iit-jee 2012 solution of paper - 2 - momentum...
TRANSCRIPT
IIT-JEE 2012 Solution of Paper - 2
PPPPARARARART I :T I :T I :T I : PHY PHY PHY PHYSICSSICSSICSSICS
SECTION I : Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct.
1. An infinitely long hollow conducti9ng cylinder with inner radius / 2R and outer radius R carries a uniform
current density along its length. The magnitude of the magnetic field, | |B
as a function of the radial
distance r from the axis is best represented by :
(A)
R/2 Rr
B
(B)
R/2 Rr
B
(C)
R/2 Rr
B
(D)
R/2 Rr
B
1.(D) Field at P :
22
04
2
RJ x
Bx
µ π
π
−
=
Qx
PR/2
x
Field at Q is 0
2
IB
x
µπ
=
2. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in
water in half-submerged state. If C
ρ is the relative density of the material of the shell with respect to water,,
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 2]
then the correct statement is that the shell is :
(A) more than half-filled if C
ρ is less than 0.5
(B) more than half-filled if C
ρ is more than 1.0
(C) half-filled if C
ρ is more than 0.5
(D) less than half-filled if C
ρ is less than 0.5
2.(D)
h/2
h/2
3. In the given circuit, a charge of 80 Cµ+ is given to the upper plate of the 4 Fµ capacitor. Then in the
steady state, the charge on the upper plate of the 3 Fµ capacitor is :
+80 µC4 µF
3 µF2 µF
(A) 32 Cµ+
(B) 40 Cµ+
(C) 48 Cµ+
(D) 80 Cµ+
3.(C)80
3 2
q q−=
4µF
80µC
3µF2µFq(80 – ) q5 240q =
48q Cµ=
4. Two moles of ideal helium gas are in a rubber balloon at 30oC . The balloon is fully expandable and can be
assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly charged
to 35oC . The amount of heat required in raising the temperature is nearly
(take 8.31 / mol.R J K= )
(A) 62 J (B) 104 J (C) 124 J (D) 208 J
4.(D) 2n =
pQ nC T∆ = ∆
52 5 208
2R J= × × =
5. Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O . The
disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown
in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously
projected at an angle towards R . The velocity of projection is in the y z− plane and is same for both
pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 3]
1
8rotation, (ii) their range is less than half the disc radius, and (iii) ω remains constant throughout. Then
Q
O
P
R
(A) P lands in the shaded region and Q in the unshaded region.
(B) P lands in the unshaded region and Q in the shaded region.
(C) Both P and Q land in the unshaded region.
(D) Both P and Q land in the shaded region.
5.(A)
6. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm.
The frequency of the tuning fork is 512 .Hz The air temperature is 038 C in which the speed of sound is
336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When thefirst resonance occurs, the reading of the water level in the column is(A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm
6.(B) 4d cm= , 512f Hz= , 38o
t C= , 336 /v m s= , 0.6 2 0.6 1.2e r cm= = × =
4l e
λ+ =
0
e
l1.2
4
vl cm
f+ =
3360.164
4 512m= =
×
16.4 1.2 15.2l cm= − =
7. Two identical discs of same radius R are rotating about their axes in opposite directions with the same
constant angular speed ω . The discs are in the same horizontal plane. At time 0t = , the points P and Q
are facing each other as shown in the figure. The relative speed between the two points P and Q is v . In
one time period ( )T of rotaion of the discs, v as a function of time is best represented by
QP
R R
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 4]
(A)
rv
O Tl
(B)
rv
O Tl
(C)
O Tl
(D)
O Tl
7.(A) ˆ ˆsin cosP
V v i v jθ θ= − −
ˆ ˆsin cosQ
V v i v jθ θ= −
ˆ2 sinQP
V v iθ=
| | | 2 sin | 2 |sin |QPV v v tθ ω= =
θ θv
θ θ
P Q
v
θ = ω t
8. A loop carrying current I lies in the x y− plane as shown in the figure. The unit vector k is coming out of
the plane of the paper. The magnetic moment of the current loop is
I
(A) 2 ˆa Ik (B) 21
2a I k
π +
(C) 21
2a I k
π − +
(D) ( ) 22 1 a I kπ +
8.(B)
2
2 14
2 4
aI A I aµ π
= = + ⋅
2 ˆ1
2I a k
π = +
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 5]
SECTION II : Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on eachparagraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for Questions 9 and 10
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre ofmass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontallyat its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin
on a horizontal frictionless plane with angular speed ω , the motion at any instant can be taken as a
combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the discthrough an instantaneous vertical axis passing through its centre of mass (as is seen from the changed
orientation of points P and Q ). Both these motions have the same angular speed ω in this case.
