IIT jee advanced 2013

2013 JEE-ADVANCED_P1_Code::0 QPaper with & Key &
Sol Solutions
Sri Chaitanya IIT Academy., India
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which ONE or MORE are correct
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Sol Solutions
PART-I_PHYSICS
Section-1
(Only one Option correct Type)
This section contains 10 Multiple Choice questions. Each Question has Four choices
(A), (B), (C) and (D). Out of Which Only One is correct
1. The work done on a particle of mass m by a force
   
3/ 2 3/22 2 2 2
ˆ ˆx y
K i j
x y x y
 
 
   
(K being a constant of appropriate dimensions), when the particle is taken from the
point  ,0a to the point  0,a along a circular path of radius a about the origin in
the x y plane is
A)
2K
a

B)
K
a

C)
2
K
a

D) 0
Ans : D
Sol : 3
kr
F
r



i.e conservative force from the origin at equal distances potentials will be
constant.
work done = U 0 
Ans : 0
2. Two rectangular blocks, having identical dimensions, can be arranged either in con-
figuration I or in configuration II as shown in the figure, One of the blocks has
thermal conductivity  and the other 2 . The temperature difference between the
ends along the x-axis is the same in both the configurations. It takes 9 s to transport
a certain amount of heat from the hot end to the cold end in the configuration I. The
time to transport the same amount of heat in the configuration II is
A) 2. 0 s B) 3.0 s C) 4.5 s D) 6.0 s
2. Ans : A
Sol :
2 1 1 2
1 2 2 1
Q kA t k A
t t k A

    
l
l l
(Q,  are same in both the configurations)

Sol Solutions
2
1
k
4t 1 13
kt 2 23
2
   
2
2
t 9s 2s
9
  
3. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2:3.
The ratio of their partial pressures, when enclosed in a vessel kept at a constant
temperature, is 4:3. The ratio of their densities is
A) 1 : 4 B) 1 : 2 C) 6 : 9 D) 8 : 9
Ans : D
Sol : A
RT
and M mN
d m
 
P
1 1 1
2 2 2
d m p 2 4 8
d m p 3 3 9
     
4. A particle of mass m is projected from the ground with an intial speed 0u at an angle
 with the horizontal. At the highest point of its trajectory, it makes a completely
inelastic collision with another identical particle, which was thrown vertically upward
from the ground with the same intial speed 0u . The angle that the composite system
makes with the horizontal immediately after the collision is
A)
4

B)
4

 C)
2

 D)
2

ANS : A
SOL :
2 2
2 0
2 0
u sin
v u 2gH but H
2g

  
0u

0u
2 0v u cos 
v
1v ucos 
H


Sol Solutions
2 2
2 0
0 0
u sin
u 2g u cos
2g
       
If we assume that we need to find the angle made by the velocity of combined mass
just after the collision
1 2
1
v
tan
v 4

       
5. A pulse of light of duration 100 ns is abosorbed completely by a small object initially
at rest. Power of ithe pulse is 30 mW and the speed of light is 8 1
3 10 ms
 . The final
momentum of the object is
A) 17 1
0.3 10 kg ms 
 B) 17 1
1.0 10 kg ms 
 C) 17 1
3.0 10 kg ms 
 D) 17 1
9.0 10 kg ms 

Ans : B
Sol : Energy of the pulse E = pt
   3 9 9
30 10 100 10 3 10 J  
     
since momentum is conserved, final momentum of the object
= initial momentum of pulse
9
8
E 3 10
C 3 10




= 17 1
1.0 10 kg ms 

6. In the Young’s double slit experiment using a monochromatic light of wavelength  ,
the path difference (in temrs of an integer n ) corresponding to any point having half
the peak intensity is
A)  2 1
2
n

 B)  2 1
4
n

 C)  2 1
8
n

 D)  2 1
16
n


Ans : B
Sol :
2max
max
I
I I cos ,
2 2

   is phase difference
1
cos
2 2

 
2
etc. but x
2 4
  
   

2
x etc. x
2 4
  
   

etc.  general expression  x 2n 1
4

 

Sol Solutions
7. The image of an object, formed by a plano-convex lens at a distance of 8 m behind
the lens, is real and is one-third the size of the object. The wavelength of light inside
the lens is
2
3
times the wavelength in free space. The radius of the curved surface of
the lens is
A) 1 m B) 2 m C) 3 m D) 6 m
Ans : C
Sol :  
1 2
1 1 1 1
1
v u R R
         
v = + 8m
| V |
u 24m | m |
| u |
      

0
m
3
2

  

