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    Master SolutionPremier Institute for the preparation of IIT-JEE /

    Date 01 05 2011 Class : 10 + 1 M.M:171 Time: 1:30 hrs

    Test Code 4001 TEST : Mathematical Tools (Master solution)

    SECTION-A (ONE OPTION CORRECT TYPE QUESTIONS)

    The value ofo2o4

    o42

    20cos20sin

    20cos20sin

    ++

    isQ.1

    (a.) 1 (b.) 2 (c.)

    2

    1

    (d.) None of these

    Ans. (a)

    o2o4

    o4o2

    20cos20sin

    20cos20sin

    ++

    =o2o22

    o2o2o2

    20cos]20cos1[20sin

    )20sin1(20cos20sin

    ++

    120cos20sin1

    20cos20sin1o2o2

    2o2

    ==

    sinP2P 20 + sinP2 P70Q.2

    (a.) 0 (b.) 1 (c.) 2 (d.) sinP2Px

    Ans. (b)

    Let p = sinP2P 20 + sinP2P 70= sinP2P20 + sinP2P(90 - 20)= 1

    The value of8

    7sin

    8

    5sin

    8

    3sin

    8sin 2222

    +

    +

    +

    isQ.3

    (a.) 1 (b.) 2 (c.)

    8

    11

    (d.)

    8

    12

    Ans. (b)= sinP2P

    8

    + sinP2

    8sin

    8

    5sin

    8

    3 22 7++

    =

    +

    +

    +

    8

    3sin

    8

    3sin

    8sin

    8sin 2222

    =

    +

    8

    3sin

    8sin2 22

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    =

    +

    82sin

    8sin2 22

    =

    +

    8cos

    8sin2 22 = 2

    The value of cos 1 cos 2 cos 3.. cos 179 is equal toQ.4

    (a.)

    2

    1

    (b.) 0 (c.)

    2

    1

    (d.) 2

    Ans. (b)

    cos1 cos2 cos3 cos 179= 0 {as cos 90 = 0

    If tan A =2

    1and tan B =

    3

    1, then the value of A + B is

    Q.5

    (a.)

    4

    (b.) 0 (c.) (d.)

    4

    Ans. (a/d)

    Tan (A + B) =Btan.Atan1

    BtanAtan

    +

    Put tan A =2

    1; tan

    3

    1B =

    1)BAtan( =+A + B = /4The maximum and minimum values of 6 sin x cos x + 4 cos 2x are:Q.6

    (a.) 5, 5 (b.) 4, 4 (c.) 7,7 (d.) 132,132

    Sol: (a)

    6 sin x cos x + 4 cos 2x = 3sin2x + 4 cos 2x

    Max = 22 43 + 5

    The greatest value of sinx. cosx isQ.7

    (a.) 1 (b.) 2 (c.)

    2

    1

    (d.) 1

    Sol: (c)

    y = sin x cos x = x2sin2

    1 greatest value =

    2

    1

    The third term of a G. P. is 4. The product of the first five terms isQ.8

    (a.) 4P3P (b.) 4P5P (c.) 4P4P (d.) 4P2P

    Sol: (b)

    TB3B = arP2

    P = 4

    Product of first five terms

    = a ar arP2P arP3P arP4

    P aP5PrP10P = (arP2P)P5P = 4P5P

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    Which term of the sequence 72, 70, 68, 66,.. is 40?Q.9

    (a.) 17 (b.) 15 (c.) 19 (d.) 16

    Sol: (a)

    72, 70, 68, 66 ., 40 A. P

    a = 72, d = 2, TB

    nB

    = 40 40 = 72 + (n 1) (2)32 = 2 (n 1) n = 17

    The 10PthP and 18PthP terms of an A. P. are 41 and 73 respectively. Find 26PthP term.Q.10

    (a.) 100 (b.) 105 (c.) 110 (d.) 108

    Sol: (b)

    aB10B = 41 and aB18B = 73

    a + 9d = 41 .(1) and a + 17 d = 73 ..(2)

