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0518/IITEQ18/Paper1/QP&Soln/Pg.1 Solution to IIT JEE 2018 (Advanced) : Paper - I PART I PHYSICS SECTION 1 (Maximum Marks: 24) This section contains SIX (06) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme: Full Mark : + If only (all) the correct option(s) is (are) chosen. Partial Marks : + If all the four options are correct but ONLY three options are chosen. Partial Marks : + If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : + If two or more options are correct but ONLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : In all other cases. For Example: If first, third and fourth are the ONLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the three correct options (either first or third or fourth option) ,without selecting any incorrect option (second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in -2 marks. 1. The potential energy of a particle of mass m at a distance r from a fixed point O is given by V(r) = kr 2 /2, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true? (A) k v R 2m (B) k v R m (C) 2 L mk R (D) 2 mk L R 2 1. (B), (C) The potential energy of the particle is, V = 2 kr 2 F = dv dr F = kr At r = R, F = kR To keep the particle in circular motion, this force must be equal to centripetal force. So, kR = 2 mv R or v = 2 kR k R m m The angular momentum of the particle is, L = mvR = m 2 k R m O r v

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  • 0518/IITEQ18/Paper1/QP&Soln/Pg.1

    Solution to IIT JEE 2018 (Advanced) : Paper - I

    PART I – PHYSICS

    SECTION 1 (Maximum Marks: 24) This section contains SIX (06) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE

    of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme:

    Full Mark : +𝟒 If only (all) the correct option(s) is (are) chosen. Partial Marks : +𝟑 If all the four options are correct but ONLY three options are

    chosen. Partial Marks : +𝟐 If three or more options are correct but ONLY two options are chosen, both of which are correct options. Partial Marks : +𝟏 If two or more options are correct but ONLY one option is chosen and it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : −𝟐 In all other cases.

    For Example: If first, third and fourth are the ONLY three correct options for a question with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the three correct options (either first or third or fourth option) ,without selecting any incorrect option (second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in -2 marks.

    1. The potential energy of a particle of mass m at a distance r from a fixed point O is given by V(r) = kr2/2, where k is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius R about the point O. If v is the speed of the particle and L is the magnitude of its angular momentum about O, which of the following statements is (are) true?

    (A) k

    v R2m

    (B) k

    v Rm

    (C) 2L mk R (D) 2mk

    L R2

    1. (B), (C) The potential energy of the particle is,

    V = 2kr

    2

    F = dvdr

    F = kr

    At r = R, F = kR To keep the particle in circular motion, this force must be equal to centripetal force.

    So, kR = 2mv

    R or v =

    2kR kR

    m m

    The angular momentum of the particle is,

    L = mvR = m 2k

    Rm

    O

    r v

  • (2) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.2

    2. Consider a body of mass 1.0 kg at rest at the origin at time t = 0. A force ˆ ˆF ( t i j) is applied on the body, where = 1.0 Ns1 and = 1.0 N. The torque acting on the body about the origin at time t = 1.0 S is Which of the following statements is (are) true?

    (A) 1

    N m3

    (B) The torque is in the direction of the unit vector k̂

    (C) The velocity of the body at t = 1 is 11 ˆ ˆv (i 2j)m s2

    (D) The magnitude of displacement of the body at t = 1 s is 1

    m6

    2. (A), (C)

    The force applied on the body is, ˆ ˆF t i j At = 1 N s1 and = 1 N

    ˆ ˆF t i j or mdv ˆ ˆt i jdt

    On integrating both the sides,

    2t ˆ ˆmv i t j2

    2t ˆ ˆv i t j [m 1 kg]2

    2dr t ˆ ˆi t j

    dt 2

    At t = 0, the body is at rest so, v 0 and r 0. On integrating both the sides,

    3 2t tˆ ˆr i j6 2

    At t = 1 s, torque acting on the body is,

    r F

    = 1 1 1 1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆi j i j k k k6 2 6 2 3

    At t = 1 s, velocity of the body is, 11 1ˆ ˆ ˆ ˆv i j i 2j ms2 2

    At t = 1 s, the magnitude of displacement of the body is,

    1 0 1 1 1 1ˆ ˆ ˆ ˆs r r i j 0 i j6 2 6 2

    2 2

    1 1 10 10| s | m

    6 2 36 6

    3. A uniform capillary tube of inner radius r is dipped vertically into a beaker filled with water. The water rises to a height h in the capillary tube above the water surface in the beaker. The surface tension of water is . The angle of contact between water and the wall of the capillary tube is . Ignore the mass of water in the meniscus. Which of the following statements is (are) true?

    (A) For a given material of the capillary tube, h decreases with increase in r. (B) For a given material of the capillary tube, h is independent of (C) If this experiment is performed in a lift going up with a constant acceleration, then

    h decreases (D) h is proportional to contact angel

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (3)

    0518/IITEQ18/Paper1/QP&Soln/Pg.3

    3. (A), (C)

    2

    ghR

    where R is the radius of meniscus.

    h = 2

    gR

    R = r

    ,cos

    where r is the radius of capillary and is the angle of contact.

    h = 2 cos

    gr

    (A) For a given material, is constant.

    Therefore, h 1

    r

    (B) h depends on as h . (C) If lift is going up with constant acceleration a, geff = g + a

    h = 2 cos

    ;(g a) r

    h decreases

    (D) h is proportional to cos . 4. In the figure below, the switches S1 and S2 are closed simultaneously at t = 0 and a current

    starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current I in the middle wire reaches its maximum magnitude Imax a x at time t = T. Which of the following statements is (are) true?

    (A) maxV

    I2R

    (B) maxV

    I4R

    (C) L

    ln 2R

    (D) 2L

    ln 2R

    4. (B), (D)

    max 2 1 maxI I I

    R/2L t (R/L)t

    2 1

    V VI I I 1 e 1 e

    R R

    = R/L t R/2L tV

    e eR

    I1 I2

  • (4) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.4

    For Imax, d

    dt(I) = 0.