2R  
1 1 3 1
1 0
8 24 2 R
              
R 3m 
8. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an
end of another horizontal thin copper wire of length L and radius R. When the
arrangement is stretched by applying forces at two ends, the ratio of the elogation in
the thin wire to that in the thick wire is
A) 0.25 B) 0.50 C) 2.00 D) 4.00
Ans : C
Sol : Tension in both the wires is same
2
1 1 2
2
2 2 1
F / A F R
y
/ AY R

      
 
l l l
l
l l l l
1
2
2 1 1
1 4 2

  

l
l
 required ratio =
2
1
2

 

l
l

Sol Solutions
9. A ray of light travelling in the direction  
1 ˆ ˆ3
2
i j is incident on a plane mirror. After
reflection, it travels along the direction  
1 ˆ ˆ3
2
i j . The angle of incidence is
A) 30° B) 45° C) 60° D) 75°
Ans : A
Sol : Angle between the incident ray and the reflected ray is
   1 1
i 3j . i 3 j
a.b 2 2cos2
ab 1 1
 
  

     
 
1
1 3
4
   
1
2
 
0
2 120 
0
60 
0
30  


2 120 
10. This diameter of a cylinder is measured using a vernier callipers with non zero error.
It is found that the zero of the vernier scale lies between 5.10 cm and 5.15 cm of the
main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th
divi-
sion of the Vernier scale exactly coincides with one of the main scale divisions. The
diameter of the cylinder is
A) 5.112 cm B) 5.124 cm C) 5.136 cm D) 5.148 cm
Ans : B
Sol : Reading = MSR + L.C  No.of divisions coincidence
=
0.05
5.10 24
50
 
one division in MS
L.C
n

= 5.10 + 0.024
= 5.124cm

Sol Solutions
Section-2
(One or More options Correct Type)
This section contains 5 multiple choice equations. Each quation has four choices (A)
(B)(C) and (D) out of which ONE or MORE are correc.
11. In the circuit shown in the figure, there are two parallel plate capacitors each of
capacitance C. The switch 1S is pressed first to fully charge the capacitor 1C and then
released. The switch 2S is then pressed to charge the capacitor 2C . After some time,
2S is released and then 3S is pressed. After some time,
02V
1C 2C
0V
1S 2S 3S
A) the charge on the upper plate of 1C is 02CV .
B) the charge on the upper plate of 1C is 0CV .
C) the charge on the upper plate of 2C is 0.
D) the charge on the upper plate of 2C is 0CV .
Ans : BD
Sol : Step : 1
potential of c1
= 2V0
and
that of c2
= 0
with copper plate is positively changed
Step : 2 Common polential
   0
0
2 0  
 

c V c
V V
c c
with upper platespositively changed.
Step : 3 Potential on c1
= V0
 change on upper phase of c1
= + cV0
potential on c2
= V0
with upper plate negatively changed
 change on upper plate of c2
= – cV0
Ans : B,D

Sol Solutions
12. A particle of mass M and positive charge Q, moving with a constant velcoity 1
1
ˆ4u ims


,
enters a region of uniorm static magnetic field normal to the x y plane. The region of
the magnetic field extends from 0x  to x L for all values of y .After passing through
this region, the particle emerges on the side after 10 milliseconds with a velocity
  1
2
ˆ ˆ2 3u i j ms
 

. The correct statement(s) is (are)
A) The direction of the magnetic field is z derection.
B) The direction of the magnetic field is z direction.
C) The magnitude of the magnetic field
50
3
M
Q

units.
D) The magnitude of the magnetic field is
100
3
M
Q

units.
Ans : AC
Sol :
Magnetic field must be along ‘–z’ direction as shown in fig.
Angle of deviation =
21
2
tan
       
y
x
V
V
1 1
tan
3

    
6

 rad
There for time sprent in the field
3 3
10 12 10
12
 
    
T
S T S
But
2 2 50
3
  
   
M M M
T B
BQ TQ Q
Ans : A,C

Sol Solutions
13. A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic
according to the equation,        1 1
, 0.01 sin 62.8 cos 628y x t m m x s t 
        . Assum-
ing 3.14, the correct statement(s) is (are)
A) The number of nodes is 5.
B) the length of the string is 0.25 m.
C) the maximum displicement of the midpoint of the string, from its equilibrium
postion is 0.01 m
D) the fundamentals frequency is 100 Hz
Ans : BC
Sol : y(x,t) = 0.01 sin(62.8) cos 628 t No.of nodes is 6 since 5 loops will form.
Length of the string L =
5 5 2
2 2 k
 
