Matematika attestatsiya №12 Aprel 8, 2022Aprel 8, 2022 da chop etilgan InfoMaster tomonidan Matematika attestatsiya №12 ga 3 fikr bildirilgan 0% 11 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 a,b>0 sonlari uchun f(a·b) = f(a) + f(b) tengligi, p – tub sonlari uchun esa f(p)=p tengligi qanoatlantirilsa, u holda f(25/11) ning qiymatini toping. A) 21 B) -1 C) 1 D) 11 2 / 40 Qfodaning qiymatini toping. A) 0,5 B) 2 C) 1 D) 0,58 3 / 40 Agar bo’lsa, ni qiymatini toping. A) 1 B) 1/25 C) 1/5 D) 5 4 / 40 m/n=p-7=7+5/p bo'lsa, m²/n² ni toping. A) 56 B) 70 C) 63 D) 54 5 / 40 y=xlnx funksiyani hosilasini toping. A) lnx/2+1 B) 2lnx-1 C) 1 D) lnx+1 6 / 40 Tenglama ildizlari yig'indisini toping. A) 2 B) 0 C) -1 D) 4 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) -4 C) -6 D) -5 8 / 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) 6700/13 B) 7551/15 C) 15 D) 7500/13 9 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 24 B) 58 C) 25 D) 6 10 / 40 Hisoblang. A) 1 B) -1 C) 2 D) 0 11 / 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 12 / 40 5-x ≤ 0,04 x ning eng kichik butun qiymatini toping. A) -1 B) 0 C) 2 D) 3 13 / 40 Hisoblang. A) 14-2b B) 24+3b C) 24-3b D) 26-3b 14 / 40 a=10,(6) , b=80,(3) , c=20,(9) bo'lsa, (a+b)c=? A) 16 B) 25 C) 9 D) 0 15 / 40 Hisoblang. A) 96 B) 32 C) 98 D) 94 16 / 40 Ayirmasi d ga teng bo'lgan arifmetik progrssiya uchun ni toping. A) 1 B) 1/2 C) 2 D) 2/3 17 / 40 (tgx+ctgx)⁴ yoyilmaning ozod hadini toping. A) 5 B) 3 C) 6 D) 4 18 / 40 Hisoblang. A) 131 B) 121 C) 135 D) 133 19 / 40 Hisoblang. A) 0 B) -1 C) 1 D) 2 20 / 40 Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping. A) 7/16 B) 5/16 C) 9/16 D) 5/13 21 / 40 Tenglamani yeching: A) 2 B) 6 C) 2,5 D) 5,5 22 / 40 |x²-2x|≤x yengsizlikni yeching. A) {0}∪[1;3] B) {0}∪[1;3) C) (-2;0)∪[1;3] D) {0}∪(1;3] 23 / 40 ABCD parallelogramm va SKBCE = 18 bo'lsa, SABCD =? A) 54 B) 62 C) 56 D) 48 24 / 40 Hisoblang. A) 6 B) 2 C) 3 D) 1 25 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) x+c B) c C) 4x+c D) 2x+c 26 / 40 Tengsizlikni yeching: 3 · 9x –10 · 21x + 7 · 49 x ≤ 0 A) [-1;0) B) [-1;0] C) (-1;3] D) [-1;1] 27 / 40 f(x) = x²-3x-2 bo'lsa. f(a+b+c)-f(abc)=? Bunda a, b, c - kvadrat funksiya koeffitsiyentlari. A) 40 B) 42 C) 38 D) 36 28 / 40 Hisoblang. A) 3 B) 1 C) 2 D) 4 29 / 40 Hisoblang. A) 3/2 B) 2/3 C) 5/3 D) 3/5 30 / 40 To'g'ri burchakli uchburchakka ichki va tashqi chizilgan aylanalar raduslari 6 va 15 ga teng bo'lsa, katetlar yig'indini toping. A) 42 B) 28 C) 35 D) 48 31 / 40 (2; 3; -5) nuqtaning OX o'qiga nisbatan simmetrik nuqtasini toping. A) (2; -3; -5) B) (-2; 3; 5) C) (2; 3; -5) D) (2; -3; 5) 32 / 40 Agar cos4x = acos⁴x+bcos2²x+1 bo'lsa, a+b=? A) 1 B) -1 C) 2 D) 0 33 / 40 Tengsizlikni yeching. A) x∈(1;1000] B) x∈[0;1000] C) x∈(10;1000) D) x∈(0;1000) 34 / 40 Hisoblang. cos²10° + cos²30° + cos²80° A) 7/4 B) 3/7 C) 7/3 D) 0 35 / 40 y = x²+1-|sinx| funksiya grafigi OX o'qidan pastda yotadimi? A) Yotmaydi B) Yotadi 36 / 40 Hisoblang. A) 0 B) 7/11 C) 9/11 D) 9/13 37 / 40 Agar bo'lsa f(x)=? A) x²+2 B) x²-2 C) x²-1 D) 2x²-1 38 / 40 Agar bo'lsa, ni toping. A) 0 B) 2,5 C) 5,2 D) 3 39 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) x+c B) 2x+c C) 4x+c D) x/2+c 40 / 40 Hisoblang. A) 1 B) 0 C) -1 D) 2 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