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2018-Jee-Advanced Question Paper-2_Key & Solutions
Sri Chaitanya IIT Academy # 304, Kasetty Heights, Ayyappa Society, Madhapur, Hyderabad – 500081
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JEE (ADVANCED) 2018 PAPER 2 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 acorrect 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.
Q.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 tdt
, 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 Key: (A, B, D)
Sol: dk tdt
as 212
k mv dk dvmv tdt dt
0 0
v t
m vdv tdt
2 2
2 2mv t
v tm
dvadt m
Constant Since F=ma F m m
m Constant
Q.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 0u . 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 0u D) The tangential (shear) stress on the plate varies linearly with the viscosity of the liquid Key: (A, C, D)
0uh
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Sol: viscous force is given by dvF Ady
since h is very small therefore; magnitude of viscous
force is given by dvF Ady
00&AuF F F u
h
1 ,F F Ah
Q.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 03 /R 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 02 /R D) The electric field is normal to the surface of the shell at all points Key: (A, B)
P
Q
O060
0120
R
z-axis
Sol: filed due to straight wire is perpendicular to the wire & radially outward. Hence 0zE length,
2 sin 60 3PQ R R according to gauss’s law
Total flux=0 0
3. inq RE ds
Q.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.)
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A)
B)
C)
D) Key: (D)
A
B
F
f
045
/ 2f
Sol: distance of a point A is 2f
Let A’ is the image of A from mirror, for this image
1 1 1
2fv f
1 2 1 1v f f f
Image of line AB should be perpendicular to the principle axis & image of F will form at infinity, therefore correct image diagram is
'A
'B
or
f-xx f-x
2
1
f hf u h
2
f f xh
f x
2h f
Q.5 Ina radioactive decay chain, 23290Th nucleus decay 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) 5N
B) 6N
C) 2N
D) 4N
Key: (A, C)
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Sol: 23290 Th is converting into
21282 Pb
Change in mass number (A) =20
Number of particle =20 54
Due to 5 particle, z will be change by 10 unit. Since given change is 8, therefore number of particle is 2 Q.6 In an experiment to measure the speed of sound by a resonating air column, a tuning fork of
frequency 500Hz 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 50.7cm and 83.9 cm. Which of the following statements is (are) true?
A) The speed of sound determined from this experiment is 332m/s B) The end correction in this experiment is 0.9cm C) The wavelength of the sound wave is 66.4cm D) The resonance at 50.7cm corresponds to the fundamental harmonic Key: (A, C) Sol: let 1n harmonic is corresponding to 50.7 cm & 2n harmonic is corresponding 83.9 cm.
(83.9 50.7)2 cm 33.2
2 cm
66.4 cm 16.64 cm
Length corresponding to fundamental mode must be close to 4
& 50.7 cm must be an odd
multiple of this length 16.6 3 49.8 cm. therefore 50.7 is 3rd harmonic If end correction is e, then
350.74
e cm Speed of sound, v f
500 66.4v cm/sec 332.000 m/s
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.
Q.7 A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass
0.4m kg is at rest on this surface. An impulse of 1.0Ns is applied to the block at time 0t so that it starts moving along the x-axis with a velocity /
0tv t v e , where
0v is a constant and 4s . The displacement of the block, in metres, at t is . Take 1 0.37e .
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Key: 6.30
J=1 m=0.4
Sol: /0
tV V e 0 2.5JVm
m/s
/0
tV V e /0
tdx V edt
/0
0 0 1
x xt x edx V e dt e dx
0
0
/
1
tx V
e
1 02.5( 4)( )X e e 2.5 4 0.37 1X 6.30X Q.8 A ball is projected from the ground at an angle of 45° with the horizontal surface. It
reaches a maximum height of 120m 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 meters, is .
Key: 30.00
045
u
1 120H m030 2H
2V u
Sol: 2 2
1sin 45 1202
uHg
2
1204ug
……..(i)
When half of kinetic energy is lost V=2
u
22
2
2
sin 302
2 16
uuH
g g
……..(ii) From (i) & (ii)
2H = 30.00 Q.9 A particle, of mass 310 kg and charge 1.0C , is initially at rest. At time 0t , the
particle comes under the influence of an electric field 0 sinE t E t i
, where
0 1.0 /E N C and 310 /rad s . Consider the effect of only the electrical force on the particle. Then the maximum speed, in m/s , attained by the particle at subsequent times is .