    R/L t R/2L t1

    e e2

    R/2L t1

    e2

    R

    t n 22L

    t = 2L

    n 2R

    When I = Imax, t =

    = 2L

    n 2R

    Imax = R 2L R 2L

    n2 n2L R 2L R

    Ve eR

    |Imax| = V 1 1

    R 4 2

    =

    V

    4R

    5. Two infinitely long straight wires lie in the xy-plane along the lines x = +R. The wire

    located at x = +R carries a constant current I1 and the wire located at x = R carries a constant current I2 . A circular loop of radius R is suspended with its centre at (0, 0, 3R ) and in a plane parallel to the xy-plane. This loop carries a constant current / in the clockwise direction as seen from above the loop. The current in the wire is taken to be

    positive if it is in the ĵ direction. Which of the following statements regarding the

    magnetic field B is (are) true?

    (A) If I1 = I2, then B cannot be equal to zero at the origin (0, 0, 0)

    (B) If I1 > 0 and I2 < 0, then B can be equal to zero at the origin (0, 0, 0)

    (C) If I1 < 0 and I2 > 0, then B can be equal to zero at the origin (0, 0, 0) (D) If I1 = I2, then the z-component of the magnetic field at the centre of the loop is

    0I

    2R

    5. (A), (B), (D)

    (A) At the origin, B 0 due to two wires if I1 = I2, hence netB at the origin is equal to B , due to loop, which is non zero.

    (B) If I1 > 0 and I2 < 0, B at the origin due to wires will be along k̂ direction and B

    due to loop is along k̂ direction, hence B can be zero at the origin.

    (0, 0, 3 R) I

    z

    y

    x R R x

    y

    O

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (5)

    0518/IITEQ18/Paper1/QP&Soln/Pg.5

    (C) If I1 < 0 and I2 > 0, B at the origin due to wires is along k̂ and also due to ring is along k̂ so, B can not be zero.

    (D) At the centre of the loop B due to wires is along x-axis. Hence the z-component of

    the magnetic field at the center of the loop is 0I k̂2R

    .

    6. One mole of a monatomic ideal gas undergoes a cyclic process as shown in the figure

    (where V is the volume and T is the temperature). Which of the statements below is (are) true?

    (A) Process I is an isochoric process (B) In process II, gas absorbs heat (C) In process IV, gas releases heat (D) Processes I and III are not isobaric

    6. (B), (C), (D) (A) Volume V is decreasing in process I. (B) U = 0, W > 0 Q > 0 Process II is isothermal expansion. (C) U = 0, W < 0 Q < 0 Process III is isothermal compression. (D) For an isobaric process TV graph must be linear.

    SECTION II (Maximum Marks:24)

    This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE.

    For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

    Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases.

    7. Two vectors A and B are defined as ˆA ai and ˆ ˆB a(cos t i sin t j), where a is a

    constant and = /6 rad s1. If A B 3 A B at time t = for the first time, the

    value of , in, seconds, is ________.

    7. [2]

    t

    A B 2a cos2

  • (6) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.6

    t

    A B 2a sin2

    So, 2a cost t

    3 2a sin2 2

    tant 1

    2 3

    t

    2 6

    t = 3

    Now, = 6

    rad s1

    t6 3

    t = 2 s

    8. Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed 1.0 m s1 and the man behind walks at a speed 2.0 ms-1. A third man is standing at a height 12 m above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency 1430 Hz . The speed of sound in air is 330 m s-1. At the instant, when the moving men are 10 m apart, the stationary man is equidistant from them. The frequency of beats in Hz, heard by the stationary man at this instant, is________.

    8. [5]

    A

    1v 330 2cos

    f f 1430 1430 1430 12cos1v 2cos 330 2cos 330

    330

    Bv 330 cos

    f f 1430 1430 1v cos 330 cos 330

    f = 14303cos

    13cos330

    = 135 5

    5Hz From CAO,cos13 13

    9. A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle 60° with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the

    ground is (2 3 ) / 10 s, then the height of the top of the inclined plane, in metres, is _______ . Take g = 10 m s2.

    2.0 m s1 1.0 m s1

    13 m 13 m 12 m

    A O B

    C

    5 m 5 m

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (7)

    0518/IITEQ18/Paper1/QP&Soln/Pg.7

    9. [0.75] Acceleration down the inclined plane is

    a = 2

    2

    g sin

    k1

    R

    R

    D

    gsina

    2

    2gsina

    3

    2R R 2h 1 gsin 4h 16h 3

    t t sin 60sin 2 2 gsin 3g 2

    2D D 2h 1 2gsin 3h 4h 1

    t t cos60sin 2 3 gsin g 2

    Since R D2 3

    t t10

    16h 4h 2 3

    3g g 10

    h 4 2 3

    2g 3 10

    [g = 10 m s2]

    2 3 3 3 3

    h h 0.75m2 44 2 3

    10. A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 N m1 and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of 2.0 m s1 collides elastically with the first block. The collision is such that the 2.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is _______ .

    10. [2.09] After collision 1v1 + 2v2 = 1 2 v1 + 2v1 = 2 ………. (i)

    e = 2 1v v

    12

    v2 v1 = 2 ……….. (ii)

    On solving v2 = 14 ms

    3 and v1 =

    2

    3 m s1

    Time period T = m 2

    2 2 2k 2

    A B

    C

    h 60

    K = 2 N m1 2 m s1

  • (8) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.8

    Displacement of 1st block in time t = T

    2

    s = 2 6.28

    2.093 3

    s = 2.09 m 11. Three identical capacitors C1, C2 and C3 have a capacitance of 1.0 F each and they are

    uncharged initially. They are connected in a circuit as shown in the figure and C1 is then filled completely with a dielectric material of relative permittivity r. The cell electromotive force (emf) V0 = 8 V. First the switch S1 is closed while the switch S2 is kept open. When the capacitor C3 is fully charged, S1 is opened and S2 is closed simultaneously. When all the capacitors reach equilibrium, the charge on C3 is found to be 5 C. The value of r = ________ .