    Solving eq (1) & (2), we get

    8d = 32 d = 4 a + 9 4 = 41 a = 41 36 = 5, a = 5

    aB26B = a + 25 d = 5 + 25 4 = 5 + 100

    aB2BB6B = 105

    One root of the equation 5xP2P + 13x + K = 0 is the reciprocal of the other, ifQ.11

    (a.) K = 0 (b.) K = 5 (c.)K =

    6

    1

    (d.) 6

    Ans. (b)5xP2P + 13x + K = 0

    Let one root be Then other root be

    1

    Product of roots ==

    1

    5

    K

    K = 5For what value of p the difference of the roots of the equation xP2P px + 8 = 0 is 2?Q.12

    (a.) 2 (b.) 4 (c.) 6 (d.) 8

    Ans. (c)xP2P px + 8 = 0

    Let its root be &

    + = p = 8( - ) = 2

    222 p2 =++

    16p222 =+

    4222 =+

    48216p2 =p = 6

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    If , are the roots of the equation xP2P p(x + 1) c = 0, then ( + 1) ( + 1) =Q.13

    (a.) c (b.) c -1 (c.) 1 c (d.) None of these

    Ans. (c)xP2P p(x + 1) c = 0xP2P px (p + c) = 0

    )cp( +=

    .1)1()1( +++=++= p c + p + 1= 1 - c

    The roots of the equation 2P2xP 10. 2PxP + 16 = 0 areQ.14

    (a.) 2, 8 (b.) 1, 3 (c.) 1, 8 (d.) 2, 3

    Ans. (b)

    2P2xP 10 2PxP + 16 = 0Let 2PxP = y

    yP2P 10y + 16 = 0yP2P 8y 2y + 16 = 0

    y(y 8) - 2(y 8) = 0(y 2) (y 8) = 0y = 2, 8

    2PxP = 2P1 P x = 12PxP = 2P3P x = 3

    The angle between the lines 2x y +3 = 0 and x + 2y + 3 = 0 isQ.15

    (a.) 90 (b.) 60 (c.) 45 (d.) 30

    Sol: (a)

    2x y + 0

    x + 2y + 3 = 0

    mB1B = + 2, mB2B =2

    1

    0

    2/5

    m.m1

    mmtan

    12

    12 =+

    =

    = /2

    The equation of the line which has Y-intercept 4 units and is parallel to the line 2x 3y 7 = 0. Will cuts

    the X-axis.

    Q.16

    (a.) (-6, 0) (b.) (-5, 0) (c.) (5, 0) (d.) (4, 0)Ans. (a)

    2x 2y 7 = 0Line parallel to above line2x 3y + k = 0 (1)y- intercept = 4y-intercept is obtained by putting x = 0

    y =3

    k

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    12k;43

    k==

    Eqn (1) become ; 2x 2y + 12 = 0It will cuts the x ax is it y = 02x = -12x = -6 pt, (-6, 0)The inclination of the straight line passing thro the point (-3, 6) and the midpoint of the line joining the

    point (4, -5) and (-2, 9) is

    Q.17

    (a.)

    4

    (b.)

    6

    (c.)

    3

    (d.)

    4

    3

    Ans. (d)

    xB1B=2

    24 yB1B=

    2

    59

    xB1B = 1 yB1B = 2xB2B = -3 yB2B = 6

    113

    26m ==

    tan =4

    3

    The ratio in which the line segment joining (2, 3) and (4, 1) is divided by the line joining (1, 2) and ( 4, 3)

    is

    Q.18

    (a.) 1 : 2 (b.) 1 :3 (c.) 2 : 3 (d.) None of these

    Sol: (b)

    eq. of CD is( ) ( )112

    121 xx

    xx

    yyyy

    =

    (4,1

    y 2 = ( 1x1423

    ) y 2 = ( )1x

    31

    (x, y)

    (4,3)

