Post on 09-Jul-2018
0518/IITEQ18/Paper2/QP&Soln/Pg.1
Solution to IIT JEE 2018 (Advanced) : Paper - II
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 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. A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving
along the x-axis. Its kinetic energy K changes with time as dK/dt = t, where is a positive
constant of appropriate dimensions. Which of the following statements is (are) true?
(A) The force applied on the particle is constant
(B) The speed of the particle is proportional to time
(C) The distance of the particle from the origin increases linearly with time
(D) The force is conservative
1. (A), (B), (D)
dK
tdt
K = 21mv
2
dK 1 dv
m 2v tdt 2 dt
dv t
vdt m
vdv t dtm
2 2v t
2 m 2
(2) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.2
v tm
(proportional to time)
a = dv
dt m
F ma m (constant)
2. Consider a thin square plate floating on a viscous liquid in a large tank. The height h of
the liquid in the tank is much less than the width of the tank. The floating plate is pulled
horizontally with a constant velocity u0. Which of the following statements is (are) true?
(A) The resistive force of liquid on the plate is inversely proportional to h
(B) The resistive force of liquid on the plate is independent of the area of the plate
(C) The tangential (shear) stress on the floor of the tank increases with u0
(D) The tangential (shear) stress on the plate varies linearly with the viscosity of the liquid
2. (A), (C), (D)
Viscous force F = Adv
dh
0uF A
h
F 1
h
Shear stress = 0
Fu
A
F
A
3. An infinitely long thin non-conducting wire is parallel to the
z-axis and carries a uniform line charge density . It pierces a thin
non-conducting spherical shell of radius R in such a way that the
arc PQ subtends an angle 120° at the centre O of the spherical
shell, as shown in the figure. The permittivity of free space is 0 .
Which of the following statements is (are) true ?
(A) The electric flux through the shell is 03R /
(B) The z-component of the electric field is zero at
all the points on the surface of the shell
(C) The electric flux through the shell is 02R /
(D) The electric field is normal to the surface of the shell at all points
3. (A), (B)
Gauss law
= 0
QE dA
where Q = L
L = PQ = 2R sin 60
L = R 3
0
R 3
Electric field lines are radially outwards, perpendicular to length of wire. Hence
component of E.F. is zero along z-axis.
u0 F
h
120
Q
P
Z
E
E
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (3)
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4. A wire is bent in the shape of a right angled triangle and is placed in front of a concave
mirror of focal length f , as shown in the figure. Which of the figures shown in the four
options qualitatively represent(s) the shape of the image of the bent wire? (These figures
are not to scale.)
(A) (B)
(C) (D)
4. (D)
If |u| = | f |
v
If | u | < | f |
Image is virtual, erect and magnified.
Let a point ‘P’ at distance ‘x’ from f
u = (f x)
1 1 1
V f x f
1 1 1 x
V f x f f (f x)
f (f x)
Vx
Magnification M = I
0
h v f
h u x
where h0 = x Ih f (independent of x)
5. In a radioactive decay chain, 23290 Th nucleus decays to 212
82 Pb nucleus. Let N and N be
the number of and particles, respectively, emitted in this decay process. Which of the
following statements is (are) true?
(A) N = 5 (B) N = 6 (C) N = 2 (D) N = 4
5. (A), (C)
Z = 90 82 = 8
A = 232 212 = 20
Number of -particle N = 20
54
Z = 5 2 = 10
Number of -particles N = 10 8 = 2
90Th232
82Pb212
f
2
f
45
x P
(4) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.4
6. In an experiment to measure the speed of sound by a resonating air column, a tuning fork
of frequency 500 Hz is used. The length of the air column is varied by changing the level
of water in the resonance tube. Two successive resonances are heard at air columns of
length 50.7 cm and 83.9 cm. Which of the following statements is (are) true?
(A) The speed of sound determined from this experiment is 332 m s1
(B) The end correction in this experiment is 0.9 cm
(C) The wavelength of the sound wave is 66.4 cm
(D) The resonance at 50.7 cm corresponds to the fundamental harmonic
6. (A), (B), (C)
For 1st resonance
0
1
3Vf
4 L e
……….. (1)
For 2nd
resonance
0
2
5Vf
4 L e
……….. (2)
1 2
3V 5V
4 L e 4 L e
2 13 L e 5 L e
2e = 3L2 5L1
e = 3 83.9 5 50.7 251.7 253.5 1.8
0.92 2 2
0
1
3Vf
4 L e
V = 500 4 49.8
20 16.6 332m / s3 100
= 0
V 332100 66.4cm
f 500
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 onscreen 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. A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass
m = 0.4 kg is at rest on this surface. An impulse of 1.0 N s is applied to the block at time
t = 0 so that it starts moving along the x-axis with a velocity v(t) = t /e
0v
, where v0 is a
constant and = 4 s . The displacement of the block, in metres, at t = is ________.
Take e1
= 0.37.
7. [6.3]
t /
0v v e F = 1 N.S.
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (5)
0518/IITEQ18/Paper2/QP&Soln/Pg.5
t /
0
dSv v e
dt
S
t/
0
0 0
dS v e dt
S = t /
0 0v e
S = 0
1v 1
e
Impulse = P = mv
v = 1
2.50.4
s = 4 2.5 1 0.37 10 0.63 6.3m/ s
8. A ball is projected from the ground at an angle of 45° with the horizontal surface. It
reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground
for the first time, it loses half of its kinetic energy. Immediately after the bounce, the
velocity of the ball makes an angle of 30° with the horizontal surface. The maximum
height it reaches after the bounce, in metres, is ________ .
8. [30 m]
H = 2 2u sin 45
2g
h = 2 2v sin 30
2g
22
22
1v
h 1 v 141H 2 u 4
u2
9. A particle, of mass 103
kg and charge 1.0 C, is initially at rest. At time t = 0, the particle
comes under the influence of an electric field E (t) = E0 sin t i , where E0 = 1.0 N C1
and = 103 rad s
1 . Consider the effect of only the electrical force on the particle. Then
the maximum speed, in m s1
, attained by the particle at subsequent times is ______.
9. [2]
Velocity is maximum when acceleration is zero.
F = 0
E = E0 sin t = 0
sin t = 0
t = 0 or
t
E = E0 sin t
F = qE = qE0 sin t
a = 0qEqEsin t
m m
a = 0qEdvsin t
dt m
v t
0
0 0
qEdv sin t
m
45
30
v
u
(6) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.6
v = t
0 0
0
qE qEcos t1 cos t
m m
replacing t =
v = 02qE
m
max 3 3
2 1 1v 2
10 10
10. A moving coil galvanometer has 50 turns and each turn has an area 2 104
m2. The
magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional
constant of the suspension wire is 104
N m rad1
. When a current flows through the
galvanometer, a full scale deflection occurs if the coil rotates by 0.2 rad. The resistance of
the coil of the galvanometer is 50 . This galvanometer is to be converted into an
ammeter capable of measuring current in the range 0 1.0 A. For this purpose, a shunt
resistance is to be added in parallel to the galvanometer. The value of this shunt
resistance, in ohms, is __________.