 
5
k


3.14
5 0.25
62.8
  
The maximum displacement of the mid point of the string which is antinode from it’s
equilibriumis 0.01
Hence : B,C
14. A solid sphere of radus R and density  is attched to end of a mass-less spring of
force constant. k . The other end of the spring is connected to another solid sphere
of radius R and density 3 . The complete arrangement is placed in a liquid of
density 2 and is allowed to reach equillibrium. The correct statement(s) is (are)
A) the net elongation of the spring is
3
4
3
R g
k
 
B) the net elongation of the spring is
3
8
3
R g
k
 
.
C) the light sphere is paratially submerged.
D) the light sphere is completely submerged.
Ans : AD

Sol Solutions
Sol : The system is in equilibrium. Gravitational pull is greater than maximum possible
buoyancy force hence lighter sphere is completely submerged. For lighter sphere.
Fb
mgke
ke + mg = Fb
ke+
3 34 4
R dg R 2dg
3 3
  
34
R dg
3e
k


Ans : A,D
15. Two non-conduction solid spheres of radii R and 2R , haveing uniform volume
charge densities 1 and 2 respectively, touch each other. The net electric field at a
distance 2R from the centre of the smaller sphere, along the line joining the centres
of the spheres, is zsero. The ratio
1
2

 can be
A) -4 B)
32
25
 C)
32
25
D) 4
Ans : BD
Sol : R 2R
For external point
 
 
 
33
1 2
2 2
4 4
k R k 2R
3 3
2R 5R
   
 = 0
1 28
4 25
 
 = 0
1
2
32
25
 



Sol Solutions
For Internal point :
 
 
33
1 2
32
4 4
k. R k 2R .R
3 3
4R 2R
   

1
2

4

Section-3
(Integer Value Correct Type)
This section contains 5 questions. The answer to each questionis a single digit
integer, ranging from 0 to 9 (both inclusive).
16. A bob of mass m, suspended by a string of length 1l , is given a minimum velocity
required to complete a full circle in the vertical plane. At the highest point, it collides
elastically with another bob of mass m suspended by a string of length 2l , which is
initially at rest, Both the strings are mass-less and inextensible. If the second bob,
after collision acquires the minimum speed required to complete a full circle in the
vertical plane, the ration
1
2
l
l
is
Ans : 5
Sol : l1
Just completes vertical circular motion hence at the highest point its velocity is
1
gl
In the elastic collision they exchange their velocities
1 2g 5gl l
1
2
5
l
l
17. A particle of mass 0.2kg is moving in one dimension under a force that delivers a
constant power 0.5W to the particle. If the initial speed  1
inms
of the particle is
zero, the speed  1
inms
after 5s is
Ans : 5

Sol Solutions
Sol : Power =
d K.E
dt
pdt K.E 
‘p’ is constant
pt =
21
mv
2
2pt 2 0.5 5
v 5m / sec
m 0.2
 
  
18. The work functions of Silver and Sodium are 4.6 and 2.3eV , respectively. The ratio
of the slope of the stopping potential versus frequency plot for Silver to that of
Sodium is
Ans : 1
Sol : shf w eV 
s
wh
V f
e e

 
slope =
h
e
which is independent of the nature of substance.
ratio of slopes = 1
19. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 3
10
disintegrations per second. Given that ln 2 0.693 , the fraction of the initial number
of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after
preparation of the sample is
Ans : 4
Sol : t
0N N e

No. of nuclei decay =  t
0 0N N N 1 e
  
fraction of decay =
 t
O
O
N 1 e
N


t
1 e 100%     
0.04
1 exp 100%     
   1 1 0.04 100% 4%   

Sol Solutions
20. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular
velocity of 1
10 ra d s 
about its own axis, which is vertical. Two uniform circular
rings, each of mass 6.25kg and radius 0.2m , are gently placed symmetrically on
the disc in such a manner that they are touching each other along the axis of the disc
and are horizontal. Assume that the friction is large enough such that the rings are at
rest relative to the disc and the system rotates about the original axis. The new
angular velocity  1
in rad s
of the system is
Ans : 8
Sol: Apply the conservation of angular momentum
D D D ringI I ' 2I '    
 
