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CODE-9
PAPER-1
Time: 3 Hours
P1-15-9
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 with the options are legible.
QUESTION PAPER FORMATAND MARKING SCHEME:
8.
The question paper consists of 3 parts (Physics, Chemistry and Mathematics). Each part consists of three
sections.
9.
Carefully read the instructions given at the beginning of each section.
10. In Section 1 contains 8 questions. The answer to each question is a single digit integer ranging from 0 to 9
(both inclusive).
Marking scheme: + 4 for correct answer, 0 if not attempted and –2 in all other cases.
11. In 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. In 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 multilate the ORS.
18. Write your name, roll number and the name of the examination centre 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.
DARKENING THE BUBBLES ON THE ORS:
19. Use a BLACK BALL POINT PEN to darken the bubbles in the upper sheet.
20. Darken the bubble COMPLETELY.
21. Darken the bubbles ONLY if you are sure of the answer.
22. The correct way of darkening a bubble is as shown her:
23. The is NO way to erase or “un-darken” a darkened bubble.
24. The marking scheme given at the beginning of each section gives details of how darkened and not darkened
bubbles are evaluated.
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-1 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
JEE-Advanced-2015
PART I : PHYSICS
SECTION 1 (Maximum Marks: 32)

This section contains EIGHT questions

The answer to each question is a SINGLE DIGIT INTEGER 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
1.
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
Sol: [3] If initially x is amount of fuel, then energy is E.
Energy required =
1
E
8
x 
Tx 
Tx 
Tx 
 3T
2
4
8
n=3
2.
A Young’s double slit interference arrangement
with slits S 1 and S 2 is immersed in water
S1
(refractive index = 4/3) as shown in the figure.
The positions of maxima on the surface of water
2
2
2 2
d
x
2
are given by x = p m  – d , where  is the
air
d
wavelength of light in air (refractive index = 1),
S2
2d is the separation between the slits and m is
water
an integer. The value of p is
d 2  x2
c
Sol: [3] t 
t 
x2 = p2m22 – d2
4 d 2  x2
3
c
 4 1
t  t   t    1 (d 2  x 2 )1/2
3 c
t 
1 2
(d  x 2 )1/2
3c
2
2
x 
t

T
9m22 = d2 + x2
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-2 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
f 2
( d  x 2 )1/2
3c
3mc = f(d2 + x2)1/2
x 
JEE-Advanced-2015
x2 = 9m22 – d2
3mc
 ( d 2  x 2 )1/ 2
f
P=3
3.
Consider the 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.
llll
llll
llll
lllllll
lllll
llllllllllllllllllllllllll
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 in a medium of refractive index
7/6, the magnification becomes M2. The
15 cm
M2
magnitude M is
1
50 cm
Sol: [7]
1 1 1
 
v u f
m
v 30

2
u 15
1
1
1


v 15 10
1
1 1 3  2 1
  

v = –30
v
10 15
30
30
For lens u = 2f = 20. So the final image is at 2f ahead of lens.
M1 = 2
....(i)
1 3 6   1  1 
1 

f   2  1   R1 R2 

1
1
 1
1 
  
10
2
 R1 R2 
10 4

f  14
1  3  1
1
   1   
10  2   R1 R2 
M2 = m1 × m2 = 2 ×
140
= 2 × 7 = 14
20
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-3 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
JEE-Advanced-2015
70
4
f
1 4
1 87 1




v 70 20 140 40
....(ii)
1
1
4


v 20 70
M2 = 14
M2
So M  7
1
4.
z
An infinitely long uniform line charge distribution of
charge per unit length  lies parallel to the y-axis in
3
a (see figure). If the
2
magnitude of the flux of the electric field through the
rectangular surface ABCD lying in the x-y plane with
the y-z plane at z 
y
O
A
B
z
1
Sol: [6]    Q
0

C
a
L
its centre at the origin is n (0 = permittivity of
o
free space), then the value of n is
L
D
L
 L
0
L
360º ––––––– 
0
1 L
60º –––––––– 6 
0
n=6
5.
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).
Sol: [2]
E
hc 1242

eV = 13.8 eV

90
K = 10.4 eV
 = E – K = 13.8 eV – 10.4 eV = 3.4 eV
n=2
6.
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 the value of N is
(ignore energy loss due to atmosphere).
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-4 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
Sol: [2]
g  g
JEE-Advanced-2015
R2
( R  h2 )
2
1
R2
 2
4 R  h2
R + h = 2R
h=R

GMm 1 2
GMm
 mv  
R
2
2R
1 2
GMm GMm
mv  

2
2R
R
v
Vs 
GM
R
2GM
V 2
R
N=2
7.
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).
A
v1 = 3 m/s
30 m
B
C
v2
27 m
D
Sol: [7]
1 1 v2 1 2
1
1 1
9
   mv  mg 30  m  9   mR 2 2
2 2 R 2
2
2 2
R
3 2
27 m
mv  30mg 
4
4
3 2
27
v  300 
4
4
...(i)
2
3 2
1
1 1
 v
v  270m  mv22    mR 2   22
4
2
2 2
 R
3 2
3
v  270  v22
4
4
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-5 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
300 
30 
JEE-Advanced-2015
27
3v 2
 270  2
4
4
27 3v22