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Key: 2.00 Sol: 310 1 0n kg q C t
0 sinE E t Force on particle will be 0 sinF qE qE t At max' 0, 0 sin 0V a F qE t 0 sinF qE t
0 sindv Eq tdt m
/
0
0 0
sinV qEdv t dt
m
/0
00 cosqEV
m t
00 cos cos0qEV
m
3 3
1 1 2 210 10
V
m/s =2.00
Q.10 A moving coil galvanometer has 50 turns and each turn has an area 4 22 10 m . The magnetic field produced by the magnet inside the galvanometer is 0.02T . The torsional constant of the suspension wire is 4 110 Nmrad . When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2rad . 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 1.0A . For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms,is…..
Key: 5.55 Sol: n= 50 turns A= 4 22 10 m B=0.02 T 410K mQ =0.2 rad 50gR AI =0-1.0A ,MB C M nIA BINA=C 4 40.02 1 50 2 10 10 0.210 gI =01.A For galvanometer, resistance is to be connected to ammeter in shunt.
0.1gI 50gR
gI I S g g gI R I I S 0.1 50 1 0.1 S
50 5.559
S
Q.11 A steel wire of diameter 0.5mm and Young’s modulus 11 22 10 /N m carries a load of mass M. The length of the wire with the load is 1.0m . 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 1.0mm , 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 1.2kg , the vernier scale division which coincides with a main scale division is . Take
210 /g m s and 3.2 .
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Key: 3.00 Sol: 0.5d mm 112 10Y 1l m
2 24 11
1.2 10 1
5 10 2 1044
Fl mglldAy y
8 11
1.2 103.2 25 10 2 104
l
=
3 3
12 12 0.30.8 25 2 10 40 10
mm
So 3rd division of a vernier scale will coincicle with main scale. Q.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 100K and the universal gas constant 1 18.0R J mol K , the decrease in its internal energy, in Joule , is .
Key: 900
Sol: iV V 8FV V For adiabatic process 5{3
for monotonic process
1 11 1 2 2.TV T V
23100 V =
23
2 8T V
2 25T k 1 100 25 12 75 9002v
FRU nc T
Joule
Q.13 In a photoelectric experiment a parallel beam of monochromatic light with power of 200W is incident on a perfectly absorbing cathode of work function6.2eV . The frequency of light is just above the thres hold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500V 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 410F n N due to the impact of the electrons. The value of n is . Mass of the electron 319 10em kg and
191.0 1.6 10eV J . Key: 24 Sol: power = nhv n= number of protons per second Since KE= 0, hv
19200 6.25 1.6 10n joule 19
2001.6 10 6.25
n
As photon just above threshold frequency maxKE is zero and they are accelerated by
potential difference of 500V. fKE q V 2
22P q V P m Vm
Since efficiency is 100%, number of electrons = number of photons per second As photon is completely absorbed force exerted = nmv
31 19
19
200 2 9 10 1.6 10 5006.25 1.6 10
254
19
3 200 10 1600 2 40 10 3 246.25 1.6 10 6.25 1.6
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Q.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 2 1n to n transition has energy 74.8eV higher than the photon emitted in the 3 2n to n transition. The ionization energy of the hydrogen atom is 13.6eV . The value of Z is .
Key: 3
Sol: 2 22 1
1 313.6 1 13.64 4
E z z
2 23 2
1 1 513.6 13.64 9 36
E z z
2 1 3 2 74.8E 2 23 513.6 13.6 74.8
4 36z z
2 3 513.6 74.84 36
z 2 9z 3z
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.
Q.15 The electric field E is measured at a point 0,0,P 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-1 LIST-II P. E is independent of d 1. A point charge Q at the origin
Q. 1Ed
2. A small dipole with point charges Q at 0,0,l and -
At 0,0, l take 2l d
R. 21E
d 3. An infinite line charge coincident with the x-axis,
with uniform linear charge density
S. 31E
d 4. Two infinite wires carrying uniform linear charge
density parallel to the x- axis.The one along 0,y z l has a charge density and the one along 0,y 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; R → 1, 4; S → 2 C) P → 5; Q → 3; R → 1, 2; S → 4 D) P → 4; Q → 2, 3; R → 1; S → 5
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Key: (B)
Sol: (i) 2 2
1KQE Ed d
(ii) dipole
23
2 1 3coskpEd
3
1Ed
for dipole
(iii) for line charge 2kE
d
1Ed
(iv) 2 2K KEd l d l
2 22 d l d lKl
d l
ˆ ˆdr t
V i jdt
2
1Ed
(V) Electric filed due to sheet 02
V is independent of r
Q.16 A planet of mass M, has two natural satellites with masses 1m and 2m . The radii of their
circular orbits are 1R and 2R respectively. Ignore the gravitational force between the
satellites. Define 1 1 1 1, ,v L K and T to be, respectively, the orbital speed, angular momentum,
kinetic energy and time period of revolution of satellite1; and 2 2 2 2, ,v L K and T to be the
corresponding quantities of satellite 2.Given 1
22m
m and 1
2
14
RR
, match the ratios in List-I to
the numbers in List-II.