D D
D ring
I
'
I 2I

 

2
D
2
2D D
ring ring
M R 1000
8
125M R
2.2M R
2
  
 
 
  

Sol Solutions
PART-II_CHEMISTRY
Section-1
This section contains 10 multiple choice questions. Each question has four choices (A),
(B), (C) and (D) out of which ONLY ONE is correct.
21. Consider the following complex ions, P, Q and R
 
3
6P FeF

 ,  
2
2 6
Q V H O

    and  
2
2 6
R Fe H O

   
The correct order of the complex ions, according to their spin - only magnetic
moment values (in B.M.) is
A) R < Q < P B) Q < R < P C) R < P < Q D) Q < P < R
Ans : B
Sol:    
3
6P FeF

has 5 unpaired e–
   
2
2 6
Q V H O

 
   ___ 3 unpaired e–
   
2
2 6
R Fe H O

 
   ___ 4 unpaired e–
order Q < R < P
22. The arrangement of X  ions around A ion in solid AX is given in the figure (not
drawn to scale). If the radius of X  is 250 pm, the radius of A is
A) 104 pm B) 125pm C) 183 pm D) 57pm
Ans : A
Sol : Fcc
2a 4r
1000pm
a
2
 = 707.21 pm
2r 2r a
 
2r 707.21 500.00 207.21pm
  
r 104pm


Sol Solutions
23. Sulfide ores are common for the metals
A) Ag, Cu and Pb B) Ag, Cu and Sn C) Ag, Mg and Pb D) Al, Cu and Pb
Ans : A
Sol: Ag, Cu and Pb occur in nature as sulphide ores Ag2
S, CuFeS2
or Cu2
S and PbS
24. The standard enthalpies of formation of  2CO g ,  2H O l and glucose (s) at 25 C
are 400 /kJ mol , 300 /kJ mol and 1300 /kJ mol respectively. The standard en-
thalpy of combustion per gram of glucose at 25 C is
A) +2900 kJ B) -2900 kJ C) -16.11 kJ D) +16.11 kJ
Ans : C
Sol : 2 2 1C O CO H 400 kJ     .......... (1)
2 2 2 2
1
H O H O H 300 kJ
2
     ........ (2)
2 2 6 12 6 36C 6H 3O C H O H 1300 kJ     ...... (3)
6 12 6 2 2 2C H O 6O 6CO 6H O   ; H ? 
     1 6 2 6 3   
per 1 mole –2900 kJ
per 1 gm ?
Ans : –16.11 kJ
25. Upon treatment with ammoniacal 2H S , the metal ion that precipitates as a sulfide is
A) Fe(III) B)Al(III) C) Mg(II) D) Zn (II)
Ans : D
Sol: With ammonical H2
S Both Fe(III) and Zn(II) precipitate as sulphide more appro
priate is Zn (II)
26. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at
25 C . For this process, the correct statement is
A) The adsorption requires activation at 25 C
B) The adsorption is accompanied by a decrease in enthalpy
C) The adsorption increases with increase of temperature.
D) The adsorption is irreversible.
Ans : B
Sol :Adsorption from solution is similar to adsorption of gas on solid
accompanied by decrease in enthalpy

Sol Solutions
27. KI in acetone, undergoes 2NS reaction with each of P, Q, R and S. The rates of the
reaction vary as
H3C Cl Cl Cl
Cl
O
P Q
R S
A) P > Q > R > S B) S > P > R > Q C) P > R > Q > S D) R > P > S > Q
Ans : B
Sol:
O
Cl
reacts with maximum rate due to participation of C = O “P”
orbitals in the stabilization of transition state i.e, SN2
T.S. chloroacetophenone is
32,000 more reactive towards KI & actone at 75o
C than 1 - chlorobutane.
Ph
C
C
H H
O Cl
Nu
T.S of “S” molecule.
Cl practically speacking allyl chloride is highly reactive than methyl
chloride towards SN2
r x n
[Ref : Jerry March page No : - 339]
due to may be above reason.
After allyl chloride methyl chloride is reactive and least reactive is isopropyl chlo
ride.
But with in the boundaries of given options, option “B” is best suitable one.

