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JEE ADVANCED 2015- PAPER 2
PART – I : PHYSICS
SECTION – 1 : (Maximum Marks : 32)




1.
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
The densities of two solid spheres A and B of the same radii R vary with radial distance r
5
r
r
ρA  r   k   and ρa  r   k   , respectively, where k is a constant. The moments of
R
R
inertia of the individual spheres about axes passing through their centres are IA and IB,
I
n
respectively, If B  , the value of n is :
IA 10
Ans. 6
Solution:
Consider a shell of radius r and thickness dr
I   dI
R
IB

IA
2 r5
2
2
o 3 k R5 . 4πr dr r
R
2 r
 3 k R 4πr dr
2
r2

6
10
o
2.
Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0,
π 2π
,
and π . When they are superposed, the intensity of the resulting wave is nl0. The
3 3
value of n is:
Ans. 3
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Solution:
I0  A2
Let amplitude of individual wave be A then amplitude of resulting wave 2A cos
π
   3A
6
I

3A

2
I  3I0
3.
For a radioactive material, its activity A and rate of change of its activity R are defined as
dN
dA
A=
and R =
, where N( τ ) is the number of nuclei at time t. Two radioactive
dt
dt
sources P (mean life τ ) and Q (mean life 2 τ ) have the same activity at t = 0.
Their rates of change of activities at t = 2 τ are Rp and RQ, respectively.
R
n
If P  , then the value of n is :
RQ e
Ans. 2
Solution:
t
t
A p  A0e τ , A0  A0e2τ
4.
t
A0 τt
A0 2τ
R p  e , R0  e
τ
2τ
at t = 2 τ
A0 2
e
Rp
2
 τ

R Q A0 e1 e
2τ
A monochromatic beam of light is incident at 60° on one face of an equilateral prism of
refractive index n and emerges from the opposite face making an angle θ (n) with the
dθ
 m. The value of m is :
normal (see the figure). For n = 3 the value of θ is 600 and
dn
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Ans. 2
Solution:
sin 60 = n sin r ……………..(1)
sin θ = n sin (60 – r)…..(2)
Differentiating eq…..(2)
dθ
dr
cos θ 
 n cos 60  r 
 sin 60  r 
dn
dn
Differentiating eq……(1)
dr
n cos r
 sinr  0
dn
dθ
  tan r 
cos θ
 n cos 60  r  
  sin 60  r 
dn
 n 
dθ
1

  cos 60  r  tan r  sin 60  r 
dn cos θ
dθ
1

 cos 30  tan 30  sin 30
dn cos 60
1 1
= 2    2
2 2
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In the following circuit, the current through the resistor R (= 2  ) is I Amperes. The value
of I is :
Ans. 1
Solution:
I
6.5
 lamp
6.5
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An electron in an excited state of Li2+ ions has angular momentum 3h/2 π .
The de-Broglie wavelength of he electron in this state is pπa0 (where a0 is the Bohr
radius). The value of p is
Ans. 2
Solution :
nh 3h
L

2π 2π
n=3
h
h
h2πr 2πr
λ 

=
p mv
3h
3
7.
n2
r  a0
Z
2π n2 2π 32
λ
a0

a0  2πa0
3
Z
3
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 I 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 r = 3 from M, the tension 1
 M 
in the rod is zero for m = 
 . The value of k is
 288 
Ans. 7
Solution.
For point mass at distance r = 3 .
GMm Gm2
 2  ma
2
3 
For point mass at distance r = 4
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GMm Gm2
 2  ma
2
4 
8.
GMm Gm2 GMm Gm2
 2 
 2
9 2
16 2
7GMm 2Gm2

2
144
7M
m=
288
The energy of a system as a function of time t is given as E(t) = A2 exp(–at), where a =
0.2s–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
Ans. 4
Solution :
E  t   A2eαt
α  0.2s1
dA
 1.25%
A
dE
?
E
log E – 2 log A – αt
dE
dA
 2
 αdt
E
A
= 21.25  0.27.5
= 2.5  1.5
= 4%
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



9.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
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 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
A parallel plate capacitor having plates of area S and plate separation d, has capacitance
C1 in air. When two dielectrics of different relative permittivities ( ε1  2 and ε2  4 )are
introduced between the two plates as shown C2. in the figure, the capacitance becomes
C
C2. The ratio 2 is
C1
(a) 6/5
Ans. (D)
Solution:
(b) 5/3
(c) 7/5
(d) 7/3
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0 A
d
s
20
2  20 s
C
d
d
2
s
4ε0
2  4ε0s
C' 
d
d
2
s
2ε0
2  0 s
C" 
d
d
C C'
4 0 s 0 s
C2 
 C" 