    (2,3)

    m n

    P

    D

    C(xB1B, yB1B)

    (1,2)

    A B

    (xB2B, yB2B)

    3(y 2) = (x 1) .(1)

    nm

    n2m4

    nm

    2n4m

    ++

    =++

    P(x,y) is x =

    nm

    n3m

    nm

    3n1my

    ++=

    ++

    =

    Put in eq. (1), we get

    ++= ++1

    nmn2m4

    12

    nmn3m3 + += + + nm

    nmn2m4nm

    n2m2n3m3

    nm31

    nm3 +=

    +

    3m + 3n = 3m + n 3m 3m = n 3n

    6

    2

    n

    m= m : n = 1 : 36m = - 2n

    Q.19 The vertices of a triangle ABC are (1, 2), (0, 1) and (2, 0) respectively. The equation of the median

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    through (1, 2) is

    (b.) (c.) (d.) None of these(a.) 5x 4y 3 = 0 x + 2y 2 = 0 4x + y 1 = 0

    Sol: (a)

    ++=

    2

    01,

    2

    20D

    =

    2

    1,1D

    B

    A CD(0,1) (1,1/2)

    (1,2)

    (2,0)

    )xx(xx

    yy1

    12

    12

    eq of BD is y yB1B =

    ( )1x11

    2

    12

    2

    1y

    = ( )1x

    22

    14

    2

    1y

    =

    ( 1x4

    5

    2

    1y2=

    ( ) 5x51y22 =)

    5x 5 4y +2 = 05x52y4 =

    5x 4y 3 = 0

    If 7 times the 7 PthP term of an A.P is equal to 11 times its 11PthP term. Then 18PthP term of the A.P. isQ.20

    (a.) Zero (b.) 7 (c.) 13 (d.) 9

    Ans. (a)

    7(a + 6d) = 11(a + 10d)

    7a + 42d = 11a + 10d

    4a + 68d = 0

    4(a + 17d) = 0

    A + 17d = 0

    18PthP term = a + 17d

    = 0

    SECTION-B (ONE OR MORE THAN ONE CORRECT TYPE QUESTIONS)5

    12Q.1If cot = , is in quadrant III then

    13

    5

    13

    12cos=

    (b.)(a.)sin =

    1312cos =

    13cossin 7=+(c.) (d.)

    Ans. (a,b)

    cot =5

    12

    sin [ ]vesinquadrantIIIrdtheinAs13

    5=

    13

    5

    12

    cos = ]quadrantIIIrdinveis[cos13

    12

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    The fourth, seventh and last term of a G. P. are 10, 80 and 2560 respectively. Which of the following is/are

    correct.

    Q.2

    Ratio of 5PthP term of series to the 4 PthP term is 2.(a.)

    (b.) The common ratio of the series is 2.

    (c.) Total number of terms of the series is 12

    (d.) The fifth term of the series is 20

    Sol: (a, b, c, d)

    aB4B = 10, aB7B = 80, aBnB = 2560

    arP3P = 10 ..(1)

    arP6P = 80 ..(2)

    Divide (2) by (1), we get

    10

    80

    ar

    ar3

    6

    = rP3P = 8 r = 2

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    a(2)P3P =10 a 8 = 10 a =4

    5

    8

    10 =4

    5a =

    24

    5 aB2B = ar = aB2B =

    2

    5

    2560)2(4

    5 1n=

    Also arPn1P = 2560

    5(2)Pn1P = 2560 4

    (2)Pn1P = 512 4

    (2)Pn1P = (2)P11P n 1 = 11 n =12

    aB5B = arP4P

    = ( )424

    5 16

    4

    5= = aB5B =20

    Q.3

    4

    1If sin = , then which of the following is/are correct?

    8

    7

    15

    1

    15

    4(c.)(b.) (d.)(a.)