    11. [1.50] Applying loop rule,

    r

    5 3 30

    1 1

    2 r

    30

    r3

    1.502

    12. In the xy-plane, the region y > 0 has a uniform

    magnetic field 1ˆB k and the region y < 0 has

    another uniform magnetic field 2ˆB k . A positively

    charged particle is projected from the origin along the positive y-axis with speed v0 = m s1 at t = 0, as shown in the figure. Neglect gravity in this problem. Let t = T be the time when the particle crosses the x-axis from below for the first time. If B2 = 4B1, the average speed of the particle, in m s1, along the x-axis in the time interval T is _______ .

    1 F V0 = 8V

    + 8 C

    8 C

    + 3 C 3 C

    +3 C 3 C

    1F

    + 5 C

    5 C

    1F

    r

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (9)

    0518/IITEQ18/Paper1/QP&Soln/Pg.9

    12. [2]

    R1 = 1

    mv

    qB

    R2 = 2

    mv

    qB

    B2 = 4B1

    R2 = 11

    R4

    Distance traveled in x direction x = 2R1 + 2R2

    x = 2R1 + 1 1R 5R

    2 2

    T1 = 1

    m

    qB

    T2 = 1

    2

    Tm

    qB 4

    Total time = 1 2T T

    2 2 = 1

    5T

    8

    Average speed V =

    1

    1

    1 1

    5R

    Rx 24

    5Tt T

    8

    R = mV

    qB

    T = 2 m

    qB

    Average 4R 4V

    speed 2T 2

    Average speed = 2 13. Sunlight of intensity 1.3 kW m2 is incident normally on a thin convex lens of focal length

    20 cm. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in kW m2 , at a distance 22 cm from the lens on the other side is ______ .

    13. [130] I = 1.3 kW/m2 f = 20 cm ABF ~ PQF

    r 2 1

    R 20 10

    a 1

    A 100

    Total energy incident E = P0 A = P a

    P = P0A

    a

    P = 1.3 100 = 130

    V0 = m/s 1ˆB k

    F V0

    2ˆB k

    x

    y

    A

    R

    B Q

    P

    r

    F

    20 cm 2 cm

  • (10) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.10

    14. Two conducting cylinders of equal length but different radii are connected in series between two heat baths kept at temperatures T1 = 300 K and T2 = 100 K, as shown in the figure. The radius of the bigger cylinder is twice that of the smaller one and the thermal conductivities of the materials of the smaller and the larger cylinders are K1 and K2 respectively. If the temperature at the junction of the two cylinders in the steady state is 200 K, then K1 / K2 = _______.

    14. [4] r2 = 2r1 A2 = 4A1

    1 1 2 2k A k AdQ

    (300 200) (200 100)dt L L

    k1A1 = k2A2

    1 2

    2 1

    k A4

    k A

    SECTION III (Maximum Marks:12) This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO

    (02) questions. Each question has FOUR options. ONLY ONE of these four options is corresponds to

    the correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : 1 In all other cases.

    Paragraph for Q. No. 15 & 16

    PARAGRAPH “X” In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, [E] and [B] stand for dimensions of electric and magnetic fields respectively, while [0] and [0] stand for dimensions of the permittivity and permeability of free space respectively. [L] and [T] are dimensions of length and time respectively. All the quantities are given in SI units.

    15. The relation between [E] and [B] is (A) [E] = [B] [L] [T] (B) [E] = [B] [L]1 [T] (C) [E] = [B] [L] [T]1 (D) [E] = [B] [L]1 [T]1

    T1 = 300 K T2 = 100 K

    T = 200 K

    k1 k2

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (11)

    0518/IITEQ18/Paper1/QP&Soln/Pg.11

    15. (C) F = qvB = qE E = vB [E] = [L] [T]1 [B] 16. The relation between [0] and [0] is (A) [0] = [0] [L]2 [T]2 (B) [0] = [0] [L]2 [T]2 (C) [0] = [0]1 [L]2 [T]2 (D) [0] = [0]1 [L]2 [T]2 16. (D)

    0 0 = 21

    c

    0 = [L]2 [T]2 [0]1 Paragraph for Q. No. 17 & 18

    PARAGRAPH “A” If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z = x/y. If the errors in x, y and z are x, y and z, respectively, then

    z z = 1

    x x x x y1 1

    y y y x y

    .

    The series expansion for

    1y

    1y

    , to first power in y/y, is 1 (y/y). The relative errors

    in independent variables are always added. So the error in z will be z = zx y

    x y

    .

    The above derivation makes the assumption that x/x

  • (12) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.12

    18. In an experiment the initial number of radioactive nuclei is 3000. It is found that 1000 40 nuclei decayed in the first 1.0 s. For |x|

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (13)

    0518/IITEQ18/Paper1/QP&Soln/Pg.13

    2. The correct statement(s) regarding the binary transition metal carbonyl compounds is (are) (Atomic numbers: Fe = 26, Ni = 28)

    (A) Total number of valence shell electrons at metal centre in Fe(CO)5 or Ni(CO)4 is 16 (B) These are predominantly low spin in nature (C) Metal-carbon bond strengthens when the oxidation state of the metal is lowered (D) The carbonyl C O bond weakens when the oxidation state of the metal is increased

    2. (B), (C) Total number of valence shell electrons at metal centre in Fe(CO)5 or Ni(CO)4 is 18. Metal carbonyl compounds are predominantly low spin in nature. MetalCarbon bond strengthens when the oxidation state of the metal is lowered.