QP
x
z
P Qy
ω
Now consider two similar systems as shown in the figure : Case (A) the disc with its face vertical and
parallel to x z− plane; Case (B) the disc with its face making an angle of 45o with x y− plane and its
horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P , and the systems are
rotated with constant angular speed ω about the z-axis.
Case (A)
x
z
P
Q
y
ω
Case (B)
x
z
P
Q
y
ω
45o
9. Which of the following statements regarding the angular speed about the instantaneous axis (passingthrough the centre of mass) is correct?
(A) It is 2 ω for both the cases
(B) It is ω for case (A); and 2
ω for case (B)
(C) It is ω for case (A); and 2 ω for case(B)
(D) It is ω for both the cases
9.(D)
10. Which of the following statements about the instantaneous axis (passing through the centre of mass) iscorrect ?
(A) It is vertical for both the cases (A) and (B)
(B) It is vertical for case (A). and is at 45o to the x z− plane and lies in the plane of the disc for case (B)
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 6]
(C) It is horizontal for case (A); and is at 45o to the x z− plane and is normal to the plane of the disc for
case (B)
(D) It is vertical for case (A); and is at 45o to the x z− plane and is normal to the plane of the disc for case
(B)10.(A)
Paragraph for Questions 11 and 12
The β -decay process, discovered around 1900 , is basically the decay of a neutron ( )n . In the laboratory,,
a proton ( )p and an electron ( )e− are observed as the decay products of the neutron. Therefore,
considering the decay of a neutron as a two-body decay process, it was predicted theoretically that thekinetic energy of the electron should be a constant. But experimentally, it was observed that the electron
kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e. e
n p e v−→ + + ,
around 1930 , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ( )e
v to
be massless and possessing negligible energy, and the neutron to be at rest, momentum and energyconservation principles are applied. From this calculation, the maximum kinetic energy of the electron is
60.8 10 eV× . The kinetic energy carried by the proton is only the recoil energy..
11. If the anti-neutrino had a mass of 23 /eV c (where c is the speed of light) instead of zero mass, what
should be the range of the kinetic energy, K , of the electron ?
(A) 60 0.8 10K eV≤ ≤ × (B) 63.0 0.8 10eV K eV≤ ≤ ×
(C) 63.0 0.8 10eV K eV≤ < × (D) 60 0.8 10K eV≤ < ×
11.(D) 60.8 10Q eV×
If υ has mass, the mass defect will be less & energy released will be less, by 3eV
60 0.8 10K eV≤ < ×
12. What is the maximum energy of the anti-neutrino?
(A) Zero (B) Much less than 60.8 10 eV×
(C) Nearly 60.8 10 eV× (D) Much larger than 60.8 10 eV×12.(C)
Paragraph for Questions 11 and 12
Most materials have the refractive index, 1n > . So, when a light ray from air enters a naturally occurring
material, then by Snell’s law, 1 2
2 1
sin
sin
n
n
θθ
= , it is understood that the refracted ray bends towards the normal.
But it never emerges on the same side of the normal as the incident ray. According to electromagnetism,
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 7]
the refractive index of the medium is given by the relation, r r
cn
vε µ
= = ± , where c is the speed of
electromagnetic waves in vacuum, v its speed in the medium, r
ε and r
µ are the relative permittivity and
permeability of the medium respectively.
In normal materials, both r
ε and r
µ are positive, implying positive n for the medium. When both r
ε and
rµ are negative, one must choose the negative root of n . Such negative refractive index materials can
now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior,
without violating any physical laws. Since n is negative, it results in a change in the direction of propagation
of the refracted light. However, simllar to normal materials, the frequency of light remains unchanged uponrefraction even in meta-materials.
13. For light incident from air on a meta-material, the appropriate ray diagram is :
(A)
θ 1
θ 2
Meta
material
Air(B)
θ1
θ 2
Meta
material
Air(C)
θ1
θ 2
Meta
material
Air(D)
θ 1
θ2
Meta
material
Air
13.(C)
14. Choose the correct statement :
(A) The speed of light in the meta-material is | |v c n=
(B) The speed of light in the meta-material is | |
cv
n=
(C) The speed of light in the meta-material is v c=
(D) The wavelength of the light in the meta-material ( )mλ is given by | |m air nλ λ= , where airλ is the
wavelength of the light in air.14.(B)
SECTION III : Multiple Correct Answer(s) Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE are correct.
15. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it.
The correct statement(s) is ( )are
(A) The emf induced in the loop is zero if the current is constant.
(B) The emf induced in the loop is finite if the current is constant.
(C) The emf in duced in the loop is zero if the current decreases at a steady rate.
(D) The emf induced in the loop is finite if the current desreases at a steady rate.
15.(A,C)
16. Six point charges are kept at the vertices of a regular haxagon of side L and centre O , as shown in the
figure. Given that 2
0
1
4
qK
Lπε= , which of the following statement (s) is ( )are correct?
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 8]
q+F L E
D2q-
q+B
2q+S T
RC
P
O
(A) The electric field at O is 6k along OD .
(B) The potential at O the line PR is same.
(C) The potential at all points on the line PR is same.
(D) The potential at all points on the line ST is same.
16.(A, B, C)
q+
D2q-
q+B
2q+
C
O
k
kk
k
2k 2k
17. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from
the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while
Q has most of its mass concentrated near the axis. Which statement (s) is ( )are correct?
(A) Both cylinders P and Q reach the ground at the same time.
(B) Cylinder P has larger linear acceleration that cylinder Q .
(C) Both cylinders reach the ground with same translational kinetic energy.
(D) Cylinder. Q reaches the ground with larger angular speed.
17.(D)
2
sinMga
IM
R
θ=
+ P Q
I I>∵P Q
a a∴ <
18. To spherical planets P and Q have the same uniform density ρ , masses pM and Q
M , and surface
areas A and 4A , respectively. A spherical planet R also has uniform density ρ and its mass is
( )P QM M+ . The escape velocities from the planets P , Q and R , are ,
P QV V and
RV , respectively..
Then
(A) Q R PV V V> > (B) R Q P
V V V> > (C) / 3R P
V V = (D) 1
/2
P QV V =
18.(B,D)
A
P
a
MP
4A
Q
2a
MQ
8QM M=PM M=
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 9]
9R
M M=
3 34 49
3 3b aπ ρ π ρ⋅ = ⋅ ⋅
( )1/ 39b a= ⋅
2e
GMv
R=∵
2P
GMv
a=
2 8 24
2Q
G M GMv
a a
⋅= = ⋅
2 / 3
1/ 3
2 9 29
9R
G M GMv
a a
⋅= = ⋅
⋅
1/ 32
9GM
a= ⋅
R Q Pv v v∴ > >
19. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on
a horizontal surface with angular speed ω and (ii) an inner disc of radius 2 R rotating anti-clockwise with
angular speed / 2ω . The ring and disc are separated by frictionless ball bearings. The system is in the
x z− plane. The point P on the inner disc is at a distance R from the origin, where OP makes an angle
of 30o with the horizontal. Then with respect to the horizontal surface,
30oP
x
z
ω/2
3R
2R
R
O
(A) the point O has a linear velocity ˆ3R iω
(B) the point P has a linear velocity 11 3 ˆˆ4 4
R i R kω ω+
(C) the point P has a linear velocity 13 3 ˆˆ4 4
R i R kω ω−
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 10]
(D) the point P has a linear velocity 3 1 ˆˆ3
4 4R i R kω ω
− +
19.(A,B)
R
2
P
3 R
300
0 0 ˆˆ ˆ3 sin30 sin302 2
P
R RV R i i k
ω ωω
= + − +
1 3 ˆˆ34 4
R i R kω ω = − +
11 3 ˆˆ4 4
R i R kω ω= +
20. In the given circuit, the AC source has 100 / .rad sω = considering the inductor and capacitor to be ideal,
the correct choice (s) is ( )are
20 V
0.5 H 50
100100 F
(A) The current through the circuit, I is 0.3 .A
(B) The current through the circuit, I is 0.3 2 .A
(C) The voltage across 100 Ω resistor = 10 2 .V
(D) The voltage across 50 Ω resister = 10 .V
20.(A,C)
100 /rad sω =
2 2
1 50 50Z = + 50 2= [ ]LX Lω=
2 2
2 100 100Z = + 100 2=
6
1
100 100 10CX −
= × ×
R
Z
V
LX
1I
tan = 1
R
Z
2I
V
cX
tan =1
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 11]
1 45oφ = & 2 45oφ =
1
20
50 2I∴ = 2
5=
2
20
100 2I =
1
5 2=
phase difference between 1I & 2I = 090
2 2
1 2I I I∴ = +
2 1
25 50= +
4 1
50
+=
1
10= 0.3A=
100 2
1x 100 10 2
5 2V I RΩ = = =
50 1
2x 50 10 2
5V I RΩ = = =
PPPPARARARART II :T II :T II :T II : CHEMISTR CHEMISTR CHEMISTR CHEMISTRYYYY
SECTION I : Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out
of which ONLY ONE is correct.
21. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are :
(A) 2O andCO respectively (B) 2O and Zn dust respectively
(C) 3HNO and Zn dust respectively (D) 3HNO and CO respectively
21.(C)
( )2 224 2Ag S NaCN NaAg CN Na S+ +
2 2 24 2Na S SO H O+ + →
2 42 4 2Na SO NaOH S+ +
( ) ( )22 42 2NaAg CN Zn Ag Na Zn CN+ → +
3Ag KNO Ag+ →
Anode - Impure Ag
Cathode - pure Ag plate
Electrolyte - 3 3AgNO HNO+
22. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus
containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the otherproduct are respectively :
(A) redox reaction; 3− and 3+ (B) redox reaction ; 3+ and 5+(C) disproporationation reaction; 3− and 5+ (D) disproporationation reaction; 3− and 3+
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 12]
22.(D) 4 24 2P NaOH H O+ + → 3 2 32 2PH Na HPO+-3 +3
23. The shape of 2 2XeO F molecule is :
(A) Trigonal bipyramidal (B) Square planer(C) tetrahedral (D) see-saw
23.(D) See-Saw but shape is see-saw because of lone pair.
Xe-O
O
F
F
24. For a diute solution containing 2.5g of a non-volatile non-electrolyte solute in 100g of water, the elevation inboiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentra-
tion of solvent, the vapour pressure (mm of Hg ) of the solution is (take 10.76
bK K kg mol
−= )
(A) 724 (B) 740 (C) 736 (D) 718
24.(A)2.5
2 0.760.1M
= ××
2.5 0.76
2M
×=
0
0
sP P n
P n N
−=
+
0
s
s
P P n
P N
−=
760 2.5 2 18
25 0.76 100
s
s
P
P
− × ×=
× ×
760 2 18 1
100 7.6 20
s
s
P
P
− ×= =
×
76020
ss
PP− =
760 20
21sP
×= 723.8= 724
25. The compound that undergoes decarboxylation most readily under mild condition is :
(A)
COOH
2CH COOH(B)
COOH
O
(C)
COOH
COOH(D)
2CH COOH
O
25.(B) β − keto acids are readily decarboxylate under mild condition because of stable carbanion is formed as an
intermediate (B) is correct :
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 13]
α
O
C OH-
O
2CO
∆− ↑
→ O
H
26. Using the data provided, calculate the multiple bond energy ( 1kJmol − ) of C C≡ bond in 2 2C H . That
energy is (take the bond energy of a C H− bond as 350 1kJ mol− )
( ) ( ) ( )2 2 22C s H g C H g+ →
1225H kJ mol−∆ =
( ) ( )2 2C s C g→
11410H kJ mol−∆ =
( ) ( )2 2H g H g→
1330H kJ mol−∆ =(A) 1165 (B) 837 (C) 865 (D) 815
26.(D) ( ) ( ) ( )2 2 22C s H g C H g+ → ;
225H kJ∆ = ..1
( ) ( )2 2C s C g→ ; 1410H kJ∆ = ...2
( ) ( )2 2H g H g→ ; 330H kJ∆ = .. 3
This can be get by substrating 1, 2 & 3
( ) ( ) ( )2 22 2C g H g C H g+ → ;
1515H kJ∆ = −
2 52
C H C H C CH H H− ≡∆ = − ∆ − ∆
( )1515 2 350C C
H ≡+ = + + ∆
815C C
H ≡∆ =27. The major product H of the given reaction sequence is :
3 2 3CH CH CO CH− − − CNG
−
→
2 495%H SO
HeatH→
(A) 3CH CH C COOH- = -
3CH(B)
3CH CH C CN- = -
3CH
(C) 3 2CH CH C COOH- - -
3CH
OH
(D) 3 2CH CH C CO NH- = - -
3CH
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 14]
27.(A)3 2CH CH C O- - -
3CH
CN
3 2CH CH C- -C N
O3
CHG
3 2CH CH C- -C N
OH3CH
3 2CH CH C- -C OH-
OH3CH
O
95%
2 4H SO( )2H O5%
Hydrolysis
β α
2H O-
Hydrolysis
3CH CH C C OH- = - -
O
3CH
28. ( ) ( )2 2 5 6 5 22 NiCl C H C H exhibits temperature dependent magnetic behaviour (paramagnetic/diamag-
netic). The coordination geometries of 2Ni + in the paramagnetic and diamagnetic states are respectively..
