3

SO‘Z BOSHI Ushbu o‘quv-uslubiy materiallar qurilish ta’lim yo‘nalishlarida sirtdan

o‘qiyotgan talabalar uchun mo‘ljallangan va «Oliy matematika» fanini o‘rganishda ular uchun ko‘rsatma vazifasini o‘taydi. U sirtqi talabalar uchun nazorat ishlarini bajarishlariga oid asosiy tavsiyalarni va shuningdek fanning «Chiziqli algebra elementlari», «Vektorli va analitik geometriya» va «Matematik analizga kirish» bo‘limlarini o‘rganish bo‘yicha uslubiy ko‘rsatmalarni o‘z ichiga oladi.

Uslubiy ko‘rsatmada «Oliy matematika» fanidan savollar, tavsiya qilinayotgan adabiyotlar ro‘yxati va nazorat ishlari uchun yigirma besh variantdan iborat topshiriiqlar keltirilgan. Nazorat ishlarining har bir topshirig‘iga oid namunaviy misol-masalalar yechib ko‘rsatilgan.

Materiallarda nazorat topshiriqlari yigirma besh variant uchun berilgan bo‘lib ular uchta qismga ajratilgan.

Ushbu qo‘llanma «Matematika va tabiiy fanlar» kafedrasi tomonidan qurilish ta’lim yo‘nalashlari sirtqi talabalarini o‘quv-uslubiy ta’minlashning tarkibiy qismlaridan biri hisoblanadi.

SIRTDAN O‘QIYOTGAN TALABALAR USHUN NAZORAT ISHLARINI BAJARISH BO‘YICHA

UMUMIY TAVSIYALAR 1. Sirtqi talaba fanni o‘rganish jarayonida oily matematikaning turli

bo‘limlaridan nazorar ishlaruini bajarishi lozim. Bu nazorat ishlari o‘qituvchi tomonidan taqriz qilinadi. Bajarilgan ishga yozilgan taqriz talabaga uning materialni o‘zlashtirganligi bo‘yicha baho berish imkonini beradi, mavjud kamchiliklarini ko‘rsatadi va keyingi ishlarini muvofiqlashtiradi va o‘qituvchining qo‘yiladigan savollarni tizimlashtirishida yordam beradi.

2. O‘rganilayotgan material bo‘yicha yetarli sondagi misol va masala yechmasdan talaba nazorat ishini bajarishga kirishmasligi lozim.

3. Har bir nazorat ishi mustaqil bajarilishi kerak. Mustaqil bajarilmagan nazorat ishi taqrizchi - o‘qituvchiga uning ishida materialni o‘zlashtirish bo‘yicha kamchiliklarni ko‘rsatishi uchun imkon bermaydi, natijada talaba kerakli bilimga ega bo‘lmasdan yakuniy nazoratni topshirish uchun tayyor bo‘lmasligi mumkin.

4. Nazorat ishi o‘z vaqtida topshirilishi lozim. Bu talabning bajarilmasligi taqrizchi - o‘qituvchiga talabaning kamchiliklarini o‘z vaqtida ko‘rsatish imkonini bermaydi va ishning taqriz qilinishi vaqtini cho‘zilishiga olib keladi.

5. Nazarat ishini bajarish va rasmiylashlashtirishda talaba quyidagi qoidalarga qat’iy amal qilishi lozim:

a) nazorat ishi alohida daftarga taqrizchi - o‘qituvchining qaydlari uchun xoshiya qoldirilgan holda bajarilshi kerak;

b) daftarning muqavasida quyidagilar qayd etilishi lozom: - oily matematikadan nazaorat ishi va uning tartib raqami;

4

- talabaniing familiyasi va ismi-sharifi, reyting daftarchasining nomeri; - fakultet, kurs, guruh; - ishning oily o‘quv yurtiga jo‘natilgan sanasi va talabaning manzili. v) masalalarning yechimi uning keltirilgan tartibida joylashtirilishi kerak; g) har bir masalani yechishdan oldin uning sharti zarur joylarda harfli ifodalar

o‘zining variantiga mos qiymatlar bilan almashtirilgan holda to‘liq ko‘chirilishi kerak;

d) masala yechimining asosiy bosqichlari qisqa va lo‘nda izohlar bilan berilishi lozim;

e) nazorat ishining oxirida foydalanilgan adabiyotlar ro‘yxati berilishi kerak. 6. Talaba reyting daftarchasi nomerining oxirgi ikki raqamiga mos variantni

bajaradi. Bunda bu ikki raqam 25 ga bo‘linadi va qoldiq talaba bajarishi kerak bo‘lgan variant nomerini bildiradi. Agar bu ikki raqam 75,50,25,00 dan iborat bo‘lsa, talaba 25- variantni bajaradi.

7. Taqriz qilingan ishni olgandan so‘ng talaba taqrizchi tomonidan ko‘rsatilgan kamchiliklarni tuzatishi va ishni qayta taqrizga jo‘natishi lozim.

8. Belgilangan tartibda taqrizdan o‘tgan va inobatga olingan (zachet qilingan) nazorat ishlarini topshirmagan talaba yakuniy nazoratga kiritilmaydi.

«OLIY MATEMATIKA» KURSIDAN SAVOLLAR

RO‘YXATI.

Chiziqli algebra elementlari Ikkinchi va uchinchi tartibli determinantlar. Determinantning xossalari.

n -tartibli determinantlarni hisoblash. Matritsa va uning turlari. Matritsalar ustida arifmetik amallar. Teskari

matritsa. Matritsaning rangi. Chiziqli algebraik tenglamalar sistemasi. Chiziqli tenglamalar sistemasini yechishning Gauss usuli. n noma’lumli m ta chiziqli tenglamalar sistemasini tekshirish va yechish. Xosmas tenglamalar sistemasini yechish. Bir jinsli chiziqli tenglamalar sistemasi. Chiziqli tenglamalar sistemasini matematik paketlarda yechish.

Vektorli algebra elementlari va analitik geometriya Vektorlar ustida chiziqli amallar. Vektorning o‘qdagi proeksyasi.

Vektorlarning chiziqli bog‘liqligi, bazis. Dekart koordinatalar sistemasida vektorlar.

Ikki vektorning skalyar ko‘paytmasi. Ikki vektorning vektor ko‘paytmasi. Uchta vektorning aralash ko‘paytmasi.

Tekislikdagi chiziq. Tekislikdagi to‘g‘ri chiziq tenglamalari. Tеkislikda ikki to‘g‘ri chiziqning o’zaro joylashishi. Nuqtadan to‘g‘ri chiziqgacha bo‘lgan masofa.

5

Ikkinchi tartibli chiziqlarning umumiy tenglamasi. Aylana va ellips. Giperbola. Parabola.

Qutb koordinatalari. Qutb koordinatalar sistemasiida chiziqlar Fazoda sirt va chiziq. Tekislik tenglamalari. Fazoda ikki tekislikning o’zaro joylashishi. Nuqtadan tekislikkacha bo‘lgan masofa. Fazodagi to‘g‘ri chiziqning tenglamalari. Fazoda ikki to‘g‘ri chiziqning o’zaro joylashishi. Fazoda to‘g‘ri chiziq bilan tekislikning o’zaro joylashishi. Nuqtadan fazodagi to‘g‘ri chiziqgacha bo‘lgan masofa.