    4

    3 Tan = Sec =Cos 2 =Sin 2 =

    Sol: (b, c, d)

    sin =4

    1

    sin 2 = 2 sin. cos = 2 4

    15

    4

    1

    sin2 =8

    15 .(1) A B

    C

    1

    4

    15

    4P2P = 1P2P + ABP2Pcos 2 = 1 2 sinP2

    P

    = 1 2 2

    4

    1

    16

    1

    8

    18 AB116 = = 1 2 =

    AB15 =

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    cos 2 =8

    7

    tan =15

    1

    Q.4 If 3 sin + 4 cos = A sin ( + ), then which of the following is/are correct?

    (d.)(a.) A = 5 (b.) A = 7 (c.)

    3

    4tan =

    4

    3tan =

    Sol: (a, c)

    3sin + 4 cos = A sin ( + )22 baA += 22 43 + 169 + 25= = = = 5

    As [ ] + cos4sin3

    =

    + cos

    5

    4sin

    5

    35

    5

    4sin,

    5

    3=cos =

    3

    5

    5

    4 tan =

    tan =3

    4

    Q.5 The equation of straight line is given as 3x 4y + 10 = 0 . Then

    4

    3

    4

    3

    (a.) (b.)Its slope is Its slope is

    3

    10

    (c.) (d.) It will pass through (6, 7)Its x-intercept is

    Ans. (b, c, d)3x 4y + 10 = 0 (1)4y = 3x + 10

    y =4

    10x

    4

    3+

    slope =4

    3

    to find x-intercept put y = 03x = -10

    x =3

    10

    Since pt (6,7) satisfy eq(1) so the eqn pass through (6,7)

    Q.6 The right bisector of segment joining (3, 4) and (-1, 2) has

    (a.) Slope 2 (b.) Slope 2

    (c.) Equation 2x + y = 5 (d.) Equation 2x + y + 5 = 0

    Ans. (a, c)A (3, 4) B = (-1, 2)

    2

    24,

    2

    13 +Pt. from which bisector pass =

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    = (1, 3)& slope of right bisector = -2

    Eqn = y 3 = -2(x -1)y 3 =-2x + 2

    2x + y -3 2 = 0

    2x + y 5 = 0Q.7 It and are the roots of 4xP2P + 3x + 7 = 0, then

    7

    3

    7

    3

    16

    47

    16

    47

    (a.) (b.) (c.) (d.) + = + = P2P + P2P = P2P + P2P =Sol: (d)

    4xP2P + 3x + 7 = 0

    4

    3=+

    4

    7=

    +=+ 2)( 222

    =4

    72

    16

    9

    2

    7

    16

    9=

    16

    47=22 +

    Q.8 A person standing on the bank of a river, observes that the angle of elevation of the top of a tree on the

    opposite bank of the river is 60 and when he retires 40 m away from the tree the angle of elevation

    becomes 30. Then

    (a.) width of the river is 20 m (b.) width of the river is 60 m

    (c.) (d.)320 3/20Height of the tree is Height of the tree is

    Ans. (a, c)

    6040 x

    30y

    o60tanx

    y= x3y =

    o30tan40x

    y=+

    3

    40xy+= 40x)x3(3 +=

    3x = x + 40 x = 20

    320y =

    Width of the river = 20 m

    3 Height of tree = 20

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    Q.9 The sequence 9, 12, 15, 18, is an A.P. Then

    (a.) Its general term is 3n + 6 (b.) Its general term is 3n 6

    (c.) Its 16PthP term is 54 (d.) Its 10PthP term is 36

    Ans. (a,c, d)

    9, 12, 15, 18 .

    a = 9d = 3general term = a +(n -1) d

    = 9 + (n -1)3= 3n + 6

    10PthP term 9 +(10 1)9 + 27 = 36

    16PthP term = 9 + 15 3= 54

    Q.10 A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3.Then

    31

    34(a.) (b.)Its y-intercept is Its y-intercept is

    Its equation is x 3y + 4 = 0 Its equation is x 3y = 4(c.) (d.)

