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Matematika attestatsiya №12

InfoMaster

Aprel 8, 2022

124 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

Tenglamani ildizlari yig’indisini toping. |x + 3| + |x −1| + |x − 4| = 6

A) -2

B) Ildizi yo’q

C) -4

D) 0

2 / 40

A(1; 3) nuqtadan 12x+5y+a–4=0 to’g’ri chiziqqacha bo’lgan masofa 2 bo’lsa, a ni toping.

A) 2

B) 4

C) 1

D) 3

3 / 40

Agar x=2t+5, y=3t-1 va f(x)=y(x) bog’lanish mavjud bo’lsa, f(5) ning qiymatini toping

A) -1

B) -2

C) 2

D) 1

4 / 40

Piramidaning asosi tomoni 4√3 va o‘tkir burchagi 45° ga teng bo’lgan rombdan iborat. Ushbu piramidaga ichki chizilgan konusning yasovchisi asos tekisligi bilan 60° li burchak tashkil etadi. Konusning hajmini toping.

A) 6√2π

B) 6√3π

C) 3π

D) 6π

5 / 40

Hisoblang.

A) 6

B) 4

C) 5

D) 3

6 / 40

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

A) Yotmaydi

B) Yotadi

7 / 40

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

A) 16

B) 9

C) 0

D) 25

8 / 40

Hisoblang.

A) 26-3b

B) 24-3b

C) 14-2b

D) 24+3b

9 / 40

Hisoblang.

A) 2

B) 3

C) 4

D) 1

10 / 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) -6

C) -5

D) -4

11 / 40

Hisoblang.

A) 32

B) 94

C) 98

D) 96

12 / 40

Tenglamani yeching.

A) 0

B) 5

C) 1

D) 2

13 / 40

Agar bo'lsa f(x)=?

A) x²-1

B) x²-2

C) x²+2

D) 2x²-1

14 / 40

(tgx+ctgx)⁴ yoyilmaning ozod hadini toping.

A) 4

B) 6

C) 5

D) 3

15 / 40

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

A) Ha

B) Yo'q

16 / 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) 4

C) 5

D) 6

17 / 40

Tangga besh martta tashlanganda 2 martta gerb chiqishi ehtimolini toping.

A) 5/16

B) 9/16

C) 7/16

D) 5/13

18 / 40

Noaniq integralni toping. ∫(sin²x+cos²x)dx

A) x/2+c

B) 2x+c

C) x+c

D) 4x+c

19 / 40

Tengsizlikni yeching.

A) x∈[0;1000]

B) x∈(10;1000)

C) x∈(0;1000)

D) x∈(1;1000]

20 / 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]

21 / 40

Hisoblang.

A) -1

B) 0

C) 1

D) 2

22 / 40

Tenglama ildizlari yig'indisini toping.

A) 2

B) -1

C) 4

D) 0

23 / 40

Hisoblang.

A) 133

B) 121

C) 135

D) 131

24 / 40

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

A) 80°

B) 60°

C) 90°

D) 75°

25 / 40

Hisoblang.

A) -2 yoki 0

B) -2 yoki 2

C) 0

D) -2 yoki -1

26 / 40

Chizmada nechta uchburchak tasvirlangan?

A) 52

B) 48

C) 50

D) 54

27 / 40

Hisoblang.

A) 9/13

B) 9/11

C) 7/11

D) 0

28 / 40

Agar bo'lsa, ni toping.

A) 3

B) 0

C) 5,2

D) 2,5

29 / 40

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

A) 62

B) 48

C) 56

D) 54

30 / 40

Hisoblang.

A) 2020

B) 2021

C) 2019

D) 1

31 / 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)

32 / 40

Tenglamani yeching:

A) 5,5

B) 6

C) 2

D) 2,5

33 / 40

Hisoblang. cos²10° + cos²30° + cos²80°

A) 3/7

B) 7/3

C) 0

D) 7/4

34 / 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) 6700/13

B) 7551/15

C) 7500/13

D) 15

35 / 40

Hisoblang.

A) 3/2

B) 5/3

C) 2/3

D) 3/5

36 / 40

|x²-2x|≤x yengsizlikni yeching.

A) (-2;0)∪[1;3]

B) {0}∪[1;3]

C) {0}∪(1;3]

D) {0}∪[1;3)

37 / 40

Tenglamani yeching.

A) 1

B) 2

C) 3

D) 4

38 / 40

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

A) 35

B) 28

C) 42

D) 48

39 / 40

Hisoblang.

A) 2

B) 6

C) 3

D) 1

40 / 40

Hisoblang.

A) 2

B) -1

C) 1

D) 0

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”