JEE-Advance – 2015 (Hints & Solutions) Paper - 1
NARAYANA IIT ACADEMY
INDIA
PAPER - I
CODE : 0
Time : 3 Hours
Maximum Marks : 264
READ THE INSTRUCTIONS CAREFULLY
GENERAL :
1.
This sealed booklet is your Question Paper. Do not break the seal till you are instructed to do so.
2.
The question paper CODE is printed on the left hand top corner of this sheet and the right hand top
corner of the back cover of this booklet.
3.
Use the Optical Response Sheet (ORS) provided separately for answering the questions.
4.
The ORS CODE is printed on its left part as well as the right part. Ensure that both these codes are
identical and same as that on the question paper booklet. If not, contact the invigilator.
5.
Blank spaces are provided within this booklet for rough work.
6.
Write your name and roll number in the space provided on the back cover of this booklet.
7.
After breaking the seal of the booklet, verify that the booklet contains 32 pages and that all the 60
questions along the with the options are legible.
QUESTION PAPER FORMAT AND MARKING SCHEME :
8.
The question paper has three parts : Physics, Chemistry and Mathematics. Each part has three
sections.
9.
Carefully read the instructions given at the beginning of each section.
10.
Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from
0 to 9 (both inclusive)
11.
Section 2 contains 10 multiple choice questions with one or more than one correct option.
Marking Scheme : +4 for correct answer, 0 if not attempted and –2 in all other cases.
12.
Section 3 contains 2 “match the following” type questions and you will have to match entries in
Column I with the entries in Column II.
Marking Scheme : for each entry in Column I, +2 for correct answer, 0 if not attempted and –1 in
all other cases.
OPTICAL RESPONSE SHEET :
13.
The ORS consists of an original (top sheet) and its carbon-less copy (bottom sheet).
14.
Darken the appropriate bubbles on the original by applying sufficient pressure. This will leave an
impression at the corresponding place on the carbon-less copy.
15.
The original is machine-gradable and will be collected by the invigilator at the end of the
examination.
16.
You will be allowed to take away the carbon-less copy at the end of the examination.
17.
Do not tamper with or mutilate the ORS.
18.
Write your name, roll number and the name of the examination center and sign with pen in the
space provided for this purpose on the original. Do not write any of these details anywhere else.
Darken the appropriate bubble under each digit of
your roll number.
NARAYANA Group of Educational Institutions
-1-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
JEE-ADVANCE SOLUTIONS
PHYSICS




1.
Ans.
Sol:
SECTION : 1 (Maximum Marks : 32)
This section contains EIGHT questions
The answer to each question is a SINGLE DIGIT NUMBER ranging from 0 to 9, both
inclusive.
For each question, darken the bubble corresponding to the correct integer in the ORS.
Marking scheme:
+4 If the bubble corresponding to the answer is darkened.
0 In all other cases.
Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm
each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An
object is placed at a distance of 15 cm from the mirror. Its erect image formed by this
combination has magnification M1. When the set-up is kept is a medium of refractive index
M2
7/6, the magnification becomes M2. The magnitude
is
M1
(7)
For mirror u  15cm f  10cm
1 1 1
  
v u f
1
1
1

 
v 15 10
1 1 1
  
v 15 10
 v = –30 cm
For lens
u  20cm f  10cm
1
1
1

 
v 20 10
1 1 1
 
v 10 20
 v = 20 cm
30 20
M1  
2
15 20
In second case, for mirror
v  30 cm
New focal length of lens
NARAYANA Group of Educational Institutions
-2-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
2.
Ans.
Sol:
3.
Ans.
3
1
1
35
f 2
10  2 10 
3 6
9
2
 1
1
2 7
7
 for lens
1
1
2


v 20 35
1 2
1
1


 
v 35 20 140
 v= 140 cm
30 140
M2  
 14
15 20
M 2 14

  7.
M1
2
A Young’s double slit interference arraignment with slits S1 and S2 is immersed in water
(refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of
water are given by x2 = p2m22 – d2, where  is the wavelength of light in air (refractive index
= 1), 2d is the separation between the slits and m is an integer. The value of p is
(3)
For maxima path difference = n 
4 
4 
 d 2  x 2   1  n
 d 2  x 2   1  n
3 
3 
2
2 2
2
 x  9n   d
p=3
Two identical uniform discs roll without slipping on two different surface AB and CD (see
figure) starting at A and C with linear speeds v1 and v2, respectively, and always remain in
contact with the surfaces. If they reach B and D with the same linear speed and v1 = 3 m/s,
then
v2
in
m/s
is
(g = 10 m/s2)
(7)
NARAYANA Group of Educational Institutions
-3-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
Sol:
4.
Ans.
Sol:
3
3
mv12  mgh1  mv 22  mgh 2
4
4
3 2
  v 2  v12   g  h1  h 2 
4
 v 22  v12  4g  40
 v 22  49
 v2 = 7 m/s.
A bullet is fired vertically upwards with velocity v from the surface of a spherical planet.
When it reaches its maximum height, its acceleration due to the planet’s gravity is 1/4th of its
value at the surface of the planet. If the escape velocity form the planet is vese = v N , then
the value of N is (ignore energy loss due to atmosphere)
(2)
At the maximum height, acceleration due to gravity is 1/4th  hmax = R (radius of planet)
1
GMm
GMm

Now COE  mv 2 
2
R
2R
 v  GM
R
Now vesc 
5.
Ans.
Sol:
2GM
v 2N2
R
Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B
 
and A emits 104 times the power emitted from B. The ratio  A  of their wavelengths A
 B 
and B at which the peaks occur in their respective radiation curves is
(2)
r2
(For spherical stars)
p 4

 
 10   400    B 
 A 
4
4
2
A
2
B
A nuclear power plant supplying electrical power to a village uses a radioactive material of
half life T years as the fuel. The amount of fuel at the beginning is such that the total power
requirement of the village is 12.5% of the electrical power available from the plant at that
time. If the plant is able to meet the total power needs of the village for a maximum period of
nT years, then the value of n is
(3)
At t = 0
Prequired = Pavailable  (12.5/100) = Pavailable/8
 Pavailable = 8 Prequired
 After first half life, Pavailable = 4 Prequired
After second half life, Pavailable = 2Prequired
After third half life, Pavailable = Prequired
 Power required by the village can be met upto 3 half lives.
 n=3

6.
Ans.
Sol:
NARAYANA Group of Educational Institutions
-4-
7.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
An infinitely long uniform line charge distribution of charge per unit length  lies parallel to
3
a (see figure). If the magnitude of the flux of the electric
the y-axis in the y-z plane at z 
2
field through the rectangular surface ABCD lying in the x-y plane with its centre at the
L
( 0  permittivity of free space), then the value of n is
origin is
n 0
z
D L C
a O
A
3
a
2
y
B
x
Ans.
(6)
From figure,
H
a/2
1
=
tan  

