Matematika attestatsiya №12

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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

f (x) = (x² )^1−log_xa funksiya uchun, f (3^x+1) = a^x−1 bo`lsa, a parametr quyidagilardan qaysi biriga teng?

A) 9

B) 1

C) 3

D) 27

2 / 40

x² - 4|x|-a - 3 = 0 tenglama ikkita yechimga ega bo‘ladigan barcha butun manfiy a lar nechta?

A) 1

B) 0

C) 2

D) 3

3 / 40

Poyezda 5 ta vagon va har bir vagonda kamida 1 ta yo’lovchi bor. Ikkita yo’lovchi bitta vagonda yoki ketma-ketma kelgan vagonda bo’lsa, qo’shni deb ataladi. Har bir yo’lovchining aniq 5 ta yoki aniq 10 ta qo’shnisi mavjud bo’lsa, poyezdda jami nechta yo’lovchi bor?

A) 20

B) 13

C) 17

D) 10

4 / 40

Hisoblang.

A) -2

B) 0

C) 1

D) 2

5 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 3

B) 5

C) 6

D) 4

6 / 40

Hisoblang.

A) 98

B) 94

C) 96

D) 32

7 / 40

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

A) 51

B) 47

C) 49

D) 48

8 / 40

Hisoblang.

A) 0

B) 4

C) 2

D) 1

9 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 5/13

B) 7/16

C) 5/16

D) 9/16

10 / 40

|x²-2x|≤x yengsizlikni yeching.

A) {0}∪[1;3]

B) (-2;0)∪[1;3]

C) {0}∪[1;3)

D) {0}∪(1;3]

11 / 40

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

A) Yotadi

B) Yotmaydi

12 / 40

Hisoblang.

A) 2

B) 0

C) -1

D) 1

13 / 40

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

A) Ha

B) Yo'q

14 / 40

Hisoblang.

A) 2020

B) 1

C) 2021

D) 2019

15 / 40

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

A) [-1;1]

B) [-1;0)

C) (-1;3]

D) [-1;0]

16 / 40

Hisoblang.

A) -2 yoki 2

B) -2 yoki -1

C) 0

D) -2 yoki 0

17 / 40

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

A) 48

B) 54

C) 62

D) 56

18 / 40

Tenglamani yeching:

A) 2,5

B) 2

C) 6

D) 5,5

19 / 40

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

A) 0

B) -1

C) 2

D) 3

20 / 40

Hisoblang.

A) -47/24

B) -47/25

C) 49/25

D) 51/29

21 / 40

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

A) 28

B) 35

C) 48

D) 42

22 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) 4x+c

B) x+c

C) 2x+c

D) c

23 / 40

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

A) 1/2

B) 1

C) 2

D) 2/3

24 / 40

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

A) 1

B) -1

C) 0

D) 2

25 / 40

y=xlnx funksiyani hosilasini toping.

A) lnx/2+1

B) lnx+1

C) 2lnx-1

D) 1

26 / 40

Hisoblang.

A) 9/11

B) 9/13

C) 7/11

D) 0

27 / 40

Tenglama ildizlari yig'indisini toping.

A) -1

B) 0

C) 4

D) 2

28 / 40

Agar bo'lsa, ni toping.

A) 2,5

B) 3

C) 0

D) 5,2

29 / 40

Hisoblang.

A) 133

B) 135

C) 131

D) 121

30 / 40

Tenglamani yeching.

A) 1

B) 0

C) 2

D) 5

31 / 40

f(x)+(x-2)·g(x)=x²+5, f(3)=10 bo'lsa, g^-1=(4) ni toping. Bunda g^-1(x) g(x) ga teskari funksiya.

A) 3

B) 6

C) 5

D) 4

32 / 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) 15

B) 7500/13

C) 7551/15

D) 6700/13

33 / 40

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

A) 40

B) 42

C) 36

D) 38

34 / 40

Hisoblang.

A) -1

B) 2

C) 1

D) 0

35 / 40

Hisoblang.

A) 1

B) 4

C) 2

D) 3

36 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) x/2+c

B) 2x+c

C) 4x+c

D) x+c

37 / 40

Hisoblang.

A) 1

B) 2

C) 6

D) 3

38 / 40

Hisoblang.

A) 4

B) 3

C) 5

D) 6

39 / 40

Tengsizlikni yeching.

A) [-7;-5) u [-1;∞)

B) [-7;-3] u [-1;∞)

C) [-7;-3) u [-1;∞)

D) [-7;3 u [-1;∞)

40 / 40

Tenglamani yeching.

A) 2

B) 1

C) 4

D) 3

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”

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An interesting discussion is worth comment. I think that you should write more on this topic, it might not be a taboo subject but generally people are not enough to speak on such topics. To the next. Cheers

Matematika attestatsiya №12

28.03.2022 da Toshkentda tushgan savollar

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