4
4
147 3v22

4
4
49  v22
v2 = 7
8.
Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A
 A 
emits 104 times the power emitted from B. The ratio    of their wavelength A and B at which the
 B
peaks occur in their respective radiation curves is
Sol: [2] RA = 400 RB
PA = 104 PB
P
...(i)
E
 AT 4
t
TA4 PA  AB PA AB

 
TB4
AA PB
PB AA
TA4
RB2
4

10

TB4
(400) 2 RB2
TA
1 1
 10  
TB
20 2
 A TB

2
 B TA
SECTION 2 (Maximum Marks: 40)

This section contains TEN questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE 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 the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 if
none of the bubbles in darkened –2 in all other cases.
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-6 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
9.
JEE-Advanced-2015
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 m and 1.0 × 10–7 m,
respectively. The electrical resistance between the two faces P and Q of the composite bar is
Q
Al
Fe
50 mm
2 mm
P
7 mm
(A)
2475

64
(B)
1875

64
(C)
1875

49
(D)
2475

132
1 1
1
Sol: [B] R  R  R
1
2
A
A
1
 1  2
R 1l1  2l2
R1 = 30 × 10–6 
R2 = 1250 × 10–6 
1 1
1
125  3
64




R 30 1250 3750 1875
R
1875

64
10. For photo-electric effect with incident photon wavelength , the stopping potential is V0. Identify the
correct variation (s) of V0 with  and 1/.
V0
V0
(A)
(B)
V0
V0
(C)
(D)
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-7 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
Sol: [A, C]
JEE-Advanced-2015
Km = E – 
eV 
hc


eV0 
hc hc

 0
when  = 0
11.
Consider a Verner callipers 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 callipers, 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 callipers, the least count of
the screw gauge is 0.01 mm.
(B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the lest 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 lest count of the Vernier
callipers, 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
capillers, the least count of the screw gauge is 0.005 mm.
Sol: [B, C]
L.C of vernier = 1 msd – 1 vsd = 0.25 mm
B.
Pitch = 0.5 m
 L.C 
C.
Pitch 1mm
 L.C 
0.5
 0.05 mm
100
1
 0.01 mm
100
12. 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 correct option (s) is (are)
(A) M  c
(B) M  G
(C) L  h
(D) L  G
Sol: [A, C, D]
h = [ML2T–1]
c = [LT–1]
G = [M–1L3T–2]
hc
M  h
G
M c
1
M
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M2 
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
L2 
hG L5T 3

c 3 L3T 3
JEE-Advanced-2015
L h
L G
L  C 3/2
13. Two independent harmonic oscillators of equal mass are oscillating about the origin with angular
frequencies 1 and 2 and have total energies E1 and E2, respectively. The variation of their momenta
p with positions x are shown in the figures. If
p
Energy = E 1
a
a
 n 2 and  n , then the correct equation(s) is (are)
b
R
p
Energy = E2
b
a
(A) E11  E2 2
Sol: [B, D]
x
R
2
2
(B)   n
1
x
E1 E2
(D)   
1
2
(C) 12  n 2
A2 R 1
Amplitudes A  a  n
1
v2 m R Rn 2
 
n
Maximum speeds
v1m b
a
vm = A
2 vm2 A1


 n2
1 vm1 A2
2
1 2
E2 vm2
E 
1
E  mvm 
 2  n2  2  1  n 2  2  1 
2
E1 vm1
E1 2
n
E1 E2

1 2
14. A ring of mass M and radius R is rotating with angular speed
 about a fixed vertical axis passing through its centre O
with two point masses each of mass M/8 at rest at O. These
masses can move radially outwards along two massless rods
fixed on the ring as shown in the figure. At some instant the
O
angular speed of the system is 8/9  and one of the masses
is at a distance of 3/5R from O. At this instant the distance
of the other mass from O is
(A)
2
R
3
(B)
1
R
3
(C)
3
R
5
(D)
4
R
5
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-9 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
JEE-Advanced-2015
Sol: [D]

By conservation of angular momentum, MR 2    MR 2  M
8


9R 2
9R 2 x2
 R2 

8
200 8
 x
2
3  M 28
R
5   8 x  9



4
R
5
15. 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)
x
+q
–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.
Sol. [C]
Using E 
3k 
r
Only positive charge will oscillate.
16. 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
S2
P
50 cm
(A) 60 cm
d
(B) 70 cm
(C) 80 cm
(D) 90 cm
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- 10 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
JEE-Advanced-2015
Sol. [B]
1 1.5 1.  1.5


V 50
10
for
S1 ,

V = 50
for S 2 , 0 

1 1.5  1

u
10
u = – 20
Therefore, d = 50 + 20 = 70
17. 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)
y
R
R
I
x
L
R

(A) If B is along zˆ, F  ( L  R

(C) If B is along yˆ, F  ( L  R )
R
L

(B) If B is along xˆ, F  0

(D) If B is along zˆ, F  0
Sol. [A, B, C]
The conductor can be replaced by a straight conductor.
18. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in 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 1/2
(D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2
Sol. [A, B, D]
A.
5
3
Average K.E. per pole  RT  RT  2 RT
2
2
B.
U mixture
U He
 mixture 
 mixtrue  He

 he
 mix
C pmixture
CV mixture
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- 11 -
Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
C p mixture
Cv mixture
7
5
R R
2  3R
2
2
5
3
R R
2  2R
2
2
Therefore,
D.
JEE-Advanced-2015
U mixture
3 3 4
6
4  1  2 1

  
3
because  He  4 and  mixture 
He
2 5 5
5
2
1
1
2
mH U M2  mHe U He
2
2
U He
mH
2
1
 U  m  4
2
H
He
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 whose layout will be similar to the one shown below.

