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  • 8/12/2019 class XII sample paper CBSE

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    MATHEMATICSClass: XII

    Sample Paper -5

    Time Allowed : 3 Hrs Maximum Marks: 100

    _______________________________________________________________1. All questions are compulsory.2. The question paper consist of 29 questions divided into three sections A, B

    and C. Section A comprises of 10 questions of one mark each, section Bcomprises of 12 questions of four marks each and section C comprises of 07questions of six marks each.

    3. All questions in Section A are to be answered in one word, one sentence oras per the exact requirement of the question.

    4. There is no overall choice. However, internal choice has been provided in 04questions of four marks each and 02 questions of six marks each. You haveto attempt only one of the alternatives in all such questions.

    5. Use of calculators is not permitted.

    SECTION A

    1.

    Evaluate 11

    sin sin3 2

    2.This graph does not represent a function . Make the required changes inthis graph, and draw the graph , so that it does represent a function.

    3. For what value of, are the vectors $ $2a i j k = + +r

    $ and $ $2 3b i j k = +r

    $

    orthogonal?

    4. Find the value of for which ( + +k) is a unit vector.

    5.Find the Obtuse angle of inclination to the Z-axis of a line that is inclined toX-axis at 45oand to Y-axis at 60o.

    6.Find the slope of the tangent to the curve 3 1y x x= + at the point where the

    curve cuts y-axis.

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    7. Evaluate:1

    1

    2 xlog dx

    2 x

    +

    .

    8.This 3 x 2 matrix gives information about the number of men and womenworkers in three factories I, II and III who lost their jobs in the last 2

    months. What do you infer from the entry in third row and second column ofthis matrix?

    Men workers Women workers

    Factory I 40 15

    Factory II 35 40

    Factory III 72 64

    9. If A and B are two matrices of the same order ,under what conditions is (A-B)(A+B) = A2- B2

    x sin cos

    10.Evaluate the determinant : sin 1

    cos 1

    x

    x

    SECTION B

    ( )

    3

    5 3

    x 511. Differentiate , w . r .t x

    7 3 8 5

    x

    x x

    +

    +

    OR

    2 '' '11.If y = a cos (log x) + b sin (log x) , prove that x 0 ,y xy y+ + =

    12. ( ) 3,g(x) = x -3; ,

    Show that (i) f is not an onto function (ii) gof is an onto function

    = + Let f x x x N

    $ $( ) $ $( )

    13.Find the distance between the parallel planes

    . 2 1 3 4and . 6 3 9 13 0r i j k r i j k + = + + =r r

    $

    2 2 2 2

    14. plane is at a distance of p units from the origin.

    It makes an intercept of a,b,c with the x , y and z axis repectively.Show that it satisfies the equation:

    1 1 1 1

    a+ + =

    A

    b c p

    15. Four cards are drawn successively with replacement from a well shuffled

    deck of 52 cards . What is the probability that

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    (i)

    All the four cards are spades?(ii)

    Only 3 cards are spades(iii) None is a spade

    21 2 1 2 516.Solve the equation:(tan ) (cot )

    8x x

    + =

    17. Find the equation of tangent to the curve given by3 3sin , cosx a t y b t= = at

    a point, where2

    t = .

    18.Evaluate:4

    0

    sinx cos xdx

    9 16 sin2x

    +

    +

    12-3x

    0

    19.Evaluate e dx as limit of sum

    20.Ifcos sinsin cos

    A

    = then prove that cos sin ,

    sin cos

    n n nA n Nn n

    =

    OR

    2

    2

    2

    is one of the cube roots of unity , evaluate the given determinant

    1

    1

    1

    =

    If

    21. Show that the function f defined by f(x) = 1-x + x , x R is continous .OR

    Show that a logarithmic function is continuous at every point in its domain.

    2

    (3 sin 2) cos22.Evaluate

    5 4sind

    cos

    OR

    11

    32

    dxEvaluate

    x x+

    Section C

    23.Show that the right circular cone of least curved surface and given volume

    has an altitude equal to 2 times the radius of the base.OR

    4 3.Find the points at which the function f given by f(x) = (x - 2 ) ( 1)

    (i)local maxima (ii) local minima (iii) point of inflexion .