    3. Based on the compounds of group 15 elements, the correct statement(s) is (are) (A) Bi2O5 is more basic than N2O5 (B) NF3 is more covalent than BiF3 (C) PH3 boils at lower temperature than NH3 (D) The N N single bond is stronger than the P P single bond 3. (A), (B), (C) Metal oxides are basic in nature while non-metal oxides are acidic in nature. Bi2O5 is more basic than N2O5. Non-metals mainly forms covalent bonds while metal forms ionic bonds. NF3 is more covalent than BiF3. Boiling point : NH3 > PH3 Bond energy P P > N N

    4. In the following reaction sequence, the correct structure(s) of X is (are)

    (A) (B) (C) (D) 4. (B)

    OHMe3 3

    22 2 HCONMeEt O COPBr NaNNaI

    Me

    Me N3

    Enantiomerically Pure 5. The reaction(s) leading to the formation of 1, 3, 5-trimethylbenzene is (are) (A) (B)

    (C) (D)

  • (14) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.14

    5. (A), (B), (D)

    6. A reversible cyclic process for an ideal gas is shown below. Here, P, V, and T are

    pressure, volume and temperature, respectively. The thermodynamic parameters q, w, H and U are heat, work, enthalpy and internal energy, respectively.

    The correct option (s) is (are) (A) qAC = UBC and WAB = P2 (V2 V1) (B) WBC = P2(V2 V1) and qBC = HAC (C) HCA < UCA and qAC = UBC (D) qBC = HAC and HCA > UCA 6. (B), (C)

    AB = Isothermal process, E or U = 0 AC = Isochoric process, v = U BC = Isobaric process, P = H

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (15)

    0518/IITEQ18/Paper1/QP&Soln/Pg.15

    SECTION II (Maximum Marks:24)

    This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE.

    For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

    Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases.

    7. Among the species given below, the total number of diamagnetic species is _______.

    H atom, NO2 monomer, 2O (superoxide), dimeric sulphur in vapour phase, MnsO4,

    (NH4)2[FeCl4], (NH4)2 [NiCl4], K2MnO4, K2CrO4 7. [1.00] K2CrO4 i.e. Cr

    +6 d0 (diamagnetic) Paramagnetic = H-atom, NO2 (monomer), 2O

    (superoxide), S2(vap), Mn3O4, (NH4)2[FeCl4],

    (NH4)2[NiCl4], K2MnO4. 8. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is

    completely used by NiCl2.6H2O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickelammonia coordination compound thus produced is _______.

    (Atomic weights in g mol1 : H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59)

    8. [2992.00]

    4 2 4(NH ) SO + Ca(OH)2 CaSO4 + 2NH3 + 2H2O 1584g = 12 mol 12 mol 24 mol

    NiCl26H2O + 6NH3 [Ni(NH3)6] Cl2 + 6H2O 952 g = 4 mol 24 mol 4 mol

    Total mass = 12 172 + 4 232 = 2992 g

    9. Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX following the sequential instructions given below. Neglect the charge balance.

    (i) Remove all the anions (X) except the central one (ii) Replace all the face centered cations (M) by anions (X)

    (iii) Remove all the corner cations (M) (iv) Replace the central anion (X) with cation (M)

    The value of number of anions

    number of cations

    in Z is ______ .

    9. [3.00] MX have NaCl type structure.

    M+ occupies all FCC positions and X occupies all octahedral voids.

  • (16) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.16

    The value of number of anions

    number of cations

    in Z = 3

    1

    10. For the electrochemical cell, Mg(s) | Mg2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu(s) the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is.

    (given, F

    R= 11500 K V1, where F is the Faraday constant and R is the gas constant, The

    value of ln(10) = 2.30) 10. [10.00] Mg(s) | Mg

    +2 (aq. 1M) || Cu+2 (aq. 1M) | Cu(s)

    E = 2

    02

    RT [Mg ]E n

    2F [Cu ]

    E = 0E 2.70 , when [Mg+2] = [Cu+2] Mg(s) | Mg

    +2 (aq, x M) || Cu+2 (aq. 1M) | Cu(s)

    Mg(s) + 2

    (aq)Cu 2(s) (aq)Cu Mg

    E = 2

    0 e 2

    RT [Mg ]E n

    2F [Cu ]

    2.67 = e300 x

    2.70 n2 11500 1

    0.03 = e300

    n (x)2 11500

    e0.03 2 11500

    n x 2.30300

    x = 10.00 11. A closed tank has two compartments A and B, both filled with oxygen (assumed to be

    ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in m3) of the compartment A after the system attains equilibrium is _______.

    = M+

    = X

    (Z)

    (i) (ii) (iii) (iv)

    = M+

    = X

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (17)

    0518/IITEQ18/Paper1/QP&Soln/Pg.17

    11. [2.22]

    nA = No. of moles in vessel A = 5 1 1

    R 400 80R

    ….(1)

    nB = No. of moles in vessel B = 1 3 1

    R 300 100R

    ….(2)

    Pressure and temperature on the both sides are must be same.

    A = No. of moles in vessel A =

    Px

    RT ….(3)

    Bn = No. of moles in vessel B = P(4 x)

    RT

    ….(4)

    Using equation 1, 2, 3 and 4, we get x = 20

    2.229

    12. Liquids A and B form ideal solution over the entire range of composition. At temperature

    T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions xA and xB, respectively, has vapour pressure of 22.5 Torr. The value of xA/xB in the new solution is ______. (given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

    12. [19.00]

    The vapour pressure of pure liquid A, AP 20 torr

    The vapour pressure of pure liquid B, BP = ?