3
3
a
2
E
+ + + + + + + + + + a/2
+ + + + + + + + + +
D 
a 
a/2
A
B
o
8.
Ans.
Sol:
G
C (3/2)a
y

30
 2 = 60o
Total flux through 360o =
q in L

x
o
o
 Flux through 60o is equal to
L 60o
L


 n=6
o
o 360
6o
Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation
of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is
10.4 eV, then the value of n is (hc = 1242 eV nm)
(2)
Energy of the incident radiation = hc/ = 1242/90 = 13.8 eV
Total energy of electron in nth orbit =
13.6
 13.8  10.4 eV
n2
13.6
 13.8  10.4  3.4
n2
13.6
n2 
4 n=2
3.4
13.6
eV
n2

NARAYANA Group of Educational Institutions
-5-




9.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
SECTION : 2 (Maximum Marks : 40)
This section contains TEN questions
Each question has FOUR options (A), (B), (C) and (D). ONE OR ORE THAN ONE of these
four option(s) is(are) correct.
For each question, darken the bubble(s) corresponding to all the correct option(s) in the
ORS.
Marking scheme:
+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.
0 If one of the bubbles is darkened. 2 In all other cases.
The figures below depict two situations in which two infinitely long static line charges of
constant positive line charge density  are kept parallel to each other. In their resulting
electric field, point charges q and –q are kept in equilibrium between them. The point
charges are confined to move in the x direction only. If they are given a small displacement
about their equilibrium positions, then the correct statement(s) is(are)


+q
Ans.
Sol:

x

q
x
(A) Both charges execute simple harmonic motion.
(B) Both charges will continue moving in the direction of their displacement.
(C) Charge +q executes simple harmonic motion while charge –q continues moving in the
direction of its displacement.
(D) Charge –q executes simple harmonic motion while charge +q continues moving in the
direction of its displacement.
(C)
In situation 1, let the distance between 2 line charges be 2a. Now if +q is shifted to the left by a
small distance x , then
2K
2K
Fnet =
(towards right)

a  x  a  x 
2x
4K
 2
x
= 2K 2
2
a  x  a  x2 
Force is opposite to direction of displacement
 4K 
x
 Fnet =   2
2
 a  x 
Now as x << a,  a2 – x2 = a2
 4K 
 Fnet =   2  x
 a 
 +q will execute SHM
Whereas in situation 2 when q is displaced it will continue to move in the direction of
displacement.
 option (C) is correct.
NARAYANA Group of Educational Institutions
-6-
10.
Ans.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of
curvature 10 cm. They are placed with the curved surfaces at a distance d as shown in the
figure, with their axes (shown by the dashed line) aligned. When a point source of light P is
placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays
emanating from it are found to be parallel to the axis inside S2. The distance d is
S1 P
S2
50 cm
d
(A) 60 cm
(B) 70 cm
(C) 80 cm
(B)
For refraction on curved surface of S1
1 = 1.5, 2 = 1, R = 10, u = 50
 2 1  2  1
 
using
, we get v = 50 cm
v u
R
(D) 90 cm
20 cm
S1
S2
P
50 cm
11.
Ans.
Sol:
Now for refraction at curved surface of S2
v =  (as rays inside S2 are parallel to the principal axis)
1 = 1, 2 = 1.5, R = +10, u = x
1.5 1 1.5  1



 x = 20
 x
10
 d = 50 + 20 = 70, so option (B) is correct.
A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a
uniform magnetic field B ^. If F is the magnitude of the total magnetic force acting on the
conductor, then the correct statement(s) is (are)
(A)
If B is along ẑ , F   L  R 
(B)
(C)
If B is along x̂ , F=0
If B is along ŷ , F   L  R 
(D)
If B is along ẑ , F=0
(A, B, C)
Fm  i dl  B


In (A) part Fm   L  R 
If B is along ẑ axis.
In (B) part Fm  0
If B is along x̂ axis.
In (C) part Fm   L  R 
If B is along ŷ axis.
All the result as according to above formula.
NARAYANA Group of Educational Institutions
-7-
12.
Ans.
Sol:
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium
equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is (are)
(A)
The average energy per mole of the gas mixture is 2RT
(B)
The ratio of speed of sound in the gas mixture to that in helium gas is 6 / 5
(C)
The ratio of the rms speed of helium atoms to that of hydrogen molecules is ½
(D)
The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/ 2
(A, B, D)
3
5
RT  RT
2
Average energy/ mole  2
 2RT
2

RT 
 mix MHe
Speed of sound in mixture
(B)


 v

M 
Speed of sound in the Helium gas
Mmix  He

 mix 
n i C P
n i CV
1
1
Average molar mass of mixture   4   2  2  1  3
2
2
(D)
6
 3/ 2  4 
 


5
 3  5 / 3 
1
vrms 
M
 vrms He 2 1


4
 vrms H
2
2
13.
In an aluminium (Al) bar of square cross section, a square hole is drilled and is filled with
iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 108  and
1.0 10 7  m , respectively. The electrical resistance between the two faces P and Q the
composite bar is
(A)
Ans.
Sol:
2475

64
(B)
1875

64
(C)
1875

49
(D)
2475

132
(B)
l  50 103m , Al  2.7 108  , Fe  1.0 107 
R Al 
Al  l 2.7 108  50 103

 3.05 105 
6
A
7

7

2

2

10


NARAYANA Group of Educational Institutions
-8-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
R Fe 
Fe  l 1.0 107  50 103

 125 105 
6
A
 7  7  2  2  10
R Fe  R Al  1875 

 
R Fe  R Al  64 
For photo-electric effect with incident photon wavelength  , the stopping potential is V0.
Identify the correct variation(s) of V0 with  and 1/ 
Now R eqn 
14.
(A)
(B)
(C)
(D)
Ans.
Sol:
(A, C)
By photoelectric equation
hf  W0  eV0
c
h  W0  eV0