10. [5.56]
Number of turns = 50
Area = 2 104
m2
B = 0.02 T
Torsional constant C = 104
At full scale deflection = 0.2 rad
Resistance R = 50
Torque = C = niAB
Full scale current i = 4
4
C 0.2 10
nAB 50 2 10 0.02
i 0.1amp
0.9 S = 0.1 R
S = R 50
9 9
S 5.56
11. A steel wire of diameter 0.5 mm and Young's modulus 2 1011
N m2
carries a load of
mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is
attached to the end of this wire. Next to the steel wire is a reference wire to which a main
scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond
to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero
of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale
division which coincides with a main scale division is _______ .
Take g = 10 m s2
and = 3.2.
11. [3]
Diameter d = 0.5 mm
Y = 2 1011
N/m2
Strain = Stress
Y
L F
L AY
G
S 0.9 A
0.1 A 1.0 A
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (7)
0518/IITEQ18/Paper2/QP&Soln/Pg.7
L =
2
3
11
FL 1.2 10 1
AY 0.5 103.2 2 10
4
L = 5
5
1230 10 0.3mm
1.6 0.25 10
L.C. of Vernier scale = 0.1 mm
Number of division which coincide with main scale = 3.
12. One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume
becomes eight times its initial value. If the initial temperature of the gas is 100 K and the
universal gas constant R = 8.0 Jmol1
K1
, the decrease in its internal energy, in Joule,
is _________ .
12. [900 J]
Adiabatic Expansion
PV = constant
T.V1
= constant
T2 = T1
1
1
2
V 5where
V 3
T2 = T1
2/3
1T125K
8 4
U = f 3
nR T 1 8 75 900J2 2
13. In a photoelectric experiment a parallel beam of monochromatic light with power of
200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The
frequency of light is just above the threshold frequency so that the photoelectrons are
emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency
is 100%. A potential difference of 500 V is applied between the cathode and the anode.
All the emitted electrons are incident normally on the anode and are absorbed. The anode
experiences a force F = n 104
N due to the impact of the electrons. The value of
n is _______ .
Mass of the electron me = 9 1031
kg and 1.0 eV = 1.6 1019
J .
13. [24]
P = 200 w
KEmax = h = 0
Energy of 1 photon E1 = h = = 6.25 eV
Number of photons incident per unit time
n = 19
19
1
P 20020 10
E 6.25 1.6 10
= 2 10
20
Change in momentum of 1 photon
P1 = 2mk 2me( v)
Force F = n(P1)
F = 2 1020
31 192 9 10 1.6 10 500
= 2 1020
489 1.6 10
= 2 12 1020
1024
= 24 104
(8) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
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14. Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In
the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has
energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The
ionization energy of the hydrogen atom is 13.6 eV. The value of Z is _______ .
14. [3]
E = 13.6 z2
2 2
1 2
1 1
n n
E1 = 13.6z2 21 3
1 13.6z4 4
E2 = 13.6z2 21 1 5
13.6z4 9 36
E1 E2 = 13.6 z2
3 5
4 36
13.6z2
2274.8
36
z2 =
74.8 189
13.6 11
z = 3
SECTION III (Maximum Marks:12) This section contains FOUR (04) questions.
Each question has TWO (02) matching lists: LIST-I and LIST-II.
FOUR options are given representing matching of elements from LIST-I and
LIST-II. ONLY ONE of these four options corresponds to a correct matching.
For each question, choose the option corresponding to the correct matching.
For each question, marks will be awarded according to the following marking
scheme :
Full Marks : +3 If ONLY the option corresponding to the correct matching 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.
15. The electric field E is measured at a point P(0,0, d) generated due to various charge
distributions and the dependence of E on d is found to be different for different charge
distributions. List-I contains different relations between E and d. List-II describes
different electric charge distributions, along with their locations. Match the functions in
List-I with the related charge distributions in List-II.
LIST-I LIST-II
P. E is independent of d 1. A point charge Q at the origin
Q. E
1
d
2. A small dipole with point charges Q at (0, 0,l) and
Q at (0, 0, l). Take 2l << d.
R. E
2
1
d
3. An infinite line charge coincident with the x-axis,
with uniform linear charge density
S. E
3
1
d
4. Two infinite wires carrying uniform linear charge
density parallel to the x- axis. The one along (y = 0,
z = l) has a charge density + and the one along
(y = 0, z = l) has a charge density . Take 2l << d
5. Infinite plane charge coincident with the xy-plane
with uniform surface charge density
(A) P 5; Q 3, 4; R 1; S 2 (B) P 5; Q 3; R1, 4; S 2
(C) P 5; Q 3; R 1, 2; S 4 (D) P 4; Q 2, 3; R 1; S 5
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (9)
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15. (B)
(P) (5), (Q) (3), (R) (1), (4); (S) (2)
(1) E.F. due to a point charge at origin.
2 2
kq 1E E
d d
(2) E.F. at any point on axis of dipole
E = 3 3
2KP 4KQL
d d
3
1E
d
(3) E.F. due to an infinite long charge
E = 2K
d
1
Ed
(4) E.F. due to two infinite long wires
1 2E E E
2 22 2
2K 2K 2k (2L) 4K LE
d L d L d Ld L
If d >> L E = 2
4K L
d
2
1E
d
(5) E.F. due to infinite plane charge
0
E
(independent of d)
16. A planet of mass M, has two natural satellites with masses m1 and m2. The radii of their
circular orbits are R1 and R2 respectively. Ignore the gravitational force between the
satellites. Define v1, L1, K1 and T1 to be, respectively, the orbital speed, angular
momentum, kinetic energy and time period of revolution of satellite 1; and v2, L2, K2 and
T2 to be the corresponding quantities of satellite 2. Given m1 / m2 = 2 and R1/R2 = 1/4,
match the ratios in List-I to the numbers in List-II.
LIST-I LIST-II
P. 1
2
v
v
1. 1
8
Q. 1
2
L
L
2. 1
R. 1
2
K
K
3. 2
S. 1
2
T
T
4. 8
(A) P 4; Q 2; R 1; S 3 (B) P 3; Q 2; R 4; S 1
(C) P 2; Q 3; R 1; S 4 (D) P 2; Q 3; R 4; S 1
16. (B)
(P) (3), (Q) (2), (R) (4), (S) (1)
1
2
m2
m
(10) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.10
1
2
R 1
R 4
(P) For orbital speed
F = 2
2
mv GMm
R R
v = GM
R
1 2
2 1
V R4 2
V R
(Q) Angular momentum
L = mVR
1 1 1 1
2 2 2 2
L m V R 12 2 1
L m V R 4
(R) Kinetic energy
K = 21mv
2
2
21 1 1
2 2 2
K m V2 (2) 8
K m V
(S) Time period of revolution
T = 2 R
V
1 1 2
2 2 1
T R V 1 1 1
T R V 4 2 8
17. One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown
schematically in the PV-diagram below. Among these four processes, one is isobaric, one
is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in
List-I with the corresponding statements in List-II.
LIST-I LIST-II
P. In process I 1. Work done by the gas is zero
Q. In process II 2. Temperature of the gas remains unchanged
R. In process III 3. No heat is exchanged between the gas and its
surroundings
S. In process IV 4. Work done by the gas is 6P0V0
(A) P 4; Q 3; R 1; S 2 (B) P 1; Q 3; R 2; S 4
(C) P 3; Q 4; R 1; S 2 (D) P 3; Q 4; R 2; S 1
M m1 m2
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (11)
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17. (C)
(P) (3), (Q) (4), (R) (1), (S) (2)
(P) Process I is adiabatic
Hence No heat is exchanged between gas and surrounding.
