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• 24

QUESTION PAPER CODE 65/3/RU

SECTION - A

1. ycotx.y1

2ydydx

2 m

Integrating factor = 2y1log y1ore 2 m

2.333

yxzzyxzyxzyx

m

= 0 m

3. order 2, degree 1 (any one correct) m

sum = 3 m

4.10

110

3z

151

157y

51

53x

m

Direction cosines are 73,

72,

76

or 73,

72,

76

m

5. a2ccbba22cba2cba2 2222 m

6cba2 m

6. ji011111kji

b a

m

unit vector is 2j

2i m

Marks

• 25

SECTION - B

7. k5j5ibc;kjiba 1 m

k4j4551

111kji

bcba

1 m

Unit vector perpendicular to both of the vectors = 2k

2j 1 m

8. let the equation of line passing through (1, 2, 4) be

kcjbiak4j2ir 1 mSince the line is perpendicular to the two given lines

3a 16 b + 7 c = 0 1 m 3a + 8 b 5 c = 0

Solving we get, 6c

3b

2aor

72c

36b

24a

1 m

Equation of line is : k6j3i2k4j2ir mOR

Equation of plane is : 011112212

z2y1x

3 m

Solving we get, x + 2y + 3z 3 = 0 1 m

9. Let x = No. of spades in three cards drawn

x : 0 1 2 3 1 m

P(x) :

6427

433

3

C0

64

274

34

132

C1

64

94

34

132

C2

64

14

34

1303

C3

2 m

x . P(x) : 0 6427

6418

643 m

• 26

Mean = 43

6448P(x)x m

OR

let p = probability of success ; q = Probability of failure

then, 9 P(x = 4) = P(x = 2)

422624

46 qpCqpC9 2 m

3pqq9p 22 1 m

Also, p + q = 1 p + 3p = 1 p = 41

1 m

10. 2xsinm

xsinm

x1em

dxdyget wex, w.r.t.atedifferenti,ey

11

1 m

dxdyx1 2 = my, Differentiate again w.r.t. x

dxdym

dxdy

x1x

dxydx1

22

22 1 m

(my)mdxdyx1m

dxdyx

dxydx1 22

22

m

0ymdxdyx

dxydx1 22

22 m

11. f (x) = 32x(x)h,1x

1x(x)g,1x 22

Differentiating w.r.t. x, we get

2(x)h,

1xx2x1(x)g,

1xx(x)f 22

2

2

1+1+1 m

5

2(x)ghf m

• 27

12.

dxxx22x1dx

21x

232dxxx22x3 2

222 2 m

cxx232

23

21x

sin89xx2

221x

2 23212

2 m

or

cxx2

32

312xsin

49xx2

212x 23212

OR

dx

2x1

53dx

1x12x

51dx

2x1x1xx

22

2

2 m

dx2x1

53dx

1x1

51dx

1x2x

51

22 m

c2xlog53xtan

511xlog

51 12 1 m

13. 4

04

4

03

dxtan x2xcos

1dx x2sin 2xcos

11 m

=

4

0

22

dxxsecxtan2xtan1

1 m

= 1

0

2

dttt1

21

Taking, tan x = t; 1 m

= 1

0

25

t52t2

21

m

= 56

522

21

m

• 28

14. dx1x1

x1

1x1xlogdx

1x1xlog 2 2 m

= dx1x1dx1

1xxlog

x 1 m

= c1)(xlogxlog1xxlog

1 m

or c1xxlog

1xxlog

15.

390002300030000

402050

15040050075250300

1003004002 m

cost incurred respectively for three villages is Rs. 30,000, Rs. 23,000, Rs. 39,000 1 m

One value : Women welfare or Any other relevant value 1 m

16.

