Iitjee 2012 Paper i Sol

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    Part I : PhysicsSection - I Single Correct Answer Type

    This section contains 10 multiple choice questions. Each question has four choice (A), (B), (C) and

    (D) out of which ONLY ONE is correct.

    Q.1 A bi-convex lens is formed with two thin plano-convex lenses as shown

    in the figure. Refractive index n of the first lens is 1.5 and that of the

    second lens is 1.2 Both the curved surfaces are of the same radius ofcurvature R = 14 cm. For this bi-convex lens, for an object distance of

    40 cm, the image distance will be

    n = 1.2n = 1.5

    R = 14 cm(A) 280.0 cm (B) 40.0 cm

    (C) 21.5 cm (D) 13.3 cm

    Ans. [B]

    [Sol.1

    f

    1= (1.2 1)

    14

    11=

    14

    2.0

    2f

    1= (1.5 1)

    1

    14

    1=

    14

    5.0

    eqf

    1=

    20

    1

    14

    7.0

    20

    1

    40

    1

    v

    1

    v = 40 cm ]

    Q.2 A thin uniform rod, pivoted at O, is rotating in the horizontal plane with constant angular speed , asshown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v with

    respect to the rod towards the other end. If reaches the end of the rod at t = T and stops. The angular

    speed of the system remains throughout. The magnitude of the torque |)(|

    on the system about O, as

    a function of time is best represented by which plot ?

    v

    O

    z

    (A)0

    Tt

    | |

    (B)0

    Tt

    | |

    (C)

    0T

    t

    | |

    (D)

    0T

    t

    | |

    Ans. [B]

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    [Sol. =dt

    dL=

    dt

    d(I + mx2) = m 2x

    dt

    dx = 2 mv2 t [as x = vt] ]

    Q.3 Three very large plates of same area are kept parallel and close to each other. They are considered as

    ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at

    temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady

    state condition is

    (A) T2

    654/1

    (B) T4

    974/1

    (C) T2

    974/1

    (D) T97

    4/1

    Ans. [C]

    [Sol.dt

    dQ= A 4

    1

    4 T)T3(

    = A 441

    )T2(T

    2T 3T

    T1

    T1

    = T2

    974/1

    ]

    Q.4 Consider a thin spherical shell of radius R with its centre at the origin, carrying uniform positive surface

    charge density. The variation of the magnitude of the electric field |)r(E|

    and the electric potential V(r)

    with the distance r from the centre, is best represented by which graph ?

    (A)

    R

    | E(r) |

    V(r)

    r0

    (B)

    R

    | E(r) |

    V(r)

    r0

    (C)

    R

    | E(r) |

    V(r)

    r0

    (D)

    R

    | E(r) |

    V(r)

    r0

    Ans. [D]

    Q.5 In the determination of Youngs modulus

    2d

    MLg4Y

    lby using Searles method, a wire of length L =

    2m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length ofthe wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively.

    They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions

    to the maximum probable error of the Y measurement

    (A) due to the errors in the measurements of d and l are the same

    (B) due to the error in the measurement of d is twice that due to the error in the measurement ofl.

    (C) due to the error in the measurement ofl is twice that due to the error in the measurement of d.

    (D) due to the error in the measurement of d is four times that due to the error in the measurement ofl.

    Ans. [A]

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    [Sol.y

    y=

    L

    L+

    l

    l+

    d

    d2

    l = 0.25, d = 0.5 ]

    Q.6 A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other end

    of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is

    stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular

    frequency = 3rad/s. Simultaneously at t = 0, a small pebble is projected with speed v from point P

    at an angle of 45 as shown in the figure. Point P is at a horizontal distance of 10 m from O. If the pebble

    hits the block at t = 1s, the value of v is (take g = 10 m/s2)

    v

    45x

    10 mO

    z

    P

    (A)50

    m/s (B)51

    m/s (C) 52 m/s (D)53

    m/s

    Ans. [A]

    [Sol. T =g

    45sinv2 ]

    Q.7 Youngs double slit experiment is carried out by using green, red and blue light, one color at a time. The

    fringe widths recorded are G,

    Rand

    B, respectively. Then,

    (A) G

    > B

    > R

    (B) B

    > G

    > R

    (C) R

    > B

    > G

    (D) R

    > G

    > B

    Ans. [D]

    [Sol. V I B G Y O R ( increases)

    =dD ]

    Q.8 A small mass m is attached to a massless string whose other end is fixed at P as

    shown in the figure. The mass is undergoing circular motion in the x-y plane with

    centre at O and constant angular speed . If the angular momentum of the

    system, calculated about O and P are denoted by0

    L

    andP

    L

    respectively,,

    then Om

    z

    P

    (A)0

    L

    andP

    L

    do not vary with time.

