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Matematika attestatsiya №6

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Yanvar 30, 2022

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Matematika fanidan attestatsiya savollari №6

1 / 40

y=kx+n funksiyaning grafigi I va III chorakdan o‘tishi uchun k va n qanday bo‘lishi kerak?

A) n = 0, k > 0

B) n > 0, k = 0

C) n = 0, k < 0

D) n < 0, k = 0

2 / 40

Kamayish tartibida joylashtiring.

a)5√6
b)12
c)4√10

A) a>b>c

B) a>c>b

C) c>a>b

D) b>a>c

3 / 40

f(x-1)+f(x+2)=2(x²+7) ekani ma‟lum bo'lsa, f(x) ko'phadni toping.

A) f(x)=x²+3x+7

B) f(x)=x²-x+5

C) f(x)=x²-4

D) f(x)+2x²-1

4 / 40

funksiyaning qiymatlar to'plamini toping.

A) [-√2;√2]

B) [-√2;0)∪(0;√2]

C) (-2√2;√2)

D) [-√2;-1)∪(-1;1)∪(1;√2]

5 / 40

Hisoblang?

A) 2

B) 4

C) 0,25

D) 0,5

6 / 40

Quyida berilgan shakillar ichidagi sonlar yig`indisi nolgateng bo`lishi uchun . Nechanchi shakilni olib tashlash kerak?

A) 2

B) 1

C) 3

D) 4

7 / 40

a,b,c -1 dan katta musbat sonlar uchun a³=b² va a⁴=c⁵ bo'lsa, log_bc ni toping.

A) 5/6

B) 1/2

C) 8/15

D) 1/3

8 / 40

Yig‘indini toping.

A) 1/12

B) 1/4

C) 1/6

D) 1/9

9 / 40

x²-n≤0 tengsizlik o‘rinli bo‘ladigan x ning 7 ta butun ildizi bor bo‘lsa. n nechta turli butun qiymat qabul qiladi?

A) 7

B) 9

C) 15

D) 12

10 / 40

To‘g‘ri burchakli uchburchakning bir burchagi 52° ga teng bo‘lsa, to‘g‘ri burchak uchidan tushirilgan balandlik va mediana orasidagi burchakni toping.

A) 14°

B) 24°

C) 7°

D) 17°

11 / 40

x²-√11x+1=0 0 bo‘lsa,

A) 11

B) 10

C) 12

D) 9

12 / 40

Asoslari 5 va 5√7 ga teng bo‘lgan trapetsiyaning yuzini teng ikkiga bo‘luvchi kesma asoslarga parallel. Shu kesma uzunligini toping.

A) 10

B) 8

C) 6√7

D) 4√7

13 / 40

Tenglamani yeching:

A) 5

B) 6; -5

C) -6

D) -6; 5

14 / 40

Agar geometrik progressiyaning umumiy hadi b_n= 3·2ⁿ bo‘lsa, ni toping.

A) 1

B) 4

C) 2

D) 3

15 / 40

Rasmdagi shakl perimetrini toping.

A) 28

B) 24

C) 30

D) 32

16 / 40

Tengsizlik nechta butun juft yechimga
ega?

A) 110

B) 115

C) 112

D) 116

17 / 40

Hisoblang:

A) cos10°

B) 1

C) sin10°

D) cos50°

18 / 40

sonlarini taqqolsang.

A) a

B) c

C) b

D) c

19 / 40

Tenglamalar sistemani yeching:

A) (9; 0), (2; 7)

B) (7; 2), (28; -1)

C) (2; 3)

D) (9; 0), (28; -1)

20 / 40

Kvadratga ikkita yarim aylana ichki chizilgan. Bo‘yalgan soha yuzini toping.

A) 8(π+2)

B) 16(π-2)

C) 32

D) 64

21 / 40

Tenglamalar sistemasining ildizlari yig‘indisini toping.

A) 12

B) 14

C) 10

D) 20

22 / 40

kasrning o‘nli kasr ko‘rinishidagi raqamlarining yig‘indisini toping.

A) 5

B) 11

C) 7

D) 10

23 / 40

Sonlarining o‘rta geometrik qiymatini toping.

A) 2√2

B) 4√3

C) 3√2

D) 2√3

24 / 40

vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping.

A) 2

B) 3

C) 4

D) 1

25 / 40

P(x+1)=x³+3x²-2x+a+3 ko‘phadi berilgan. P(x+2) ko‘phadining koeffitsiyentlari yig‘indisini 8 ga teng bo‘lsa, a nechaga teng?

A) -4

B) -6

C) -3

D) 5

26 / 40

Ifodaning qiymatini toping.

A) 0,(04)

B) 0,0(2)

C) 0,04

D) 0,0(4)

27 / 40

A(4;6), B(2;1), C(6;1) nuqtalarni tutashtirishdan hosil bo‘ladigan uchburchak yuzini toping.

A) 20

B) 15

C) 10

D) 8

28 / 40

ABC to‘g‘ri burchakli uchburchakning og‘irlik markazi G nuqta. Bunda AB ⊥ BC, AG ⊥ GB va AG = 8 bo‘lsa, BG ni toping.

A) 4√2

B) 3√2

C) 4√3

D) 6

29 / 40

b ning qanday qiymatlarida M(2;1) nuqtadan 4x - 3y + b = 0 to‘g‘ri chiziqqacha masofa 2 ga teng bo‘ladi?

A) 4 yoki –12

B) 3 yoki –8

C) 6

D) 5 yoki –15

30 / 40

Ifodaning qiymatini quyidagi sonlardan qaysi biriga teng.

3¹⁹·2⁵
3¹⁶·2⁸
3¹³·2⁸
32⁹·2⁸

A) 1

B) 4

C) 2

D) 3

31 / 40

a_n - arifmetik progressiyaning umumiy hadi bo‘lsa, quyidagi nisbatni toping:

A) 6

B) 7

C) 8

D) 5

32 / 40

Tenglikdan foydalanib a ni toping. (a+b+c+d)·(a-b-c+d)=(a-b+c-d)·(a+b-c-d)

A) 1

B) bc/d

C) bd/c

D) cd/b

33 / 40

Kvadratlarning yuzlari yig‘indisini toping.

A) 22

B) 121

C) berilganlar yetarli emas

D) 11

34 / 40

tenglamaning ildizlari yig‘indisini (agar ildizi bitta bo‘lsa, o‘zini) toping.

A) 8

B) 5

C) 3

D) -2

35 / 40

√3

A) 5/5

B) 7/3

C) 3/2

D) 21/10

36 / 40

P(x)=x¹ºº ko‘phadni x³-3x+2 ga bo‘lganda qoladigan qoldiqni toping.

2¹ºº-1
(2¹ºº-1)x+2(2⁹⁹-1)
(2¹ºº-1)x-2(2⁹⁹-1)
2¹ººx-3·2¹⁰⁰

A) 3

B) 1

C) 4

D) 2

37 / 40

bo‘lsa,

A) 3/2

B) 2/3

C) 2

D) 1

38 / 40

ABC o‘tkir burchakli uchburchakning BC asosiga AD balandlik, AC yon tomoniga BE balandlik o‘tkazilgan. Bunda CE =2AE = 8, DC = 6 bo‘lsa, BD ni toping.

A) 8

B) 12

C) 6

D) 10

39 / 40

Chizmadan foydalanib α ni toping.