1. society
2. crime
3. property crime
4. usury

CODE-9
PAPER-2
Time: 3 Hours
P2-15-9
Maximum Marks: 240
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 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. 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 and 0 in all other cases.
11. In Section 2 contains 8 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 “paragraph” type questions. Each paragraph describes an experiment, a situation or
a problem. Two multiple choice questions will be asked based on this paragraph. One or more than one
option can be correct.
Marking scheme: +4 for correct answer, 0 if not attempted and –2 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.
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-1 -
Physics, Chemistry & Mathematics (Paper-2) 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 monochromatic beam of light is incident at 60º on one face of an equilateral prism of refractive index
n and emerges from the opposite face making an angle (n) with the normal (see figure). For n  3
the value of  is 60º and
d
 m. The value of m is
dn
Sol: [2] r1 + r2 = A
r2 = (A – r1)
A = 60º
sin 60º
x
sin r1
3
x
2sin r1
sin r1 
sin r
3
2x
sin r2 1

sin  x
cos r1  1 
3
4 x2
 sin  

  sin r2
 n 
 sin  

  sin( A  r1 )
 n 
sin  = n(sin A cos r1 – cos A sin r1)

3
3
sin   n  sin A 1  2  cos A 

4n
2n 



4n 2  3
3
sin    (sin A)
 cos A 
cos A
2
2


C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-2 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
sin A  8n
 d 
cos    
 dn  2  2 4n 2  3
 d   sin 60º 2  3
cos 60º   
9
 dn 
1  d 
3
3
 2
 
2  dn  2
3
 d 
 2
 dn 
2.
In the following circuit, the current through the resistor R (= 2) is I Amperes. The value of I is
6.5V
Sol: [1]
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-3 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
i = 1 Amperes
3.
An electron in an excited state of Li2+ ion has angular momentum 3h/2. The de Broglie wavelength of
the electron of the electron in this state is p0 (where 0 is the Bohr radius). The value of p is
Sol: [2]

4.
h
h
3 ; n3
2
2
3h
mv 
(2) r
p  mvr  n

h
h k (2r )


p mv
3h

2r
3
r
a0 n 2
a (3) 2
r  0
x
(3)

2
 3a0  p  2
3
A large spherical mass M is fixed at one position and two identical point masses m are kept on a line
passing through the centre of M (see figure). The point masses are connected by a rigid massless rod
of length l and this assembly is free to move along the line connecting them. All three masses interact
only through their mutual gravitational interaction. When the point mass nearer to M is at a distance
 M 
r = 3l from M, the tension in the rod is zero for m  k 
 . The value of k is
 288 
m
M
r
Sol: [7]
m
l
F1  F '  ma
F2  F '  ma
F1  F '  F2  F '
F1  F2  2 F '
GMm GMm 2Gm  m


(3 x) 2 (4l ) 2
l2
M × 7 = 2; M = 7
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-4 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
5.
JEE-Advanced-2015
The energy of a system as a function of time t is given as E(t) = A2 exp (–t), where a = 0.2 s–1. The
measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the
percentage error in the value of E(t) at t = 5 s is
Sol: [4] E(t) = A2e–t
 = 0.2 s–1
dE A2 de t  e t dA2

E
E
dE A2 de t e t dA2
 2 t  2 t
E
Ae
Ae
dE de t
dA2
e t dt 2 A dA
dt
dA



  .t  2
 0.2  1.5  2  1.25  1.5  2.5  4%
t
2 =
t
2
dt dt e
A
e
A
t
A
6.
The densities of two solid spheres A and B of the same radii R vary with radial distance r as
5
r
r
 A (r )  k   and  B (r )  k   , respectively, where k is a constant. The moments of inertia of the
R
R
IB n
individual spheres about axes passing through their centres are IA and IB, respectively. If I  10 , the
A
value of n is
r
Sol: [6]  A (r )  k  
R
r
 B (r )  k  
R
5
r

2
2
2
2
k  .r 4  dr
dm1r 2

e
dV
r
e
4

r
dr
r

IA 5
 1 1  1
R 



5
2
2
2
2
IB
 e2 4r dr r   k r 5 .r 4  dr
dm1r 2  e2 dV2 r

5
 R

R
 r6 
r5
R 6 R5


dr

10
 6R 0
 R9

 6R

R
10
r
R
6
 10 
 R5 dr 10r R5  10 R

0
IB 6

I A 10
n=6
7.
Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, /3, 2/3 and
. When they are superposed, the intensity of the resulting wave is nI0. The value of n is
2
Sol: [3] Irms = I0 + I0 + 2 I 0 cos  / 3  2 I 0  2 I 0 
1
 3I 0
2
n=3
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-5 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
8.
JEE-Advanced-2015
For a radioactive material, its active A and rate of change of its activity R are defined as A  
dN
and
dt
dA
, where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life ) and Q
dt
(mean life 2) have the same activity at t = 0. Their, rates of change of activities at t = 2 and RP and
RP n
RQ, respectively. If R  e , then the value of n is
Q
R
Sol: [2]
dN
dt
P
dA
dt
Q
A
t=0
t = 2
R