(A)
P
Q
R
S
T
(B)
P
Q
R
S
T
(C)
P
Q
R
S
T
(D)
P
Q
R
S
T
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 none of the bubbles is darkened. – 1 In all other cases.
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19. Match the nuclear processes 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
B.
Fission in a nuclear reactor
(Q)
C.
-decay
D.
-ray emission
235
92
U
60
27
Co nucleus
(R) Energy production in stars via hydrogen
conversion to helium
(S) Heavy water
(T) Neutrino emission
Sol. [A-(R); B-(P, S); C-(Q, T); D-(Q)]
20. 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.
2
U  x 
U1 ( x )  0 1    
2   a  
B.
U 2 ( x) 
C.
2
  x 2 
U0  x 
U 3 ( x) 
exp     
2  a 
  a  
D.
U 4 ( x) 
U0  x 
 
2 a
U0
2
(P)
The force acting on the particle is zero at x = a
(Q)
The force acting on the particle is zero at x = 0
(R)
The force acting on the particle is zero at x = – a.
(S)
The particle experiences an attractive force towards
2
 x 1  x 3 
    
 a 3  a  
x = 0 in the region | x | < a.
(T)
The particle with total energy
U0
can oscillate about
4
the point x = – a
Sol. [A-(P),(Q),(R),(T); B-(Q),(S), C-(P),(Q),(R),(S); D-(P),(R),(T)]
f 

dU
dx
 x2 
F1  2U o x 1  2 
 a 
F2  
U0
2x
a2
x2
 x x3   2
F3  U 0  2  4  e a
a 
a
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F4  
JEE-Advanced-2015
 3x2 
U0  1 x2 
d 2U 2
 3

(
x


a
)


1
U
 2U 0 1   is positive, therefore U is a


0 1 
2 
1
2  a a3  and dx 2
a
 2


manima.
 1 3x 2 
d 2U 3
u 2

U
  02 is negative therefore is a maxima.
0 2 
(at x = –a) dx 2
4 
a 
a
a
U x
d 2U 4
  03 is +ve, therefore is a minima.
2
dx
a
PART II : CHEMISTRY
SECTION 1 (Maximum Marks: 32)

This section contains EIGHT questions

The answer to each question is a SINGLE DIGIT INTEGER 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. Not consider 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
Sol: [9] n = 3, l = 0, 1, 2
l = 0, m = 0
l = 1 m = 1–1, 0, + 1
l = 2, m = –2, –1, 0 + 1, +2
For H– degeneracy will be 9 in second excited state
22. All the energy released from the reaction X Y, rG0 = – 193 kJ mol–1 is used for oxidizing M+ as
M+  M3+ +2e–, E0 = –0.25V.
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]
Sol: [4] X Y, G0 = –193 kJ mol–1
As G0 = –nFE0Cell
= –2 Ú 96500 × (0.25) J
= Total available energy
= –193 × 103 J
+
No. of moles of M oxidised = –193 × 103
––––––––––––– = 4
–2 × 96500 (0.25)
23. 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 –0.0558°C, the number of chloride(s) in the coordination
sphere of the complex is [Kf of water = 1.86 K kg mol–1]
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Sol: [3] Tf = 0.0538
Kf = 1.86
m = 0.01
As Tf = i × Kf × m
i
Tf
Kf  m
1 .8 6
3
0 .0 5 5 8  0 .0 1
Possible structure will be [Co(NH3)3Cl]Cl2
So only one Cl– is within co-ordination sphere.

24. The total number of stereoisomers that can exist for M is
CH3
H3 C
H3 C
M
H3C
O
CH3
Chiral centres
Sol: [4]
H3C
O
There are two chiral centres so possible optical isomers = 2 + 2 = 4
No geometrical isomerism is possible
25. The number of resonance structures for N is
OH
NaOH

N
Sol: [9]
26. The total number of lone pairs of electrons in N2O3 is
Sol: [8] Lone pair = 8
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27. For the octahedral complexes of Fe3+ in SCN– (thiocyanato-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]
Sol: [2] SCN– is a weak ligand but CN– is a strong ligand. Difference between spin only magnetic movements
in B.M. is approximately 2.
28. Among the triatomic molecules/ions, BeCl2, N3 , N2O, NO2 , O3, SCl2, ICl2 , I3 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]
Sol: [4]
BeC l 2 , N 3 , N 2 O, NO 2
SECTION 2 (Maximum Marks: 40)