1
 V0


15.
Consider a Vernier calipers in which each 1 cm on the main scale is divided into 8 equal
divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier calipers, 5
divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw
gauge, one complete rotation of the circular scale moves it by two divisions on the linear
scale. Then :
(A)
If the pitch of the screw gauge is twice the least count of the Vernier calipers, the least
count of the screw gauge is 0.01 mm.
(B)
If the pitch of the screw gauge is twice the least count o the Vernier calipers, the least
count of the screw gauge is 0.005 mm
(C)
If the least count of the linear scale of the screw gauge is twice the least count of the
Vernier calipers, the least count of the screw gauge is 0.01 mm
(D)
If the least count of the linear scale of the screw gauge is twice the least count of the
Vernier calipers, the least count of the screw gauge is 0.005 mm
(B, C)
L.C. of vernier caliper = 1 MSD – 1 VSD
Ans.
Sol:
NARAYANA Group of Educational Institutions
-9-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
16.
Ans.
Sol.
1
1
 cm 
cm
8
25
10  8 2
1



cm  0.025cm  0.25 mm
80
80 40
pitch
L.C. of screw gauge 
total no.of division on circular scale
Pitch = linear distance moved on main scale in one rotation of circular scale.
Planck’s constant h, speed of light c and gravitational constant G are used to form a unit of
length L and a unit of mass M. Then the correct option(s) is (are)
(A) M  c
(B) M  G
(C) L  h
(D) L  G
(A,C, D)
[h] = ML2T-1
...(i)
[G] = M-1L3T-2 .....(ii)
[C]= LT-1
...(iii)
From (i) and (ii)
[hG] = L5T-3 = L2 =L2(L3 T-3)
L2 [LT-1]3
 hG 
 L2   3 
C 
L G
L h
hc  ML3T 2
hc ML3T 2

G M 1L3T 2
hc
 m2
G
hc
M
G
m c
17.
Option (A), (C) and (D) are correct.
Two independent harmonic oscillators of equal mass are oscillation about the origin with
angular frequencies 1 and 2 and have total energies E1 and E2, respectively. The
a
a
variations of their momenta p with positions x are shown in the figure. If  n 2 and  n,
b
R
then the correct equations (s) is (are)
p
p
Energy = E 1
b
a
Energy = E 2
x
x
R
NARAYANA Group of Educational Institutions
-10-
(A) E11  E 22
Ans.
Sol.
(B)
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
E
E
(C) 12  n 2
(D) 1  2
1 2
2
 n2
1
(B,D)
a
a
 n2 ,  n
n
R
2
p
1
E
 kx 2
2m 2
Given
p
p
Energy = E 1
b
a
Energy = E 2
x
x
R
From figure
m12 2
E1 
a
2
m
E 2  22 R 2
2
2
2
E   a   
 1   1      1  .n 2 .....(i)
E 2  2   R   2 
p12
2
2
E1 2m  p1   b 
 2    
Also,
p2  p2   R 
E2
2m
2


E1  a 
1

 2 .....(ii)
Or,

E2  n2  a 
n

n
From (i) and (ii)
2
2
 1 
1
1
1
 2
   4
n
2 n
 2 

 2  n2
1
E

And 1  1
E 2 2
E E
 1 2
1 2
Hence, (B) and (D) are correct options
NARAYANA Group of Educational Institutions
-11-
18.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
A ring of mass M and radius R is rotating with angular speed  about a fixed vertical axis
M
passing through its centre O with two point masses each of mass
at rest at O. These
8
masses can move radially outwards along two massless rods fixed on the ring as shown in the
8
figure. At some instant the angular speed of the system is  and one of the masses is at a
9
3
distance of R from O. At this instant the distance of the other mass from O is
5
O
(A)
Ans.
Sol.
2
R
3
(B)
1
R
3
(D)
Using conservation of angular momentum
We have
 MR 2     MR 2  M8  259 R 2  M8 x 2   98 
9
9R 2 x 2
 R2  R2 

8
200 8
1 2
9 2 x2
 R 
R 
8
200
8
16
 x2 
R2
25
4
x R
5
(C)
3
R
5
4
(D) R
5
O
NARAYANA Group of Educational Institutions
-12-









19.
Ans.
Sol.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
SECTION : 3 (Maximum Marks : 16)
This section contains TWO questions.
Each question contains two columns, Column I and Column II.
Column I has four entries (A), (B), (C) and (D).
Column II has five entries (P), (Q), (R), (S) and (T).
Match the entries in column I with the entries in column II.
One or more entries in column I may match with one or more entries in column II.
The ORS contains a 4  5 matrix.
For each entry in column I, darken the bubbles of all the matching entries. For example, if
entry (A) in column I matches with entries (Q), (R) and (T), then darken these three bubbles
in the ORS. Similarly, for entries (B), (C) and (D).
Marking scheme:
For each entry in column I.
+2 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened.
0 If one of the bubbles is darkened.
1 In all other cases.
2 In all other cases.
Match the nuclear process given in column I with the appropriate option(s) in column II.
Column I
Column II
(A)
Nuclear fusion
(P)
Absorption of thermal neutrons by
235
92 U
60
(B)
Fission in a nuclear reactor
(Q)
27 Co nucleus
(C)
 -decay
(R)
(D)
  ray emission
(S)
(T)
Energy production in stars via
hydrogen conversion to helium
Heavy water
Neutrino emission
AR,T; BP,S,T; CQ,T; DR,T
Based on theory
NARAYANA Group of Educational Institutions
-13-
20.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
A particle of unit mass is moving along the x-axis under the influence of a force and its total
energy is conserved. Four possible forms of the potential energy of the particle are given in
column I (a and U0 are constants). Match the potential energies in column I to the
corresponding statement (s) in column II.
Column I
Column II
2
(A)
(P)
The force acting on the particle is zero
2
U  x 
at x = a
U1  x   0 1    
2   a  
2
(B)
(Q) The force acting on the particle is zero
U x
U2  x   0  
at x = 0
2 a
2
(C)
(R) The force acting on the particle is zero
  x 2 
U0  x 
U3  x  
exp



at x = -a
 
 
2 a
  a  
3
(D)
(S)
The particle experiences an attractive
U0  x 1  x  
U4  x  
    
force towards x = 0 in the region
2  a 3  a  
x  a.
(T)
The particle with total energy
U0
can
4
oscillate about the point x = -a
Ans.
Sol.
AP,Q,R,T; BQ,S; CP,Q,R,S; DP,R,T.
du1 2U0 x  x 2 
F1  x   
 2 1  2 
dx
a  a 
 U 0 2x
F2  x  
 2
2
a
x2
U0  2x  a 2 x 2  x 2 /a 2 2x 
F3  x  
 2 
 e  2 e
2  a 2
a
a 
x2
 2
2U
 F3  2 0 xe a
a
F4  
U0
2
 x2 
1  2 
 a 
 1 1 3x 2 
 a  3  a3 


NARAYANA Group of Educational Institutions
-14-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
PART II: CHEMISTRY