LIST–I LIST–II
P. 1
2
vv
1. 18
Q. 1
2
LL
2.1
R. 1
2
KK
3.2
S. 1
2
TT
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
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Key: (B)
2m1R
M
1
2
2mm
1
2
14
RR
1R
1m
Sol: 2
1 1 121 1
GMm mVR R
2 21 2
1 2
,GM GMV VR R
2
1 22
2 1
4V RV R
(P) 1
2
2VV
(Q) L mvR 1 1 1 1
2 2 2 2
12 2 14
L m v RL m v R
(R) 212
K mv 2
21 1 12
2 2 2
2 2 8K m vK m v
(S) 2 /T R V 1 1 2 1 2
2 1 2 2 1
1 1 14 2 8
T R v R vT v R R v
Q.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-1 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 0 06PV
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 Key: (C)
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Sol: process-I is an adiabatic process
Q U W 0Q W U
Volume of gas is decreasing 0W 0U
Temperature of gas increases
No heat is exchanged between the gas and surrounding.
Process –II is an isobaric process (pressure remain constant)
0 0 0 0 03 3 6W P V P V V PV
Process-III is an isochoric process ( volume remain constant)
Q U W Q U
Process-IV is an isothermal process (temperature remains constant)
Q U W 0U
Q.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 t i t j 1. p
Q. cos sinr t t i t j 2. L
R. cos sinr t t i t j 3. K
S. 2
2r t t i t j 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
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Key: (A)
Sol: (P) ˆ ˆr t ti tj
ˆ ˆdr t
V i jdt
{Constant}
0dvadt
P mv
(Remain constant)
212
k mv {Remain constant} ˆ ˆ 0U UF i ix y
U Constant E= K+ U
0dL r Fdt
L
=constant
(Q) ˆ ˆcos sinr t i t j
2 ˆ ˆcos sindva t i t jdt
2a r 2U r
U depends on r hence it will change with time
Total energy remain constant because force is central
(R) ˆ ˆ( ) cos sinr t ti t j
( ) ˆ ˆ( ) sin cosdr tv t t i t jdt
2 ˆ ˆcos sina t t i t j
(S) 2ˆ ˆ2
r ti t j ˆdva j
dt
Constant F ma
Constant
0
ˆ ˆ ˆ. .t
U F dr m j i tj dt
212
E k U m [Remain constant]
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PART-2:CHEMISTRY SECTION -1 (Maximum Marks: 24)
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 acorrect 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+
Key. ABD
Sol. A) [Co(en)(NH3)3(H2O)]+3 complex is type of [M(AA)b3c] have two G.I.
B) If (en) is replaced by two cynide ligand, complex will be type of [Ma3b2c] and have 3 G.I.
C) [Co(en)(NH3)3(H2O)]3+ have d6 configuration (t 62g) on central metal with SFL therefore it
is dimagnetic in nature.
D)Complex [Co(en)(NH3)3(H2O)]3+ have lesser CFSE ( O) value than [Co(en)(NH3)4]3+
therefore complex [Co(en)(NH3)3(H2O)]+ absorbs longer wavelength for d–d transition.
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)
Key. BD
Sol. A) Cu+2 and Mn+2 both gives green colour in flame test and cannot distinguished. B) Cu+2 belongs to group-II of cationic radical will gives ppt. of CuS in acidic medium. C) Cu+2 and Mn+2 both form ppt. in basic medium. D) Cu+2/Cu = +0.34 V (SRP)
Mn+2/Mn = – 1.18 V (SRP)
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3. Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4) at 288 K to give P (51%), Q
(47%) and R (2%). The major product(s) the following reaction sequence is (are) :-
22 22 3 2
233 22
1) /1) ,2) /2) ,
3) , /273 2783)4)4) , /273 2798
5) ,
Sn HClAc O pyridineBr H O excessBr CH CO H
NaNO HCl KH OH PONaNO HCl K
EtOH
R S major product s
A)
Br
Br
Br
Br B)
Br
Br
Br
Br C)
Br
Br Br D)
Br
BrBr
Br
Key. D
Sol.
