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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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1 / 40

Tenglamaning haqiqiy ildizlari o‘rta arifmetigini toping:

A) 4

B) 1,(6)

C) 1/2

D) 1

2 / 40

Bir nechta natural sonlardan tuzilgan S to‘plam berilgan. Bu sonlarning o‘rta
arifmetigi eng kattasini hisobga olmaganda 32 ga, eng kichigini hisobga olmaganda 40 ga, eng kattasini ham eng kichigini ham hisobga olmaganda esa 35 ga teng. Agar eng katta son eng kichigidan 72 ga ortiq bo‘lsa, u holda S to‘plamdagi barcha sonlarning o‘rta arifmetigini toping.

A) 36,8

B) 36,4

C) 36,2

D) 36,6

3 / 40

Berilgan chizmadan foydalanib α + β ning qiymatini toping:

A) 60°

B) 30°

C) 75°

D) 45°

4 / 40

P(x)=x²+3ax+c ko‘phadi x + a ga qoldiqsiz bo‘linsa, P( x) ni x - a ga bo‘lganda qoladigan qoldiqni aniqlang.

A) 0

B) 6a²

C) a²

D) 2a²

5 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) 2x+c

B) 4x+c

C) x+c

D) x/2+c

6 / 40

Hisoblang.

A) 121

B) 133

C) 135

D) 131

7 / 40

Agar bo'lsa, ni toping.

A) 0

B) 2,5

C) 3

D) 5,2

8 / 40

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

A) 1/2

B) 2/3

C) 2

D) 1

9 / 40

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

A) 48

B) 42

C) 35

D) 28

10 / 40

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

A) 1

B) 0

C) 2

D) -1

11 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) c

B) x+c

C) 2x+c

D) 4x+c

12 / 40

Hisoblang.

A) 3

B) 2

C) 1

D) 4

13 / 40

Hisoblang.

A) 1

B) 0

C) 2

D) 4

14 / 40

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

A) 3√7

B) 4√5

C) 2√7

D) √7

15 / 40

Hisoblang. cos²10° + cos²30° + cos²80°

A) 3/7

B) 0

C) 7/4

D) 7/3

16 / 40

Chizmada nechta uchburchak tasvirlangan?

A) 52

B) 48

C) 50

D) 54

17 / 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) 32 va 22

B) 30 va 24

C) 34 va 20

D) 36 va 18

18 / 40

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

A) Yotmaydi

B) Yotadi

19 / 40

Tenglamani yeching.

A) 2

B) 5

C) 0

D) 1

20 / 40

|x²-2x|≤x yengsizlikni yeching.

A) {0}∪[1;3]

B) (-2;0)∪[1;3]

C) {0}∪[1;3)

D) {0}∪(1;3]

21 / 40

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

A) -1

B) 2

C) 0

D) 3

22 / 40

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

A) Ha

B) Yo'q

23 / 40

Tengsizlikni yeching.

A) x∈(0;1000)

B) x∈[0;1000]

C) x∈(10;1000)

D) x∈(1;1000]

24 / 40

Tengsizlikni yeching.

A) [-7;-3) u [-1;∞)

B) [-7;-3] u [-1;∞)

C) [-7;3 u [-1;∞)

D) [-7;-5) u [-1;∞)

25 / 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) -3

B) -6

C) -5

D) -4

26 / 40

f(x) = x²-3x-2 bo'lsa. f(a+b+c)-f(abc)=? Bunda a, b, c - kvadrat funksiya koeffitsiyentlari.

A) 42

B) 38

C) 40

D) 36

27 / 40

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

A) (-1;3]

B) [-1;0)

C) [-1;1]

D) [-1;0]

28 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 3

B) 5

C) 4

D) 6

29 / 40

Hisoblang.

A) 2

B) 1

C) 3

D) 6

30 / 40

Hisoblang.

A) 2

B) 0

C) 1

D) -1

31 / 40

Hisoblang.

A) 32

B) 94

C) 96

D) 98

32 / 40

Hisoblang.

A) 0

B) -2 yoki 0

C) -2 yoki -1

D) -2 yoki 2

33 / 40

Tenglama ildizlari yig'indisini toping.

A) 0

B) 4

C) 2

D) -1

34 / 40

Hisoblang.

A) 51/29

B) 49/25

C) -47/25

D) -47/24

35 / 40

Hisoblang.

A) 2021

B) 2019

C) 2020

D) 1

36 / 40

y=xlnx funksiyani hosilasini toping.

A) 1

B) lnx+1

C) lnx/2+1

D) 2lnx-1

37 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 9/16

B) 7/16

C) 5/16

D) 5/13

38 / 40

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

A) 21

B) 18

C) 20

D) 23

39 / 40

Uchburchak tamonlari 10, 24 va 26 ga teng bo'lsa, katta burchagi necha gradus?

A) 75°

B) 90°

C) 80°

D) 60°

40 / 40

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

A) 25

B) 58

C) 24

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”;

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

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- 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”