Monksays StudMonk The App Facebook Monksays StudMonk – The App Facebook 5 : (D) Given, x) x

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  • 1 Monksays StudMonk.com StudMonk The App Facebook

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  • 2 Monksays StudMonk.com StudMonk The App Facebook

    1 : (B)

    Here,

    E(x) = nP = 5 and Var (x) = npq = 2.5

    2.5 1

    5 2q

    1

    2p and n = 10

    Now, ( 1) ( 0)P x P x

    10

    10

    0

    1( 1)

    2P x C

    2 : (D)

    Correct option is (D)

    3: (B)

    As required circle touches y-axis at the origin.

    Let Centre of the circle is d (a, 0) and radius is a

    Equation of circle will be,

    2 2 2( ) ( 0)x a y a

    2 2 2 22x ax a y a

    2 2 2 0x y ax (i)

    By differentiating above equation w.r.t. x, we get

    101

    ( 1)2

    P x

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  • 3 Monksays StudMonk.com StudMonk The App Facebook

    2 2 2 0dy

    x y adx

    2 2 2dy

    a x ydx

    (ii)

    From (i) and (ii),

    2 2 2 2 0dy

    x y x y xdx

    2 2 22 2 0dy

    x y x xydx

    2 2 2 0dy

    x y xydx

    4 : (B)

    Here,

    aij is stands for element of matrix A as ith row and jth column, and Aij stands for co-factor of element aij of

    matrix A.

    11 121, 1a a and 13 0a

    And

    2 1

    21

    1 0( 1) 1

    2 1A

    2 2

    22

    1 0( 1) 1

    1 1A

    2 3

    23

    1 1( 1) 1

    1 2A

    Therefore

    11 21 12 22 13 23 1 ( 1) 1 (1) 0 ( 1) 0a A a A a A

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  • 4 Monksays StudMonk.com StudMonk The App Facebook

    5 : (D)

    Given, ( ) (sin cos )xf x e x x

    '( ) [cos sin ] [sin cos ]x xf x e x x x x e

    '( ) 2 sinxf x e x

    To verify Rolles Theorem.

    '( ) 0f c

    2 sin 0ce c

    sin 0c

    c

    6 : (C)

    As given, both line passes through (0, 0), and 1 2,6 3

    Equation of first line is.

    0 tan ( 0)6

    y x

    1

    3y x

    3 0x y (i)

    Equation of second line is

    0 tan ( 0)3

    3

    y x

    y x

    3 0x y (ii)

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  • 5 Monksays StudMonk.com StudMonk The App Facebook

    Hence, joint equation of these line is

    ( 3 )( 3 ) 0x y x y

    2 23 3 3 0x xy xy y

    2 23 3 4 0x y xy

    7 : (B)

    As given, 1 12 tan (cos ) tan (2 cos )x ec x

    1 1 1tan (cos ) tan (cos ) tan (2 cos )x x ec x

    1 1

    2

    2 costan tan (2 cos )

    1 cos

    xec x

    x

    2

    2 cos2 cos

    sin

    xec x

    x

    2 cot 2x

    cot 1x

    4x

    Hence,

    sin cos sin cos4 4

    x x

    1 1

    2 2

    = 2

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  • 6 Monksays StudMonk.com StudMonk The App Facebook

    8 : (A)

    Option (A) is correct.

    9 : (D)

    Let 28 2

    dxI

    x x

    29 2 1

    dxI

    x x

    2 23 ( 1)

    dxI

    x

    1 1sin3

    xI c

    10 : (A)

    As given,

    3 2( ) 5 7 9f x x x x

    (1.1) 8.6f

    11 : (B)

    As given,

    1, 0 5

    ( ) 5

    0,

    xf x

    otherwise

    Now, probability of waiting time not more than 4 is 1

    4 0.85

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  • 7 Monksays StudMonk.com StudMonk The App Facebook

    12 : (B)

