Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 31 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 Agar bo‘lsa, x ni toping. A) 5 B) -2 C) 2 D) -5 2 / 40 ni 9 ga bo’lgandagi qoldiqning kvadratini toping. A) 9 B) 25 C) 16 D) 4 3 / 40 Agar va bo'lsa f(4) ning qiymatini toping. A) 35 B) 11 C) 12 D) 4 4 / 40 Agar a>0 bo'lsa, funksiyaning vertikal asimptotasini toping. A) x=a B) y=1-a C) x=-a D) y=-a 5 / 40 funksiyaning [2; 3] kesmadagi eng katta qiymatini A) 7,5 B) 4 C) 3 D) 4,5 6 / 40 Tenglama ildizlari ayirmasining modulini toping. A) √5+5 B) √5 C) 5 D) 2√5 7 / 40 Hisoblang? A) 0,25 B) 0,5 C) 4 D) 2 8 / 40 Quyida berilgan tizimda yonma - yon shakillardagisonlar ustida pasdagi ( EKUB yoki EKUK topish) amali bajariladi , natija amal pastigi shakilga yoziladi Shunga ko`ra 1 , 2 va 3 amallar EKUB yoki EKUK dan qaysi biri bo`lishi kerak A) 1-EKUB 2-EKUK 3-EKUB B) 1-EKUB 2-EKUB 3-EKUK C) 1-EKUB 2-EKUB 3-EKUB D) 1-EKUB 2-EKUK 3-EKUK 9 / 40 Hisoblang 200−199+198−197+⋯+4−3 A) 98 B) 100 C) 101 D) 99 10 / 40 f(x)+f(3x)+f(5x)+...+f(39x)=(3x+10)4 bo'lsa, f'(0) ni toping. A) 25 B) 10 C) 30 D) 0 11 / 40 Tenglamalar sistemasidan foydalanib x + y + z ni toping: A) berilganlar yetarli emas B) 10 C) 8 D) 6 12 / 40 Integralni hisoblang: A) 2 B) 4 C) 3 D) 1 13 / 40 Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d) A) bd/c B) bc/d C) cd/b D) 1 14 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) 121 B) berilganlar yetarli emas C) 11 D) 22 15 / 40 Tenglamalar sistemasining ildizlari yig‘indisini toping. A) 20 B) 10 C) 14 D) 12 16 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 35 B) 14 C) 23 D) 24 17 / 40 ABC uchburchakning A burchgi 30° ga, B burchagi 75° ga teng. B uchidan AC tomonga BD kesma o‘tkazilgan. ABD burchak 45° ga teng bo‘lsa, quyidagilardan qaysi biri noto‘g‘ri? A) DC < AD B) AB = BC C) BC > AD D) BD = BC 18 / 40 Hisoblang: A) 20/21 B) 19/20 C) 9/10 D) 10/11 19 / 40 √3 A) 21/10 B) 7/3 C) 3/2 D) 5/5 20 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 32 B) 64 C) 16(π-2) D) 8(π+2) 21 / 40 P(x)=x¹ºº ko‘phadni x³-3x+2 ga bo‘lganda qoladigan qoldiqni toping. 2¹ºº-1 (2¹ºº-1)x+2(299-1) (2¹ºº-1)x-2(299-1) 2¹ººx-3·2100 A) 2 B) 3 C) 4 D) 1 22 / 40 ABC to‘g‘ri burchakli uchburchakning og‘irlik markazi G nuqta. Bunda AB ⊥ BC, AG ⊥ GB va AG = 8 bo‘lsa, BG ni toping. A) 4√3 B) 6 C) 4√2 D) 3√2 23 / 40 Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping. A) 6√7 B) 8 C) 4√7 D) 10 24 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 4 B) 3 C) 1 D) 2 25 / 40 Tengsizlik nechta butun juft yechimga ega? A) 116 B) 112 C) 115 D) 110 26 / 40 P(x+1)=x³+3x²-2x+a+3 ko‘phadi berilgan. P(x+2) ko‘phadining koeffitsiyentlari yig‘indisini 8 ga teng bo‘lsa, a nechaga teng? A) 5 B) -4 C) -6 D) -3 27 / 40 Rasmdagi shakl perimetrini toping. A) 30 B) 24 C) 28 D) 32 28 / 40 Tenglamalar sistemani yeching: A) (9; 0), (2; 7) B) (2; 3) C) (9; 0), (28; -1) D) (7; 2), (28; -1) 29 / 40 Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng. 319·25 316·28 313·28 329·28 A) 3 B) 4 C) 2 D) 1 30 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 2√3 B) 3√2 C) 2√2 D) 4√3 31 / 40 sonlarini taqqolsang. A) a B) c C) c D) b 32 / 40 1+sin 2x = 7(sin x + cos x) tenglamani yeching. A) Ø B) 0 C) -π/4+πk, k∈Z D) π/4+πk, k∈Z 33 / 40 x²-√11x+1=0 0 bo‘lsa, A) 10 B) 12 C) 11 D) 9 34 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 13 B) 14 C) 15 D) 12 35 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 44 B) 43 C) 45 D) 42 36 / 40 b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi? A) 6 B) 5 yoki –15 C) 4 yoki –12 D) 3 yoki –8 37 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 3 B) 4 C) 2 D) 1 38 / 40 To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping. A) 42 B) 48 C) 32 D) 64 39 / 40 Hisoblang: A) 1/32 B) -1/2 C) 0 D) -√3/2 40 / 40 Hisoblang: A) cos50° B) 1 C) cos10° D) sin10° O'rtacha ball 0% 0% Testni qayta ishga tushiring Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
Istaklar ro'yxatiga qo'shildiIstaklar ro'yxatidan olib tashlandi 13 Matematika fanidan attestatsiya savollari №16