    Also find the (iv)local maximum value and the (v) local m

    +x has

    inimum value

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    24. Obtain the inverse of the following matrix using elementary operations.

    0 1 2

    A = 1 2 3

    3 1 1

    2 2

    2 2

    2 2

    2 2

    x y25. Calculate the area (i) between the curves + 1,and the x-axis between x = 0 to x =a

    a b

    x y(ii) Triangle AOB is in the first quadrant of the ellipse + 1,where OA = a and OB = b.

    a b

    Fi

    =

    =

    nd the area enclosed between the chord AB and the arc AB of the ellipse

    (iii) Find the ratio of the two areas found .

    OR2 2 2 2Find the smaller of the two areas in which the circle x y 2a is divided by the parabola y ax+ = =

    $( ) $( )26.Find the equation of a plane that is parallel to the X-axis and passes through the lin

    common to two intersectiing planes r. i+j+k 1 0and r. 2i+3j-k 4 = = r r

    $ $ $

    27.In a bank, principal increases continuously at the rate of 5% per year. Inhow many years will Rs 1000 double itself?

    28. Two trainee carpenters A and B earn Rs 150 and Rs 200 per day respectively. A can

    make 6 frames and 4 stools per day while B can make 10 frames and 4 stools per day. many days shall each work, if it is desired to produce atleast 60 frames and 32 stools aminimum labour cost? Solve the problem graphically.

    29. In answering a question on a MCQ test with 4 choice per question, astudent knows the answer, guesses or copies the answer. Let be theprobability that he knows the answer be the probability that he guessesand that he copies it. Assuming that a student, who copies the answer, willbe correct with the probability , what is the probability that student knowsthe answer, given that he answered it correctly?

    Arjun does not know that answer to one of the questions in the test. Theevaluation process has negative marking. Which value would Arjun violate ifhe resorts of unfair means? How would an act like the above hamper hischaracter development in the coming years?

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    Solutions

    Section A

    1 11. sin sin sin3 2 3 6

    sin sin 1 (1 Mark)3 6 2

    =

    = + = =

    2. If a vertical line intersects the graph of a relation in two or more points,then the relation is nota function.

    Graph should have no vertical lines

    3.Two vectors anda br r

    are orthogonal/perpendicular, if a.b 0=r r

    $ $(2i j k).(i 2j 3k) 0

    2 2 3 0

    5

    2

    + + + =

    + =

    =

    $ $ $ $

    (1 Mark)

    4. Magnitude of : ( + + k )=1 , for it to be a unit vector.

    2 2 2 2.( i + j+k ) 3 3 1

    1

    3

    = + + = = =

    =

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    (1 mark)

    o o

    2 2 2 2

    o

    1 15.Let cos45 , m cos60

    22

    1 1 1 1

    Also, m n 1 n 1 n .2 4 4 21

    Let n cos2

    1For to be obtuse , we take the negative value of n

    2

    120 [1mark]

    = = = =

    + + =

    = =

    =

    = =

    =

    =

    l

    l

    3

    2

    2

    x 0x 0

    6.y x x 1

    dy3x 1

    dx

    At a point where the curve cuts y-axis, x = 0

    dy3x 1 1 (1 Mark)

    dx ==

    = +

    =

    = =

    1

    1

    1

    1

    2 x7.I log dx ...(1)

    2 x

    2 xLet f(x) log and

    2 x

    2 x 2 xf( x) log log f(x).

    2 x 2 xf(x) is an odd function.