C  C'
3 d
d
7 0 s
=
3 d
C2 7

C1 3
C1 
10. An ideal monoatornicgas is confined in a horizontal cylinder by a spring loaded piston (as
shown in the figure). Initially the gas is at temperature!,, 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 distance x.
Ignoring the friction between the piston and the cylinder, the correct statement(s) is(are)
1
P1V1
4
(b) If V2 = 2V1 and T2 = 3T, then the change in internal energy is 3P1V1
7
(c) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is P1V1
3
(a) If V2 = 2V1 and T2 = 3T, then the energy stored in the spring is
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(d) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is
17
P1V1
6
Ans. (A, B, C)
Solution:
(A) P  P1 
Kx
A
3
V
P2  P1
x 1

2
A
3P1
Kx
 P1 
2
A
PA
Kx  1
2
Energy of spring
1 2 P1A
PV
kx 
x 1 1
Ans. A
2
4
4
f
(B) U   P2 V2  P2 V1 
2
= 3 P1V1
Ans. B
(C) Pf 
4P1
3

KX 
Wgas   WPatm  Wspring

P1
A
3
X
2V1
A
1


=  P1 Ax  Kx,x 
2


2V 1 P A 2V 

=   P1 A. 1  . 1 . 1 
A 2 3 A 

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P1V1 7P1V1

3
3
(D) Q  W  U
= 2P1V1 
7P1V1 3
  P2V2  P1V1 
3
2
7P V 3  4

= 1 1   P1 .3V1  P1V1 
3
2 3

7P V 9
41P1V1
= 1 1  P1V1 
3
2
6
11. A fission reaction is given by
=
Considering
236
92
236
92
140
94
U 54
Xe 38
Sr  x  y, where x and y are two particles.
U to be at rest, the kinetic energies of the products are denoted by
Kse , Ksr , K x (2MeV) and Ky (2MeV), respectively. Let the binding energies per nucleon of
236
92
94
U, 140
54 Xe and 38Sr be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different
conservation laws, the correct option (s) is (are)
(a) x = n, y = n, Ksr = 129 MeV, Kxe = 86 MeV
(b) x = p, y = e–, Ksr = 129 MeV, Kxe = 86 MeV
(c) x = p, y = n, Ksr = 129 MeV, Kxe = 86 MeV
(d) x = n, y = n, Ksr = 86 MeV, Kxe = 129 MeV
Ans. (A)
Solution:
a4
236
140
1
1
92 U 54  38 Sr 0 x 0 y
x=y=n
Q = 236  7.5 – (140  8.5 + 94  8.5)
= 1770 – (1190 + 799)
= 219 MeV
In A and D energy and charge conservation is followed
So
Q  K xe  Ksr  K x  K y
= 129 + 86 + 4
= 219
In D,
pxe  psr  px  py
so conservation of momentum will not hold.
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12. Two spheres P and Q of equal radii have densities ρ1 and ρ2 , respectively. The spheres are
connected by a massless string and placed in liquids L1 and L2 of densities ς1 and ς2 and
viscosities η1 and η2 , respectively. They float in equilibrium with the sphere P in L1 and
sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal
velocity V p and Q alone in L1 has terminal velocity V Q , then
(a)
Vp
VQ

η1
η2
(b)
Vp
VQ

η2
η1
(c) V p .V Q  0
(d) V p .V Q  0
Ans. (A, D)
Solution:
For floating
ρ1  ρ2  V  ς1  ς2  V
ρ1  ρ2  ς1  ς2
since strings in taut so
ρ1  ς1
ρ2  ς2
2
2 r  ς2  ρ1  g
Vp 
9
η2
Vς 
2  ς1  ρ2  g
9
η1
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since ς2  ρ1    ς1  ρ2 
VP η1

VQ η2
V p . V Q  0 because Vp and VQ are opposite
13. In terms of potential difference V, electric current I, permittivity ε0 ,
permeability μ 0 and speed of light c, the dimensionally correct equations(s)
is(are)
(a) μ0I2  ε0V 2
(b) ε0I  μ0V
(c) I  ε0V
(d)
μ0CI  ε0V
Ans. (AC)
Solution.
1 2 1 M0N2A 2
(A) Energy of inductor LI 
I
2
2
1
1 A
Energy of capacitor = CV 2  0 V 2
2
2 d
A
A
μ0 I2
& 0 V 2
have same dimension
d
So μ0I2 & 0 V 2 have same dimension
(C) Q = CV
Q CV

