3603 pavyzdys 1 x y x y 2 x y x y 3 4 1 2 1 2 1 2 · taikomoji matematika. pvz. antram testui...

TAIKOMOJI MATEMATIKA. Pvz. Antram testui serija3603

variantasPAVYZDYS

Funkcija g(x, y) apibrėžta formule g(x, y) = 2x4y4 + 3 cos (5x− 5y).

1 ∂g

∂x=

1© 8x5y4+15 cos (5x−5y); 2© 8x3y4+15 sin (5x−5y);3© 8x5y4−15 cos (5x−5y); 4© 8x3y4−3 sin (5x−5y);5© 8x3y4−15 sin (5x−5y); 6© 8x3y4+3 sin (5x−5y).

2 ∂g

∂y=

1© 8x4y5−15 cos (5x−5y); 2© 8x4y3+3 sin (5x−5y);3© 8x4y3+15 sin (5x−5y); 4© 8x4y3−3 sin (5x−5y);5© 8x4y3−15 sin (5x−5y); 6© 8x4y5+15 cos (5x−5y).

w(x) =43x2 − 50

44x

3 w′(−4) = 1© 2932 ; 2© − 369

88 ; 3© − 369352 ; 4© 43

44 ; 5© 0; 6© − 2932 ; 7© 369

352 ; 8© − 4344 .

4 w′′(−2) = 1© 0; 2© − 25176 ; 3© 25

44 ; 4© − 2511 ; 5© 25

88 ; 6© − 2544 ; 7© − 25

88 ; 8© 25176 .

5 limx→0

x7 lnx =1© ∞; 2© riba neegzistuoja;3© ln 7; 4© 0;5© 7; 6© 1.

6 limx→∞

x15

9x =1© 5

3 ; 2© 15;3© riba neegzistuoja; 4© 0;5© 9; 6© 1

9 .

Funkcija f(x) apibrėžta formule f(x) = 4x5 − 4x.

7 Raskite funkcijos f(x) išvestinę f ′(x).1© f ′(x)=20x5−4; 2© f ′(x)=20x4−4;3© f ′(x)=16x6−4; 4© f ′(x)=16x5−4;5© f ′(x)=20x6−4; 6© f ′(x)=16x4−4.

8 f ′(2) = 1© 1276; 2© 1020; 3© 252; 4© 316; 5© 508; 6© 636.

9 Užrašykite funkcijos f(x) liestinės,einančios per tašką A

(2, f(2)

), lygtį.

1© y=−316x−512; 2© y=−120x−316;3© y=316x−512; 4© y=−316x+752;5© y=316x+752; 6© y=120x+316.

Tarkime, kad z(x) = 3√1 + 20x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© z(x)≈1− 20

3 x; 2© z(x)≈− 203 x;

3© z(x)≈1+20x; 4© z(x)≈ 203 x;

5© z(x)≈1− x3 ; 6© z(x)≈1+ x

3 ;7© z(x)≈1−20x; 8© z(x)≈1+ 20

3 x.

11 z(− 1

9

)≈ 1© 8

9 ; 2© 2827 ; 3© 47

27 ; 4© 727 ; 5© 10

9 ; 6© − 2027 ; 7© 20

27 ; 8© 2627 .

f(x) = 2x3 − 18x2 − 42x+ 98

12 maxx∈[−4,3]

f(x) = 1© −155; 2© 137; 3© −150; 4© 120; 5© 131; 6© −392; 7© −136.

13 minx∈[−4,3]

f(x) = 1© 131; 2© −155; 3© −392; 4© 120; 5© 137; 6© −136; 7© −150.

14 maxx∈[−4,3]

|f(x)| = 1© 392; 2© 150; 3© 120; 4© 155; 5© 137; 6© 131; 7© 51; 8© 136.

f(x) = 24 sin(5x)− 53x2

15 f ′′(0) =1© −106; 2© 53; 3© 173; 4© 120; 5© −173; 6© −53; 7© 106; 8© 0; 9© −120.

16 f ′′′(0) =1© 3000; 2© 106; 3© −706; 4© −3000; 5© −106; 6© 0; 7© 600; 8© 706; 9© −600.

Funkcijos y =−6x2 + 15x

−13x− 29grafiko asimptotė, kai x→∞, yra tiesė y = qx+ r.

17 q = 1© − 613 ; 2© 29

15 ; 3© 136 ; 4© 15

29 ; 5© 613 ; 6© − 29

15 ; 7© − 136 ; 8© − 15

29 .

18 r = 1© 4325 ; 2© 369

169 ; 3© 2543 ; 4© − 169

369 ; 5© 169369 ; 6© − 43

25 ; 7© − 2543 ; 8© − 369

169 .

19Nustatykite funkcijosy = 5e−10x

2

grafiko iškilumoaukštyn sritį.

1©(−∞;− 1

2√

5

)∪(

12√

5;+∞

); 2© {0}; 3©

(− 1

2√

5; 12√

5

);

4© (−2√5;0); 5© (−∞;+∞); 6© (−∞;0);

7© (0;+∞); 8© (−∞;0)∪(0;+∞); 9© (−2√5;2√5);

0© (0;2√5).

20 (sin4

(3x9))′

=

1© 67x10 sin5(3x9) cos(3x9); 2© 108x8 sin3(3x9) cos(3x9);3© 67x8 sin3(3x9) cos(3x9); 4© −67x8 sin3(3x9) cos(3x9);5© −108x10 sin5(3x9) cos(3x9); 6© −108x8 sin3(3x9) cos(3x9);7© −67x10 sin5(3x9) cos(3x9); 8© 108x10 sin5(3x9) cos(3x9).

21(

7√6 + ln (x+ 8)

)′=

1© 7(x+8)7√

(6+ln(x+8))6; 2© 1

7√

(6+ln(x+8))6;

3© 1

7 7√

6+ln(x+8); 4© 1

7(x+8) 7√

6+ln(x+8);

5© 7√

(6+ln(x+8))6

7(x+8); 6© 7

7√

6+ln(x+8);

7© 1

7(x+8)7√

(6+ln(x+8))6; 8© 1

7√

6+ln(x+8).

22 s q

b dp c

· ( 1 0 10 1 0

)=

1©

s q sp b pc p c

; 2©

s c sp q ps p c

; 3©

s q sb d bp c p

; 4©

q s qd b dc p c

.

23 Jei rc− pb = 1, tai(

r pb c

)−1=

1©(

c pb r

); 2©

(−r −p−b −c

); 3©

(c −b−p r

); 4©

(c bp r

); 5©

(c −p−b r

).

Išspręskite tiesinių lygčių sistemą{−17z1 −8z2 = 114

8z1 −11z2 = −157 .

24 z1 = 1© −6; 2© 9; 3© 7; 4© −4; 5© −10; 6© 0.

25 z2 = 1© 6; 2© 7; 3© 4; 4© 9; 5© −4; 6© −8.

Išspręskite tiesinių lygčių sistemą

x1 +12x2 +6x3 = 882x1 −3x2 −7x3 = −62−2x1 +13x2 +16x3 = 158

.

26 x1 = 1© 3; 2© 1; 3© −7; 4© 7; 5© 8; 6© −8.

27 x2 = 1© 9; 2© −8; 3© 1; 4© 6; 5© −9; 6© 10.

28 x3 = 1© −2; 2© −10; 3© −3; 4© 4; 5© 6; 6© 2.

29 Apskaičiuokite

∣∣∣∣∣∣−4 −2 54 −1 −3−5 −1 −4

∣∣∣∣∣∣.1© −199; 2© 7; 3© 132; 4© 61; 5© 9; 6© −111.

30 Matricos

y u vd c ft w p

adjunktas A32 =

1© yp− ft ;2© vd− yf ;3© yf − vd ;4© yp− fu .

31 Matricos(−73 −72−49 −12

)adjunktas A12 =

1© −49 ; 2© −72 ; 3© 72 ; 4© 49 ; 5© −73 ; 6© −12 ; 7© 12 ; 8© 73 .


2x1 − 4x2 − 5x3 = 2

−3x1 − 2x2 + 4x3 = 1

2x1 − 2x2 − 2x3 = −5Kramerio metodu.

Pažymėkime:

A – sistemos matrica; D = detA;

x1

x2

x3

=

(D1

D

D2

D

D3

D

)T

– sistemos sprendinys.

32 D = 1© 45 ; 2© −15 ; 3© 34 ; 4© −34 ; 5© −45 .

33 D1 = 1© −97 ; 2© 0 ; 3© 156 ; 4© −39 ; 5© −80 .

34 D2 = 1© 24 ; 2© −25 ; 3© 81 ; 4© −12 ; 5© −1 .

35 D3 = 1© 94 ; 2© 96 ; 3© 73 ; 4© 47 ; 5© −18 .

36 x1 + x2 + x3 = 1© 934 ; 2© − 24

17 ; 3© − 4017 ; 4© − 21

17 ; 5© − 22734 .

37 Matricos A =

−4 −2 −24 −4 −13 5 −2

determinantas yra lygus

1© −126 ;2© 126 ;3© 250 ;4© −251 ;5© −218 ;6© 90 .

38 Adjunktas A12 = 1© 14 ; 2© 5 ; 3© −9 ; 4© −15 .

39 Adjunktas A31 = 1© −12 ; 2© −6 ; 3© 13 ; 4© 3 .

40 Adjunktas A33 = 1© −8 ; 2© −43 ; 3© −37 ; 4© 24 .

41 Adjunktas A11 = 1© 27 ; 2© 13 ; 3© −5 ; 4© −11 .

42 A−1 = 1©

717

1134

817

117

217

617

417 − 1

34717

; 2©

− 13126 − 5

126 − 1663

19 − 1

9 − 19

121

221 − 4

21

; 3©

− 13126

19

121

− 5126 − 1

9221

− 1663 − 1

9 − 421

.


variantas001

Funkcija u(x, y) apibrėžta formule u(x, y) = 2x4y7 + 3 cos (3x− 4y).

1 ∂u

∂x=

1© 8x3y7−9 sin (3x−4y); 2© 8x5y7+9 cos (3x−4y);3© 8x5y7−9 cos (3x−4y); 4© 8x3y7−3 sin (3x−4y);5© 8x3y7+9 sin (3x−4y); 6© 8x3y7+3 sin (3x−4y).

2 ∂u

∂y=

1© 14x4y8+12 cos (3x−4y); 2© 14x4y6−3 sin (3x−4y);3© 14x4y6+3 sin (3x−4y); 4© 14x4y6−12 sin (3x−4y);5© 14x4y8−12 cos (3x−4y); 6© 14x4y6+12 sin (3x−4y).

h(x) =23x2 − 26

40x

3 h′(−4) = 1© − 171320 ; 2© − 197

320 ; 3© 2340 ; 4© − 197

80 ; 5© 0; 6© 197320 ; 7© 171

320 ; 8© − 2340 .

4 h′′(−5) = 1© 132500 ; 2© − 13

10 ; 3© − 131250 ; 4© − 13

250 ; 5© 13250 ; 6© 13

1250 ; 7© − 132500 ; 8© 0.

5 limx→0

x17 lnx =1© 17; 2© 1;3© riba neegzistuoja; 4© 0;5© ∞; 6© ln 17.

6 limx→∞

x16

21x =1© riba neegzistuoja; 2© 16;3© 1

21 ; 4© 0;5© 21; 6© 16

21 .



8 f ′(2) = 1© 90; 2© 250; 3© 186; 4© 506; 5© 122; 6© 378.


(2, f(2)

), lygtį.

1© y=−122x+296; 2© y=−122x−192;3© y=52x+122; 4© y=122x−192;5© y=122x+296; 6© y=−52x−122.

Tarkime, kad h(x) = 5√1 + 3x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© h(x)≈1+ 3

5x; 2© h(x)≈1−3x;3© h(x)≈− 3

5x; 4© h(x)≈1+3x;5© h(x)≈1− 3

5x; 6© h(x)≈ 35x;

7© h(x)≈1− x5 ; 8© h(x)≈1+ x

5 .

11 h(16

)≈ 1© 11

10 ; 2© 3130 ; 3© 9

10 ; 4© 2930 ; 5© 7

6 ; 6© 110 ; 7© 5

6 ; 8© − 110 .

f(x) = 2x3 − 54x+ 42

12 maxx∈[0,8]

f(x) = 1© 640; 2© −66; 3© 634; 4© 150; 5© −76; 6© 653; 7© 42.

