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11-sinf Matematika 4-chorak

InfoMaster

Aprel 20, 2021

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11-sinf Matematika 4-chorak

Testni Salomov Sardor tayyorladi.

1 / 25

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

A) 84

B) 60

C) 74

D) 70

2 / 25

Funksiya hosilasini toping

f(x)=(2x+4)(3x+1)

A) 3x+1

B) 12x+14

C) 16x+14

D) 12x

3 / 25

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

A) y=3x-2,2

B) y=2x

C) y=2x-2,25

D) y=2x+3

4 / 25

$f(x)=x^3-2x^2+3x-2$ funksiya grafigiga x=4 abssissali nuqtada o`tkazilgan urinma tenglamasini tuzing

A) y=35x+30

B) y=10x

C) y=35x

D) y=20x+20

5 / 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) 80°

B) 90°

C) 15°

D) 52°

6 / 25

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

A) y=-x

B) y= 2x

C) y=x

D) y=1/2

7 / 25

Murakkab funksiyaning hosilasini toping. y=x²sinx

A) y’=2xsinx+x²cosx

B) y’=x²cosx-2xsinx

C) y’=-2xsinx+x²

D) y’=2xsinx-x²

8 / 25

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

A) (-1/3;+∞)

B) (-∞;-1)

C) (-1;1/3)

D) [-1;-1/3]

9 / 25

f(x)=2x+3 funksiyasi uchun A(1;5) nuqtadan o’tuvchi boshlang’ich funksiyasini toping.

A) F(x)=x²-3x+1

B) F(x)=x²+3x+1

C) F(x)=x²-3x-1

D) F(x)=x²+3x-1

10 / 25

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

A) 0

B) 36

C) 12

D) 16

11 / 25

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

A) 0,01

B) 0,6

C) 0,0269

D) 0,5

12 / 25

f(x)=(3x+2)² funksiyasining barcha boshlang’ich funksiyalarini toping.

A) F(x)=6x³+6x+4

B) F(x)=6x³+3x²-4

C) F(x)=6x³+6x²+4

D) F(x)=6x³+3x²+4

13 / 25

Havo shariga t $\in$ [0;10] minut oralig’ida V(t)=t³+13t²+t+20 (m³) havo purkamoqda. t=10 minutdagi havo hajmini toping.

A) 2330

B) 2300

C) 2025

D) 1023

14 / 25

F(x)=0.2sin(5x+12) funksiyasi uchun f(x) ni toping

A) f(x)=0.5cos(5x+12)

B) f(x)=-cos(5x+12)

C) f(x)=-0,5sin(5x+12)

D) f(x)=cos(5x+12)

15 / 25

f(x)=e^xcosx funksiyasi uchun f ’(π/2) ni aniqlang.

A. e^-0.5π B. e^0,5π C. e^π D. e^-π

A) e^(0.5π)

B) e^(-0.5π)

C) e^(π)

D) e^(-π)

16 / 25

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

A) 124 dm²

B) 124 dm

C) 135 dm²

D) 224 dm²

17 / 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) (1√3)/2

B) √3/2

C) (8√3)/2

D) (81√3)/2

18 / 25

Funksiyaning aniqlanish sohasini toping : y=log_x-2(x²+7x-8)

A) (2;3)U(3;+∞)

B) (8;+∞)

C) (-∞;-1)U(8;+∞)

D) (-∞;-8)U(2;3)U(3;+∞)

19 / 25

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

A) log₄e

B) ln3

C) ln4

D) log₃e

20 / 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) 1

D) 2√2

21 / 25

f(x)=(sinx)^cosx bo’lsa, f '(5π/6) ni toping

A) (ln2+3)/2)*(√3^√3)

B) (ln2+3)/2)*(√2^√2)

C) (ln2-3)/2)*(√2^√3)

D) (ln2+3)/2)*(√2^√3)

22 / 25

integralni hisoblang

A) 5ln2e

B) 5ln4e

C) 5

D) 5ln3e³

23 / 25

Uchburchakli muntazam prizma asosining radiusi $\sqrt{3}$ ga va balandligi 2 ga teng bo’lgan silindrga ichki chizilgan. Prizma yon sirtining yuzini toping

A) 3

B) 4

C) 335

D) 36

24 / 25

Ixtiyoriy kvadratga tashqi va ichki chizilgan aylanalar yuzlari nisbati a ga teng. Ixtiyoriy aylanaga ichki va tashqi chizilgan kvadratlar yuzlari nisbati b ga teng bolsa, a/b ni toping

A) 4

B) 335

C) 1

D) 1/2

25 / 25

(x²+23x+23)(x²+x+23)=23x² tenglama haqiyqiy ildizlari yig’indisini toping

A) 24

B) -23

C) -24

D) 23