A
RP
2
A
RQ
dN
 N 0 e t  N
dt
dA
dN
R
 
  2 N
dt
dt
R =  2N
N = N0e–t = 
2
Q2  P 1 2 2
RP  2P N P Q N 01  e 2 1

 
  

RQ  2Q N Q 2P
N 02 e 1 e = 2P Q e e e




9.
SECTION 2 (Maximum Marks: 32)
This section contains EIGHT 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.
An ideal monatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the
figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed
state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this
process the piston moves out by a distances x. Ignoring the friction between the piston and the cylinder,
the correct statement (s) is (are)
(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is
1
PV
1 1
4
(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-6 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is
7
PV
1 1
3
(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is
17
PV
1 1
6
Sol: [A,B,C] Let area of cross section of piston = A
For A and B case
P2 
3P1
2
(P2 – P1)A = kx
...(i)
V2 – V1) = Ax
...(ii)
From (i) and (ii), stored energy in spring =
3
( PV
2 2  PV
1 1 ) = 3P1V1
2
Internal energy, U = nCv T =
For C and D case
P2 
1 2 PV
kx  1 1
2
4
4
P1
3
(P2 – P1)A = kx
...(iii)
V2 – V1) = Ax
...(iv)
From (iii) and (iv), stored energy in spring U sp 
Internal energy U = nCv T =
PV
1 1
3
3
9
( PV
PV
2 2  PV
1 1) =
1 1
2
2
Applying work-energy-theorem, Wgas  Wsp  Wother side gas  0
Wgas 
PV
1 1
 2 PV
1 1 0
3
Wgas 
7 PV
1 1
3
Q = U + Wgas + Usp =
10. A fission reaction is given by
Considering
236
92
236
92
43
PV
1 1
6
94
, where x and y are two particles.
U 
 140
54 Xe + 38Sr + x + y
U to be at rest, the kinetic energies of the products are denoted by K Xe , K Sr K x (2 MeV )
236
and K y (2 MeV ) , respectively. Let the binding energies per nucleon of 92
and 94
be 7.5
U , 140
54 Xe
38 Sr
MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s)
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-7 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
is (are)
(A) n  n y  n, K sr  129 MeV, K Xe  86 MeV
(B) x  p, y  e, K Sr  129 MeV , K Xe  86 MeV
(C) x  p, y  n, K Sr  129 MeV , K Xe  86 MeV
(D) x  n, y  n, K Sr  86 MeV , K Xe  129 MeV
Sol: [A]
Energy released = 8.5× (40 + 94) – 7.5 × 236
= 234 – 15 = 219 MeV
219 = 1219 + 86 + 2 + 2
The heavier particle Xe will have lower kientic energy threfore option A is correct and option D is
incorrect.
11.
Two spheres P and Q of equal radii have densities 1 and 2 , respectively. The spheres are connected
by a massless string and placed in liquids L1 and L2 of densities 1 and 2 and viscosities 1 and 2 ,
respectively. They float in equilibrium with the sphere P and L1 and sphere Q in L2 and the string being

taut (see figure). If sphere P alone in L2 has terminal velocity VP and Q alone in L1 has terminal

velocity VQ , then

| VP | 1
(A) | V |  
2
Q
Sol: [A,D]
L1
P
L2
Q

| VP | 2
(B) | V |  
1
Q
 
(C) V p .VQ  0
 
(D) V p .VQ  0
P will rise while Q will go down
Due to equilibrium
     
for P;
4 3
r (1 – 2) = 62rvP
3
for Q;
4 3
r (1 – 2) = 61rvQ
3
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-8 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
2 vP
dividing; 1 =  v
1 Q
12. In terms of potential difference V, electric current I, permittivity o, permeability 0 and speed of light
c, the dimensionally correct equation(s) is (are)
(A) 0 I 2  0V 2
Sol: [A,C]
(B)  0 I   0V
(C) I   0 cV
(D)  0 cI   0V
Informative
13. Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution,
a spherical cavity of radius R2, centred at P with distance OP = a = R1 – R2 (see figure) is made. If the
 
electric field inside the cavity at position r is E ( r ) , then the correct statement (s) is (are)
R2
P
a
R1
O
Strain

(A) E is uniform, its magnitude is independent of R2 but its direction depends on r

(B) E is inform, its magnitude depends on R2 and its direction depends on r

(C) E is uniform, its magnitude is independent of a but its direction depends on a