    (B)0

    L

    varies with time whileP

    L

    remains constant

    (C)0

    L

    remains constant whileP

    L

    varies with time.

    (D)0

    L

    andP

    L

    both vary with time

    Ans. [C]

    [Sol. About point P, angular momentum changes in direction, but about O angular momentum does not change

    either in direction or in magnitude. ]

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    Q.9 A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40

    amu) is kept at 300 K in a container. The ratio of the r.m.s. speeds

    )on(argv

    )helium(v

    rms

    rmsis

    (A) 0.32 (B) 0.45 (C) 2.24 (D) 3.16

    Ans. [D]

    [Sol. vrms

    =

    M

    RT3

    Ratio =He

    ar

    M

    M= 10 ]

    Q.10 Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage

    source of potential difference X. A proton is released at rest midway between the two plates. It is found

    to move at 45 to the vertical JUST after release. Then X is nearly

    (A) 1 105 V (B) 1 107 V (C) 1 109 V (D) 1 1010 V

    Ans. [C]

    [Sol.d

    qV = mg

    mg

    qE

    V =q

    mgd= 19

    227

    106.1

    1010106.1

    ]

    Section - II : Multiple Correct Answer(s) Type

    This section contains 5 multiple choice questions. Each question has four choice (A), (B), (C) and (D)

    out of which ONE or MORE may be correct.

    Q.11 A cubical region of side a has its centre at the origin. It encloses three fixed point charges, q at

    0,4

    a,0 , + 3q at (0, 0, 0) and q at

    0,4

    a,0 . Choose the correct option(s).

    y

    x

    z

    a

    q

    3q

    q

    (A) The net electric flux crossing the plane2

    ax is equal to the net electric flux crossing the plane

    2

    ax

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    (B) The net electric flux crossing the plane2

    ay is more than the net electric flux crossing the plane

    2

    ay

    (C) The net electric flux crossing the entire region is0

    q

    (D) The net electric flux crossing the plane2az is equal to the net electric flux crossing the plane

    2

    ax

    Ans. [A,C,D]

    [Sol. = charge enclosed /0

    A and D because of symmetry. ]

    Q.12 For the resistance network shown in the figure, choose the correct option(s)

    I1444

    22

    2S

    TQ

    P I2

    12V

    1 1(A) the current through PQ is zero

    (B) I1

    = 3A

    (C) The potential at S is less than that at Q

    (D) I2

    = 2A

    Ans. [A,B,C,D]

    [Sol. By input output symmetry, the current in each of the 2 resistance is the same. Similarly, the current ineach of the 4 resistance is the same. So the current in each of the 1 resistance is zero.

    So, I2

    =6

    12= 2A

    I4

    =12

    12= 1 AA

    I1 = I2 + I4 = 3ATaking potential at negative terminal of the battery to be zero, Potential at S = 4 Volt,

    Potential at Q = 8 Volt ]

    Q.13 A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle with thehorizontal. A horizontal force of 1N acts on the block through its centre of mass as shown in the figure.

    The block remains stationary if (take g = 10 m/s2)

    Q

    1N

    O P

    (A) = 45(B) > 45 and a frictional force acts on the block towards P(C) > 45 and a frictional force acts on the block towards Q(D) < 45 and a frictional force acts on the block towards Q

    Ans. [A,C]

    [Sol. Fnet

    = 1 cos 1 sin ]

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    Q.14 Consider the motion of a positive point charge in a region where are simultaneous uniform electric and

    magnetic field jEE0

    and jBB0

    . At time t = 0, this charge has velocity v

    in the x-y plane, making

    an angle with the x-axis. Which of the following option(s) is(are) correct for time t > 0?(A) If = 0, the charge moves in a circular path in the x-z plane(B) If = 0, the charge undergoes helical motion with constant pitch along the y-axis(C) If = 10, the charge undergoes helical motion with its pitch increasing with time, along the y-axis(D) If = 90, the charge undergoes linear but accelerated motion along the y-axis

    Ans. [C,D]

    [Sol.

