Matematika attestatsiya №12 Aprel 8, 2022Aprel 8, 2022 da chop etilgan InfoMaster tomonidan Matematika attestatsiya №12 ga 3 fikr bildirilgan 0% 134 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 Agar ning qiymatini toping. A) 12 B) 12,5 C) 6 D) 8 2 / 40 Soddalashtiring: ctg2α-tgα A) -1/sin2α B) -1/cos2α C) 1/sin2α D) 1/cos2α 3 / 40 DC-B markazli aylana yoyi, AC-urinma, AB-kesuvchi va S₁=2S₂ ekanligi maʼlum boʻlsa, tgα/α ni toping. A) 3 B) 5/2 C) 1 D) 3/2 4 / 40 (a, b∈ N) va 3a – 4b = 5 bo‘lsa, 3a+4b ning eng kichik qiymatini toping. A) 13 B) 9 C) 37 D) 12 5 / 40 Tengsizlikni yeching: 3 · 9x –10 · 21x + 7 · 49 x ≤ 0 A) [-1;0) B) [-1;0] C) (-1;3] D) [-1;1] 6 / 40 To'g'ri burchakli uchburchakning yuzi 96 ga, perimetri esa 48 ga teng bo'lsa, gipatenuza uzunligini toping. A) 21 B) 18 C) 23 D) 20 7 / 40 Tenglama ildizlari yig'indisini toping. A) 2 B) 0 C) 4 D) -1 8 / 40 Agar bo'lsa, ni toping. A) 5,2 B) 0 C) 3 D) 2,5 9 / 40 Hisoblang. A) 0 B) 1 C) 2 D) -1 10 / 40 Hisoblang. A) 4 B) 1 C) 3 D) 2 11 / 40 Uchburchak tamonlari 10, 24 va 26 ga teng bo'lsa, katta burchagi necha gradus? A) 75° B) 80° C) 90° D) 60° 12 / 40 Hisoblang. A) 1 B) 2 C) 4 D) 0 13 / 40 Agar bo'lsa f(x)=? A) 2x²-1 B) x²-1 C) x²+2 D) x²-2 14 / 40 Hisoblang. A) 26-3b B) 24+3b C) 14-2b D) 24-3b 15 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 58 B) 25 C) 6 D) 24 16 / 40 Hisoblang A) 12 B) 6 C) 36 D) 9 17 / 40 Hisoblang. A) 131 B) 133 C) 121 D) 135 18 / 40 Tenglamani yeching. A) 2 B) 1 C) 0 D) 5 19 / 40 Hisoblang. A) 2/3 B) 3/2 C) 5/3 D) 3/5 20 / 40 (tgx+ctgx)⁴ yoyilmaning ozod hadini toping. A) 6 B) 3 C) 5 D) 4 21 / 40 Hisoblang. A) 4 B) 5 C) 6 D) 3 22 / 40 Hisoblang. A) 1 B) 6 C) 3 D) 2 23 / 40 Agar va ular orasidagi burchak 120° bo'lsa, ni toping. A) √7 B) 4√5 C) 3√7 D) 2√7 24 / 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) 30 va 24 D) 36 va 18 25 / 40 Tenglamani yeching: A) 2 B) 2,5 C) 5,5 D) 6 26 / 40 Hisoblang. A) -2 yoki 0 B) -2 yoki -1 C) 0 D) -2 yoki 2 27 / 40 Hisoblang. A) -47/24 B) 51/29 C) -47/25 D) 49/25 28 / 40 Hisoblang. A) 9/11 B) 0 C) 9/13 D) 7/11 29 / 40 Hisoblang. A) 1 B) 2021 C) 2019 D) 2020 30 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) x+c B) 2x+c C) 4x+c D) x/2+c 31 / 40 |x²-2x|≤x yengsizlikni yeching. A) {0}∪[1;3] B) {0}∪(1;3] C) (-2;0)∪[1;3] D) {0}∪[1;3) 32 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) 2x+c B) c C) x+c D) 4x+c 33 / 40 f(x) = x²-3x-2 bo'lsa. f(a+b+c)-f(abc)=? Bunda a, b, c - kvadrat funksiya koeffitsiyentlari. A) 36 B) 38 C) 42 D) 40 34 / 40 ABCD parallelogramm va SKBCE = 18 bo'lsa, SABCD =? A) 54 B) 62 C) 56 D) 48 35 / 40 Hisoblang. cos²10° + cos²30° + cos²80° A) 0 B) 7/3 C) 3/7 D) 7/4 36 / 40 Chizmada nechta uchburchak tasvirlangan? A) 48 B) 54 C) 52 D) 50 37 / 40 To'g'ri burchakli uchburchakning gipotenuzasiga o'tkazilgan bissiktriksasi uni 20 va 30 ga teng kesmalarga ajratadi. To'g'ri burchakli uchburchak yuzini toping. A) 7500/13 B) 6700/13 C) 15 D) 7551/15 38 / 40 5-x ≤ 0,04 x ning eng kichik butun qiymatini toping. A) -1 B) 3 C) 0 D) 2 39 / 40 Tenglamani yeching. A) 1 B) 4 C) 2 D) 3 40 / 40 Tengsizlikni yeching. A) x∈(1;1000] B) x∈[0;1000] C) x∈(0;1000) D) x∈(10;1000) 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