Matematika attestatsiya №12 Aprel 8, 2022Aprel 8, 2022 da chop etilgan InfoMaster tomonidan Matematika attestatsiya №12 ga 3 fikr bildirilgan 0% 133 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 Birinchi va oxirgi raqami juft bo‘lgan va 1000 ga bo‘linmaydigan barcha to‘rtxonali sonlar nechta? A) 960 B) 1600 C) 1996 D) 1800 2 / 40 n ning nechta butun qiymatida kasr butun son bo‘ladi? A) 8 B) 9 C) 3 D) 10 3 / 40 f(x) va g(x) funksiyalar f(x)+(x-2)·g(x)=x²+5 bog’lanishda va f(3)=10 bo’lsa, g-1 (4) ning qiymatini toping.(g-1 (x)-g(x) ga teskari funksiya) A) 3 B) 2 C) 4 D) 6 4 / 40 360 ning juft va tub bo‘luvchilarining yig‘indisini toping. A) 1102 B) 1112 C) 960 D) 1092 5 / 40 Hisoblang. A) 5 B) 6 C) 4 D) 3 6 / 40 f(x) = x²-3x-2 bo'lsa. f(a+b+c)-f(abc)=? Bunda a, b, c - kvadrat funksiya koeffitsiyentlari. A) 38 B) 42 C) 40 D) 36 7 / 40 y = x²+1-|sinx| funksiya grafigi OX o'qidan pastda yotadimi? A) Yotadi B) Yotmaydi 8 / 40 Hisoblang. A) 51/29 B) -47/25 C) 49/25 D) -47/24 9 / 40 Hisoblang A) 6 B) 12 C) 9 D) 36 10 / 40 Agar bo'lsa, ni toping. A) 5,2 B) 3 C) 0 D) 2,5 11 / 40 Hisoblang. A) 133 B) 135 C) 131 D) 121 12 / 40 a=10,(6) , b=80,(3) , c=20,(9) bo'lsa, (a+b)c=? A) 0 B) 9 C) 16 D) 25 13 / 40 Hisoblang. A) 7/11 B) 9/11 C) 0 D) 9/13 14 / 40 5-x ≤ 0,04 x ning eng kichik butun qiymatini toping. A) -1 B) 0 C) 3 D) 2 15 / 40 Hisoblang. A) 3/5 B) 3/2 C) 5/3 D) 2/3 16 / 40 Hisoblang. A) 0 B) 1 C) -1 D) 2 17 / 40 Hisoblang. A) 14-2b B) 24-3b C) 24+3b D) 26-3b 18 / 40 Tengsizlikni yeching. A) x∈(10;1000) B) x∈(0;1000) C) x∈[0;1000] D) x∈(1;1000] 19 / 40 To'g'ri burchakli uchburchakka ichki va tashqi chizilgan aylanalar raduslari 6 va 15 ga teng bo'lsa, katetlar yig'indini toping. A) 48 B) 35 C) 28 D) 42 20 / 40 (tgx+ctgx)⁴ yoyilmaning ozod hadini toping. A) 6 B) 5 C) 3 D) 4 21 / 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) 22 / 40 Ayirmasi d ga teng bo'lgan arifmetik progrssiya uchun ni toping. A) 2/3 B) 1 C) 1/2 D) 2 23 / 40 Tengsizlikni yeching: 3 · 9x –10 · 21x + 7 · 49 x ≤ 0 A) [-1;0) B) [-1;0] C) [-1;1] D) (-1;3] 24 / 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) 7551/15 B) 6700/13 C) 7500/13 D) 15 25 / 40 Hisoblang. A) 1 B) 2 C) 4 D) 3 26 / 40 Hisoblang. A) 94 B) 98 C) 96 D) 32 27 / 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) 3 B) 5 C) 6 D) 4 28 / 40 Tenglamani yeching. A) 3 B) 1 C) 2 D) 4 29 / 40 Tengsizlikni yeching. A) [-7;-5) u [-1;∞) B) [-7;-3] u [-1;∞) C) [-7;3 u [-1;∞) D) [-7;-3) u [-1;∞) 30 / 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) -5 B) -4 C) -6 D) -3 31 / 40 Hisoblang. A) 2 B) 1 C) -1 D) 0 32 / 40 Agar cos4x = acos⁴x+bcos2²x+1 bo'lsa, a+b=? A) 0 B) 2 C) 1 D) -1 33 / 40 (tgx-2ctgx)⁴ yoyilmaning ozod hadini toping. A) 6 B) 25 C) 58 D) 24 34 / 40 Hisoblang. cos²10° + cos²30° + cos²80° A) 3/7 B) 7/4 C) 7/3 D) 0 35 / 40 Hisoblang. A) 3 B) 1 C) 6 D) 2 36 / 40 Agar bo'lsa f(x)=? A) x²-2 B) x²+2 C) 2x²-1 D) x²-1 37 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) 4x+c B) c C) x+c D) 2x+c 38 / 40 Tenglamaning [0;π] kesmadagi ildizlari soni 3 tadan ko'pmi? A) Ha B) Yo'q 39 / 40 Noaniq integralni toping. ∫(sin²x+cos²x)dx A) x/2+c B) x+c C) 4x+c D) 2x+c 40 / 40 Hisoblang. A) -2 yoki 0 B) -2 yoki 2 C) -2 yoki -1 D) 0 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