    2 xThus, log dx 0 (

    2 x

    =

    +

    =

    +

    + = = =

    +

    =

    +

    1 Mark)

    8. 64 Women lost their jobs in the Factory III in the last two months. (1Mark)

    9. (A-B)(A+B) = A2

    +AB BA - B2

    == A2

    - B2

    iff AB = BA (1 Mark)

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    2

    3 2 2

    3 2 2 3 3

    x sin cos

    10. sin x 1 =x(-x -1 ) - sin ( x sin cos ) cos ( sin x cos )

    cos 1 x

    x x x sin sin cos sin cos xcos

    x x x(sin cos ) x x x x (1 Mark)

    + +

    = + + +

    = + + = + =

    Section B

    ( )

    ( ) ( ) ( )

    ( )

    ( ) ( )

    3

    5 3

    x 5 x11.Let y =

    7 3x 8 5x

    Taking log on both sides , we get

    1 1log y = 3 log x + log 5 x 5log 7 3x log 8 5x (1 Mark)

    2 31 dy 1 1 1 1 1 1

    . 3. . .1 5. 3 . .5y dx x 2 5 x 7 3x 3 8 5x

    1 dy 3 1 15 5. (1 M

    y dx x 2 5 x 7 3x 3 8 5x

    +

    +

    + +

    = + + +

    = + + + +

    ( ) ( )

    ( ) ( ) ( )

    3

    5 3

    ark)

    dy 3 1 15 5y

    dx x 2 5 x 7 3x 3 8 5x

    dy x 5 x 3 1 15 5 (2 Marks)

    dx x 2 5 x 7 3x 3 8 5x7 3x 8 5x

    = + +

    + +

    + = + +

    + + +

    OR

    ( ) ( )

    ( ) ( )'

    '' '

    y = a cos (log x) + b sin (log x)

    dy 1 1a sin logx . b. cos logx . (1 Mark)

    dx x x

    dyx xy a sin logx b. cos logx (1 Mark)

    dx

    Differentiating,

    xy y a cos

    = +

    = = +

    + = ( ) ( )

    ( ) ( )2 ''

    2 ''

    '

    '

    1 1logx . b. sin logx . (1 Mark)

    x xx y xy a cos logx b. sin logx y

    x y xy y 0 (1 Mark)

    +

    + = + =

    + + =

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    12.f(x) x 3,x N,

    Domainof f {1,2,3,...}

    Range {4,5,6,...} Codomain of f ={1,2,3,...}

    f is not an onto function (2 Marks)

    f(x)=x+3g(x) = x -3

    g[f(x)] x 3 -3

    = +

    =

    =

    = + =x

    Domainof gof {1,2,3,...}

    Range {1,2,3,4,5,6,...} Codomain of gof ={1,2,3,...}

    gof is an onto function (2 Marks)

    =

    = =

    $( )$( )

    $( )

    13. Distance between the parallel planes

    d-kisgiven by (1 Mark)

    n

    r. 6i 3j 9k 13 0

    13r. 2i j 3k (1 Mark)

    3

    the distance between the given parallel planes

    r. 2i 1j 3k

    + + =

    + =

    +

    r

    r$ $

    r$ $

    r$ $( )

    ( )22 2

    134 and r. 2i j 3k

    3

    134-

    3is (1 Mark)

    2 1 3

    13 254+ 3 253= (1 Mark)4 1 9 14 3 14

    = + =

    + +

    = =+ +

    r$ $

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    2 2

    14. The equation of the plane in the intercept form is

    x y z1

    a b c

    1 1 1x y z 1 0

    a b c

    Plane is at a distance of p units from the origin

    1 1 10 0 0 1

    a b c

    1 1 1

    a b

    + + =

    + + =

    + +

    + +

    2

    2 2 2

    2 2 2 2

    =p (2 Marks)

    c

    1 1 1 1

    p a b c

    1 1 1 1 (2 Marks)

    a b c p

    = + +

    + + =

    15.This is a case of bernoulli trials.