    Let 2 2 2 2( ) cos ( ) sin

    2 2

    c cX a b a b

    As we know,

    ( )( )sin

    2

    c s a s b

    ab

    and

    ( )cos

    2

    c s s c

    ab

    Where 2

    a b cs

    By substituting these value on above equation we will get

    2X c

    13 : (B)

    Let 1 log (sec tan )y

    21 1 (sec tan sec )sec tan

    dya

    d

    1 sec [sec tan ]

    [sec tan ]

    dy

    d

    1 secdy

    d

    .(i)

    Now,

    Let 2 secy

    2 sec tandy

    d

    (ii)

    1

    2

    seccot

    sec tan

    dy

    dy

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  • 8 Monksays StudMonk.com StudMonk The App Facebook

    1

    24

    cot 14

    dy

    dy

    14 : (B)

    We can say that both line passes through point (5, 3) and makes angle 45o and 135o with x axis

    Equation of first line is ,

    3 tan 45 ( 5)oy x

    3 5y x

    2 0y x (i)

    Similarly,

    3 tan 135 ( 5)oy x

    3 1( 5)y x

    8 0y x .(ii)

    Joint equation of line is

    ( 2)( 8) 0y x y x

    2 2 10 6 16 0x y x y

    15 : (A)

    As given, required point is on the

    Curve 36 2y x

    Therefore, only point (4, 11)

    Satisfy the given equation,

    Hence, option (A) is Correct.

    16 : (A)

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  • 9 Monksays StudMonk.com StudMonk The App Facebook

    1sin 0

    ( )

    0

    x for xf x x

    k for x

    0

    1lim sinx

    x kx

    0 k

    17 : (C)

    Given, 1sinm xy e

    1sin

    2

    1

    1

    m xdy e mdx x

    2 2 2

    21

    dy m y

    dx x

    2

    2 2 2(1 )dy

    x m ydx

    2A m

    18 : (B)

    Let 4 25

    2 5

    x

    x

    eI dx

    e

    10 25 6

    2 5

    x x

    x

    e eI dx

    e

    5(2 5) 6

    2 5

    x x

    x

    e eI dx

    e

    65

    2 5

    x

    x

    eI dx

    e

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  • 10 Monksays StudMonk.com StudMonk The App Facebook

    5 3 log(2 5)xI x e C

    5A and 3B

    19 : (B)

    By solving we will get,

    1 1

    1 1

    tan ( 3) sec ( 2) 4

    1 5cos ( 2) cos

    2ec

    20 : (C)

    As given, 2log(1 2 )sin

    0( )

    0

    x xfor x

    f x x

    k for x

    Is continuous at x = 0

    20

    log(1 2 ) sinlimx

    x xk

    x

    0 0

    sin2 log(1 2 ) 180

    lim lim2 180

    180

    x x

    xx

    xx

    = k

    2180

    k

    90k

    21 : (A)

    Given, 2 2

    10 2 2log 2

    x y

    x y

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  • 11 Monksays StudMonk.com StudMonk The App Facebook

    2 2

    2 2100

    x y

    x y

    2 2 2 2

    2 2

    100 1000

    x y x y

    x y

    2 2

    2 2

    99 1010

    x y

    x y

    2 299 101 0x y

    (2 99) (2 101) 0dy

    x ydx

    99

    101

    dy x

    dx y

    22 : (D)

    Let

    /2

    /2

    2 sinlog

    2 sin

    xI dx

    x

    As given function is odd.

    0I

    23 : (C)

    By using anti differentiation method,

    We will get to know that, Option (C) is correct.

    24 : (B)

    Degree 3

    Order 2

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  • 12 Monksays StudMonk.com StudMonk The App Facebook

    25 : (B)

    Acute angle is 12

    sin3

    26 : (B)

    2

    2

    0

    (2 )A x x dx

    23

    2

    03

    xA x

    84

    3A

    4

    3A sq unit.

    27 : (A)

    Given ( )