Sol Solutions
28. In the reaction,
P Q R S  
the time taken for 75% reactio of P is twice the time taken for 50% reaction of P. The
concentration of Q varies with reaction time as shown in the figure. The overall order
of the reaction is
 Q
 0
Q
Time
A) 2 B) 3 C) 0 D) 1
Ans : D
Sol : 75%
2.303 100
t log
k 25

50%
2.303 100 0.693
t log
k 50 k
 
first order
29. Concentrated nitric acid, upon long standing, turns yellow - brown due to the forma-
tion of
A) NO B) 2NO C) 2N O D) 2 4N O
Ans : B
Sol: Conc HNO3
dissociate slowly
2HNO3  H2
O + 2NO2
+ O2
Due to dissolution of NO2
, Conc HNO3
upon long standing turns yellow - brown.
30. The compound that does NOT liberate 2CO , on treatment with aqueous sodium
bicarbonate solution, is
A) Benzoic acid B) Benzenesulphonic acid
C) Salicylic acid D) Carbolic acid (Phenol)
Ans : D
Sol:
OH
+ NaHCO3
(aq)  No carbondioxide liberation

Sol Solutions
Section-2
(One or More Options correct Type)
This section contains 5 multiple choice questions. Each question has four choices (A),
(B), (C) and (D) out of which ONE or MORE are correct.
31. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is
1/100th of that of a strong acid  ,1 ,HX M at 0
25 .C The aK of HA is
A) 4
1 10
 B) 5
1 10
 C) 6
1 10
 D) 3
1 10

Ans : A
  2
H 10 

  aH k .c

4
ak 10

32. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are
due to
A) p  (empty) and *
  electron delocalisations
B) *
  and   electron delocalisations
C) p  (filled) and   electron delocalisations
D)   *
p filled  and *
  electron delocalisations
Ans : A
Sol:
CH3
CH3
C+
H3C & CH3
– CH = CH – CH3
stabilities can be explained as follows.
In case of tertiary butyl carbocation, stabilization is due to transfer of " " elec
trons to empty “p” orbitals. So, the delocalization is p  (empty).
C
H
H
H
H
HH
H
H H
C+
C
C
Stability of 2-butene can be explained as follows.
C = C  In the double bonded region two “p” orbitals of two carbons are
giving two molecular orbitals those are & *  orbitals. In the hyper conjugative
stabili zation of double bonded region,  – bonded electrons are confined to  –
molecular orbital, and C H " " bonded electrons are entering into * –mo
lecular orbital.

Sol Solutions
33. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is
(are)
A)  3 25
Cr NH Cl Cl   and  3 24
Cr NH Cl Cl  
B)  3 24
Co NH Cl

   and    3 22
Pt NH H O Cl

  
C)  
2
2 2CoBr Cl

and  
2
2 2PtBr Cl

D)    3 33
Pt NH NO Cl   and  3 3
Pt NH Cl Br  
Ans : B & D
Sol: a
     3 2 3 24 2
B Co NH Cl and Pt NH H O Cl
 
   
      
exhibit geometrical isomers.
     3 3 33 3
D Pt NH NH Cl and Pt NH Cl Br   
      
exhibit ionisation isomers
34. Among P,Q,R and S, the aromatic compound (s) is/are
A) P B) Q C) R D) S
3AlCl
P
Cl
N aH
Q
R0
100 115 C
 4 32
NH CO
O O
H C l
S
O

Sol Solutions
Ans : ABCD
Sol:
35. Benzene and naphthalene form an ideal solution at room temperature. For this pro-
cess, the true statement(s) is (are)
A) G is positive B) systemS is positive C) 0surroundingsS  D) 0H 
Ans : B, C, D
Sol : No heat absorbed or released

Sol Solutions
Section-3
(Integer Value correct Type)
This section contains 5 questions. The answer to each question is a single digit
integer, ranging from 0 to 9 (both inclusive)
__________________________________________________________________
36. The atomic masses of He and Ne are 4 and 20 a.m.u., respectively. The value of the
de Broglie wavelength of He gas at 0
73 C is “M” times that of the de Broglie wave-
length of Ne at 0
727 C . M is
Ans : 5
Sol :
h
mv
 
h
2mK.E
  ,
1 2 2
2 1 1
m T
m T

 

,
M 20 1000
1 4 200
 
M = 5
37. 4
EDTA  is ethylenediaminetetraacetate ion. The total number of N Co O  bond angles
in  
1
Co EDTA

 
  complex ion is
Ans : 8
Sol:
6 are 90o
2 are 180o
Total are 8
More appropriate answer = 8