SECTION : 1 (Maximum Marks : 32)
This section contains EIGHT questions
The answer to each question is a SINGLE DIGIT NUMBER ranging from 0 to 9, both
inclusive.
For each question, darken the bubble corresponding to the correct integer in the ORS.
Marking scheme:
+4 If the bubble corresponding to the answer is darkened.
0 In all other cases.
21.
If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia
complex (which behaves as a strong electrolyte) is –0.0558oC, the number of chloride(s) in
the coordination sphere of the complex is
Key: (1)
Sol.
Tf  i K f m
0.0558  i 1.86  0.01
0.0558
i
3
0.0186
CoCl3 .x NH3
 If i = 3, then total ions = 3
 No. of Cl– ions in the coordination sphere = 1
The complex could be Co  NH3  x Cl  Cl2
22.
All the energy released from the reaction X  Y, G o  193kJ mol1
 M 3  2e , E o  0.25 V .
is used for oxidizing M+ as M  
Under standard conditions the number of moles of M+ oxidized when one mole of X is
converted to Y is [ F = 96500 C mol-1]
Key: (4)
Sol. Energy released when one mole of X converts into Y = 193 kJ.
Energy needed for oxidation of 1 mole of M+ under standard condition = nFE o
 2  96500   0.25 J
 48.25 kJ
193
4
48.25
23.
For the octahedral complexes of Fe3+ in SCN– (thicyanato-S) and in CN– ligand
environments, the difference between the spin-only magnetic moments in Bohr magnetons
(when approximated to the nearest integer) is [Atomic number of Fe = 26]
Key: 4
Sol. Fe3+ with SCN– (weak filed ligand)
High spin octahedral complex
 No. of moles of M+ oxidized 
 Fe  SCN 6 
3
NARAYANA Group of Educational Institutions
-15-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
3
Fe  tg eg
3
2
I  5  5  2  35 BM
Fe3 with CN  (strong field ligand)
Low spin octahedral complex
 Fe  CN 6 
3
Fe3  t 2 g 5
II  11  2  3 BM
I  II  35  3
 5.91  1.73  4.18  4 BM
24.
The total number of lone pairs of electrons in N 2 O3 is
Key: (8)
O
N
N
O
Sol.
N
O
O
N
O
O
(symmetric)
(Asymmetric)
Total number of lone pairs are 8 in N 2 O3 molecule.
25.
Among the triatomic molecules/ions, BeCl2, N 3 , N 2 O , NO 2 , O3 , SCl2 , ICl 2 , I 3 and XeF2,
the total number of linear molecule(s)/ion(s) where the hybridization of the central atom
does not have contribution from the d-orbital(s) is [Atomic number : S = 16, Cl = 17, I = 53
and Xe = 54]
Key: (4)
BeCl2 :
sp
Sol.
N 3
:
sp
N2O
:
sp
NO 2
:
sp
Other linear molecules/ions involve d orbitals.
26.
Key.
Sol.
Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H
atom is 9, while the degeneracy of the second excited state of H is
(3)
The possible excited states are
1s12 p1x
1
1s1 2p y
1s2 2p1z
NARAYANA Group of Educational Institutions
-16-
27.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
The total number of stereoisomers that can exist for M is
H3C
CH3
H3C
O
M
Key.
Sol.
(2)
9
H3C
5
7
CH3
8
4
3
6
H3C
1
M
2
O
The compound M has two stereogenic carbons, hence 1R4R, 1S4S, 1R4S and 1S4R
configurations are possible. Due to cage like structure it is not possible to have R configuration at
C1 and S-configuration at C4. So total number of stereoisomers are 2.
28.
The number of resonance structures for N is
OH
NaOH

 N
Key.
Sol.
(9)
O


O
O
O
 

 
O
 
O
O
O
 
O
NARAYANA Group of Educational Institutions
-17-




29.
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
SECTION : 2 (Maximum Marks : 40)
This section contains TEN questions
Each question has FOUR options (A), (B), (C) and (D). ONE OR ORE THAN ONE of these
four option(s) is(are) correct.
For each question, darken the bubble(s) corresponding to all the correct option(s) in the
ORS.
Marking scheme:
+4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened.
0 If one of the bubbles is darkened.
2 In all other cases.
The major product of the reaction is
H3C
CO 2H
H3C
NH2
H3C
(A) H3C
OH
(B) H3C
H3C
OH
CO 2H
NH2
H3C
OH
CO 2H
H3C
Key.
Sol.
NaNO 2 , aqueous HCl

0o C
H3C
OH
NH2
OH
CO 2H
NN
1
N2
CO 2H
H3C
30.
(D) H3C
H3C
H3C
CO 2H
H3C
CO 2H
(C) H3C
(B)
H3C
H3C
NH2
H3C
Key.
Sol.
NaNO 2 , aqueous HCl

0o C
H 2O
H 
H3C
H3C
COOH

OH
The correct statement(s) about Cr2+ and Mn3+ is(are)
[Atomic numbers of Cr = 24 and Mn = 25]
(A) Cr2+ is a reducing agent
(B) Mn3+ is an oxidizing agent
(C) Both Cr2+ and Mn3+ exhibit d4 electronic configuration
(D) When Cr2+ is used as a reducing agent, the chromium ion attains d5 electronic configuration
(A), (B), (C)
Cr 2
t 2g3eg1

Cr 3
t 2g3 (more stable)
NARAYANA Group of Educational Institutions
-18-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
3
Mn  e
3d
31.
Key.
Sol.
32.
Key.
Sol.