Ikkinchi tartibli sirtlarning umumiy tenglamalari. Sfera va ellipsoidlar. Giperboloidlar. Konus sirtlar. Paraboloidlar. Silindrik sirtlar. Ikkinchi tartibli sirtlarning to’g’ri chiziqli yasovchilari.

Matematik analizga kirish

To‘plam. Sonli to’plamlar. Matematik mantiq elementlari. Haqiqiy sonlar. Kompleks son tushunchasi va tasviri. Kompleks sonlarning yozilish shakllari.

Kompleks sonlar ustuda amallar Sonli ketma-ketliklar. Cheksiz katta va cheksiz kichik ketma-ketliklar. Ketma-ketlikning limiti. Yaqinlashuvchi ketma-ketliklar. e soni. Bir o‘zgaruvchining funksiyasi. Asosiy elementar funksiyalar. Teskari funksiya. Murakkab funksiya. Elementar funksiyalar sinfi. Giperbolik funksiyalar. Oshkormas va parametrik ko’rinishda berilgan funksiyalar. Funksiyaning limiti. Cheksiz kichik funksiyalar.

Funksiya uzluksizligining ta’riflari. Uzluksiz funksiyalarning xossalari. Funksiyaning uzilish nuqtalari. Tekis uzluksizlik.

TAVSIYA QILINAYOTGAN ADABIYOTLAR

1. Sh.R. Xurramov. Oliy matematika. Darslik, I-jild, T., “Tafakkur”. 2018. 2. Sh.R. Xurramov. Oliy matematika. Darslik, 2-jild, T., “Tafakkur”. 2018.

3. Sh.R. Xurramov. Oliy matematika (masalalar to‘plami, nazorat topshiriqlari). Oliy ta’lim muassasalari uchun o‘quv qo‘llanma. 1-qism. –T.: «Fan va texnologiya», 2015, 408 bet. 4. А.П.Рябушко и др. Сборник задач индивидуальных заданий по высшей математике. Ч. 2– Минск, Высшая школа, 1991. 5. О.В Зимина, А.И.Кириллов, Т.А. Сальникова, Высшая математика. М.: Физматлит, 2001. 6. П.С. Данко, А.Г.Попов, Т.Я.Кожевникова. Высая математика в упражнениях и задачах. Ч.1. –М.: 2003. 7. К.Н.Лунгу, Е.В.Макаров. Высшая математика. Руководство к решению задач. Ч.1 – М.: Физматлит, 2007.

8. Черненко В.Д. Высшая математика в примерах и задачах. 1том. СПб. “Политехника”, 2003.

6

NAZORAT ISHINI BAJARISH BO‘YICHA USLUBIY KO‘RSATMALAR

Uslubiy ko‘rsatmaning ushbu bandida nazorat ishlarining namunaviy

masalalari yechib ko‘rsatilgan. Masalalarning yechimi talaba nazorat ishini bajarishi jarayonida o‘rganishi kerak bo‘lgan mavzular bo‘yicha keltirilgan. Masalalarning yechimi talaba o‘zining variantini bajarishida faodalanilishi mumkin bo‘lgan formula va tushunchalarni o‘z ichiga olgan. Ta’kidlash joizki, bu formula va nazariy tushunchalar faqat amaliy mashg‘ulotlarda va nazorat ishlarini bajarilishida qo‘llanilishi mumkin. Ular yakuniy nazoratni topshirish uchun yetarli emas.

1- MAVZU. CHIZIQLI ALGEBRA ELEMENTLARI

1-masala. 1. Berilgan determinantni hisoblang: a) i satr elementlari bo‘yicha

yoyib; b) j ustun elementlari bo‘yicha yoyib; c) j ustundagi bittadan boshqa elementlarni nolga aylantirib va shu ustun elementlari bo‘yicha yoyib.

.2,2,

3212110332120214

ji

Yechish. a) Determinantni 2i satr elementlari bo‘yicha yoyamiz. 24242323222221212424232322222121 AaMaMaMaAaAaAaAa .

322113024

1321110021

2

212103214

3312103014

2

)020023(2 )940020(2)18800412(

.361830186)640620(3

b)Determinantni 2j ustun elementlari bo‘yicha yoyamiz:

42423232222212124242323222221212 AaMaMaMaAaAaAaAa

113322024

10322113024

1322113322

1

)18800412()18461846(

.3618180)41200188(

7

c) Determinantni 2j ustundagi bittadan boshqa elementlarni nolga aylantirib va shu ustun elementlari bo‘yicha yoyib hisoblaymiz.

Buning uchun: – 1-satr elementlarini 2- satrning mos elementlariga qo‘shamiz; – 1-satr elementlarini )1( ga ko‘paytirib 4-satrning mos elementlariga qo‘shamiz; – determinantni 2-ustun elementlari bo‘yicha yoyamiz

.306113342

306113342

)1(1

3006110334020214

21

Uchinchi tartibli determinantda 2 ustunning 2 satri elementidan boshqa elementlarini nolga aylantiramiz. Bunda 32a element nolga teng bo‘lgani uchun faqat 12a elementni nolga aylantiramiz. Buning uchun 1-satrga )4( ga ko‘paytirilgan 2-satrni qo‘shamiz, hosil bo‘lgan determinantni 2 ustun elementlari bo‘yicha yoyamiz va kelib chiqqan ikkinchi tartibli determinantni hisoblaymiz:

.3636110

)1(13061131010

22

2-masala. BA, matritsalar va , sonlari berilgan. BA , AB , 1A matritsalarni toping.

,121642414

A ,

211052110

B ,4 4 .

Yechish. a) BA matritsani topish uchun A matritsa elementlarini ga, B matritsa elementlarini ga ko‘paytiramiz va hosil qilingan A va B matritsalarning mos elementlarini qo‘shamiz:

211052110

4121642414

)4(BA

8440208440

4842416816416

8

12402436020816

844844024201688416)4(4016

.

b) AB martitsani matritsalarni ko‘paytirish qoidasi asosida topamiz:

211052110

121642414

AB

.183

14162432

201110114012026202680804454420

c) A matritsa determinantini hisoblaymiz:

0562481616616121642414

||

A .

ijA algebraik to‘ldiruvchilarni topamiz:

,81264

11

A ,81162

12 A ,8

2142

13

A

,71241

21

A ,01144

22

A ,72114

23 A

,106441

31

A ,326244

32

A .184214

33 A

Bundan

.

289

81

71

28160

71

285

81

71

187832081078

561

||1

332313

322212

3121111

AAAAAAAAA

AA

3-masala. Tenglamalar sistemalarini birgalikda bo‘lish-bo‘lmasligini Kroniker-Kapelli teoremasi bilan tekshiring. Birgalikda bo‘lgan sistemani Kramer formulalari orqali, matritsalar va Gauss usullari bilan yeching.