    13 1p = P(Success) = P(getting a spade in a single draw ) =

    52 4

    1 3q P(Failure) 1 p 1 (1 Mark)

    4 4

    (i)All the

    =

    = = = =

    4

    4 4 04

    4 3 13

    4

    4 0 40

    1 1four cards are spades =P(X =4) = C p q (1 Mark)

    4 256

    12 3(ii)Only 3 cards are spades=P(X=3)= C p q (1 Mark)

    256 643 81

    (iii)None is a spade=P(X=0) = C p q4 256

    = =

    = =

    = =

    (1 Mark)

    16.Here,2

    1 2 1 2 5(tan ) (cot )

    8x x

    + =

    21 1 2 1 1

    2 21 1

    2 21 1

    2 21 1

    5(tan x cot x) 2 tan x.cot x

    8

    52 tan x.cot x (1mark)

    2 8

    52 tan x.cot x4 8

    (2 5) 32 tan x.cot x (1mark)

    8 8

    + =

    =

    =

    = =

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    21 1

    21 1 2

    1 2 1 2

    32tan x. tan x 0

    2 8

    3tan x 2(tan x) 0

    8

    16(tan x) 8 tan x 3 0

    + =

    + =

    =

    (1 mark)

    2 21

    21

    1

    1

    8 64 4 16 3tan x

    2 16

    8 256 8 16tan x

    32 32

    3tan x and

    4 4

    x so tan x2 2 4

    x tan4

    x 1

    + =

    = =

    =

    < < =

    =

    =

    Q

    (1 mark)

    17.Here, 3 3sin , cosx a t y b t= = (1)

    Differentiating (1) w.r.t. t

    2dx

    3asin t.cos tdt = and

    2dy 3bcos t.sintdt

    = (1 Mark)

    2

    2

    dydy 3bcos t.sint bdt cott

    dxdx a3asin t.costdt

    = = =

    (1Mark)

    Slope of the tangent at2

    t =

    2

    dy bcot 0

    dx a 2

    = =

    ( 1 Mark)

    Hence, equation of tangent is given by

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    ( ) ( ) ( ) ( )1

    2-3. 0 2-3. 0+h 2-3. 0+2h 2-3. 0+(n-1)h2-3x

    h 00

    2 2-3h 2-6h 2-3.(n-1)hh 0

    h

    119. Here a =0, b=1 , b-a =1-0 = 1 h (1/2Mark)

    n

    e dx limh e e e ... e

    limh e e e ... e (1Mark)

    lim

    =

    = + + + +

    = + + + +

    =

    ( )

    ( )

    ( )

    ( )

    2 -3h -6h -3.(n-1)h

    0

    n-3h

    2

    -3hh 0

    -3

    2

    -3hh 0

    -3

    2

    -3hh 0

    -32

    -3hh 0

    he 1 e e ... e

    1 elimhe ,but nh = 1 (1Mark)

    1 e

    1 elimhe

    1 e

    1 e

    e lim 1 eh

    1 eelim

    -3 1 e

    3h

    + + + +

    =

    =

    =

    =

    ( )

    ( )

    2-3

    -3h3h 0

    2 2 1-3

    (1/2Mark)

    e 11 e lim

    -3 e 1

    3h

    e e e1 e (1Mark)-3 3

    =

    = =

    20: We shall prove by principle of mathematical induction

    Here, letcos sin

    sin cos

    n n n

    An n

    =

    ( )cos sin

    : ,sin cos

    n n n

    P n An n

    =

    ( mark )

    ( )

    1 cos sinSo, Asin cos

    1 is true.P

    =

    ( mark )Assuming result to be true for n = k i.e P(k) to be true

    ( )cos sin

    :sin cos

    k k k

    P k Ak k

    =

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    ( mark)

    We have to prove ( )1P k+ is true,

    ( ) 1 1

    1

    1 :

    cos sin cos sin

    sin cos sin cos

    k k

    k

    P k A A A

    k kA

    k k

    +

    +

    + =

    =

    (mark )

    ( ) ( )

    ( ) ( )

    ( ) ( )( ) ( )

    cos cosk sin sink cos sink sin cosk

    sin cosk cos sink sin sink cos cosk

    cos k sin k

    sin k cos k

    cos k 1 sin k 1sin k 1 cos k 1

    + =

    +

    + + =

    + +

    + + = + +

    ( )1P k + is true.