4
3d5 (stable electronic configuration)
Copper is purified by electrolytic refining of blister copper. The correct statement(s) about
this process is (are)
(A)
Impure Cu strip is used as cathode
(B)
Acidified aqueous CuSO4 is used as electrolyte
(C)
Pure Cu deposits at cathode
(D)
Impurities settle as anode mud
(B, C, D)
Anodes are of impure Cu and Cu-strip is cathode. The electrolyte is acidified CuSO4 and the net
process is,
Anode:
Cu  Cu 2  2e
Cu 2  2e  Cu  s 
Cathode:
Impurities from the blister Cu deposited as anode mud.
Fe3+ is reduced to Fe2+ by using
(A)
H2O2 in presence of NaOH
(B)
Na2O2 in water
(C)
H2O2 in presence of H2O4
(D)
Na2O2 in presence of H2SO4
(A, B)
In acidic medium,
H2O2  2FeSO4  H2SO4  Fe2 SO4 3  2H2O
In Basic medium,
33.
Mn
2
H2O2  2K3 Fe  CN 6   2KOH  2K 4 Fe  CN 6   2H 2O  O2
H2O2  2Fe  2OH  2Fe2  2H2O  O2
Na 2O2  2H2O  2NaOH  H2O2
Since,
Therefore, Na2O2 in water also represent same reaction as that of basic H2O2.
The % yield of ammonia as a function of time in the reaction
N 2  g   3H 2  g 
2NH3  g  , H  0
At (P, T1) is given below
If this reaction is conducted at (P, T2), with T2 > T1, the % yield of ammonia as a function of
time is represented by
NARAYANA Group of Educational Institutions
-19-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
(A)
(B)
(D)
Key.
(C)
(B)
Sol.
N 2  g   3H 2  g 
2NH3  g  , H  0
With referenced to the graph given at )P, T1)
Initially at high temperature yield will be more but after some interval it will be constant with type
34.
Key.
Sol.
If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m
fraction of octahedral holes occupied by aluminium ions and n fraction of tetrahedral holes
occupied by magnesium ions, m and n, respectively, are
1 1
1
1 1
1 1
,
1,
,
,
(A)
(B)
(C)
(D)
2 8
4
2 2
4 8
(A)
In cubic close packing unit cell = fcc
O2– form ccp lattice (Z=4)
Al3+ occupy m fraction of OV = 4m
Mg2+ occupy n fraction of TV = 8n
Charge should be neutral
Hence,
4m ×(3) + 8n × (2) = 4 × (2)
12 m + 16 n = 8
3 m + 4n =
1
1
That is satisfied by when m  , n 
2
8
NARAYANA Group of Educational Institutions
-20-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
Compound(s) that on hydrogenation produce(s) optically inactive compound (s) is (are)
35.
Key.
Sol.
(A)
(B)
(C)
(D)
(B, D)
Both (B) (D) will form optically inactive product.
Hydrogenation
Optically inactive
Hydrogenation
Optically inactive
36)
The major product of the following reaction is
O
CH 3
i. KOH , H 2O
ii. H  , heat
O
CH 3
CH 3
O
O
(A)
(B)
O
CH3
O
CH 3
(C)
(D)
KEY: (A)
SOL:
O
O
CH3
OH
O
O
O
OH
CH3
O
CH3
CH3
O
H
+
CH3
O
NARAYANA Group of Educational Institutions
-21-
37)
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
In the following reaction, the major product is
CH3
1 equivalent HBr
CH 2
H2C
CH 3
CH 3
CH3
H 3C
H 2C
Br
(A)
Br
(B)
CH 3
CH 3
(C)
H 2C
Br
(D)
H2C
Br
KEY: (D)
SOL:
CH 3
C
CH 2
CH 2
H+
CH3
CH 3
CH2
C
+
+
CH2
C
C
CH 3
–
CH3
CH3
Br 
CH 2 Br
C
CH3
C
CH3
CH 3
CH 2
Br
C
H
Thermodynamically
more stable product
Br
Since temperature is not given, therefore, option (B) may also be correct.
NARAYANA Group of Educational Institutions
-22-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
38)
The structure of D-(+) glucose is
CHO
H
HO
H
H
OH
H
OH
OH
CH 2OH
The structure of L- (-) glucose is
CHO
(A)
HO
H
HO
HO
CHO
CHO
H
OH
(B)
H
H
CH 2OH
H
HO
H
HO
OH
H
OH
H
CH 2OH
(C)
HO
HO
H
HO
H
H
OH
H
CH 2OH
CHO
(D)
HO
HO
HO
H
H
H
H
Oh
CH2OH
KEY: (A)
SOL: L (–) glucose should be mirror image of D(+) glucose hence
The correct answer should be
CHO
HO
H
HO
HO
H
OH
H
H
CH 2OH
Hence correct answer is (A)
NARAYANA Group of Educational Institutions
-23-









JEE-Advance – 2015 (Hints & Solutions) Paper - 1
SECTION : 3 (Maximum Marks : 16)
This section contains TWO questions.
Each question contains two columns, Column I and Column II.
Column I has four entries (A), (B), (C) and (D).
Column II has five entries (P), (Q), (R), (S) and (T).
Match the entries in column I with the entries in column II.
One or more entries in column I may match with one or more entries in column II.
The ORS contains a 4  5 matrix.
For each entry in column I, darken the bubbles of all the matching entries. For example, if
entry (A) in column I matches with entries (Q), (R) and (T), then darken these three bubbles
in the ORS. Similarly, for entries (B), (C) and (D).
Marking scheme:
For each entry in column I.
+2 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened.
0 If one of the bubbles is darkened.
1 In all other cases.
2 In all other cases.
39)
Match the anionic species given in Column I that are present in the ore(s) given in column II
(A)
(B)
(C)
(D)
Column I
Carbonate
Sulphide
Hydroxide
Oxide
(P)
(Q)
(R)
(S)
(T)
Column II
Siderite
Malachite
Bauxite
Calamine
Argentite
KEY: AP,Q,S; BT; CR,Q, DR
SOL:
(P)
Siderite -
FeCO3
(Q)
Malachite
-
(R)
Bauxite -
AlOx(OH)3-2x
(S)
Calamine
-
ZnCO3
(T)
Argentite
-
Ag2S
CuCO3.Cu(OH)2
NARAYANA Group of Educational Institutions
-24-
40)
(A)
(B)
(C)
(D)
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
Match the thermodynamics process given under Column I with the expressions given under
Column II
Column I
Freezing of water at 273 K and 1 atm
Expansion of 1 mol of an ideal gas into
a vacuum under isolated conditions
Mixing of equal volumes of two ideal
gases at constant temperature and
pressure in an isolated container
Reversible heating of H2(g) at 1 atm
from 300 K to 600 K, followed by
reversible cooling to 300 K at 1 atm
(P)
(Q)
Column II
Q=0
w=0
(R)
Ssys  0
(S)
U  0
(T)
G  0
KEY: AR,T; BP,Q,S; CP,Q,S; DP,Q,S,T
SOL:
(A)
For freezing of water at 273K and 1 atm
S  0, G  0 (it is at equilibrium)
w is not zero as there will be expansion in freezing
(B)
Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions
Q = 0, w = 0, U  0
(C)
Mixing of equal volume of two ideal gases at constant temperature and pressure in an
isolated container
Q = 0, w = 0, U  0
(D)
Reversible heating of H2(g) at 1 atm from 300 K to 600 K, followed by reversible cooling
to 300 K at 1 atm
Q = 0, w = 0, U  0 , G  0
NARAYANA Group of Educational Institutions
-25-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
MATHEMATICS
x2 
41.
Let F  x  

x
Key.
6
 1
2 cos 2 tdt for all x  R and f : 0,    0,   be a continuous function. For
 2
 1
a   0,  , if F'  a   2 is the area of the region bounded by x = 0, y = 0, y = f(x) and x = a,
 2
then f(0) is
(3)
x2 
Sol.