.443,105

,332

321

321

321

xxxxxxxxx

9

Yechish. Sistemaning kengaytirilgan matritsasi ustida elementar almashtirishlar bajaramiz:

)(33)( CrAr

Demak, sistema Kroniker-Kapelli teoremasiga ko‘ra aniq sistema. 1) Sistemani Kramer formulalari bilan yechamiz.

Sistemaning determinantini va yordamchi determinantlarni hisoblaymiz:

;51143151312

;511441510313

1

x

;1021431101332

2

x ;51443

1051312

3

x

Tenglamaning yechimini Kramer formulalari bilan topamiz:

;151511

1

xx ;2

511022

2

xx

.151513

3

xx

2) Sistemani matritsalar usuli bilan yechamiz. Sistema uchun .51

~ 3

2

41433312

10151

C

414310151

3312

~

~

34419017511010151

~ 11:

3441901117

11510

10151

~

~ 19

1151:

11001117

11510

10151

.

1151

115100

1117

11510

10151

~

10

Sistema determinantining algebraik to‘ldiruvchilarini topamiz:

11415

11

A ; 41311

12

A ; 194351

13

A ;

;111431

21 A ;71332

22 A ;54312

23 A

;141531

31 A ;5

1132

32 A .11

5112

33 A

U holda

.11519574

14111

5111

A

Tenglamaning yechimini BAX 1 formula bilan topamiz:

.121

51102

51

511

445057207012561103

511

410

3

11519574

14111

5111

BAX

Demak, 11 x , 22 x , 13 x .

3) Sistemani Gauss usuli bilan yechamiz. Gauss usulining 1-bosqichi yuqorida sistemani tekshirishda uning

kengaytirilgan matritsasida bajarildi va quyidagi ko‘rinish hosil qilindi: Gauss usulining 2-bosqichini bajaramiz:

10)1(5

,1117)1(

115

,1

1

,1117

115

,105

21

2

3

3

32

321

xx

x

x

x

xx

xxx

.1,2,1

1125,2,1

3

2

1

21

2

3

xxx

xxx

11001117

11510

10151

.

11

000731

36160.

2

1880731

36160 ~ A

5

3

313731115

~

4-masala. Bir jinsli tenglamalar sistemasini yeching.

.033,073,05

321

321

321

xxxxxxxxx

Yechish a) Sistema matritsasi ustida elementar almashtirishlar bajaramiz:

.,3,2)( nrnAr Demak, sistema cheksiz ko’p yechimga ega. Ularni topamiz:

.73

,5073,05

321

321

321

321

xxx

xxxxxxxxx

,163115

,4371

33

31 xx

xx

.3671

53

3

32 xx

xx

,4

311

xxx

4

9 322

xxx

.

Erkin noma’lumni kx 43 ( k ixtiyoriy son) deb, sistemaning umumiy

yechimini topamiz: .4,9, 321 kxkxkx

2- MAVZU. VEKTORLI ALGEBRA ELEMENTLARI

VA ANALITIK GEOMETRIYA

5-masala. ABC uchburchak uchlarining koordinatalari berilgan: a) C uchdan tushirilgan balandlik tenglamasini tuzing va uning uzunligini toping; b) B uchdan o‘tkazilgan mediana tenglamasini tuzing va uchburchak medianalarining kesishish nuqtalarini toping; c) A burchakning radian qiymatini hisoblang va uning bissektrisasi tenglamasini tuzing. ),20( A ),10;5(B ).1;4(C Yechish. a) AB tomon tenglamasini berilgan ikki nuqtadan o‘tuvchi to‘g‘ri chiziq tenglamasi formulasidan topamiz:

,102

10505

yx 010512 yx ).(AB

Bundan

,25

12 xy .

512

1 k

12

CM balandlik AB tomonga perpendikular bo‘lib, C nuqtadan o‘tadi (1-shakl). Shu sababli uning tenglamasi

),4(1 xky ),4(111

xk

y ),4(1251 xy

08125 yx ).(CM

CM balandlik uzunligi C nuqtadan AB to‘g‘ri chiziqqacha bo‘lgan masofaga teng.

Demak,

.)..(1363

512|1015412|||

22buСM

b) AC tomon o‘rtasi );( yxN nuqtada bo‘lsin. U holda kesmaning o‘rtasi koordinatalarini topish formulasiga ko‘ra:

,22

40 x 21

212 y yoki .

21;2

N

BN mediana tenglamasini tuzamiz:

,10

21

10525

yx 0523 yx ).(BN

Uchburchak medianalarining xossasiga ko‘ra medianalarning kesishish

nuqtasi );( yxK da 212

||||

KNBK bo‘ladi.

U holda

;31

21225

x 3

2121210

y

yoki .3;31

K

c) AC tomon tenglamasini tuzamiz:

,212

040

yx 0843 yx ).(AC

AB va AC tomonlar orasida burchak A bo‘lsin. Uni ikki to‘g‘ri chiziq

orasidagi burchak formulasidan foydalanib hisoblaymiz:

6516

)4(3512)4(5312cos

2222

yoki .3134,06516arccos

A burchak bissektrisasi CB tomon bilan );( yxL nuqtada kesishsin (1-shakl).

A

1-shakl.

K

x

y

C

L

B

N

.

.

.

.

.

. 5 4 2

10

13

Uchburchak bissektrisasining xossasiga ko‘ra

.||||

||||

ABAC

LBCL

5)21()04(|| 22 AC va 13)210()05(|| 22 AB ekanidan

135

||||

LBCL .

U holda

,23

1351

)5(1354

x

27

1351

101351

y yoki .

27;

23

L

Ikki nuqtadan o‘tuvchi to‘g‘ri chiziq tenglamasidan topamiz:

227

2

023

0

yx

yoki 06311 yx ).(AL

6-masala. 1. Har bir );( yxM nuqtasidan berilgan );( 11 yxA va );( 22 yxB nuqtalargacha bo‘lgan masofalar nisbati a ga teng bo‘lgan chiziq tenglamasini tuzing.

),2;3( A )6;4(B , .53

a

Yechish. Ikki nuqta orasidagi masofa formulasidan topamiuz:

,)2()3(|| 22 yxAM .)6()4(|| 22 yxBM

Misolning shartiga ko‘ra

aBMAM

|||| yoki .

53

)6()4()2()3(

22

22

yxyx

Bu tenglikda almashtirishlar bajaramiz: ),3612168(9)4496(25 2222 yyxxyyxx

,46810897293251002515025 2222 yyxxyyxx ,143208167816 22 yyxx

,143138

3916 22

yyxx

,2

131639

16143

213

2132

1639

16392

2222

22

yyxx

14

.16

65152

131639

222

yx

Bu tenglama markazi

213;

1639 nuqtada joylashgan va radiusi

166515 ga

teng bo‘lgan aylanani aniqlaydi.

2. Har bir );( yxM nuqtasidan berilgan );( 11 yxA nuqtagacha va bx to‘g‘ri chiziqqacha bo‘lgan masofalar nisbati m ga teng bo‘lgan chiziq tenglamasini tuzing.

