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Matematika attestatsiya №12

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Aprel 8, 2022

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OMAD YOR BO'LSIN!

Matematika fanidan attestatsiya savollari №12

28.03.2022 da Toshkentda tushgan savollar

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Testda 76% va undan yuqori natija oling va sertifikatni yuklab oling.

1 / 40

f(g(x))=x² + 4x - 1, g(x)=x + a va f′(0)=1 bo'lsa, α ni toping.

A) 2

B) 1,5

C) 3,5

D) 3

2 / 40

A(1; 3) nuqtadan 12x+5y+a–4=0 to’g’ri chiziqqacha bo’lgan masofa 2 bo’lsa, a ni toping.

A) 3

B) 4

C) 2

D) 1

3 / 40

Rasmda tasvirlangan favvora ko‘rinishi y=-x(x+6) ko'rinishda bo'lsa, favvora eng ko’pi bilan qaysi balandlikga ko’tarila oladi. (Koordinata Oy o’qi , suv yuzasida joylashgan)

A) 6

B) Hisoblab bo'lmaydi

C) 9

D) 8

4 / 40

0, 1, 2, 3, 4, 5 raqamlaridan nechta 3 xonali tuzish mumkin?

A) 180

B) 200

C) 125

D) 100

5 / 40

y=xlnx funksiyani hosilasini toping.

A) lnx+1

B) 1

C) 2lnx-1

D) lnx/2+1

6 / 40

Hisoblang

A) 36

B) 9

C) 12

D) 6

7 / 40

Tengsizlikni yeching.

A) x∈[0;1000]

B) x∈(1;1000]

C) x∈(10;1000)

D) x∈(0;1000)

8 / 40

Chizmada nechta uchburchak tasvirlangan?

A) 48

B) 52

C) 50

D) 54

9 / 40

Hisoblang.

A) 24-3b

B) 14-2b

C) 24+3b

D) 26-3b

10 / 40

Tengsizlikni yeching.

A) [-7;-3) u [-1;∞)

B) [-7;-5) u [-1;∞)

C) [-7;3 u [-1;∞)

D) [-7;-3] u [-1;∞)

11 / 40

Ayirmasi d ga teng bo'lgan arifmetik progrssiya uchun ni toping.

A) 1/2

B) 2/3

C) 1

D) 2

12 / 40

Agar va ular orasidagi burchak 120° bo'lsa, ni toping.

A) 2√7

B) 3√7

C) 4√5

D) √7

13 / 40

Hisoblang.

A) -2 yoki 2

B) -2 yoki -1

C) -2 yoki 0

D) 0

14 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 7/16

B) 9/16

C) 5/13

D) 5/16

15 / 40

Hisoblang.

A) 5/3

B) 3/5

C) 3/2

D) 2/3

16 / 40

Tenglamani yeching.

A) 5

B) 0

C) 1

D) 2

17 / 40

Hisoblang. cos²10° + cos²30° + cos²80°

A) 3/7

B) 0

C) 7/3

D) 7/4

18 / 40

Tengsizlikni yeching: 3 · 9^x–10 · 21^x+ 7 · 49 ^x ≤ 0

A) (-1;3]

B) [-1;0)

C) [-1;0]

D) [-1;1]

19 / 40

Hisoblang.

A) 96

B) 94

C) 32

D) 98

20 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) 2x+c

B) x/2+c

C) x+c

D) 4x+c

21 / 40

y = x²+1-|sinx| funksiya grafigi OX o'qidan pastda yotadimi?

A) Yotadi

B) Yotmaydi

22 / 40

Tenglama ildizlari yig'indisini toping.

A) 2

B) 0

C) 4

D) -1

23 / 40

Hisoblang.

A) 131

B) 133

C) 135

D) 121

24 / 40

Hisoblang.

A) 9/13

B) 7/11

C) 0

D) 9/11

25 / 40

Hisoblang.

A) 2

B) 1

C) 0

D) -1

26 / 40

Hisoblang.

A) 3

B) 1

C) 2

D) 6

27 / 40

Hisoblang.

A) 1

B) 2020

C) 2019

D) 2021

28 / 40

Tenglamaning [0;π] kesmadagi ildizlari soni 3 tadan ko'pmi?

A) Ha

B) Yo'q

29 / 40

Agar cos4x = acos⁴x+bcos2²x+1 bo'lsa, a+b=?

A) 0

B) 1

C) 2

D) -1

30 / 40

(2; 3; -5) nuqtaning OX o'qiga nisbatan simmetrik nuqtasini toping.

A) (2; 3; -5)

B) (2; -3; 5)

C) (2; -3; -5)

D) (-2; 3; 5)

31 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 3

B) 5

C) 6

D) 4

32 / 40

Tamonlarini ayirmasi 14 bo'lgan parallelogramning o'rmas burchagidan katta diagonaliga tushgan balandlik uni 12 va 30 ga teng kesmalarga ajratadi. Tamonlar uzunligini toping.

A) 34 va 20

B) 36 va 18

C) 30 va 24

D) 32 va 22

33 / 40

To'g'ri burchakli uchburchakning yuzi 96 ga, perimetri esa 48 ga teng bo'lsa, gipatenuza uzunligini toping.

A) 21

B) 23

C) 20

D) 18

34 / 40

Hisoblang.

A) -1

B) 0

C) 2

D) 1

35 / 40

To'g'ri burchakli uchburchakka ichki va tashqi chizilgan aylanalar raduslari 6 va 15 ga teng bo'lsa, katetlar yig'indini toping.

A) 42

B) 48

C) 28

D) 35

36 / 40

Arimetik progerssiyada s₁₂=354, P=a₁+a₂+...+a₁₁, Q=a₂+a₄+...+a₁₂ va P/Q=32/27 ga teng bo'lsa, d ni toping.

A) -5

B) -3

C) -6

D) -4

37 / 40

Agar bo'lsa, ni toping.

A) 2,5

B) 5,2

C) 0

D) 3

38 / 40

Hisoblang.

A) 1

B) 2

C) 4

D) 3

39 / 40

ABCD parallelogramm va S_KBCE= 18 bo'lsa, S_ABCD=?

A) 48

B) 56

C) 62

D) 54

40 / 40

(tgx-2ctgx)⁴ yoyilmaning ozod hadini toping.

A) 25

B) 24

C) 58

D) 6

Baholash mezoni

- 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) saqlansin”;

- 75 foizdan past ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi)ga tushirilsin”;

- 75 foiz va undan yuqori ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, oliy malaka toifasi (bosh o’qituvchi lavozimi) berilsin”;

- 74 foizdan 65 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, birinchi malaka toifasi (yetakchi o’qituvchi lavozimi) saqlansin”;

- 65 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi)ga tushirilsin”

- 64 foizdan 60 foizgacha ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tdi, ikkinchi malaka toifasi (katta o’qituvchi lavozimi) saqlansin”;

- 60 foizdan kam ko’rsatgichga ega bo’lsa - “Attestatsiyadan o’tmadi, mutaxassis (oliy yoki o’rta maxsus, kasb-hunar ma'lumotli o’qituvchi) lavozimiga tushirilsin”