Barcha fanlardan online testlar

11-sinf Matematika 4-chorak

InfoMaster

Aprel 20, 2021

441 Ko'rishlar 83 izohlar

SaqlashSaqlanganOlib tashlandi 6

OMAD YOR BO'LSIN!

Tomonidan yaratilgan InfoMaster

11-sinf Matematika 4-chorak

Testni Salomov Sardor tayyorladi.

1 / 25

$y=6^{ln(x^{2}+5)}$ funksiyasining kamayish oralig’ini toping

A) (-√5;√5)

B) (0;+∞)

C) ∅

D) (-∞;0]

2 / 25

Tomoni 4 ga teng bo’lgan teng tomonli uchburchak ichiga uchburchakning tomonlariga urinuvchi uchta teng doira joylashtirilgan. O’zaro urunuvchi doiralar yoylaridan hosil bo’lgan egri chiziqli uchburchakning yuzini toping

A) (2-√3)(2√3-π)

B) (4-√3)(2√3-π)

C) (2-√3)(2-π)

D) 2√3-π

3 / 25

2cosx+sinx=-2 tenglamaning [-π;π] kesmada nechta ildizi bor

A) 2

B) 1

C) 3

D) Ø

4 / 25

Tekislikdagi 1 m masofada yotgan nuqtadan ikkita teng og’ma o’tkazilgan. Agar og’malar perpendikulyar va tekislikka o’tkazilgan perpendikulyar bilan 60⁰ ga teng burchaklar tashkil etsa, og’malarning asoslari orasidagi masofani toping

A) 2√3

B) 4√2

C) 2√2

D) 1

5 / 25

Bir burchagi 60⁰ bo’lgan to’g’ri burchakli uchburchakka tomoni 6 ga teng romb shunday ichki chizilganki, 60⁰ li burchak ular uchun umumiy, rombning barcha uchlari uchburchakning tomonlarida yotadi. Uchburchak yuzini toping

A) (81√3)/2

B) (8√3)/2

C) (1√3)/2

D) √3/2

6 / 25

x=1, y=2^x va y=2^-x chiziqlari bilan chegaralangan sohaning yuzini toping

A) ln3

B) log₄e

C) log₃e

D) ln4

7 / 25

$\sqrt[3]{131}$ sonini taqriban hisoblang.

A) 5,01

B) 3,08

C) 4,08

D) 5,08

8 / 25

To’g’riburchakli parallelepiped asosining tomonlari 7:24 nisbatda, diagonal kesimining yuzi 50dm² ga teng. Yon sirtining yuzini toping.

A) 124 dm

B) 124 dm²

C) 224 dm²

D) 135 dm²

9 / 25

Ushbu funksiyasining boshlang’ich funksiyasini toping

A) lnIx-1I

B) lnI(x-2)(x-3)I

C) (x+2)(x+1)

D) ln⁡(|x+2|*|x+1|)

10 / 25

tengsizlikning yechimi bo’lgan eng kichik natural sonni toping

A) 1

B) 4

C) 3

D) 5

11 / 25

funksiya hosilasining x₀=-2 nuqtadagi qiymatini hisoblang.

A) -38/961

B) 38/31

C) 38/961

D) -38/31

12 / 25

$f(x)=x^3+2x^2+x+3$ funksiyasining kamayish oralig’ini toping.

A) (-∞;-1)

B) (-1;1/3)

C) [-1;-1/3]

D) (-1/3;+∞)

13 / 25

Koordinatalar boshidan y=x²-4x+3 parabolaning simmetriya o’qigacha bo’lgan masofani toping.

A) 1

B) 4

C) 2

D) 3

14 / 25

Uchlari A(2;3;4) va B(-4;-5;-6) nuqtalarda bo’lgan AB kesmasining o’rtasining koordinatalarini toping.

A) (-1;-1;-1)

B) (-1;-1;-4)

C) (-1;-3;-1)

D) (-2;-4;-1)

15 / 25

$f(x)=x^2-3x+4$ funksiya uchun y=2x-1 to`g`ri chiziqqa parallel bo`lgan urinma tenglamasini yozing

A) y=2x-2,25

B) y=2x+3

C) y=3x-2,2

D) y=2x

16 / 25

Funksiya hosilasini toping: $f(x)=7x^3-5x^2+4$

A) 14x+12

B) 21x²-10x

C) 12x+12

D) 12x²-12

17 / 25

A(-4;3;6) nuqtaning Oxz tekisligiga nisbatan simmrtrik nuqtasini toping.

A) (-4;-3;9)

B) (-4;-3;6)

C) (-4;-6;6)

D) (-8;-3;6)

18 / 25

tgx funksiya hosilasini toping

A) sinx

B) 1/sinx

C) cosx

D) 1/cos²x

19 / 25

$\lim_{x\rightarrow -1}(5-2x)$ limitni hisoblang

A) 7

B) 5

C) 6

D) 0

20 / 25

Agar f(x)=x² va bo`lsa ,f(g(2)) ni toping

A) 36

B) 0

C) 16

D) 12

21 / 25

y=-3x³+2x²-4x+5 funksiyaga x₀=-1 nuqtada o’tkazilgan urinma tenglamasining burchak koeffitsiyentini toping.

A) -17

B) 15

C) -13

D) 17

22 / 25

Moddiy nuqta S(t)=5t³+3t²+2t+4 qonuniyat bilan harakatlanayotgan jismning t=2 sekunddagi tezligini toping.

A) 60

B) 70

C) 74

D) 84

23 / 25

A(1;-2;2), B(1;4;0), C(-4;1;1) va D(-5;-5;3) nuqtalar berilgan. AC va BD vektorlar orasidagi burchakni toping

A) 15°

B) 52°

C) 80°

D) 90°

24 / 25

Nargiza 4minutda 260ta so`zni terib, 7ta imloviy xatoga yo`l qo`ydi. Nargizaning matn terish sifatini aniqlang

A) 0,5

B) 0,6

C) 0,01

D) 0,0269

25 / 25

$f(x)=x^3-2x^2$ funksiya grafigiga x=1 abssisssali nuqtadao`tkazilgan urinma tenglamasini yozing

A) y=x

B) y=-x

C) y=1/2

D) y= 2x

Baholash mezoni

86%-100% 5 baho

71%-85% 4 baho

56%-70% 3 baho

55% va kamiga 2 baho

Author: InfoMaster

Foydali bo'lsa mamnunmiz

SaqlashSaqlanganOlib tashlandi 6

Oldingi

11-sinf Matematika 3-chorak

Keyingi

Boshlang'ich sinf ATTESTATSIYA №1

Tags: 11-sinf matematika

InfoMaster

Foydali bo'lsa mamnunmiz

Tegishli Maqolalar

Istaklar ro'yxatiga qo'shildiIstaklar ro'yxatidan olib tashlandi 27

Javob qoldiring

Sharhlash uchun siz saytga kirishingiz lozim.

Aprel 2021
Du	Se	Ch	Pa	Ju	Sh	Ya
	1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30

Elementlarni Solishtiring

Jami (0)

Solishtiring

11-sinf Matematika 4-chorak

Baholash mezoni

86%-100% 5 baho

71%-85% 4 baho

56%-70% 3 baho

55% va kamiga 2 baho

Author: InfoMaster

11-sinf Matematika 3-chorak

Boshlang'ich sinf ATTESTATSIYA №1

6-sinf Matematika 1-chorak

8-sinf Matematika 4-chorak №2

5-sinf Matematika 4-chorak

11-sinf Matematika 3-chorak

Javob qoldiring

Elementlarni Solishtiring