13 minx∈[0,8]

f(x) = 1© 634; 2© 653; 3© 42; 4© −76; 5© 640; 6© 150; 7© −66.

14 maxx∈[0,8]

|f(x)| = 1© 634; 2© 653; 3© 42; 4© 76; 5© 640; 6© 66; 7© 140; 8© 150.

f(x) = 16 sin(4x)− 95x2

15 f ′′(0) =1© −159; 2© 190; 3© 159; 4© 64; 5© −95; 6© −64; 7© −190; 8© 95; 9© 0.

16 f ′′′(0) =1© 1024; 2© −190; 3© 0; 4© −446; 5© −256; 6© −1024; 7© 446; 8© 256; 9© 190.

Funkcijos y =−10x2 − 3x

−4x+ 25grafiko asimptotė, kai x→∞, yra tiesė y = kx+ a.

17 k = 1© 52 ; 2© − 2

5 ; 3© 25 ; 4© − 25

3 ; 5© 253 ; 6© 3

25 ; 7© − 325 ; 8© − 5

2 .

18 a = 1© − 8131 ; 2© 8

131 ; 3© 1140 ; 4© − 131

8 ; 5© 4011 ; 6© − 11

40 ; 7© − 4011 ; 8© 131

8 .

19Nustatykite funkcijosy = −20e−3x2


1© (−√6;0); 2© (−∞;0); 3© (−

√6;√6);

4© {0}; 5©(−∞;− 1√

6

)∪(

1√6;+∞

); 6© (−∞;+∞);

7©(− 1√

6; 1√

6

); 8© (−∞;0)∪(0;+∞); 9© (0;

√6);

0© (0;+∞).

20 (sin7

(9x8))′

=

1© 504x7 sin6(9x8) cos(9x8); 2© −504x9 sin8(9x8) cos(9x8);3© 741x7 sin6(9x8) cos(9x8); 4© 741x9 sin8(9x8) cos(9x8);5© −741x7 sin6(9x8) cos(9x8); 6© 504x9 sin8(9x8) cos(9x8);7© −741x9 sin8(9x8) cos(9x8); 8© −504x7 sin6(9x8) cos(9x8).

21(

10√

7 + ln (x+ 9))′

=

1© 10√

(7+ln(x+9))9

10(x+9); 2© 1

10 10√

7+ln(x+9);

3© 1010√

7+ln(x+9); 4© 10(x+9)

10√

(7+ln(x+9))9;

5© 110√

(7+ln(x+9))9; 6© 1

10(x+9)10√

(7+ln(x+9))9;

7© 110√

7+ln(x+9); 8© 1

10(x+9) 10√

7+ln(x+9).

22 v u

q tb z

· ( 0 1 01 0 1

)=

1©

b z qb u vz b z

; 2©

v z vb u bv b z

; 3©

u v ut q tz b z

; 4©

v u vb q bz b z

.

23 Jei xr − sw = 1, tai(

x sw r

)−1=

1©(

r sw x

); 2©

(r −s

−w x

); 3©

(r ws x

); 4©

(r −w−s x

); 5©

(−x −s−w −r

).

Išspręskite tiesinių lygčių sistemą{

15z1 −6z2 = −81−8z1 +7z2 = 28

.

24 z1 = 1© 9; 2© −7; 3© −3; 4© −4; 5© −10; 6© −1.

25 z2 = 1© −3; 2© 5; 3© −4; 4© 7; 5© −7; 6© 6.


−t1 +10t2 −12t3 = −1510t1 +15t2 +7t3 = 47−16t1 +13t2 +10t3 = 237

.

26 t1 = 1© −10; 2© 8; 3© −1; 4© 1; 5© −7; 6© 2.

27 t2 = 1© 0; 2© 1; 3© −6; 4© −3; 5© 5; 6© 10.

28 t3 = 1© 3; 2© −9; 3© 9; 4© 0; 5© −3; 6© 6.

29 Apskaičiuokite

∣∣∣∣∣∣4 4 −3−6 −4 −3−6 1 −1

∣∣∣∣∣∣.1© −5; 2© 4; 3© 166; 4© −247; 5© −65; 6© −161.

30 Matricos

d q zw a hv b f

adjunktas A12 =

1© df − hv ;2© wf − hv ;3© df − hq ;4© hv − wf .

31 Matricos(−82 −25−70 45

)adjunktas A22 =

1© 70 ; 2© 25 ; 3© −25 ; 4© −82 ; 5© 82 ; 6© −70 ; 7© 45 ; 8© −45 .


−5x1 − 5x2 + 4x3 = −15x1 + 4x2 − 4x3 = 1

2x1 − 5x2 + 2x3 = 1

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© −42 ; 2© 18 ; 3© −22 ; 4© −18 ; 5© 22 .

33 D1 = 1© 67 ; 2© −96 ; 3© 6 ; 4© 18 ; 5© −27 .

34 D2 = 1© 96 ; 2© 54 ; 3© 75 ; 4© 65 ; 5© 0 .

35 D3 = 1© 52 ; 2© −99 ; 3© 51 ; 4© −30 ; 5© 3 .

36 x1 + x2 + x3 = 1© 19 ; 2© 1

2 ; 3© 116 ; 4© − 77

18 ; 5© −4 .

37 Matricos A =

3 4 24 2 −4−3 −1 5


1© −15 ;2© 10 ;3© −10 ;4© 18 ;5© 28 ;6© 17 .

38 Adjunktas A12 = 1© 11 ; 2© −8 ; 3© 7 ; 4© −14 .

39 Adjunktas A33 = 1© −12 ; 2© −25 ; 3© −10 ; 4© 13 .

40 Adjunktas A21 = 1© −20 ; 2© −22 ; 3© −13 ; 4© 40 .

41 Adjunktas A22 = 1© 39 ; 2© 25 ; 3© 21 ; 4© −10 .

42 A−1 = 1©

− 956 − 1

7128

− 128 − 1

7 − 314

− 114

314

114

; 2©

− 35

115 2

45 − 21

10 −2− 1

5910 1

; 3©

− 35

45 − 1

5115 − 21

10910

2 −2 1

.


variantas002

Funkcija u(x, y) apibrėžta formule u(x, y) = 5x3y6 + 3 cos (5x+ 2y).

1 ∂u

∂x=

1© 15x2y6+3 sin (5x+2y); 2© 15x2y6−15 sin (5x+2y);3© 15x2y6−3 sin (5x+2y); 4© 15x4y6−15 cos (5x+2y);5© 15x4y6+15 cos (5x+2y); 6© 15x2y6+15 sin (5x+2y).

2 ∂u

∂y=

1© 30x3y5+6 sin (5x+2y); 2© 30x3y7+6 cos (5x+2y);3© 30x3y5−6 sin (5x+2y); 4© 30x3y5−3 sin (5x+2y);5© 30x3y7−6 cos (5x+2y); 6© 30x3y5+3 sin (5x+2y).

z(x) =27x2 − 2

15x

3 z′(−5) = 1© − 95 ; 2© − 677

75 ; 3© 673375 ; 4© 9

5 ; 5© − 673375 ; 6© 0; 7© 677

375 ; 8© − 677375 .

4 z′′(5) = 1© 0; 2© − 4375 ; 3© 4

375 ; 4© − 415 ; 5© 2

1875 ; 6© − 41875 ; 7© − 2

1875 ; 8© 41875 .

5 limx→0

x13 lnx =1© ∞; 2© 13;3© ln 13; 4© 0;5© 1; 6© riba neegzistuoja.

6 limx→∞

x6

15x =1© 6; 2© 1

15 ;3© 15; 4© 0;5© riba neegzistuoja; 6© 2

5 .



8 f ′(3) = 1© 319; 2© 724; 3© 2182; 4© 2911; 5© 238; 6© 967.


(3, f(3)

), lygtį.

1© y=−228x−319; 2© y=−319x+1185;3© y=319x−729; 4© y=319x+1185;5© y=−319x−729; 6© y=228x+319.

Tarkime, kad v(x) = 3√1 + 7x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© v(x)≈ 7

3x; 2© v(x)≈1+ 73x;

3© v(x)≈1− 73x; 4© v(x)≈1+ x

3 ;5© v(x)≈1−7x; 6© v(x)≈− 7

3x;7© v(x)≈1− x

3 ; 8© v(x)≈1+7x.

11 v(18

)≈ 1© 31

24 ; 2© 724 ; 3© 25

24 ; 4© 78 ; 5© − 7

24 ; 6© 1724 ; 7© 23

24 ; 8© 98 .

f(x) = 2x3 + 6x2 − 144x+ 91

12 maxx∈[−1,5]

f(x) = 1© −229; 2© 739; 3© 251; 4© −277; 5© 255; 6© 239; 7© −261.

13 minx∈[−1,5]

f(x) = 1© −277; 2© 251; 3© 739; 4© 255; 5© −229; 6© 239; 7© −261.

14 maxx∈[−1,5]

|f(x)| = 1© 739; 2© 239; 3© 277; 4© 255; 5© 261; 6© 229; 7© 359; 8© 251.

f(x) = 12 sin(4x)− 61x2

15 f ′′(0) =1© 122; 2© 61; 3© 48; 4© −109; 5© −122; 6© −61; 7© 109; 8© 0; 9© −48.

16 f ′′′(0) =1© −192; 2© −768; 3© 0; 4© −314; 5© 122; 6© 192; 7© 314; 8© 768; 9© −122.


6x+ 7grafiko asimptotė, kai x→∞, yra tiesė y = qx+ a.

17 q = 1© 712 ; 2© − 7

12 ; 3© 12 ; 4© 12

7 ; 5© − 127 ; 6© −2; 7© − 1

2 ; 8© 2.

18 a = 1© 5381 ; 2© − 53

81 ; 3© 8153 ; 4© 3

14 ; 5© − 143 ; 6© − 81

53 ; 7© − 314 ; 8© 14

3 .


grafiko iškilumožemyn sritį.

1© (−∞;− 12 )∪(

12 ;+∞); 2© {0}; 3© (0;2);

4© (−2;2); 5© (−∞;0); 6© (−∞;0)∪(0;+∞);7© (0;+∞); 8© (−2;0); 9© (−∞;+∞);0© (− 1

2 ;12 ).

20 (sin5

(2x3))′

=

1© −30x4 sin6(2x3) cos(2x3); 2© 15x2 sin4(2x3) cos(2x3);3© −15x4 sin6(2x3) cos(2x3); 4© 15x4 sin6(2x3) cos(2x3);5© 30x2 sin4(2x3) cos(2x3); 6© −15x2 sin4(2x3) cos(2x3);7© 30x4 sin6(2x3) cos(2x3); 8© −30x2 sin4(2x3) cos(2x3).

21(

10√

7 + ln (x+ 8))′

=

1© 1

10(x+8) 10√

7+ln(x+8); 2© 10

√(7+ln(x+8))9

10(x+8);

3© 110√

(7+ln(x+8))9; 4© 1

10 10√

7+ln(x+8);

5© 1010√

7+ln(x+8); 6© 1

10√

7+ln(x+8);

7© 1

10(x+8)10√

(7+ln(x+8))9; 8© 10(x+8)

10√

(7+ln(x+8))9.

22 q p

r ta x

· ( 0 1 01 0 1

)=

1©

p q pt r tx a x

; 2©

a x ra p qx a x

; 3©

q p qa r ax a x

; 4©

q x qa p aq a x

.

23 Jei as− yx = 1, tai(

a yx s

)−1=

1©(

s −x−y a

); 2©

(s xy a

); 3©

(−a −y−x −s

); 4©

(s −y−x a

); 5©

(s yx a

).

Išspręskite tiesinių lygčių sistemą{−13z1 −2z2 = 84−6z1 +5z2 = 98

.

24 z1 = 1© −1; 2© −8; 3© −4; 4© 8; 5© 10; 6© −3.

25 z2 = 1© −6; 2© 7; 3© 10; 4© −4; 5© −1; 6© −7.


−x1 +10x2 +6x3 = 79−10x1 +9x2 +13x3 = 147−12x1 +11x2 +6x3 = 139

.

26 x1 = 1© 6; 2© 9; 3© −10; 4© 0; 5© −5; 6© −7.