t
t
A V
I 0
t
A
have unit of speed
t
So
I 0 cV
14. 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)
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(a) E is uniform, its magnitude is independent of R2 but its direction depends on r
(b) E is uniform, its magnitude in depends of 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 depends on a
Ans. (D)
Solution:
Electric field in cavity
E
ρOP
3 0
OP  R1  R2
=
ρa
3 0
15. In plotting stress versus strain curves for the 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)
(a) P has more tensile strength than Q
(b) P is more ductile than Q
(c) P is more brittle than Q
(d) The Young's modulus of P is more than that of Q.
Ans. (A,B)
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Solution:
Breaking stress of P is more than Q so P is more ductile
1
strain = stress
Y
Yp  YQ
16. 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)
P  r  3R / 4  63
(a) P(r = 0) = 0
(b)

P  r  2R /3 80
(c)
P  r  3R /5 16

P  r  2R /3 21
(d)
P  r  R / 2 20

P  r  R /3 27
Ans. (B,C)
Solution:
Gravitative field at a distance r
4
GP πr3
4Gρπr
3
=

2
r
3
Consider a small patch of width dr at a distance r
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4
dP. A  Gπρr A dr.ρ
3
P
r
4G ρ2π
O dP R 3 rdr
4Gρ2π 2
r  R 2 
P 
32
2Gρ2π 2 2
R  r 
P
 P  c  R 2  r2 
3
 2 9R 2 
3R
7R 2
r
P1  c  R 
 =c
4
16
16


2R
r
3
 2 4R 2 
5R 2
P2  c  R 
 =c
9
9


P1 63

P2 80
3R
r
9
16R 2
9 2

P3  c  R 2  R  = c
25 
25

2R
5

21R 2
4R 2 
c
=
R 4  c  R2 

25 
25

r
P3 16

P4 21
 2 R 2  3R 2
r = R P5  c  R   
G 
9

r

R
R 2  8R 2
P6  c  R 2   
3
9 
9

Ps 21

Pr 32
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
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SECTION – 3 : (Maximum Marks : 16)
This section contains TWO paragraphs
Based on each paragraph, there will be TWO questions.
Each equation 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 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
17 to 18
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 I, w
and d, respectively.
A uniform magnetic field E is applied on the strip 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 is appearance of equal and
opposite charges on the face opposite to PORS. A potential difference along the zdirection 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) to the same material. Their lengths are
the same, width 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 1 flowing through them in a given magnetic field
strength B, the correct statement(s) is (are)
(a) If w1 = w3 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
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(d) If w1 = 2w2 and d1 = d2, then V2 = V1
Ans. (A, D)
Solution:
q  Vm  Vk 
qvB 
v velocity of electrons
w
Vm  Vk  wvB .
I  neAv  ne  wd  v
wv 
I
ned
1
B
ned
(A) w1 = w2, d1 = 2d2
 V2 = 2V1
(D) w1 = 2w2, d1 = d2
 V1 = V2
18. Consider two different metallic strips (1 and 2) of same dimensions (length I, width
wand 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 differences developed between K and
M in 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 = 2n3, then V2 = V1
(c) If B1 = 2B3 and n1 = n2, then V2 = 0.5V1
(d) If B1 = 2B3 and n1 = n3, then V2 = V1
Ans. (A, C)
Solution :
IB
VM  VK 
ned
(A) n1 = 2n2 ;
B1 = B2
(correct)
 V2 = 2V1
(C) B1 = 2B2, n1 = n2
V1  2V2  V2  0.5 V , correct
Vm  Vk 
PARAGRAPH 19 to 20
Light guidance in an optical fiber can be understood by considering a structure
comprising of thin solid glass cylinder of refractive index n, 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 n3 as shown in
the figure. All rays with the angle of incidence i less than a particular value im are
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confined in the medium of refractive index n1. The numerical aperture (NA) of the
structure is defined 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
16
refractive index
3 15
16
(b) NA of S1 immersed in liquid of refractive index
is that as that of S2 immersed in
15
water
(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
Ans. (A, C)
Solution:
ns sin im  n1 sin  90  c 
sin c 
n2
n1
Where ns : Refractive index of surroundings
NA  sin im 
n1
ns
1
n22 1

n12  n22
2
n1 ns
For S1 in Air
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1 45 9 3
 