(D) E is uniform and both its magnitude and direction depend on a
Sol: [D] The Electric field is uniform and equal to Electric field at P when the cavity was absent, in the
direction of a .
14. In plotting stress versus string curves for two materials P and Q, a student by mistake puts strain on the
y-axis and stress on the x-axis as shown in the figure. Then the correct statement (s) is (are)
P
Q
Stress
(A) P has more tensile strength then Q
(B) P is more ductile than Q
(C) P is more brittle then Q
(D) The young’s modulus of P is more than that of Q
Sol: [A, B]
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
-9 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
P can with stand more maximum stress
P can undergo more strain
15. A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own
gravity. If P(r) is the pressure at r(r < R), then the correct option(s) is (are)
(A) P (r  0)  0
P(r  3R / 4) 63
(B) P(r  2 R / 3)  80
P( r  3R / 5) 16
(C) P(r  2 R / 5)  21
P( r  R / 2) 20
(D) P (r  R / 3)  27
Sol: [B, C]
(A) =
dF
A
=
G  Mx 3
R3
4x 2 dx
x2
R
GMl
xdx
=
R3 x
(B)
(C)
GMl 2
 R  x2 
2 R3
P
=
P1
P2
7 R2
9R2
16  63
16
=
2 =
5R 2 80
4R
R2 
9
9
P1
P2
16 R 2
9R2
16
25
25
=
2 =
2 =
21R
4R
21
R2 
25
25
R2 
R2 
R2
27
4
(D)
=
2 =
R
32
R2 –
9
16. A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air.
When two dielectrics of deferent relative primitivities (1 = 2 and 2 = 4) are introduced between the
P1
P2
R2 –
C2
two plates as shown in the figure, the capacitance becomes C2. The ratio C is
1
d/2
+
–
S/2
d
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 10 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
(A) 6/5
(B) 5/3
JEE-Advanced-2015
(C) 7/5
(D) 7/3
Sol: [D]
2 4
7
C1 + C1 = C1
24
3
 C2 =
SECTION 3 (Maximum Marks : 16)

This section contains TWO paragraphs

Based on each paragraph, there will be TWO questions

Each question contains FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of
these four option(s) is(are) correct

For each quesion, 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 none of the bubbles is darkened
– 2 In all other cases
Pragraph-1
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in
the figure. The length, width and thickness of the strip are l, w and d, respectively.
l
y
K
w
I
S
R
d
I
M
x
z
P
Q

A uniform magnetic field B is applied on the stop along the positive y-direction. Due to this, the charge
carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers
on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A
potential difference along the z-direction is thus developed. Charge accumulation continues until the
magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on
the cross section of the strip and carried by electrons.
17. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths
are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically
located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences
between K and M in strips 1 and 2 respectively. Then, for a given current I flowing through them in a
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 11 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
given magnetic field strength B, the correct statement (s) is (are)
(A) If w1 = w2 and d1 = 2d2, then V2 = 2V1
(B) If w1 = w2 and d1 = 2d2, then V2 = V1
(C) If w1 = 2w2 and d1 = d2, then V2 = 2V1
(D) If w1 = 2w2 and d1 = 2d2, then V2 = V1
Sol: [A,D]
eE = evB
V
 vB  V  

18. Consider two different metallic strips (1 and 2) of the same dimensions (length l, width w and thickness
d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is
placed in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential deferences
developed between K and M is strips 1 and 2, respectively. Assuming that the current I is the same for
both the strips, the correct option (s) is (are)
(A) If B1 = B2 and n1 = 2n2, then V2 = 2V1
(B) If B1 = B2 and n1 = 2n2, then V2 = V1
(C) If B1 = 2B2 and n1 = n2, then V2 = 0.5V1
(D) If B1 = 2B2 and n1 = 2n2, then V2 = V1
Sol: [A,C] V  B and V  vd
I = nevdA
therefore vd 
1
n
V
1
n
Pragraph-2
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid
glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light
guidance in the structure takes place due to successive total internal reflections at the interface of the
media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value
im and confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is
define as sin im.
19. For two structures namely S1 with n1  45 / 4 and n2  3 / 2 , and S2 with n1 = 8/5 and n2 = 7/5 and
taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is (are)
(A) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index
16
3 15
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 12 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
(B) NA of S1 immersed in liquid of refractive index
JEE-Advanced-2015
6
is the same as that of S2 immersed in water
15
(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index
4
15
(D) NA of S1 placed in air is the same as that of S2 placed in water.
Sol: [A,C]
3
45
and n2 = , sin 1 =
2
4
for n1 =
for n1 =
[A]
and
6
 sin2 =
45
8
7
7
and n2 = , sin 1 =
 sin 2 =
5
5
8
4
sin im1 =
3
[C] For S1
15
=
8
1 sin im 
sin im 
For S2
15
8
3
3
9
=
 sin im1 =
45
4
16
45

4
16
8
sin im1 =
3 15
5
3
45
9
15
 sin im\2 =
16
5
45 1

4
5
3
4
4
 8  15
sin im    
5
5 8
sin im 
3
4
20. If two structures of same cross-sectional area, but different numerical apertures NA1 and NA2 (NA2
< NA1) are joined longitudinally, the numerical aperture of the combined structure is
NA1 NA2
(A) NA  NA
1
2
(B) NA1 + NA2
(C) NA 1
(D) NA 2
Sol: [D]
The lesser of the two.
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 13 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
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. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar
0
0
conductivity of a solution of a weak acid and HY (0.10 M). If X  X  , the difference in their pKa
values, pKa(HX) – pKa(HY), is (consider degree of ionisation of both acids to be << 1).
Sol: [3]
  c (HX) 
K a (HX)  0.01  mc

  m (HX) 
2
  c (HX) 
K a (HY)  0.1   m0

  m (HX) 
2
K a (HX) 1   cm (HX) 
 