3 2 4/HNO HSO
Br
2ACOPy
2 3 2/Br CHCOH
BrBrBr Br
2NaNO HCl
00 5 C 3 2HPO
4. The Fischer presentation of D-glucose is given below.
CHO
OHH
H
H
OH
OH
HHO
2CHOH
D-glucose The correct structure(s) of -glucopyranose is (are) :-
A)
O
H
H
HH
H
HO
OH
OH
HO
2CHOH
B)
O
HH
HH
H
HO
OH
OH2CHOH
OH C)
O
H
H
H
H
H
HOOH
OH
HO
2CHOH
D)
O
H
H
H
H
H
HO
OH
OHHO
2CHOH
Key. D
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Sol.
CH=O
OHH
H
H
OH
OH
HHO
2CHOH
D-glucose
O
H
H
H
H
H
HO
OH
OHHO
2CHOH
L glucopyranose
OOH
OH
OH
OH
OH H
D 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) Time
0ln 3 tP P
B) 0A
13
t
C) Time
0ln tP P
D) 0A
Rat
e co
nsta
nt
Key. AD
Sol. 2A B C
0
0
02
t Pt t P P P P
0 2 tP P P 0 0
000
1 1ln ln
2t
P PKP Pt P P t P
00 0
0
21ln ln 2 ln 33 t
t
PK Kt P P Pt P P
and 01
03
1 1ln ln 3 tan/ 3
Pt cons tK P K
Rate constant does not depends on concentration
6. For a reaction, ,A P the plots of [A] and [P] with time at temperatures 1T and 2T are given
below.
If T2 > T1, the correct statement(s) is (are)
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(Assume H and S are independent of temperature and ratio of lnK at T1 to lnK at T2 is
greater than 2
1
TT
. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium
constant, respectively.)
A) 0, 0H S B) 0, 0G H
C) 0, 0G S D) 0, 0G S
Key. AC
Sol. A P given 2 1T T
1 2
2 1
lnln
K TK T
1 1 2 2ln lnT k T k 0 0
1 2G G
1 2H T S H T S 1 2T S T S 0S
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 total number of compounds having at least one bridging oxo group among the molecules
given below is__.
2 3 2 5 4 6 4 7 4 2 5 5 3 10 2 2 3 2 2 5, , , , , , ,N O N O P O P O H P O H PO H S O H S O
Key. 5 or 6
Sol:
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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 furnance such that the
contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is
______ .
(Atomic weights in g mol–1 : O = 16, S = 32, Pb = 207)
Key. 6.47
Sol: 2 2PbS O Pb SO 1000 1000 207
32 32mol gm
Mol of Pb=mol of 2O 100032
mol
1000 20732
mass of Pb g 207 6.4732
kg kg
9. To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted
to KMnO4 using the reaction,
MnCl2 + K2S2O8 + H2OKMnO4 + H2SO4 + HCl (equation not balanced).
Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic
acid(225 g) was added in portions till the colour of the permanganate ion disappeard. The
quantity of MnCl2 (in mg) present in the initial solution is _____.
(Atomic weights in g mol–1 : Mn = 55, Cl = 35.5)
Key. 126
Sol: 2 2 2 8 2 4 2 4amole amole
MnCl K S O H O KMnO H SO HCl
2 4 4 2
2 4 4
2 0.225 / 90 51 55 71
126
H
eq eq
C O MnO COm of C O m of MnO
aa
mg
10. For the given compound X, the total number of optically active stereoisomers is____.
Key. 7
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11. In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is____.