1 16 4 4
6
For S1 in ns 
15
6
ns 
15
NA 
15 45 9 3 15
 
6
16 4
24
For S1 in water
NA 
NA 
1
 4 /3
45 9 3  3  9
 

16 4 4  4  16
For S2 in Air
1 64 49
15


1 25 25
5
For S2 in water
NA 
NA 
1
64 49 3 15


 4 /3 25 25 4 5
For S2 in ns 
ns 
16
3 15
16
3 15
15 9

5
16
4
For S2 sin ns 
15
4
ns 
S2
15
NA 
3 15
16
15 64 49
15 15 3



4
25 25
4 5
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
NA1NA2
(a)
(b) NA1 + NA2
(c) NA1
(d) NA2
NA1  NA2
NA 
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Ans. (D)
Solution:
NA 
1
n12  n22
ns
NA2 < NA1
Therefore the numerical aperture of combined structure is equal to the lesser of the two
numerical aperture, which is NA2
PART – II : CHEMISTRY
Atomic masses : [H = 1, D = 2, Li = 7, C = 12, N = 14, O = 16, F = 19, Na = 23, Mg = 24, Al =
27, Si = 28, P = 31, S = 32, Cl = 35.5, K = 39, Ca = 40, Cr = 52, Mn = 55, Fe = 56, Cu = 63.5,
Zn = 65, As = 75, Br = 80, Ag = 108, I = 127, Ba = 137, Hg = 200, Pb = 207]
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. In the complex acetyl bromidodic arbonyl b is (triethyl phosphine)iron(II), the number
of Fe-C bond(s) is
Ans. 3
Solution: [Fe(CH3CO)Br(CO)2{P(C2H5)3}2]
Θ
CH3  C Ligand 
One Fe  C bond
||
O
 two Fe-C bond
CO Ligand 
22. Among the complex ions, [Co(NH2–CH2–CH2–NH2)3Cl2]+, [CrCl2(C2O4)2]3–,
[Fe(H2O)4(OH)2]+, [Co(NH3)4(H2O)CI]2+, the number of complex ion(s) that show (s) cistrans isomerism is
Ans. (6)
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Solution:
All the complexes given show cis-trans isomerism
23. Three moles or B2H6 are completely reacted with methanol. The number of moles of
boron containing product formed is
Ans. 6
Solution:
B2He + 6CH3OH 
 2B (OCH3)3 + 6H2
For 3 moles of B2H6 mole of B containing product formed = 6
24. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than
the molar conductivity of a solution of a weak acid HY (0.10 M). If λ0X  λ0Y , the
difference in their pKa values, pKa (HX) – pKa (HY), is (consider degree of ionization of
both acids to be <<1)
Ans. 3
Solution :
λoX  λoY
 λoH  λoX  λoH  λoY  λoHX  λoHY
Also
λm
 α,
λ mo
So
λ m  HX   λ mo α1
(1)
and
λ m  HY   λ mo α2
(where α 1 and α 2 are degrees of dissociation of HX and HY respectively.)
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Now, Given that
λ m  HY   10λ m  HX 
λom α2  10  λom α1

α2  10 α1
Ka 
(2)
Cα 2
, but a  1, therefore K a  Cα 2 .
1α
K a  HX  0.01α12 0.01  1 
1


  
.

2
K a  HY  0.1α 2
0.1  10  1000
2
 log  Ka  HX    log  Ke  HY    3.
 pKa  HX   pKa  HY   3.
25. A closed vessel with rigid walls contains 1 mol of
Considering complete decay of
238
92
238
92
U and 1 mol of air at 298 K.
U to 206
82 Pb, the ratio of the final pressure to the initial
pressure of the system at 298 K is
Ans. 9
Solution:
Initial moles of gases = 1
238
92
206
U 
82 Pb  824 He  61 eo
Initial moles
1moles
Moles after
8 mole
decomposition
Total gaseous moles after decomposition = 8 + 1 = 9 moles
P n
Ratio of pressures f  f  9
Pi ni
26. In dilute aqueous H2SO4, the complex diaquodioxalatoferrate (II) is oxidized by MnO4 .
For this reaction, the ratio of the rate of change of [H+] to the rate of change of [ MnO4 ] is
Ans. 8
Solution:
8H  MnO4  [Fe  H2O2 OX2 ]2 
Mn3  Fe3  4CO2  6H2O
rate of change of [H ]
8
rate of change of[MnO4 ]
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27. The number of hydroxyl group(s) in Q is
Ans. 4
Solution:
28. Among the following the number of reaction (s) that produce (s) benzaldehyde is
Ans. 4
Solution:
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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 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
29. Under hydrolytic conditions, the compounds used for preparation of linear polymer and
for chain termination, respectively, are
(a) CH3SiCl3 and Si(CH3)4
(b) (CH3)2SiCl3 and (CH3)3SiCl
(c) (CH3)SiCl2 and CH3SiCl3
(d) SiCl4 and (CH3)3SiCl
Ans. (B)
Solution :
For linear polymer, we need R2SiCl2 and for termination, we need R3SiCl.
30. 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)
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(a) O2 is physisorbed