    E0

    B0

    V

    ]

    Q.15 A person blows into open-end of a long pipe. As a result, a high-pressure pulse of air travels down the

    pipe. When this pulse reaches the other end of the pipe.

    (A) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open

    (B) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open

    (C) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed

    (D) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closedAns. [B,D]

    [Sol. Phase change of from denser medium and no phase charge from a rarer medium. For sound, the closeend of the pipe is a rarer medium and open end is a denser medium. ]

    Section - III : Integer Answer Type

    This section contains 5 questions. The answer to each question is a single digit integer, ranging from

    0 to 9 (both inclusive).

    Q.16 An infinitely long solid cylinder of radius R has a uniform volume charge density. It has a spherical

    cavity of radius R/2 with its centre on the axis of the cylinder, as shown in the figure. The magnitude of theelectric field at the point P, which is at a distance 2R from the axis of the cylinder, is given by the

    expression0

    k16

    R23

    . The value of k is

    P

    y

    R/2

    z

    x

    R

    Ans. 6

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    [Sol. E1 2 (2R) l =

    0

    2

    R l

    01 4

    RE

    E2 4 (2R)2 =

    0

    3

    2

    R

    3

    4

    02 424

    RE

    0021 244

    R23

    24

    11

    4

    REE

    k = 6 ]

    Q.17 A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a shown in the figure. Both the

    cylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the

    magnitude of the magnetic field at the point P is given byaJ

    12

    N

    0

    , then the value of N is

    aP

    2aAns. 5

    [Sol.

    2

    a32

    2

    a

    J

    2

    JaB

    2

    00

    P

    12

    Ja5

    6

    11

    2

    Ja00

    N = 5 ]

    Q.18 A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and

    radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and

    P is IO

    and IP, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio

    O

    P

    I

    I

    to the nearest integer is

    O2R

    2R

    P

    Ans. 3

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    [Sol. I0

    =2

    1(2R)2(2R)2

    2222 RRRR2

    1=

    44R

    2

    13

    2

    38R

    IP

    =2

    3(2R)2 (2R)2

    2222 R5RRR2

    1

    2

    R37

    2

    1124R

    44

    3~I

    I

    O

    P]

    Q.19 A circular wire loop of radius R is placed in the x-y plane centered at the origin O. A square loop of side

    a (a

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    PART II: CHEMISTRY

    SECTION -I : Single Correct Answer TypeThis section contains 10 multiple choice questions. Each question has four choices(A), (B), (C) and (D) out of which ONLY ONE is correct.

    Q.21 As per IUPAC nomenclature, the name of the complex [Co(H2O)

    4(NH

    3)2]Cl

    3is

    (A) Tetraaquadiaminecobalt (III) chloride

    (B) Tetraaquadiamminecobalt (III) chloride

    (C) Diaminetetraaquacoblat (III) chloride(D) Diamminetetraaquacobalt (III) chloride

    Ans. (D)

    Sol. H2Oaqua

    NH3ammine

    Q.22 In allene (C3H

    4), the type(s) of hybridisation of the carbon atoms is (are)

    (A) sp and sp3 (B) sp and sp2 (C) only sp2 (D) sp2 and sp3

    Ans. (B)

    Sol.C = C = C

    H

    H

    H

    Hsp

    2sp sp

    2

    Q.23 For one mole of a Vander waals gas when b = 0 and T = 300 K, the PV vs 1/V plot is shown below.