K a (HY) 10   0m (HX) 
K a (HX) 1  1 
  
K a (HY) 10  10 
2
2
K a (HX)
1

K a (HY) 1000
log Ka(HX) – log Ka(HY) = log 10–3
–log Ka(HX) + logKa(HY) = – log 10–3
pKa(HX) – pKa(HY) = 3
22. A closed vessel with rigid walls contains 1 mol of
complete decay of
238
92
U to
206
82
238
92
U and 1 mol of air at 298 K. Considering
Pb, the ratio of the final pressure to the initial pressure of the system at
298 K is
Sol: [9]
238
92
286
U 
82 Pb  8 42He  6
0
1
e
No. of moles of gas before decay = 1
No. of moles of gas after decay = 1 + 8 = 9
23. In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by MnO4 . For this reaction,
the ratio of the rate of change of [H+] to the rate of change of [MnO4 ] is
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 14 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
Sol: [8]
MnO4  8H   5e 
 Mn 2  4H 2O

d[MnO 4 ]
1 d[H  ]

dt
8 dt
d[H  ]
8
dt
 8

d[MnO 4 ] 1
dt
24. The number of hydroxyl group(s) in Q is

H
aqueous dilute KMnO 4 (excess)


 P 
Q
heat
0 C
H
HO
H3C
CH3
Sol: [4]
OH
H
heat
aq.dil. KMnO 4 (excess)




0 C
H
HO
H3C
CH3
CH3
H3C
OH
HO
CH3
OH
CH3
25. Among the following, the number of reaction(s) that produce(s) benzaldehyde is
CO, HCl


Anhydrous AlCl3 /CuCl
I.
CHCl 2
H 2O


100 C
II.
COCl
H2


Pd  BaSO 4
III.
CO2 Me
IV.
DIBAL  H

Toluene78
H 2O
Sol: [4]
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 15 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
26. In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe–C bond(s) is
Sol: [2]
27. Among the complex ions, [Co(NH2 CH2 CH2 NH 2) 2Cl2 ]+, [CrCl2(C 2 O4 )2 ]3– , [Fe(H2O) 4 (OH)2] +,
[Fe(NH3)2(CN)4]–, [Co(NH2CH2CH2NH2)2(NH3)Cl]2+ and [Co(NH3)4(H2 O)Cl]2+, the number of
complex ion(s) that show(s) cis-trans isomerism is
Sol: [6] All compelx ion shows cis-trans isomerism.
28. Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing
product formed is
Sol: [6]
SECTION 2 (Maximum Marks: 32)

This section contains EIGHT 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. When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The TRUE
statement(s) regarding this adsorption is (are)
(A) O2 is physisorbed
(B) heat is released
*
(C) occupancy of  2 p of O2 is increased
(D) bond length of O2 is increased
Sol: [B, C, D]
30. One mole of monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The
relationship of interatomic potential V(r) and interatomic distance r for the gas is given by
(A)
(B)
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 16 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
(C)
JEE-Advanced-2015
(D)
Sol: [A]
31. In the following reactions, the product S is
H3C
i. O 3
NH 3

 R 
S
ii. Zn , H 2O
H3C
H3C
N
(A)
N
(B)
N
N
(C)
(D)
H3C
H3C
Sol: [A]
O
H3C
H3C
i. O 2


ii. Zn ,H 2O
O
H O
H3C
H NH
2
NH 3


H
(R)
OH
H
OH
H3C
NH
OH
2H 2O
H3C
N
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 17 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
32. The major product U in the following reactions is

CH 2 CH CH 3 , H
radical initiator,O 2

 T 
U
high pressure heat
H
O
H3C
O
CH3
(A)
CH3
O
O
(B)
H
H
CH2
O
O
O
CH2
(C)
(D)
CH3
H3C
Sol: [B]
CH 2 CHCH3 , H 
high pressure heat

CH3
H
O
CH3
O
O
radical initiator


H
33. In the following reactions, the major product W is
OH
NH2
, NaOH
NaNO2 , HCl

 V 
W
0 C
OH
(A)
N
N
OH
(B)
N
N
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 18 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
HO
N
N
(C)
CH3
Sol: [A]
NaNO 2 , HCl
0 C


