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Matematika attestatsiya №12

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Aprel 8, 2022

48 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

Hisoblang.

A) 1.5

B) 3

C) √3

D) 1

2 / 40

{a_n} - ketna-ket berilgan, bunda a₁=40 va a_k+1=a_k-k, k∈N a₈ ni toping.

A) 13

B) 35

C) 12

D) 32

3 / 40

Agar bo’lsa, (x+y)²-z² ning qiymatini toping.

A) 1/2

B) 11/12

C) 1/6

D) 23/12

4 / 40

Hisoblang.

A) 0

B) -2

C) 2

D) 1

5 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 5

B) 3

C) 6

D) 4

6 / 40

Hisoblang.

A) 49/25

B) -47/24

C) 51/29

D) -47/25

7 / 40

Hisoblang.

A) 4

B) 3

C) 5

D) 6

8 / 40

Hisoblang.

A) 3/5

B) 3/2

C) 2/3

D) 5/3

9 / 40

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

A) 51

B) 48

C) 49

D) 47

10 / 40

Tengsizlikni yeching.

A) [-7;-5) u [-1;∞)

B) [-7;3 u [-1;∞)

C) [-7;-3) u [-1;∞)

D) [-7;-3] u [-1;∞)

11 / 40

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

A) 36

B) 38

C) 40

D) 42

12 / 40

Tenglamani yeching:

A) 2,5

B) 2

C) 6

D) 5,5

13 / 40

Tengsizlikni yeching.

A) x∈[0;1000]

B) x∈(10;1000)

C) x∈(0;1000)

D) x∈(1;1000]

14 / 40

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

A) Ha

B) Yo'q

15 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) 4x+c

B) x/2+c

C) x+c

D) 2x+c

16 / 40

Hisoblang.

A) 2

B) 4

C) 3

D) 1

17 / 40

Hisoblang.

A) -1

B) 1

C) 0

D) 2

18 / 40

y=xlnx funksiyani hosilasini toping.

A) 2lnx-1

B) 1

C) lnx+1

D) lnx/2+1

19 / 40

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

A) 90°

B) 60°

C) 80°

D) 75°

20 / 40

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

A) 0

B) 3

C) -1

D) 2

21 / 40

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

A) [-1;0)

B) (-1;3]

C) [-1;1]

D) [-1;0]

22 / 40

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

A) Yotmaydi

B) Yotadi

23 / 40

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

A) 2

B) 2/3

C) 1

D) 1/2

24 / 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

25 / 40

Hisoblang. cos²10° + cos²30° + cos²80°

A) 3/7

B) 0

C) 7/4

D) 7/3

26 / 40

Tenglamani yeching.

A) 1

B) 4

C) 3

D) 2

27 / 40

Hisoblang.

A) 1

B) -1

C) 0

D) 2

28 / 40

Hisoblang.

A) -2 yoki 2

B) 0

C) -2 yoki 0

D) -2 yoki -1

29 / 40

Hisoblang.

A) 133

B) 121

C) 135

D) 131

30 / 40

Hisoblang.

A) 2

B) 0

C) 1

D) 4

31 / 40

Hisoblang.

A) 1

B) 2

C) 0

D) -1

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) 36 va 18

B) 34 va 20

C) 32 va 22

D) 30 va 24

33 / 40

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

A) 18

B) 21

C) 23

D) 20

34 / 40

Agar bo'lsa f(x)=?

A) x²-2

B) 2x²-1

C) x²+2

D) x²-1

35 / 40

Chizmada nechta uchburchak tasvirlangan?

A) 52

B) 48

C) 54

D) 50

36 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) x+c

B) 4x+c

C) c

D) 2x+c

37 / 40

Hisoblang.

A) 1

B) 2

C) 3

D) 6

38 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 5/13

B) 9/16

C) 5/16

D) 7/16

39 / 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) -3

C) -6

D) -5

40 / 40

Hisoblang.

A) 98

B) 96

C) 32

D) 94

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”