    (1mark )

    Thus by principle of mathematical inductioncos sin

    sin cos

    n n n

    An n

    =

    for all n N

    (1 mark )OR

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    2

    2

    2

    1 1 2 3

    2 2

    2 2

    2

    2

    1

    1

    1

    C C C C

    11 1 (1Mark)

    1 1

    But, is one of the cube roots of unity

    1 0

    =

    + +

    + + = + +

    + +

    + + =

    2

    2

    (1Mark)

    0

    0 1 (1mark)

    0 1

    Since I column of the determinant is zero therefore, value of the given determinant

    =

    21. f(x) = 1-x + x , x R

    Let g(x) =1-x+ x ,x R

    h(x) x ,x R

    h[g(x)]=h[1-x+ x]

    1-x + x f(x) (1Mark)

    1 x,being a polynomial function is continuous

    x ,being a modulus function is

    =

    = =

    continuous

    g(x) =1-x+ x ,x R is a continuous function (1Mark)

    Again , h(x) = x ,is a continuous function (1Mark)

    f(x)is continuous (1Mark)

    OR

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    ( )

    ( )

    ( )

    ( )

    x c x c

    h 0

    h 0

    h 0 h 0

    h 0

    Domainoflogx is(0, )

    f(x) logx,x (0, )

    Let c (0, )

    limf(x) limlogx

    limlog c h ,c 0 (1Mark)

    hlimlog c 1 ,c 0

    c

    hlimlog c limlog 1

    c

    hlog c limlog 1

    c

    log c lim

    =

    =

    = + >

    = + >

    + +

    = + +

    = +

    ( )

    ( ) ( )

    h 0

    h 0 h 0

    x 0

    h 0

    h 0

    hlog 1

    hc. (1Mark)

    h cc

    hlog 1

    hclog c lim .lim

    h c

    c

    log(1 x)Butlim 1

    x

    hlog 1

    clim 1

    hc

    hlog c 1.lim log c 1.0 logc (1Mark)

    c

    f(c)

    logarithmic function

    +

    +

    = +

    +=

    +

    =

    = + = + =

    =

    is continuous at every point in its domain (1Mark)

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    ( )

    2

    2

    2 2

    2

    (3 sin 2)cos22. Let I = d

    5 cos 4sin

    Let t sin dt cos d

    (3 t 2)I dt

    5 (1 t ) 4t

    (3 t 2) (3 t 2)I dt dt (1Mark)

    t 4t 4 (t 2)

    Using the method of Partial fractions

    (3 t 2) A

    t 2(t 2)

    = =

    =

    = =

    +

    = +

    ( )

    ( )

    2

    2

    2

    BA 3,B 4 (1Mark)

    (t 2)

    3 4I dt

    t 2 (t 2)

    3 4I dt dt (1Mark)

    t 2 (t 2)

    43log sin 2 C (1Mark)

    sin 2

    = =

    = +

    = +

    = +

    OR

    ( )

    ( )

    ( ) ( )

    11

    32

    1 1

    3 6

    16 56

    5 3

    2

    3

    3

    2

    dxLet I

    x x

    dx

    x 1 x

    Let x t x t dx 6t dt (1Mark)6t dt 6t dt

    I(1 t)t (1 t)

    t 1 1 dt6

    (1 t)

    t 1 dt dt6 6 (1Mark)

    (1 t) (1 t)

    t 1 t t 1 dt dt6 6

    (1 t) (1

    =

    +

    =

    +

    = = =

    = =++

    + =

    +

    +=

    + +

    + +=

    + +

    ( )2

    3 2

    1 1 1

    3 6 6

    t)

    dt6 t t 1 dt 6 (1Mark)

    (1 t)

    t t6 t 6log1 t

    3 2

    2 x 3x 6x 6log 1 x C (1Mark)

    = + +

    = + +

    = + + +

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    Section C

    23.