    Ans. (b,c)

    3

    1 Slope of required line =

    3

    1Eqn of line = y 2 = (x 2)

    3y 6 = x 2 x 3y + 4 = 0

    3

    4

    Its y-intercept =SECTION-C (SUBJECTIVE TYPE QUESTIONS)Q.1

    =

    +

    2tan

    2tan

    3 sin = 5 sin , then

    Sol: [4]

    = sin5sin3

    3

    5

    sin

    sin=

    35

    35

    sinsin

    sinsin

    +=+

    2sin.

    2cos2

    2cos.

    2sin2

    +

    +

    = 4

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    2.tan

    2tan

    +

    = 4

    Q.2 Find the sum of the given series up to 2 decimal places

    ..........321

    161

    81

    41

    211 +++

    Sol: [0.66 or 0.67]

    As it is infinite G.P. series; usingr1

    aS= =

    2

    11

    1=

    3

    2= 0.67

    Q.3 Find the slope of the equation of the perpendicular bisector of the line segment joining the points A(2, 3)

    and B( 6, 5)

    Sol: [0.5 ]

    Sol: 0.5

    12

    12

    xx

    yy

    4

    826

    35

    mB1B = 2= =Slope of AB =

    Using mB1B mB2B = 1

    -2 mB2B = 1

    mB2B =2

    1 mB2B = 0.5

    Q.4 If one root of the equation (x 1) (7 x) = m is three times the other, then m is equals to

    Ans. [5](x 1) (7 x) = m

    Let one root = Other root = 37x xP2P -7 + x = mxP2P 8x + 7 + m = 0

    4 = 8 = 27 + m = 3P2Pm + 7 = 12m = 5

    Q.5 The x-intercept of equation of a line perpendicular to x 2y + 3 = 0 and passing through the pt(3, -2) is

    Ans. [2]

    Slope of required line = -2

    Eqn of line y + 2 = -2(x 3)y + 2 = -2x + 62x + y 4 = 0

    For its x-intercept put y = 0x = 2

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    SECTION-D (MATCH THE COLUMN TYPE QUESTIONS)(Items of Column- I can match one or more than one Items of column- II with individual marking scheme)

    Q.1 In the following options symbols have their usual meaning. Match the column I with column-II

    Column-I Column- II

    r1a

    (i.) Sum of n terms in arithmetic progression (a.)

    ( )r1

    r1a n

    (ii.) nPthP term of arithmetic progression (b.)

    (iii.) nPthP term of geometric progression (c.) a + (n 1)d

    arPn1P(iv.) Sum of infinite series of geometric progression (d.)

    ( )[ ]d1na22

    n+

    (e.)

    Sol: (i) (e), (ii) (c), (iii) (d), (iv) (a)

    Column-I Column- IIQ.2

    (a.) 0(i.) If sinB1B + sinB2B + sinB3B = 3 then1

    cos B1B + cos B2B + cos B3B =

    (b.) 1(ii.) tan(180 + ) . tan(90 + ) =

    (c.) 2(iii.) cotA + tan (180 + A) + tan(90 + A) +

    tan(360 - A) =

    (iv.) (d.) 3

    o

    o

    o

    o

    70cot

    20tan

    36tan

    54cot+ =

    (e.) -1

    Ans. (i a), (ii e), (iii a), (iv c)(i) sin B1B + sinB2B + sinB3B = 3

    2

    So, B1B = B2B = B3B =

    So, cosB1B + cosB2B + cosB3B

    = 0

    (ii) tan(180 + ) . tan(90 + )

    =tan.[-cot ] = -1

    (iii) cotA + tan(180 + A) + tan(90 + A) +tan(360 - A)

    = cotA + tanA + [-cotA) tanA

    = 0

    o

    o

    o

    o

    70cot

    20tan

    36tan

    54cot+(iv)

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    =oo

    o

    o

    oo

    2090cot(

    20tan

    36tan

    )3690cot(

    +

    = 1 +1 = 2