Fx 


6
2 cos 2 t dt
x
a2 
Fa  


6
2 cos 2 t dt
a


F'  a   2 cos 2  a 2   .2a  2 cos 2 a.1
6



F'  a   2  4a cos 2  a 2    2 cos 2 a  2
6



= 4a cos 2  a 2    2sin 2 a
6

a

2 2
2
0 f  x  dx  4a cos  a  6   2sin a

 

 


f  a   4  a 2cos  a 2   sin  a 2   .2a  cos 2  a 2   .1  4sin a cos a
6 
6
6 



 f  0  3
42.
A cylindrical container is to be made from certain solid material with the following
constraints: It has a fixed inner volume of V mm3, has a 2 mm thick solid wall and is open at
the top. The bottom of the container is a solid circular disc of thickness 2 mm and is of
radius equal to the outer radius of the container.
If the volume of the material used to make the container is minimum when the inner radius
V
of the container is 10 mm, then the value of
is
250
Ans. (4)
Sol. Let inner radius = r
Height of cylinder = h
Volume of inner cylinder = v
v  r 2 h
v
r 2
Now, Material used
2
2
M    r  2  h  r 2 h    r  2  .2
h
NARAYANA Group of Educational Institutions
-26-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
43.
Ans.
Sol.
v
2
2

=    r  2 . 2  v  2  r  2 
r


v  4 4
2

M =   1   2  v   2  r  2  

  r r

M is the function of r
v 
4 8

f '  r      0  2  3   4  r  2   0
r r 
 

when r = 10,
v  40
8 


  48  0
  1000 1000 
48v
 48
1000
v
4
250
Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way
that all the girls stand consecutively in the queue. Let m be the number of ways in which 5
boys and 5 girls can stand in a queue in such a way that exactly four girls stand
m
consecutively in the queue. Then the value of
is
n
(5)
Clearly n  6!5!
Now, to calculate value of m
We make three cases
2
Case 1 : first four places filled by girls fifth by boy. This can be done in 5 P4 .5  5!  5.  5!
Case 2 : Four girls are in a row having both the side a boy this can be done in
2
5  5 P4  5 P2  4!  20.  5!
Case 3 : Last four places are filled by girls and 6th place by a boy this can be done in
2
5
P4 .5.5!  5.  5!
m  30.  5!
2
m
5
n
The minimum number of times a fair coin needs to be tossed, so that the probability of
getting at least two heads is at least 0.96, is
(8)
Let coin is tossed ‘m’ times, then probability of getting at least 2 heads is
m
m
 m  1
1
1
1     m C1    1 
2m
2
2
 m  1  0.96
Given that 1 
2m
m 1
1
 0.04 
m
2
25
So,
44.
Ans.
Sol.
NARAYANA Group of Educational Institutions
-27-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
m 1
2m
3
1
5
6
f 1  1, f  2   , f  3  , f  4   , f  5  
,
4
2
16
32
7
8
9
10
f  6  , f 7  
, f 8  
, f 9  
,....
64
128
256
512
So minimum value of m is 8
If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are
2
2
tangents to the circle  x  3   y  2   r 2 , then the value of r2 is
Let f  m  
45.
Ans.
Sol.
(2)
y2  4x , end points of latus rectum are 1,  2 
Equation of normal at (1, 2) is
y  x  3 , it touches the circle
So
46.
3   2   3
11
 r  r  2  r2  2
 x  , x  2
,
Let f : R  R be a function defined by f (x)  
0, x  2
Where [x] is the greatest integer less than or equal to x. If I = 
2
-1
Key.
Sol.
xf(x 2 )
dx, then the value
2+ f(x+1)
of (4I – 1) is
(0)
2
xf(x 2 )
I=
dx,
-1 2+ f(x+1)
0
1
I =  0dx+  0dx  
0
-1
2
1
2
x
dx +
20

3
2
4
0dx +  0dx
3
2
1 ( x )1
1
1
I= .
 2  1 
2
2
4
4

4I -1 = 0.
47.
The number of distinct solutions of the equation
5
cos 2 2 x  cos 4 x  sin 4 x  cos 6 x  sin 6 x  2
4
In the interval 0, 2 is
Key.
(8)
5
cos 2 2 x  1  2sin 2 x cos 2 x  1  3sin 2 x cos 2 x  2
4
5
1
3
cos 2 2 x  sin 2  2 x   sin 2  2 x   0
4
2
4
2
2
cos 2 x  sin 2 x  0
cos 4x  0
Since, x 0, 2
Hence no. of distinct solution = 8.
Sol.
NARAYANA Group of Educational Institutions
-28-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
48.
Key.
Sol.
Let the curve C be the mirror image of the parabola y 2  4 x with respect to the line
x+ y  4  0. If A and B are the points of intersection of C with the line y = -5, then the
distance between A and B is
(4)
Let a Point P(t 2 , 2t) on y 2  4 x
x - t 2 y  2t
Image of 'P' w.r.t. the line x  y  4  0 can be given as:

   t 2  2t  4 
1
1
2
y  t  4.
x  2t  4;
Hence, Curve C becomes
2
 x+ 4 
 4  - y.
4
Since, it intersect with the line y = -5
( x  4) 2  16  20

49.
( x  4) 2  4
x  4  2
 x  4  2 OR x  4  2
A = (-2, -5) ; B = (-6, -5)
 AB = 4.
Let P and Q be distinct points on the parabola y 2  2 x such that a circle with PQ as
diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the
area of the triangle OPQ is 3 2 , then which of the following is (are) the coordinates of P ?

(A) 4, 2 2
Key.
Sol.


(B) 9,3 2

1 1 
(C)  ,

4 2

(D) 1, 2

(A, D)
P  at12 , 2 at1 
O
Q  at 22 , 2 at 2 
t1 t 2  4 (Since, PQ is the diameter of circle)

0
1 2
area of OPQ = at1
2 2
at 2
3 2  t1 
2at 2 1
4
t1

t1  2, 2 2

P  1, 2 , 4, 2 2

0 1
2at1 1

(Since P lies in first Quadrant)

NARAYANA Group of Educational Institutions
-29-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1


50.
Let y(x) be a solution of the differential equation 1  e x y'  ye x  1. If y(0) = 2, then which of
Key.
the following statements is (are) true?
(A) y (4)  0
(B) y (2)  0
(C) y ( x) has a critical point in the interval (1, 0)
(D) y ( x) has no critical point in the interval (1, 0)
(A,C)
 y.e x  1
1  e x  dy
dx
d
y  e x  1  1

dx
y  e x  1  x  c

Sol.



y (0)  2
2(2)  c  c = 4
Hence, y (e x  1)  x  4 .

y(4)  0
x4
y x
e 1
x
x
dy  e  1   x  4  e

2
dx
 ex  1
For critical point
ex  x  3  1
1
x 3
1 1
Now, g  1    0
e 2
1
And g  0   1   0
3
Hence, g  x   0 must has one real root in (-1,0).
Let:
51.
g  x   ex 
Consider the family of all circles whose centers lie on the straight line y = x. If this family of
circles is represented by the differential equation Py  Qy  1  0 , where P, Q are functions
of x, y and y (here y 
Key.
Sol.
dy
d2 y