(A tomic weight in g mol–1: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis)
Key. 495 Sol:
12. The surface of copper gets tarnished by the formation of copper oxide. 2N gas was passed to
prevent the oxide formation during heating of copper at 1250 K. However, the 2N gas contains
1 mole % of water vapour as impurity. The water vapour oxidizes copper as per the reaction
given below: 2 2 22Cu s H O g Cu O s H g 2Hp is the minimum partial pressure of 2H
(in bar) needed to prevent the oxidation at 1250K. The value of 2HIn P is___
(Given : total pressure=1 bar, R(universal gas constant)= 1 18JK mol , In (10)=2.3. Cu s and
2Cu O s are mutually immiscible. At 1250K:
12 22 1/ 2 ; 78,000Cu s O g Cu O s G J mol
12 2 21/ 2 ; 1,78,000H g O g H O g G Jmol ; G is the Gibbs energy)
Key. -14.6
Sol:
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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
1mol ). If the pre-exponential factor of the forward reaction is 4 times that of the reverse
reaction, the absolute value of G (in 1mol ) for the reaction at 300K is____
(Given ; ln (2) = 0.7, RT = 2500 J mol–1 at 300 K and G is the Gibbs energy)
Key. -8500
Sol:
14. Consider an electrochemical cell: A(s) 2, 2 ,1n nA aq M B aq M B s . The value of H for
the cell reaction is twice that of G at 300K. If the emf of the cell is zero, the S
(in 1 1JK mol ) of the cell reaction per mole of B formed at 300K is___
(Given : ln (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.)
Key. -11.62
Sol:
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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. Match each set of hybrid orbitals from LIST-I with complex (es) given in LIST-II LIST-I LIST-II P. 2dsp 1. 4
6FeF Q. 3sp 2. 2 33
Ti H O Cl
R. 3 2sp d 3. 3
3 6Cr NH
S. 2 3d sp 4. 24FeCl
5. 4Ni CO
6. 2
4Ni CN
The correct option is A) 5; 4,6; 2,3; 1P Q R S B) 5,6; 4; 3; 1,2P Q R S C) 6; 4,5; 1; 2,3P Q R S D) 4,6; 5,6; 1, 2; 3P Q R S Key. C Sol:
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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: ary1>alky1>hydrogen)
Ph
Ph
O
OH
Me X LIST-I LIST-II
p. 1.
q. 2.
r. 3. Fehling solution
s. 4. HCHO, NaOH 5. NaOBr The correct option is
A) P 1; Q2,3; R1,4; S2,4 B) P1,5; Q3,4; R 4,5; S3 C) P1,5; Q3,4; R5; S2,4 D) P1,5; Q 2,3; R1,5; S2,3 Key. D Sol: Preparations of aromatic acids.
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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 product in LIST-II and choose the correct option. A) P1,5; Q2; R3; S4
B) P1,4; Q2; R4; S3
C) P1,4; Q1,2; R3,4; S4
D) P4,5; Q4; R4; S3,4
Key. B Sol:
P.
Q.
R.
S. 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)
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LIST-I LIST-II P. (10 mL of 0.1 M NaOH+20mL of 0.1 M
acetic acid) diluted to 60ML 1. the value of H does not change on
dilution
Q. (20mL of 0.1 M NaOH+20mL of 0.1 M
acetic acid) diluted to 80mL 2. the value of H changes to half of its
initial value on dilution
Q. (20mL of 0.1 M HCl+20mL of 0.1 M
ammonia solution) diluted to 80ML 3. the value of H changes to two times
of its initial value on dilution
S. 10mL saturated solution of 2Ni OH in
equilibrium with excess solid 2Ni OH is
diluted to 20mL (solid 2Ni OH is still
present after dilution)
4. the value of H changes to 12
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) P4; Q2; R3; S1 B) P4; Q3; R2; S3
C) P1; Q4; R5; S3 D) P1; Q5; R4; S1
Key. D Sol:
S. Because of dilution solubility does not change so [H+] = constant
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PART-2:MATHEMATICS SECTION -1 (Maximum Marks: 24)
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 acorrect 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
Q1. For any positive integer n, define : 0,nf as 1
1
1tan1 1
n
nj
f xx j x j
for
all 0,x
(Here, the inverse trigonometric function 1tan x
assume values in ,2 2
.) Then, which of the following statement(s) is (are) TRUE?
A) 5
21
1tan 0 55
jf
B) 10
21
11 ' 0 sec 0 10j
jf f
C) For any fixed positive integer n, 1lim tan nxf x
n
D) For any fixed positive integer n, 2limsec 1nxf x
Key: D
Sol:
1
1
1tan
1 1
n
nj
x j x jf x
x j x j
1 1
1
tan tan 1n
nj
f x x j x j
1 1tan tannf x x n x
1 1tan tan tan tannf x x n x
tan1n
x n xf x
x x n
2tan1n
nf xx nx
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2 2sec 1 tann nf x f x
2
22limsec lim1 1
1nx x
nf xx nx
2. Let T be the line passing through the points P(-2, 7) and Q(2, -5). Let 1F be the set of all pairs of circles 1 2,S S such that T is tangents to 1S at P and tangent to 2S at Q, and also such that 1S and 2S touch each other at a point, say, M. Let 1E be the set representing the locus of M as the pair 1 2,S S varies in 1F . Let the set of all straight line segments joining a pair of distinct points of 1E and passing through the point R(1, 1) be 2F . Let 2E be the set of the mid-points of the line segments in the set 2F . Then , which of the following statement(s) is (are) TRUE?