    Here, Volume V of the cone is1 3V2 2

    V r h r3 h

    = =

    ...(1)

    (1 mark)

    Surface area S

    2 2S rl r h r= = +

    ...(2)Where h= height of the cone

    r= radius of the cone

    l= Slant height of the cone

    (1 mark)

    ( )2 2 2 2 2S r h r= + by (2)Let 2

    1 =S S, then by (1)

    23 V 3V 9V2

    S h 3 Vh1 2

    h h h

    = + = +

    Differentiating S1with respect to h, we get

    2213 9

    3

    = +

    dS

    V V

    dh h

    (1 mark)

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    dS1 0 for maxima/minimadh

    223 V 9V 0

    3h

    223 V 9V3

    h

    6V3h

    =

    + =

    =

    =

    (1 mark)2 2d S 54V1

    42hdh

    2d S 6V31

    0 at h2 2

    dh

    =

    > =

    Therefore curved surface area is minimum at 3h V6

    =

    3h 1 2 2 2

    Thus, r h h 2r6 3

    h 2r

    = =

    =

    Hence for least curved surface the altitude is 2 times radius.(2 marks)

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    4 3

    ' 4 2 3 3

    3 2

    3 2

    3 2

    '

    23. f(x) = (x - 2 ) (x 1)

    f (x) =3(x - 2 ) (x 1) 4(x 1) (x - 2 )

    (x - 2 ) (x 1) 3(x - 2 ) 4(x 1)

    (x - 2 ) (x 1) 3x-6 4x 4

    (x - 2 ) (x 1) 7x-2

    f (x) =0 (x

    +

    + + +

    = + + +

    = + + +

    = +

    3 2

    '

    '

    2- 2 ) (x 1) 7x-2 x 1, ,2 (1Mark)

    7

    Let us examine the behaviour of f (x) , slightly to the left and right of each of these three val

    (i) x = -1 :

    When x

    + =

    '

    '

    '

    0

    When x>-1; f (x)>0

    x =-1 is neither a point of local maxima nor minima (1Mark)

    It may be a point of inflexion

    2(ii) x

    7 2When x< ; f (x)>0

    72

    When x> ; f (x)

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    1 2 3 0 1 0

    0 1 2 1 0 0 A

    3 1 1 0 0 1

    =

    .(1 mark)

    Applying 3 3 1R R -3R

    1 2 3 0 1 0

    0 1 2 1 0 0 A

    0 5 8 0 3 1

    =

    .. (1/2 mark)

    Applying 1 1 2R R - 2R

    1 0 1 2 1 0

    0 1 2 1 0 0 A

    0 5 8 0 3 1

    =

    . (1/2mark)

    Applying 3 3 2R 5R +R

    1 0 1 2 1 0

    0 1 2 1 0 0 A

    0 0 2 5 3 1

    =

    (1/2 mark)

    Applying 3 31

    R R2

    1 0 1 2 1 0

    0 1 2 1 0 0 A

    0 0 1 5 3 1

    2 2 2

    =

    (1 mark)

    Applying 1 1 3R R + R

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    1 1 1

    1 0 0 2 2 2

    0 1 2 1 0 0

    0 0 1 5 3 1

    2 2 2

    =

    .(1/2 mark)

    Applying 2 2 3R R -2R

    1 1 1

    1 0 0 2 2 2

    0 1 0 4 3 1 A

    0 0 1 5 3 1

    2 2 2

    =

    .(1/2 mark)

    Hence1

    1 1 1

    2 2 2

    A 4 3 1

    5 3 1

    2 2 2

    =

    . (1 mark)25.