27 x2 = 1© 0; 2© −2; 3© −5; 4© 5; 5© −4; 6© −6.

28 x3 = 1© 3; 2© 7; 3© 6; 4© 4; 5© −3; 6© 8.

29 Apskaičiuokite

∣∣∣∣∣∣1 −3 15 5 42 6 3

∣∣∣∣∣∣.1© 49; 2© 2; 3© 32; 4© −6; 5© 8; 6© −63.

30 Matricos

t x pb y wf z q

adjunktas A23 =

1© tq − wf ;2© tq − wx ;3© xf − tz ;4© tz − xf .

31 Matricos(

37 −7431 23

)adjunktas A12 =

1© −37 ; 2© 37 ; 3© −31 ; 4© 74 ; 5© 23 ; 6© −74 ; 7© −23 ; 8© 31 .


−4x1 − x2 + 5x3 = 5

x1 + 2x2 − 3x3 = −43x1 + 4x2 + 2x3 = −5

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© 72 ; 2© −9 ; 3© −72 ; 4© −63 ; 5© 63 .

33 D1 = 1© 93 ; 2© 61 ; 3© 27 ; 4© 74 ; 5© −21 .

34 D2 = 1© 29 ; 2© −80 ; 3© −99 ; 4© −31 ; 5© 72 .

35 D3 = 1© −91 ; 2© −27 ; 3© −33 ; 4© −59 ; 5© 47 .

36 x1 + x2 + x3 = 1© 57 ; 2© − 43

63 ; 3© − 2621 ; 4© − 8

7 ; 5© − 6763 .

37 Matricos A =

5 4 −5−2 2 −33 −5 −2


1© −167 ;2© 167 ;3© −159 ;4© 321 ;5© 4 ;6© 218 .

38 Adjunktas A11 = 1© −30 ; 2© −19 ; 3© 8 ; 4© −28 .

39 Adjunktas A13 = 1© 4 ; 2© −13 ; 3© −8 ; 4© 6 .

40 Adjunktas A23 = 1© 36 ; 2© −56 ; 3© 73 ; 4© 37 .

41 Adjunktas A32 = 1© 25 ; 2© 42 ; 3© −41 ; 4© 9 .

42 A−1 = 1©

19167 − 33

1672

16713167 − 5

167 − 25167

− 4167 − 37

167 − 18167

; 2©

23 12 −7−6 −3 213 7 −4

; 3©

19167

13167 − 4

167− 33

167 − 5167 − 37

1672

167 − 25167 − 18

167

.


variantas003

Funkcija h(x, y) apibrėžta formule h(x, y) = −3x3y5 − 2 cos (−5x+ 5y).

1 ∂h

∂x=

1© −9x2y5+2 sin (−5x+5y); 2© −9x2y5−2 sin (−5x+5y);3© −9x2y5+10 sin (−5x+5y); 4© −9x4y5−10 cos (−5x+5y);5© −9x2y5−10 sin (−5x+5y); 6© −9x4y5+10 cos (−5x+5y).

2 ∂h

∂y=

1© −15x3y6+10 cos (−5x+5y); 2© −15x3y4+10 sin (−5x+5y);3© −15x3y4−10 sin (−5x+5y); 4© −15x3y6−10 cos (−5x+5y);5© −15x3y4+2 sin (−5x+5y); 6© −15x3y4−2 sin (−5x+5y).

v(x) =21x2 − 47

25x

3 v′(−2) = 1© 131100 ; 2© − 21

25 ; 3© 37100 ; 4© − 131

50 ; 5© − 37100 ; 6© 21

25 ; 7© 0; 8© − 131100 .

4 v′′(−5) = 1© − 943125 ; 2© 94

3125 ; 3© 94625 ; 4© 0; 5© − 94

625 ; 6© − 9425 ; 7© − 47

3125 ; 8© 473125 .

5 limx→0

x9 lnx =1© 0; 2© 1;3© ∞; 4© ln 9;5© 9; 6© riba neegzistuoja.

6 limx→∞

x8

14x =1© 14; 2© riba neegzistuoja;3© 1

14 ; 4© 47 ;

5© 8; 6© 0.



8 f ′(4) = 1© 3066; 2© 2298; 3© 570; 4© 762; 5© 9210; 6© 12282.


(4, f(4)

), lygtį.

1© y=−762x−2304; 2© y=762x+3792;3© y=744x+762; 4© y=762x−2304;5© y=−744x−762; 6© y=−762x+3792.


10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© h(x)≈1−19x; 2© h(x)≈1− 19

3 x;3© h(x)≈1− x

3 ; 4© h(x)≈1+ x3 ;

5© h(x)≈1+ 193 x; 6© h(x)≈− 19

3 x;7© h(x)≈1+19x; 8© h(x)≈ 19

3 x.

11 h(19

)≈ 1© 46

27 ; 2© 827 ; 3© 8

9 ; 4© − 1927 ; 5© 19

27 ; 6© 2627 ; 7© 28

27 ; 8© 109 .

f(x) = 2x3 − 21x2 + 36x+ 82

12 maxx∈[−2,2]

f(x) = 1© 99; 2© 103; 3© 108; 4© −26; 5© −90; 6© 86; 7© −94.

13 minx∈[−2,2]

f(x) = 1© −94; 2© 86; 3© −26; 4© 103; 5© 99; 6© −90; 7© 108.

14 maxx∈[−2,2]

|f(x)| = 1© 103; 2© 26; 3© 90; 4© 108; 5© 86; 6© 94; 7© 169; 8© 99.

f(x) = 47 sin(5x)− 96x2

15 f ′′(0) =1© 192; 2© −331; 3© 0; 4© −96; 5© −192; 6© 331; 7© 235; 8© −235; 9© 96.

16 f ′′′(0) =1© 5875; 2© −1175; 3© 192; 4© −1367; 5© −5875; 6© 0; 7© 1367; 8© 1175; 9© −192.


−11x− 29grafiko asimptotė, kai x→∞, yra tiesė y = mx+ k.

17 m = 1© 1411 ; 2© 29

7 ; 3© − 297 ; 4© − 11

14 ; 5© 729 ; 6© − 14

11 ; 7© − 729 ; 8© 11

14 .

18 k = 1© 195224 ; 2© − 121

483 ; 3© 483121 ; 4© − 224

195 ; 5© 224195 ; 6© − 483

121 ; 7© − 195224 ; 8© 121

483 .



1©(− 1√

6; 1√

6

); 2© (−∞;0)∪(0;+∞); 3©

(−∞;− 1√

6

)∪(

1√6;+∞

);

4© (−∞;+∞); 5© {0}; 6© (0;√6);

7© (−√6;0); 8© (0;+∞); 9© (−

√6;√6);

0© (−∞;0).

20 (sin6

(8x7))′

=

1© 226x8 sin7(8x7) cos(8x7); 2© 226x6 sin5(8x7) cos(8x7);3© −336x8 sin7(8x7) cos(8x7); 4© −226x6 sin5(8x7) cos(8x7);5© −336x6 sin5(8x7) cos(8x7); 6© 336x8 sin7(8x7) cos(8x7);7© 336x6 sin5(8x7) cos(8x7); 8© −226x8 sin7(8x7) cos(8x7).

21(

3√7 + ln (x+ 5)

)′=

1© 1

3(x+5) 3√

7+ln(x+5); 2© 1

3√

(7+ln(x+5))2;

3© 3√

(7+ln(x+5))2

3(x+5); 4© 1

3 3√

7+ln(x+5);

5© 33√

7+ln(x+5); 6© 1

3√

7+ln(x+5);

7© 3(x+5)3√

(7+ln(x+5))2; 8© 1

3(x+5)3√

(7+ln(x+5))2.

22 y b

t ac s

· ( 1 0 10 1 0

)=

1©

y b yc t cs c s

; 2©

b y ba t as c s

; 3©

y b yt a tc s c

; 4©

y s yc b cy c s

.

23 Jei tr − dw = 1, tai(

t dw r

)−1=

1©(−t −d−w −r

); 2©

(r −w−d t

); 3©

(r −d

−w t

); 4©

(r wd t

); 5©

(r dw t

).


15w1 −14w2 = −158−14w1 +13w2 = 147

.

24 w1 = 1© −7; 2© −4; 3© 4; 4© −9; 5© 5; 6© −1.

25 w2 = 1© 3; 2© −9; 3© −4; 4© −6; 5© 7; 6© −10.


−u1 −8u2 −14u3 = 6014u1 −7u2 +15u3 = −16814u1 +3u2 +2u3 = −27

.

26 u1 = 1© −3; 2© −2; 3© −8; 4© 6; 5© −6; 6© −9.

27 u2 = 1© 0; 2© −1; 3© 6; 4© 5; 5© 8; 6© −7.

28 u3 = 1© 0; 2© −5; 3© −3; 4© −7; 5© 1; 6© −2.

29 Apskaičiuokite

∣∣∣∣∣∣−6 −4 −2−1 −1 −3−2 5 3

∣∣∣∣∣∣.1© −156; 2© −8; 3© −82; 4© 147; 5© −94; 6© 3.

30 Matricos

r z sb x ya f w

adjunktas A23 =

1© za− rf ;2© rw − ya ;3© rf − za ;4© rw − yz .

31 Matricos(

55 −64−12 −90

)adjunktas A11 =

1© −64 ; 2© 55 ; 3© 64 ; 4© 12 ; 5© −12 ; 6© −55 ; 7© 90 ; 8© −90 .


5x1 − x2 − 4x3 = −2−3x1 + 2x2 + 5x3 = 1

4x1 + x2 + 5x3 = 1

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© −34 ; 2© −6 ; 3© −2 ; 4© 34 ; 5© 6 .

33 D1 = 1© −6 ; 2© 89 ; 3© 61 ; 4© −91 ; 5© 85 .

34 D2 = 1© −42 ; 2© 39 ; 3© −61 ; 4© −83 ; 5© 40 .

35 D3 = 1© 20 ; 2© −27 ; 3© −47 ; 4© 82 ; 5© −13 .

36 x1 + x2 + x3 = 1© 3317 ; 2© − 99

34 ; 3© − 317 ; 4© − 14

17 ; 5© 32 .

37 Matricos A =

3 5 −21 2 23 1 1


1© 80 ;2© 35 ;3© −75 ;4© 10 ;5© 71 ;6© 42 .

38 Adjunktas A13 = 1© −13 ; 2© −1 ; 3© −3 ; 4© −5 .

39 Adjunktas A11 = 1© 0 ; 2© 4 ; 3© −2 ; 4© −5 .

40 Adjunktas A31 = 1© −26 ; 2© 1 ; 3© −24 ; 4© 14 .

41 Adjunktas A12 = 1© 2 ; 2© −4 ; 3© 5 ; 4© 4 .

42 A−1 = 1©

0 17 − 1

7− 1

5935

1235

25 − 8

35135

; 2©

− 914 − 8

7 − 1314

12 1 1

237

37

27

; 3©

0 − 15

25

17

935 − 8

35− 1

71235

135

.


variantas004

Funkcija z(x, y) apibrėžta formule z(x, y) = −2x6y7 + 2 cos (2x+ 5y).

1 ∂z

∂x=

1© −12x7y7−4 cos (2x+5y); 2© −12x5y7+4 sin (2x+5y);3© −12x5y7−2 sin (2x+5y); 4© −12x5y7+2 sin (2x+5y);5© −12x7y7+4 cos (2x+5y); 6© −12x5y7−4 sin (2x+5y).

2 ∂z

∂y=

1© −14x6y6−2 sin (2x+5y); 2© −14x6y6+2 sin (2x+5y);3© −14x6y6+10 sin (2x+5y); 4© −14x6y8−10 cos (2x+5y);5© −14x6y6−10 sin (2x+5y); 6© −14x6y8+10 cos (2x+5y).

g(x) =24x2 − 47

31x

3 g′(4) = 1© 337496 ; 2© − 337

496 ; 3© 431124 ; 4© − 24

31 ; 5© 2431 ; 6© − 431

496 ; 7© 431496 ; 8© 0.

4 g′′(5) = 1© 94775 ; 2© 0; 3© − 47

3875 ; 4© − 9431 ; 5© − 94

775 ; 6© 473875 ; 7© 94

3875 ; 8© − 943875 .

