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Matematika abituriyent testi №1

InfoMaster

Aprel 5, 2022

210 Ko'rishlar 1 izoh

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Matematika abituriyentlar uchun №1

1 / 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) 10

C) 7

D) 8

2 / 30

Ag ar tgα=2 bo’lsa, u holda ni hisoblang

A) 10/3

B) -10/27

C) 10/27

D) -10/3

3 / 30

Tenglamaning nechta ildizi bor? |x+1|=|2x-1|

A) 2

B) 1

C) 3

D) 4

4 / 30

Hisoblang:

A) 3

B) 1

C) -1

D) 2

5 / 30

Teng yonli trpetsiyaning asoslari 15 va 25 ga balandligi esa 15 ga teng trapetsiyaning dioganalini toping

A) 25

B) 30

C) 28

D) 20

6 / 30

funktsiyaning aniqlanish sohasini toping.

A) (1;2)

B) [1;2]

C) (-∞;1]v[2;∞)

D) (-∞;1)v(2;∞)

7 / 30

m ning qanday eng katta butun qiymatida y=2x-mx-5+m funksiyaning grafigi 1,3,4 –choraklarda yotadi?

A) 2

B) 5

C) 1

D) 3

8 / 30

Tomoni 6 ga teng bo`lgan teng tomonli uchburchakga tashqi chizilgan doiraning yuzini toping.

A) 10 π

B) 6 π

C) 12π

D) 7 π

9 / 30

Soddalashtiring.

A) 2018a/a+1

B) 2019

C) 2018

D) a+1

10 / 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) (5√3+3)π/3

B) 64√2π/3

C) 3√2π/4

D) 8√2π/5

11 / 30

y= funktsiyaning aniqlanish sohasini toping.

A) (2;∞)

B) (-∞;2)

C) [2;∞)

D) (-∞;2)v(2;∞)

12 / 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) 1/√3

B) 1

C) 0,5

D) 1/√2

13 / 30

Musobaqada 5 ta ishtirokchidan 3 tasiga 1, 2, 3-o’rinlarni necha xil usulda berish mumkin?

A) 47

B) 60

C) 120

D) 18

14 / 30

Radiuslari orasidagi burchagi 36^o va radiusi 5 ga teng bo`lgan sektor yoyining uzunligini toping.

A) π

B) 2π

C) 2π/3

D) π/2

15 / 30

Quyidagi 2x-y+3z=2018 va x+5y+z=2019 tekisliklarning holatini aniqlang.

A) o’zaro parallel

B) ayqash

C) o’zaro perpendikulyar

D) aniqlab bo’lmaydi

16 / 30

a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping.

A) 4

B) 12

C) 8

D) 6

17 / 30

Hisoblang

A) 1

B) 0

C) 2

D) -1

18 / 30

Perimetri 60 ga teng bo’lgan parallelogrammning tomonlari nisbati 2:3 ga, o’tkir burchagi esa 30⁰ ga teng. Parallelogrammning yuzini toping.

A) 54

B) 52√3

C) 108

D) 48√3

19 / 30

Soat 3:37 bo’lganda minut va soat millari orasidagi burchakni toping.

A) 112,5°

B) 113°

C) 113,5°

D) 100°

20 / 30

Agar va b-a=4 bo’lsa, a+b ni toping.

A) 2

B) 3,5

C) 4

D) 3

21 / 30

Tengsizlik nechta butun yechimga ega?

A) 4

B) cheksiz ko’p

C) 3

D) 1

22 / 30

A(2;-2,5) nuqtadan y= - 4x parabolagacha bo’lgan eng qisqa masofani toping.

A) 1

B) 1,5

C) √5/2

D) √3/2

23 / 30

Quyidagilardan qaysi biri barcha lar uchun ma’noga ega (aniqlangan)?

A) (512-1/2⁻⁹)°

B) 4k+1/1/8:7-1/56

C) ²ᴷ⁺⁴v2k+1/k²+1

D) ⁴ᴷ⁺³√-√2k+1

24 / 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) 48

B) 108

C) 52

D) 54

25 / 30

Fazoda (1;2;3) nuqtalardan o’tuvchi to’g’ri chiziq tenglamasini tuzing.

A) -x-y+z=0

B) x-1/2=y-2/3=z-3/4

C) x=y/3=z/2

D) 2x+3y+z=0

26 / 30

a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega

A) a≠12

B) a ning bunday qiymati yo’q

C) a≠5

D) a=12

27 / 30

|x²-5x-14|+20≥5|x+2|+4|x-7| tengsizlikni yeching.

A) (-∞;-6]v{2}v[12;∞)

B) [-2;4]v{6}

C) [1;6]

D) [2;4]v{2}v[3;∞)

28 / 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) 6

B) 4

C) 10

D) 5

29 / 30

y= $\sqrt{2x^{2}-4x}$ funksiyaning aniqlanish sohasini toping

A) (-∞;0])v[2;∞)

B) (2;∞)

C) (0,2)

D) (-∞;0)v(2;∞)

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(1+2cosα/4sin²)

B) πa²/4cos²α

C) πa²/4sin²α

D) πa²(1-2sinα/4sin²)