    PTotal = A A B BP X P X

    45 = B1 1

    20 P2 2

    90 = B20 P BP 70 torr

    For new solution, PTotal = 22.5 = 0A A B BP X P X

    22.5 = 20x + 70(1 x) x = 0.95 = XA XB = 1 x = 0.05

    AB

    X 0.9519.00

    X 0.05

    1 m3, 5 bar 400 K

    3 m3, 1 bar 300 K

    Old Partition

    Fig. 1

    A

    A B

    New Partition (Conducting)

    Fig. 2

  • (18) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.18

    13. The solubility of a salt of weak acid (AB) at pH 3 is Y 103 mol L1 . The value of Y is _______. (Given that the value of solubility product of AB (Ksp) = 2 1010 and the value of ionization constant of HB (Ka) = 1 108)

    13. [4.47]

    Solubility = sp[n ]

    K 1Ka

    3

    3 10 5

    8

    10Y 10 2 10 1 2 10

    10

    3 6 3Y 10 20 10 20 10 Y 20 4.47

    14. The plot given below shows P T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.

    On addition of equal number of moles of a non-volatile solute S in equal amount (in kg)

    of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is _______.

    14. [0.05]

    b X NaCl b X( T ) i (K ) m 362 360 2

    b Y NaCl b Y( T ) i (K ) m 368 367 1

    b Xb Y

    (K )2

    (K )

    On addition of equal no. of moles of a non-volatile solute, S in equal amount (in kg) of these solvent. SoluteS undergo dimerisation in solventX and Y

    Van’t Hoff factor = i = 12

    ( = degree of ionization)

    iY = 0.7

    1 1 0.35 0.652

    iX = ?

    b X b Y( T ) 3( T ) X b X Y b Yi (K ) m i (K ) m

    XB X b Y1 (K ) m (0.65) (K ) m

    2

    X = 0.05

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (19)

    0518/IITEQ18/Paper1/QP&Soln/Pg.19

    SECTION III (Maximum Marks:12) This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO

    (02) questions. Each question has FOUR options. ONLY ONE of these four options corresponds to the

    correct answer. For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : 1 In all other cases. Paragraph for Q. No. 15 & 16

    PARAGRAPH “X” Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with moist KOH furnishes Y as the major product. Reaction of X with H2/Pd-C, followed by H3PO4 treatment gives Z as the major product.

    15. The compound Y is (A) (B)

    (C) (D)

    16. The compound Z is (A) (B) (C) (D) 15. (C), 16. (A)

    3

    CO, HCl

    anhy. AlCl /CuCl

    C H

    O

    2Ac O/NaOAc

    CH

    O

    CH COOH

    C C HCH CH COOH

    2 2 3Br , Na CO 473 K moist KOH

    2 2CH CH COOH

    3 4H PO2H / Pd C

    (X)

    Cinnamic Acid

    (Y)(X)

    (Z)

  • (20) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.20

    Paragraph for Q. No. 17 & 18

    PARAGRAPH “A”

    An organic acid P (C11H12O2) can easily be oxidized to a dibasic acid which reacts with ethyleneglycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S.

    23 2

    2 2 2

    3 2 24 3

    1) H /Pd C2) NH / 1) H Pd C 1) HCl3) Br /NaOH 2) SOCl 2) Mg/Et O

    4) CHCl , KOH, 3) MeMgBr, CdCl 3) CO (dry ice)4) NaBH/Pd C 4) H O

    S P Q

    R

    17. The compound R is (A) (B) (C) (D)

    18. The compound S is (A) (B) (C) (D) 17. (A), 18. (B)

    Organic Acid (C11H12O2) [O] dibasic acid ethylene glycol polymer (dacron)

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (21)

    0518/IITEQ18/Paper1/QP&Soln/Pg.21

    COOH

    COOH

    H2 2

    ethylene glycol

    H O CH CH OH dacron

    2 2O CH CH O C Cn

    O

    O

    +

    terephthalic acid

    3

    2

    O

    Zn/H Oalkene

    (P) aliphatic ketone + other products.

    C OH

    O

    2H /Pd C

    C OH

    O

    2SOCl

    C Cl

    O

    Me Mg Br, CdCl2

    Me

    O

    4NaBHMe

    OH

    HClMe

    Cl

    (P)

    (Q) Mg|Et2O CO2 (dryice) H3O+

    Me

    COOH

    (R)

    2H /Pd C

    COH

    O

    OH

    O

    3NH /NH

    2

    O

    NH2

    3CHCl , KOH

    NC

    NH CH3

    Br2/NaOH

    2H /Pd C

    (P)

    (S)

  • (22) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.22

    PART III – MATHEMATICS SECTION 1 (Maximum Marks:24)

    This section contains SIX (06) questions. Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE

    of these four option(s) is (are) correct option(s). For each question, choose the correct option(s) to answer the question. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is (are) chosen. Partial Marks : +3 If all the four options are correct but ONLY three options are chosen. Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,

    both of which are correct options. Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

    it is a correct option. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : 2 In all other cases. For Example: If first, third and fourth are the ONLY three correct options for a question

    with second option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect option (second option in this case), will result in +2 marks. Selecting only one of the three correct options (either first or third or fourth option) , without selecting any incorrect option (second option in this case), will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without selection of any correct option(s) will result in -2 marks.

    1. For a nonzero complex number z, let arg (z) denote the principal argument with < arg (z) . Then, which of the following statement (s) is (are) FALSE ?

    (A) arg (1 i) = 4

    , where i = 1

    (B) The function f : R (, ], defined by f(t) = arg (1 + it) for all t R, is continuous at all points of R, where i = 1

    (C) For any two nonzero complex numbers z1 and z2, arg ( 12

    z

    z

    arg(z1) + arg(z2) is an

    integer multiple of 2. (D) For any three given distinct complex numbers, z1, z2 and z3, the locus of the point

    z satisfying the condition arg 1 2 3

    3 2 1

    (z z ) (z z )

    (z z ) (z z )

    = , lies on a straight line

    1. (A) (B) (D)

    (A) arg (1 i) = 3

    4

    (B) f(t) is discontinuous at t = 0

    (C) arg 1

    2

    z

    z

    = arg(z1) arg (z2) + 2k

    Hence, it is true.