This section contains TEN questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE 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 the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 if
none of the bubbles in darkened –2 in all other cases.
29. The % yield of ammonia as a function of time in the reaction
 2NH 3 (g), H  0 at (P, T1) is given below
N 2 (g)  3H 2 (g) 
If this reaction is conducted at (P, T2), and T2 > T1, the % yield of ammonia as a function of time is
represented by
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(A)
(B)
(C)
(D)
JEE-Advanced-2015
Sol. [C] According to Lechatellier Principle formation of Ammonia is formed at high P and low T, as well it
is an exothermic process.
30. 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
(A)
1 1
,
2 8
(B) 1,
1
4
(C)
1 1
,
2 2
(D)
1 1
,
4 8
Sol. [A]
31. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is(are)
Br
Br
H
(A)
H3C
(B) H2C
CH3
H
CH3
Br
H
H2C
H
CH3
(C)
(D) H2C
Br
CH3
CH3
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Sol. [B, D]
After hydrogenations
H3C
(A) will form
H3C
CH2
(B)
H
H
H
C
C
C
H
H
CH3
Br
presence of chiral C make of optically active.
H
C
CH2
Br
CH3
No chiral C-atom so optically inactive
H
H3C
CH C
(C)
H3C
Br
CH3
Presence of C-carbon make it optically active
(D)
H3C
H
H
H
C
C
C
H
Br H
CH3
No chiral C-atom optically inactive
32. The major product of the following reaction is
O
i. KOH , H 2O


ii. H  , heat
CH3
O
CH3
CH3
O
O
(A)
(B)
O
O
CH3
CH3
(C)
(D)
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Sol. [A, D]
O O
CH3 will first go under aldol condensation then elimination of water will take
place.
So possible answer (A) and (D)
33. In the following reaction, the major product is
CH3
CH2
H2C
1 equivalent HBr


CH 3
CH3
CH 3
(A) H 2 C
(B) H3C
Br
Br
CH3
CH3
(C)
(D)
H2C
Br
Br
H3C
Sol. [D]
34. The structure of D=(+)-glucose
CHO
H
OH
HO
H
H
OH
H
OH
CH 2 OH
The structure of L(–)-glucose is
CHO
HO
H
(A)
CHO
H
H
HO
OH
HO
H
HO
H
CH2 OH
(B)
H
HO
CHO
CHO
OH
HO
H
HO
H
H
HO
H
HO
H
HO
H
OH
(C)
H
HO
H
CH2 OH
OH
H
CH2 OH
(D)
H
OH
CH2 OH
Sol. [A]
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35. The major product of the reaction is
H3C
NaNO 2 , aqueous HCl
CO 2 H 
0 C
CH3
H3C
NH2
NH2
H3C
(A)
CH2 OH
(B)
CH3
OH
H3C
CH3
CH2 OH
OH
H3C
(C)
NH2
(D)
CH3
OH
CH3
OH
O
H3C
OH
Sol. [B]
CH3
OH
36. 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
Sol. [A, B, C]
37. 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
Sol. [B, C, D] Factual
38. Fe3+ is reduced to Fe2+ by using
(A) H2O2 in presence of NaOH
(B) Na2O2 in water
(C) H2O2 in presence of H2SO4
(D) Na2O2 in presence of H2SO4
Sol. [C, D]
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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 whose layout will be similar to the one shown below.

(A)
P
Q
R
S
T
(B)
P
Q
R
S
T
(C)
P
Q
R
S
T
(D)
P
Q
R
S
T
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 none of the bubbles is darkened. – 1 In all other cases.
39. Match the anionic species given in Column I that are present in the ore(s) given in Column II.
Column I
Column II
(A) Carbonate
(P) Siderite
(B) Sulphide
(Q) Malachite
(C) Hydroxide
(R) Bauxite
(D) Oxide
(S) Calamine
(T) Argentite
Sol. A–(P), (Q), (S); B–(T); C–(Q), (R); D–(R)
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40. Match the thermodynamic process given under Column I with the expressions given under Column II.
Column I
Column II
(A) Freezing of water at 273 K and 1 atm
(P) q = 0
(B) Expansion of 1 mol of an ideal gas into a
(Q)  = 0
vacuum under isolated conditions
(R) Ssys < 0
(C) Mixing of equal volumes of two ideal gases
at constant temperature and pressure in an
isolated container
(S) 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
(T) G = 0
Sol. A–(R), (T); B–(P), (Q), (S); C–(P), (Q), (S); D–(P), (Q), (S), (T)
PART III : MATHEMATICS
SECTION 1 (Maximum Marks: 32)

This section contains EIGHT questions

The answer to each question is a SINGLE DIGIT INTEGER 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
 x  , x  2
41. Let f :    be a function defined by f ( x )  
,
 0, x  2
2
where [x] is the greatest integer less than or equal to x. If I 
xf ( x 2 )
dx , then the value of

1 2  f ( x  1)
(4I – 1) is
[ x], x  2
Sol: [0] f (x) = 
 0, x  2
2
I
x f ( x2 )
dx
=
2

f
(
x

1)
1

2
x f ( x2 )
=
dx 
2  f ( x  1)
1

2
x f (x2 )
dx
2  f ( x  1)
2


0
0
2


x.0

=
20
1
0
1
2
x f (x )
x f ( x2 )
dx 
dx 
2  f ( x  1)
2  f ( x  1)
0


2

1
x f ( x2 )
dy
2  f ( x  1)
0
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
2