    The value of the Vander waal's constant a (atm. litre2mol2) is

    (Graph not to scale)

    24.6

    23.1

    21.6

    20.1

    02.0 3.0

    1V

    (mol liter )1

    PV

    (liter-atmm

    ol

    1)

    (A) 1.0 (B) 4.5 (C) 1.5 (D) 3.0

    Ans. (C)

    Sol. RTVV

    aP

    2

    V

    aRTPV

    Slope =a = 5.12

    3

    02

    6.246.21

    a = 1.5

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    Q.24 The number of optically active products obtained from the complete ozonolysis of the given

    compound is

    CH CH=CHCCH=CHCCH=CHCH3 3

    CH3

    CH3H

    H

    (A) 0 (B) 1 (C) 2 (D) 4

    Ans. (A)

    Sol. CH CH=CHCCH=CHCCH=CHCH3 3

    CH3

    CH3H

    H

    Zn

    O3 2CH3CH = O +

    CH3

    CH3H

    H

    OHC CCHO + OHC C CHO

    All are optically inactive

    Q.25 A compound MpX

    qhas cubic close packing (ccp) arrangement of X. Its unit cell structure is shown

    below. The empirical formula of the compound is

    M=

    X =

    (A) MX (B) MX2

    (C) M2X (D) M

    5X

    14

    Ans. (B)

    Sol. Face centre = = 4For M,

    edge centre = 144

    1

    centre = 1

    M2X

    4

    i.e. MX2

    Q.26 The number of aldol reaction (s) that occurs in the given transformation is

    CH2CHO + 4HCHO

    NaOH.aq.conc

    OHOH

    OHHO

    (A) 1 (B) 2 (C) 3 (D) 4

    Ans. (C)

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    Sol. CH3CHO + 4HCHO conc. OH

    OHOH

    OHHO

    OO||||

    HCHCHCH3 )1(aldol

    HO

    OH|

    HCCHCH||

    O

    22 )2(Aldol

    OH/HCHO CHCH

    HOCH2

    HOH2

    C

    O

    HCHO/OH

    Aldol(3)

    OHOH

    OHHO

    OHcannizaro

    C CH

    O

    OH

    HO

    OH

    Number of aldol reaction = 3

    Q.27 The colour of light absorbed by an aqueous solution of CuSO4

    is

    (A) Orange-red (B) Blue-green (C) Yellow (D) VioletAns. (A)

    Sol. aqueos solution of CuSO4

    is blue so absorbed colour is orange-red

    VI

    BGY

    R

    O

    OR

    Q.28 The carboxyl functional group(COOH) is present in

    (A) picric acid (B) barbituric acid (C) ascorbic acid (D) aspirinAns. (D)

    Sol.

    OCCH3

    COOH

    O

    Aspirin

    Q.29 The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0

    is Bohr radius]

    (A) 20

    2

    2

    ma4

    h

    (B) 202

    2

    ma16

    h

    (C) 202

    2

    ma32

    h

    (D) 202

    2

    ma64

    h

    Ans. (C)

    Sol. mvr = nh/2p

    v =a4m2

    h2

    z

    nr

    2

    4

    1

    x

    a

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    Q.33 Which of the following molecules, in pure from, is (are) unstable at room temperature ?

    (A) (B) (C) (D)

    Ans. (BC)

    Sol. Anti aromatic are unstable at room temperature

    and are anti-aromatic with 4 electron system.

    While is non-aromatic and is aromatic with 6.

    Q.34 Which of the following hydrogen halides react(s) with AgNO3(aq) to give a precipitate that dissolves in

    Na2S

    2O

    3(aq) ?

    (A) HCl (B) HF (C) HBr (D) HIAns. (ACD)

    Sol. AgF is soluble in water

    AgNO3

    + HX (X = Cl, Br, I) AgX 322OSNa

    [Ag(S2O

    3)2]3 (soluble)

    Q.35 For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state

    Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are)

    correct ? [take S as change in entropy and w as work done]

    (A) Sx z

    = Sx y

    + Sy z

    (B) Wx z

    = Wx y

    + Wy z

    (C) Wx y z

    = Wx y

    (D) Sx y

    z

    = Sx y

    Ans. (AC)

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    SECTION -III: Integer Answer TypeThis section contains 5 questions. The answer to each question is a single-digit integer,

    ranging from 0 to 9 (both inclusive)

    Q.36 The substituents R1

    and R2

    for nine peptides are listed in the table given below. How many of these

    peptides are positively charged at pH = 7.0 ?