),0;6(A 23x , .2m

Yechish. Ikki nuqta orasidagi masofa va nuqtadan to‘g‘ri chiziqqacha bo‘lgan masofa formulalari bilan topamiz:

,)0()6(|| 22 yxAM .23|| xBM

Misolning shartiga ko‘ra

mBMAM

|||| yoki .2

23

)6( 22

x

yx

Bundan

.234)6(

222

xyx

Bu tenglikda almashtirishlarni bajaramiz:

,49343612 222

xxyxx

,91243612 222 xxyxx ,273 22 yx

.1279

22

yx

Bu tenglama fokuslari Ox o‘qida joylashgan va yarim o‘qlari 33,3 ba ga teng bo‘lgan giperbolani aniqlaydi.

7-masala. ABCD piramidaning uchlari berilgan: a) AB qirra tenglamasini tuzing; b) ABC yoq tenglamasini tuzing; c) D uchdan ABC yoqqa tushirilgan balandlik tenglamasini tuzing va uning uzunligini toping; d) C uchdan o‘tuvchi AB qirraga parallel to‘g‘ri chiziq tenglamasini tuzing; e) D uchdan o‘tuvchi AB qirraga perpendikular tekislik tenglamasini tuzing; f) AD qirra bilan ABC yoq orasidagi burchak sinusini toping; g) ABC va ABD yoqlar orasidagi burchak kosinusini toping.

),7;1;2(A ),6;3;3(B ),9;3;2( C ).5;2;1(D

15

Yechish. a) AB qirra tenglamasini berilgan ikki nuqtadan o‘tuvchi to‘g‘ri chiziq tenglamasidan foydalanib tuzamiz:

767

131

232

zyx yoki

17

21

12

zyx ).(AB

b) ABC yoq tenglamasini berilgan uchta nuqtadan o‘tuvchi tekislik tenglamasi bilan tuzamiz:

.0240121712

zyx

Bundan 0152 zy ).(ABC

c) D uchdan tushirilgan DE balandlik ABC yoqqa perpendikular bo‘ladi. Shu sababli DE to‘g‘ri chiziqning yo‘naltiruvchi vektori };;{ rqps sifatida ABC yoqning normal vektori }2;1;0{1 n

ni olish mumkin. U holda to‘g‘ri chiziqning kanonik tenglamasi formulasiga ko‘ra

25

12

01

zyx ).(DE

Nuqtadan tekislikkacha bo‘lgan masofa formulasidan topamiz:

222 210|15522110|||

DE yoki .)..(5

53|| buDE

d) C uchdan o‘tuvchi CF to‘g‘ri chiziq AB qirraga parallel bo‘gani sababli CF to‘g‘ri chiziq va AB qirraning yo‘naltiruvchi vektori }1;2;1{21 ss

bo‘ladi. U holda

19

23

12

zyx ).(CF

e) D uchdan o‘tuvchi tekislik AB qirraga perpendikular bo‘lgani uchun AB to‘g‘ri chiziqning yo‘naltiruvchi vektori }1;2;1{1 s

ni izlanayotgan tekislikning normal vektori };;{2 CBAn

deb olish mumkin. Tekislik tenglamasini berilgan nuqtadan o‘tuvchi va berilgan vektorga perpendikular tekislik tenglamasi bilan topamiz:

0)5()1()2(2)1(1 zyx yoki .02 zyx

f) AD qirra tenglamasini tuzamiz:

27

11

12

zyx ).(AD

16

AD qirra bilan ABC yoq orasidagi burchak sinusini to‘g‘ri chiziq bilan tekislik orasidagi burchak formulasidan topamiz:

54,065

3)2(1)1(210

)2(211)1(0sin222222

g) ABD yoq tenglamasini tuzamiz:

0211121712

zyx

yoki 06 zyx ).(ABD

ABC va ABD yoqlar orasidagi burchak kosinusini ikki tekislik orasidagi burchak formulasidan foydalanib topamiz:

77,035

3)1()1(1210

)1(2)1(110cos222222

.

8-masala. Sirt turini aniqlang va shaklini chizing.

a) ;1829 22 zyx b) .1234 22 yx Yechish. a) Sirt tenglamasini kanonik shaklga keltiramiz:

,1829 22 yzx ),9(29 22 yzx ).9(2

92

22

yzx

Bu tenglama elliptik paraboloidni aniqlaydi (2-sahkl).

b) Berilgan tenglamada .0z Bunda berilgan sirt yasovchilari Oz o‘qqa parallel silindrik sirtdan iborat bo‘ladi.

3-shakl.

3

z

x

O y

2-shakl.

9

z

y

32

2

x

17

1234 22 yx tenglamadan topamiz:

143

22

yx .

Bu tenglama giperbola tenglamasi bo‘ladi. Demak, berilgan tenglama giperbolik silindrni aniqlaydi (3-shakl).

3- MAVZU. MATEMATIK ANALIZGA KIRISH 9-masala. Sonli ketma-ketliklarning limitini toping.

a) ;)2)(1( 232 nnnnxn b) .))34(...951(!)!1(2)!2(

nnnnxn

Yechish. a)

))2)(1(lim(lim 232 nnnnxn

nn

)2)(1(

22lim232

2355

nnnnnnnn

n

)2)(1(22lim

232

23

nnnnnn

n

.)00)(01(0

00221111

212lim

33

3

nnnn

nnn

b)

nnn

nnnnn

nnxn

2341!

)22)(1(!))34(951(!

)!1(2)!2(

.)12(

)4)(1(

nnnn

Bundan

.21

02)01)(01(

12

4111lim

)12()4)(1(limlim

n

nnnn

nnxnnnn

10-masala. Limitlarni toping:

a) ;13

2lim 3442

xxxxx

x b) ;

12530lim 3

2

5

xxx

x c) ;

231lim

38

xx

x

d) ;coscos3lim 30 xx

xtgxx

e) )).12ln()32)(ln(2(lim

xxxx

Yechish.

a) .113

121

113

121

132

4

32

44

234

34

42

xx

xx

xxx

xxx

xxxxx

18

U holda

132lim 34

42

xxxxx

x.

31

003001

113

121

113

121lim

4

32

xx

xxx

b) .7511

2556lim

)255)(5()6)(5(lim

12530lim 25253

2

5

xxx

xxxxx

xxx

xxx

c)

3142

)42)(2()31)(31(lim

231lim

33 2

33 23838 xxx

xxxxx

xx

xx

.233

4)2(2)2(31

42lim31

42)8()8(lim

233 2

8

33 2

8

x

xxx

xxxx

xx

d) xxx

xxxxx

xxxx 2020 sincos3cos

3sinlim)cos1(cos3cos

3sinlim

3113

sinlim

33sinlim

31sin

33sin3

limcos3cos

1lim 2

0

0

22

2

00

xxx

x

xx

xx

xx

xxx

x

xx.

e)

12

32ln)2(lim))12ln()32)(ln(2(limxxxxxx

xx

)2(12

4

4122

1241lnlim

1232lnlim

xxx

x

x

x xxx

.21284limlnlim 12

84

xxe

x

xx

x

11-masala. Funksiyani uzluksizlikka tekshiring va grafigini chizing.