(Q) Process II is isobaric.
w = PV = 3P0 (3V0 V0) = 6P0V0
(R) Process is isochoric
w = 0
(S) Process is isothermal
T = constant
18. In the List-I below, four different paths of a particle are given as functions of time. In
these functions, and are positive constants of appropriate dimensions and . In
each case, the force acting on the particle is either zero or conservative. In List-II, five
physical quantities of the particle are mentioned: p is the linear momentum, L is the
angular momentum about the origin, K is the kinetic energy, U is the potential energy and
E is the total energy. Match each path in List-I with those quantities in List-II, which are
conserved for that path.
LIST-I LIST-II
P. r(t) = ˆˆt t t j 1. p
Q. r(t) = cos t i + sin t j 2. L
R. r(t) = (cos t i + sin t j ) 3. K
S. r(t) = 2ˆ ˆt i t j
2
4. U
5. E
(A) P 1, 2, 3, 4, 5; Q 2, 5; R 2, 3, 4, 5; S 5
(B) P 1, 2, 3, 4, 5; Q 3, 5; R 2, 3, 4, 5; S 2, 5
(C) P 2, 3, 4; Q 5; R 1, 2, 4; S 2, 5
(D) P 1, 2, 3, 5; Q 2, 5; R 2, 3, 4, 5; S 2, 5
18. (A)
(P) (1), (2), (3), (4), (5); (Q) (2), (5); (R) (2), (3), (4), (5); (S) (5)
(P) ˆ ˆr t i t j
dr ˆ ˆv i jdt
(constant)
a 0
F 0
Angular momentum L m r v
ˆ ˆL m t k t k 0(conserved)
I adiabatic
II isobaric
III isochoric
IV isothermal
(12) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.12
(Q) ˆ ˆr a cos t i sin t j
Particle is moving on an elliptical path.
Angular momentum about origin and total energy remains conserved.
(R) ˆ ˆr a cos t i sin t j
Particle moves on a circular path
radius = a
angular velocity =
speed v = a
but v constant
hence P constant
Angular momentum 2 ˆL m a kisconstant
Kinetic energy E = 21mv
2 (constant)
P.E. U = constant
E = constant
(S) 2tˆ ˆr t i j
2
dr ˆ ˆv i t jdt
(depends on t)
hence P mv also depends on t.
dv ˆa jdt
ˆF ma m j (constant)
2
2 tˆ ˆL m r v m t k ( k)2
2m t ˆL k
2
(depends on t)
Only total energy remains conserved.
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (13)
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PART II-CHEMISTRY
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. The correct option(s) regarding the complex [Co(en)(NH3)3(H2O)]3+
(en = H2NCH2CH2NH2) is (are)
(A) It has two geometrical isomers
(B) It will have three geometrical isomers if bidentate ‘en’ is replaced by two cyanide
ligands
(C) It is paramagnetic
(D) It absorbs light at longer wavelength as compared to [Co(en)(NH3)4]3+
1. (A, B, D)
Complex [Co(en) (NH3)3 (H2O)]+3
has two geometrical isomers.
[Co(NH3)3 (H2O) (CN)2]+ has three geometrical isomers.
Co
NH3
NH3
NH3
H2O
Co
NH3
H2O
NH3
NH3
(Fac) (Mer)
en en
(14) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.14
Co
NH3
NH3
NH3
H2O
NC
CN
Co
NH3
NH3
NH3
NC
OH2
NC
Co
NH3
H2O
NH3
NC
NH3
CN
(Fac) (Fac) (Mer)
[Co(en) (NH3)3 (H2O)]+3
is diamagnetic
Co = [Ar]18 4s2 3d
7
Co+3
= [Ar]18 3d6
en is strong field ligand.
_ _ t2g
[Co(en) (NH3)3 (H2O)]+3
[Co (en) (NH3)4]+4
0 = hc
0 1
[Co (en) (NH3)3 (H2O)]+3
absorbs light at longer wavelength as compared to [Co(en)
(NH3)4]+3
2. The correct option(s) to distinguish nitrate salts of Mn2+
and Cu2+
taken separately is (are)
(A) Mn2+
shows the characteristic green colour in the flame test
(B) Only Cu2+
shows the formation of precipitate by passing H2S in acidic medium
(C) Only Mn2+
shows the formation of precipitate by passing H2S in faintly basic medium
(D) Cu2+/
Cu has higher reduction potential than Mn2+
/Mn (measured under similar
conditions)
2. (B), (D)
Both Cu+2
and Mn+2
gives green colour in the flame test.
Ksp (CuS) < Ksp (MnS)
Only Cu+2
shows the formation of precipitate by passing H2S in acidic medium.
2 20 0
Cu /Cu Mn /CuE E
Less electro positive metals have higher reduction potential.
3. Aniline reacts with mixed acid (conc.HNO3 and conc.H2SO4) at 288K to give
P(51 %),
Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are)
1) Ac2O, pyridine 1) Sn/HCl
2) Br2, CH3CO2H 2)Br2/H2O (excess)
R S major product(s)
3) H3O+
3) NaNO2, HCl/273278 K
4) NaNO2, HCl/273 278 K 4) H3PO2
5) EtOH,
eg
0
0
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (15)
0518/IITEQ18/Paper2/QP&Soln/Pg.15
(A) (B)
(C) (D)
3. (D)
NH2
NO2
2Ac O, pyridine
NH
NO2
C CH3
O
2
3
Br
CH COOH
NH
NO2
AC
Br
3H O
NH2
NO2
Br
2NaNO , HCl
273 278 k
Et OH
NO2
Br
Sn/HCl
NH2
Br
2 2Br /H O
excess
NH2
Br
Br
BrBr
2NaNO , HCl
273 278
3 2H PO
Br
Br
BrBr
(R)
4. The Fischer presentation of D-glucose is given below.
D-glucose
The correct structure(s) of β-L-glucopyranose is (are)
(A) (B) (C) (D)
(16) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.16
4. (D)
CH2OH
CHO
H OH
OH H
H OH
H OH
CH2OH
CHO
HOH
OHH
HOH
HOH
D-Glucose L-Glucose
C
HCH2OH
H
OH
H
H
O C
HOH
H
OH
OH
H
H
CH2OH
H
O
O
H
OH
OH
H
H
CH2OH
H
H
OH
HO
HO
HO
L-Glucose
HO ....
HO
-L-Glucopyranose
5. For a first order reaction A(g) → 2B(g) + C(g) at constant volume and 300 K, the total
pressure at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A
is present with concentration [A]0, and t1/3 is the time required for the partial pressure of
A to reach 1/3rd
of its initial value. The correct option(s) is (are)
(Assume that all these gases behave as ideal gases)
(A) (B)
(C) (D)
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (17)
0518/IITEQ18/Paper2/QP&Soln/Pg.17
5. (A, D)
A(g) 2 B(g) + C(g)
oe
o t
2P1k n
t 3P P
Kt = 0 o tn(2P ) n(3P P )
o t on(3P P ) n(2P ) kt
At t = t1/2
1/3
n3t
k
t1/3 is independent of o[A]
For Ist
order Rxn, rate constant is independent of initial concentration of reaction.
