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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

Agar bo'lsa ifodaning qiymatini toping.

A) 0

B) 7

C) 13

D) 9

2 / 40

Qavariq oltiburchakning 3 ta uchidan kesib olsak qanday shakl hosil bo’ladi ?

A) 14

B) 12

C) 6

D) 9

3 / 40

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

A) 0

B) 3

C) 2

D) 1

4 / 40

tenglamani yeching.

A) 5

B) log(5/3)2

C) log(3/5)2

D) 3

5 / 40

Quyidagi berilgan qoldiqli bo`lishlardan qaysi biri to`g`ribajarilgan ?
a)76 : 9 = 7 (qoldiq 13);
b)20 : 7 = 2 (qoldiq 6);
d)54 : 15 = 4 (qoldiq 1)

A) b

B) a,b

C) )a,b,d

D) b,d

6 / 40

x ni toping ?

A) 3

B) 6

C) 4

D) 5

7 / 40

Integralni hisoblang:

A) 4

B) 3

C) 1

D) 2

8 / 40

Uchburchakning 3 va 4 ga teng bo‘lgan tomonlariga o‘tkazilgan medianalar o‘zaro perpendikulyar bo‘lsa, bu uchburchakning uchinchi tomonini toping.

A) 2,5

B) √5

C) √6

D) 2,4

9 / 40

f(x)+f(3x)+f(5x)+...+f(39x)=(3x+10)⁴ bo'lsa, f'(0) ni toping.

A) 30

B) 10

C) 0

D) 25

10 / 40

Raqamlari yig‘indisi 10 bo‘lgan nechta turli 3 xonali son bor?

A) 0,216

B) 0,432

C) 0,288

D) 0,144

11 / 40

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

A) bd/c

B) 1

C) bc/d

D) cd/b

12 / 40

Agar α=60°, β=70°, γ=50° bo‘lsa, tgα+ tgβ+ tgγ yig‘indi quyidagilardan qaysi biriga teng?

A) 4 √3tg50° tg70°

B) √3tg50° tg70°

C) 2√3tg50° tg70°

D) √3ctg40° tg70°

13 / 40

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

A) 10

B) 15

C) 20

D) 8

14 / 40

vekorning Oxy tekislikdagi proyeksiyasi bo‘lgan vektorni toping.

A) 4

B) 2

C) 3

D) 1

15 / 40

Rasmdagi shakl perimetrini toping.

A) 32

B) 24

C) 28

D) 30

16 / 40

Barcha ikki xonali sonlar ko‘paytmasi 4 ning qanday eng katta darajasiga bo‘linadi?

A) 43

B) 44

C) 42

D) 45

17 / 40

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

A) 8

B) 10

C) 4√7

D) 6√7

18 / 40

Hisoblang:

A) 1/32

B) -√3/2

C) -1/2

D) 0

19 / 40

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

A) 8

B) 5

C) 7

D) 6

20 / 40

sonlarini taqqolsang.

A) c

B) c

C) b

D) a

21 / 40

√3

A) 21/10

B) 7/3

C) 3/2

D) 5/5

22 / 40

sonlari geometrik progressiyaning ketma-ket hadlari bo‘ladigan barcha n larning yig‘indisini (agar bitta qiymati bo‘lsa, o‘zini) toping.

A) 24

B) 23

C) 14

D) 35

23 / 40

Hisoblang:

A) sin10°

B) 1

C) cos10°

D) cos50°

24 / 40

bo‘lsa,

A) 1

B) 2/3

C) 3/2

D) 2

25 / 40

Tengsizlik nechta butun juft yechimga
ega?

A) 110

B) 116

C) 115

D) 112

26 / 40

Tenglamani yeching:

A) 6; -5

B) -6; 5

C) 5

D) -6

27 / 40

To‘g‘ri to‘rtburchakning 16 ga teng diagonali yon tomoni bilan 15° li burchak tashkil etadi. To‘rtburchak yuzini toping.

A) 64

B) 32

C) 48

D) 42

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) 6

B) 4√2

C) 4√3

D) 3√2

29 / 40

1+sin 2x = 7(sin x + cos x) tenglamani yeching.

A) -π/4+πk, k∈Z

B) π/4+πk, k∈Z

C) 0

D) Ø

30 / 40

3x3 o‘lchamli kvadratning tugunlarida 16 ta nuqta belgilanib, ularning o‘ng tomondan eng yuqorisidagi A bilan belgilangan. Bir uchi A nuqtada, qolgan uchlari qolgan 15 ta nuqtada orasidan tanlanadigan uchburchaklarning sonini toping.

A) 96

B) 25

C) 105

D) 100

31 / 40

Chizmadan foydalanib α ni toping.

A) 50°

B) 40°

C) 30°

D) 20°

32 / 40

ABC uchburchakning A burchgi 30° ga, B burchagi 75° ga teng. B uchidan AC tomonga BD kesma o‘tkazilgan. ABD burchak 45° ga teng bo‘lsa, quyidagilardan qaysi biri noto‘g‘ri?

A) DC < AD

B) AB = BC

C) BD = BC

D) BC > AD

33 / 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) 5

C) -6

D) -3

34 / 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) 12

B) 6

C) 10

D) 8

35 / 40

x²-√11x+1=0 0 bo‘lsa,