=
1
I
x .1
dx 
2  f ( x  1)
2

1
 x2 
x
dx   
2
 4 1
JEE-Advanced-2015
2
1
1
[2  1]   4I –1 = 0
4
4
=
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
of the container is 10 mm, then the value of
V
is
250
Sol: [4] V = r2h
M
= h[(r + 2)2 – r2] + (l + 2)22
= h(2r + 2) . 2 + 2(r + 2)2
=
V
(2r + 2) . 2 + 2(r + 2)2
r 2
4 4 
= V   2  + 2(r + 2)
r r 
dM
 4 8
 V   2  3   4( r  2)
dr
r 
 r
when r = 10, we get
8 
 4
V 

 + 48 = 0
 100 1000 
 V = 1000 H
V

=4
250
x2 
43. Let F ( x) 

x

6
 1
 1
2cos 2 t dt for all x   and f :  0,   0,   be a continuous function. For a   0,  ,
 2
 2
if F´(a) + 2 is the area of the region bounded by x = 0, y = 0, y = f(x) and x = a, then f(0) is
x2 
Sol: [3] Let F(x) =


6
2cos 2  dt  x  R
x
 1
and f : 0,   [0, ] be a continuous in for a F
 2
 1
0, 2  , if F´(a) +2 in the area
a
F´(a) + 2 =
 f ( x)dx
0
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 2 
Now F´(x) = 2cos2  x   2x – 2cos2x
6

a

 f ( x)dx  4cos
0
x=a

2
2
 a   a  2cos a  2
6

2
diff. w.r.t. a
 2 2 
 2   2  
f (a) = 4  cos  a    a.2cos  a   sin  a   2a  + 2. 2 cos a sin a
6
6 
6 




3
f (0) = 4   = 3
4
44. The number of distinct solutions of the equation:
5
cos 2 2 x  cos 4 x  sin 4 x  cos6 x  sin 6 x  2
4
in the interval [0, 2] is
Sol: [8]
5
cos22x + cos4x + sin4x + sin6x + cos6x = 2
4

5
cos22x + (1 –2 sin2x cos2x) + (1 – 3 sin2x cos2x) = 2
4

5
1
3
cos22x + 1 –
sin2x + 1 – sin22x = 2
4
2
4

5
5
cos22x –
sin22x =0
4
4

tan22x = 1  2x =
 3 5 7  9 11 13 15
, , , , ,
,
,
=8
4 4 4 4 4 4 4
4
45. Let the curve C be the mirror image of the parabola y2 = 4x 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
Sol: [4] x = 1 intersits y2 = 4x at y = ± 2 so A (1, –2) and B (1, 2)
Hence AB = 4
y2 = 4 x
(1, –4)
(0, –5)
y = –5
(1, –5)
46. 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
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Sol: [8] Let the coin be tossed n times.
Now p =
1
,q=
2
n
n
1 n 1
1 – C0    C1    0.96
2
2
n

1–
(n  1)
 0.96
2n
n 1  1

   0.04 
n
2
 25

By taking different values of n

8
1

, not true
128 25
9
1

n = 8,
, True
256 25
so n = 8
take n = 7,
47. 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 consecutively in the queue. Then the value
of
m
is
n
Sol: [5] B1, B2, B3, B4, B5, G1, G2, G3, G4, G5
we take five girls as one unit
so n = 6! × 5!
we divide 5 girls into two groups of four girls in one unit and one girl in second unit
× B1 × B2 × B3 × B4 × B5 ×
5!
m = 5! × 6C2 × 4! × 2! × 4!.1!
m 5! 15  4!  2  5
so n  6!  5!  6  5 = 5
48. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are tangents to the
circle (x – 3)2 + (y + 2)2 = r2, then the value of r2 is
A = (1, 2)
dy 2
Sol: [2] y2 = 4x, dx  y on A, slope of tan = 1
so equation of normal y –2 = –1 (x –1)
 x+y–3=0
As it is tangent to (x –3)2 + (y + 2)2 = r2
3 2 3
so r =
2
2
B
2
=2
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
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SECTION 2 (Maximum Marks: 40)

This section contains TEN questions

Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE 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 the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 if
none of the bubbles in darkened –2 in all other cases.

 

49. Let f ( x )  sin  sin  sin x   for all x   and g ( x)  sin x for all x   . Let ( f  g )( x ) denote
2

6 2
f ( g ( x )) and ( g  f )( x ) denote g(f(x)). Then which of the following is (are) true ?
 1 1
(A) Range of f is   , 
 2 2
(B) Range of f  g is   1 , 1 
 2 2 
f ( x) 

(C) lim
x 0 g ( x)
6
(D) There is an x   such that ( g  f )( x)  1
Sol: [A,B,C]
 

Let f (x) = sin  sin  sin x   for all x R

6 2
g (x) =

sin x for all x R
2
 

(A) Hence   sin x 
2 2
2

 

–1  sin  sin  sin x    1

 2

1
1
  f ( x) 
2
2


(B) (fog) (x) = f (g (x)) = f  sin x 
2

  

= sin  6 sin  2 sin  2 sin x   




  

–  1   sin 2  sin  2 sin x     1

 

1
1
Thus   fog ( x ) 
2
2
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
 

sin  sin  sin x  

6 2
lim
(C) Also x 0 
 

sin x sin  sin x 
2
6 2

(D)
JEE-Advanced-2015
 


sin  sin x  =
6 2
6

 


sin  sin  sin x   = 1

2
6 2

  