    (D) arg 1

    3

    z z

    z z

    = arg 3 21 2

    Z Z

    Z Z

    is locus of a point lying on circle.

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (23)

    0518/IITEQ18/Paper1/QP&Soln/Pg.23

    2. In a triangle PQR, let PQR = 30 and the sides PQ and QR have lengths 10 3 and 10, respectively. Then, which of the following statement (s) is (are) TRUE?

    (A) QPR = 45 (B) The area of the triangle PQR is 25 3 and QRP = 120 (C) The radius of the incircle of the triangle PQR is 10 3 15 (D) The area of the circumcircle of the triangle PQR is 100 2. (B),(C),(D)

    cos (30) = 2 2(10 3) 10 (PR)

    2.10.10 3

    PR = 10 QPR = 30 and QRP = 120

    Area = 1

    2.10.10 3 sin(30) = 25 3 sq.units

    r = S

    = 10 3 15 units

    Area of circumcircle = R2 = 2

    abc

    4

    = 100 Sq units 3. Let P1 : 2x + y z = 3 and P2 : x + 2y + z = 2 be two planes. Then, which of the following

    statement (s) is (are) TRUE? (A) The line of intersection of P1 and P2 has direction ratios 1, 2, 1

    (B) The line 3x 4 1 3y z

    9 9 3

    is perpendicular to the line of intersection of P1 and P2

    (C) The acute angle between P1 and P2 is 60 (D) If P3 is the plane passing through the point (4, 2, 2) and perpendicular to the line of

    intersection of P1 and P2, then the distance of the point (2, 1, 1) from the plane P3 is 2

    3

    3. (C), (D) Let a, b, c be D.C’s of line of intersect Then 2a + b c = 0 a + 2b + c = 0

    a b

    3 3

    =

    c

    3 a : b : c = 1 : 1 : 1

    Hence, (A, B) are incorrect.

    (C) cos () = 2 2 1

    4 1 1 1 4 1

    = 1

    2

    = 60 (D) Eq. of required plane is 1(x 4) 1(y 2) + 1(z + 2) = 0 x y + z = 0 Distance of (2, 1, 1) from x y + z = 0

    is 2 1 1

    3

    =

    2

    3

    R P

    30 10 10 3

    Q

  • (24) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.24

    4. For every twice differentiable function f : R [2, 2] with (f(0))2 + (f(0))2 = 85, which of the following statement (s) is (are) TRUE?

    (A) There exist r, s R, where r < s, such that f is oneone on the open interval (r, s) (B) There exists x0 (4, 0) such that |f(x0)| 1 (C)

    xlim f (x)

    = 1

    (D) There exists a (4, 4) such that f(a) + f(a) = 0 and f(a) 0 4. (A), (B), (D) (A) f : R [2, 2] (f(0))2 + (f (0))2 = 85 …(i) As co-domain [2, 2] and from (1) we can say function is not constant. f(x) is increasing (or decreasing) in some small interval.

    There exist r, s R where r < s, such that ‘f’ is 11 on the (r, s) (B) Apply LMVT in [4, 0]

    f [x0] = f (0) f ( 4)

    4

    , x0 (4, 0)

    | f (x0)| = f (0) f ( 4)

    4

    1

    (C) Let f(x) = sin 85 x f(0) = 0 f (x) = cos ( 85 x) ; 85 f (0) = 85 So,

    xlim

    f(x) = xlim

    sin 85 x = does not emits Option (C) not correct. (D) Let g(x) = f2(x) + (f (x))2 g(x) = 2f(x) f (x) + 2 f (x) f '' (x) = 2 f (x) [f(x) + f '' (x)] Now from option (B) | f (x0)| 1 in (4, 0)

    LMVT for (0, 4) f (x) f (0)

    4

    = '0f ' (x )

    '0| f '(x ) | 1 g(x0) 5 x0 (4, 0) g '0(x ) 5

    '0x (0, 4)

    As g(0) = (f(0))2 + (f (0))2 = 85 g(x) has max in (4, 4) g() = 0 for some (4, 4) g() = 2 f () [f() + f '' ()] 0 = 2 f () [f() + f '' ()] As f () 0 f() + f '' () = 0 for some (4, 4). 5. Let f : R R and g : R R be two nonconstant differentiable functions. If

    f(x) = (f (x) g(x))e g '(x) for all x R, and f(1) = g(2) = 1, then which of the following statement (s) is (are) TRUE?

    (A) f(2) < 1 loge 2 (B) f(2) > 1 loge 2 (C) g(1) > 1 loge 2 (D) g(1) < 1 loge 2 5. (B), (C)

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (25)

    0518/IITEQ18/Paper1/QP&Soln/Pg.25

    f (x) = f (x)

    g(x)

    e

    e g(x)

    ef(x) f (x) eg(x) g(x) = 0 f (x)e f '(x) dx g(x)e g '(x) dx = 0 ef(x) + eg(x) = K As given f(1) = g(2) = 1 ef(1) + eg(1) = K = ef(2) + eg(2) e1 + eg(1) = ef(2) + e1

    ef(2) + eg(1) = 2

    e

    So, ef(2) < 2

    e eg(1) <

    2

    e

    f (2) < n 2 1 g(1) < n 2 1 f(2) > 1 n 2 g(1) > 1 n 2

    6. Let f : [0, ) R be a continuous function such that f(x) = 1 2x + x x t

    0e f (t)dt for all

    x [0, ). Then, which of the following statement (s) is (are) TRUE? (A) The curve y = f(x) passes through the point (1, 2) (B) The curve y = f(x) passes through the point (2, 1)