5 limx→0

x11 lnx =1© 11; 2© 0;3© ln 11; 4© riba neegzistuoja;5© ∞; 6© 1.

6 limx→∞

x10

17x =1© 10; 2© 0;3© riba neegzistuoja; 4© 17;5© 1

17 ; 6© 1017 .



8 f ′(3) = 1© 2912; 2© 320; 3© 968; 4© 2183; 5© 725; 6© 239.


(3, f(3)

), lygtį.

1© y=−231x−320; 2© y=−320x−729;3© y=320x+1191; 4© y=320x−729;5© y=231x+320; 6© y=−320x+1191.

Tarkime, kad w(x) = 9√1 + 19x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© w(x)≈1− 19

9 x; 2© w(x)≈1−19x;3© w(x)≈− 19

9 x; 4© w(x)≈ 199 x;

5© w(x)≈1− x9 ; 6© w(x)≈1+ 19

9 x;7© w(x)≈1+19x; 8© w(x)≈1+ x

9 .

11 w(− 1

10

)≈ 1© 109

90 ; 2© 1990 ; 3© 91

90 ; 4© 1110 ; 5© − 19

90 ; 6© 8990 ; 7© 71

90 ; 8© 910 .

f(x) = 2x3 − 3x2 − 12x+ 62

12 maxx∈[1,6]

f(x) = 1© 49; 2© 322; 3© 314; 4© 69; 5© 42; 6© 327; 7© 23.

13 minx∈[1,6]

f(x) = 1© 23; 2© 69; 3© 314; 4© 42; 5© 322; 6© 49; 7© 327.

14 maxx∈[1,6]

|f(x)| = 1© 23; 2© 332; 3© 42; 4© 69; 5© 322; 6© 49; 7© 314; 8© 327.

f(x) = 26 sin(5x)− 35x2

15 f ′′(0) =1© −35; 2© 35; 3© −130; 4© 130; 5© −165; 6© 0; 7© 165; 8© −70; 9© 70.

16 f ′′′(0) =1© 0; 2© −650; 3© 3250; 4© 720; 5© 650; 6© −720; 7© 70; 8© −70; 9© −3250.


−9x+ 21grafiko asimptotė, kai x→∞, yra tiesė y = mx+ d.

17 m = 1© 2117 ; 2© 14

9 ; 3© 914 ; 4© − 17

21 ; 5© − 2117 ; 6© 17

21 ; 7© − 914 ; 8© − 14

9 .

18 d = 1© 167297 ; 2© 149

27 ; 3© 27149 ; 4© 297

167 ; 5© − 27149 ; 6© − 149

27 ; 7© − 167297 ; 8© − 297

167 .



1© (−∞;− 16 )∪(

16 ;+∞); 2© (−∞;0); 3© (− 1

6 ;16 );

4© (−6;6); 5© (0;+∞); 6© (−6;0);7© (−∞;0)∪(0;+∞); 8© {0}; 9© (−∞;+∞);0© (0;6).

20 (sin5

(3x7))′

=

1© 105x6 sin4(3x7) cos(3x7); 2© 105x8 sin6(3x7) cos(3x7);3© 67x8 sin6(3x7) cos(3x7); 4© −105x8 sin6(3x7) cos(3x7);5© −67x8 sin6(3x7) cos(3x7); 6© 67x6 sin4(3x7) cos(3x7);7© −67x6 sin4(3x7) cos(3x7); 8© −105x6 sin4(3x7) cos(3x7).

21(

4√8 + ln (x+ 3)

)′=

1© 4√

(8+ln(x+3))3

4(x+3); 2© 4(x+3)

4√

(8+ln(x+3))3;

3© 14√

(8+ln(x+3))3; 4© 4

4√

8+ln(x+3);

5© 1

4 4√

8+ln(x+3); 6© 1

4√

8+ln(x+3);

7© 1

4(x+3) 4√

8+ln(x+3); 8© 1

4(x+3)4√

(8+ln(x+3))3.

22 p d

w vb x

· ( 1 0 10 1 0

)=

1©

d p dv w vx b x

; 2©

p d pb w bx b x

; 3©

p d pw v wb x b

; 4©

p x pb d bp b x

.

23 Jei xb− ys = 1, tai(

x ys b

)−1=

1©(

b sy x

); 2©

(−x −y−s −b

); 3©

(b −y−s x

); 4©

(b ys x

); 5©

(b −s−y x

).

Išspręskite tiesinių lygčių sistemą{−13w1 +12w2 = −9−10w1 −3w2 = 42

.

24 w1 = 1© −9; 2© 9; 3© 1; 4© −2; 5© −3; 6© −7.

25 w2 = 1© 7; 2© −10; 3© −1; 4© −5; 5© 6; 6© −4.


−y1 +4y2 +12y3 = 71−8y1 +9y2 +17y3 = 28−16y1 +17y2 −12y3 = −308

.

26 y1 = 1© 9; 2© −1; 3© 1; 4© −8; 5© −6; 6© 5.

27 y2 = 1© −4; 2© −1; 3© −9; 4© −6; 5© 1; 6© −3.

28 y3 = 1© −10; 2© −1; 3© −6; 4© 8; 5© 10; 6© 5.

29 Apskaičiuokite

∣∣∣∣∣∣2 2 6−2 1 3−2 2 −5

∣∣∣∣∣∣.1© −66; 2© 6; 3© −53; 4© −3; 5© −74; 6© 130.

30 Matricos

f x ut c qh d r

adjunktas A12 =

1© fr − qx ;2© tr − qh ;3© qh− tr ;4© fr − qh .

31 Matricos(−79 4674 14

)adjunktas A11 =

1© −74 ; 2© 14 ; 3© −14 ; 4© −46 ; 5© 79 ; 6© 46 ; 7© 74 ; 8© −79 .


−3x1 + 2x2 + 4x3 = −5−5x1 − 5x2 − 3x3 = 5

−3x1 − 3x2 − 4x3 = 3

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© −67 ; 2© −55 ; 3© 67 ; 4© −73 ; 5© 55 .

33 D1 = 1© 91 ; 2© 87 ; 3© 35 ; 4© 73 ; 5© −33 .

34 D2 = 1© 65 ; 2© −51 ; 3© 70 ; 4© 88 ; 5© −28 .

35 D3 = 1© −60 ; 2© −39 ; 3© 0 ; 4© −23 ; 5© −57 .

36 x1 + x2 + x3 = 1© − 5355 ; 2© −1 ; 3© − 73

55 ; 4© 7455 ; 5© − 6

11 .

37 Matricos A =

2 1 4−4 3 −2−4 5 5


1© −43 ;2© −68 ;3© 46 ;4© 39 ;5© −42 ;6© −31 .

38 Adjunktas A33 = 1© −19 ; 2© 11 ; 3© 21 ; 4© 10 .

39 Adjunktas A21 = 1© −21 ; 2© 15 ; 3© −11 ; 4© 33 .

40 Adjunktas A31 = 1© −14 ; 2© 28 ; 3© −24 ; 4© −22 .

41 Adjunktas A11 = 1© 23 ; 2© 25 ; 3© −42 ; 4© −44 .

42 A−1 = 1©

2546

1546 − 7

231423

1323 − 6

23− 4

23 − 723

523

; 2©

− 12

16 0

0 29 − 1

312 − 1

1813

; 3©

2546

1423 − 4

231546

1323 − 7

23− 7

23 − 623

523

.


variantas005

Funkcija h(x, y) apibrėžta formule h(x, y) = 2x5y4 − 3 cos (3x− 5y).

1 ∂h

∂x=

1© 10x4y4+9 sin (3x−5y); 2© 10x4y4+3 sin (3x−5y);3© 10x6y4+9 cos (3x−5y); 4© 10x6y4−9 cos (3x−5y);5© 10x4y4−9 sin (3x−5y); 6© 10x4y4−3 sin (3x−5y).

2 ∂h

∂y=

1© 8x5y5−15 cos (3x−5y); 2© 8x5y3+15 sin (3x−5y);3© 8x5y3−15 sin (3x−5y); 4© 8x5y3+3 sin (3x−5y);5© 8x5y5+15 cos (3x−5y); 6© 8x5y3−3 sin (3x−5y).

z(x) =45x2 − 19

6x

3 z′(−5) = 1© − 55375 ; 2© − 572

75 ; 3© 57275 ; 4© − 572

15 ; 5© 152 ; 6© 0; 7© 553

75 ; 8© − 152 .

4 z′′(4) = 1© 19192 ; 2© − 19

384 ; 3© 0; 4© 19384 ; 5© − 19

48 ; 6© 1948 ; 7© − 19

192 ; 8© − 193 .

5 limx→0

x9 lnx =1© riba neegzistuoja; 2© 0;3© ln 9; 4© ∞;5© 9; 6© 1.

6 limx→∞

x17

10x =1© 10; 2© riba neegzistuoja;3© 17; 4© 17

10 ;5© 0; 6© 1

10 .



8 f ′(4) = 1© 16382; 2© 4094; 3© 3070; 4© 12286; 5© 1022; 6© 766.


(4, f(4)

), lygtį.

1© y=−1022x−3072; 2© y=1022x−3072;3© y=−1022x+5104; 4© y=−1016x−1022;5© y=1016x+1022; 6© y=1022x+5104.


10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© h(x)≈1− 16

3 x; 2© h(x)≈1−16x;3© h(x)≈1− x

3 ; 4© h(x)≈1+ x3 ;

5© h(x)≈1+ 163 x; 6© h(x)≈1+16x;

7© h(x)≈− 163 x; 8© h(x)≈ 16

3 x.

11 h(− 1

8

)≈ 1© 9

8 ; 2© 2524 ; 3© 5

3 ; 4© 78 ; 5© 23

24 ; 6© 23 ; 7© − 2

3 ; 8© 13 .

f(x) = 2x3 − 33x2 + 144x+ 48

12 maxx∈[−2,7]

f(x) = 1© −388; 2© 241; 3© −396; 4© 112; 5© 244; 6© 237; 7© 125.

13 minx∈[−2,7]

f(x) = 1© 125; 2© 112; 3© 241; 4© −388; 5© −396; 6© 237; 7© 244.

14 maxx∈[−2,7]

|f(x)| = 1© 125; 2© 112; 3© 396; 4© 244; 5© 388; 6© 241; 7© 237; 8© 85.

f(x) = 27 sin(5x)− 28x2

15 f ′′(0) =1© −135; 2© −163; 3© 0; 4© 135; 5© −28; 6© 56; 7© 28; 8© −56; 9© 163.

16 f ′′′(0) =1© 0; 2© 56; 3© 3375; 4© −675; 5© 731; 6© 675; 7© −3375; 8© −731; 9© −56.


3x+ 31grafiko asimptotė, kai x→∞, yra tiesė y = kx+ l.

17 k = 1© 83 ; 2© 4

31 ; 3© 38 ; 4© − 4

31 ; 5© − 83 ; 6© 31

4 ; 7© − 314 ; 8© − 3

8 .

18 l = 1© − 159185 ; 2© 159

185 ; 3© − 185159 ; 4© − 9

260 ; 5© 9260 ; 6© 260

9 ; 7© − 2609 ; 8© 185

159 .


2


1© (0;2√3); 2© {0}; 3©

(−∞;− 1

2√

3

)∪(

12√

3;+∞

);

4© (−∞;+∞); 5© (0;+∞); 6© (−∞;0)∪(0;+∞);7© (−2

√3;2√3); 8© (−2

√3;0); 9©

(− 1

2√

3; 12√

3

);

0© (−∞;0).

20 (sin9

(8x7))′

=

1© 301x6 sin8(8x7) cos(8x7); 2© 301x8 sin10(8x7) cos(8x7);3© −301x6 sin8(8x7) cos(8x7); 4© −301x8 sin10(8x7) cos(8x7);5© −504x8 sin10(8x7) cos(8x7); 6© 504x6 sin8(8x7) cos(8x7);7© 504x8 sin10(8x7) cos(8x7); 8© −504x6 sin8(8x7) cos(8x7).

21(

6√5 + ln (x+ 3)

)′=

1© 1

6 6√

5+ln(x+3); 2© 1

6(x+3) 6√

5+ln(x+3);

3© 6√

(5+ln(x+3))5

6(x+3); 4© 1

6(x+3)6√

(5+ln(x+3))5;

5© 66√

5+ln(x+3); 6© 1

6√

5+ln(x+3);

7© 6(x+3)6√

(5+ln(x+3))5; 8© 1

6√

(5+ln(x+3))5.