6. For a reaction, A ⇌ P, the plots of [A] and [P] with time at temperatures T1 and T2 are
given below.
If T2 > T1, the correct statement(s) is (are)
(Assume ΔHƟ and ΔS
Ɵ are independent of temperature and ratio of lnK at T1 to lnK at
T2 is greater than T2/T1. Here H, S, G and K are enthalpy, entropy, Gibbs energy and
equillibrium constant, respectively.)
(A) ΔHƟ < 0, ΔS
Ɵ < 0 (B) ΔG
Ɵ < 0, ΔH
Ɵ > 0
(C) ΔGƟ < 0, ΔS
Ɵ < 0 (D) ΔG
Ɵ < 0, ΔS
Ɵ > 0
Slope = k
t
o tn(3P P )
R
ate
const
ant
o[A]
o[A]
t1/3
(18) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.18
6. (A), (C)
From the graph (1) and (2), we get H = ve i.e. exothermic.
Given : 1 2
2 1
n K T
nK T
1 2
1
2
S h 1
R R T T
TS h 1
R R T
2 1
S(T T ) 0
R
S 0
SECTION 2 (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 onscreen
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 total number of compounds having at least one bridging oxo group among the
molecules given below is ____.
N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5
7. [5.00 or 6.00]
2 3N O : NO
N OO N N OO
O
Symmetrical
or
Unsymmetrical
2 5N O : N O NO O
O
4 6P O :O
OP
OPO
P
O
P
O
7 7P O :O
O
P
OP
O
P
O
P
OO
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (19)
0518/IITEQ18/Paper2/QP&Soln/Pg.19
4 2H P : P O PO OH H
H H
O O
2 3 10H P : P O PO OH P
OH OH
O H
OH
O O O
2 2 3H S : S O HOH
O
S
2 2 5H S : S S OHOH
O
O
O
8. Galena (an ore) is partially oxidized by passing air through it at high temperature. After
some time, the passage of air is stopped, but the heating is continued in a closed furnace
such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of
O2 consumed is ____.
(Atomic weights in g mol1
: O = 16, S = 32, Pb = 207)
8. [6.47]
PbS + O2 Pb + SO2
2 bO Pn n [Equal moles]
b b2 2
2 b
p po o
O p
w ww w
M M 32 207
Let, 2ow = 1 kg
bPw =
207
32 = 6.47 kg
9. To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely
converted to KMnO4 using the reaction,
MnCl2 + K2S2O8 + H2O KMnO4 + H2SO4 + HCl (equation not balanced).
Few drops of concentrated HCl were added to this solution and gently warmed. Further,
oxalic acid (225 mg) was added in portions till the colour of the permanganate ion
disappeared. The quantity of MnCl2 (in mg) present in the initial solution is ____.
(Atomic weights in g mol−1
: Mn = 55, Cl = 35.5)
9. [126.00]
2 2 2 8 2 4 2 4MnCl K S O H O KMnO H SO HCl
(ON) 5
(ON) 2
24 2 2 4 2KMnO H C O Mn CO
(ON) 5
(ON) 2
(20) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.20
No. of equivalents of MnCl2 = No. of eq. of KMnO4 = No. of eq. of H2C2O4
5 2MnCln = 2
2 2 4H C On
5 w
126 = 2
225
90
w = 2 225 126
90 5
= 126 mg
10. For the given compound X, the total number of optically active stereoisomers is ____.
10. [7.00]
Total number of stereomers of compound X = 8
X has 7 optically active stereisomers
X has 1 optically inactive
11. In the following reaction sequence, the amount of D (in g) formed from 10 moles of
acetophenone is ____.
(Atomic weights in g mol1
: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%)
corresponding to the product in each step is given in the parenthesis)
11. [495.00]
NaOBr
H3O+
COOH
NH3, Br/KOH
CONH2
NH2
Br2
CH3COOH
NH2
BrBr
Br
A (60%)
6 moles
B (50%)
3 moles
D
C (50%)
1.5 moles
(3 eq)
D (100%)
1.5 moles
10 moles
No. of moles of D = 1.5
(C6H4NBr3) = 1.5 330 = 495 gm
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (21)
0518/IITEQ18/Paper2/QP&Soln/Pg.21
12. The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed
to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas
contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per
the reaction given below:
2Cu(s) + H2O(g) Cu2O(s) + H2(g)
2HP is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at
1250 K. The value of ln2H(p ) is ____.
(Given: total pressure = 1 bar, R (universal gas constant) = 8 JK−1
mol−1
, ln(10) = 2.3.
Cu(s) and Cu2O(s) are mutually immiscible.
At 1250 K: 2Cu(s) + ½ O2(g) Cu2O(s); ΔGƟ = − 78,000 J mol
−1
H2(g) + ½ O2(g) H2O(g); ΔGƟ = − 1,78,000 J mol
−1; G is the Gibbs energy)
12. [14.60]
(s) 2 (g) 2 (s) 2 (g)2Cu H O Cu O
At 1250 K (s) 2 (g) 2 (s)
12Cu O Cu O G 78000
2 J/mol = 78 kJ/mol
2 (g) 2 (g) 2 (g)
1H O H O
2 G° = 178000 J/mol = 178 kJ/mol
(s) 2 (g) 2 (s) 2 (g)2Cu H Cu O G 100 kJ/mol
G = G° + RT n K
0 = 100 + RT 2
2
(pH )n
(pH O)
100 = 2
2
pH81250 n
1000 pH O
Given : pH2O = 1 % of total pressure (1 bar)
= 1
1100
100 = 22
pH81250 n
1000 10
10 = 22n pH n10
10 = 2n pH 2 n10
10 = 2n pH 2(2.3)
10 = 2n pH 4.6
2n pH 14.60 bar
13. Consider the following reversible reaction,
A(g) + B(g) AB(g).
The activation energy of the backward reaction exceeds that of the forward reaction by
2RT (in J mol−1
). If the pre-exponential factor of the forward reaction is 4 times that of the
reverse reaction, the absolute value of ΔGƟ (in J mol
−1) for the reaction at 300 K is ____.
(Given; ln(2) = 0.7, RT = 2500 J mol−1
at 300 K and G is the Gibbs energy)
13. [8500.00]
(g) (g) (g)A B AB
Eb Ea = 2RT J/mol ; Af = 4Ab
RT = 2500 J/mol ; T = 300 K
(22) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.22
G° = RT n K
a
b
E /RTf f
E /RTb b
K A eK
K A e
=
b aE E
RTF
b
Ae
A
= 4
2RT
2RTe 4e
G° = RT 2n 4e
= 2500 2n 4e
= 2500 2n 4 n e
= 2500 (2 n 2 + 2)
= 2500 2(0.7) 2
= 2500 3.4
= 8500
Absolute value of G° = 8500.00
14. Consider an electrochemical cell: A(s) | An+
(aq, 2 M) || B2n+
(aq, 1 M) | B(s). The value of
ΔHƟ for the cell reaction is twice that of ΔG
Ɵ at 300 K. If the emf of the cell is zero, the
ΔSƟ (in J K
−1 mol
−1) of the cell reaction per mole of B formed at 300 K is ____.
