Boshlang’ich, oraliq va yakuniy nazorat ishlar to’plami I bosqich talabalari uchun boshlang’ich, oraliq va yakuniy nazorat ishlar

Javob:

2-misol.

Elementar funksiya hosilasini :

a) (e^6x – sinx)=6e^6x -cosx b) (cosx – lnx)=-sinx-

3-misol.

f (4) va f (5) ni toping.

Yechish:

a) (2x³ +3x²-12x –6)=6x²+6x-12 b) (3x⁴ – 4x³ –12x² +8)=12x³-12x²-24x

f (4)= 6^.4²+6^.4-12=108 f (4)=12^.4³-12^.4²-24^.4=480

f (5)= 6^.5²+6^.5-12=168 f (5)=12^.5³-12^.5²-24^.5=1080

Javob: a) f (4)=108; f (5)=168; b) f (4)=480; f (5)=1080

4-misol.

f(x)=2x ⁴-8x²+5 funksiyaning [-3; 4] kesmadagi eng katta va eng

kichik qiymatini toping.

Yechish:

f(-3)= 2^.(-3)⁴-8^. (-3)²+5=95

f(4)=2^.4⁴-8^.4²+5=385

f (x)=2x ⁴-8x²+5=8x³-16x

f (-3)=8^.(-3)³-16^.(-3)=-216+48=-168

f (4)=8^.4³-16^.4=512-24=488

Hosil bo’lgan qiymatlar: 95; 385; -168; 488

Bu qiymatlar ichida eng katta qiymati 488 va eng kichik qiymat -168.

Javob: eng katta qiymati 488 va eng kichik qiymat -168.

II-variant.

1. Kesmaning bir uchi A(2;3;-1) va uning o’rtasi C(1;1;1) berilgan.

Kesmaning ikkinchi uchi B(x;y;z) ni toping.

Yechish :

AC=CB => AC vektorni yasaymiz: a₁=1-2=-1

a₂=1-3=-2

a₃=1-(-1)=2

Demak, AC(-1;-2;2)

CB vektorni yasaymiz : b₁=x-1

b₂=y-1

b₃=z-1.

Masala shartidan foydalanib, B vektorning koordinatalarini topamiz:

x-1=-1 => x=0

y-1=-2 => y=-1

z-1=2 => z=3 Demak, B(0;-1;3).

Javob: B(0;-1;3).

2. Elementar funksiya hosilasini toping.

a) e ^–6x +log₃x b) sinx – 9e ^5x

Yechish:

a) (e ^–6x +log₃x)=-6 e ^–6x + b) (sinx – 9e ^5x)=cosx-45e ^5x

3. f (2) va f (3) ni toping.

Yechish:

a) (2x⁵ +3x³-12x² –6)=10x⁴+9x²-24x b) (3x⁴ – 4x² –12x³ +8)=12x³-8x-36x²

f (2)=10^.2⁴+9^.2²-24^.2=148 f (2)=12^.2³-8^.2-36^.2²=-64

f (3)=10^.3⁴+9^.3²-24^.3=819 f (3)=12^.3³-8^.3-36^.3²=-24

Javob: a) f (2)= 148 f (3)=819 b) f (2)=-64 f (3)=-24

4. f(x)=4x⁴-6x²+8 funksiyaning [-2; 3] kesmadagi eng katta va eng

kichik qiymatini toping.

Yechish:

f(-2)=4^.(-2)⁴-6^.(-2)²+8=80

f(3)=4^.3⁴-6^.3²+8=278

f (x)=(4x⁴-6x²+8)=16x³-12x

f (-2)=16^.(-2)³-12^.(-2)=-104

f (3)=16^.3³-12^.3=396

Hosil bo’lgan qiymatlar: 80; 278; -104; 396

Bu qiymatlar ichida eng katta qiymati 396 va eng kichik qiymat -104.

Javob: eng katta qiymati 396 va eng kichik qiymat -104.

Oraliq nazorat ish –3.

I-variant.

1-masala.

1. Berilgan:

To’g’ri burchakli parallelepiped Yechish:

a=7, b=6, h=6, d=? d²=7²+6²+6²=49+36+36=841

d=29

Javob: 11

2. Berilgan: ABCDA₁B₁C₁D₁

muntazam to’rtburchakli prizma B₁ C₁

B D = 8 sm DC₁ = 7 sm A₁ D₁

ABCD kvadrat.

B₁D topilsin B

C

A D

Yechilishi

Asosdagi ABCD kvadratning tomoni a bilan, prizmaning yon qirrasini AA₁=h deb belgilaymiz. So’ngra ABD, DCC₁, BB₁D to’g’ri burchakli uchburchaklardan Pifagor teoremasiga asosan quyidagilarni topamiz:

Δ ABD : BD²=a²+a² ΔDCC₁: C₁D²=h²+a² ΔBB₁D: B₁D²= h²+BD²

8²=2a² 7²=h²+32 B₁D²=17+8²

a²=32 h²=49-32 B₁D²=17+64=81

h²=17 B₁D=9

3. 9 ta qirrasi bo’lgan ko’pyoq - asoslari uchburchak bo’lgan prizma









II-variant.

1-masala.

To’g’ri burchakli parallelepipedning uchta o’lchovlari berilgan bo’lsa, uning dioganalini toping: 12, 21, 16.