.1,4

,12,)1(,2,3

)(3

2

xxxxxx

xf

Yechish. Funksiya );( x da aniqlangan. ),2;( )1;2( , );1( oraliqlarda funksiya uzluksiz. ,2x 1x nuqtalarda funksiya analitik berilishni o‘zgartiradi. Shu sababli, bu nuqtalarda funksiya uzilishga ega bo‘lishi mumkin. 4-shakl.

y

x

.

1

4

3

1 O

2

19

2x nuqtada: ,1)3(lim)02(02

xfx

,1)1(lim)02( 202

xfx

.132)2( f Bundan ).2()02()02( fff Demak, 2x nuqtada funksiya uzluksiz.

1x nuqtada: ,4)1(lim)01( 1

2

01Axf

x

.3)4(lim)01( 2

3

01Axf

x

Demak, 1x sakrash nuqtasi va bu nuqtada funksiya birinchi tur uzilishga ega. Funksiyaning sakrashi 1|43|12 AA (4-shakl).

12-masala. Funksiyani berilgan nuqtalarda uzluksizlikka tekshiring

.3,4;5)( 2143

xxxf x Yechish. 41 x nuqtada:

,05lim)04( 43

04

x

xf .5lim)04( 4

3

04

x

xf

Demak, 41 x nuqtada funksiya ikkinchi tur uzilishga ega. 32 x nuqtada:

1255lim)03( 43

03

x

xf , ,1255lim)03( 4

3

03

x

xf

.1255)3( 433

f Demak, 32 x nuqtada funksiya uzluksiz.

NAZORAT ISHINI BAJARISH UCHUN TOPSHIRIQLAR

1-masala. 1. Berilgan determinantni hisoblang: a) i satr elementlari bo‘yicha

yoyib; b) j ustun elementlari bo‘yicha yoyib; c) j ustundagi bittadan boshqa elementlarni nolga aylantirib va shu ustun elementlari bo‘yicha yoyib.

1.1. .2,1,

1234214334124321

ji 1.2. .2,3,

0232013232213211

ji

1.3. .4,3,

0443121111233022

ji 1.4. .2,2,

2114331110221106

ji

20

1.5. .1,3,

2104312111233011

ji 1.6. .4,2,

1111021412112405

ji

1.8. .4,1,

2023173540233281

ji 1.9. .4,2,

3413120323241432

ji

1.10. .3,4,

3431221031241140

ji 1.11. .2,4,

12384511032847120

ji

1.12. .4,3,

4923641220231735

ji 1.13. .2,1,

2114214332205114

ji

1.14. .3,2,

2321101221430212

ji 1.15. .1,3,

3321015432112023

ji

1.16. .3,1,

5214111142143213

ji 1.17. .2,3,

1231312216050213

ji

1.18. 4,2,

4215122101422353

ji 1.19. 2,1,

4310320105345023

ji

21

1.20. .3,2,

45036242147541026

ji 1.21. .3,4,

1213412260321421

ji

1.22. 1,4,

1532460152630211

ji . 1.23. 3,3,

6024312009363102

ji .

1.24. .4,4,

3102421311324021

ji 1.25. .2,3,

4405121311210214

ji

1.25. .3,2,

1105214034121234

ji

2-masala. BA, matritsalar va , sonlari berilgan. BA , AB , 1A matritsalarni toping.

2.1. ,501423245

A ,

221173545

B ,1 4 .

2.2. ,574153013

A ,

203581201

B ,3 5 .

2.3. ,014507485

A ,312121551

B ,5 1 .

2.4. ,014507485

A ,312121551

B ,3 1 .

22

2.5. ,353421121

A ,

321135157

B ,1 3 .

2.6. ,121201213

A ,

173112210

B ,1 .1

2.7. ,122013376

A ,

734214502

B ,1 3 .

2.8. ,221413432

A ,

291260133

B ,2 2 .

2.9. ,210351243

A ,

214231441

B ,5 1 .

2.10. ,173232201

A ,

231713103

B ,1 4 .

2.11. ,230494371

A ,

254291256

B ,3 2 .

2.12. ,110231162

A ,

323504234

B ,1 2 .

2.13. ,7110111496

A ,

250343111

B ,5 2 .

2.14. ,812713301

A ,

465103453

B ,5 2 .

23

2.15.

148131215

A , ,061217553

B ,2 2 .

2.16. ,434603521

A ,

121332111

B ,1 2 .

2. 17. ,243678312

A ,121453212

B .2,1

2.18. ,722234013

A ,

722234013

B ,2 5 .

2.19. ,332154043

A ,

112620171

B ,1 3 .

2.20. ,105321343

A ,

211145022

B ,4 5 .

2.21. ,113342653

A ,

354013582

B ,3 2 .

2.22. ,544133012

A ,

202361203

B ,2 .3

2.23. ,172394412

A ,

147465400

B ,5 1 .

2.24. ,011351158

A ,

210123674

B 1 , 2 .

24

2.25. ,101112112

A

321642063

B , ,3 5 .

3-masala. Tenglamalar sistemalarini birgalikda bo‘lish-bo‘lmasligini Kroniker-Kapelli teoremasi bilan tekshiring. Birgalikda bo‘lgan sistemani Kramer formulalari orqali, matritsalar va Gauss usullari bilan yeching.

3.1.

.632,723

,123

321

321

321

xxxxxxxxx

3.2.

.344,42,322

321

321

321

xxxxxxxxx

3.3.

.2324,4235,623

321

321

321

xxxxxxxxx

3.4.

.1642,825,113

321

321

321

xxxxxxxxx

3.5.

.532,72,13

321

321

321

xxxxxxxxx

3.6.

.132,1234

,82

321

321

321

xxxxxxxxx

3.7.

.453,43,232

321

321

321

xxxxxxxxx

3.8.

.1223,52,2324

321

321

321

xxxxxxxxx

3.9.

.23,701325,2752

31

321

321

xxxxxxxx

3.10.

.4,15638

,634

321

321

321

xxxxxxxxx

3.11.

.443,754

,03

321

321

321

xxxxxx

xxx 3.12.

.343,1223,132

321

321

321

xxxxxxxxx

3.13.

.1653,132

,174

31

321

21

xxxxx

xx 3.14.

.1542,67,175

321

31

321

xxxxxxxx

3.15.

.554,923,623

321

321

321

xxxxxxxxx

3.16.

.2165,484,1132

21

31

321

xxxxxxx

25

3.17.

.1252,363,233

31

21

321

xxxx

xxx 3.18.

.63,1154

,742

321

321

321

xxxxxxxxx

3.19.

.234,32

,153

321

321

321

xxxxxxxxx

3.20.

.1672,232,725

31

321

321

xxxxxxxx

3.21.

.15243,205,932

321

321

321

xxxxxxxxx

3.22.