(c) occupancy of π2p
of O2 is increased
(b) heat is released
(d) bond length of O2 is increased
Ans. (A)
Solution:
31. One mole of a 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
Ans. (C)
Solution :
P(V–b) = RT

PV – Pb = RT

PV Pb

1
RT RT
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Pb
RT
Hence Z > 1 at all pressures.
This means, repulsive tendencies will be dominant when interatomic distance are small.
This means, interatomic potential is never negative but becomes positive at small
interatomic distances.
Hence answer is (C)
P(V– b) = RT
PV Pb

1
 PV – Pb = RT 
RT RT
Pb
 Z 1
RT
32. In the following reactions, the product S is

Z 1
Ans. (A)
Solution:
33. The major product U in the following reactions is
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Ans. (B)
Solution:
34. In the following reactions, the major product W is
Ans. (A)
Solution:
35. The correct statement(s) regarding, (i) HCIO, (ii) HCIO2, (iii) HClO3 and (iv) HClO4
is(are)
(a) The number of CI=O bonds in (ii) and (iii) together is two
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(b) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
(c) The hybridization of Cl in(iv)is sp3
(d) Amongst (i) to (iv), the strongest acid is (i)
Ans. (BC)
Solution:
*Number of lone pairs on Cl in (ii) & (iii) together is 3
* Hybridisation of Cl in (iv) is sp3
*Strongest acid is HClO4 (iv)
36. 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+
Ans. (CD)
Solution :
Precipitate is formed on passing H2S in acidic medium i.e. ion must be of group II.
Cu2+ , Pb2+, Hg2+ , Bi3+
SECTION – 3 : (Maximum Marks : 16)
 This section contains TWO paragraphs.
 Based on each paragraph, there will be TWO questions.
 Each equation 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 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 :
i. B2H6
PdBaSO4
C8H6 
C8H6 
X
H2
ii. H2O2 , NaOH, H2O
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i. EtMgBr, H2O
C8H8O 
Y
ii. H , heat
37. Compound X is
Ans. (C)
38. The major compound Y is
Ans. (D)
Solution:
PARAGRAPH – 2
When 100 mL of 1.0 M HCI 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. 1). 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
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acetic acid (Ka = 2.0  10–5) was mixed with 100 mL of 1.0 M NaOH (under identical
conditions to 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
Ans. (A)
Solution:
Let the heat capacity of insulated beaker be C.
Mass of aqueous content in expt. 1 = (100 + 100)  1
= 200g
 Total heat capacity = (C + 200  4.2) J/K
Moles of acid, base neutralised in expt. 1 = 0.1  1 = 0.1
 Heat released in expt. 1 = 0.1  57 = 5.7 KJ
 5.7  1000 = (C + 200  4.2)   T.
5.7  1000 = (C + 200 + 4.2)  5.7
 (C +200  4.2) =1000
In second experiment,
nCH3COOH  0.2, nNaOH  0.1
Total mass of aqueous content = 200 g
 Total heat capacity = (C +200  4.2) = 1000
 Heat released = 1000  5.6 = 5600 J.
Overall, only 0.1 mol of CH2COOH undergo neutralization.
5600
 Hneutralization of CH3COOH 
 56000 J / mol
0.1
= – 56000 J/mol
 Hionization of CH3COOH  57  56  1 KJ / mol
40. The pH of the solution after Expt. 2 is
(a) 2.8
(b) 4.7
(c) 5.0
(d) 7.0
Ans. (B)
Solution:
Final solution contain 0.1 mole of CH3COOH and CH3COONa each.
Hence it is a buffer solution.
CH3COO 
pH  pK a  log
CH3COOH
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= 5  log 2  log
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0.1
 4.7
0.1
PART- III : MATHEMATICS
SECTION – 1 : (Maximum Marks : 2)
 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
41. The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3)…………(1 + x100) is
(1 + x) (1 + x2) (1 + x3) …..(1 + x100)
Ans. 8
Solution:
9 = (0, 9) (1, 8), (2, 7), (3, 6), (4, 5) # 5 cases
9 = (1,2,6), (1,3,5), (2, 3, 4} # 3 cases
total = 8
x2 y 2
42. Suppose that the foci of the ellipse
  1 and (f1, 0) where f1 > 0 and f2 < 0. Let P1
9 5
and 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 (f1 ,0). If m1 is the slope of T1 and m2 is the slope of
 1