–sin 30°  sin sin  sin  6 sin  2 sin x     sin 30°


 
so L.H.S. can notbe equal to 1


   
 

50. 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 |2 
 | a | 12
(A)
2
   
(C) | a  b  c  a | 48 3

| c |2

 | a | 30
(B)
2

(D) a.b  72
Sol: [A,C,D]



| a | 12,| b |  4 3 and b .c  24


Clearly a  b  c  0 ; b  c   a

 

 | b |2  | c |2 2b .c | a |2
  

 48 + | c |2 2b .c  | a |2 | c |2  48  144

 | c |2  48
R
a
b
P
c
Q

| c |2

 | a | 24  12  36
2

Also a  b  c


 

 | a |2  | b |2 2(a. b ) | c |2

 144  48  2( a.b )  48


 2(a.b )  –144  a.b  –72
  
 
Also a  b  a | – a – c | = a  c  c  a
   
 
so | a  b  c  a | 2(c  a )
 
  
Also | c  a | 2 | c |2 | a |2 | c .a |2
Now
= 144 × 48 – 72 × 72
= 72 × 24 = (24)2. 3
   
Thus | a  b  c  a | 48 3
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51. 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 ?
(A) Y3Z4 – Z4Y3
(B) X44 + Y44
(C) X4Z3 – Z3X4
(D) X23 + Y23
Sol. [C,D]


X, Y are non-zero skew symmetry matrices and Z is a symmetric matrix
So, X´ = –X,
Y´ = – Y,
Z´ = Z
23
23
23
23
23
23
(X + Y )´ = (X )´ + (Y )´ = (X´) + (Y´) = (–X)23 + (–Y)23 = –(X23 + Y23)  (D)
(X44 + Y44)´ = (X´)44 + (Y´)44 = X44 + Y44  (B) is wrong
Again (Y3Z4 – Z4Y3)´ = (Y3Z4)´ –(Z4Y3)´ = (Z4)´ (Y3)´ – (Y3)´ (Z4)´
= –Z4Y3 – (–Y3) (Z4)
= – Z4Y3 + Y3Z4
It is symmetric (A)
Now (X4Z3 – Z3X4)´ = (X4Z3)´ –(Z3 X4)´
= (Z3)´ (X4)´ – (X4)´ (Z3)´
= Z3X4 – X4Z3 = – (X4Z3 – Z3X4)
Skew symmetric (C)
(1  ) 2
52. Which of the following values of a satisfy the equation (2  ) 2
(1  2) 2
(2  2 )2
(1  3) 2
(2  3) 2  648 ?
(3  ) 2
(3  2) 2
(3  3)2
(A) –4
(B) 9
(C) –9
(D) 4
Sol. [B,C]
(1  )2
(1  2) 2
(1  3) 2
(2  ) 2
(2  2)2
(2  3)2
(3  ) 2
(3  2) 2
(3  3) 2
= – 648
Applying R3  R3 – R2, R2  R2 – R1
L. H. S. =
(1  ) 2 (1  2)2
(3  2) (3  4)
5  2
5  4
(1  3)2
(3  6)
5  6
Applying C3  C3 – C2, C2  C2 – C1
L.H.S.
(1  ) 2  (2  3) 2(2  5)
(3  2)
2
2
(5  2)
2
2
Applying C3  C3 – C2
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
L.H.S.
(1  ) 2
3  2
JEE-Advanced-2015
(2  3) (2  5)
2
2
2
0
0
= 2[42 + 63 – 42 – 103]
= – 83 = – 648 (given)
3 = 81 = 9 or – 9
53. In  3 , consider the planes P1 : y = 0 and P2 : x + z = 1. Let P3 be a plane, different from P1 and P2,
which passes through the intersection of P1 and P2. If the distance of the point (0, 1, 0) from P3 is 1 and
the distance of a point (, , ) from P3 is 2, then which of the following relations is (are) true ?
(A) 2    2  2  0
(B) 2    2  4  0
(C) 2    2  10  0
(D) 2    2   8  0
Sol: [B,D]
Let plane P3 be P2 + P1 = 0
 (x + z –1) +  y = 0
 x – y + z –1 = 0
  1
2  2

1
2 + 2 + 1 = 2 + 2  =
1
2
Thus P3: 2x – y + 2z – 2 = 0
Now
2    2  2
=2
3

2 –  + 2  –2 = 6 or –6

2 –  + 2  – 8 = 0 (D)
or
2 –  + 2  + 4 = 0 (B)
54. In 3 , let L be a straight line passing through the origin. the origin. Suppose that all the points on L are
at a constant distance from the two planes P1 : x + 2y – z + 1 = 0 and P2 : 2x – y + z – 1 = 0. Let M be
the locus of the feet of the perpendiculars drawn from the points on L to the plane P1. Which of the
following points lie(s) on M ?
5 2

(A)  0,  ,  
6 3

Sol: [B]
 1 1 1
(B)   ,  , 
 6 3 6
 5 1
(C)   , 0, 
 6 6
 1 2
(D)   , 0, 
 3 3
x y z
 
a b c
| x  2 y  z  1| | 2 x  y  z  1|

=
6
6
x + 2y – z + 1 = 2x – y + z –1
x –3y + 2z – z = 0
.....(i)
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
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x y z
  
