Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 221 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 8 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 Ifodani soddalashtiring. [ln(ln x)-ln(loge10)].log10e A) ln(ln x) B) lg(ln x) C) ln(lg x) D) lg(lg x) 2 / 30 Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 300 ga teng. Parallelogrammning yuzini toping. A) 52√3 B) 54 C) 48√3 D) 108 3 / 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 yoki 50 B) 50 C) 2 yoki 50 D) 10 4 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln|x-1/x+4|+c B) ln(|x+4*|x-1|)+c C) ln|x+4/x-1|+c D) ln(|x+4+|x-1|)+c 5 / 30 Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang. A) o’zaro parallel B) aniqlab bo’lmaydi C) o’zaro perpendikulyar D) ayqash 6 / 30 x(t)=t2+6t+5 qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakat boshlangandan necha sekund o’tgach boshlang’ich nuqtaga nisbatan 77 metr masofaga siljiydi? A) 6 B) 8 C) 7 D) 10 7 / 30 Soddalashtiring. (0<m<7) A) 2m-7 B) 7-2m C) m D) 7 8 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) juft funksiya B) juft ham emas, toq ham emas funksiya C) bunday funksiya mavjud emas D) toq funksiya 9 / 30 sistemada xy ning qiymatini toping. A) 75 B) 80 C) 60 D) 64 10 / 30 Hisoblang: A) -1 B) 3 C) 2 D) 1 11 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusidan katta bo’lgan uchburchaklar sonini toping. A) 2019 B) 0 C) 1 D) cheksiz ko’p 12 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 10 B) 12 C) 11 D) √46 13 / 30 Bankda qo`yilgan pul bir yildan kegin foydasi bilan 2600 so`m bo`ldi; Agar bank yillik 30% foyda to`lasa, boshida qancha pul qo`yilgan bo`ladi ? A) 1900 B) 2000 C) 2200 D) 2100 14 / 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) 3 C) √32 D) 6 15 / 30 |x2-5x-14|+20≥5|x+2|+4|x-7| tengsizlikni yeching. A) [1;6] B) [-2;4]v{6} C) [2;4]v{2}v[3;∞) D) (-∞;-6]v{2}v[12;∞) 16 / 30 Rombning yuzi 96 ga, dioganallaridan biri 16 ga teng. Romb tomonini toping. A) 11 B) 12 C) 9 D) 10 17 / 30 To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi? A) o‘zgarmaydi B) 6,25% ga ortadi C) 6,25% ga kamayadi D) 2,5% ga ortadi 18 / 30 Agar bank qo`yilgan pulga 40% yillik foyda bersa, qo`yilgan 5000 so`m pul bir yildan keyin qancha bo`ladi ? A) 7200 B) 7000 C) 6200 D) 6900 19 / 30 Radiuslari orasidagi burchagi 36o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping. A) 2π B) 2π/3 C) π D) π/2 20 / 30 A) √6 B) 2 C) 1 D) 0 21 / 30 Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing. A) x-1/2=y-2/3=z-3/4 B) 2x+3y+z=0 C) x=y/3=z/2 D) -x-y+z=0 22 / 30 Tenglamaning ildizlari yig`indisini toping. A) 4 B) 3 C) 6 D) 5 23 / 30 Tomoni 25 ga diagonallaridan biri 4 ga teng bo’lgan rombning yuzini toping. A) 16 B) 8√5 C) 32 D) 24√5 24 / 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) 3π/2 C) 4π/3 D) 2π/3 25 / 30 a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring. A) a>b>c>d>e B) b>a>c>d>e C) b>c>a>d>e D) e>b>a>d>c 26 / 30 Agar x,y sonlar (x+5)2+(y-12)2=142 tenglikni qanoatlantirsa, x2+y2 ifodaning eng kichik qiymatini toping. A) 2 B) √2 C) √3 D) 1 27 / 30 Quyidagi va to’g’ri chiziqlarning o’zaro holatini aniqlang. A) o’zaro kesishadi B) ayqash to’gri chiziqlar C) o’zaro parallel D) o’zaro perpendikulyar 28 / 30 Agar f(x)=ax3-5x2+b va bo’lsa, a ni toping. A) 3 B) 0 C) 2 D) 1 29 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 12 B) 6 C) 8 D) 4 30 / 30 Ifodani soddalashtiring. 2 cos55o.cos40o.sin55o+cos110o.sin40o A) 0 B) 1 C) 2 D) 0,5 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