Sol Solutions
38. The total number of carboxylic acid groups in the product P is
O
O
3
3
2 2
1.H O ,
2.O
3.H O

 PO
O
O
Ans : 2
Sol:
39. A tetrapeptide has COOH group on alanine. This produces glycine (Gly), valine
(val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this
tetrapeptide, the number of possible sequences (primary structures) with 2NH group
attached to a chiral center is
Ans : 4

Sol Solutions
Sol : 2 2H N CH COOH Glycine  
3
|
NH2
CH C H COOH Alanine  
2
|
NH2
Ph CH C H COOH Phenyl Alanine   
3
| |
CH NH3 2
CH CH CH COOH Valine   
For, the given tetrapeptide – COOH on alanine is fixed. This means alanine is
fixed as, “C” terminus. Remaining aminoacids are glycine which is without chiral
carbon, and valine and phenyl alanine which are chiral. Then the number of pos
sible sequences (primary structures) with - NH2
group attached to a chiral center
is four. Because we have left with only two aminoacids. Which are chiral and
which can act as N-terminal ends. By fixing one chiral amino acid at N-terminus
we can obtain two peptide sequence by inter changing internal two aminoacids.
Above phenomenon can be repeated even for second chiral aminoacid. As a
whole ultimately we will be having total four peptide sequences.
40. The total number of lone-pairs of electrons in melamine is
Ans : 6
Sol: Melamine’s structure is
N N
N
NH2
H2N NH2
Total no.of lone pairs in the melamine is “6”

Sol Solutions
PART-III_MATHEMATICS
Section-1
This Section contains 10 multiple choice Questions. Each question has four choices (A),(B),(C)
& (D) out of which ONLY ONE is correct.
41. For 0,a b c   the distance between  1,1 and the point of intersection of the lines
0ax by c   and 0bx ay c   is less than 2 2.Then
A) 0a b c   B) 0a b c   C) 0a b c   D) 0a b c  
Ans : A (or) C
Sol :The point of intersection of two lines is
c c
,
a b a b
       
distance d < 2 2
2
d 8 
2
c
2 1 8
a b
     
c
1 2
a b
  

a b c 0    and also a - b + c > 0 (since a > b > c > 0)
42. The area enclosed by the curves sin cosy x x  and cos siny x x  over the interval
0,
2
 
 
 
is
A)  4 2 1 B)  2 2 2 1 C)  2 2 1 D)  2 2 2 1
Ans : B
Sol : 1 2y 2 sin x , y 2 sin x
4 4
                
put x
4

  
1 2
3
and y 2 sin , y 2 cos
4 4
 
      
 required area =    
3
2 4
4 2
2 sin cos d 2 sin cos d
 
 
        
=  2 2 2 1

Sol Solutions
43. The number of points in  ,  , for which 2
sin cos 0x x x x   , is
A) 6 B) 4 C) 2 D) 0
Ans : C
Sol :Let      2
f x x xsin x cos x f ' x x 2 cos x     
y
x
O
     f ' x 0 x 0, f ' x 0 x 0 and f ' x 0 for x 0        
44. The value of
23
1
1 1
cot cot 1 2
n
n k
k
 
  
  
  
  is
A)
23
25
B)
25
23
C)
23
24
D)
24
23
Ans : B
Sol :Let  
23 n 23
1 1
n 1 k 1 n 1
n 1 n
cot 1 2k tan
1 n n 1
 
  
                  
  
  
23
1 1
n 1
tan n 1 tan n 

   =
1
tan 24
4
 

23
tan
25
  
45. A Curve passses through the point 1, .
6
 
 
 
Let the slope of the curve at each point
 ,x y be sec , 0
y y
x
x x
 
  
 
. Then the equation of the curve is
A)
1
sin log
2
y
x
x
 
  
 
B) cos log 2
y
ec x
x
 
  
 
C)
2
s log 2
y
ec x
x
 
  
 
D)
2 1
cos log
2
y
x
x
 
  
 
Ans : A

Sol Solutions
Sol :
dy y y
sec
dx x x
     
2
xdy ydx y dx
cos
x x x
             
y y dx
cos d
x x x
            
y
sin ln x c
x
     
but  y 1
6


1
c
2
 
46. Let
1
: ,1
2
f R
 
 
 
(the set of all real numbers) be a positive, non constant and differen-
tiable function such that    1
2f x f x and
1
1
2
f
 
 
 