22 a u

y qz t

· ( 1 0 10 1 0

)=

1©

a u ay q yz t z

; 2©

a t az u za z t

; 3©

u a uq y qt z t

; 4©

a u az y zt z t

.

23 Jei yd− wp = 1, tai(

y wp d

)−1=

1©(

d pw y

); 2©

(d −p

−w y

); 3©

(−y −w−p −d

); 4©

(d wp y

); 5©

(d −w−p y

).

Išspręskite tiesinių lygčių sistemą{−11u1 −10u2 = 135−4u1 +9u2 = −52 .

24 u1 = 1© 9; 2© −7; 3© 8; 4© 6; 5© −5; 6© −10.

25 u2 = 1© 6; 2© −10; 3© −6; 4© −8; 5© −7; 6© −9.


−z1 −12z2 +2z3 = −10112z1 −7z2 +3z3 = −185−4z1 +15z2 +2z3 = 142

.

26 z1 = 1© 1; 2© −9; 3© 9; 4© 7; 5© −2; 6© 10.

27 z2 = 1© 8; 2© 2; 3© −10; 4© 10; 5© 3; 6© 9.

28 z3 = 1© 4; 2© −6; 3© 0; 4© 9; 5© −5; 6© −7.

29 Apskaičiuokite

∣∣∣∣∣∣2 −4 5−3 −4 −4−4 5 1

∣∣∣∣∣∣.1© −199; 2© −252; 3© 5; 4© −4; 5© −5; 6© 351.

30 Matricos

r q cy a wf t x

adjunktas A23 =

1© rt− qf ;2© rx− wf ;3© rx− wq ;4© qf − rt .

31 Matricos(−12 −8037 −13

)adjunktas A21 =

1© −12 ; 2© −13 ; 3© −37 ; 4© 80 ; 5© 12 ; 6© 13 ; 7© 37 ; 8© −80 .


3x1 + x2 − x3 = 5

−2x1 − 3x2 + 4x3 = 1

4x1 − 2x2 + 4x3 = 1

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© −4 ; 2© −9 ; 3© 4 ; 4© 6 ; 5© 9 .

33 D1 = 1© −8 ; 2© −81 ; 3© −21 ; 4© −34 ; 5© −95 .

34 D2 = 1© 72 ; 2© 126 ; 3© 32 ; 4© 96 ; 5© −37 .

35 D3 = 1© −6 ; 2© 95 ; 3© −26 ; 4© 45 ; 5© 83 .

36 x1 + x2 + x3 = 1© 5 ; 2© − 814 ; 3© −47 ; 4© − 13

4 ; 5© − 314 .

37 Matricos A =

−4 3 4−5 1 −4−3 4 −1


1© −107 ;2© −134 ;3© 107 ;4© 123 ;5© 210 ;6© −181 .

38 Adjunktas A31 = 1© 11 ; 2© 10 ; 3© −16 ; 4© 15 .

39 Adjunktas A13 = 1© 2 ; 2© 3 ; 3© 35 ; 4© −17 .

40 Adjunktas A21 = 1© −29 ; 2© 16 ; 3© 19 ; 4© 20 .

41 Adjunktas A22 = 1© −22 ; 2© 16 ; 3© −33 ; 4© −11 .

42 A−1 = 1©

− 15107 − 7

10717107

− 19107 − 16

107 − 7107

16107

36107 − 11

107

; 2©

−1 − 34 0

12 − 1

812

− 12 − 5

812

; 3©

− 15107 − 19

10716107

− 7107 − 16

10736107

17107 − 7

107 − 11107

.


variantas006

Funkcija h(x, y) apibrėžta formule h(x, y) = 3x3y7 + 2 cos (5x+ 5y).

1 ∂h

∂x=

1© 9x4y7−10 cos (5x+5y); 2© 9x2y7−2 sin (5x+5y);3© 9x2y7+2 sin (5x+5y); 4© 9x4y7+10 cos (5x+5y);5© 9x2y7−10 sin (5x+5y); 6© 9x2y7+10 sin (5x+5y).

2 ∂h

∂y=

1© 21x3y6−10 sin (5x+5y); 2© 21x3y8+10 cos (5x+5y);3© 21x3y6+2 sin (5x+5y); 4© 21x3y8−10 cos (5x+5y);5© 21x3y6+10 sin (5x+5y); 6© 21x3y6−2 sin (5x+5y).

g(x) =20x2 − 42

45x

3 g′(−5) = 1© 49 ; 2© − 4

9 ; 3© − 4581125 ; 4© 542

1125 ; 5© 4581125 ; 6© − 542

1125 ; 7© 0; 8© − 542225 .

4 g′′(−2) = 1© − 760 ; 2© 7

60 ; 3© 715 ; 4© − 28

15 ; 5© − 715 ; 6© 0; 7© 7

30 ; 8© − 730 .

5 limx→0

x8 lnx =1© 0; 2© ln 8;3© 1; 4© 8;5© ∞; 6© riba neegzistuoja.

6 limx→∞

x17

8x =1© 0; 2© 1

8 ;3© riba neegzistuoja; 4© 8;5© 17; 6© 17

8 .



8 f ′(2) = 1© 954; 2© 186; 3© 234; 4© 378; 5© 762; 6© 474.


(2, f(2)

), lygtį.

1© y=234x−384; 2© y=−84x−234;3© y=−234x−384; 4© y=84x+234;5© y=234x+552; 6© y=−234x+552.

Tarkime, kad z(x) = 7√1 + 20x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© z(x)≈1−20x; 2© z(x)≈1− 20

7 x;3© z(x)≈1− x

7 ; 4© z(x)≈1+ 207 x;

5© z(x)≈ 207 x; 6© z(x)≈− 20

7 x;7© z(x)≈1+ x

7 ; 8© z(x)≈1+20x.

11 z(16

)≈ 1© 10

21 ; 2© 3121 ; 3© 43

42 ; 4© 1121 ; 5© − 10

21 ; 6© 56 ; 7© 41

42 ; 8© 76 .

f(x) = 2x3 − 27x2 + 48x+ 30

12 maxx∈[−3,4]

f(x) = 1© −290; 2© 68; 3© −411; 4© −82; 5© 55; 6© 53; 7© −429.

13 minx∈[−3,4]

f(x) = 1© 55; 2© −411; 3© −429; 4© 68; 5© −290; 6© 53; 7© −82.

14 maxx∈[−3,4]

|f(x)| = 1© 55; 2© 290; 3© 453; 4© 82; 5© 68; 6© 411; 7© 53; 8© 429.

f(x) = 29 sin(3x)− 65x2

15 f ′′(0) =1© 152; 2© 0; 3© 65; 4© −87; 5© −152; 6© −65; 7© 130; 8© −130; 9© 87.

16 f ′′′(0) =1© −261; 2© −391; 3© 783; 4© 0; 5© 130; 6© 261; 7© 391; 8© −130; 9© −783.


−13x− 31grafiko asimptotė, kai x→∞, yra tiesė y = rx+ c.

17 r = 1© 813 ; 2© − 13

8 ; 3© − 1331 ; 4© 13

8 ; 5© 1331 ; 6© 31

13 ; 7© − 3113 ; 8© − 8

13 .

18 c = 1© − 417169 ; 2© 417

169 ; 3© 169417 ; 4© − 303

55 ; 5© 55303 ; 6© 303

55 ; 7© − 169417 ; 8© − 55

303 .



1© (−∞;0)∪(0;+∞); 2©(−∞;− 1√

34

)∪(

1√34

;+∞); 3© {0};

4© (0;√34); 5© (−

√34;0); 6© (0;+∞);

7© (−√34;√34); 8© (−∞;+∞); 9© (−∞;0);

0©(− 1√

34; 1√

34

).

20 (sin2

(9x8))′

=

1© 144x7 sin1(9x8) cos(9x8); 2© −144x9 sin3(9x8) cos(9x8);3© 100x9 sin3(9x8) cos(9x8); 4© 100x7 sin1(9x8) cos(9x8);5© 144x9 sin3(9x8) cos(9x8); 6© −144x7 sin1(9x8) cos(9x8);7© −100x7 sin1(9x8) cos(9x8); 8© −100x9 sin3(9x8) cos(9x8).

21(

6√9 + ln (x+ 3)

)′=

1© 16√

(9+ln(x+3))5; 2© 6(x+3)

6√

(9+ln(x+3))5;

3© 1

6(x+3)6√

(9+ln(x+3))5; 4© 1

6√

9+ln(x+3);

5© 66√

9+ln(x+3); 6© 1

6 6√

9+ln(x+3);

7© 6√

(9+ln(x+3))5

6(x+3); 8© 1

6(x+3) 6√

9+ln(x+3).

22 p a

v rw b

· ( 1 0 10 1 0

)=

1©

a p ar v rb w b

; 2©

p a pw v wb w b

; 3©

p b pw a wp w b

; 4©

p a pv r vw b w

.

23 Jei px− sd = 1, tai(

p sd x

)−1=

1©(

x −d−s p

); 2©

(x −s−d p

); 3©

(x ds p

); 4©

(x sd p

); 5©

(−p −s−d −x

).

Išspręskite tiesinių lygčių sistemą{−7x1 −16x2 = −11716x1 +3x2 = 66

.

24 x1 = 1© 3; 2© 5; 3© −3; 4© 0; 5© −2; 6© 1.

25 x2 = 1© 6; 2© 1; 3© −5; 4© 0; 5© 5; 6© 10.


−s1 +2s2 −14s3 = −105−16s1 +17s2 +5s3 = −186−4s1 +9s2 +10s3 = −32

.

26 s1 = 1© 5; 2© 3; 3© −3; 4© −5; 5© 10; 6© 0.

27 s2 = 1© 2; 2© −7; 3© −8; 4© −4; 5© 4; 6© −1.

28 s3 = 1© 4; 2© −8; 3© −1; 4© 5; 5© 6; 6© 3.

29 Apskaičiuokite

∣∣∣∣∣∣4 −4 −1−1 −3 55 4 5

∣∣∣∣∣∣.1© −307; 2© 7; 3© 9; 4© −186; 5© 315; 6© −271.

30 Matricos

w c zh b rf y x

adjunktas A23 =

1© wy − cf ;2© wx− rc ;3© cf − wy ;4© wx− rf .

31 Matricos(−15 10−86 75

)adjunktas A22 =

1© 10 ; 2© 15 ; 3© 75 ; 4© −75 ; 5© 86 ; 6© −15 ; 7© −86 ; 8© −10 .


4x1 + 3x2 + x3 = −2−x1 − x2 − 4x3 = 2

−x1 + 5x2 + 5x3 = 2

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© 81 ; 2© −69 ; 3© −81 ; 4© −133 ; 5© 69 .

33 D1 = 1© 65 ; 2© 37 ; 3© −72 ; 4© 69 ; 5© −82 .

34 D2 = 1© 54 ; 2© −49 ; 3© 68 ; 4© −19 ; 5© −22 .

35 D3 = 1© −36 ; 2© −98 ; 3© −12 ; 4© 30 ; 5© 70 .

36 x1 + x2 + x3 = 1© 4481 ; 2© − 50

81 ; 3© − 23 ; 4© − 5

81 ; 5© − 1027 .

37 Matricos A =

2 −1 −43 2 −45 1 5


1© −136 ;2© −173 ;3© 81 ;4© −132 ;5© −24 ;6© 91 .

38 Adjunktas A13 = 1© −1 ; 2© −7 ; 3© 18 ; 4© −6 .

39 Adjunktas A22 = 1© 30 ; 2© 38 ; 3© 41 ; 4© −28 .

40 Adjunktas A11 = 1© 4 ; 2© −21 ; 3© 14 ; 4© 27 .

41 Adjunktas A23 = 1© −7 ; 2© 2 ; 3© 16 ; 4© −14 .

42 A−1 = 1©

213 − 5

13 − 113

191

3091 − 1

131291 − 4

91113

; 2©

213

191

1291

− 513

3091 − 4

91− 1

13 − 113

113

; 3©

1136 − 1

9 − 518

49

19 − 2

91336 − 2

9 − 118

.


variantas007

Funkcija u(x, y) apibrėžta formule u(x, y) = 4x7y6 + 5 cos (4x+ 4y).