(Given: n (2) = 0.7, R(universal gas constant) = 8.3 J K−1
mol−1
. H, S and G are enthalpy,
entropy and Gibbs energy, respectively.)
14. [11.62]
n 2n+
s aq, 2M aq,1M sA A B B
Oxidation
nA A ne …. (1)
Reduction
B2n
+ + 2ne B …(2)
multiply equation (1) by (2) and add both
2n n2A B B
Given E.M.F. of the cell is zero; the system is in equilibrium , G = 0
G = RT n k n 2
2n
[A ]G RT n k
[B ]
2G RT n(2)
G H S
G 2 G S (Given H = 2G)
G = TS
G
ST
= RT n4
T
= 2R n 2
= 8.3 0.7 2
= 11.62 J k1
mol1
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (23)
0518/IITEQ18/Paper2/QP&Soln/Pg.23
SECTION 3 (Maximum Marks: 12)
This section contains FOUR (04) questions.
Each question has TWO (02) matching lists: LIST‐I and LIST‐II.
FOUR options are given representing matching of elements from LIST‐I and LIST‐II.
ONLY ONE of these four options corresponds to a correct matching.
For each question, choose the option corresponding to the correct matching.
For each question, marks will be awarded according to the following marking scheme:
Full Marks : +3 If ONLY the option corresponding to the correct matching 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.
15. Match each set of hybrid orbitals from LIST–I with complex(es) given in LIST–II.
LIST–I LIST–II
P dsp2 1. [FeF6]
4−
Q sp3 2. [Ti(H2O)3Cl3]
R sp3d
2 3. [Cr(NH3)6]
3+
S d2sp
3 4. [FeCl4]
2−
5. Ni(CO)4
6. [Ni(CN)4]2−
The correct option is
(A) P → 5; Q → 4, 6; R → 2, 3; S → 1
(B) P → 5, 6; Q → 4; R → 3; S → 1, 2
(C) P → 6; Q → 4, 5; R → 1; S → 2, 3
(D) P → 4, 6; Q → 5, 6; R → 1,2; S → 3
15. (C)
P : dsp2
(6)
2
4[Ni (CN) ]
Ni 4S2 3d
8
Ni2+
CN is a strong ligand Pairing takes place.
dsp2 hybridization square planar
Q : sp3 (4), (5)
(4) [FeCl4]2
Fe 4s2 3d
6
2Fe
Cl Weak ligand No pairing takes place.
sp3 hybridization Tetrahedral structure.
4s 4p 3d
4s 4p 3d
(24) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.24
(5) Ni(CO)4
Ni 4s2 3d
8
CO strong ligand pairing from electrons of s orbital.
sp3 hybridization Tetrahedral
R ; (1)
(1) [Fe F6]4
Fe 4s2 3d
6 ; Fe
2+ 3d
6
F weak field ligand No pairing
sp3d
2 hybridization octahedral
S; d2sp
3 2, 3
(2) [Ti (H2O)3Cl3]
Ti 4s2 3d
2
Ti3+
3d1
Weak ligands H2O, Cl
d2sp
3 octahedral
(3) 3
3 3 6Cr(NH )
Cr : 4s1 3d
5 ; Cr
3+ 4s
0 3d
3
d2sp
3 octahedral
soln. (C)
16. The desired product X can be prepared by reacting the major product of the reactions in
LIST-I with one or more appropriate reagents in LIST-II.
(given, order of migratory aptitude: aryl > alkyl > hydrogen)
4s 3p 3d
4s 4p 3d
4s 4p 3d 4d
4s 4p 3d
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (25)
0518/IITEQ18/Paper2/QP&Soln/Pg.25
LIST I LIST II
P.
1. I2, NaOH
Q.
2. [Ag(NH3)2]OH
R.
3. Fehling solution
S.
4. HCHO, NaOH
5. NaOBr
The correct option is
(A) P → 1; Q → 2,3; R → 1,4; S → 2,4
(B) P → 1,5; Q → 3,4; R → 4,5; S → 3
(C) P → 1,5; Q → 3,4; R → 5; S → 2,4
(D) P → 1,5; Q → 2,3; R → 1,5; S → 2,3
16. (D)
OH
Ph
Ph
Me
OHMe
2 4H SO 2I , NaOHX
NaOBr
X
(P) :
Iodo form reaction
P (1), (5)
NH2
Ph
Ph
H
OHMe
2HNO 2N
Ph
Ph
H
OHMe
X
(Q) :
(2) Ag (NH3)2 OH
(3) Fehlings solution
Gives Tollens & Fehlings Test
Q (2), (3)
(26) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.26
(R)OH
Ph
OHMe
Me
Ph H2SO
4 Me
CH3
O
Ph
Ph
NaOH
X
Me
CH3
O
Ph
Ph
[Ag(NH3)
2]OH
FehlingsX
+
(1) I2 ,
(2) NaOBr
+ AgNO3
(1)
(2)
(S)
Haloform reaction
P 1, 5 ; Q 2, 3 ; R 1, 5 ; S 2, 3
17. LIST-I contains reactions and LIST-II contains major products.
LIST I LIST II
P.
1.
Q.
2.
R.
3.
S.
4.
5.
Match each reaction in LIST-I with one or more products in LIST-II and choose the
correct option.
(A) P → 1,5; Q → 2; R → 3; S → 4 (B) P → 1,4; Q → 2; R → 4; S → 3
(C) P → 1,4; Q → 1,2; R → 3,4; S → 4 (D) P → 4,5; Q → 4; R → 4; S → 3,4
17. (P) (1), (4)
ONa BrElimination
OH+ +
(Q)
OMeSubstitution
O
Me
HBr
Br+ H++
+ MeOH
(2)
(R)
Brbase
NaOMe
(4)
R (4)
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (27)
0518/IITEQ18/Paper2/QP&Soln/Pg.27
(S)
ONaSubstitution
OMeMeBr
-NaBr+
(3)
Soln. (B)
P 1, 4 ; Q 2 ; R 4 ; S 3
18. Dilution processes of different aqueous solutions, with water, are given in LIST-I. The
effects of dilution of the solutions on [H+] are given in LIST-II.
(Note: Degree of dissociation () of weak acid and weak base is << 1; degree of
hydrolysis of salt <<1; [H+] represents the concentration of H
+ ions)
LIST I LIST II
P. (10 mL of 0.1 M NaOH + 20 mL of
0.1M acetic acid) diluted to 60
mL
1. the value of [H+] does not change
on dilution
Q. (20 mL of 0.1 M NaOH + 20 mL of
0.1 M acetic acid) diluted to 80 mL
2. the value of [H+] changes to half of
its initial value on dilution
R. (20 mL of 0.1 M HCl + 20 mL of
0.1 M ammonia solution) diluted to
80 mL
3. the value of [H+] changes to two
times of its initial value on
dilution
S. 10 mL saturated solution of
Ni(OH)2 in equilibrium with excess
solid Ni(OH)2 is diluted to 20 mL
(solid Ni(OH)2 is still present after
dilution).