N
N
(D)
OH
OH
N
, NaOH

OH
N
34. The correct statement(s) regarding (i) HClO, (ii) HClO2, (iii) HClO3 and (iv) HClO4, is(are)
(A) The number of Cl = O bonds on (ii) and (iii) together is two
(B) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
(C) The hybridisation of Cl in (iv) is sp3
(D) Amongst (i) to (iv), the strongest acid is (i)
Sol: [B, C]
O
(i)
O
H
Cl
(ii) HO Cl O
(iii) H
O
Cl
O
(iv) H
O
Cl
O
O
O
35. The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute
HCl, is (are)
(A) Ba2+, Zn2+
(B) Bi3+, Fe3+
(C) Cu2+, Pb2+
(D) Hg2+, Bi3+
Sol: [C, D]
36. Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain
termination, respectively, are
(A) CH3SiCl2 and Si(CH3)4
(B) (CH3)2SiCl2 and (CH3)3SiCl
(C) (CH3)2SiCl2 and CH3SiCl3
(D) SiCl4 and (CH3)3SiCl
Sol: [B]
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 19 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
SECTION 3 (Maximum Marks : 16)

This section contains TWO paragraphs

Based on each paragraph, there will be TWO questions

Each question contains FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of
these four option(s) is(are) correct

For each quesion, 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 none of the bubbles is darkened
– 2 In all other cases
PARAGRAPH 1
In the following reactions
Pd  BaSO 4
i. B2 H 6
C8 H6 
 C8 H8 
X
H2
ii. H 2 O 2 , NaOH, H 2 O
H 2O
HgSO 4 , H 2SO 4
i. EtMgBr, H 2O
C8 H8O 
Y
ii. H  , heat
37. Compound X is
O
OH
CH3
(A)
CH3
(B)
OH
CHO
(C)
(D)
CH
CH2
Pd/BaSO4


H2
Sol: [C]
OH
i. B2 H6


ii. H 2O2 , NaOH, H2O
(X)
C8 H 6
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 20 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
38. The major compound Y is
CH3
CH3
(A)
(B)
CH3
CH2
CH3
(C)
CH3
(D)
Sol: [D]
CH3
CH
O
CH3
H2 O


HgSO 4 , H 2SO 4
C8 H 6
HO
CH3
i. CH 3CH 2 MgBr, H 2O


C8 H 8 O

ii.H , 

CH3
H3C
(Y)
PARAGRAPH 2
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant
pressure, a temperature increase of 5.7°C was measured for the beaker and its contents (Expt. I). Because
the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1), this experiment
could be used to measure the calorimeter constant.
In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of
1.0 M NaOH (under identical conditions in Expt. 1) where a temperature rise of 5.6°C was measured.
(Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1).
39. Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the Expt. 2 is
(A) 1.0
(B) 10.0
(C) 24.5
(D) 51.4
Sol: [A] Q = mst
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 21 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
= 200 × 4.2 × 0.1 = 84 J
Heat evolved during neutralisation of 0.1 mole CH3COOH is 84 J
H neutralisation 
84
 840 mol1
0.1
= 0.84 kJ mol–1  1.0 kJ mol1
(Because of common ion effect the Hneutralisation would be greater than 0.84 kJ mol–1)
40. The pH of the solution after Expt. 2 is
(A) 2.8
(B) 4.7
(C) 5.0
(D) 7.0
Sol: [B] CH3COOH  NaOH 
 CH3COONa  H 2O
100 ml
100 ml
2.0 M
1.0
0.2 mol
0.1 mole
0
0
– 0.1 mol
– 0.1 mol
0.1 mol
0.1 mol
––––––––––––––––––––––––––––––––––––––––
0.1 mol
0.0 mol
pH  pK a  log
 407  log
0.1 mole 0.1 mol
[salt]
[acid]
(0.1)
(0.1)
pH = 4.7
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
1
41. If
 12  9 x 2 
    e9 x  3 tan 1  
 dx where tan–1 x takes only principal values, then the value of
 1  x2 
0
3 

 log e |1   |   is

4 
1
Sol: [9]  =

0
e(9 x 3tan
1
x)
 12  9 x 2 
dx

2 
 1 x 
Let tan–1x = 
dx
 d
1  x2
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 22 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
/ 4

(9tan  3 )
(12  9 tan 2 )
(9 tan  3 )
(3  9sec 2 )d 
e
=
0
/ 4
=
e
0
= [e9tan + 3]0/4

 9  3 
4 1

=  e


(1 + ) = e(  3 / 4)
Thus log (1 + ) 
3
=9
4
42. Let f :    be a continuous odd function, which vanishes exactly at one point and f (1)  1 .
2
x
x
Suppose that F ( x) 

f (t ) dt for all x  [ 1, 2] and G ( x )   t | f ( f (t )) | dt for all x [ 1, 2] . If
1
1
lim
x 1
F ( x) 1
 , then the value of f  1  is
G ( x ) 14
2
Sol: [7] f is continuous odd function on R and f (x) = 0 at only one point, Also f (1) =

Now lim
x 1

 lim
x 1
1
2
F ( x) 1

G ( x) 14
f ( x)
1

x | f ( f ( x )) | 14
 |f (f(1))| = 7
1
1
 |f   | = 7 so f   = 7
 2
 2
 


43. Suppose that p, q and r are three non-coplanar vectors in 3 . Let the components of a vector s
 


along p, q and r be 4, 3 and 5, respectively. If the components of this vector s along
  
  
  
  p  q  r  ,   p  q  r  and   p  q  r  are x, y and z, respectively, then the value of 2x + y + z
is