.573,3862

,104

31

21

321

xxxx

xxx

3.23.

.642,325,123

321

321

321

xxxxxxxxx

3.24.

.164,1053

,3432

321

321

321

xxxxxxxxx

3.25.

.12432,74,3357

321

31

321

xxxxx

xxx

4-masala. Bir jinsli tenglamalar sistemasini yeching.

4.1.

.044,025,032

321

32

321

xxxxxxxx

4.2.

.0742,033,024

321

321

321

xxxxxxxxx

4.3.

.032,05112,052

321

321

321

xxxxxxxxx

4. 4.

.055,023,035

321

321

321

xxxxxxxxx

4.5.

.02,025,034

321

321

321

xxxxxxxxx

4. 6.

.01010,0232,045

321

321

321

xxxxxxxxx

4. 7.

.034,0323,032

321

321

321

xxxxxxxxx

4. 8.

.0582,032,024

321

321

321

xxxxxxxxx

26

4. 9.

.025,02,034

321

321

321

xxxxxxxxx

4. 10.

.047,025,032

321

321

321

xxxxxxxxx

4. 11.

.0432,024,023

321

321

321

xxxxxxxxx

4. 12.

.0134,032,053

321

321

321

xxxxxxxxx

4. 13.

.0714,023,0362

321

321

321

xxxxxxxxx

4. 14.

.0232,0597,023

321

321

321

xxxxxxxxx

4. 15.

.082,053,032

321

321

321

xxxxxxxxx

4 .16.

.05,053,023

321

321

321

xxxxxxxxx

4. 17.

.01411,0433,0542

321

321

321

xxxxxxxxx

4. 18.

.038,023,045

321

321

321

xxxxxxxxx

4. 19.

.0332,043,025

321

321

321

xxxxxxxxx

4. 20.

.01433,0944,0455

321

321

321

xxxxxxxxx

4. 21.

.0,034,023

321

31

321

xxxxxxxx

4. 22.

.07,02,043

21

321

321

xxxxxxxx

4.23.

.034,05112,053

321

321

321

xxxxxx

xxx 4. 24.

.0,02,032

321

321

31

xxxxxxxx

4. 25.

.06102,032,034

321

321

321

xxxxxxxxx

27

5-masala. ABC uchburchak uchlarining koordinatalari berilgan: a) C uchdan tushirilgan balandlik tenglamasini tuzing va uning uzunligini toping; b) B uchdan o‘tkazilgan mediana tenglamasini tuzing va uchburchak medianalarining kesishish nuqtalarini toping; c) A burchakning radian qiymatini hisoblang va uning bissektrisasi tenglamasini tuzing.

5.1. ),2;1(A ),8;9(B ).14;6(C 5.2. ),3;2( A ),9;3(B ).0;6(C

5.3 ),2;1( A ),4;7(B ).10;4(C 5.4. ),1;1( A ),5;9(B ).11;6(C

5.5. ),4;1( A ),8;4(B ).1;5( C 5.6. ),1;1(A ),7;7(B ).13;4(C

5.7. ),2;5( A ),2;8(B ).3;7(C 5.8. ),4;2( A ),1;14(B ).1;2( C

5.9. ),0;6(A ),4;9(B ).5;6(C 5.10. ),2;8(A ),7;4(B ).10;14(C

5.11. ),6;1( A ),6;6(B ).3;3( C 5.12. ),1;4( A ),5;7( B ).4;8(C

5.13. ),0;12(A ),5;0(B ).8;18(C 5.14. ),2;1( A ),3;11(B ).6;7(C

5.15. ),4;3(A ),9;15(B ).7;1(C 5.16. ),2;1(A ),8;7(B ).14;4(C

5.17. ),1;1(A ),7;9(B ).13;6(C 5.18. ),6;14( A ),1;26( B ).2;20(C

5.19. ),1;2( A ),5;10(B ).11;7(C 5.20. ),3;5( A ),2;17(B ).0;1(C

5.21. ),1;2(A ),7;6(B ).13;3(C 5.22. ),1;2( A ),4;10(B ).7;8(C

5.23. ),1;1( A ),5;7(B ).11;4(C 5.24. ),6;2( A ),1;10( B ).3;6( C

5.25. ),7;3( A ),5;2(B ).4;7( C

6-masala. 1. (1-11) Har bir );( yxM nuqtasidan berilgan );( 11 yxA va );( 22 yxB nuqtalargacha bo‘lgan masofalar nisbati a ga teng bo‘lgan chiziq tenglamasini tuzing. 2. (11-25) Har bir );( yxM nuqtasidan berilgan );( 11 yxA nuqtagacha va bx to‘g‘ri chiziqqacha bo‘lgan masofalar nisbati m ga teng bo‘lgan chiziq tenglamasini tuzing.

6.1. ),1;4(A )1;2( B , .4a 6.2. ),7;5(A )1;2(B , .4a

6.3. ),3;3(A )1;5(B , .31

a 6.4. ),4;2( A )5;3(B , .32

a

6.5. ),6;1(A )2;4( B , .2a 6.6. ),2;3( A )6;4(B , .53

a

6.7. ),0;6(A )3;0( B , .2a 6.8. ),0;4(A )0;0(B , .3a

6.9. ),2;4( A )6;1(B , .2a 6.10. ),1;2(A )2;2(B , .4a

28

6.11. ),3;3(A )1;5(B , .3a 6.12. ),1;6(A 5x , .31

m

6.13. ),2;1(A 9x , .41m 6.14. ),0;1(A 8x , .

51

m

6.15. ),5;0(A 3x , .21m 6.16. ),1;2(A 5x , .3m

6.17. ),4;3(A 3x , .3m 6.18. ),0;2(A 25x , .

54

m

6.19. ),0;2(A 58

x , 45

m . 6.20. ),0;1(A 4x , .21m

6.21. ),0;4(A 2x , .21m 6.22. ),0;3(A

29x , .

32

m

6.23. ),3;1(A 6x , .21m 6.24. ),5;1(A 1x , .

41m

6.25. ),0;3(A )0;6(B , .21a

7-masala. ABCD piramidaning uchlari berilgan: a) AB qirra tenglamasini tuzing; b) ABC yoq tenglamasini tuzing; c) D uchdan ABC yoqqa tushirilgan balandlik tenglamasini tuzing va uning uzunligini toping; d) C uchdan o‘tuvchi AB qirraga parallel to‘g‘ri chiziq tenglamasini tuzing; e) D uchdan o‘tuvchi AB qirraga perpendikular tekislik tenglamasini tuzing; f) AD qirra bilan ABC yoq orasidagi burchak sinusini toping; g) ABC va ABD yoqlar orasidagi burchak kosinusini toping.