    HRRH

    OCOHCNHCOHCNHCOHCNHCOHCNH

    21

    ||||3

    3242

    2422

    222

    242242

    2222

    24222

    2

    3

    CHNH)CH(IX

    NH)CH(OHCHVIII

    CONHCHCOOHCHVII

    NH)CH(NH)CH(VI

    CONHCHCONHCHV

    NH)CH(CONHCHIV

    HCOOHCHIII

    CHHII

    HHI21

    RRPeptide

    Ans. 4

    Sol. Any amino acid will exist in a cationic form in solution of pH < pI (isoelectric point)

    pH = 7 is less than pI given which implies pI > 7 basic amino acid.

    Given peptide will have basic nature if number of basic sites are greater than acidic sites.

    Peptide IV R1

    = CH2CONH

    2R

    2= (CH

    2)4NH

    2

    Peptide VI R1

    = (CH2)4NH

    2R

    2= (CH

    2)4NH

    2

    Peptide VIII R1

    = CH2OH R

    2= (CH

    2)4NH

    2

    peptide IX R1

    = (CH2)4NH

    2R

    2= CH

    3

    Ans. 4

    Q.37 The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes

    a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic

    table ?

    XH2n6HCu 11

    1

    0

    1

    1

    63

    29

    Ans. 8

    Sol. X52261

    1

    4

    2

    1

    0

    1

    1

    63

    29 H2n6HCu

    Ans. 8th group.

    1 2 3 4 5 6 7 8

    Kr Cr Sc Ti V Cr mn Fe

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    Q.38 When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its

    pyranose form is

    OHCH|CHOH|

    CHOH|CHOH

    |

    CH|CHO

    2

    2

    Ans. 8

    Sol.

    OHCH|CHOH|CHOH|

    CHOH|

    CH|CHO

    2

    2

    1

    2

    3

    4

    5

    (=)

    CHO

    CH2

    CHOH

    CHOH

    CH OH2

    H OH

    Given is D configuration for pyranose. Cyclization will happen by OH of fifth carbon.

    CH

    CH2

    CH2OH

    CHOH

    CHOH

    OH

    H O

    1*

    3*

    4*

    5

    2

    Configuration at C-5 is D(glucose) this will not change. New chiral generates at C-1 and total numberof chiral center which may have D or L continue are 3. Hence total number of stereoisomers= 23 = 8

    Ans. 8

    Q.39 29.2% (w/w) HCl stock solution has a density of 1.25 g mL1. The molecular weight of HCl is 36.5 g

    mol1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is

    Ans. 8

    Sol. V 1.25 5.364.01000

    200

    100

    2.29

    V = ml

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    Q.40 An organic compound undergoes first-order decomposition. The time taken for its decomposition to

    1/8 and 1/10 of its initial concentration are t1/8

    and t1/10

    respectively. What is the value of

    10]t[

    ]t[

    10/1

    8/1 ? [Take log10

    2 = 0.3]

    Ans. 9

    Sol.xa

    aln

    K

    1t

    33K

    1

    8/1

    1ln

    K

    1

    8

    1

    1K

    1

    10/1

    1ln

    K

    1

    10

    1

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    PART III: MATHEMATICSSECTION -I : Single Correct Answer Type

    This section contains 10 multiple choice questions. Each question hasfour choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

    Q.41 The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line

    4x5y = 20 to the circle x2 + y2 = 9 is

    (A) 20(x2 + y2)36x + 45y = 0 (B) 20(x2 + y2) + 36x45y = 0

    (C) 36(x2 + y2)

    20x + 45y = 0 (D) 36(x2 + y2) + 20x

    45y = 0Ans. [A]

    [Sol. Equation of chord of contact with mid-point of (h, k)

    yk =k

    h(xh) hx + ky = h2 + k2

    Equation of chord of contact for point

    5

    204,

    x + y

    5

    204= 9

    h

    =

    5

    204

    k

    = 9kh

    22

    4x5y = 20

    (h, k) (0, 0)

    x + y = 92 2

    5

    204,

    = 22kh

    h9

    ;

    5

    204 = 22

    kh

    k9

    22 kh

    h6

    20 =

    22 kh

    k45

    20 (h2 + k2)36h + 45k = 0 Locus is 20(x2 + y2)36x + 45y = 0. Ans.]

    Q.42 The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that

    each person gets atleast one ball is

    (A) 75 (B) 150 (C) 210 (D) 243

    Ans. [B]

    [Sol. Number of ways =!2!3!1!1

    !3!5

    !2!2!2!1

    !3!5

    = 90 + 60 = 150. Ans.]