3 1 0
x + 2y – z + 1 = –2x + y – z + 1
3x + y = 0
.......(ii)
x y z

 
1 3 2
x  3 y   z  0


 t1
1
2
1
x = 3 + t1, y =  + 2t1, z = –t1
3 + t1 + 2 + 4t1 + t1 + 1 = 0
5 + 6t1 + 1 = 0
1  5
6
1  5
1  5
1  5
x = 3 –
y=–
,z=
6
3
6
13  1
1  2
1  5
x=
,y=
,z=
6
3
6
For  = 0, we get (B).
t1 =
 1 1 1
  , , 
 6 3 6
55. Let P and Q be distinct points on the parabola y2 = 2x 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 
(B)
1 1 
(C)  ,

4 2
 9,3 2 
(D)
1, 2 
Sol. [A,D]
1 2 
1 2 
P  t1 , t1  and Q  t2 , t2 
2

2

P
1 2 
1 2

Circle  x  t1  x  t2  + (y – t1) (y – t2) = 0
2 
2 

1 22
t1 t2  t1t2  0
4
t1t2 + 4 = 0; t1t2 = –4
O
4
1
 8
P  t12 , t1  , Q  2 ,   , O (0, 0)
t1 
2
  t1
1 2
t1
2
8
2
A = t1
0
t1

y2 = 2x
Q
1
4
1
t1
0
1
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
3 2
JEE-Advanced-2015
1 
8
 2t1  
2 
t1 
8
2t1 + t = ± 6 2
1
t12 + 4 = ± 3 2t1
t12  3 2t1  4  0
t1 = 2 2, 2

A (4, 2 2 ) and D(1,
2)
56. Let y(x) be a solution of the differential equation (1 + ex)y´ + yex = 1. If y(0) = 2, then which of 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)
Sol. [A,D]
(1 + en)
dy
+ yen = 1
dx
dy
ex
1


x
dx 1  e 1  e x
ex
I.F. =  1 e x dx
e
1
= e
 t dt  e log t  t
= ex +1
y . (ex + 1) =
1
 1 e
x
 (e x  1) dx  c
y(ex + 1) = x + c
x = 0, y = 2
2×2=0+cc=4
y(ex + 1) = x + 4
y (x) =
x4
ex  1
y (–4) = 0, y (–2)  0
y '( x ) 
(e x  1).1  ( x  4)e x
(e x  1)2
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)

JEE-Advanced-2015
Let y(x) = 1 – ex (3 + x)
y(0) h (–1) < 0
critical point in (–1, 0)
57. 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´ =
dy
d2y
, y´´ 2 ), then which of the following statemtns is (are) ture ?
dx
dx
(A) P = y + x
(B) P = y – x
(C) P + Q = 1 – x + y + y´ + (y´)2
(D) P– Q = x + y – y´ – (y´)2
Sol. [B,C]
(x –)2 + (y –)2 = r2
2(x –) + 2 (y –) y´= 0
x –  + yy´ – y´ = 0
–(y´ + 1) = –x – yy´
=
x  yy '
y ' 1
( y ' 1)[1  yy '' ( y ') 2 ]  ( x  yy ')( y '')
=
( y ' 1) 2
0 = y´ + yy´yy´´ + (y´)3 + 1 + yy´´ + (y´)2 – xy´´ – yy´y´´
y´´(y –x) + (y´)3 + (y´)2 + y´ + 1 = 0
y´´(y –x) + y´[(y´)2 + (y´) + 1] + 1 = 0
P = y – x; Q = (y´)2 + y´ + 1
Option (B) and (C) is correct
58. If g :    be a differentiable function with g(0) = 0, g´(0) = 0 and g´(1)  0. Let
 x
g ( x),

f ( x)  | x |
 0,

x0
x0
and h(x) = e|x| for all x   . Let ( f  h)( x ) denote f(h(x)) and (h  f )( x ) denote h(f(x)). Then which
of the following is (are) ture ?
(A) f is differentiable at x = 0
(C)
(B) h is differentiable at x = 0
f  h is differentiable at x = 0
(D) h  f is differentiable at x = 0
Sol. [A,D]
Lf ´(0) = lim
h 0
f ( h)  f (0)
h
(1) g ( h)
h 0
h
= lim
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g ( h)
h 0 h
So f is diff. at x = 0 (A)
Rf ´(0) = lim
eh  1
 1
h 0  h
Again Lf ´(0) lim
eh  1
1
h0
h
 h is non differentiable at x = 0 (B)
Rh´(0) = lim

 g '(1) for x  0
Similarly f ´(h(x)) = 
  g '(1) for x  0
foh is non-differentiable at x = 0
| g ( h) |
and Left
h 0
h
Now for differentiability of (hof) at x = 0 me have right hand derivative = lim
hand derivative = lim
h 0
| g ( h) |
and they are equal as g´(0) = 0 (D)
h
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 whose layout will be similar to the one shown below.


