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Matematika abituriyent testi №1

InfoMaster

Aprel 5, 2022

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Matematika abituriyentlar uchun №1

1 / 30

Bir vaqtning o’zida 9,13, . . . ,405 va 15,21, . . . ,255 ketma–ketliklarning hadlari bo’lgan sonlarning eng kattasi va eng kichigining ayirmasini toping

A) 228

B) 147

C) 150

D) 231

2 / 30

Agar f(x)=ax³-5x²+b va bo’lsa, a ni toping.

A) 2

B) 3

C) 0

D) 1

3 / 30

Soddalashtiring.

A) a+1

B) 2019

C) 2018

D) 2018a/a+1

4 / 30

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

A) ayqash

B) o’zaro parallel

C) aniqlab bo’lmaydi

D) o’zaro perpendikulyar

5 / 30

Tomoni 25 ga diagonallaridan biri 4 ga teng bo’lgan rombning yuzini toping.

A) 8√5

B) 24√5

C) 16

D) 32

6 / 30

Tenglamani yeching.

|x²-11x+10|=x²-11x+10

A) (-∞;1]v[10;∞)

B) (-∞;1]

C) [10;∞)

D) 1; 10

7 / 30

a(x+2)=2x+1 tenglama a ning qanday qiymatida yechimga ega emas?

A) (2;∞)

B) (-∞;2)

C) (-∞;2)v(2;∞)

D) (∞;∞)

8 / 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) 52

C) 108

D) 54

9 / 30

sistema yagona yechimga ega bo’ladigan a ning barcha qiymatlari to’plamini toping.

A) {-1;3}

B) {2;4}

C) {1;2}

D) {-1;2}

10 / 30

Soddalashtiring. (0<m<7)

A) m

B) 7-2m

C) 2m-7

D) 7

11 / 30

sistema a ning qanday qiymatida cheksiz ko’p yechimga ega?

A) (-∞;6)

B) 6

C) (2;∞)

D) (-∞;6])v[6;∞)

12 / 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 ortadi

B) 2,5% ga ortadi

C) o‘zgarmaydi

D) 6,25% ga kamayadi

13 / 30

sistemada xy ning qiymatini toping.

A) 80

B) 64

C) 60

D) 75

14 / 30

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

A) ⁴ᴷ⁺³√-√2k+1

B) ²ᴷ⁺⁴v2k+1/k²+1

C) (512-1/2⁻⁹)°

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

15 / 30

(-3;4) nuqtaga absissa, ordinata o’qlariga va koordinata boshiga nisbatan simmetrik bo’lgan nuqtalarni tutashtirishdan hosil bo’lgan uchburchakning eng kata tomonini toping.

A) 10

B) 24

C) 12

D) 14

16 / 30

O’qishni bilmaydigan bola alifbening A,A,A, N,N, S- 6 ta harflarini ixtiyoriy ravishda terib chiqadi. Bunda ANANAS so’zining hosil bo’lish ehtimolini toping.

A) 1/60

B) 6/720

C) 5/24

D) 5/720

17 / 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+2cosα/4sin²)

C) πa²/4sin²α

D) πa²(1-2sinα/4sin²)

18 / 30

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

A) 3

B) 1

C) 4

D) 2

19 / 30

Teng yonli uchburchakning tomonlari 5, 5 va 6 ga teng. Bu uchburchakning bissektiritsalari va medianalari kesishgan nuqtalar

A) 1/2

B) 1

C) 1/6

D) 1,2

20 / 30

Agar va b-a=4 bo’lsa, a+b ni toping.

A) 3

B) 4

C) 3,5

D) 2

21 / 30

Ushbu arifmetik progressiyaning manfiy hadlari yig’indisini toping.

A) -1

B) -0,75

C) -0,5

D) -0,25

22 / 30

Hisoblang:

A) 2

B) 1

C) -1

D) 3

23 / 30

bo’lsa, ni x orqali ifodalang.

A) 2/x

B) 25/x

C) x/25

D) 2-x

24 / 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) 2

C) 374389

D) aniqlab bo’lmaydi

25 / 30

ABCD to’gri to’rtburchak ichidan olingan O nuqtadan A, B, C, D uchlarigacha bo’lgan masofalar mos ravishda 1; 2; 1,5; 1,2 ga teng bo’ladigan barcha to’g’ri to’rtburchaklar sonini toping

A) 1

B) cheksiz ko’p

C) 2

D) 0

26 / 30

Arifmetik progressiyada a₁₇=33 va a₄₅=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping.

A) √2

B) 4

C) 2

D) 2√2

27 / 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

28 / 30

integralning qiymatini toping.

A) 0

B) -π/2

C) π/2

D) π/4

29 / 30

Agar f(x)=4x³-6x²-2x+3^x ^.log₃e bo’lsa, u holda ni toping.

A) e

B) 1

C) √2e

D) 3e

30 / 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) 3√2π/4

B) 64√2π/3

C) 8√2π/5

D) (5√3+3)π/3