7.1. ),3;5;3(A ),4;7;8(B ),4;10;5(C ).8;7;4(D

7.2. ),5;6;6(A ),5;9;4(B ),11;6;4(C ).3;9;6(D

7.3. ),2;2;3(A ),2;3;5( B ),1;3;5( C ).7;3;2( D

7.4. ),5;4;0(A ),1;2;3( B ),6;5;4(C ).2;3;3( D

7.5. ),3;1;1( A ),8;5;6(B ),8;5;3(C ).1;4;8(D

7.6. ),7;2;1( A ),10;2;4(B ),5;3;2( C ).7;3;5(D

7.7. ),7;2;4(A ),0;2;1(B ),7;5;3(C ).5;3;2( D

7.8. ),5;3;2(A ),7;3;5( B ),7;2;1(C ).0;2;5(D

7.9. ),7;3;5(A ),5;3;2(B ),7;2;4(C ).7;2;1( D

29

7.10. ),4;1;3(A ),1;6;1(B ),6;1;1(C ).1;4;0( D

7.11. ),2;1;3( A ),1;0;1(B ),3;7;1(C ).8;5;9(D

7.12. ),4;5;3(A ),4;8;5(B ),2;2;1( C ).2;3;1(D

7.13. ),3;4;2(A ),5;1;1(B ),3;9;4(C ).7;6;3(D

7.14. ),5;5;9(A ),1;7;3(B ),8;7;5(C ).2;0;6(D

7.15. ),6;9;2(A ),2;8;2(B ),6;8;9(C ).3;9;7(D

7.16. ),1;5;2( A 14

52

3

zyx .

7.17. ),6;8;1(A ),2;2;5(B ),6;7;5(C ).1;8;4( D

7.18. ),1;7;0(A ),5;1;2( B ),3;6;1(C ).8;9;3( D

7.19. ),4;5;5(A ),4;1;1( B ),1;5;3(C ).3;8;5( D

7.20. ),1;1;6(A ),6;6;1(B ),0;2;4(C ).6;2;1(D

7.21. ),3;5;7(A ),4;4;9(B ),7;5;4(C ).6;9;7(D

7.22. ),2;8;6(A ),7;4;5(B ),2;8;2(C ).7;3;7(D

7.23. ),5;2;4(A ),1;6;0(B ),7;2;0(C ).0;4;1(D

7.24. ),9;4;4(A ),3;10;7(B ),4;8;2(C ).9;6;9(D

7.25. ),5;6;4(A ),4;9;6(B ),5;3;2(C ).9;5;7(D

8-masala. Sirt turini aniqlang va shaklini chizing.

8.1. a) 035 222 zyx ; b) 42 22 yz .

8.2. a) ;064 22 yzx b) 1234 22 zx .

8.3. a) ;03248 222 zyx b) 623 22 zy .

8.4. a) ;0301056 222 zyx b) 645 22 zx .

8.5. a) ;362 22 zyx b) 1863 22 zx .

8.6. a) ;030532 222 zyx b) 623 2 xz .

8.7. a) ;01246 222 zyx b) .632 22 zx

8.8. a) ;093 22 xyz b) 1553 22 zx .

8.9. a) ;34 22 xzy b) 44 22 zx .

30

8.10. a) 0453 22 zyx ; b) 2045 22 zx .

8.11. a) 0724129 222 zyx ; b) 1234 22 yx .

8.12. a) 0915910 222 zyx ; b) zzy 22 2 .

8.13. a) 018236 222 xyz ; b) 2054 22 zy .

8.14. a) ;02793 222 zyx b) .104 22 zx

8.15. a) ;024 22 yzx b) 32 xy .

8.16. a) ;362 22 xzy b) .42 xz

8.17. a) ;0243124 222 zyx b) 303 22 zx .

8.18. a) ;0542 222 zyx b) .3557 22 zx

8.19. a) ;042627 222 zyx b) 44 22 zx .

8.20. a) 03694 222 zyx ; b) 1232 2 xy .

8.21. a) 020544 222 zyx ; b) 3649 22 yx .

8.22. a) 030655 222 zyx ; b) 34 22 yz .

8.23. a) 024234 222 zyx ; b) yyx 222 .

8.24. a) ;03228 222 zyx b) zzx 12632 22 .

8.25. a) 010522 222 zyx ; b) 12 22 zxx .

9-masala. Sonli ketma-ketliklarning limitini toping.

9.1. a) ;652 nnnxn b) .)!4()!3()!2(

n

nnxn

9.3. a) ;6262 22 nnnnxn b) .22)12(531

2

nn

nxn

9.4. a) nnxn 28533 ; b) ).321(12 nn

xn

9.5. a) ;23 44 nnxn b) .52642 n

nnxn

9.6. a) ;)1( nnnxn b) .632

3613

65

n

nn

nx

31

9.7. a) )285(3 3 nnnxn ; b) .321

3 6 nnnxn

9.8. a) ;33 3 nnxn b) .5)32(27452

nnnxn

9.9. a) );23( nnnxn b) .!)!1(!

nnnxn

9.10. a) );34(2 nnnxn b) .)1(1

321

211

nnxn

9.11. a) ;3 32 nnnxn b) .3232

11

nnnn

nx

9.12. a) ;)3)(2( nnnxn b) .)13)(23(1

741

411

nnxn

9.13. a) ;32)2( 2 nnnnxn b) .)52)(12(1

931

711

nnxn

9.14. a) ;)3)(2( nnnnnxn b) .18321

2

n

nxn

9.15. a) ;)5(8 25 nnnnxn b) .

21

21

211

31

31

311

2

2

n

n

nx

9.16. a) );35( 3 33 32 nnnxn b) .22)12(4321

2nnnxn

9.17. a) ;)2()2( 3 23 2 nnxn b) .)35(127223 2

n

nnxn

9.18. a) ;1242 nnnxn b) .)12(5313 23 3

n

nnxn

9.19. a) ;12 22 nnnnxn b) .)12)(12(1

531

311

nnxn

9.20. a) ;32 44 nnxn b) .)1(!3)!13()!13(

nn

nnxn

9.21. a) ;4 22 nnnxn b) .32)13(852

313

nnnxn

32

9.22. a) ;13 224 nnnxn b) .1)23(741

24

nn

nxn

9.23. a) );)1(( 33 23 nnnnxn b) .1)32(753

2

nn

nxn

9.24. a) );12(8 3 23 33 nnnxn b) .2525

1 nn

nn

nx

9.25. a) ;832 3 3nnxn b) .1025

1000133

10029

107

n

nn

nx

10-masala. Limitlarni toping:

10.1. .35324563lim 2

2

5

xxxx

x .

1122lim

2

2

0

xx

x

.5

coscoslim 22

0 xxx

x

.3212lim

12

x

x xx

10.2. .5423lim 2

3

1

xxxx

x .

423143lim

2

1

xxxx

x

.4

5cos1lim 20 xx

x

.95lim

4x

x xx

10.3. .181142lim 2

3

2

xxxx

x .

8314lim

32

xx

x

.5sin3

4lim0 x

xtgx

.3532lim

2x

x xx

10.4. 1223lim 4

24

1

xxxx

x. .

273lim33

x

xxx

.3

2sin2lim 20 xxxtg

x

.75lim

32

x

x xx

10.5. .483274lim 2

2

2

xxxx

x .

5162lim

4

xx

x

.sin

4cos1lim0 xx

xx

.1525lim

x

x xx

10.6. .12

4133lim 22

4

xxxx

x .

