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Matematika attestatsiya №12

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Aprel 8, 2022

103 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

ABCD trapetsiyaning yuzi 24 ga teng asoslari DC = 6, AB = 2, BC tomondan E nuqta olingan bo‘lib, BE = 2EC bo‘lsa, ADE uchburchakning yuzini toping.

A) 12

B) 14

C) 16

D) 21

2 / 40

Soddalashtiring:

A) sinα

B) 0

C) 1

D) cosα

3 / 40

Hisoblang.

A) √2+1

B) √2-1

C) 1-√2

D) √2+√3

4 / 40

Quyidagi keltirilgan ko‘paytirish amalida qatnashgan harflar turli qiymatlarga ega bo‘lsa, A+ B+C = ?

A) 10

B) 13

C) 12

D) 11

5 / 40

Tenglamani yeching:

A) 5,5

B) 6

C) 2

D) 2,5

6 / 40

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

A) -1

B) 3

C) 2

D) 0

7 / 40

Hisoblang.

A) 2/3

B) 5/3

C) 3/5

D) 3/2

8 / 40

Tenglamani yeching.

A) 5

B) 2

C) 0

D) 1

9 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 5/13

B) 9/16

C) 5/16

D) 7/16

10 / 40

Hisoblang.

A) -1

B) 1

C) 0

D) 2

11 / 40

y=xlnx funksiyani hosilasini toping.

A) lnx+1

B) 1

C) 2lnx-1

D) lnx/2+1

12 / 40

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

A) 23

B) 21

C) 18

D) 20

13 / 40

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

A) -1

B) 2

C) 0

D) 1

14 / 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) -5

B) -3

C) -4

D) -6

15 / 40

Tengsizlikni yeching.

A) x∈(1;1000]

B) x∈(10;1000)

C) x∈(0;1000)

D) x∈[0;1000]

16 / 40

Hisoblang.

A) 0

B) -2 yoki 0

C) -2 yoki -1

D) -2 yoki 2

17 / 40

Tengsizlikni yeching.

A) [-7;3 u [-1;∞)

B) [-7;-3) u [-1;∞)

C) [-7;-3] u [-1;∞)

D) [-7;-5) u [-1;∞)

18 / 40

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

A) 48

B) 49

C) 47

D) 51

19 / 40

Chizmada nechta uchburchak tasvirlangan?

A) 54

B) 48

C) 52

D) 50

20 / 40

Hisoblang.

A) 2019

B) 2021

C) 1

D) 2020

21 / 40

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

A) Ha

B) Yo'q

22 / 40

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

A) 80°

B) 60°

C) 75°

D) 90°

23 / 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) 34 va 20

C) 36 va 18

D) 32 va 22

24 / 40

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

A) 25

B) 58

C) 6

D) 24

25 / 40

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

A) Yotmaydi

B) Yotadi

26 / 40

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

A) 16

B) 0

C) 25

D) 9

27 / 40

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

A) 56

B) 54

C) 62

D) 48

28 / 40

Agar bo'lsa f(x)=?

A) x²-2

B) x²+2

C) x²-1

D) 2x²-1

29 / 40

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

A) 1

B) 1/2

C) 2/3

D) 2

30 / 40

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

A) 4√5

B) 3√7

C) √7

D) 2√7

31 / 40

|x²-2x|≤x yengsizlikni yeching.

A) {0}∪[1;3]

B) (-2;0)∪[1;3]

C) {0}∪[1;3)

D) {0}∪(1;3]

32 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) x/2+c

B) 4x+c

C) x+c

D) 2x+c

33 / 40

Hisoblang.

A) -47/24

B) 49/25

C) -47/25

D) 51/29

34 / 40

Hisoblang.

A) 2

B) 1

C) 0

D) -1

35 / 40

Hisoblang.

A) 3

B) 6

C) 5

D) 4

36 / 40

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

A) 42

B) 40

C) 36

D) 38

37 / 40

Hisoblang.

A) 32

B) 96

C) 94

D) 98

38 / 40

Tenglamani yeching.

A) 1

B) 2

C) 3

D) 4

39 / 40

Hisoblang.

A) 14-2b

B) 26-3b

C) 24+3b

D) 24-3b

40 / 40

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

A) (-1;3]

B) [-1;0]

C) [-1;0)

D) [-1;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”