4. the value of [H+] changes to 1/ 2
times of its initial value on dilution
5. the value of [H+] changes to 2
times of its initial value on dilution
Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct
option is
(A) P → 4; Q → 2; R → 3; S → 1
(B) P → 4; Q → 3; R → 2; S → 3
(C) P → 1; Q → 4; R → 5; S → 3
(D) P → 1; Q → 5; R → 4; S → 1
18. [D]
P : 3 3 2CH COOH NaOH CH COONa H O
10 ml of 0.1m20 ml of 0.1m
The value of H+ does not change on dilution for buffer solution P(1).
Q : 3 3 2NaOH CH COOH CH COONa H O
20 ml of 0.1M20ml of 0.1M
C1 = 0.05 M
C2 = 0.025 M on dilution to 80 ml
M1V1 = M2V2
KH
OH
(28) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.28
n
a
KOH K C C
K
aK KH
C
1
HC
1 2
12
H C 0.025
C 0.05H
2
1
H 0.052
0.025H
2 1H 2 H
R : 4 4 2NH OH HCl NH Cl H O
20 ml of 0.1M20 ml of 0.1M
C1 = 0.05 M
C2 = 0.025 M
nH K C
H C
1 1
22
H C
CH
2 2
11
H C 0.025
C 0.05H
1
2
2 1
1H H
2
R (4)
S : The value of H does not change on dilution.
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (29)
0518/IITEQ18/Paper2/QP&Soln/Pg.29
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 any positive integer n, define fn : (0, ) → as
n 1
n j 1
1f (x) tan
1 (x j)(x j 1)
for all x (0, )
(Here, the inverse trigonometric function tan1
x assumes values in ,2 2
)
Then, which of the following statement(s) is (are) TRUE?
(A) 5 2jj 1
tan f (0) 55
(B) 10 2j jj 1
1 f (0) sec (f (0)) 10
(C) For any fixed positive integer n, nx
1lim tan(f (x))
n
(D) For any fixed positive integer n, 2
nxlim sec (f (x)) 1
1. (A, B, D) n
1
n
j 1
(x j) (x j 1)f (x) tan
1 (x j) (x j 1)
= n
1 1
j 1
tan (x j) tan (x j 1)
= 1 1 1 ntan (x n) tan (x) tan
1 x(x n)
n 2
ntan(f (x))
1 x nx
(30) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
0518/IITEQ18/Paper2/QP&Soln/Pg.30
nxlim tan(f (x)) 0
2
nxlim sec (f (x)) 1
Also, 2 2
jtan (f (0)) j
5 5
2 2
j
j 1 j 1
tan (f (0)) j 55
Also (B) option is correct.
and (A), (D) are correct.
2. Let T be the line passing through the points P(2, 7) and Q(2, 5). Let F1 be the set of all
pairs of circles (S1, S2) such that T is tangent to S1 at P and tangent to S2 at Q, and also
such that S1 and S2 touch each other at a point, say, M. Let E1 be the set representing the
locus of M as the pair (S1, S2) varies in F1. Let the set of all straight line segments joining
a pair of distinct points of E1 and passing through the point R(1, 1) be F2. Let E2 be the
set of the mid-points of the line segments in the set F2. Then, which of the following
statement(s) is (are) TRUE?
(A) The point (2, 7) lies in E1 (B) The point 4 7
,5 5
does NOT lie in E2
(C) The point 1
,12
lies in E2 (D) The point
30,
2
does NOT lie in E1
2. (D)
Let M (h, k).
PMQ = 90 (prove yourself)
k 5 k 7
1h 2 h 2
E1 = Locus of M is
x2 + y
2 2y 39 = 0
Equation of chord of circle whose midpoint (h, k) is
T = S1
xh + yk (y + k) 39 = h2 + k
2 2k 39
Since this locus passes through (1, 1).
E2 h2 + k
2 h 2k + 1 = 0
Options (B) & (C) are incorrect.
(A) is also incorrect as in that case one circle will be a point option (D) is correct.
3. Let S be the set of all column matrices
1
2
3
b
b
b
such that b1, b2, b3 and the system of
equations (in real variables)
x + 2y + 5z = b1
2x 4y + 3z = b2
x 2y + 2z = b3
has at least one solution. Then, which of the following system(s) (in real variables) has
(have) at least one solution for each
1
2
3
b
b
b
S ?
(A) x + 2y + 3z = b1, 4y + 5z = b2 and x + 2y + 6z = b3
(B) x + y + 3z = b1, 5x + 2y + 6z = b2 and 2x y 3z = b3
(C) x + 2y 5z = b1, 2x 4y + 10z = b2 and x 2y + 5z = b3
(D) x + 2y + 5z = b1, 2x + 3z = b2 and x + 4y 5z = b3
P
(2,7)
Q
(2,5)
M
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (31)
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3. (A),(C), (D)
For atleast one solution, either 0 or = 1 = 2 = 3 = 0.
=
1 2 5
2 4 3 0
1 2 2
1 =
1
2
3
b 2 5
b 4 3
b 2 2
= 0 b1 + 7b2 13b3 = 0
2 =
1
2
3
1 b 5
2 b 3 0
1 b 2
b1 + 7b2 13b3 = 0
Also, 3 = 0
For option (A), =
1 2 3
0 4 5 12 0
1 2 6
, so unique solution.
For option (B), =
1 1 3
5 2 6 0
2 1 3
, 1 = 0, 2 =
1
2
3
1 b 3
5 b 6
2 b 3
= 3 (b1 + b2 + 3b3) 0
So no solution.
For option (C), =
1 2 5
2 4 10 0
1 2 5
Also, 1 = 2 = 3 = 0. So, infinitely many solution.
For option (D), =
1 2 5
2 0 3
1 4 5
= 54 0, so unique solution.
Hence (A), (C), (D) are correct.
4. Consider two straight lines, each of which is tangent to both the circle 2 2 1x y
2 and
the parabola y2 = 4x. Let these lines intersect at the point Q. Consider the ellipse whose
center is at the origin O(0, 0) and whose semimajor axis is OQ. If the length of the
minor axis of this ellipse is 2 , then which of the following statement(s) is (are) TRUE?
(A) For the ellipse, the eccentricity is 1
2and the length of the latus rectum is 1
(B) For the ellipse, the eccentricity is 1
2and the length of the latus rectum is
1
2
(C) The area of the region bounded by the ellipse between the lines 1
x2
and x = 1 is
1( 2)
4 2
(D) The area of the region bounded by the ellipse between the lines 1
x2
and x = 1 is
1( 2)
16
(32) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
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4. (A, C)
Eq. of tangent to y2 = 4x is
y = mx + 1
m …(1)
Eq. of tangent to x2 + y
2 =
21
2
is
y = mx 21 m
2
…(2)
Comparing (1) and (2), we get m2 = 1 m = 1
Q (1, 0)
Eq. of ellipse is 2 2
2
x y
11
2
= 1
Eccentricity = 1
12
= 1
2
Length of latus rectum = 1
22
= 1
Area = 2
1 2
1
2
1 xdx
2
=
12 1
1
2
x 12 1 x sin (x)
2 2
= ( 2)
4 2
5 Let s, t, r be non-zero complex numbers and L be the set of solutions
z = x + iy (x, y , i = 1 ) of the equation sz tz r 0 and z x iy. Then,
which of the following statement(s) is (are) TRUE?