 
  
  
  
Sol: [9] Here s  4 p  3q  5r = x(  p  q  r )  y ( p  q  r )  z ( p  q  r )
  
As p, q , r are non-coplanar so
–x+y–z=4
x–y–z=3
x+y+z=5
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 23 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
9
7
On solving, we get x = 4, y  , z  ,
2
2
 2x + y + z = 8 +
9 7
 =9
2 2
 k 
 k 
44. For any integer k, let  k  cos   + i sin   , where i  1 . The value of the expression
 7 
 7 
12
| 
k 1
 k |
is
k 1
3
| 
4 k 1
  4k 2 |
k 1
Sol: [4] ak = cos
k
k
 i sin
7
7
12
| a
k 1
 ak |
k 1
3
| a
4 k 1
 a4 k  2 |
k 1
 



ak + 1 – ak =  2i sin  cos(2k  1)  i sin(2k  1) 
14  
14
14 

 |ak + 1 – ak| = 2 sin
so required =


and |a4k –1 – a4k –2| = 2 sin
14
14
12
=4
3
45. Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the
sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in
between 130 and 140, then the common difference of this A.P. is
Sol: [9]
S7 6

S11 11 , 130 < a7 < 140
7
[2a  6d ]
6
2
2a  6d 6



11
11
[2a  10d ]
2a  10d 7
2

14a + 42 d = 12a + 60 d

2a = 18d  a = 9 d
Again 130 < 15d < 140  8
2
1
d 9
3
3
d=9
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 24 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
46. The coefficient of x9 in the expansion of (1 + x)(1 + x2)(1 + x3)......(1 + x 100) is
Sol: [5] I can be made by (1, 8), (2, 7), (3, 6), (4, 5) and (0, 9)
Total coefficient = 5
x2 y 2

 1 are (f1, 0) and (f2, 0) where f1 > 0 and f2 < 0. Let P1 and
9
5
P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2, 0), respectively. Let
T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through
47. Suppose that the foci of the ellipse
 1
2
(f1, 0). if m1 is the slope of T1 and m2 is the slope of T2, then the value of  m2  m2  is
 1

Sol: [4] e for ellipse =
1
5 2

9 3
2
=2
3
(f1, 0 )  (2, 0) and (2f2, 0)  (–4, 0)
ae = 3 ×
2
1
T1 : y = m1x + m passes through (–4, 0)  m1 =
2
1
4
T2 : y = m2x + m passes through (2, 0)  m2 =  2
2
1
 m22 = 4
2
m1
n
48. Let m and n be two positive integers greater than 1. If
 cos( a )  e 
e
lim  e
    ,
m
x 0
2
a


then the value of
m
is
n
Sol: [2]
 e cos
lim
a0
a
an
e
m
  e
2
n
ecos a ( sin a n )na n1 e

 lim
a0
2
ma m1
n
 necos a a 2 n m e


m
2

m
2
n
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 25 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
SECTION 2 (Maximum Marks: 32)

This section contains EIGHT 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)  7 tan 8 x  7 tan 6 x  3 tan 4 x  3tan 2 x for all x    ,  . Then the correct expression(s)
 2 2
is(are)
/ 4

(A)
xf ( x) dx 
0
Sol: [A,B]
/ 4
1
12
/ 4
f ( x )dx  0

(B)
xf ( x ) dx 

(C)
0
0
1
6
/ 4
(D)

f ( x) dx  1
0
f (x) = 7 tan6x sec2x – 3 tan2x sec2x

 /4
0
f ( x) dx
=

 /4
0
7 tan 6 x sec2 xdx 

/4
0
3tan 2 x sec2 x dx
tan x = t  sec2x dx = dt
1
1
= 7 t 6 dt  3 t 2 dt


0
0
1
1
= 7.  3.  0
7
3

 /4
0
xf ( x )dx
= 7

 /4
0
(B)
x.(tan 6 x sec 2 x) dx  3

  tan 7 x  /4
=   x. 7  

0


 /4
0
1.
=

=

 /4
0
 /4
0

 /4
0
0
x.(tan 2 x sec 2 x )dx
tan 7 x 
dx 
7


  tan 3 x  /4
– 3   x.
 
3 0




=   0 
4


 /4

 /4
0
1.
tan 3 x 
dx 
3




tan 7 xdx   0  
4



 /4
0
tan 3 xdx
tan 3 x(1  tan 1 x) dx
tan 3 x(1  tan 2 x)sec2 xdx
= tan x = t  sec2xdx = dt
1
=
3
 t (1  t
0
2
) dt 
1
 (t
0
3
 t5 ) =
1 1 32 1
 

4 6
12 12
(A)
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 26 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
50. Let f ´( x) 
JEE-Advanced-2015
1
192 x 3
10
m

f
for
all
with
.
If
x




 f ( x)dx  M , then the possible values
2  sin 4 x
2
1/ 2
of m and M are
1
1
(B) m  , M 
4
2
(A) m = 13, M = 24
(C) m  11, M  0
(D) m  1, M  12
Sol: [D]
51. Let S be the set of all non-zero real numbers  such that the quadratic equation x2 – x +  = 0 has two
distinct real roots x1 and x2 satisfying the inequality |x1 – x2| < 1. Which of the following intervals is(are)
a subset(s) of S ?
 1 1 
(A)   2 ,

