so‘z boshitaqi.uz/pdf/sirtqi/oliy_matematika.pdf · 2018. 9. 28. · 1- mavzu. chiziqli algebra...
TRANSCRIPT
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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;
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- 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.
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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.
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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(
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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
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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
~
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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
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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
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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
.2
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
.3
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
.21lim
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
.2/sin1lim 2
2xx
x
.
534lim
6x
x xx
10.21 .132347lim 2
2
1
xxxx
x .
657132lim
26
xxxx
x
.24lim
0 xarctgxtgx
x
.113lim
11
1
x
x xx
-
35
10.22. .1
154lim 224
1
xxx
x .
42lim
4
16 xx
x
.cos1
lim 222
xx
x
.)32(lim 13
1
x
x
xx
10.23. .352
27lim 23
3
xxx
x .
548143lim
2
1 xxxx
x
.1
sincoslim4
ctgxxx
x
.)53(lim 42
2
2
x
x
xx
10.24. .3423lim 2
3
1
xxxx
x .
211132lim
2
2 xxxx
x
.sin3sin
2sinlim0 xx
xx
)].12ln()12)[ln(13(lim
xxxx
10.25. .1
1lim 244
1
xxxx
x .
410932lim
2
3 xxxx
x
.2sin
coscoslim5
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