Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 134 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 4 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Radiusi 1 ga teng aylana uchta yoyga bo`lingan. Ularga mos markaziy burchaklar 1, 2 va 6 sonlariga proporsional. Yoylardan eng kattasining uzunligini toping. A) 3π/4 B) 2π/3 C) 4π/3 D) 3π/2 2 / 30 y= funktsiyaning aniqlanish sohasini toping. A) (2;∞) B) (-∞;2)v(2;∞) C) (-∞;2) D) [2;∞) 3 / 30 sonning oxirgi raqamini toping. A) 4 B) 6 C) 2 D) 8 4 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) 6,25% ga kamayadi B) o‘zgarmaydi C) 2,5% ga ortadi D) 6,25% ga ortadi 5 / 30 Agar f(x)=sin2x va g(x)=cos2x bo’lsa, u holda f(g(x)) funksiyaning hosilasini toping. A) 4sin2x*cos(cos2x) B) 4sin2x*cos(2cos2x) C) -4sin2x*cos(cos2x) D) -4sin2x*cos(2cos2x) 6 / 30 a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega A) a≠5 B) a ning bunday qiymati yo’q C) a=12 D) a≠12 7 / 30 A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping. A) 1 B) √3/2 C) √5/2 D) 1,5 8 / 30 Agar geometrik progressiyaning ketma–ket dastlabki uchta hadining yig’indisi 62 ga, ularning o’nli logarifmlari yig’indisi 3 ga teng bo’lsa, shu geometrik progressiyaning birinchi hadini toping. A) 2 yoki 50 B) 50 C) 10 yoki 50 D) 10 9 / 30 To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping A) 6 B) 4 C) 5 D) 7 10 / 30 tenglamalar sistemasini yeching A) (3;–4) B) (4;–3) C) (–4;3) D) (4;3) 11 / 30 7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin? A) 5 B) 3 C) 4 D) 6 12 / 30 Hisoblang A) 1 B) -1 C) 0 D) 2 13 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) 2√2 B) 4 C) 2 D) √2 14 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) juft funksiya B) toq funksiya C) bunday funksiya mavjud emas D) juft ham emas, toq ham emas funksiya 15 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) √2e B) 1 C) 3e D) e 16 / 30 Konsert zalining birinchi qatorida 40 ta o’rindiq bor. Har bir keyingi qatordagi o’rindiqlar soni oldingi qatordan 4 ga ko’p. Agar konsert zalida jami 40 ta qator bo’lsa, u holda shu zaldagi barcha o’rindiqlar sonini toping. A) 4720 B) 4680 C) 4760 D) 4716 17 / 30 sistemada xy ning qiymatini toping. A) 75 B) 64 C) 80 D) 60 18 / 30 To’rtburchakli muntazam piramidaning yon qirrasidagi ikki yoqli burchak 120 ga teng. Diagonal kesimining yuzasi S ga teng bo’lsa, uning yon sirtini toping. A) 0,5S B) 4S C) 2S D) 3S 19 / 30 Agar f(x)=ax3-5x2+b va bo’lsa, a ni toping. A) 3 B) 0 C) 2 D) 1 20 / 30 sin2x-cos2x=1 tenglama [-π; 2π] oraliqda nechta ildizga ega? A) 6 B) 9 C) 7 D) 10 21 / 30 Hisoblang A) √3 B) 2 C) √3/2 D) 1/2 22 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 10 π B) 7 π C) 12π D) 6 π 23 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) √3 B) 1 C) √2 D) 2 24 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 2π B) 3π C) 2π/3 D) 4π 25 / 30 Teng yonli uchburchakning tomonlari 5, 5 va 6 ga teng. Bu uchburchakning bissektiritsalari va medianalari kesishgan nuqtalar A) 1/6 B) 1/2 C) 1 D) 1,2 26 / 30 Soddalashtiring. (0<m<7) A) 7-2m B) 7 C) 2m-7 D) m 27 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) 6 B) (-∞;6) C) (-∞;6])v[6;∞) D) (2;∞) 28 / 30 Ostki asosining yuzi 16π ga va ustki asosining yuzi 4π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda sharning hajmini toping. A) 8√2π/5 B) (5√3+3)π/3 C) 64√2π/3 D) 3√2π/4 29 / 30 To’g’ri to’rtburchakning perimetri 50 ga teng. Bir tomoni boshqa tomonidan 5 ga ko’p. To’g’ri to’rtburchakning yuzini toping. A) 150 B) 50 C) 225 D) 60 30 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) b>c>a>d>e B) e>b>a>d>c C) b>a>c>d>e D) a>b>c>d>e 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
