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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

Agar a∈[-3;1] va b∈[-2;2]bo‘lsa, a² + b³ qanday oraliqda bo‘ladi?

A) [0;17]

B) [-17;8]

C) [-8;0]

D) [-8;17]

2 / 40

Quyidagi chizmada barcha 11 ta kesmaning uzunligi 2 ga teng bo‘lsa, ABCDE beshburchakning yuzini toping.

A) √11+√12

B) Aniqlab bo‘lmaydi

C) 2√13

D) 11√+√13

3 / 40

Hisoblang

A) 0

B) 1

C) 0,5

D) 0,1

4 / 40

Soddalashtiring:

A) 2√b-2√a

B) -2√a

C) 2√a-2√b

D) -2√b

5 / 40

Chizmada nechta uchburchak tasvirlangan?

A) 50

B) 52

C) 48

D) 54

6 / 40

Tenglama ildizlari yig'indisini toping.

A) 4

B) 2

C) -1

D) 0

7 / 40

Hisoblang.

A) 135

B) 131

C) 121

D) 133

8 / 40

1, 2, 3, 4 raqamlaridan foydalanib 400 dan kichik kechta uc xonali son yasash mumkin?

A) 49

B) 47

C) 48

D) 51

9 / 40

Hisoblang.

A) -1

B) 0

C) 2

D) 1

10 / 40

Agar bo'lsa f(x)=?

A) x²+2

B) 2x²-1

C) x²-1

D) x²-2

11 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) 4x+c

B) x+c

C) x/2+c

D) 2x+c

12 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 5

B) 6

C) 3

D) 4

13 / 40

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

A) [-1;0)

B) (-1;3]

C) [-1;0]

D) [-1;1]

14 / 40

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

A) Yotadi

B) Yotmaydi

15 / 40

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

A) √7

B) 2√7

C) 4√5

D) 3√7

16 / 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)

17 / 40

a=1^0,(6), b=8^0,(3) , c=2^0,(9) bo'lsa, (a+b)^c=?

A) 9

B) 0

C) 25

D) 16

18 / 40

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

A) 1

B) 0

C) -1

D) 2

19 / 40

Hisoblang.

A) 1

B) 2

C) -1

D) 0

20 / 40

Tengsizlikni yeching.

A) [-7;-3] u [-1;∞)

B) [-7;3 u [-1;∞)

C) [-7;-5) u [-1;∞)

D) [-7;-3) u [-1;∞)

21 / 40

Tenglamani yeching.

A) 2

B) 5

C) 1

D) 0

22 / 40

Hisoblang.

A) 3

B) 5

C) 4

D) 6

23 / 40

Tenglamani yeching.

A) 2

B) 4

C) 1

D) 3

24 / 40

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

A) Yo'q

B) Ha

25 / 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

26 / 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) -4

B) -6

C) -3

D) -5

27 / 40

Hisoblang.

A) -2 yoki 0

B) -2 yoki 2

C) -2 yoki -1

D) 0

28 / 40

Hisoblang.

A) 5/3

B) 3/5

C) 2/3

D) 3/2

29 / 40

Hisoblang.

A) 2

B) 0

C) 1

D) -1

30 / 40

Hisoblang.

A) 2020

B) 2021

C) 1

D) 2019

31 / 40

Agar bo'lsa, ni toping.

A) 0

B) 2,5

C) 3

D) 5,2

32 / 40

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

A) 6

B) 58

C) 24

D) 25

33 / 40

Tenglamani yeching:

A) 6

B) 2,5

C) 5,5

D) 2

34 / 40

5^-x≤ 0,04 x ning eng kichik butun qiymatini toping.

A) 2

B) 3

C) 0

D) -1

35 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 5/16

B) 5/13

C) 7/16

D) 9/16

36 / 40

|x²-2x|≤x yengsizlikni yeching.

A) {0}∪[1;3)

B) {0}∪[1;3]

C) (-2;0)∪[1;3]

D) {0}∪(1;3]

37 / 40

Hisoblang.

A) 9/11

B) 0

C) 7/11

D) 9/13

38 / 40

Hisoblang

A) 6

B) 12

C) 36

D) 9

39 / 40

Hisoblang.

A) 96

B) 94

C) 32

D) 98

40 / 40

y=xlnx funksiyani hosilasini toping.

A) 1

B) lnx/2+1

C) 2lnx-1

D) lnx+1

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”