1 ∂u

∂x=

1© 28x6y6+5 sin (4x+4y); 2© 28x8y6−20 cos (4x+4y);3© 28x6y6−5 sin (4x+4y); 4© 28x8y6+20 cos (4x+4y);5© 28x6y6+20 sin (4x+4y); 6© 28x6y6−20 sin (4x+4y).

2 ∂u

∂y=

1© 24x7y5+5 sin (4x+4y); 2© 24x7y7−20 cos (4x+4y);3© 24x7y5−20 sin (4x+4y); 4© 24x7y5−5 sin (4x+4y);5© 24x7y5+20 sin (4x+4y); 6© 24x7y7+20 cos (4x+4y).

h(x) =41x2 − 6

16x

3 h′(5) = 1© − 1031400 ; 2© 1031

400 ; 3© − 1019400 ; 4© 1019

400 ; 5© 103180 ; 6© 0; 7© 41

16 ; 8© − 4116 .

4 h′′(2) = 1© 364 ; 2© − 3

64 ; 3© 316 ; 4© − 3

16 ; 5© 332 ; 6© − 3

32 ; 7© − 34 ; 8© 0.

5 limx→0

x22 lnx =1© 0; 2© riba neegzistuoja;3© 22; 4© 1;5© ln 22; 6© ∞.

6 limx→∞

x16

18x =1© 18; 2© 16;3© riba neegzistuoja; 4© 1

18 ;5© 8

9 ; 6© 0.



8 f ′(2) = 1© 188; 2© 124; 3© 92; 4© 60; 5© 44; 6© 28.


(2, f(2)

), lygtį.

1© y=−24x−44; 2© y=24x+44;3© y=44x−64; 4© y=−44x+112;5© y=−44x−64; 6© y=44x+112.

Tarkime, kad w(x) = 7√1 + 11x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© w(x)≈1− 11

7 x; 2© w(x)≈1− x7 ;

3© w(x)≈ 117 x; 4© w(x)≈1−11x;

5© w(x)≈1+ x7 ; 6© w(x)≈1+ 11

7 x;7© w(x)≈1+11x; 8© w(x)≈− 11

7 x.

11 w(− 1

6

)≈ 1© 53

42 ; 2© 3142 ; 3© 5

6 ; 4© 1142 ; 5© − 11

42 ; 6© 4342 ; 7© 7

6 ; 8© 4142 .

f(x) = 2x3 − 12x2 − 72x+ 53

12 maxx∈[−3,4]

f(x) = 1© 107; 2© −308; 3© 140; 4© −379; 5© 135; 6© −299; 7© 133.

13 minx∈[−3,4]

f(x) = 1© −299; 2© −308; 3© −379; 4© 135; 5© 140; 6© 133; 7© 107.

14 maxx∈[−3,4]

|f(x)| = 1© 308; 2© 299; 3© 133; 4© 494; 5© 135; 6© 107; 7© 140; 8© 379.

f(x) = 38 sin(5x)− 97x2

15 f ′′(0) =1© −287; 2© 194; 3© −97; 4© 97; 5© 190; 6© −190; 7© −194; 8© 0; 9© 287.

16 f ′′′(0) =1© 4750; 2© −950; 3© −194; 4© −4750; 5© 950; 6© 194; 7© −1144; 8© 0; 9© 1144.


7x− 15grafiko asimptotė, kai x→∞, yra tiesė y = dx+ n.

17 d = 1© − 67 ; 2© − 8

15 ; 3© − 76 ; 4© 7

6 ; 5© 158 ; 6© − 15

8 ; 7© 67 ; 8© 8

15 .

18 n = 1© 8187 ; 2© 187

8 ; 3© − 1878 ; 4© 49

146 ; 5© − 49146 ; 6© − 146

49 ; 7© − 8187 ; 8© 146

49 .



1© (−∞;0); 2© (0;+∞); 3© (−∞;0)∪(0;+∞);4© (−

√42;√42); 5©

(−∞;− 1√

42

)∪(

1√42

;+∞); 6© {0};

7© (−∞;+∞); 8© (0;√42); 9©

(− 1√

42; 1√

42

);

0© (−√42;0).

20 (sin6

(8x7))′

=

1© 372x6 sin5(8x7) cos(8x7); 2© −336x6 sin5(8x7) cos(8x7);3© −336x8 sin7(8x7) cos(8x7); 4© −372x6 sin5(8x7) cos(8x7);5© 372x8 sin7(8x7) cos(8x7); 6© 336x8 sin7(8x7) cos(8x7);7© 336x6 sin5(8x7) cos(8x7); 8© −372x8 sin7(8x7) cos(8x7).

21(

3√4 + ln (x+ 7)

)′=

1© 3√

(4+ln(x+7))2

3(x+7); 2© 1

3(x+7) 3√

4+ln(x+7);

3© 1

3(x+7)3√

(4+ln(x+7))2; 4© 3(x+7)

3√

(4+ln(x+7))2;

5© 33√

4+ln(x+7); 6© 1

3 3√

4+ln(x+7);

7© 13√

4+ln(x+7); 8© 1

3√

(4+ln(x+7))2.

22 z a

q vw x

· ( 1 0 10 1 0

)=

1©

z a zq v qw x w

; 2©

z a zw q wx w x

; 3©

a z av q vx w x

; 4©

z x zw a wz w x

.

23 Jei bq − rp = 1, tai(

b rp q

)−1=

1©(−b −r−p −q

); 2©

(q −r−p b

); 3©

(q −p−r b

); 4©

(q pr b

); 5©

(q rp b

).

Išspręskite tiesinių lygčių sistemą{−17r1 −4r2 = 111−10r1 +5r2 = 80

.

24 r1 = 1© −6; 2© 9; 3© −8; 4© 6; 5© −4; 6© −7.

25 r2 = 1© 5; 2© 2; 3© 8; 4© −6; 5© −1; 6© −10.


w1 −12w2 −6w3 = 3014w1 +11w2 −15w3 = −2010w1 −7w2 +6w3 = 112

.

26 w1 = 1© −10; 2© −2; 3© −1; 4© 9; 5© −5; 6© 6.

27 w2 = 1© −9; 2© −4; 3© 7; 4© 3; 5© −6; 6© −2.

28 w3 = 1© 1; 2© 4; 3© 6; 4© 9; 5© 8; 6© −4.

29 Apskaičiuokite

∣∣∣∣∣∣4 −1 5−5 −6 21 −2 −4

∣∣∣∣∣∣.1© 210; 2© 148; 3© −16; 4© 6; 5© −5; 6© −7.

30 Matricos

x b qs c ew a h

adjunktas A12 =

1© ew − sh ;2© xh− eb ;3© xh− ew ;4© sh− ew .

31 Matricos(

89 −65−81 66

)adjunktas A11 =

1© −65 ; 2© 89 ; 3© 81 ; 4© −66 ; 5© 65 ; 6© −81 ; 7© −89 ; 8© 66 .


−5x1 + x2 − 4x3 = −13x1 + x2 + 5x3 = −5−2x1 + 3x2 − 3x3 = −3

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© 45 ; 2© −7 ; 3© 4 ; 4© −45 ; 5© −4 .

33 D1 = 1© 94 ; 2© −47 ; 3© 37 ; 4© −91 ; 5© 36 .

34 D2 = 1© −73 ; 2© −34 ; 3© 37 ; 4© −95 ; 5© −19 .

35 D3 = 1© −66 ; 2© −52 ; 3© −10 ; 4© −9 ; 5© 62 .

36 x1 + x2 + x3 = 1© − 8945 ; 2© 2

15 ; 3© 115 ; 4© 14

15 ; 5© 745 .

37 Matricos A =

2 2 −1−3 −1 24 −1 5


1© 72 ;2© −57 ;3© 43 ;4© 29 ;5© 33 ;6© 2 .

38 Adjunktas A33 = 1© 4 ; 2© 11 ; 3© −10 ; 4© 6 .

39 Adjunktas A32 = 1© −1 ; 2© −2 ; 3© 6 ; 4© 3 .

40 Adjunktas A11 = 1© −2 ; 2© 11 ; 3© −3 ; 4© −11 .

41 Adjunktas A31 = 1© −5 ; 2© −11 ; 3© −8 ; 4© 3 .

42 A−1 = 1©

− 111 − 3

11111

2333

1433 − 1

33733

1033

433

; 2©

47

197

137

57

157

117

17

107

57

; 3©

− 111

2333

733

− 311

1433

1033

111 − 1

33433

.


variantas008

Funkcija u(x, y) apibrėžta formule u(x, y) = 3x3y4 + 5 cos (4x− 4y).

1 ∂u

∂x=

1© 9x4y4−20 cos (4x−4y); 2© 9x4y4+20 cos (4x−4y);3© 9x2y4+5 sin (4x−4y); 4© 9x2y4+20 sin (4x−4y);5© 9x2y4−20 sin (4x−4y); 6© 9x2y4−5 sin (4x−4y).

2 ∂u

∂y=

1© 12x3y3+5 sin (4x−4y); 2© 12x3y3−5 sin (4x−4y);3© 12x3y3+20 sin (4x−4y); 4© 12x3y3−20 sin (4x−4y);5© 12x3y5−20 cos (4x−4y); 6© 12x3y5+20 cos (4x−4y).

w(x) =45x2 − 42

2x

3 w′(4) = 1© 33916 ; 2© 0; 3© − 381

16 ; 4© − 452 ; 5© 45

2 ; 6© 38116 ; 7© − 339

16 ; 8© 3814 .

4 w′′(−2) = 1© − 218 ; 2© 21

8 ; 3© − 214 ; 4© −42; 5© 21

2 ; 6© − 212 ; 7© 0; 8© 21

4 .

5 limx→0

x11 lnx =1© 11; 2© 1;3© ln 11; 4© riba neegzistuoja;5© 0; 6© ∞.

6 limx→∞

x4

8x =1© 0; 2© 8;3© 4; 4© 1

2 ;5© riba neegzistuoja; 6© 1

8 .



8 f ′(4) = 1© 6138; 2© 8186; 3© 2042; 4© 1530; 5© 378; 6© 506.


(4, f(4)

), lygtį.

1© y=−506x+2512; 2© y=506x+2512;3© y=−506x−1536; 4© y=488x+506;5© y=−488x−506; 6© y=506x−1536.

Tarkime, kad g(x) = 9√1 + 2x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© g(x)≈1− x

9 ; 2© g(x)≈ 29x;

3© g(x)≈1+ 29x; 4© g(x)≈1−2x;

5© g(x)≈− 29x; 6© g(x)≈1− 2

9x;7© g(x)≈1+2x; 8© g(x)≈1+ x

9 .

11 g(− 1

7

)≈ 1© 65

63 ; 2© 6263 ; 3© − 2

63 ; 4© 6163 ; 5© 2

63 ; 6© 87 ; 7© 6

7 ; 8© 6463 .

f(x) = 2x3 − 27x2 − 60x+ 84

12 maxx∈[2,6]

f(x) = 1© −1216; 2© −128; 3© 115; 4© −818; 5© −119; 6© −816; 7© −111.

13 minx∈[2,6]

f(x) = 1© −128; 2© −111; 3© −818; 4© −119; 5© −816; 6© −1216; 7© 115.

14 maxx∈[2,6]

|f(x)| = 1© 115; 2© 111; 3© 154; 4© 816; 5© 119; 6© 1216; 7© 818; 8© 128.

f(x) = 26 sin(3x)− 58x2

15 f ′′(0) =1© 116; 2© 136; 3© −136; 4© −78; 5© −58; 6© 58; 7© 0; 8© 78; 9© −116.

16 f ′′′(0) =1© −702; 2© 234; 3© 116; 4© 0; 5© 350; 6© −116; 7© 702; 8© −350; 9© −234.


−7x− 21grafiko asimptotė, kai x→∞, yra tiesė y = kx+ n.

17 k = 1© 712 ; 2© 2

3 ; 3© − 712 ; 4© − 12

7 ; 5© 127 ; 6© − 2

3 ; 7© 32 ; 8© − 3

2 .