318tan

1x1x11 x 1 xtan 11 2 m

318

x22x

2 4x2 + 31x 8 = 0 1 m

,41x 8 (Rejected) 1 m

OR

L.H.S. =

zx1

xztanyz1zytan

xy1yxtan 111 2 m

RHS0xtanztanztanytanytanxtan 111111

2 m

• 29

17.ccbbabba

cacaabc

cbcbabacbaba

cacbca

22

22

22

Taking a, b & c common from C1 , C

2 and C

31 m

ccbcbabba

caccaabc2

3211 C CCC and taking 2 common from C1 1 m

bcbcbbbba

0ccaabc2

133 CCC 1 m

bcbcb0cca0cca

abc2

322 RRR m

Expand by C3, = 2 abc ( b) ( ac c2 ac + c2) = 4a2 b2 c2 m

663

2+1 m

100010001

27366636

663

122212221

1 m

19. f (x) = 1x1x

L

2

1x1x2lim

1x21x1xlim(1)f

1)(x1)(x

1 m

• 30

R 01x0lim

1x21x1xlim(1)f

1)(x1)(x

1 m

(x)f02 is not differentiable at x = 1

L 01x

0lim1x

21x1xlim(1)f 1x1x

1 m

R 21x1x2lim

1x21x1xlim(1)f

1x1x

1 m

(x)f20 is not differentiable at x = 1

SECTION - C

20. correct figure 1 m

ar (ABDOA) 3

1612

dy41

4

0

34

0

2

yy ......(i) 1 m

ar (OEBDO) 4

0

32

34

0

24

0 12xx

34dx

4xdxx2

316

316

332

.........(ii) 1 m

ar (OEBCO) 3

1612xdxx

41

4

0

34

0

2

................(iii) 1 m

From (i), (ii) and (iii) we get ar (ABDOA) = ar (OEBDO) = ar (OEBCO)

21.1

xy

xy

dxdy

xxyy

dxdy

2

2

2

, Hence the differential equation is homogeneous 1 m

Put y = 1dx

dxgetwe,dxdx

dxdyandx

2

vvvvvvv 1+1 m

• 31

1

1dxdx

2

vvv

vvv

1 m

cxloglogdxx1d1 vvvv

v1 m

cylog

xyor,cxlog

xylog

xy

1 m

OR

Given differential equation can be written as 21

2 y1ytanx

y11

dydx

1 m

Integrating factor =

dyy1eytanex:issolutionande 2

ytan1ytanytan

111

1+1 m

x t)ytan(wherec1)y(tanecetedttee 11ytantttytan11

1 m

x = 1, y = 0 c = 2 21)y(taneex 1ytanytan11

1 m

ytan1 1e21ytanxor

22. E1 : Bolt is manufactured by machine A

E2 : Bolt is manufactured by machine B

E3 : Bolt is manufactured by machine C

A : Bolt is defective

;10020)P(E;

10050)P(E;

10030)P(E 321

1001)P(A/E;

1004)P(A/E;

1003)P(A/E 321 3 m

• 32

P(E2/A) = 3120

2020090200

1001

10020

1004

10050

1003

10030

1004

10005

2 m

P( E2/A) = 1 P(E2/A) = 3111

1 m

23. Equation of line through A and B is (say)6

5z1

4y13x

2 m

General point on the line is 564,3, 1 m

If this is the point of intersection with plane 2x + y + z = 7

then, 2756432 1 m

Point of intersection is (1, 2, 7) 1 m

Required distance = 7742413 222 1 m

24. Let the two factories I and II be in operation for x and y

days respectively to produce the order with the minimum cost

then, the LPP is :

Minimise cost : z = 12000 x + 15000 y 1 m

Subject to :

50x + 40y > 6400 or 5x + 4y > 640

50x + 20y > 4000 or 5x + 2y > 400 2 m

30x + 40y > 4800 or 3x + 4y > 480

x, y > 0

• 33

correct graph 2 m

Vertices are A (0, 200) ; B (32, 120)

C (80, 60) ; D (160, 0) m

z (A) = Rs. 30,00,000; z (B) = Rs. 21,84,000;

z (C) = Rs. 18,60,000 (Min.); z (D) = Rs. 19,20,000;

On plotting z < 1860000

or 12x + 15y < 1860, we get no

point common to the feasible region

Factory I operates for 80 days m

Factory II operates for 60 days

25. f : 5

35y54(y)f;96x5xf(x);)9,R 12

y95

35y546

535y54

5(y)fof2

1

3 m

x

5396x5x554

(x)f of2

1

2 m

Hence f is invertible with 5

35y54(y)f 1

m

OR

(i) commutative : let x, y then1R

x * y = x + y + xy = y + x + yx = y * x * is commutative 1 m

(ii) Associative : let x, y, z then1R

x * (y * z) = x * (y + z + yz) = x + (y + z + yz) + x (y + z + yz)

• 34

= x + y + z + xy + yz + zx + xyz 1 m

(x * y) * z = (x + y + xy) * z = (x + y + xy) + z + (x + y + xy) . z

= x + y + z + xy + yz + zx + xyz 1 m

x * (y * z) = (x * y) * z * is Associative

(iii) Identity Element : let 1Raaa*ee*athatsuch1Re m

a + e + ae = a e = 0 m

(iv) Inverse : let a * b = b * a = e = 0 ; a, b 1R m

a + b + ab = 0 a1aaor

a1ab 1

m

26. Solving the two curves to get the points of intersection 8,p3 1 m

m1 = slope of tangent to first curve = 9p2x

1 m

m2 = slope of tangent to second curve = p2x

1 m

curves cut at right angle iff 1p

2x9p2x

m

)p3x(Put4x9p 22

p)(949p2

p = 0 ; p = 4 1 m