(A)
P
Q
R
S
T
(B)
P
Q
R
S
T
(C)
P
Q
R
S
T
(D)
P
Q
R
S
T
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 none of the bubbles is darkened. – 1 In all other cases.
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
59.
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Column I
Column II
(A) In  2 , if the magnitude of the projection vector of the vector
iˆ  ˆj on
3iˆ  ˆj is
(P)
1
(Q)
2
(R)
3
(S)
4
(T)
5
3 and if   2  3 , then possible
values(s) of || is (are)
(B) Let a and b be real numbers such that the function
3ax 2  2, x  1
f ( x)  
2
 bx  a , x  1
is differentiable for all x   . Then possible value(s) of a is (are)
(C) Let   1 be a complex cube root of unity. If
(3  3  22 )4 n 3  (2  3  32 )4 n 3  (3  2  32 )4 n 3  0
then possible values(s) of n is (are)
(D) 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)
Sol: [A-(P),(Q); B-(P),(Q); C-(P),(Q,(S),(T); D-(Q),(T)]
A.
Projection vector of iˆ  ˆj on
given
3iˆ  ˆj
(iˆ  ˆj ).( 3iˆ  ˆj )
 3
3 1
 | 3   | 2 3
Also  = 2 +
So
3 
3
2
3
2 3
 |4 – 2| = 6
 = 2 or –1  || = 2 or 1
Thus A- (P), (Q)
B.
3ax 2  2 , x  1
f (x) = 
2
, x 1
 bx  a
f(x) is continuous at x = 1
so –3a –2 = b + a2
Also f (x) is diff. at x = 1, so –6a = b
So a2 – 6a = –3a –2
 a2– 3a + 2 = 0  (a –1) (a –2) = 0  a = 1, 2
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C.
D.
JEE-Advanced-2015
Thus B-(P), (Q)
(3 –3 + 22)4n + 3 + (2 + 3 – 32)4n + 3 + (–3 +2 + 32)4n + 3 = 0
8n + 6 (2 + 3 – 32)+ (2 +3 – 32)4n + 3 + 4n+ 3 (2 + 3 – 32) = 0
(8n + 6 + 4n+ 3 + 1) (2 + 3 – 32) = 0
 n = 1, 2, 4, 5 are solutions
Thus C-(P),(Q),(S),(T)
2ab
 4 , a, s, q, b  AP
ab
 2a(15 –2a) = 4a + 4(15 – 2a) b = 15 – 2a
 30 a – 4a2 = 4a + 60 – 8a
 4a2 – 34a + 60 = 0
 2a2 – 17a + 30 = 0
 2a2 – 12a – 5a + 30 = 0
 (2a – 5) (a – 6) = 0 a = 5/2, a = 6
5 15
 , |q – a| = 5
2 2
q = 4, |q –a| = 2
Clearly value of |q –a| are 2
Thus D-(Q),(T)
q=5+
60.
Column I
Column II
(A) In a triangle XYZ, let a, b and c be the lengths of the sides
(P)
1
(Q)
2
(R)
3
(S)
5
(T)
6
opposite to the angles X, Y and Z, respectively. If 2(a2 – b2) = c2
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
and  
opposite to the angles X, Y and Z, respectively.
If 1 +cos2X – 2cos2Y = 2sinX sinY, then possible value(s) of
a
b
is (are)
(C) In  2 , let 3iˆ  ˆj , iˆ  3 ˆj and iˆ  (1  ) ˆj be the position
vectors of X, Y and Z with respect to the origin O, respectively.

If the distance of Z from the bisector of the acute angle of OX
3

with OY is , then possible value(s) of || is are
2
(D) Suppose that F() denotes the area of the region bounded
by x = 0, x = 2, y2 = 4x and y = |x – 1| + |x – 2| + ax,
where  {0,1} . Then the value(s) of F () 
8
2 , when
3
 = 0 and  = 1, is (are)
Sol: [A-(P),(R),(S); B-(P); C-(P),(Q); D-(S),(T)]
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
A.
JEE-Advanced-2015
2( a 2  b 2 )  c2

x
sin( x  y )
sin z
b
c
cos(n)  0

sin x cos y  cos x sin y
sin z
y
=
a
b
cos y  cos x
c
c
=
a cos y  b cos x a  a 2  c 2  b 2  b  c 2  b 2  a 2 
 .
 c

c
c 
2ac
2cb



=
2a 2  2b 2 1

2c 2
2
z
a
 n 
 cos    0  ( P), ( R) ( S )
  
B.
2cos2 x  2(2cos 2 y  1)  2sin x sin y
2cos 2 x  4cos 2 y  2  2sin x sin y
4  2sin 2 x  4  4sin 2 y  2sin x sin y
42
sin 2 x
sin x
2
2
sin y
sin y
 2

a2 a

b2 b
a2 a
a 1  9
 20 
 2 &1
2
b
b
b
2
B-(P)
C.
The vector along angle bisector is
 3iˆ  ˆj iˆ  3 ˆj 

= 

 2
2 
Cleanly the equation of < bisector is y = x
1  
Distance =


3
y=3
x=0
2
3–x
 |  | 2 &1
C-(P) and (Q)
D.
x=2
When  = 0
F() = Area between the line y = 3, x = 0, x = 2
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Physics, Chemistry & Mathematics (Paper-1) Code-9 (24/05/2015)
and y2 = 4x is = 6 
F ( ) 
JEE-Advanced-2015
8 2
3
8 2
6
3
and when  = 1
F (1) 
8 2
= Area between the lines y = 3 – x, y = x + 1 adn parabola and x = 0 and x = 0 which
3
is = 5.
D-(S) and (T)

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