    Q.43 Let f(x) = ,Rx,

    0x,0

    0x,x

    cosx2

    then f is

    (A) differentiable both at x = 0 and at x = 2

    (B) differentiable at x = 0 but not differentiable at x = 2

    (C) not differentiable at x = 0 but differentiable at x = 2

    (D) differentiable neither at x = 0 nor at x = 2

    Ans. [B]

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    [18]

    [Sol. f(x) = ,Rx,

    0x,0

    0x,x

    cosx2

    At x = 0

    LHD =h

    )0(f)h0(fLim

    0h

    =

    h

    hcosh

    Lim

    2

    0h

    = 0

    RHD =h

    )0(f)h0(fLim

    0h

    =h

    hcosh

    Lim

    2

    0h

    = 0

    f(x) =

    h2x2,x

    cosx

    2xh2,x

    cosx

    2

    2

    f '(x) =

    h2x2,x

    sinx

    cosx2

    2xh2,x

    sinx

    cosx2

    LHD = ; RHD = f(x) is differentiable at x = 0 but not derivable at x = 2. Ans.]

    Q.44 The function f : [0, 3] [1, 29], defined by f(x) = 2x315x2 + 36x + 1, is(A) one-one and onto. (B) onto but not one-one.

    (C) one-one but not onto. (D) neither one-one nor onto.

    Ans. [B][Sol. f(x) = 2x315x2 + 36x + 1

    f '(x) = 6x230x + 36 = 6 (x 2) (x3)

    Sign of f ' 0 2 3

    + +

    sign of f ' changes in [0 3]

    function is not one-one

    O

    (0, 1)

    (3, 28)

    (2, 29)

    f(0) = 1 ; f(2) = 29 and f(3) = 28

    Range = [1, 29] Function is onto but not one-one. Ans.]

    Q.45 If

    bax

    1x

    1xxLim

    2

    x= 4, then

    (A) a = 1, b = 4 (B) a = 1, b =4 (C) a = 2, b = 3 (D) a = 2, b = 3

    Ans. [B]

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    [20]

    [Sol. Let P =

    333231

    232221

    131211

    aaa

    aaa

    aaa

    det.P =

    333231

    232221

    131211

    aaa

    aaa

    aaa

    then det Q =

    336

    325

    314

    235

    224

    213

    134

    123

    112

    a2a2a2

    a2a2a2

    a2a2a2

    = 222324

    332

    3231

    232

    2221

    132

    1211

    a2a2a

    a2a2a

    a2a2a

    = 29 2 22

    333231

    232221

    131211

    aaa

    aaa

    aaa

    = 212 det P = 213. Ans.]

    Q.49 The integral dx

    xtanxsec

    xsec29

    2

    equals (for some arbitrary constant K)

    (A) Kxtanxsec71

    11

    1

    xtanxsec

    1 2211

    (B)

    Kxtanxsec7

    1

    11

    1

    xtanxsec

    1 2211

    (C)

    Kxtanxsec7

    1

    11

    1

    xtanxsec

    1 2211

    (D)

    Kxtanxsec7

    1

    11

    1

    xtanxsec

    1 2211

    Ans. [C][Sol. y = sec x + tan x

    sec2xtan2x = 1

    sec xtan x =y

    1

    2sec x = y +y

    1 sec x =

    y2

    1y2

    dy = (sec x tan x + sec2x) dx dy = y sec x dx dx =xsecy

    dy

    I =

    29

    2

    y

    xsecy

    dyxsec

    =

    dyy

    y2

    1y

    211

    2

    =

    dyy

    1y

    2

    1213

    2

    = dyyy21 21329

    I =

    211

    y

    27

    y

    2

    121127

    + K

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    [21]

    I = 21127 )xtanx(sec11

    1

    xtanxsec7

    1

    + K

    I =

    7

    xtanxsec

    11

    1

    xtanxsec

    12

    211+ K. Ans.]