, y  2 ), then which of the following statements is (are) true?
dx
dx
(B)
P=y–x
(A)
P=y+x
(C)
P  Q  1  x  y  y   y
2
(D)
P  Q  x  y  y   y
2
(B, C)
Equation of family of circles is
x2  y 2  2 x  2 y    0
NARAYANA Group of Educational Institutions
-30-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
dy
   x 
dx
2
d2y
dy
 y    2     1
dx
 dx 
 y  
……..(i)
…….(ii)
yy "  y '  1
2

y"
Putting the value of  in (i)
2
2

y " y   y '  1  y " y   y '  1
y ' y 
x

y"
y"




or y " y  x   y ' y '  y '  1  1  0
2
P yx
Q  y '  y '  1
2
P  Q  1  x  y  y '  y '
2
be a differentiable function with g(0) =0, g  0   0 and g 1  0 . Let

52.
Let g :
Key.
 x
gx, x  0

f  x   | x |

0,
x0

|x|
And h  x   e for all x  . Let  f o h  x  denote f  h  x   and  h of  x  denote h  f  x   .
Then which of the following is (are) true?
(A)
f is differentiable at x = 0
(B)
h is differentiable at x = 0
(C)
f o h is differentiable at x = 0
(D)
h o f is differentiable at x = 0
(A, D)
Sol.
f '  0   lim
f  0  h   f  0
h 0
 lim
h
h
g  h  0
h
h 0
h
g h
 lim
0
h0
h
(As g '  0  exists and g  0   0 )


lim
g  0  h  g  0
h 0
h
g h
0
0
h
f  0  h   f  0
lim
h 0
f '  0   lim
h 0
 lim
h
h
g  h  0
h
h 0
h
 g  h 
 lim  

h 0
 h 
NARAYANA Group of Educational Institutions
-31-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
  1  0  0
 function f is differentiable at x = 0
e  x x  0
h x   x
x0
e
 e  x x  0
h ' x    x
x  .0
 e
h '  0   1 ;
h '  0   1
h is not differentiable at x = 0
 
f  h  x   f e x

e
x
e
x
g  e| x |   g  e | x | 
f  h  x   is not differentiable at x =0 as g  1  0
  x

h  g  x   x  0
h  f  x     x


h
0
x0



 x g  x
 x
x0
 e
 1
x0
e g  x  x  0

x0
 1
h  f  x   is continuous at x = 0
Also h  f  x   is differentiable at x = 0.
53.

 

Let f  x   sin  sin  sin x   for all x  and g  x   sin x for all x  . Let  f o g  x 
2
2

6
denote f  g  x   and  g o f  x  denote g  f  x   . Then which of the following is (are) true?
 1
Range of f is   ,
 2
f x 
lim

(C)
x 0 g  x 
6
(A, B, C)



(A)
Key.
Sol.
1
2 
(B)
 1
Range of f o g is   ,
 2
(D)
There is an x 
1
2 
such that  g of  x   1
f  x   Sin  sin  sin x  
2

6
 xR 

2


sin x 

2
2



1  sin  sin x   1
2


  
 
  sin  sin x  
6 6 2
 6
NARAYANA Group of Educational Institutions
-32-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
1
 
 1
  sin  sin  sin x   
2
2
 2
6
1 1
Range of f is   , 
 2 2






f  g  x    f  sin x   sin  sin  sin  sin x   
2

2

2
6
Range of f  g  x   is   , 
 2 2



1 1
sin  sin  sin x  
6
2
  
lim 
 
x 0
6 6



 6 sin  2 sin x  






g  f  x    g  sin  sin x  
2

6




 sin  sin  sin x  
2
2

6

1

 1
  sin  sin  sin x   
2
2
 2
6



4
for no x  R ;
54.



 
sin  sin  sin x   
2
2
 4
6
g  f  x   1
Let PQR be a triangle. Let a  QR , b  RP and c  PQ . If a  12 , b  4 3 and b.c  24 ,
then which of the following is (are) true?
c
 a  12
2
a  b  c  a  48 3
(A)
(C)
Key.
Sol.
2
c
(B)
(D)
2
 a  30
2
a .b  72
(A, C, D)
a  QR , b  RP , c  PQ
a  12 , b  4 3 ,
b .c  24
a b c 0
As
b  c  a
2
b  c  2b .c  a
2
2
or 48  c  48  144
2
c  48
2
c
2
2
2
c
2
a 
48
 12  12
2
a 
48
 12  36
2
NARAYANA Group of Educational Institutions
-33-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
a  b  c
2
a  b  2a.b  c
2
2
 
 a . b   72
a  b  c  a  2a  b 
2 a . b  48  144  48

 2 a b  2 a b  a .b
2
2

2
 2 144  48  72  72
 48 3
55.
Key.
Sol.
Let X and Y be two arbitrary, 3×3, non zero, skew-symmetric matrices and Z be an
arbitrary 3×3, non zero, symmetric matrix. Then which of the following matrices is (are)
skew symmetric?
Y 3 Z4  Z4 Y 3
X 44  Y 44
X 4 Z3  Z3X 4 (D)
X 23  Y 23
(A)
(B)
(C)
(C, D)
X T  X
Y T  Y
ZT  Z
Y Z
3
 Z 4Y 3   Y 3 Z 4    Z 4Y 3 
T
4
T
T
  Z T   Y T   Y T   Z T 
4
3
3
4
 Z 4  Y 3    Y   Z   Y 3 Z 4  Z 4Y 3
3
X
 Y 44    X T   Y T 
T
44
44
44
   X T    Y T 
44
   X    Y 
44
X
4
44
44
 X 44  Y 44
4
Z3  Z3X
   X Z   Z X 
 Z   X    X  Z 
4 T
4
3 T
T 3
4 T
3
T 4
T 4
T 3
  Z   x    X   Z 
3
4
4
3
 Z 3 X 4  X 4Z 3   X 4Z 3  Z 3 X 4 
X
23
 Y 23    X T   Y T 
T
23
23
   X    Y 
23
56.
   X 23  Y 23 
Which of the following values of  satisfy the equation
1    1  2  1  3 
2
2
2
 2     2  2   2  3 
2
2
2
 3     3  2   3  2 
2
Key
23
(A) –4
(B, C)
2
2
(B) 9
 648 
(C) –9
NARAYANA Group of Educational Institutions
(D) 4
-34-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
Sol.
1   2  2
4   2  4
9   2  6
1  4 2  4
4  4 2 8
9  4 2  12
1  9 2  c
4  9 2  72
9  9 2  18
Then by simplifying
1  2 2
2 2
3
5
3 2  2
2
2
1
6
0
2 2 6 10  4  2 2  8 3  648
 2  81
  9
57.
In R2 consider the planes P1 : y = 0 and P2 : x+z =1. Let P3 be a plane different from P1 and
P2 passes through the intersection of P1 and P2. If the distance of the point (0,1,0) from P2 is 1
and the distance of a point  ,  ,   from P3 is 2, then which of the following relations is
Key
Sol
(are) true ?
(A) 2    2  2  0
(B) 2    2  4  0
(C) 2    2  10  0
(D) 2    2  8  0
(B, D)
Given
P1 = y = 0 and x + z – 1 = 0
P2 = x + z– 1 = 0
P3 = P1 +  P2 = 0
 x  z   y 1  0
The perpendicular distance from (0, 1, 0) = 1