(A) If L has exactly one element, then | s | | t |
(B) If | s | = | t |, then L has infinitely many elements
(C) The number of elements in L ∩ {z : | z 1 + i | = 5 } is at most 2
(D) If L has more than one element, then L has infinitely many elements
5. (A), (B), (C) & (D)
Let, s = 1 2s is
t = 1 2t it (s1, s2, t1, t2, r1, r2 R)
r = r1 + ir2
sz tz r 0
1 1 2 2 1 2 2 1 1 2(s t )x (t s )y r i (s t )x s t y r 0
1 1 2 2 1s t x t s y r 0
And 2 2 1 1 2s t x s t y r 0
i) If 1 1 2 2
2 2 1 1
s t t s,
s t s t
then locus is two coincident lines.
ii) If 1 1 2 2
2 2 1 1
s t t s,
s t s t
then liens are intersecting and hence locus is a point.
If |s| |t|, L has exactly one point.
And if |s| \ |t|, L has infinitely many elements as it’s a straight line.
Q
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (33)
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6. Let f : (0, ) be a twice differentiable function such that
2
t x
f (x) sin t f (t)sin xlim sin x
t x
for all x (0, )
If f6 12
, then which of the following statement(s) is (are) TRUE?
(A) f4 4 2
(B) 4
2xf (x) x
6 for all x (0, )
(C) There exists (0, ) such that f() = 0
(D) f f 02 2
6. (B, C, D)
2
t x
f (x) sin(t) f (t) sin(x)lim sin (x)
t x
2
t x
f (x) cos(t) f (t) sin(x)lim sin (x)
1
2f (x) cos x f (x)sin(x) sin (x)
d f (x)
1dx sin x
f(x) = x sin(x) ( C 0 )
For f(x) = x sin(x), option (B), (C), (D) are correct.
SECTION 2 (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 the integral
1
2
10 2 6 4
1 3dx
(x 1) (1 x)
is _____ .
7. [2.00] 1/2
2/4 6/4
0
( 3 ) dx
(x 1) (1 x)
=
1/2
1/2
0 2
( 3 1) dx
x 1(1 x)
1 x
Let x 1
t1 x
(34) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
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= 3
1
( 3 1) dt
2t
2
dx dt
(1 x) 2
= ( 3 1) ( 3 1) 2
8. Let P be a matrix of order 3 3 such that all the entries in P are from the set {1, 0, 1}.
Then, the maximum possible value of the determinant of P is _____ .
8. (4.00)
=
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
= (a1b2c3 + a3c2b1 + a2c1b3) (a1b3c2 + a2c3b1 + a3c1b2)
{ai , bi, ci} {1, 0, 1} i = 1, 2, 3
Maximum value can be 6, but that’s not possible.
(For = 6, 3 entries should be 1 and 6 entries should be 1 in first bracket and 3 entries
should be 1 and 6 entries should be 1 in second bracket. That’s a contradiction !)
5. (Even if one element is 0)
max = 4 which is possible.
Ex :
1 1 0
1 1 1
1 1 1
9. Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If is the
number of one-one functions from X to Y and is the number of onto functions from Y
to X, then the value of 1
5!( ) is _________ .
9. (119.00)
= 7C5 5! = 21 5!
= 7C4 3 2
5C2 3! +
7C3 5 4!
5!
= 119.00
10. Let f : → be a differentiable function with f(0) = 0. If y = f(x) satisfies the
differential equation
dy
(2 5y)(5y 2)dx
then the value of xlim f (x)
is ________ .
10. [0.40]
dydx
(2 5y) (5y 2)
1 (2 5y) (5y 2)dy dx
4 (2 5y) (5y 2)
e
1 5y 2log x c
5 4 5y 2
20x c5y 2e
5y 2
20x2 5y
e2 5y
( f (0) 0)
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (35)
0518/IITEQ18/Paper2/QP&Soln/Pg.35
20x
20x
2 (1 e )y
5 (1 e )
x
2lim f (x)
5
11. Let f : → be a differentiable function with f(0) = 1 and satisfying the equation
f(x + y) = f(x) f(y) + f(x) f(y) for all x, y
Then, the value of loge(f(4)) is _____.
11. (2.00)
f(0) = 1
Put x = 0, y = 0
f(0) = f(0) f (0) + f (0) f (0) f(0) = 1
2
put y = 0
f(x) = f(x) f (0) + f (x) f(0)
f(x) = f(x) 1
2 + f (x)
f (x) = f (x)
2
f '(x)
dxf (x) =
1dx
2
loge (f(x)) = x
2+ c
f(x) =
x
2e (C = 0 since f(0) = 1)
loge f(4) = loge (e
2) = 2
12. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the
line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in
the plane x + y = 3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R
is the image of P in the xy-plane, then the length of PR is _____ .
12. (8.00)
Let P (x0, y0, z0)
0 0 0x x y y z z
1 1 0
= 0 0
2 2
2(x y 3)
1 1
x = 3 y0
y = 3 x0
Since, mirror images lies on zaxis
3 y0 = 0, 3 x0 = 0
x0 = 3, y0 = 3
Also, 2 20 0z y = 5 z0 = 4
PR = 4 (4) = 8
13. Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the
x-axis, yaxis and z-axis, respectively, where O(0, 0, 0) is the origin. Let S1 1 1
, ,2 2 2
be the centre of the cube and T be the vertex of the cube opposite to the origin O such
(36) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
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that S lies on the diagonal OT. If p SP , q SQ, r SR and t ST, then the value
of (p q) r t) is __________
13. (0.5)
SP = p = 1
i2
1 1ˆ ˆj k2 2
SQ = q = 1
i2
+ 1 1ˆ ˆj k2 2
SR = r = 1
i2
1 1ˆ ˆj k2 2
ST = t = 1
i2
+ 1 1ˆ ˆj k2 2
ˆ ˆ ˆi j k
1 1 1p q
2 2 2
1 1 1
2 2 2
= ˆ ˆi j
2
ˆ ˆ ˆi j k
1 1 1r t
2 2 2
1 1 1
2 2 2
= ˆ ˆi j
2
ˆ 1k
| (p q) (r t ) |22
14. Let 2 2 2 2
10 10 10 101 2 3 10X C 2 C 3 C ......... 10 C
where 10
Cr, r {1, 2, …., 10} denote binomial coefficients. Then, the value of
1X is ____.
1430
14. (646)
X = 10 2
10r
r 0
r C
X = 10 2
1010 r
r 0
10 r C
2X = 10 2
10r
r 0
10 C
X = 20105. C
X646.00
1430
R
z
y Q(0,1,0)
P(0,1,0)
x
S 1 1 1
, ,2 2 2
T
O
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (37)
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SECTION 3 (Maximum Marks: 12)
This section contains FOUR (04) questions.
Each question has TWO (02) matching lists: LIST‐I and LIST‐II.
FOUR options are given representing matching of elements from LIST‐I and
LIST‐II. ONLY ONE of these four options corresponds to a correct matching.
For each question, choose the option corresponding to the correct matching.
For each question, marks will be awarded according to the following marking
scheme:
Full Marks : +3 If ONLY the option corresponding to the correct matching 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.
15. Let 1
xE {x : x 1and 0}
x 1
and 12 1 e
xE {x E :sin log is a real number}
x 1
.