    Q.50 The point P is the intersection of the straight line joining the points Q (2, 3, 5) and R (1,1, 4) with the

    plane 5x

    4y

    z = 1. If S is the foot of the perpendicular drawn from the point T (2, 1, 4) to QR, thenthe length of the line segment PS is

    (A)2

    1(B) 2 (C) 2 (D) 22

    Ans. [A]

    [Sol. Line QR1

    4z

    4

    1y

    1

    1x

    = r

    Let coordinates of point P are (r + 1, 4r1, r + 4) which lies on the plane

    5x4yz = 1

    5r + 54 (4r1)(r + 4) = 1

    12r + 5 = 1 12r = 4 r =3

    1

    P

    R (1,1, 4)

    T (2, 1, 4)

    Q (2, 3, 5)

    S

    P

    3

    13,

    3

    1,

    3

    4

    Let coordinate of S are ( + 1, 41, + 4) ; ST QR ( + 12) 1 + (411) 4 + ( + 44) 1 = 0

    1 + 16 + = 0 =2

    Coordinate of S are

    2

    9,1,

    2

    3

    Now, length PS =222

    2

    9

    3

    131

    3

    1

    2

    3

    3

    4

    =36

    1

    9

    4

    36

    1 =

    2

    1. Ans.]

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    [22]

    SECTION -III: Integer Answer TypeThis section contains 5 questions. The answer to each question is a single-digit integer,ranging from 0 to 9 (both inclusive)

    Q.51 Let , [0, 2] be such that

    2 cos (1sin ) = sin2

    2cot

    2tan cos 1,

    tan(2) > 0 and 1 < sin e

    1

    e

    1

    e

    11As

    Again S = 1

    21

    x21

    0

    x dxedxe22

    < 1

    21

    2121

    0

    dxedx1

    S <

    2

    11

    e

    1

    2

    1

    Also,

    e

    11

    4

    1

    e

    11 =

    e4

    1

    e

    1

    4

    3 =

    e4

    e4e3 > 0

    S >

    e

    11

    4

    1. Ans.]

    Q.53 A ship is fitted with three engines E1, E

    2and E

    3. The engines function independently of each other with

    respective probabilities2

    1,

    4

    1and

    4

    1. For the ship to be operational at least two of its engines must

    function. Let X denote the event that the ship is operational and let X1, X

    2and X

    3denotes respectively

    the events that the engines E1, E

    2and E

    3are functioning. Which of the following is(are) true?

    (A) 16

    3X|XP c

    1 (B) P [Exactly two engines of the ship are functioning | X] =

    8

    7

    (C) P [X | X2] =

    16

    5(D) P [X | X

    1] =

    16

    7

    Ans. [B, D]

    [Sol. P(X1) =

    2

    ; P(X

    2) =

    4

    1and P(X

    3) =

    4

    1

    P(X) = P (atleast 2 of X1, X

    2, X

    3happens)

    = P(X1 X

    2) + P(X

    2 X

    3) + P(X

    3 X

    1)2P(X

    1 X

    2 X

    3)

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    [24]

    =4

    1

    4

    1

    2

    12

    2

    1

    4

    1

    4

    1

    4

    1

    4

    1

    2

    1 =

    4

    1

    (A)

    X

    XP

    c1 =

    )X(P

    XXPc1

    =

    )X(P

    XXXP23

    c1

    =

    4

    14

    1

    4

    1

    2

    1

    =8

    1

    (B)X

    )happensX,X,Xof2Exactly(P321

    =)X(P

    X)XXXof2Exactly(P

    1

    321 =

    4

    14

    1

    4

    1

    2

    1

    4

    1

    =

    8

    7

    (C)

    2X

    XP =

    )X(P

    )XX(P

    2

    2

    =)X(P

    )XXX(P)XXX(P)XXX(P

    2

    c3123

    c12321

    =

    4

    14

    3

    2

    1

    4

    1

    4

    1

    2

    1

    4

    1

    4

    1

    42

    1

    =8

    5

    (D)

    1X

    XP =

    )X(P

    )XX(P

    1

    1

    =)X(P

    )XXX(P)XXX(P)XXX(P

    1

    c3213

    c21321

    =

    2

    14

    3

    4

    1

    2

    1

    4

    1

    4

    3

    2

    1

    4

    1

    4

    1

    2

    =16

    7. Ans.]

    Q.54 Tangents are drawn to the hyperbola 14y

    9x

    22

    , parallel to the straight line 2xy = 1. The points

    of contact of the tangent on the hyperbola are

    (A)

