Uy » Attestatsiya testlar » Matematika attestatsiya » Matematika attestatsiya №5 Matematika attestatsiya Matematika attestatsiya №5 InfoMaster Yanvar 24, 2022 33 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0% 2 ovozlar, 1 avg 0 12345678910111213141516171819202122232425262728293031323334353637383940 Matematika fanidan attestatsiya savollari №5 1 / 40 y=4cos 2x+cos8x funksiyaning hosilasini toping A) 16sin3x*cos5x B) -16cos3x*sin5x C) 16sin3x*sin5x D) 16cos3x*cos5x 2 / 40 Hisoblash natijasida hosil bo’lgan sondan 4 marta katta sonning natural bo’luvchilari soni a bo’lsa , a ning butun bo’luvchilari sonini toping. A) 8 B) 16 C) 24 D) 32 3 / 40 ABCD qavariq to‘rtburchakka aylana ichki chizilgan. AB = 8, BC = 12 bo‘lsa, CD – AD ayirmani toping. A) aniqlab bo‘lmaydi B) 4 C) 3 D) 2 4 / 40 f(x)=3x-2 funksiyaning qiymatlar sohasini toping. A) [-1; ∞) B) (-2; ∞) C) (-1; ∞) D) (0; ∞) 5 / 40 y=f(x) funksiya uchun tenglik o'rinli bo'lsa, f(π/4)=? A) 1/5 B) 1/3 C) 4 D) 1/4 6 / 40 Bir odam shunday vasiyat qildi: “Naqd 10 dirham pulim bor. Bir kishiga qarz ham berganman. Qarzning miqdori o'g'lim oladigan merosga teng. Ikkala o'g'lim barobar meros olishsin. Ukamga jami merosning 0,2 qismini va yana 1 dirham beringlar”. U kishining o'g'illari necha dirhamdan meros olishgan? A) 6 B) 35/6 C) 25/3 D) 8 7 / 40 Quyida berilgan holatda stol va uni atrofida qo’yiladigan stullarning holati berilgan va raqamlangan k-holatda stullarning raqamlari yig`indisi 351 taga teng bo`lsa, k ni toping ? A) 12 B) 13 C) 11 D) 10 8 / 40 Bir idishda 32/5 kg, ikkinchisida esa unga qaraganda 16/5 kg ortiq yog` bor. Ikkala idishda qancha yog` bor? A) 64/5 B) 48/5 C) 16 D) 10 9 / 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-EKUB 3-EKUB B) 1-EKUB 2-EKUB 3-EKUK C) 1-EKUB 2-EKUK 3-EKUK D) 1-EKUB 2-EKUK 3-EKUB 10 / 40 Dastlabki n ta hadining yig‘indisi formula bilan aniqlanadigan arifmetik progressiyaning 6-hadini toping. A) 6 B) 9 C) 8 D) 10 11 / 40 cos75° ·cos 45° ·cos15° ni hisoblang. A) √3/8 B) √2/8 C) 1/8 D) √3/4 12 / 40 Agar bo'lsa, ni m orqali ifodalang. A) 2/m B) (m+1)/2 C) (m+4)/4 D) (4-m)/4 13 / 40 b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi? A) 5 yoki –15 B) 6 C) 3 yoki –8 D) 4 yoki –12 14 / 40 Agar geometrik progressiyaning umumiy hadi bn= 3·2n bo‘lsa, ni toping. A) 4 B) 2 C) 1 D) 3 15 / 40 tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping. A) 8 B) 3 C) 5 D) -2 16 / 40 Hisoblang: A) 0 B) 1/32 C) -1/2 D) -√3/2 17 / 40 Tenglamani yeching: A) -6; 5 B) -6 C) 6; -5 D) 5 18 / 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) 4 C) 1 D) 3 19 / 40 Hisoblang: A) cos10° B) cos50° C) sin10° D) 1 20 / 40 Tenglamalar sistemani yeching: A) (7; 2), (28; -1) B) (9; 0), (28; -1) C) (9; 0), (2; 7) D) (2; 3) 21 / 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) 6 B) 3√2 C) 4√2 D) 4√3 22 / 40 sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping. A) 14 B) 23 C) 24 D) 35 23 / 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) AB = BC B) BC > AD C) BD = BC D) DC < AD 24 / 40 Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d) A) 1 B) cd/b C) bd/c D) bc/d 25 / 40 Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping. A) 16(π-2) B) 8(π+2) C) 64 D) 32 26 / 40 Chizmadan foydalanib α ni toping. A) 20° B) 30° C) 50° D) 40° 27 / 40 sonlarini taqqolsang. A) a B) b C) c D) c 28 / 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) 10 D) 4√7 29 / 40 Agar α=60°, β=70°, γ=50° bo‘lsa, tgα+ tgβ+ tgγ yig‘indi quyidagilardan qaysi biriga teng? A) √3tg50° tg70° B) 4 √3tg50° tg70° C) 2√3tg50° tg70° D) √3ctg40° tg70° 30 / 40 an - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping: A) 7 B) 6 C) 5 D) 8 31 / 40 Sonlarining o‘rta geometrik qiymatini toping. A) 4√3 B) 3√2 C) 2√2 D) 2√3 32 / 40 6-(5:3)·(-3)²-(-3)³+15:(-3) hisoblang. A) 13 B) 14 C) 15 D) 12 33 / 40 Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi? A) 42 B) 45 C) 43 D) 44 34 / 40 ABC o‘tkir burchakli uchburchakning BC asosiga AD balandlik, AC yon tomoniga BE balandlik o‘tkazilgan. Bunda CE =2AE = 8, DC = 6 bo‘lsa, BD ni toping. A) 10 B) 6 C) 8 D) 12 35 / 40 vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping. A) 2 B) 3 C) 4 D) 1 36 / 40 Kvadratlarning yuzlari yig‘indisini toping. A) berilganlar yetarli emas B) 22 C) 11 D) 121 37 / 40 Tengsizlik nechta butun juft yechimga ega? A) 115 B) 112 C) 110 D) 116 38 / 40 √3 A) 7/3 B) 5/5 C) 3/2 D) 21/10 39 / 40 3x3 o‘lchamli kvadratning tugunlarida 16 ta nuqta belgilanib, ularning o‘ng tomondan eng yuqorisidagi A bilan belgilangan. Bir uchi A nuqtada, qolgan uchlari qolgan 15 ta nuqtada orasidan tanlanadigan uchburchaklarning sonini toping. A) 100 B) 25 C) 105 D) 96 40 / 40 Hisoblang: A) 10/11 B) 9/10 C) 20/21 D) 19/20 O'rtacha ball 0% 0% Testni qayta ishga tushiring Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
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