A) The point (-2, 7) lies in 1E B) The point 4 7,5 5
does NOT lie in 2E
C) The point 1 ,12
lies in 2E D) The point 30,2
does NOT lie in 1E Key. D Sol:
AP AQ AM
Locus of M is a circle having PQ as its diameter
Hence, 1 : 2 2 7 5 0E x x y y and 2x
Locus of B (midpoint) is a circle having RC as its diameter 22 : 1 1 0E x x y
Now, after checking the options, we get (D)
3. Let S be the of all column matrices1
2
3
bbb
such that 1b , 2b , 3b and the system of equations (in
real variables) Has at least one solution. Then, which of the following system(s) (in real
variables) has (have) at least one solution of each1
2
3
bb Sb
?
A) 1 22 3 ,4 5x y z b y z b and 32 6x y z b B) 1 23 ,5 2 6x y z b x y z b and 32 3x y z b C) 1 22 5 ,2 4 10x y z b x y z b and 32 5x y z b D) 1 22 5 ,2 3x y z b x z b and 34 5x y z b
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Key. AD
Sol. We find D=0 & since no pair of planes are parallel, so there are infinite number of solution s.
Let 1 2 3P P P 1 2 37 13P P P 1 2 37 13b b b A) 0D unique solution for any 1 2 3, ,b b b B) 0D but 1 2 37 13P P P C) 0D Also 2 1 3 12 ,b b b b but 1 2 37 13b b b hence two simultaneous equations in 1 2 3, ,b b b can be considered as two different planes hence line of intersection has only common solutions other solutions are not common hence C is Not Correct D) 0D
4. Consider two straight lines, each of which is tangent to both the circle 2 2 12
x y and the
parabola 2 4y x . Let these lines intersect at the point Q. Consider the ellipse whose center is at the origin O(0, 0) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is 2 , then the which of the following statement(s) is (are) TRUE?
A) For the ellipse, the eccentricity is 12
and the length of the latus rectum is 1
B) For the ellipse, the eccentricity is 12
and the length of the latus rectum is 12
C) The area of the region bounded by the ellipse between the lines 12
x and 1x is
1 24 2
D) The area of the region bounded by the ellipse between the lines 12
x and 1x is
1 216
Key. AD Sol.
Let equation of common tangent is 1y mxm
4 2
2
10 0 1 2 0 121
m m m mm
Equation of common tangents are 1y x and 1y x
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Point Q is 1,0
Equation of ellipse is 2 2
11 1/ 2x y
A) 1 112 2
e and 22 1bLR
a
C)
Area 12
2 2 1
1/ 21/ 2
1 12. . 1 2 1 9 sin2 22xx dx x x
1 1 22 24 4 8 8 4 4 2
Correct answer are (A) and (D) 5. Let s, t, r be the non-zero complex numbers and L be the set of solutions
, , 1z x iy x y i of the equation 0sz t z r , where z x iy .Then, which of the
following statements(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 : 1 5L z z i is at most 2
D) If L has more than one element, then L has infinitely many elements
Key. ACD Sol. Given
0sz tz r ….(1)
On taking conjugate 0s z tz r ……..(2)
From (1) and (2) eliminating z
2 2z s t rt rs
A) If s t then z has unique value
B) If s t then rt rs may or may not be zero so L may be empty set
C) locus of z is null set or singleton set or a line in all cases it will intersect given circle at most
two points
D) In this case locus of z is a line so L has infinite elements
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6. Let : 0,f be a twice differentiable function such that 2sin sinlim sint
f x t f t xx
t x
for all 0,x . If 6 12
f
, then which of the following statement (s) is (are) TRUE?