(Here, the inverse trigonometric function sin1
x assumes values in ,
2 2
)
Let f : E1 be the function defined by f(x) = loge
x
x 1
and g : E2 be the function defined by g(x) = sin1
e
xlog
x 1
LIST I LIST II
P The range of f is 1. 1 e, ,1 e e 1
Q The range of g contains 2. (0, 1)
R The domain of f contains 3. 1 1,
2 2
S The domain of g is 4. (, 0) (0, )
5. e,e 1
6. (, 0)
1 e,
2 e 1
The correct option is:
(A) P → 4; Q → 2; R → 1; S → 1
(B) P → 3; Q → 3; R → 6; S → 5
(C) P → 4; Q → 2; R → 1; S → 6
(D) P → 4; Q → 3; R → 6; S → 5
15. (A)
P 4, Q 2, R 1, S 1
E1 : x
0x 1
x (, 0) (1, )
E2 : x
0x 1
and 1 log x
x 1
1
(38) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
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x (, 0) (1, ) 1 x
e x 1
e
0 x
x 1
1
e
x
x 1 e 0
0 (e 1) x 1
e (x 1)
x(e 1) e
x 1
0
x 1
,1 e
(1, ) x (, 1) e
,e 1
Intersection x 1
,1 e
e
,e 1
So, domain of g is x 1 e
, ,1 e e 1
Range of x
x 1is R
+ {1}
Range of f is R {0} or (, 0) (0, )
Range of g is , {0}2 2
16. In a high school, a committee has to be formed from a group of 6 boys
M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5
(i) Let 1 be the total number of ways in which the committee can be formed such
that the committee has 5 members, having exactly 3 boys and 2 girls.
(ii) Let 2 be the total number of ways in which the committee can be formed such
that the committee has at least 2 members, and having an equal number of boys
and girls.
(iii) Let 3 be the total number of ways in which the committee can be formed such
that the committee has 5 members, at least 2 of them being girls.
(iv) Let 4 be the total number of ways in which the committee can be formed such
that the committee has 4 members, having at least 2 girls and such that both M1
and G1 are NOT in the committee together.
LIST I LIST II
P. The value of 1 is 1. 136
Q. The value of 2 is 2. 189
R. The value of 3 is 3. 192
R. The value of 4 is 4. 200
5. 381
6. 461
The correct option is:
(A) P → 4; Q → 6; R → 2; S → 1
(B) P → 1; Q → 4; R → 2; S → 3
(C) P → 4; Q → 6; R → 5; S → 2
(D) P → 4; Q → 2; R → 3; S → 1
16. (C)
(i) 6C3
5C2 = 200 (P 4)
(ii) 6C1
5C1 +
6C2
5C2 +
6C3
5C3 +
6C4
5C4 +
6C5
5C5
= 461 (Q 6)
(iii)5C2
6C3 +
5C3
6C2 +
5C4
6C1 +
5C5
6C0 = 381
(R 5)
(iv) (5C2
6C2
4C1
5C1) + (
5C3
6C1
4C2
5C0) +
5C4
6C0
= 189 (S 2)
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (39)
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17. Let H : 2 2
2 2
x y1,
a b where a > b > 0, be a hyperbola in the xy-plane whose conjugate
axis LM subtends an angle of 60° at one of its vertices N. Let the area of the triangle
LMN be 4 3 .
LIST I LIST II
P The length of the conjugate axis of H is 1. 8
Q The eccentricity of H is 2. 4
3
R The distance between the foci of H is 3. 2
3
S The length of the latus rectum of H is 4. 4
The correct option is:
(A) P → 4; Q → 2; R → 1; S → 3
(B) P → 4; Q → 3; R → 1; S → 2
(C) P → 4; Q → 1; R → 3; S → 2
(D) P → 3; Q → 4; R → 2; S → 1
17. (B)
LMN = 4 3
1
2ab 4 32
ab = 4 3
a
a . 4 33
2a 12 a 2 3
obtan30
a
b = 2 3
23
Length of conjugate axis 2b = 4.
4 1 2
e 1 112 3 3
SS = 2 ae = 2
2 2 3 83
L.R. = 22b 2 4 4
a 2 3 3
18. Let 1 2f : f : ,2 2
2
3 4, f : 1, e 2 and f :
be functions
defined by
(i) 2x
1f (x) sin 1 e
(ii) 12
| sin x |if x 0
f (x) ,tan x
1 if x 0
where the inverse trigonometric function of tan1
x
30 a N(a, 0)
M(0, b)
(0, 0)
L(0, b)
y
x
(40) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution
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assume values in ,2 2
,
(iii) f3(x) = [sin(loge(x + 2))], where, for t , [t] denotes the greatest integer less
than or equal to t,
(iv)
2
4
1x sin if x 0
f (x) x
0 if x 0
LIST I LIST II
P The function f1 is 1. NOT continuous at x = 0
Q The function f2 is 2. continuous at x = 0 and NOT differentiable at x = 0
R The function f3 is 3. differentiable at x = 0 and its derivative is NOT
continuous at x = 0
S The function f4 is 4. differentiable at x = 0 and its derivative is
continuous at x = 0
The correct option is:
(A) P → 2; Q → 3; R → 1; S → 4
(B) P → 4; Q → 1; R → 2; S → 3
(C) P → 4; Q → 2; R → 1; S → 3
(D) P → 2; Q → 1; R → 4; S → 3
18. (D)
(P) 1f (x) =
2 2
2
x x
x
x e .cos 1 e
1 e
1f (0) does not exists.
(Q) 2f (x) = 1
| sin x |, x 0
tan x
1, x 0
=
1
1
sin x, x 0
tan x
sin x, x 0
tan x
1, x 0
f2(0+) = 1
f2(0) = 1
Not continuous at x = 0
(R) f3(x) = esin log (x 2 , f3 : (1, 2e 2
) R
Given 1 < x + 2 < e/2
0 < elog (x 2)2
f3 (x) = 0
Always continuous and differentiate.
IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (41)
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(S) 4f (x) =
2 1x sin , x 0
x
0, x 0
Obviously, f4(x) continuous and differentiable at x = 0.
4f (x) =
2
2
1 1 12x sin x cos , x 0
x x x
0, x 0
=
1 12x sin cos , x 0
x x
0, x 0
4f (0 ) = h 0
1 12h sin cos 0
h hlim
h
= h 0
1cos
1 hlim 2sinh h
= does not exists.
4f (x) is not differentiate at x = 0.
For continuity 4f (x)
4h 0
1 1f (0 ) lim 2hsin cos
h h
does not exists.
4f (0 ) not continuous at x = 0.
Either n(A)
S1 S3 S5 S2 S4
S1 S4 S2 S5 S3
S2 S4 S1 S3 S5
S2 S4 S1 S5 S3
S2 S5 S3 S1 S4
S2 S5 S4 S1 S3
S3 S5 S1 S4 S2
S3 S5 S2 S4 S1
S3 S1 S4 S2 S5
S3 S1 S5 S2 S4
S4 S1 S3 S2 S5
S4 S2 S5 S1 S3
S5 S1 S3 S2 S4
S5 S2 S4 S1 S3
END OF THE QUESTION PAPER