. Then the value of  
1
1/2
f x dx lies
in the interval
A)  2 1,2e e B)  1,2 1e e  C)
1
, 1
2
e
e
 
 
 
D)
1
0,
2
e  
 
 
Ans : D
Sol :   2xd
e f x 0
dx


 2x 1
e f x
e

 
  2x 1
f x e 
 
 
1 1
2x 1
1/2 1/2
0 f x dx e dx
     
1
1/2
e 1
0 f x dx
2

  
47. Ler ˆˆ ˆ3 2PR i j k  

and ˆˆ ˆ3 4SQ i j k  

determine diagonals of a parallelogram PQRS
and ˆˆ ˆ2 3PT i j k  

be another vector. Then the volume of the parallelepiped
determined by the vectors PT

, PQ

and PS

is
(A) 5 (B) 20 (C) 10 (D) 30
Ans : C

Sol Solutions
Sol: From the given data PR PQ PS 
  
but ˆ ˆ ˆPQ PS i 3j 4k   
 
ˆ ˆ ˆ ˆ ˆ ˆPQ 2i j 3k and PS i 2j k      
 
 volume of the parallelepiped is v PT PQ PS 10    
  
48. Perpendiculars are drawn from ponits on the line
2 1
2 1 3
x y z 
 

to the plane
3x y z   . The feet of perpendiculars lie on the line
(A)
1 2
5 8 13
x y z 
 

(B)
1 2
2 3 5
x y z 
 

(C)
1 2
4 3 7
x y z 
 

(D)
1 2
2 7 5
x y z 
 

Ans : D
Sol :Let  2 2k, 1 k, 3k    be a point on the given line and its projection on the given
plane is
x 2 2k y 1 k z 3k 3 4k 3
1 1 1 3
               
hence the equation of required line is
x y 1 z 2
2 7 5
 
 

49. Four persons independently solve a certain problem correclty with probabilities
1 3 1 1
, , , .
2 4 4 8
Then the probability that the problem is solved correctly by at least one of
them is
(A)
235
256
(B)
21
256
(C)
3
256
(D)
253
256
Ans : A
Sol :Required propability =
1 1 3 7 235
1 . . .
2 4 4 8 256
 
50. Let complex numbers  and
1

lie on circles    
2 2 2
0 0x x y y r    and
   
2 2 2
0 0 4x x y y r    , respectivley. If 0 0 0z x iy  satisfies the equation
2 2
02 2,z r  then  
(A)
1
2
(B)
1
2
(C)
1
7
(D)
1
3

Sol Solutions
Ans : C
Sol :
1
,

lie on 0 0| z z | r and | z z | 2r   
respectivelty, 2
1
R
 
 
 
(say). Here | | R 
0 02
| z | r and z 2r
R

     
2
0 0| R z | 2 z    
squaring and rewriting their equation
 4 4 2 2
0 01 4R 4R 7R 2 1 7R z z 0       .......... (1)
But by hypohesis we have
2 2
02| z | r 2 
  0 0 0 02z z z z 2      
2
0 0 0 0z z R z z 2      ...... (2)
substituting (2) in (1)
  2
0 01 7R 1 z z 0  
1
R
7
 
Section-2
(One or More Options Correct Type)
This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C)
and (D) out of which ONE or MORE are correct.
51. A line l passing through the origin is perpendicular to the lines
     1
ˆˆ ˆ: 3 1 2 4 2 ,l t i t j t k t         
     2
ˆˆ ˆ: 3 2 3 2 2 ,l s i s j s k s        
Then, the coordinate(s) of the point(s) on l2
at a distance of 17 from the point of
intersection of l and l1
is (are)
A)
7 7 5
, ,
3 3 3
 
 
 
B)  1, 1,0  C) (1,1,1) D)
7 7 8
, ,
9 9 9
 
 
 
Ans : BD

Sol Solutions
Sol :Let line passing through the origin and perpendicular to given lines is
x y
m
z

n

l
2m 2n 0   l
2 2m n 0  l
2m 2n 0 m n
2 2m n 0 2 3 2
   
 
    
l l
l
 equation of the line l is
x y z
2 3 2
 
 
and we have line 1
x 3 y 1 z 4
is
1 2 2
  
 l
Point of intersection of 1l and l is (2, –3, 2)
By given condition, 17 is distance between (2s + 3, 2s + 3, s + 2) and (2, –3, 2)
2
9s 28s 20 0   
10
s 2, s
9
    
