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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

x, y, z sonlari uchun x > y > z va x + y + z + 18 munosabatlar o‘rinli. Bu 3 ta son bir-biridan 1, 2 va 3 sonlariga farq qiladi. Shunga ko‘ra x ning mumkin bo‘lgan barcha qiymatlari yig‘indisini
toping.

A) 14

B) 16

C) 9

D) 15

2 / 40

Agar biror oyda 5 ta payshanba bo‘lsa, shu oyda … bo‘la olmaydi?

A) 5 ta juma

B) 5 ta shanba

C) 5 ta seshanba

D) 5 ta yakshanba

3 / 40

EKUB(a;b)=3 va EKUK(a;b) =60 bo’lsa, a+b yig’indining eng katta qiymatini toping.

A) 54

B) 60

C) 72

D) 63

4 / 40

(0!+1!+ 2022!)¹²⁵ sonning birlar xonasida qaysi raqam turadi?

A) 6

B) 4

C) 2

D) 3

5 / 40

Tenglamani yeching.

A) 1

B) 4

C) 3

D) 2

6 / 40

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

A) 90°

B) 75°

C) 80°

D) 60°

7 / 40

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

A) 24

B) 25

C) 6

D) 58

8 / 40

Tenglamani yeching:

A) 5,5

B) 2

C) 2,5

D) 6

9 / 40

Hisoblang.

A) 6

B) 3

C) 1

D) 2

10 / 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

11 / 40

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

A) 3√7

B) 4√5

C) 2√7

D) √7

12 / 40

Hisoblang.

A) 131

B) 121

C) 135

D) 133

13 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 6

B) 4

C) 3

D) 5

14 / 40

Hisoblang. cos²10° + cos²30° + cos²80°

A) 7/3

B) 7/4

C) 3/7

D) 0

15 / 40

Hisoblang.

A) 51/29

B) -47/24

C) 49/25

D) -47/25

16 / 40

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

A) -1

B) 1

C) 2

D) 0

17 / 40

Agar bo'lsa, ni toping.

A) 0

B) 5,2

C) 3

D) 2,5

18 / 40

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

A) 62

B) 48

C) 56

D) 54

19 / 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)

20 / 40

Agar bo'lsa f(x)=?

A) x²-1

B) x²-2

C) x²+2

D) 2x²-1

21 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 7/16

B) 5/16

C) 5/13

D) 9/16

22 / 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

23 / 40

Hisoblang.

A) -2 yoki 2

B) -2 yoki -1

C) -2 yoki 0

D) 0

24 / 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) -6

B) -4

C) -3

D) -5

25 / 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) 28

D) 35

26 / 40

Tenglama ildizlari yig'indisini toping.

A) 0

B) 4

C) 2

D) -1

27 / 40

Hisoblang.

A) 3/5

B) 3/2

C) 5/3

D) 2/3

28 / 40

Hisoblang.

A) 1

B) 2

C) -1

D) 0

29 / 40

Hisoblang.

A) 96

B) 32

C) 98

D) 94

30 / 40

Hisoblang.

A) 24-3b

B) 26-3b

C) 24+3b

D) 14-2b

31 / 40

Tengsizlikni yeching.

A) x∈(0;1000)

B) x∈(10;1000)

C) x∈(1;1000]

D) x∈[0;1000]

32 / 40

Hisoblang.

A) 2

B) 1

C) 3

D) 4

33 / 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

34 / 40

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

A) 2

B) -1

C) 3

D) 0

35 / 40

Hisoblang.

A) 0

B) -1

C) 1

D) 2

36 / 40

Hisoblang.

A) 0

B) 1

C) -1

D) 2

37 / 40

Hisoblang.

A) 0

B) 9/13

C) 9/11

D) 7/11

38 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) c

B) 4x+c

C) 2x+c

D) x+c

39 / 40

Hisoblang.

A) 2021

B) 1

C) 2019

D) 2020

40 / 40

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

A) 9

B) 0

C) 16

D) 25

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”;

- 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”