    (C) The area of the region {(x, y) [0, 1] R : f(x) y 21 x } is 2

    4

    (D) The area of the region {(x, y) [0, 1] R : f(x) y 21 x } is 1

    4

    6. (B), (C)

    ex f(x) = (1 2x)ex + x

    t

    0

    e f (t) dt

    diff. ex f (x) ex f(x) = (1 2x) ex 2ex + ex f(x) f (x) 2 f(x) = 3 + 2x I.F. = e2x

    Soln. y e2x = C + 2xe (2x 3) dx = C + 2 e2x x dx 3 e2x dx

    = C + 2 2x 2xe e

    x2 4

    +

    3

    2e2x

    y = C e2x x 1

    2 +

    3

    2

    y = Ce2x + 1 x As f(0) = 1 1 = C + 1 C = 0 y = 1 x f(x) = 1 x It passes (2, 1) (B)

    A = 2

    4

    11 1

    2

    = 1

    4 2

    =

    2

    4

    (C)

    (0, 1)

    (1, 0)

  • (26) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.26

    SECTION II (Maximum Marks:24)

    This section contains EIGHT (08) questions. The answer to each question is a NUMERICAL VALUE.

    For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 6.25, 7.00, -0.33, -.30, 30.27, -127.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

    Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct numerical value is entered as answer. Zero Marks : 0 In all other cases.

    7. The value of 2 2 41 1

    2 log (log 9) log 72(log 9) 7 is _______ .

    7. [8]

    7(log 9)2 log 4log 2 12 22log 9 7 =

    2log 92

    log 2

    2log 9 7log 27

    = 4 2 = 8 8. The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3,

    4, 5} and the repetition of digits is allowed, is ______ . 8. [625] Divisible by 4 means last two digits 12, 24, 32, 44, 52

    So,

    No of ways 5 5 5

    Total ways = 5 5 5 5 = 54 = 625. 9. Let X be the set consisting of the first 2018 terms of the arithmetic progression 1, 6, 11,

    ….., and Y be set consisting of the first 2018 terms of the arithmetic progression 9, 16, 23, ….., Then, the number of elements in the set X Y is _______ .

    9. [3748] X = 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, …… 10086. Y = 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, …… 14128 X Y = 16, 51, 86, 121, …… (A.P. d = 35, a = 16)

    So, n(X Y) = t So, 16 + (t 1) 35 10086 t 288.7 t = 288

    n(X Y) = n(X) + n(Y) n(X Y) = 2018 + 2018 288 = 3748

    1 2

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (27)

    0518/IITEQ18/Paper1/QP&Soln/Pg.27

    10. The number of real solutions of the equation

    i1 i 1

    i 1 i 1

    xsin x x

    2

    =

    i1 i

    i 1 i 1

    xcos ( x)

    2 2

    lying in the interval 1 1

    ,2 2

    is _______ .

    (Here, the inverse trigonometric functions sin1 x and cos1x assume values in ,2 2

    and [0, ], respectively.) 10. [2]

    i 1

    i 1

    x

    = x2 + x3 + x4 + …… =

    2x

    1 x

    i

    i 1

    x

    2

    = x

    2x

    12

    =

    x

    2 x

    i

    i 1

    x

    2

    = x

    2x

    12

    = x

    2 x

    ii 1

    x

    = x

    1 ( x)

    = x

    1 x

    As given sin1 i i

    ii 1 1

    i 1 i 1 i 1 i 1

    x xx x cos x

    2 2

    =

    2

    i i

    ii 1

    i 1 i 1 i 1 i 1

    x xx x x

    2 2

    2x x x x

    x1 x 2 x 2 x 1 x

    x2 1 1 1 1

    x1 x 2 x 1 x 2 x

    x 1 1

    (1 x)(2 x) (1 x) (2 x)

    x3 + 2x2 + 4x 2 = 0 x = 0 Let f(x) = x3 + 2x2 + 4x 2

    f(0) < 0, one root lies between 1

    0,2

    f (x) = 3x2 + 4x + 4 D < 0 f (x) is increasing So, only two solutions

  • (28) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.28

    11. For each positive integer n, let yn =

    1

    n1 ((n 1) (n 2) ... (n n))n

    .

    For x R, let [x] be the greatest integer less than or equal to x. If nlim

    yn = L, then the

    value of [L] is _______ . 11. [1]

    n yn = 1 n 1 n 2 n n

    log ....n n n n

    = n

    r 1

    1 rlog 1

    n n

    nlim

    n yn = n

    n r 1

    1 rlim log 1

    n n

    n nnlim y = 1

    0

    log (1 x) dx

    n (L) = 1

    1

    00

    xlog (1 x) x dx

    1 x

    = log 2 1 10 0

    x n |1 x |

    = log 2 1 + [ n 2 0]

    = n 4

    e

    L = 4

    e

    [L] = 1

    12. Let a and b be two unit vectors such that a . b 0 For some x, y R, let c x a y b (a b) . If | c |= 2 and the vector c is inclined at the same angle a to both a

    and b , then the value of 8 cos2 a is _______ . 12. [3]

    | a | = | b | = 1

    a b = 0, | c | = 2 a c = 2 cos b c = 2 cos c = x a + y b + (a b) c a = x, c b = y So, x = y = 2 cos dot product with c

    4 = x (a c) + y(b c) + a b c 4 = x2 + y2 + [a b c]

    4 = x2 + x2 + 24 2x

    4 2x2 = 24 2x

    Solving x2 = 2 or x2 = 3

    2

    [a b c] = a a a b a c

    b a b b b c

    c a c b c c

    =

    1 0 x

    0 1 x

    x x 4

    = 4 2x2

    [abc] = 24 2x defined if x2 < 2

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (29)

    0518/IITEQ18/Paper1/QP&Soln/Pg.29

    But x2 2, as x2 > 2 (2 cos )2 = 3

    2

    8 cos2 = 3

    13. Let a, b, c be three nonzero real numbers such that the equation 3 a cos x + 2 b sin

    x = c, x ,2 2

    , has two distinct real roots a and with a + =

    3

    . Then, the value of

    b

    a is _______ .

