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Matematika abituriyent testi №1

InfoMaster

Aprel 5, 2022

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Matematika abituriyentlar uchun №1

1 / 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) 2

B) 412967

C) 374389

D) aniqlab bo’lmaydi

2 / 30

To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping

A) 4

B) 7

C) 5

D) 6

3 / 30

Markazi nuqtada bo‘lgan aylanaga va urinmalar o‘tkazilgan bo’lib, va nuqtalar urinish nuqtalari bo’lsin. Aylanadagi Q nuqtadan o‘tkazilgan uchinchi urinma va kesmalarni X va Y nuqtalarda kesib o‘tadi. Agar uchburchakning perimetri 48 va aylana radiusi 7 ga teng bo‘lsa, u holda kesma uzunligini toping

A) 25

B) 12

C) 30

D) 15

4 / 30

Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping.

A) e⁵⁰⁵⁰

B) e⁻⁴⁹⁵⁰

C) e⁻⁵⁰⁵⁰

D) e⁴⁹⁵⁰

5 / 30

sonning oxirgi raqamini toping.

A) 2

B) 8

C) 4

D) 6

6 / 30

Sin9x =4sin3x tenglamani yeching

A) π/3+πn, n€Ζ

B) πn, n€Ζ

C) πn/3, n€Ζ

D) π/2+πn, n€Ζ

7 / 30

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

A) (-∞;6])v[6;∞)

B) (2;∞)

C) 6

D) (-∞;6)

8 / 30

Soddalashtiring. (0<m<7)

A) 7

B) 7-2m

C) m

D) 2m-7

9 / 30

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

A) 4

B) 2

C) 2√2

D) √2

10 / 30

Tenglamani yeching.

|x²-11x+10|=x²-11x+10

A) (-∞;1]

B) [10;∞)

C) 1; 10

D) (-∞;1]v[10;∞)

11 / 30

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

A) {1;2}

B) {-1;3}

C) {2;4}

D) {-1;2}

12 / 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) 6/720

B) 5/24

C) 5/720

D) 1/60

13 / 30

Ifodani soddalashtiring.

2 cos55^o.cos40^o.sin55^o+cos110^o.sin40^o

A) 0

B) 1

C) 2

D) 0,5

14 / 30

Agar bank qo`yilgan pulga 40% yillik foyda bersa, qo`yilgan 5000 so`m pul bir yildan keyin qancha bo`ladi ?

A) 7000

B) 6900

C) 7200

D) 6200

15 / 30

Ikki burchagi graduslari yig’indisi uchinchi burchagi gradusidan katta bo’lgan uchburchaklar sonini toping.

A) cheksiz ko’p

B) 2019

C) 1

D) 0

16 / 30

Tengsizlikni yeching.

A) to'g'ri javob yo'q

B) (-1/³ √2 ;0)

C) (0;+∞)

D) 0

17 / 30

sin2x-cos2x=1 tenglama [-π; 2π] oraliqda nechta ildizga ega?

A) 6

B) 10

C) 9

D) 7

18 / 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

19 / 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) 4716

C) 4760

D) 4680

20 / 30

funksiya uchun quyidagi mulohazalardan qaysi biri o’rinli?

A) bunday funksiya mavjud emas

B) juft ham emas, toq ham emas funksiya

C) juft funksiya

D) toq funksiya

21 / 30

y=cos(2sinx) funksiyaning qiymatlar sohasini toping.

A) [0;cos2]

B) [-1;1]

C) [cos2;1]

D) [0;1]

22 / 30

To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping

A) 12

B) 10

C) √46

D) 11

23 / 30

bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya.

A) -12

B) -4

C) -14

D) -10

24 / 30

Rombning yuzi 96 ga, dioganallaridan biri 16 ga teng. Romb tomonini toping.

A) 10

B) 9

C) 11

D) 12

25 / 30

integralning qiymatini toping.

A) π/4

B) -π/2

C) 0

D) π/2

26 / 30

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

A) 2

B) 0

C) 3

D) 1

27 / 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(2cos2x)

C) 4sin2x*cos(cos2x)

D) 4sin2x*cos(2cos2x)

28 / 30

tenglamalar sistemasini yeching

A) (4;4)

B) (-4;4)

C) (4;–4)

D) (-4;-4)

29 / 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) 10 yoki 50

C) 2 yoki 50

D) 50

30 / 30

a=sin 1; b=sin 2; c=sin 3; d=sin 4 va e=sin 5 sonlarni kamayish tartibida joylashtiring.

A) e>b>a>d>c

B) b>a>c>d>e

C) a>b>c>d>e

D) b>c>a>d>e