Matematika attestatsiya №12 Aprel 8, 2022Aprel 8, 2022 da chop etilgan InfoMaster tomonidan Matematika attestatsiya №12 ga 3 fikr bildirilgan 0% 4 12345678910111213141516171819202122232425262728293031323334353637383940 OMAD YOR BO'LSIN! Matematika fanidan attestatsiya savollari №12 28.03.2022 da Toshkentda tushgan savollar DIQQAT! Endi siz o'z bilmingizni sinab ko'rish bilan birga sertifikatga ham ega bo'lishingiz mumkin. Sertifikat olish uchun barcha ma'lumotlarni to'g'ri kiriting! e-mail manzilini to'g'ri kriting, barcha ma'lumotlar sizga yuboriladi. Testda 76% va undan yuqori natija oling va sertifikatni yuklab oling. 1 / 40 Hisoblang. A) 1.5 B) √3 C) 3 D) 1 2 / 40 {an} - ketna-ket berilgan, bunda a1=40 va ak+1=ak-k, k∈N a8 ni toping. A) 12 B) 13 C) 32 D) 35 3 / 40 BCD-A markazli aylananing yoyi, ED // BA, DE⟘EF, EC=ED=7 va BF=1 boʻlsa, AF ning qiymatini toping. A) 12 B) 13 C) 15 D) 15 4 / 40 0, 1, 2, 3, 4, 5 raqamlaridan nechta 3 xonali tuzish mumkin? A) 180 B) 100 C) 200 D) 125 5 / 40 1, 2, 3, 4 raqamlaridan foydalanib 400 dan kichik kechta uc xonali son yasash mumkin? A) 49 B) 48 C) 51 D) 47 6 / 40 ABCD parallelogramm va SKBCE = 18 bo'lsa, SABCD =? A) 54 B) 48 C) 62 D) 56 7 / 40 Arimetik progerssiyada s12=354, P=a1+a2+...+a11 , Q=a2+a4+...+a12 va P/Q=32/27 ga teng bo'lsa, d ni toping. A) -3 B) -5 C) -4 D) -6 8 / 40 y=xlnx funksiyani hosilasini toping. A) 1 B) 2lnx-1 C) lnx/2+1 D) lnx+1 9 / 40 Chizmada nechta uchburchak tasvirlangan? A) 54 B) 50 C) 52 D) 48 10 / 40 Hisoblang. A) 1 B) 4 C) 3 D) 2 11 / 40 Hisoblang. A) 2020 B) 1 C) 2021 D) 2019 12 / 40 Hisoblang. A) 0 B) 9/11 C) 7/11 D) 9/13 13 / 40 Agar va ular orasidagi burchak 120° bo'lsa, ni toping. A) 2√7 B) √7 C) 4√5 D) 3√7 14 / 40 Tenglamani yeching: A) 6 B) 2,5 C) 2 D) 5,5 15 / 40 Hisoblang. A) 24+3b B) 14-2b C) 24-3b D) 26-3b 16 / 40 Hisoblang. A) 133 B) 131 C) 135 D) 121 17 / 40 Hisoblang. A) 3/2 B) 5/3 C) 2/3 D) 3/5 18 / 40 (tgx+ctgx)⁴ yoyilmaning ozod hadini toping. A) 5 B) 4 C) 3 D) 6 19 / 40 Tengsizlikni yeching. A) [-7;-3] u [-1;∞) B) [-7;-3) u [-1;∞) C) [-7;-5) u [-1;∞) D) [-7;3 u [-1;∞) 20 / 40 Agar bo'lsa, ni toping. A) 3 B) 0 C) 2,5 D) 5,2 21 / 40 Tenglamaning [0;π] kesmadagi ildizlari soni 3 tadan ko'pmi? A) Yo'q B) Ha 22 / 40 f(x)+(x-2)·g(x)=x²+5, f(3)=10 bo'lsa, g-1=(4) ni toping. Bunda g-1(x) g(x) ga teskari funksiya. A) 6 B) 3 C) 5 D) 4 23 / 40 Tamonlarini ayirmasi 14 bo'lgan parallelogramning o'rmas burchagidan katta diagonaliga tushgan balandlik uni 12 va 30 ga teng kesmalarga ajratadi. Tamonlar uzunligini toping. A) 32 va 22 B) 34 va 20 C) 36 va 18 D) 30 va 24 24 / 40 Hisoblang. cos²10° + cos²30° + cos²80° A) 0 B) 7/3 C) 7/4 D) 3/7 25 / 40 Tengsizlikni yeching. A) x∈[0;1000] B) x∈(0;1000) C) x∈(1;1000] D) x∈(10;1000) 26 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 24 B) 25 C) 6 D) 58 27 / 40 Agar bo'lsa f(x)=? A) 2x²-1 B) x²-2 C) x²-1 D) x²+2 28 / 40 Tenglama ildizlari yig'indisini toping. A) 0 B) 4 C) -1 D) 2 29 / 40 Hisoblang. A) -1 B) 0 C) 1 D) 2 30 / 40 Hisoblang. A) 49/25 B) -47/25 C) 51/29 D) -47/24 31 / 40 Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping. A) 7/16 B) 9/16 C) 5/16 D) 5/13 32 / 40 Hisoblang. A) 32 B) 94 C) 96 D) 98 33 / 40 Hisoblang A) 6 B) 9 C) 36 D) 12 34 / 40 |x²-2x|≤x yengsizlikni yeching. A) {0}∪[1;3) B) {0}∪(1;3] C) {0}∪[1;3] D) (-2;0)∪[1;3] 35 / 40 Hisoblang. A) 2 B) 4 C) 1 D) 0 36 / 40 Hisoblang. A) 3 B) 2 C) 6 D) 1 37 / 40 a=10,(6) , b=80,(3) , c=20,(9) bo'lsa, (a+b)c=? A) 0 B) 25 C) 9 D) 16 38 / 40 Uchburchak tamonlari 10, 24 va 26 ga teng bo'lsa, katta burchagi necha gradus? A) 60° B) 75° C) 90° D) 80° 39 / 40 Ayirmasi d ga teng bo'lgan arifmetik progrssiya uchun ni toping. A) 1/2 B) 2/3 C) 1 D) 2 40 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) 4x+c B) 2x+c C) x/2+c D) x+c 0% Testni qayta ishga tushiring Baholash mezoni - 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) saqlansin”; - 75 foizdan past ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi)ga tushirilsin”; - 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) berilsin”; - 74 foizdan 65 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi) saqlansin”; - 65 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi)ga tushirilsin” - 64 foizdan 60 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi) saqlansin”; - 60 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, mutaxassis (oliy yoki o’rta maxsus, kasb-hunar ma'lumotli o’qituvchi) lavozimiga tushirilsin” Fikr-mulohaza yuboring Author: InfoMaster Matematika attestatsiya