5

Sol: [A,D]
 1 
,0 
(B)  
5 

 1 
(C)  0,

5

 1 1
, 
(D) 
 5 2
1
 4 1
a2
|x1 – x2| =
1
1  5a 2

5

0,
0
 2
a
a2
 5a2 – 1 > 0
1 
 1
 
,     , 
a 

5
 5  
and also
1
40  a 
a2
 1   1
  2 ,0    0, 2 
1  4 
6
52. If   3sin 1   and   3cos   , where the inverse trigonometric functions take only the principal
9
 11 
values, then the correct options(s) is(are)
(A) cos   0
(B) sin   0
Sol: [B,C,D]  = 3 sin–1
3sin 1
(C) cos(  )  0
(D) cos   0
6
11
1
3
   3sin 1
2
2


2
 = 3cos–1
 3cos–1
4
9
1
4
 3cos 1  3cos 1 0
2
9
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 27 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
1
 < 3cos
JEE-Advanced-2015
4 3

9 2
 sin  < 0, cos  < 0, cos  < 0, sin  > 0  cos() > 0
53. Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie along the
x-axis and the y-axis, respectively. let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3
2 2
. If e1 and
3
e2 are the eccentricities of E1 and E2, respectively, then the correct expression(s) is(are)
touches the curves S, E1 and E2 at P, Q and R, respectively. Suppose that PQ  PR 
2
2
(A) e1  e2 
43
40
(B) e1e2 
7
5
8
2
2
(C) | e1  e2 |
2 10
(D) e1e2 
3
4
Sol: [A,B]
54. Consider the hyperboal H : x2 – y2 = 1 and a circle S with center N(x2, 0). Suppose that H and S touch
each other at a point P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects
the x-axis at point M. If (l, m) is the centroid of the triangle PMN, then the correct expression(s) is
(are)
(A)
dl
1
 1  2 for x1 > 1
dx1
3 x1
(B)
dm

dx1 3
(C)
dl
1
 1  2 for x1 > 1
dx1
3x1
(D)
dm 1
 for y1 > 0
dy1 3
x1

x12  1

for x1 > 1
Sol: [A,B,D] Let circle be (x – x2)2 + y2 = r2
( x  x ) dy
x
dy
 1 2
 1 (from hyperbola)
 dx
y1
dx ( x1 , y1 ) y1
( x1 , y1 )
1


 x1  2 x1  x y 
1

, 1
 2x1 = x2 (l, m)  
3
3




x1
 equation of tangent is y – y1 = y ( x  x1 )
1
1 
 M  x ,0 
 1 
1
dl
1
 l = x1 + 3x  dx  1  2
3x1
1
1
m=
dm 1
y1
 dy  3
3
1
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 28 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
x12  1
m=
3
JEE-Advanced-2015
x1
dm

 dx
1
3 x12  1
55. The option(s) with the values of a and L that satisfy the following equation is(are)
4
t
6
t
6
 e (sin
at  cos 4 at ) dt
0

 L?
 e (sin
4
at  cos at ) dt
0
(A) a  2, L 
e4  1
e  1
(B) a  2, L 
e4  1
e  1
(C) a  4, L 
e4  1
e  1
(D) a  4, L 
e4  1
e  1
Sol: [A,C]
56. Let f, g : [–1, 2]   be continuous functions which are twice differentiabel on the interval (–1, 2).
Let the values of f and g at the points –1, 0 and 2 be as given in the following table
x=–1
x=0
x=2
f(x)
3
6
0
g (x )
0
1
–1
In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)´´ never vanishes. Then the correct
statement(s) is (are)
(A) f ´(x) – 3g´(x) = 0 has exactly three solutions in ( 1,0)  (0, 2)
(B) f ´(x) – 3g´(x) = 0 has exactly one solutions in (1, 0)
(C) f ´(x) – 3g´(x) = 0 has exactly one solutions in (0, 2)
(D) f ´(x) – 3g´(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in (0, 2)
Sol.: (B, C)
SECTION 3 (Maximum Marks : 16)

This section contains TWO paragraphs

Based on each paragraph, there will be TWO questions

Each question contains FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of
these four option(s) is(are) correct

For each quesion, darken the bubble(s) corresponding to all the correct option(s) in the ORS

Marking scheme:
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 29 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
+ 4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened
0 If none of the bubbles is darkened
– 2 In all other cases
PARAGRAPH 1
Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number
of red and black balls, respectively, in box II.
57. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of
this box. the ball was found to be red. if the probability that this red ball was drawn from box II is
then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are)
(A) n1  3, n2  3, n3  5, n4  15
(b)
n1  3, n2  6, n3  10, n4  50
(C) n1  8, n2  6, n3  5, n4  20
(d)
n1  6, n2  12, n3  5, n4  20
Sol: [A,B]
Box -I: Red balls – n1; Black balls – n2.
Box -II: Red balls – n3; Black balls – n4.
red balls from be the enert that box 1 and box 2 E be the enert tha redball is drawn.
 E 
P   P( E2 )
E 
 E2 
P 2  
 E 
E
 E 
P   P( E2  P   P ( E1 )
 E2 
 E1 
1
n3

