Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 154 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 0 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 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(cos2x) C) -4sin2x*cos(2cos2x) D) 4sin2x*cos(2cos2x) 2 / 30 sistema a ning qanday qiymatida cheksiz ko’p yechimga ega? A) 6 B) (-∞;6) C) (-∞;6])v[6;∞) D) (2;∞) 3 / 30 Radiusi 25 bo’lgan doirada 48 ga teng vatar o’tkazilgan. Doira markazidan shu vatargacha masofani toping. A) 10 B) 9 C) 7 D) 8 4 / 30 Hisoblang A) 0 B) 2 C) -1 D) 1 5 / 30 bo’lsa, ni x orqali ifodalang. A) 2/x B) 25/x C) 2-x D) x/25 6 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 48√3 B) 52√3 C) 108 D) 54 7 / 30 Agar f(x)=4x3-6x2-2x+3x .log3e bo’lsa, u holda ni toping. A) 3e B) √2e C) e D) 1 8 / 30 Tomoni 25 ga diagonallaridan biri 4 ga teng bo’lgan rombning yuzini toping. A) 16 B) 8√5 C) 32 D) 24√5 9 / 30 Agar bank qo’yilgan pulga 40% yillik bersa, qo’yilgan 4500 so’m pul bir yildan so’ng qancha bo’ladi? A) 6200 B) 6300 C) 6100 D) 6000 10 / 30 Anvar tub son o’yladi va o’ylagan sonini 5 ga ko’paytirib, 8 ni ayirgan edi, yana tub son hosil bo’ldi. Anvar qanday son o’ylagan? A) 412967 B) 374389 C) 2 D) aniqlab bo’lmaydi 11 / 30 ABC muntazam uchburchak ichidan ixtiyoriy P nuqta olinib, undan BC, CA va AB tomonlarga mos ravishda PD, PE va PF perpendikulyarlar tushirilgan bo’lsa,ni toping. A) 0,5 B) 1/√2 C) 1/√3 D) 1 12 / 30 n ning qanday qiymatida ushbu 81 .82 .83 .….8n=51222 tenglik o’rinli bo’ladi? A) 14 B) 12 C) 11 D) 10 13 / 30 Agar va b-a=4 bo’lsa, a+b ni toping. A) 2 B) 3 C) 3,5 D) 4 14 / 30 a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas? A) (∞;∞) B) (-∞;2) C) (-∞;2)v(2;∞) D) (2;∞) 15 / 30 Yuzasi 10 ga teng bo’lgan kvadratning ketma-ket ikki uchidan o’tuvchi aylana chizilgan. Uchinchi uchidan aylanaga urunma o’tkazilgan. Urunma tomondan ikki marta katta bo’lsa, aylana radiusini toping. A) 5 B) 4 C) 6 D) 10 16 / 30 ABCD kvadrat ichidan olingan O nuqtadan A, B, C uchlarigacha bo’lgan masofalar mos ravishda 3, 4, 5 ga teng bo’lsa, u holda OD kesma uzunligini toping. A) √37 B) 6 C) 3 D) √32 17 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) 2 B) 1 C) √3 D) √2 18 / 30 Soddalashtiring. A) 2018a/a+1 B) a+1 C) 2018 D) 2019 19 / 30 Tengsizlik nechta butun yechimga ega? A) 3 B) cheksiz ko’p C) 1 D) 4 20 / 30 Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin? A) 47 B) 60 C) 18 D) 120 21 / 30 Arifmetik progressiyada a17=33 va a45=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping. A) 2√2 B) 2 C) √2 D) 4 22 / 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) 10 B) 2 yoki 50 C) 50 D) 10 yoki 50 23 / 30 Tenglamaning ildizlari yig’indisi va ko’paytmasining yig’indisini toping. |x+1|.|x-4|=5 A) 9 B) -3 C) -6 D) 6 24 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 67,5√3 B) 135√3/4 C) 67,5 D) 48√3 25 / 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) b>a>c>d>e C) e>b>a>d>c D) a>b>c>d>e 26 / 30 Bekzodda 50 so’m va Sobirda 70 so’m pul bor edi. Anvar Bekzodga o’z pulining 10 foizini bergandan so’ng, Bekzod Sobirga pulining yarmini berdi. So’ng Sobir Anvarga pulining 10 foizini berdi. Anvar o’zidagi pullarini hisoblab, pullari dastlabki holdagi puli bilan teng ekanligini bildi. Anvarda qancha pul bo’lgan? A) 100 B) 400 C) 127 D) 375 27 / 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) 3S B) 4S C) 2S D) 0,5S 28 / 30 A) 2 B) 1 C) 0 D) √6 29 / 30 Tenglamani yeching. (x+2)+(x+4)+(x+6)+…+(x+100)=2800 A) 3 B) 5 C) 4 D) 7 30 / 30 Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping. A) -0,5 B) -1 C) -0,25 D) -0,75 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