    2

    1,

    22

    9(B)

    2

    1,

    22

    9(C) 22,33 (D) 22,33

    Ans. [A, B]

    [Sol. If line y = mx + c is tangent at (x1, y

    1) then contact points is

    c

    b,

    c

    ma22

    Equation of tangent of slope m is y = mx 222 bma

    point of contact

    2

    1,

    22

    9;

    2

    1,

    22

    9. Ans.]

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    [25]

    Q.55 If y (x) satisfies the differential equation y' y tan x = 2x sec x and y(0) = 0, then

    (A)284

    y2

    (B)

    184'y

    2

    (C)93

    'y2

    (D)

    33

    2

    3

    4

    3'y

    2

    Ans. [A, D]

    [Sol. )x(secx2xtanydxdy

    I.F. = dxxtane =

    xcosnel = cos x

    Solution is ycos x = dxxcosxsecx2ycos x = x2 + C ; y(0) = 0 C = 0 y = x2 sec x and y' = 2x sec x + x2 sec x tan x

    4

    y =28

    2;

    4

    'y =282

    2

    3

    y =9

    22

    ;

    3

    'y =33

    2

    3

    4 2

    . Ans.]

    SECTION -III: Integer Answer TypeThis section contains 5 questions. The answer to each question is a single-digit integer,ranging from 0 to 9 (both inclusive)

    Q.56 Let f : R R be defined as f (x) = | x | + | x21 |. The total number of points at which f attains eithera local maximum or a local minimum is

    [Ans. 5][Sol. f : R R

    f (x) = | x | + | (x1) (x + 1) | =

    1x,1xx

    1x0,1xx

    0x1,1xx

    1x,1xx

    2

    2

    2

    2

    f ' (x) =

    1x,x211x0,x21

    0x1,1x2

    1x,1x2

    1 1/2 1/2O

    1

    Graph of f(x) Total number of points = 5. Ans.]

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    [26]

    Q.57 The value of

    ......

    23

    14

    23

    14

    23

    14

    23

    1log6

    2

    3 is

    [Ans. 4]

    [Sol. Let x = ......23

    14

    23

    14

    23

    14

    23

    1

    x = x423

    1 18x2 = 4x

    18x2 + x4 = 0

    x =36

    441811 =

    36

    171 =

    36

    16=

    9

    4=

    2

    3

    2

    Hence, 6 +

    2

    2

    33

    2log

    = 62 = 4 Ans.]

    Q.58 Let p (x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum

    at x = 3. If p(1) = 6 and p(3) = 2, then p'(0) is

    [Ans. 9]

    [Sol. p ' (x) = k(x1)(x3) = k(x24x + 3)

    p'(x) =

    x3

    2

    x4

    3

    xk

    23

    +

    p(1) = 6

    323

    1k + = 6 6

    3

    k4

    p(3) = 2

    k(918 + 9) + = 2 = 2

    3

    k4= 4 k = 3

    p'(x) = 3(x24x + 3) p'(0) = 9 Ans.]

    Q.59 If candb,a

    are unit vectors satisfying 9accbba222

    , then c5b5a2

    is

    [Ans. 3]

    [Sol.222

    accbba

    = 62 ba

    = 9 ba

    =2

    3

    2cba

    = 0ba2a 2

    0cba

    acb

    c5b5a2

    = a5a2

    = 3. Ans.]

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    Q.60 Let S be the focus of the parabola y2 = 8x and let PQ be the common chord of the circle

    x2 + y22x 4y = 0 and the given parabola. The area of the triangle PQS, is

    [Ans. 3]

    [Sol. Parabola y2 = 8x focus = (2, 0)circle x2 + y22x4y = x(x2) + y(y4) = 0

    S(2,0)

    Q(2,4)

    P(0,0)

    circle with diametric ends (0, 0) and (2, 4)which are vertex and one end of Latus rectum.

    Area of PQS

    = 4221 = 4 Ans.

    Aliter: Since parabola and circle both passes through origin.

    one common point P (0, 0).

    Solving above two equations simultaneously, x2 + 6x8 x2 = 0

    By hit and trail, x = 2 satisfy above equation.

    So another common point Q(2, 4)

    Area of PQS = 422

    1 = 4 Ans.]