Uy » Abituriyent » Matematika abituriyent » Matematika abituriyent testi №1 Matematika abituriyent Matematika abituriyent testi №1 InfoMaster Aprel 5, 2022 218 Ko'rishlar 1 izoh SaqlashSaqlanganOlib tashlandi 0 8 Vaqtingiz tugadi! Tomonidan yaratilgan InfoMaster Matematika abituriyentlar uchun №1 1 / 30 a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping. A) 6 B) 4 C) 8 D) 12 2 / 30 Tenglamaning nechta ildizi bor? |x+1|=|2x-1| A) 3 B) 1 C) 2 D) 4 3 / 30 To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping A) 11 B) √46 C) 10 D) 12 4 / 30 Tenglamani yeching. (x+2)+(x+4)+(x+6)+…+(x+100)=2800 A) 7 B) 5 C) 4 D) 3 5 / 30 sistemadan x+y ning qiymatini toping. A) -12 B) 35/4 C) 12 D) 6 6 / 30 Parallelogrammning tomonlari nisbati 3:5 kabi. Agar parallelogrmmning perimetri 48 ga burchaklaridan biri 1200 ga teng bo’lsa, uning yuzini toping. A) 135√3/4 B) 48√3 C) 67,5 D) 67,5√3 7 / 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 8 / 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) 4760 C) 4680 D) 4716 9 / 30 Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusiga teng bo’lgan uchburchak qanday uchburchak deyiladi? A) o’tmas burchakli uchburchak B) to’gri burchakli uchburchak C) teng tomonli uchburchak D) o’tkir burchakli uchburchak 10 / 30 sistemadan x+y+z ning qiymatini toping. A) -139/41 B) 140/41 C) 150/41 D) 139/41 11 / 30 Tomoni 2 ga teng kvadratga tashqi chizilgan aylana uzunligini toping. A) 2π/3 B) 3π C) 2π D) 4π 12 / 30 x2-5|x|-6=0 tenglama ildizini toping A) ±6 B) -1;-6 C) ±;±6 D) -1;6 13 / 30 Tengsizlikni yeching. A) (-4;2)v(2;3) B) (-3;2)v(2;4) C) (2;4) D) (-3;2) 14 / 30 Ushbu funksiyaning boshlang’ich funksiyasini toping. A) ln(|x+4+|x-1|)+c B) ln|x-1/x+4|+c C) ln(|x+4*|x-1|)+c D) ln|x+4/x-1|+c 15 / 30 Parallelogrammning yuzi 213 ga, tomonlaridan biri 7 ga va o’tkir burchagi 600 ga teng bo’lsa, ikkinchi tomonini toping A) 6 B) 8 C) 4 D) 5 16 / 30 Ostki asosining yuzi 32π va ustki asosining yuzi 18π ga teng bo‘lgan kesik konus berilgan. Agar kesik konusga shar ichki chizilgan bo‘lsa, u holda sharning sirtini toping. A) 72π B) 56π C) 100π D) 96π 17 / 30 sistemada xy ning qiymatini toping. A) 60 B) 64 C) 80 D) 75 18 / 30 Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping. A) 7 π B) 12π C) 6 π D) 10 π 19 / 30 Agar sinx+cosx=a bo’lsa, ning qiymatini toping. A) 1/3 B) -1/2 C) 1/2 D) 2/5 20 / 30 Uchburchakning balandligi 12 ga teng bo’lib, u asosni 5:16 nidbatda bo’ladi. Agar asosning uzunligi 21 ga teng bo’lsa, uchburchakning perimetrini toping A) 108 B) 54 C) 48 D) 52 21 / 30 Muntazam uchburchakli piramidaning yon qirrasi asos tekisligi bilan 45o li burchak tashkil etgan bo‘lsa, u holda piramidaning yon sirti yuzining uning asosi yuziga nisbatini toping. A) 2√5 B) 2√3 C) 3√3 D) 4 22 / 30 Hisoblang A) 3√3 B) 2√3 C) √3 D) 1 23 / 30 y=cos(2sinx) funksiyaning qiymatlar sohasini toping. A) [-1;1] B) [cos2;1] C) [0;cos2] D) [0;1] 24 / 30 Teng yonli uchburchakning tomonlari 5, 5 va 6 ga teng. Bu uchburchakning bissektiritsalari va medianalari kesishgan nuqtalar A) 1 B) 1/2 C) 1/6 D) 1,2 25 / 30 Ushbu (y6+y3+1)(y3+1)(y3-1)-y6+y3+1 ifodani soddalashtish natijasida ko’phad hosil qilindi. Uning nechta hadi bor? A) 4 B) 3 C) 2 D) 1 26 / 30 funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli? A) juft ham emas, toq ham emas funksiya B) toq funksiya C) juft funksiya D) bunday funksiya mavjud emas 27 / 30 bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya. A) -14 B) -10 C) -12 D) -4 28 / 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) 3S C) 2S D) 4S 29 / 30 Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping. A) e⁻⁵⁰⁵⁰ B) e⁻⁴⁹⁵⁰ C) e⁴⁹⁵⁰ D) e⁵⁰⁵⁰ 30 / 30 O’q kesimining diagonallari o’zaro perpendikulyar bo’lgan kesik konus yasovchisi va asos tekisligi orasidagi burchak ga teng. Agar o’q kesimining diagonali ga teng bo’lsa, kesik konus asosining yuzini toping. A) πa²/4cos²α B) πa²(1-2sinα/4sin²) C) πa²/4sin²α D) πa²(1+2cosα/4sin²) 0% Testni qayta ishga tushiring Baholash mezoni To'g'ri javob uchun 3,1 ball. Fikr-mulohaza yuboring Author: InfoMaster Foydali bo'lsa mamnunmiz