T2, then the value of  2  m22  is.
 m1

Ans. 4
Solution:
b2
5 4
2
e 1 2 1 
a
9 9
2
e  foci (2, 0) (–2, 0)
3
p1 : y 2  8x,
y  m1x 
2
m1
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0  4m1 
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2
m1
 4m12  2
 m1  
1
2
p2 : y 2  16x
 y  m2x 
4
m2
 0  2m2 
4
m2
 2m22  4
1
 m22  2  2  4
2
m1
 ecosα n e 
e

43. Let m and n be two positive integers greater than 1. If lim 


 2  , then the
α 0 
αm 
 


m
value of
is
n
Ans. 2
Solution:
m  2 and n  2
cos  an 1
e  e
 1   cos a n  1  2n
 
a
= lim 
2
a 0
n

 am
cos a n  1
a


 ecosan 1  1 
 cos a n  1 
  lim 
  lim a2nm
= e  lim 
2n
n
a 0  cos a
 a0
a
 1  a0 



1
= e  1    lim a2nm
2 a0
Now lim a2nm must be equal to 1.
   
 
 
 
 
a 0
i.e., 2n – m = 0
m
2
n
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1

44. If α   e
0
9x 3 tan 1 x

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 12  9x2 
dx where tan–1 x takes only principal values, then the value

2 
 1 x 
3π 

of  log e 1  α 
 is
α 

Ans. 9
Solution :
1
α   e9x 3tan
1
x
0

 12  9x2 
dx

2 
1

x


 α  e9x 3 tan
3π
9
4
 α e
1
x

1
0
1
 n 1  α   9 
3π
4
Alter :
1
9x3tan x   12  9x2 
1
α e
0
dx

2 
 1 x 
 12  9x2 
3 

 9
dx  dt  
dx  dt
2 
1  x2 
1

x



93π/4

 α
 
et dt  et
0
93π/4
0
 e93π/4  1
Now loge 1  α  3π / 4  loge e
93π/4
 3π / 4  9
45. Let f: R  R be a continuous odd function, which vanishes exactly at one point and f(1) =
1
. Suppose that F(x) =
2
x
 f  t dt for all x  [1,2] and G(x) =
1
Fx 1
 , then the value of
x 1 G  x 
14
x
  t |f  f  t   |dt for all x  [1,2]. If lim
1
1
f   is.
2
Ans. 7
Solution:
x
x
1
1
F  x    f  t dt   f  t dt
x
x
1
1
G  x    t |f  f  t   |dt   t |f  f  t   |dt
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lim
x 1
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Fx
Gx
L’ hospitals lim
x 1
f x
x|f  f  x  

1
14
1
1
2

 1  14
1f 
2
1
f   7
2
46. Suppose that p, q and r are three non-coplanar vectors in R3. 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
Ans. BONUS
Solution :
This question in seem to be wrong but examiner may think like this
S  4p  3q  5r

 
 
S  x p  q  r  y p  q  r  z p  q  r
–x+y–z=4
x–y–z=3
x + y + z=5
add (1) and (2)
2z  7


….(1)
….(2)
….(3)
z
7
2
 x=4
2x  8
2x + y + z = 2(4) + 1 = 9
 kπ 
 kπ 
47. For any integer k, let α k  cos 
  i sin 
 , where i  1 . The value of the
 7 
 7 
12
expression
 α k 1  α k
k 1
3
is
 α 4k 1  α 4k 2
k 1
Ans. 4
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Solution:
2kπ
2kπ i 2kπ
α k  cos
 i sin
 e 14
14
14
12
e
Now
i2 k 1 π
14
k 1
3
e
i 4k 1 π
14
k 1
e
e
i2kπ
14
12
i2π
 e 14  1
i 4k 2 π
14

k 1
3
e
k 1
2π
14

1
12
4
3
48. 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
Ans. 9
Solution :
S7
6

S11 11
7
[2a  6d]
6
2

11
[2a  10d] 11
2
7  a  3d 
6

11  a  5d  11
7a + 21d = 6a + 30d
a = 9d
Given
130 < a + 6d < 140

130 < 15d < 140
Hence d = 9
a = 81
Hence d = 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 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
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49. Let f, g : [–1, 2]  R be continuous function which are twice differentiable 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
3
6
0
f(x)
0
g(x)
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 solution in (–1, 0)
(c) f’(x) – 3g'(x) = 0 has exactly one solution in (0, 2)
(d) f’(x) – 3g'(x) = 0 has exactly two solutions in (–1, 0) and exactly two solutions in
(0, 2)
Ans. (B, C)
Solution :
Let h(x) = f(x) – 3g(x)
h  1 
  h'  x  = 0 has atleast one root in (–1, 0) and atleast one root in (0, 2)
h  0  3
h(2) = 3
But since h”(x) = 0 has no root in (–1, 0) & (0, 2) therefore h’(x) = 0 has exactly 1 root in
(–1, 0) & exactly 1 root in (0,2)
 π π
50. Let f(x) = 7 tan8x +7tan6x – 3tan4x – 3tan2x for all x    ,  . Then the correct
 2 2
expression(s) is (are)
π/4