2134lim

5

xx

x

33

.3cos1coscoslim

3

0 xxx

x

.3lim

32

2

2

x

x xx

10.7. .22

54lim 2324

1

xxxxx

x .

3223lim

7

xx

x

.4cos18cos1lim

0 xx

x

.

4313lim

14

x

x xx

10.8. .23

168lim 2324

1

xxxxx

x .

2529lim

38 xx

x

.cos1lim2

0 tgxxx

x

.

2325lim

3

x

x xx

10.9. .103

43lim 223

2

xxxx

x .

9102372lim

29

xxxx

x

.1sin

1lim0

tgxxx .

1414lim

3x

x xx

10.10. .124

123lim 233

1

xxxxx

x .

134353lim

21

xxxx

x

.2cos1

1sin1lim2

0 xx

x

.

2321lim

x

x xx

10.11. .125315112lim 2

2

3

xxxx

x .

626lim

2

2

xxxx

x

.2

)1(lim1

xtgxx

.285lim

4

x

x xx

10.12. .434023lim 2

2

4

xxxx

x .

410212lim

2

3 xxxx

x

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sin1lim2

xx

x

.

1412lim

12

x

x xx

10.13. 35192

145lim 22

7

xxxx

x. .

64204lim

34

xx

x

.4

2sin1lim4

xx

x .)54(lim 1

3

1

2

x

x

xx

34

10.14. .1

2lim 32

1

xxx

x .

547173lim

25

xxxx

x

.3

2lim0 xtg

xarctgx

.)34(lim 11

2

x

x

xx

10.15. .372673lim

2

2

3

xxxx

x .

1111lim

330 xxxx

x

.arcsin

3sinsinlim0 x

xxx

].ln)2)[ln(32(lim xxxx

10.16. .1092

8lim 23

2

xxx

x .

743548lim

21

xxxx

x

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sin2sinlim 222

0 xxx

x

.)32(lim 23

2

x

x

xx

10.17. 1

2lim 223

1

xxxxx

x. .

2912823lim

2

2 xxxx

x

.2cos1lim2

0 arctgxxx

x

)].32ln()31)[ln(12(lim xxx

x

10.18. 103

4lim 22

2

xxx

x. .

22312lim

4

xx

x

.1

sincoslim4

tgxxx

x

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13

x

x xx

10.19. .122036112lim 2

2

6

xxxx

x .

158612lim

25

xxxx

x

.arcsincos1lim

2

0 xxx

x

.

131lim

12

x

x xx

10.20. .27

6lim 32

3

xxx

x .

2795lim

38 xx

x

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2xx

x

.

534lim

6x

x xx

10.21 .132347lim 2

2

1

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x .

657132lim

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xxxx

x

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0 xarctgxtgx

x

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11

1

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35

10.22. .1

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1

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4

16 xx

x

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lim 222

xx

x

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1

x

x

xx

10.23. .352

27lim 23

3

xxx

x .

548143lim

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1 xxxx

x

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sincoslim4

ctgxxx

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2

2

x

x

xx

10.24. .3423lim 2

3

1

xxxx

x .

211132lim

2

2 xxxx

x

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2sinlim0 xx

xx

)].12ln()12)[ln(13(lim

xxxx

10.25. .1

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1

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410932lim

2

3 xxxx

x

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0 xxxx

x

.)94(lim 2

5

2

xx

xx

11-masala. Funksiyani uzluksizlikka tekshiring va grafigini chizing.

11.1.

.2,2,20,0,0,1

)(xxxxx

xf 11.2.

.3,7,30,1,0,3

)(xxxxxx

xf

11.3.

.1,3,11,2,1,4

)( 2

xxxx

xxxf 11.4.

.2,2,20,0,0,

)(

2

xxxxx

xf

11.5.

.3,1,31,,1),1(2

)( 2

xxxx

xxxf 11.6.

.1,1,10,,0,

)( 3

xxxxxx

xf

11.7.

.1,1,12,1,2,

)(2 xx

xxxx

xf 11.8.

.,1,0,cos,0,1

)(

xxxxx

xf

11.9.

.2,2,20,4,0,3

)( 2

xxxxxx

xf 11.10.

.,3,0,sin,0,1

)(

xxxxx

xf

36

11.11.

.2,2,20,)1(

,0,)( 2

xxxx

xxxf 11.12.

.3,3,31,2,1,1

)(

2

xxxxxx

xf

11.13.

.3,5,31,1,1,

)(

3

xxxx

xxxf 11.14.

.,2

,2

,0

,2

,cos

)(

xx

x

xx

xf

11.15.

.1,3

,11,1,1,2

)( 2

xxxx

xxxf 11.16.

.3,1,30,)1(,0,

)( 23

xxxxxx

xf

11.17.

.3,1,31,,1),1(2

)(2

2

xxxx

xxxf 11.18.

.1,1,10,,0,

)( 2

xxxxxx

xf

11.19.

.1,1,12,1,2,

)(2 xx

xxxx

xf 11.20.

.,1,0,sin,0,1

)(

xxxxx

xf

11.21.

.2,2,20,4,0,13

)( 2

xxxxxx

xf 11.22.

.,4,0,cos,0,1

)(

xxxxx

xf

11.23.

.1,4,11,2,1,21

)( 2

xxxx

xxxf 11.24.

.2,2,20,0,0,3

)(2 xxxx

xxx

xf

11.25.

.3,82

,31,1,1,

)(

2

xxxx

xxxf

12-masala. Funksiyani berilgan nuqtalarda uzluksizlikka tekshiring

12.1. .2,3;25)( 21

xx

xxxf 12.2. .5,4;2)( 214

1

xxxf x

37

12.3. .5,3;5

4)( 21 xx

xxxf 12.4. .2,1;3)( 212

2

xxxf x

12.5. .2,1;1

2)( 212 xx

xxxf 12.6. .4,2;7)( 213

4

xxxf x

12.7. .2,1;4)( 211 xxxf xx

12.8. .3,2;4

3)( 212 xx

xxxf

12.9. .4,3;5)( 2131

xxxf x 12.10. .3,2;6)( 2131

xxxf x

12.11. .3,2;25)( 21

xx

xxxf 12.12. .2,3;8)( 212

4

xxxf x

12.13. .1,2;8

)( 213 xx

xxxf 12.14. .3,4;5)( 214

3

xxxf x

12.15. .5,3;5

3)( 21 xx

xxxf 12.16. .2,1;6)( 211

2

xxxf x

12.17. .2,1;4

2)( 212 xx

xxxf 12.18. .5,2;5)( 213

4

xxxf x

12.19. .2,1;4)( 211 xxxf xx

12.20. .3,2;4

3)( 212 xx

xxxf

12.21. .4,3;4)( 2141

xxxf x 12.22. .3,5;5)( 2151

xxxf x

12.23. .3,1;15)( 212

xx

xxxf 12.24. .2,4;7)( 212

4

xxxf x

12.25. .2,3;21)( 21

2

xxxxxf