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Matematika attestatsiya №12

InfoMaster

Aprel 8, 2022

119 Ko'rishlar 1 izoh

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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

Ushbu {1, 2, 3, 4, …, 37} to‘plamdan shunday ikkita son tanlanganki, ularning ko‘paytmasi qolgan 35 ta sonning yig‘indisiga teng. Shu ikki sonning farqini toping.

A) 8

B) 10

C) 5

D) 9

2 / 40

Tengsizlikni yeching:

A) (2;6)∪(6;∞)

B) (-∞;-2) ∪(6;∞)

C) (2;6)

D) (6;∞)

3 / 40

AB masofani toping ?

A) √27

B) √30

C) √29

D) √24

4 / 40

Agar f(x)=x^sin2x bo'lsa, f'(π/4) toping.

A) 1

B) π/2

C) π/4

D) 0

5 / 40

Hisoblang.

A) 0

B) 4

C) 1

D) 2

6 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) 4x+c

B) c

C) x+c

D) 2x+c

7 / 40

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

A) 0

B) -1

C) 2

D) 1

8 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 5/16

B) 5/13

C) 9/16

D) 7/16

9 / 40

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

A) 48

B) 56

C) 62

D) 54

10 / 40

Hisoblang.

A) 1

B) 3

C) 2

D) 4

11 / 40

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

A) 25

B) 6

C) 24

D) 58

12 / 40

Hisoblang.

A) 96

B) 94

C) 98

D) 32

13 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) 4x+c

B) x/2+c

C) 2x+c

D) x+c

14 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 4

B) 6

C) 3

D) 5

15 / 40

Tengsizlikni yeching.

A) [-7;-3) u [-1;∞)

B) [-7;3 u [-1;∞)

C) [-7;-3] u [-1;∞)

D) [-7;-5) u [-1;∞)

16 / 40

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

A) -1

B) 3

C) 0

D) 2

17 / 40

To'g'ri burchakli uchburchakning gipotenuzasiga o'tkazilgan bissiktriksasi uni 20 va 30 ga teng kesmalarga ajratadi. To'g'ri burchakli uchburchak yuzini toping.

A) 7551/15

B) 7500/13

C) 6700/13

D) 15

18 / 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) -5

C) -6

D) -4

19 / 40

Hisoblang.

A) -2 yoki 0

B) -2 yoki 2

C) 0

D) -2 yoki -1

20 / 40

Hisoblang

A) 9

B) 12

C) 6

D) 36

21 / 40

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

A) Yotadi

B) Yotmaydi

22 / 40

Hisoblang.

A) -47/24

B) 51/29

C) 49/25

D) -47/25

23 / 40

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

A) Yo'q

B) Ha

24 / 40

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

A) √7

B) 2√7

C) 4√5

D) 3√7

25 / 40

Chizmada nechta uchburchak tasvirlangan?

A) 54

B) 52

C) 50

D) 48

26 / 40

Agar bo'lsa, ni toping.

A) 5,2

B) 0

C) 2,5

D) 3

27 / 40

Tenglama ildizlari yig'indisini toping.

A) -1

B) 2

C) 4

D) 0

28 / 40

Hisoblang.

A) 3

B) 6

C) 4

D) 5

29 / 40

|x²-2x|≤x yengsizlikni yeching.

A) {0}∪[1;3)

B) {0}∪(1;3]

C) (-2;0)∪[1;3]

D) {0}∪[1;3]

30 / 40

y=xlnx funksiyani hosilasini toping.

A) lnx/2+1

B) 2lnx-1

C) lnx+1

D) 1

31 / 40

Hisoblang.

A) -1

B) 2

C) 1

D) 0

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) 30 va 24

B) 36 va 18

C) 34 va 20

D) 32 va 22

33 / 40

Hisoblang.

A) 135

B) 133

C) 121

D) 131

34 / 40

Hisoblang.

A) 0

B) 9/11

C) 7/11

D) 9/13

35 / 40

Hisoblang.

A) 26-3b

B) 14-2b

C) 24-3b

D) 24+3b

36 / 40

Hisoblang. cos²10° + cos²30° + cos²80°

A) 7/3

B) 3/7

C) 0

D) 7/4

37 / 40

Hisoblang.

A) 1

B) -1

C) 0

D) 2

38 / 40

Tenglamani yeching.

A) 0

B) 2

C) 1

D) 5

39 / 40

Tengsizlikni yeching.

A) x∈(1;1000]

B) x∈[0;1000]

C) x∈(10;1000)

D) x∈(0;1000)

40 / 40

Tenglamani yeching:

A) 5,5

B) 2,5

C) 6

D) 2

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”