    13. [0.5]

    3 a cos x + 2b sin x = c

    3 cos x + 2b

    asin x =

    c

    a

    As , are roots 3 cos + 2b

    a sin =

    c

    a …(1)

    3 cos + 2b

    asin =

    c

    a …(2)

    (1) (2) 3(cos cos ) + 2b

    a(sin sin ) = 0

    3 2 sin sin2 2

    + 2b

    2 cos sina 2 2

    = 0

    Given + = 3

    1

    3 2 sin2 2

    + 2b 3

    2 sina 2 2

    = 0

    1 = 2b

    a

    b

    a =

    1

    2

    14. A farmer F1 has a land in the shape of a triangle with vertices at P(0, 0), Q (1, 1) and R(2,

    0). From this land, a neighbouring farmer F2 takes away the region which lies between the side PQ and a curve of the form y = xn (n > 1). If the area of the region taken away by the farmer F2 by the farmer F2 is eaxtly 30% of the area of PQR, then the value of n is ______ .

    14. [4]

    1

    n

    0

    (x x ) dx = 3

    10

    12

    0

    x

    2

    1n 1

    0

    x

    n 1

    =

    3

    10

    1

    2

    1

    n 1 =

    3

    10

    1 1 3

    n 1 2 10

    n + 1 = 5 n = 4

    (0, 0)

    P

    R(2,0)

    y = xn

    Q(1, 1)

  • (30) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.30

    SECTION III (Maximum Marks:12)

    This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions.

    Each question has FOUR options. ONLY ONE of these four options corresponds to the correct answer.

    For each question, choose the option corresponding to the correct answer. Answer to each question will be evaluated according to the following marking scheme: Full Marks : +3 If ONLY the correct option is chosen. Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered). Negative Marks : 1 In all other cases. Paragraph for Q. No. 15 & 16

    PARAGRAPH “X”

    Let 5 be the circle in the xy-plane defined by the equation x2 + y2 = 4.

    15. Let E1, E2 and F1F2 be the chords of S passing through the point P0 (1, 1) and parallel to

    the x-axis and the y-axis, respectively. Let G1G2 be the chord of S passing through P0 and having slope 1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3, and G3 lie on the curve

    (A) x + y = 4 (B) (x 4)2 + (y 4)2 = 16 (C) (x 4) (y 4) = 4 (D) xy = 4 15. (A)

    E1 ( 3,1) , E2 ( 3,1) , F1 (1, 3) , F2 (1, 3) , G1 (0, 2), G2 (2, 0)

    Tangent at E2 ( 3,1) is x 3 + y = 4

    E1 ( 3,1) is x 3 + y = 4 Solving (0, 4) = E3 Similarly (4, 0) = F3

    (2, 2) = G3 E3, F3, G3 satisfy option (A)

    E3

    G3

    F3

    F2

    G1 F1 E2

    P0(1,1)

    G2 (2,0)

    E1

  • IIT JEE 2018 Advanced : Question Paper & Solution (Paper – I) (31)

    0518/IITEQ18/Paper1/QP&Soln/Pg.31

    16. Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

    (A) (x + y)2 = 3xy (B) x2/3 + y2/3 = 24/3 (C) x

    2 + y2 = 2xy (D) x2 + y2 = x2 y2

    16. (D) Tangent x cos + y sin = 2

    M2

    , 0cos

    , N

    20,

    sin

    Let P(h, k) mid point of M and N

    P(h, k) =

    2 20 0

    cos sin,2 2

    cos = 1

    h, sin =

    1

    k

    cos2 + sin2 = 1

    2 2

    1 1

    h k = 1

    h2 + k2 = h2k2 Locus x2 + y2 = x2y2

    Paragraph for Q. No. 17 & 18

    PARAGRAPH “A”

    There are five students S1, S2, S3, S4 and S5 in a music class and for them there are five seats R1, R2, R3, R4 and R5 arranged in a row, where initially the seat Ri is allotted to the student Si, i = 1 , 2 , 3 , 4 , 5 . But, on the examination day, the five students are randomly allotted the five seats.

    17. The probability that, on the examination day, the student S1 gets the previously allotted seat R1, and NONE of the remaining students gets the seat previously allotted to him/her is

    (A) 3

    40 (B)

    1

    8 (C)

    7

    40 (D)

    1

    5

    17. (A) n(s) = 5! = 120

    event A = the student S1 gets the previously allotted seat R1 and None of the remaining students gets the seat previously allotted.

    n(A) = D4 = 4! 1 1 1 1

    11! 2! 3! 4!

    = 9

    P(A) = 9

    120 =

    3

    40

    18. For i = 1, 2, 3, 4, let Ti denote the event that the students Si and Si+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1 T2 T3 T4 is

    (A) 1

    15 (B)

    1

    10 (C)

    7

    60 (D)

    1

    5

  • (32) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution

    0518/IITEQ18/Paper1/QP&Soln/Pg.32

    18. (C)

    P(T1 T2 …… T4) = 1 2 3 4Total n(T T T T )

    n(s)

    = 1 1 2 1 2 35! { n (T ) n (T T ) n (T T T )....}

    5!

    = 4 3 3 4 21 1 1 1 1120 C 4! 2! ( C 3! 2! C 2! 2! 3!) ( C 2! 2! C 2! 2!) 2

    120

    = 120 {192 108 24 2}

    120

    = 120 106

    120

    =

    14

    120 =

    7

    60