A) 4 4 2
f
B) 4
2
6xf x x for all 0,x
C) There exists 0, such that ' 0f
D) " 02 2
f f
Key. BCD
Sol. 2sin sin
lim sint x
f x t f t xx
t x
By using L’Hopital
1
2cos sinlim sin
1t x
f x t f t xx
2cos sin sinf x x f x x x 1
2
sin cos1
sinf x x f x x
x
1
sinf x
dx
sinf x
x cx
Put &6 6 12
x f
0 sinc f x x x
A) 1
4 4 2f
B) sinf x x x
as 3 4
2sin , sin6 6x xx x x x x
42 0,
6xf x x x
C) 1 sin cosf x x x x
1 0 tanf x x x there exist 0, for which 1 0f
D) 11 2cos sinf x x x x 11
2 2f
,
2 2f
11 02 2
f f
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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
12
12 60 4
1 3
1 1x x
Key. 2
Sol.
12
1/42 60
1 3
1 1
dx
x x
12
1/460 2
6
1 3
11
1
dx
xx
x
Put 2
1 21 1
x dxt dtx x
1/31/3
6/41
1 3 1 3 2 1 3 3 1 22 2
dtI
t t
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_______.
Key. 4
Sol: 1 2 3
1 2 3 1 2 3 2 3 1 3 1 2 3 2 1 2 1 3 1 3 2
1 2 3x y
a a ab b b a b c a b c a b c a b c a b c a b cc c c
Now if 3x and 3y thecan be maximum 6 But it is not possible as 3x each term of 1x and 3y each term of 1y
3
1
1i i ii
a b c
and3
1
1i i ii
a b c
Which is contradiction. So now next possibility is 4
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 form X to Y and is the number of onto functions from Y to X,
then the value of 15!
is ______.
Key. 119
Sol. 5n X 7n Y
Number of one-one function 75 5!C
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Number of onto function Y to X
1,1,1,1,3 1,1,1,2,2
7 7 73 3 33
7! 7!5! 5! 3. 5! 4 5!3!4! 2! 3!
C C C
7 73 54 4 35 21 119
5!C C
10. Let :f be a differentiable function with 0 0f . If y f x satisfies the differential
equation 2 5 5 2dy y ydx
, then the value of limx
f x
is____.
Key. 0.4
Sol. 225 4dy ydx
So, 225 4dy dxy
Integrating,
21 1 5ln2 225 2
5 5
yx c
y
5 2ln 20
5 2y x cy
Now, 0c as 0 0f Hence 205 25 2
xy ey
205 25 2
x
x x
f xlet let e
f x
Now, RHS = 0 5 2 0xlet x x
25x
let f x
11. Let :f be a differentiable function with 1f O and satisfying the equation
' 'f x y f x f y f x f y for all ,x y. Then, then value of log 4e f is
__________
Key. 2
Sol. , : ' ' .P x y f x y f x f y f x f y x y R
0,0 : 0 0 ' 0 ' 0 0P f f f f f 1 2 ' 0f
,0 : . ' 0 ' . 0P x f x f x f f x f 1 '2
f x f x f x
1'2
f x f x 12
xf x e ln 4 2f
12. Let P be a point in the first octant, whose image Q in the plane 3x y (that is, the line segment PQ is perpendicular to the plane 3x y and the mid-point of PQ lies in the plane 3x y ) 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_____
Key. 8
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Sol. Let , ,P 0,0,Q & , ,R
Now, ˆ ˆ ˆ ˆ ˆ ˆ|| ||PQ i j i j i j
Also, mid point of PQ lies on the plane 3 6 32 2
Now, distance of point P from X-axis 2 2 5 2 2 225 16 as 3 as 4 Hence PR = 2 8 13. Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-
axis and z-axis, respectively, where O(0, 0, 0) is the origin. Let 1 1 1, ,2 2 2
S
be the centre of the
cube and T be the vertex of the cube opposite to the origin O such 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 _____
Key. 0.5
Sol.
1 1 1 1 ˆˆ ˆ, ,2 2 2 2
p SP i j k
1 1 1 1 ˆˆ ˆ, ,2 2 2 2
q SQ i j k
1 1 1 1 ˆˆ ˆ, ,
2 2 2 2r SR i j k
1 1 1 1 ˆˆ ˆ, ,2 2 2 2
t ST i j k
ˆ ˆˆ ˆ ˆ ˆ
1 11 1 1 1 1 14 4
1 1 1 1 1 1
i j k i j kp q r t
ˆ1 1ˆ ˆ ˆ ˆ2 2 2 2
16 2 2ki j i j
14. Let 2 2 2 210 10 10 101 2 3 102 3 ...... 10X C C C C , where 10 , 1, 2,.....10rC r denote binomial
coefficients. Then, the value of 11430
X is_____
Key. 646
Sol: 2
0. ; 10
nn
rr
X r C n
11
0. .
nn n
r rr
X n C C
11
0. .
nn n
n r rr
X n C C
2 11. ; 10n
nX n C n
19910.X C
199
1 .1430 143
X C = 646
2018-Jee-Advanced Question Paper-2_Key & Solutions
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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 : 1 01
xE X x andx
and 1
2 1 : sin log1e
xE X is a real numberx
.
