?· z x y x y z x y z x y z

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  • 24

    QUESTION PAPER CODE 65/3/RU

    EXPECTED ANSWERS/VALUE POINTS

    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

    18. Adj A = 27A;366636

    663

    2+1 m

    A. Adj A = 3IA

    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