2 n3  n4
=
1 n3
1 n1

2 n3  n4 2 n1  n2
n3
1
 (given)
n1 (n3  n4 3
n3 
n1  n2
=
Now (A) gives
(B) gives
(C) gives
5
20
53
6
10
60
10  3
9
5
25
58
14



5 1

(A)
15 3
10 1

30 3
(B)
70 1

270 3
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 30 -
1
,
3
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
(D) given
5
25
56
18

JEE-Advanced-2015
15 1

40 3
58. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball
from box I, after this transfer, is
1
, then the correct option(s) with the possible values of n1 and n2 is
3
(are)
(A) n1 = 4 and n2 = 6
Sol: [C,D]
(B) n1 = 2 and n2 = 3
(C) n1 = 10 and n2 = 20 (D) n1 = 3 and n2 = 6
Let E1 be the enert that transferred ball is R and E2 be the enert that transferred ball is B
P (E)drrawing a red ball from bag I)
= P(E  E1) + P(E  E2)
= P(E1) P(E/E1) + P(E2) P(E/E2)
n1
n1  1
n2
n1
= n  n n  1 n  1  n  n . ( n  n  1)
1
2 1
2
1
2
1
2
n1 ( n1  1)  n2 n1
1
= (n  n )( n  n  1)  3 (given)
1
2
1
2
4  3  24 36 1
Now (A)  (10)  (9)  90  3
(B) 
2 1  6 8 1


5 4
20 3
(C) 
10  9  200
290
1


30  29
30  29 3
(D) 
3  2  18 24 1


98
72 3
PARAGRAPH 2
Let F :    be a thrice differentiable function. Suppose that F(1) = 0, F(3) = –4 and F´(x) < 0 for
all x  (1/ 2,3) . Let f(x) = xF(x) for all x   .
59. The correct statement(s) is(are)
(A) f´(1) < 0
(B) f(2) < 0
(C) f´(x)  0 for any x  (1,3)
(D) f´(x) = 0 for some x  (1,3)
Sol: [A,B,C] F is thrice differentiable function
F(1) = 0, (F(3) = –4 and F´(x) < 0 for
1
< x < 3.
2
Here f (x) = x F(x) for all x R
Here f (x) = x F(x)
 f ´(x) = F(x) + x F´(x)
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 31 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
 f ´(1) = 0 + F´(1) < 0 as F´(x) < 0
f ´(1) < 0
(A)
Here F´(x) < 0  F(x) is differentiable function for
1
< x < 3.
2
Now F(1) = 0
 F(x) < 0 for 1 < x < 3.
 f (2) = 2F(2) < 0
(B)
Again if 1 < x < 3
f ´ (x) = F(x) + x F´(x) = (–ve) + (–ve) = –ve
So f ´(x)  0 for 1 < x < 3
60. If
(C)
3
3
2
 x F´( x)dx  12 and
 x F´´( x)dx  40 , then the correct expression(s) is(are)
1
1
3
3
(A) 9 f ´(3)  f ´(1)  32  0
(B)
 f ( x)dx  12
1
3
(C) 9 f ´(3)  f ´(1)  32  0
(D)
 f ( x)dx  12
1
3
2
 x F '( x)dx  12
Sol: [C,D]
1
3
x

2
1
xf '( x)  f ( x)
dx  12
x2
3
3
3
 xf ( x)1   f ( x)dx   f ( x)dx  12

1
3
2
1
3
 x F ( x)   2 f ( x )dx  12

1


1
3

9F(3) – F(1) – 2
 f ( x)dx  12
1
3

–36 –2
 f ( x)dx  12
1
3

 f ( x)dx  24
–2
1
3

 f ( x)dx  12
(D)
1
3
Again
3
 x F ''( x)dx  40
1
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 32 -
Physics, Chemistry & Mathematics (Paper-2) Code-9 (24/05/2015)
JEE-Advanced-2015
3
2
 ( x f ''( x)  2 xf '( x )  2 f ( x)) dx  40

1
3
 x2f ´(x)|13 –
 4 xf '( x)  24  40
1
3


3
4
xf
(
x
)
|

f ( x) dx  = 64

1
 [9f ´(1) – f ´(1)] –4[xf(x)|1 –
1


3

 9f ´(1) – f ´(1) –4[–36 + 12] =64
 9f ´(1) – f´(1) + 32 = 0 (C)
C. O.: Ambition, 383, Bank Street, Munirka, New Delhi - 110067, Tel.: 011-26177182/282, 09818185251
E-mail : info@ambitiongroup.org, visit us at : ambitiongroup.org
- 33 -