(a)
0
π/4
1
xf  x dx 
12
 f  x dx  0
(b)
0
π/4
(c)

0
1
xf  x dx 
6
π/4
(d)
 f  x  dx  1
0
Ans. (A,B)
Solution:
f(x) = (7tan6x – 3tan2x).sec2x
π
4
1
  f  x  dx   7t 6  3t 2 dt   t 7  t 3   0
0
1
0
0
π
4
1
0
0
Now  xf  x  dx  
7t
6
Ans. (B)

 3t 2 tan 1 t
dt
II
I
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
= tan t  t  t
1
=

0
=
1
7

t3 1  t 4
1 t
2
3
1
    t
 dt 
1
0
7
 t3
0
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 1 1t dt
2
1
 t 1  t  dt
3
2
0
1 1 1
 
4 6 12
1
192x3
1
51. Let f'(x)=
for all x  R with f   =0. If m   f  x dx  M, then the possible
4
2  sin πx
2
1/2
values of m and M are
(a) m = 13, M = 24
(b) m =
1
1
,M 
4
2
(c) m = – 11, M = 0
(d) m = 1, M = 12
Ans. (D)
Solution:
192x3
1
f ' x  
x  R;f    0
4
2  sin  πx 
2
1 
Now 64x3  f ' x   96x3 x   ,1
2 
3
1 
So 16x 4  1  f  x   24x 4  x   ,1
2
2 
1
16 31 1
24 31 3
.
   f  x dx  .

5 32 2 1/2
5 32 4
1
26
78
  f  x dx 

10 1/2
20
Hence (D)
52. Let S be the set of all non-zero real numbers a such that the quadratic equation
α x2 – x + α = 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
 1 
 1 
,0 
(a)   , 
(b)  
(c)  0,


5
5
 2
 5 

Ans. (A, D)
Solution :
 x1  x 2 
2
 1 1
, 
(d) 
 5 2
 4x1x2  1
1
41
α2
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1
0
α2
5α 2  1
0
α2
 5
1   1


a    , 
, 

5  5 

D>0
1  4α2  0
 1 1
α   , 
 2 2
(1) & (2)
 1 1   1 1 
α ,
,


 2 5  5 2
..(1)
..(2)
 6 
4
53. If α  3 sin 1   and β  3cos1   where the inverse trigonometric functions take
 11 
9
only the principal values, then the correct option(s) is(are)
(a) cos β  0
(b) sin β  0
(c) cos  α  β  0
(d) cos α  0
Ans. (B, C, D)
Solution :
6
6
4
4
α  3sin 1
 3 sin 1
and β  3cos1  3 cos1
11
12
9
8
π
βπ
&
 α
2
3π
 α β 
2
54. 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 touches the curves S, E1 and E2at P,Q and R, respectively. Suppose
2 2
. If e1 and e2 are the eccentricities of E1 and E2, respectively, then the
3
correct expression(s) is (are)
that PQ = PR =
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(a) e12  e22 
43
40
(b) e1  e2 
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7
2 10
(c) |e12  e22 | 
5
8
(d) e1e2 
3
4
Ans. (A,B)
Solution:
x2 y 2
E1  2  2  1
a
b
x2 y 2
E2  2  2  1
A
B
Now as x + y = 3 is a tangent
a2 + b2 = A2 + B2 = 9
Now point P is
x2 + (2 – x)2 = 2
2x2 – 4x + 2 = 0
x=1
so P is (1, 2)
5 4 1 8
points Q & R are  ,  &  , 
3 3 3 3
25
16
5 4
Now  ,  lies on E1 so 2 
1
2
9a
3
3
9
9

a


 
 225 – 25a2 + 16a2 = 9a2(9 – a2)
 225 – a2 = a2(9 – a2)
 a4 – 10a2 + 25 = 0
1
e12 
5
1 8
Now  ,  lies on E2
3 3
1
64