1( si, .2
n ,2
Here the inverse trigonometric function assumes valu inx es
Let 1:f E be the function defined by log1e
xf xx
and 2:g E be the function
defined by 1sing x log .1
xx
LIST-I LIST-II
P. The range of f is 1. 1, ,1 1
ee e
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. ,1
ee
6. 1,0 ,2 1
ee
The correct options 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
Key. A
Sol. 1 : 01
xEx
1 : ,0 1,E x
2 : 1 11
xE nx
11
x ee x
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Now 1 01
xx e
1 1
01
e xe x
1, 1,
1x
e
also 01
x ex
10
1e x e
x
,1 ,
1ex
e
So 21: , ,
1 1eE
e e
as Range of1
xx
is 1R
Range of f is 0R or ,0 0,
Range of g is , \ 0 ,0 0,2 2 2 2
or
Now 4, 2, 1, 1P Q R S Hence a is correct 16. In a high school, a committee has to be formed form a group of 6 boys
1 2 3 4 5 6, , , , ,M M M M M M and 5girls 1 2 3 4 5, , , ,G G G G G . 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 number of ways in which the committee can be formed such that the committee
has 5 numbers, 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 lest 2 girls and such that both 1M and 1G 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
S. The value of 4 is 4. 200 5. 381 6. 461
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The correct options 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 Key. C
Sol. (1) 1
6 5200
3 2
so 4P
(2) 2
6 5 6 5 6 5 6 5 6 51 1 2 2 3 3 4 4 5 5
111
5
46! So 6Q
(3) 3
5 6 5 6 5 6 5 62 3 3 2 4 1 5 0
11 5 6 5 65 0 5 1 4
= 381 So 5R
(4) 2
5 6 4 5 5 6 4 1 5189
2 2 1 1 3 1 2 1 4
So 2S
17. Let 2 2
2 2: 1x yHa b
, where 0a b , 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. 43
R. The distance between the foci of H is 3. 23
S. The length of the latus rectum of H is 4. 4 The correct options 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 Key. B
Sol.
tan 30 ba
3a b
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Now are of 1 .. . 32
LMN b b 24 3 3b 2& 2 3b a
P. Length of conjugate axis = 2b = 4 So 4P
Q. Eccentricity 23
e So 3Q
R. distance between foci = 2ae 22 2 3 83
So 1R
S. Length of latus rectum 22 2 22 42 3 3
ba
So 2S
18. Let 21 2 3: , : , , 1, 2
2 2f f f e
and 4 :f be functions defied by
(i) 2
1 sin 1 xf x e
(ii) 12
sin0
tan1 0
xif xf x xif x
, where the inverse trigonometric function 1tan x assumes
values in ,2 2
(iii) 3 sin log 2ef x x , where for ,t t denotes the greatest integer less than or equal to t,
(iv)
2
4
1sin 0
0 0
x if xx
f x if x
LIST-I LIST-II P. the function 1f is 1. NOT continuous at x = 0 Q. The function 2f is 2. Continuous at x = 0 and NOT differentiable at x = 0 R. The function 3f is 3. Differentiable at x = 0 and is derivative is NOT continuous at x = 0
S. The function 4f is 4. Differentiable at x = 0 and its derivative is continuous at x = 0 The correct options 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
Key. D
Sol. 2
sin 1 xi f x e
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2 2
2
'1
1cos 1 . 0 . 22 1
x x
xf x e e x
e
at 0x '1f x does not exist
So. 2P
12
sin, 0
tan0 0
xxii f x xx
10
sinlim 1tanx
x xx x
2f x does not continuous at x = 0 So 1Q 3 sin 2 0iii f x n x /21 2x e
0 22
n x
0 sin 2 1n x 3 0f x So 4R
2
4
1sin , 0
0, 0
x xiv f x x
x
So 3S