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    OR

    ( ) ( )

    2 2 2

    The circle is x y 2a C(O, 2a)12 2

    The parabola is y ax, a 0 y 4 ax, a 04

    2 2Their point of intersection is given by :x ax 2a

    2 2x ax 2a 0

    x 2a x a 0

    x a, 2a

    x a (1Mark)

    2 2y a y a (1Mark)

    sh

    + =

    = > = >

    + =

    + =

    +

    =

    =

    = =

    ( )

    ( )

    ade region is the smaller of the two areas in which the circle is divided by the parabola1/2a 2a 2 2

    A=2 axdx 2a x dx (1Mark)0 0

    a2a

    3/2 21/2x x 2a x2 2 12 a 2 2a x sin

    3 2 2 2a0

    2 0

    4 3a a

    3

    +

    = + +

    =

    ( )2a

    1/2 x/2 2 2 2 1x 2a x 2a sin

    2a 0

    24a 2a sq units (3Marks)

    3

    + +

    = +

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    26.r. i+j+k 1 0and r. 2i+3j-k 4

    x y z 1 0 and 2x 3y z 4 0

    The required plane passes through the line common to two intersectiing planes

    x y z 1 k(2x 3y z 4) 0

    = =

    + + = + + =

    + + + + + =

    r r$ $ $ $

    (1Mark)x(1 2k) y(1 3k) z(1 k) ( 1 4k) 0......(1)

    The required plane is parallel to the X-axis whose d.cs are 1,0,0(1Mark)

    1.(1 2k) 0.(1 3k) 0.(1 k) 0 (1Mark)

    (1 2k

    + + + + + + =

    + + +

    + =

    +1

    ) 0 k (1Mark)2

    Substituting in (1),we get

    1 1 1 1x(1 2 ) y(1 3 ) z(1 ) ( 1 4 ) 0

    2 2 2 2

    1 3

    x(0) y( ) z( ) 3 02 2y 3z 6 0 y 3z 6

    = =

    + + + + + + =

    + + =

    + = + = 0 (2Marks)

    27.Let P be the principal. According to the given problem,5

    100 20

    dP PP

    dt

    = =

    (i)

    Separating the variables in equation (i), we get

    20

    dP dt

    P=

    (1 mark)

    1 1 1

    20 20 20

    Integrating both sides, we get

    20

    1log

    20

    ...(2)t k t t

    k

    dP dt

    P

    P t k

    P e e e Ce+

    =

    = +

    = = =

    (2marks)

    1

    20

    Now, 1000, when 0

    Substituting the values of P and t in (2), we get C = 1000

    1000t

    P t

    P e

    = =

    =

    (1mark)

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    1

    20

    1

    20

    Let years be the time required to double the principal. Then

    2000 1000

    2

    20 log 2

    t

    t

    e

    t

    e

    e

    t

    =

    =

    =

    .(2 marks)

    28.Let the two carpenters work for x days and y days respectively,then our problem is to minimise the objective functionC= 150 x + 200 ySubject to the constraints

    6x + 10y 60 3x +5y30

    4x+4y 32 x+ y 8

    And

    x0,y0..(2 Mark)

    Feasible region is shown shaded ..

    (2 Marks)

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    This region is unbounded.corner points objective funcA( 10,0) 1500

    E(5,3) 1350.D(0,8) 1600

    The labour cost is the least,when carpenter A works for 5 days and carpenter B works for 3 days,

    (2 Marks)

    29.

    Let A be the event that he knows the answer, B be the event that he guesses

    and C be the event that he copies.

    ( ) ( ) ( )

    ( ) ( ) ( )

    1 1 1Then, P A ,P B and P C2 4 4

    Let X be the event that he has answered correctly. (2 Marks)

    31Also, P X / A 1, P X /B and P X /C4 4

    = = = = = =

    Thus, Required probability = P(A/X) =

    ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

    P X / A P A

    P X / A P A P X /B P B P X / C P C

    + +

    (1 Mark)

    11

    21 1 1 1 3

    12 4 4 4 4

    =

    + +

    (1 Mark)

    1

    21 1 3

    2 16 16

    =

    + +

    23

    = (1 Mark)

    If Arjun copies the answer, he will be violating the value of honesty in hischaracter. He should not guess the answer as well because that may fetchhim negative marking for a wrong guess. He should accept the question theway it is and leave it unanswered as cheating may get him marks in this

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    exam but this habit may not let him develop integrity of character in the longrun. (1 Mark)