 1
1
2  2
1

2
P3  2 x  y  2 z  2  0
The distance from  ,  ,  from P3 = 2
2    2  2
2
3
2    2  4  0 and 2    2  8  0
In R3, let L be a straight line passing through the origin. Suppose that all the points an L are
at a constant distance from the two planes P1 : x + 2y – z + 1 = 0: 2x – y + z –2 = 0. Let M be
the locus of the feet of the perpendicular drawn from the points on L to the plane P1. Which
of the following points lie(s) of M ?
5 2

 1 1 1
 5 1
 1 2
(A)  0,  ,  
(B)   ,  , 
(C)   , 0, 
(D)   , 0, 
6 3
 3 3

 6 3 6
 6 6

58.
NARAYANA Group of Educational Institutions
-35-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
Key
Sol
(A, B)
Equation of angle bisectors
x  2 y  z 1
2x  y  z 1

6
6
x  3 y  2 z  2  0 and 3x  y  0

Hence plane 3x+y = 0 taken, as passes through origin. Line through origin on 3x + y = 0 and
parallel to
i j k
x
y
z


   any point on this line is   ,  3,  5 
1 2 1  iˆ  3 ˆj  5kˆ is
1 3 5
2 1 1
Foot of perpendicular from   ,  3,  5  to x + 2y –z + 1 = 0
x   y  3 z  5  1



1
2
1
6
1
1
1

Point is    ,  3  ,  5   , which is satisfied by A, B
6
3
6

59.
(A)
Column – I
Column – II
In
if the magnitude of the projection vector of the vector (P)
1
 iˆ   ˆj on 3 iˆ  ˆj is 3 and if   2  3  , then
R2,
possible value(s) of |  | is (are)
(B)
Let a and b real numbers such that the function
(Q)
2
(C)
3ax  2, x  1
f  x  
is differential for all x  R. Then
2
 bx  a , x  1
possible value (s) of a is (are)
Let   1 be a complex cube root of unity. If
(R)
3
(S)
4
(T)
5
2
3  3  2 
 2  3  3 
2 4 n 3
2 4 n 3
(D)

  3  2  3 2 
4 n 3
 0, then possible
value(s) of n is (are)
Let the harmonic mean of two positive real numbers a and
b be 4. If q is a positive real number such that a, 5, q, b is
an arithmetic progression, then the value(s) of | q  a | is
(are)
Key
(A-P,Q, B-PQ, C-P,Q,S,T, D-Q,T)
Sol
(A)
 i   j  . 
3 iˆ  ˆj
2

3
3    2 3
NARAYANA Group of Educational Institutions
-36-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1


3
  2 , compare with   2  3 
  0,    3
Then  1, 2
(B)
6ax
f ' x  
 b
x  1 for differentiable function
.... 1
6a  b
For continuous function, 3a  2  6a  a 2 ….(2)
From equation (1) & (2) a  2,1
(C)
3  3  2    2  3  3    3  2  3 
  3  2  31  

0
2 4 n 3
2 4 n 3
4 n 3
2
2 4 n 3
8n6
3 2  2  3  0


(D)
1   n   2n  0
Hence n is not multiple of 3
n = 1, 2, 4, 5
2ab
 4,
Given a  b
a, 5, q, b are in A.P.
q  10  a , q 
…(1)
b5
2
a  10  q
b  2q  5
…(2)
From (1) and (2)
2q 2  23q  60  0
15
 for q = 4, a = 6
2
Then q = 15/2, a = 5/2
| q  a |  2,5
q  4,
NARAYANA Group of Educational Institutions
-37-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
60.
(A)
Column – I
In a triangle XYZ , let a, b and c be the lengths of the sides
Column – II
(P) 1
opposite to the angles X, Y and Z respectively. 2  a 2  b 2   c 2
and  
sin  X  Y 
, then possible values of n for which
sin Z
cos  n     0 is (are)
(B)
In a triangle XYZ , let a, b and c be the lengths of the sides
(Q) 2
opposite to the angles X, Y and Z, respectively. If
1  cos 2 X  2 cos 2Y  2sin X sin Y , then possible value(s) of
a
b
is (are)
(C)
(D)
In R2, let
3 iˆ  ˆj , iˆ  3 ˆj and  iˆ   1    ˆj be the position
(R) 3
vectors of X,Y and Z with respect to the origin O,
respectively. If the distance of Z from the bisector of the acute
3
angle of OX and OY is
, then possible value(s) of |  | is
2
(are)
Suppose that F(x) denotes the area of the region bounded by
(S)
2
x = 0, x = 2, y = 4x and y |  x  1|  |  x  2 |  x, where
 {0,1}. Then the value(s) of F   
5
8
2, when   0 and
3
  1, is (are)
(T)
Key
(A-P,R,S; B – P; C- P; D – S,T)
Sol
(A)
Given 2  a 2  b 2   c 2 and  
6
sin  X  Y 
,
sin Z
2  sin 2 X  sin 2 Y   sin 2 Z
2 sin  X  Y   sin Zz
 2 1 sin Z  0
1
2
 n
cos 
 2


(B)

0

n = odd number
Given 1  cos 2 X  2 cos 2Y  2sin X sin Y
1  1  2sin 2 X  2  4sin 2 Y  2sin X sin Y
NARAYANA Group of Educational Institutions
-38-
JEE-Advance – 2015 (Hints & Solutions) Paper - 1
2b  a  ab
2
2
2
a a
   2  0
b b
a
  1
b
(C)
3 iˆ  ˆj  iˆ  3 ˆj
2
Internal angular bisector is
Angular bisector is y  x  0
3
2
The perpendicular distance   ,1    is
(D)
|1  2 | 3

, |  |  1 or 2
2
2
for   0 , y |  x  1|  |  x  2 |  x
y=3
y=3
(2,2 2)
x=0
x=L
(2,–2 2)


F    3  2 2  2 
2 2

0
 6
y2
dy
4
8 2
3
F   
8 2
6
3
 x  3 0  x  1
 x 1 1  x  2
For   1 , y |  x  1|  |  x  2 |  x  
y=2 x
y=2
x=1
F    6 
F   
x=2
8 2
8 2
1  5 
3
3
8
2 5
3
NARAYANA Group of Educational Institutions
-39-