18 n = 1© 269145 ; 2© − 7

50 ; 3© − 145269 ; 4© − 269

145 ; 5© 750 ; 6© 50

7 ; 7© 145269 ; 8© − 50

7 .



1© (0;4); 2© (− 14 ;

14 ); 3© (0;+∞);

4© (−∞;0); 5© {0}; 6© (−∞;0)∪(0;+∞);7© (−4;0); 8© (−∞;− 1

4 )∪(14 ;+∞); 9© (−∞;+∞);

0© (−4;4).

20 (sin8

(2x3))′

=

1© −69x4 sin9(2x3) cos(2x3); 2© −48x2 sin7(2x3) cos(2x3);3© −69x2 sin7(2x3) cos(2x3); 4© 69x2 sin7(2x3) cos(2x3);5© 48x4 sin9(2x3) cos(2x3); 6© −48x4 sin9(2x3) cos(2x3);7© 69x4 sin9(2x3) cos(2x3); 8© 48x2 sin7(2x3) cos(2x3).

21(

4√3 + ln (x+ 9)

)′=

1© 4(x+9)4√

(3+ln(x+9))3; 2© 1

4(x+9) 4√

3+ln(x+9);

3© 1

4(x+9)4√

(3+ln(x+9))3; 4© 4

√(3+ln(x+9))3

4(x+9);

5© 14√

3+ln(x+9); 6© 1

4 4√

3+ln(x+9);

7© 44√

3+ln(x+9); 8© 1

4√

(3+ln(x+9))3.

22 w b

x pc s

· ( 1 0 10 1 0

)=

1©

w b wc x cs c s

; 2©

w b wx p xc s c

; 3©

w s wc b cw c s

; 4©

b w bp x ps c s

.

23 Jei tc− xp = 1, tai(

t xp c

)−1=

1©(−t −x−p −c

); 2©

(c −p−x t

); 3©

(c px t

); 4©

(c xp t

); 5©

(c −x−p t

).


9y1 −14y2 = 732y1 −11y2 = 32

.

24 y1 = 1© −10; 2© 8; 3© 5; 4© −7; 5© −6; 6© 7.

25 y2 = 1© 3; 2© −2; 3© −1; 4© 9; 5© −3; 6© −7.


r1 +10r2 +12r3 = 3514r1 +11r2 +7r3 = −8916r1 −5r2 +10r3 = −37

.

26 r1 = 1© 6; 2© −4; 3© 9; 4© 3; 5© 4; 6© −7.

27 r2 = 1© −9; 2© −3; 3© 5; 4© −1; 5© −7; 6© 3.

28 r3 = 1© 6; 2© −3; 3© −10; 4© 10; 5© 1; 6© −4.

29 Apskaičiuokite

∣∣∣∣∣∣2 6 −63 −4 −65 −3 4

∣∣∣∣∣∣.1© 10; 2© 1; 3© −277; 4© −226; 5© −386; 6© 218.

30 Matricos

s v hf y tx r e

adjunktas A12 =

1© tx− fe ;2© se− tv ;3© fe− tx ;4© se− tx .

31 Matricos(

17 −8794 29

)adjunktas A11 =

1© 17 ; 2© 94 ; 3© −87 ; 4© −94 ; 5© 87 ; 6© −29 ; 7© −17 ; 8© 29 .


5x1 − 3x2 + 5x3 = 4

−3x1 − 2x2 + 3x3 = 1

−4x1 + x2 − 2x3 = 1

Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© −4 ; 2© 8 ; 3© −17 ; 4© −8 ; 5© 4 .

33 D1 = 1© 4 ; 2© −68 ; 3© 18 ; 4© −66 ; 5© 47 .

34 D2 = 1© 72 ; 2© 76 ; 3© −92 ; 4© −71 ; 5© 19 .

35 D3 = 1© −56 ; 2© 47 ; 3© 8 ; 4© −35 ; 5© 63 .

36 x1 + x2 + x3 = 1© −36 ; 2© 134 ; 3© 4 ; 4© 65

4 ; 5© 14 .

37 Matricos A =

−5 −2 −14 −5 33 −1 3


1© 27 ;2© 55 ;3© −98 ;4© −73 ;5© −43 ;6© −80 .

38 Adjunktas A22 = 1© 22 ; 2© −4 ; 3© −12 ; 4© 3 .

39 Adjunktas A12 = 1© −3 ; 2© 4 ; 3© −2 ; 4© −7 .

40 Adjunktas A33 = 1© 33 ; 2© 17 ; 3© −5 ; 4© 13 .

41 Adjunktas A23 = 1© −8 ; 2© 2 ; 3© 15 ; 4© −11 .

42 A−1 = 1©

− 11144

1372 − 1

161372

136 − 1

8− 1

24 − 112 − 1

8

; 2©

− 1255

755 − 1

5− 3

55 − 1255

15

15 − 1

535

; 3©

− 1255 − 3

5515

755 − 12

55 − 15

− 15

15

35

.


variantas009

Funkcija u(x, y) apibrėžta formule u(x, y) = −5x3y4 + 4 cos (−3x+ 5y).

1 ∂u

∂x=

1© −15x2y4+4 sin (−3x+5y); 2© −15x2y4−4 sin (−3x+5y);3© −15x4y4+12 cos (−3x+5y); 4© −15x4y4−12 cos (−3x+5y);5© −15x2y4−12 sin (−3x+5y); 6© −15x2y4+12 sin (−3x+5y).

2 ∂u

∂y=

1© −20x3y3−20 sin (−3x+5y); 2© −20x3y3−4 sin (−3x+5y);3© −20x3y5−20 cos (−3x+5y); 4© −20x3y3+20 sin (−3x+5y);5© −20x3y5+20 cos (−3x+5y); 6© −20x3y3+4 sin (−3x+5y).

v(x) =26x2 − 45

49x

3 v′(5) = 1© − 2649 ; 2© 139

245 ; 3© − 139245 ; 4© − 121

245 ; 5© 0; 6© 121245 ; 7© 139

49 ; 8© 2649 .

4 v′′(3) = 1© 5147 ; 2© − 5

147 ; 3© − 10147 ; 4© 10

147 ; 5© 1049 ; 6© − 90

49 ; 7© − 1049 ; 8© 0.

5 limx→0

x9 lnx =1© ∞; 2© 9;3© 0; 4© riba neegzistuoja;5© 1; 6© ln 9.

6 limx→∞

x14

16x =1© 1

16 ; 2© 16;3© riba neegzistuoja; 4© 14;5© 7

8 ; 6© 0.



8 f ′(4) = 1© 382; 2© 1022; 3© 254; 4© 94; 5© 62; 6© 1534.


(4, f(4)

), lygtį.

1© y=−94x+496; 2© y=94x+496;3© y=−94x−256; 4© y=94x−256;5© y=−120x−94; 6© y=120x+94.

Tarkime, kad y(x) = 7√1 + 2x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© y(x)≈1−2x; 2© y(x)≈ 2

7x;3© y(x)≈1− 2

7x; 4© y(x)≈1+ x7 ;

5© y(x)≈− 27x; 6© y(x)≈1− x

7 ;7© y(x)≈1+ 2

7x; 8© y(x)≈1+2x.

11 y(

110

)≈ 1© 36

35 ; 2© 135 ; 3© 34

35 ; 4© 1110 ; 5© − 1

35 ; 6© 6970 ; 7© 9

10 ; 8© 7170 .

f(x) = 2x3 + 3x2 − 120x+ 29

12 maxx∈[−4,2]

f(x) = 1© −200; 2© −275; 3© 454; 4© 429; 5© 431; 6© 446; 7© −183.

13 minx∈[−4,2]

f(x) = 1© −200; 2© −275; 3© 429; 4© −183; 5© 446; 6© 431; 7© 454.

14 maxx∈[−4,2]

|f(x)| = 1© 183; 2© 446; 3© 269; 4© 429; 5© 454; 6© 200; 7© 431; 8© 275.

f(x) = 8 sin(5x)− 57x2

15 f ′′(0) =1© −40; 2© −57; 3© 114; 4© −114; 5© 40; 6© 97; 7© 57; 8© −97; 9© 0.

16 f ′′′(0) =1© −200; 2© −114; 3© −1000; 4© −314; 5© 314; 6© 200; 7© 1000; 8© 0; 9© 114.


7x− 21grafiko asimptotė, kai x→∞, yra tiesė y = lx+ d.

17 l = 1© 12 ; 2© 21

11 ; 3© − 1121 ; 4© − 21

11 ; 5© − 12 ; 6© 2; 7© −2; 8© 11

21 .

18 d = 1© − 537 ; 2© 53

7 ; 3© 7724 ; 4© − 24

77 ; 5© 2477 ; 6© − 77

24 ; 7© − 753 ; 8© 7

53 .


2


1© (0;2√6); 2©

(− 1

2√

6; 12√

6

); 3© (−2

√6;2√6);

4© (−∞;0)∪(0;+∞); 5©(−∞;− 1

2√

6

)∪(

12√

6;+∞

); 6© {0};

7© (−∞;0); 8© (−2√6;0); 9© (0;+∞);

0© (−∞;+∞).

20 (sin4

(5x6))′

=

1© −116x5 sin3(5x6) cos(5x6); 2© 116x5 sin3(5x6) cos(5x6);3© 120x5 sin3(5x6) cos(5x6); 4© 116x7 sin5(5x6) cos(5x6);5© −120x7 sin5(5x6) cos(5x6); 6© 120x7 sin5(5x6) cos(5x6);7© −116x7 sin5(5x6) cos(5x6); 8© −120x5 sin3(5x6) cos(5x6).

21(

5√3 + ln (x+ 6)

)′=

1© 5(x+6)5√

(3+ln(x+6))4; 2© 1

5(x+6) 5√

3+ln(x+6);

3© 1

5(x+6)5√

(3+ln(x+6))4; 4© 1

5 5√

3+ln(x+6);

5© 5√

(3+ln(x+6))4

5(x+6); 6© 1

5√

3+ln(x+6);

7© 15√

(3+ln(x+6))4; 8© 5

5√

3+ln(x+6).

22 w r

u yd a

· ( 1 0 10 1 0

)=

1©

r w ry u ya d a

; 2©

w r wd u da d a

; 3©

w r wu y ud a d

; 4©

w a wd r dw d a

.

23 Jei ct− qs = 1, tai(

c qs t

)−1=

1©(

t sq c

); 2©

(t qs c

); 3©

(−c −q−s −t

); 4©

(t −q−s c

); 5©

(t −s−q c

).


15h1 −6h2 = −12−10h1 −13h2 = 144

.

24 h1 = 1© 2; 2© 3; 3© −4; 4© 8; 5© −6; 6© 5.

25 h2 = 1© −7; 2© 10; 3© 8; 4© −2; 5© 0; 6© −8.


x1 +6x2 −12x3 = 9216x1 +3x2 +17x3 = −1534x1 −15x2 −14x3 = 223

.

26 x1 = 1© 7; 2© −10; 3© 1; 4© 8; 5© 2; 6© 0.

27 x2 = 1© 8; 2© −9; 3© −4; 4© 6; 5© −5; 6© 1.

28 x3 = 1© −3; 2© −4; 3© −9; 4© −5; 5© 2; 6© −10.

29 Apskaičiuokite

∣∣∣∣∣∣−1 1 43 6 52 −5 1

∣∣∣∣∣∣.1© −132; 2© 61; 3© 5; 4© 173; 5© 229; 6© 4.

30 Matricos

f p et c xr a v

adjunktas A12 =

1© xr − tv ;2© tv − xr ;3© fv − xr ;4© fv − xp .

31 Matricos(

54 −4644 −11

)adjunktas A11 =

1© −11 ; 2© 54 ; 3© −44 ; 4© 46 ; 5© −54 ; 6© 11 ; 7© −46 ; 8© 44 .


−3x1 − x2 + x3 = 4

−x1 − x2 + 3x3 = 5

−3x1 + 2x2 − 2x3 = −3Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© 18 ; 2© 34 ; 3© 20 ; 4© −18 ; 5© −20 .

33 D1 = 1© 44 ; 2© −89 ; 3© −51 ; 4© −53 ; 5© −10 .

34 D2 = 1© 73 ; 2© 45 ; 3© −97 ; 4© −79 ; 5© −23 .

35 D3 = 1© 19 ; 2© −49 ; 3© 98 ; 4© −50 ; 5© −56 .

36 x1 + x2 + x3 = 1© 6118 ; 2© − 7

9 ; 3© 299 ; 4© − 13

18 ; 5© − 5918 .