52. Let   sin , 0f x x x x   . Then for all natural numbers n, ( )f x vanishes at
A) a unique point in the interval
1
,
2
n n
 
 
 
B) a unique point in the interval
1
, 1n n
2
 
  
 
C) a unique point in the interval (n,n+1)
D) two points in the interval (n,n+1)
Ans : BC
Sol : 1 sin x 1 x xsin x x       and
1
f n 0, n N
2
      

Sol Solutions
53. Let  
 1
4 2
1
1
k k
n
n
k
S


  k2
. Then Sn
can take value(s)
A) 1056 B) 1088 C) 1120 D) 1332
Ans : AD
Sol : 2 2 2 2 2 2 2 2
nS 1 2 3 4 5 6 7 8 . ..........upto 4n terms        
   2 2 2 2 2 2 2 2 2 2
3 1 7 5 11 9 ........... 4 2 8 6 .....

          
   2 3 1 7 5 ..... 4 2 1 4 3 . ....

        
2 2 2
8n 8n 4n 16n 4n    
ns 1056 for n 8 and 1332 for n 9   
54. For 3 3 matrices M and N, which of the following statement(s) is (are) NOT
correct?
A) NT
MN is symmetric or skew symmetric, according as M is symmetric or skew
symmetric
B) MN-NM is skew symmetric for all symmetric matrices M and N
C) M N is symmetric for all symmetric matrices M and N
D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N
Ans : CD
(A)
Sol :If  
TT T T T T
M M then N MN N M N N MN  
If  
TT T T T T
M M then N MN N M N N MN  
(B)
If      
T T TT T
M M and N N then MN NM MN NM    
=  T T T T
N M M N N M MN MN NM      
(C)
If T T
M M and N N  then  
T T T
MN N M NM 
(D)
Adj (MN) = adj (N) adj (M)

Sol Solutions
55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8:15
is converted into an open rectangular box by folding after removing squares of equal
area from all four corners. If the total area of removed squares is 100, the resulting
box has maximum volume. Then the lengths of the sides of the rectangular sheet are
A) 24 B) 32 C) 45 D) 60
Ans: AC(Approximately)
Sol: Question not clear
Section-3
(Integer Value correct Type)
This section contains 5 questions. The answer to each question is a single digit integer,
ranging from 0 to 9 (both inclusive)
56. Consider the set of eight vectors   ˆˆ ˆ : , , 1,1V ai bj ck a b c     .Three non coplanar
vectors can be chosen from V in 2P ways. Then p is.
Ans : 5
Sol :If    a, b, c 1, 1, 1   
the no. of ways in which 3 vectors are coplanar is 24
        1, 1, 1 , 1, 1, 1 . .....6ways, 1, 1, 1 , 1, 1, 1 .....6 ways     
        1, 1, 1 , 1, 1, 1 . .....6ways, 1, 1, 1 , 1, 1, 1 .....6 ways     
but total ways = 56
 non-coplanar vectors = 32
57. Of the three independent events 1E , 2E and 3E the probability that only E1
occurs is
 , only E2
occurs is  and only E3
occurs is  . Let the probability p that none of
events E1
,E2
or E3
occurs satisfy the equations    2 3 2p and p         .
All the given probabilities are assumed to lie in the interval (0,1).
Then
1
3
Probability of occurrence of E
Probability of occurrence of E

Ans : 6
58. The coefficients of three consecutive terms of  
5
1
n
x

 are in the ratio 5:10:14.
Then n =
Ans : 6
Sol:
    5 24 15 14
n 5 11 n 6
100 70

    


Sol Solutions
59. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards
are removed from the pack and the sum of the numbers on the remaining cards is
1224. If the smaller of the numbers on the removed cards is k, then k-20 =
Ans : 5
Sol :Let the removed numbers are k, k + 1
  2n n 1
1224 2k 1 n n 4k 2450
2

       
by observation n = 50, k = 25
60. A vertical line passing through the point (h,0) intersects the ellipse
2 2
1
4 3
x y
  at the
points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If
 h = area of the triangle PQR,    1 2
1/2 11/2 1
max min
hh
h and h
  
      , then 1 2
8
8
5
   
Ans : 9
Sol :Here the tangents at P and Q to the ellipse are intersect at Q on its major axis
 area
2
3 4 h
h 1
2 h 4
      
here  is decreasing from
1
to 1
2
1 2
45 9
5,
8 2
    
1 2
8
8 9
5
    

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