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Matematika abituriyent testi №1

InfoMaster

Aprel 5, 2022

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Matematika abituriyentlar uchun №1

1 / 30

Ushbu funksiyaning boshlang’ich funksiyasini toping.

A) ln(|x+4+|x-1|)+c

B) ln|x+4/x-1|+c

C) ln(|x+4*|x-1|)+c

D) ln|x-1/x+4|+c

2 / 30

Hisoblang.

A) 2/17

B) 17/34

C) 2/34

D) 15/34

3 / 30

ifodaning qiymatini toping.

A) 0,5

B) 0

C) -2

D) -0,5

4 / 30

ABCD to’gri to’rtburchak ichidan olingan O nuqtadan A, B, C, D uchlarigacha bo’lgan masofalar mos ravishda 1; 2; 1,5; 1,2 ga teng bo’ladigan barcha to’g’ri to’rtburchaklar sonini toping

A) cheksiz ko’p

B) 1

C) 0

D) 2

5 / 30

Sin9x =4sin3x tenglamani yeching

A) πn, n€Ζ

B) π/2+πn, n€Ζ

C) πn/3, n€Ζ

D) π/3+πn, n€Ζ

6 / 30

To‘g‘ri to‘rtburchakning eni 25% ga orttirildi, bo‘yi esa 25% ga kamaytirildi. Natijada uning yuzi qanday o‘zgardi?

A) 6,25% ga ortadi

B) o‘zgarmaydi

C) 2,5% ga ortadi

D) 6,25% ga kamayadi

7 / 30

To’rtburchakning uchi M (0, 4) ; N(-4,0) ; P(-3;2) uchlari berilgan. Agar bo’lsa, Q uchining koordinatalarini toping.

A) (-7; 1)

B) (-7;-1)

C) (7; 1)

D) (4; -3)

8 / 30

To’g’ri to’rtburchakning perimetri 50 ga teng. Bir tomoni boshqa tomonidan 5 ga ko’p. To’g’ri to’rtburchakning yuzini toping.

A) 60

B) 150

C) 225

D) 50

9 / 30

Agar f(x)=ax³-5x²+b va bo’lsa, a ni toping.

A) 1

B) 3

C) 2

D) 0

10 / 30

Radiusi 1 ga teng aylana uchta yoyga bo`lingan. Ularga mos markaziy burchaklar 1, 2 va 6 sonlariga proporsional. Yoylardan eng kattasining uzunligini toping.

A) 3π/4

B) 4π/3

C) 2π/3

D) 3π/2

11 / 30

qonuniyat bo’yicha harakatlanayotgan moddiy nuqta harakatning 200- metrida qanday tezlikka (m/s) erishadi?

A) 42

B) 36

C) 32

D) 38

12 / 30

tenglamalar sistemasini yeching

A) (4;3)

B) (4;–3)

C) (–4;3)

D) (3;–4)

13 / 30

Tomoni 12 ga teng bo`lgan teng tomonli uchburchakga ichki chizilgan aylana uzunligini toping.

A) 2√3π

B) 12π

C) 3π

D) 4√3π

14 / 30

a ning qanday qiymatlarida ushbu 7x-a-13=(a-5)(x+7) tenglama yagona yechimga ega

A) a≠5

B) a ning bunday qiymati yo’q

C) a≠12

D) a=12

15 / 30

sistema a ning qanday qiymatida cheksiz ko’p yechimga ega?

A) 6

B) (2;∞)

C) (-∞;6])v[6;∞)

D) (-∞;6)

16 / 30

To’g’ri burchakli uchburchakning yuzi 24 ga, katetlaridan biri 6 ga teng bo’lsa, gipotenuzasini toping

A) 11

B) 12

C) 10

D) √46

17 / 30

Agar x,y sonlar (x+5)²+(y-12)²=14² tenglikni qanoatlantirsa, x²+y² ifodaning eng kichik qiymatini toping.

A) √2

B) √3

C) 1

D) 2

18 / 30

Agar geometrik progressiyaning ketma–ket dastlabki uchta hadining yig’indisi 62 ga, ularning o’nli logarifmlari yig’indisi 3 ga teng bo’lsa, shu geometrik progressiyaning birinchi hadini toping.

A) 50

B) 10

C) 10 yoki 50

D) 2 yoki 50

19 / 30

x²-5|x|-6=0 tenglama ildizini toping

A) -1;6

B) -1;-6

C) ±6

D) ±;±6

20 / 30

Arifmetik progressiyada a₁₇=33 va a₄₅=89. Progressiyaning birinchi hadi hamda ayirmasining o’rta geometrigini toping.

A) 2√2

B) √2

C) 2

D) 4

21 / 30

bo’lsa ni toping. Bunda funksiya f(x) ga teskari funksiya.

A) -10

B) -4

C) -12

D) -14

22 / 30

Hisoblang

A) 2

B) 0

C) -1

D) 1

23 / 30

a ning 7x-a-13=(a-5)(x+7) tenglama yechimga ega bo’lmaydigan qiymatining natural bo’luvchilar sonini toping.

A) 12

B) 6

C) 4

D) 8

24 / 30

To’g’ri burchakli uchburchakning gipotenuzasi 13 ga, katetlaridan biri 52 ga teng. Gipotenuzaga tushirilgan balandlik uzunligini toping

A) 7

B) 5

C) 6

D) 4

25 / 30

Agar funksiya berilgan bo’lsa, u holda M=ning qiymatini toping.

A) e⁴⁹⁵⁰

B) e⁻⁴⁹⁵⁰

C) e⁻⁵⁰⁵⁰

D) e⁵⁰⁵⁰

26 / 30

tenglamani yeching.

A) 2019

B) 0

C) 2018

D) 2017

27 / 30

Hisoblang

A) 2√3

B) 1

C) 3√3

D) √3

28 / 30

Rombning balandligi 8 ga dioganallarining ko’paytmasi 80 ga teng. Rombning perimetrini toping

A) 16

B) 24

C) 20

D) 32

29 / 30

7 sonini uchta natural sonlar yig’indisi ko’rinishida necha xil usulda yozish mumkin?

A) 5

B) 4

C) 6

D) 3

30 / 30

(-3;4) nuqtaga absissa, ordinata o’qlariga va koordinata boshiga nisbatan simmetrik bo’lgan nuqtalarni tutashtirishdan hosil bo’lgan uchburchakning eng kata tomonini toping.

A) 12

B) 24

C) 10

D) 14