37 Matricos A =

2 −3 5−3 −1 −24 4 5


1© −80 ;2© −52 ;3© −55 ;4© 55 ;5© 113 ;6© 46 .

38 Adjunktas A11 = 1© 2 ; 2© 3 ; 3© −11 ; 4© −2 .

39 Adjunktas A22 = 1© 14 ; 2© −6 ; 3© −10 ; 4© −17 .

40 Adjunktas A23 = 1© 17 ; 2© −20 ; 3© −3 ; 4© 29 .

41 Adjunktas A33 = 1© −5 ; 2© −6 ; 3© −11 ; 4© 7 .

42 A−1 = 1©

112 − 7

1849

16 − 4

929

14

16 − 1

3

; 2©

− 355 − 7

11 − 15

− 755

211

15

855

411

15

; 3©

− 355 − 7

55855

− 711

211

411

− 15

15

15

.


variantas010

Funkcija v(x, y) apibrėžta formule v(x, y) = 2x5y6 + 2 cos (2x− 4y).

1 ∂v

∂x=

1© 10x6y6+4 cos (2x−4y); 2© 10x4y6+4 sin (2x−4y);3© 10x4y6+2 sin (2x−4y); 4© 10x6y6−4 cos (2x−4y);5© 10x4y6−2 sin (2x−4y); 6© 10x4y6−4 sin (2x−4y).

2 ∂v

∂y=

1© 12x5y5+8 sin (2x−4y); 2© 12x5y7+8 cos (2x−4y);3© 12x5y5+2 sin (2x−4y); 4© 12x5y5−2 sin (2x−4y);5© 12x5y7−8 cos (2x−4y); 6© 12x5y5−8 sin (2x−4y).

y(x) =3x2 − 29

47x

3 y′(4) = 1© 77752 ; 2© 19

752 ; 3© 0; 4© 77188 ; 5© − 3

47 ; 6© 347 ; 7© − 19

752 ; 8© − 77752 .

4 y′′(3) = 1© − 5847 ; 2© 58

423 ; 3© − 581269 ; 4© 58

1269 ; 5© 0; 6© − 291269 ; 7© − 58

423 ; 8© 291269 .

5 limx→0

x12 lnx =1© ∞; 2© 12;3© ln 12; 4© 0;5© riba neegzistuoja; 6© 1.

6 limx→∞

x21

18x =1© 1

18 ; 2© riba neegzistuoja;3© 18; 4© 0;5© 21; 6© 7

6 .



8 f ′(3) = 1© 4858; 2© 1618; 3© 14578; 4© 3886; 5© 1294; 6© 11662.


(3, f(3)

), lygtį.

1© y=1618x−3888; 2© y=−1618x+5820;3© y=966x+1618; 4© y=1618x+5820;5© y=−966x−1618; 6© y=−1618x−3888.

Tarkime, kad v(x) = 5√1 + 2x.

10 Taikydami diferencialą užrašykite apytikslę formulę, kai x→ 0.1© v(x)≈1− 2

5x; 2© v(x)≈− 25x;

3© v(x)≈1+ x5 ; 4© v(x)≈1−2x;

5© v(x)≈1− x5 ; 6© v(x)≈1+2x;

7© v(x)≈1+ 25x; 8© v(x)≈ 2

5x.

11 v(− 1

8

)≈ 1© 39

40 ; 2© 78 ; 3© 9

8 ; 4© − 120 ; 5© 19

20 ; 6© 120 ; 7© 21

20 ; 8© 4140 .

f(x) = 2x3 + 6x2 − 90x+ 50

12 maxx∈[−3,−1]

f(x) = 1© −112; 2© 124; 3© 144; 4© 400; 5© 320; 6© 330; 7© 325.

13 minx∈[−3,−1]

f(x) = 1© 124; 2© 144; 3© 400; 4© −112; 5© 325; 6© 320; 7© 330.

14 maxx∈[−3,−1]

|f(x)| = 1© 325; 2© 124; 3© 400; 4© 477; 5© 144; 6© 320; 7© 112; 8© 330.

f(x) = 46 sin(4x)− 70x2

15 f ′′(0) =1© −184; 2© 254; 3© 140; 4© −70; 5© −140; 6© −254; 7© 70; 8© 0; 9© 184.

16 f ′′′(0) =1© 2944; 2© −736; 3© 140; 4© 736; 5© 876; 6© 0; 7© −140; 8© −2944; 9© −876.


16x+ 25grafiko asimptotė, kai x→∞, yra tiesė y = kx+ l.

17 k = 1© 85 ; 2© 5

3 ; 3© − 58 ; 4© − 3

5 ; 5© − 53 ; 6© − 8

5 ; 7© 58 ; 8© 3

5 .

18 l = 1© 1285 ; 2© 206

191 ; 3© 5128 ; 4© − 128

5 ; 5© − 5128 ; 6© 191

206 ; 7© − 191206 ; 8© − 206

191 .



1© (− 12 ;

12 ); 2© (−∞;− 1

2 )∪(12 ;+∞); 3© (0;+∞);

4© (−2;0); 5© (−2;2); 6© (0;2);7© (−∞;0)∪(0;+∞); 8© (−∞;+∞); 9© (−∞;0);0© {0}.

20 (sin9

(4x3))′

=

1© −108x2 sin8(4x3) cos(4x3); 2© −100x2 sin8(4x3) cos(4x3);3© 108x2 sin8(4x3) cos(4x3); 4© 100x4 sin10(4x3) cos(4x3);5© 108x4 sin10(4x3) cos(4x3); 6© 100x2 sin8(4x3) cos(4x3);7© −100x4 sin10(4x3) cos(4x3); 8© −108x4 sin10(4x3) cos(4x3).

21(

5√10 + ln (x+ 8)

)′=

1© 1

5(x+8) 5√

10+ln(x+8); 2© 1

5(x+8)5√

(10+ln(x+8))4;

3© 1

5 5√

10+ln(x+8); 4© 5(x+8)

5√

(10+ln(x+8))4;

5© 55√

10+ln(x+8); 6© 1

5√

10+ln(x+8);

7© 5√

(10+ln(x+8))4

5(x+8); 8© 1

5√

(10+ln(x+8))4.

22 d z

t vx a

· ( 1 0 10 1 0

)=

1©

d a dx z xd x a

; 2©

z d zv t va x a

; 3©

d z dt v tx a x

; 4©

d z dx t xa x a

.

23 Jei ac− pr = 1, tai(

a pr c

)−1=

1©(

c rp a

); 2©

(c pr a

); 3©

(c −r−p a

); 4©

(−a −p−r −c

); 5©

(c −p−r a

).


7z1 +12z2 = −25−16z1 +9z2 = 130

.

24 z1 = 1© −7; 2© 9; 3© −6; 4© 2; 5© −2; 6© 6.

25 z2 = 1© −8; 2© 5; 3© −3; 4© −10; 5© −6; 6© 2.


x1 −12x2 +8x3 = −1622x1 +7x2 +11x3 = −15

−10x1 +13x2 +2x3 = 165.

26 x1 = 1© 10; 2© −1; 3© −7; 4© −8; 5© −6; 6© 8.

27 x2 = 1© 1; 2© 0; 3© 4; 4© 3; 5© 10; 6© 9.

28 x3 = 1© −6; 2© 7; 3© 8; 4© −5; 5© 2; 6© −1.

29 Apskaičiuokite

∣∣∣∣∣∣−2 −5 31 5 63 5 −2

∣∣∣∣∣∣.1© −48; 2© −50; 3© 42; 4© 97; 5© −5; 6© 0.

30 Matricos

q p us h at v z

adjunktas A32 =

1© qz − ap ;2© qa− us ;3© qz − at ;4© us− qa .

31 Matricos(−90 −23−59 −60

)adjunktas A12 =

1© 90 ; 2© −59 ; 3© −23 ; 4© −60 ; 5© 23 ; 6© 60 ; 7© 59 ; 8© −90 .


2x1 + 4x2 + 2x3 = 2

5x1 + 3x2 − 3x3 = 5

5x1 + 2x2 − x3 = −5Kramerio metodu.

Pažymėkime:


x1

x2

x3

=

(D1

D

D2

D

D3

D

)T


32 D = 1© −19 ; 2© 37 ; 3© 44 ; 4© −37 ; 5© −44 .

33 D1 = 1© 11 ; 2© 79 ; 3© 98 ; 4© −48 ; 5© 136 .

34 D2 = 1© −160 ; 2© −12 ; 3© 71 ; 4© 36 ; 5© 90 .

35 D3 = 1© 68 ; 2© 94 ; 3© −34 ; 4© 140 ; 5© 98 .

36 x1 + x2 + x3 = 1© − 94 ; 2© 83

44 ; 3© 711 ; 4© − 29

11 ; 5© 6944 .

37 Matricos A =

1 −1 3−4 1 2−5 −2 −5


1© 86 ;2© −142 ;3© −22 ;4© 22 ;5© 68 ;6© −57 .

38 Adjunktas A21 = 1© 20 ; 2© −11 ; 3© −24 ; 4© 26 .

39 Adjunktas A11 = 1© −1 ; 2© −3 ; 3© −2 ; 4© 5 .

40 Adjunktas A22 = 1© −20 ; 2© 10 ; 3© 14 ; 4© 5 .

41 Adjunktas A13 = 1© 20 ; 2© −27 ; 3© 29 ; 4© 13 .

42 A−1 = 1©

− 738 − 9

38 − 1776

119

419

819

− 338

738

976

; 2©

− 168 − 11

68 − 568

− 1534

534 − 7

341368

768 − 3

68

; 3©

− 168 − 15

341368

− 1168

534

768

− 568 − 7

34 − 368

.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 210 5 3 7 5 4 4 2 4 3 8 4 4 7 2 1 4 5 8 3 2 71 1 6 6 6 4 4 2 5 4 1 1 3 7 1 7 6 1 8 5 1 62 2 3 7 6 4 4 4 1 3 2 1 6 7 5 5 2 6 8 0 5 73 5 2 1 2 1 6 2 4 4 5 1 1 6 8 5 5 1 6 3 7 84 6 5 7 8 2 2 3 2 4 6 7 3 4 7 8 9 2 2 3 1 85 1 3 3 7 2 5 4 5 2 5 8 6 4 5 8 7 5 6 9 6 46 5 1 4 7 1 1 5 3 1 4 2 6 2 6 8 9 1 1 2 1 37 6 3 2 6 1 6 6 5 3 6 2 7 1 2 7 4 1 6 5 7 38 5 3 6 8 5 1 5 6 6 3 4 2 5 4 9 1 5 8 2 8 39 6 1 2 3 3 6 6 4 4 7 1 4 4 4 4 3 7 1 2 3 3

10 6 1 1 3 4 4 4 2 1 7 5 5 2 6 5 8 3 3 1 3 2

22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 420 3 5 5 2 6 4 4 6 2 4 4 3 2 2 5 1 2 2 4 2 31 3 2 2 3 5 5 6 3 4 4 2 3 5 5 2 3 2 3 2 3 22 1 4 2 3 5 4 4 3 3 3 4 3 5 2 4 1 2 1 4 1 13 3 3 2 5 2 4 4 5 1 8 4 1 1 1 4 2 4 1 4 3 34 3 3 5 6 1 1 4 1 3 2 2 5 4 3 2 3 4 2 1 2 15 1 5 5 4 2 1 6 1 4 4 1 3 2 5 3 1 3 4 3 2 36 4 2 1 1 1 3 5 6 3 6 1 3 1 1 3 6 2 1 3 1 27 1 2 6 2 6 2 2 1 1 8 1 5 1 2 1 5 1 1 3 4 18 2 5 3 2 6 2 1 5 1 8 5 1 3 1 1 2 3 1 1 4 29 3 4 3 6 5 5 6 1 1 1 1 5 5 1 2 3 2 3 2 3 2

10 3 5 1 6 5 6 1 2 4 7 5 5 1 4 4 5 2 1 2 4 2

3603 pavyzdys 1 x y x y 2 x y x y 3 4 1 2 1 2 1 2 · taikomoji matematika. pvz. antram testui...

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