -13 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