9
A2 9  A2


225 – 9a2 = 9a2(9 – a2)

a2 = 5 so b2 = 4

9 – A2 + 64A2 = 9A2 (9 – A2)
 A2 = 1 so B2 = 8
1 + 7A2 = A2 = 9A2 – A4  A4 – 2A2 + 1 = 0
7
e22 
8
55. Consider the hyperbola 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 
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PMN, then the correct expression(s) is(are)
dm
x1
dl
1
(a)
(b)

for x1  1
 1  2 for x1  1
dx1 3 x2  1
dx1
3x1
1

dl
1
 1  2 for x1  1
dx1
3x1
(c)
(d)

dm 1
 for y 1  0
dy 1 3
Ans. (A, B, D)
Solution:
Equation tangent to H at P is xx1 – yy1 = 1
1
x1  x 2 
x12  1
y1
2
 x2  2x1

, m 
3
3
3
1
So,  x1 
3x1
d
1 dm 1 dm 1
x1
1  2 ,
 ,
 .
dx1
3x1 dy 1 3 dx1 3 x12  3
56. The option(s) with the values of a and L that satisfy the following equation is(are)
4π
 e  sin
t

6
at  cos4 at dt
6
at  cos4 at dt
0
π
 e  sin
t

0
e4π  1
(a) a = 2, L = π
e 1
e4π  1
(c) a  4, L  π
e 1
Ans. (A, C)
Solution:
e4π  1
(b) a = 2, L = π
e 1
e4π  1
(d) a = 4, L = π
e 1
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π


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2π


3π


I1   e sin at  cos at dt   e sin at  cos at dt   et sin6 at  cos4 at dt 
t
6
4
0
6
π
4π
  sin
t
6
4
2π

at  cos4 at dt
3π
π
= 1  eπ  e2π  r3π   et  sin6 at  cos4 at  dt
0
I1
e4π  1
π
2π
3π

1 e e e  π
I2
e 1
SECTION - 3 : (Maximum Marks : 16)
 This section contains TWO paragraphs
 Based on each paragraph, there will be TWO questions.
 Each equation 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 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
1
ball was drawn from box II is , then the correct option(s) with the possible values of
3
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
Ans. (A, B)
Solution :
Red  n1
Red  n3
Box  I 
Box  II 
Black  n2
Black  n 4
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n3
1
n1
1
PR   .
 .
2 n1  n2 2 n3  n4
n3
n3
1
.
2 n3  n 4
n3  n 4
R  II / R  

n3
n3
1
n1
1
n1
.
 .

2 n1  n 2 2 n3  n 4 n 1  n 2 n 3  n 4
n1 = 3, n2 = 3, n3 = 5, n4 = 15
58. A ball is drawn at random from box I and transferred to box II. If the probability of
1
drawing a red ball from box I, after this transfer, is , then the correct option(s) with
3
the possible values of n1 and n2 is(are)
(a) n1 = 4 and n2 = 6
(b) n1 = 2 and n2 = 3
(c) n1 = 10 and n2 = 20
(d) n1 = 3 and n2 = 6
Ans. (C,D)
Solution:
n1
n1  1
n2
n1
1
Given
.

.

n1  n2 n1  n2  1 n1  n2 n1  n2  1 3


3 n12  n1  n1n2   n1  n2  1
3n1  n1  n2  1  n1  n2  n1  n2  1
2n1 = n2
PARAGRAPH 2
Let F : R  R 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  R.
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)
Ans. (A, B, C)
Solution:
f ' x   xf ' x   f  x 

f '1  f ' 1  f 1  f ' 1  0
 (A)
 (B)
f(2) = 2f(2) < 0
for x  (1, 3) , f’(x) = xf’(x) + f(x) < 0  (C)
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3
IIT – JEE /AIPMT/NTSE/Olympiads Classes
3
60. If  x F'  x  dx  12 and  x3F"  x  dx  40 , then the correct expression (s) is (are)
2
1
1
3
(b)  f  x  dx  12
(a) 9f’(3) + f’(1) – 32 = 0
1
3
(d)  f  x  dx  12
(c) 9f’ (3) – f’(1) + 32 = 0
1
Ans. (C, D)
Solution:
3
3
3
3
2
 x f "  x  dx  40  x f ' x 1   3x f ' x  dx  40
3
1
1
3
  x2f ' x   xf  x   3 12  40
1
 9f’(3) – 3f(3) – f’(1) + f(1) = 4
 9f’(3) + 36 – f’(1) + 0 = 4  9f’(3) – f’(1) + 32 = 0
3

2
 x f ' x  dx  12
1
3
1
(C)
  x2f  x     2xf  x dx  12
1
1
3
 36  2 f  x dx  12

3
3

 f  x dx  12

(D)
1
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