Berilgan:

To’g’ri burchakli parallelepiped Yechish:

a=12, b=21, h=16, d=? d²=12²+21²+16²=144+441+256=841

d=11

Javob: 11

2-masala. S

B erilgan:

SABC uchburchakli piramida

ΔABC-teng yonli

B DAC, AC=12sm B

AB=BC=10 sm A

A S=BS=CS=13 sm. D

SO=? C

Yechish:

AS=CS=BS => ularning proyeksiyalari ham o’zaro teng, AO=BO=CO. O nuqta ΔABC ga tashqi chizilgan aylananing markazi va AO=R ushbu aylananing radiusi va uni quyidagi formula bilan topamiz: . Uchburchaklarning radiusini geron formulasi yordamida topamiz:

. U holda

To’g’ri burchakli ΔAOS dan Pifagor teoremasi yordamida topamiz: SO²=AS²-AO²=13²-

SO= Javob:

3. 8 ta qirrasi bo’lgan ko’pyoq – asosi to’rt burchak bo’lgan piramida.

Nazorat ish – 4.

I-variant.

1. Funksiya hosilasini toping. y=( )=

2. Berilgan funksiyaning boshlang’ich funksiyasini toping.

e^8x- sin3x

3. Quyidagi chiziqlar bilan chegaralangan figuralarning yuzlarini toping.

y=6x-x² parabola va y=x+4 to’g’ri chiziq

y =6x-x² va y=x+4 funksiya grafigini yasaymiz.

x y x y

1 5 1 5

0 0 2 6

3 9 3 7

4 8 4 8

5 5 5 9

6 0

S=

Parabola va to’g’ri chiziqlarning kesishgan nuqtalarining x o’qi bilan kesishgan nuqtalarini integral uchun chegara deb olamiz:

II-variant

1. Funksiya hosilasini toping.

y=

2. Berilgan funksiyaning boshlang’ich funksiyasini toping.

e^2x- cos5x

3. Quyidagi chiziqlar bilan chegaralangan figuralarning yuzlarini toping.

y= 4-x² parabola va y=x+2 to’g’ri chiziq

x y x y

1 3 1 3

2 0 2 4

-1 3 0 2

-2 0 -1 1

0 4 -2 0

S=

y=4-x² va y= x+2 funksiya grafigini yasaymiz.

x y x y

1 3 1 3

2 0 2 4

-1 3 0 2

-2 0 -1 1

0 4 -2 0

Nazorat ish-5.

I-variant.

1-masala.

A gar 1 m² tomni bo’yashga 0,12 kg bo’yoq ketsa, asosining diametric 10m va balandligi 12 m bo’lgan konus shaklidagi tunuka tomni bo’yash uchun necha kilogramm bo’yoq ketadi? B

Berilgan: Yechish:

konus shaklidagi tom 2R=10 => R=5m

1m² => 0,12 kg bo’yoq l h ΔAOB dan AB=l=

AC=10 m AB=l= m

h=12 m A O C S_yon=Rl=^.5^.13=65=204,1(m²).

bo’yoq=? 204,1^. 0,12=24,492 (kg).

Javob: tomni bo’yash uchun 24,492 kg bo’yoq kerak.

2-masala.

Diametri 25 sm bo’lgan koptok uchun necha kvadrat metr rezina sarf bo’lgan?

Berilgan: Yechish:

D =25sm bo’lgan koptok D=25sm=0,25m

yasash uchun qancha rezina S_sh=4R²=D²

kerak? S_sh=3,14^. 0,25²=0,196 (m²)

Javob: koptok yasash uchun 0,196 m² rezina kerak bo’ladi.

3-masala.

37 xillik kartadan iborat dasturdan 5 tasini necha xil usulda olish mumkin?

II-variant.

1-masala.

Konusning balandligi 28 m va asosining radiusi 10 m. Konusning yon sirti topilsin.

B

Berilgan: Yechish:

konus. l S_{yon k}=Rl=10l

h=28m h ΔAOB dan l=

R =10m l= (m)

S_{yon k}=? A C S_{yon k}=10 ^. 29 =290 (m²).

Javob: konus yon sirti 290 m²

2-masala.

Silindr to’la sirtining yuzi 62 sm², yon sirtining yuzi 30 sm² bo’lsa, silindrning balandligi topilsin.

Berilgan: Yechish:

A B₁ – silindr natijada

S_t=62 sm²

S_yon=30 sm^{2 => =>}

H=? =>

Javob: silindr balandligi .

3-masala.

41 xillik kartadan iborat dasturdan 6 tasini necha xil usulda olish mumkin?

1.

Oraliq nazorat ish – 1.

I-variant

1. Agar A to’plam x²-7x+6=0 tenglamaning yechimlari to’plami va

B={1;6} bo’lsa, A=B bo’lishini isbotlang.

2. Hisoblang:

a) b) (50000-1397,3):(20,4+33,603)

3. 175 % i 78,75 ni tashkil qiladigan sonni toping.

II-variant

1. Agar A={3;4;5} va B to’plam x²-7x+12=0 tenglamaning yechimlari to’plami bo’lsa, BA

bo’lishini isbotlang.

2. Hisoblang:

a) b) (2779,6+8024,4):(1,98+2,02)

3. 46,6 soni 11,65 ning necha foizini tashkil qiladi?