(A) tetrahedral and tetrahedral (B) square planar and square planar(C) tetrahedral and square planar (D) square planar and tetrahdral
28.(C) [ ]2 83Ni Ar d+ →
When it is paramagnetic
Hybridisation → 3sp
Shape → tetrahedral
When it is diamagnetic
Hybridisation → 2dsp
Shape →square planar
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 15]
SECTION II : Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on eachparagraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph : 1In the following reaction sequence, the compound J is an intermediate.
( ) ( )( )( )
3 2 2
3 2
3
, /
.
CH CO O i H Pd C
CH COONa ii SOCl
iii anhyd AlCl
I J K→ →
( )9 8 2J C H O gives effervescence on treatment with 3NaHCO and a positive Baeyer’s test.
29. The compound K is
(A) (B) O
(C)
O
(D) O
30. The compound I is
(A)
O H
(B)
OH
(C)
O 3CH
(D)
HH
H
29 & 30∵ J ( )9 8 2C H O gives effervescences with
3NaHCO & positive Baeyer’s test ∴ it has C OH- -
Ogroup & unsaturation.
2CHH
3 3CH C O C CH- - - -
3CH C ONa- -
O O
O
CH CH C OH= - -
O
Reduce onlydouble bond
2 | |H Pd C
2 2CH CH C O- - =
OH
2SOCl
NPSR→
2 2CH CH C O- - =
Cl
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 16]
F.C.acylation ( )3 .AlCl Anhyd
2 2CH CH C O- - =
OK
29.(C)30.(A)Paragraph : 2
The electrochemical cell shown below is concentration cell.
2|M M+ (saturated solution of a sparingly soluble salt,
2MX ) || ( )2 30.001 |M mol dm M+ −
The emf of the cell depends on the difference in concentration of 2M
+ ions at the two electrodes.
The emf of the cell of 298 K is 0.059 V.
31. The value of ( )1G kJ mol
−∆ of the given cell is (take 11 96500F C mol−= )
(A) 5.7− (B) 5.7
(C) 11.4 (D) 11.4−31.(D) G nFE∆ = −
2 96500 0.059= − × × 111387J Mol−= −
111.387kJ Mol−= − 111.4 kJ mol
−−
32. The solubility product (3 9;spK mol dm ) of 2MX at 298 K based on the information available for the given
concentration cell is ( take 2.303 298 / 0.059R F V× × = )
(A) 151 10−× (B) 154 10−× (C) 121 10−× (D) 124 10−×
32.(D)2 2| |
C aM M
+ +→
0 0Cell
E = ( Assuming concentration cell)
2
2
|0.059log
2 |
a
C
mE
m
+
+
−=
( )1/3
4
3
/0.0590.059 log
2 10
spK
−
−=
1/3
9
0.0590.059 log
2 4 10
spK−
−= ×
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 17]
( )32
910
4 10
spK−
−=
×154 10spK
−= ×
Paragraph : 3Bleaching powder and bleach solution are produced on a large scale and used in several house holdproducts. The effectiveness of bleach solution is often measured by iodometry.
33. Bleaching powder contians a salt of an oxoacid as one of its components. The anhydride of that oxoacid is:
(A) 2Cl O (B) 2 7Cl O (C) 2ClO (D) 2 6Cl O
33.(A) Bleaching powder → ( )Ca OCl Cl
acid of ClO− is HClO
and anhydride of that acid is 2Cl O
34. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4N acetic acid. In
the titration of the liberated iodine, 48 mL of 0.25 N 2 2 3Na S O was used to reach the end point. The molarity
of the household bleach solution is :(A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 M
34.(C) 3 2. .B S CH COOH Cl+ →
2 2I Cl I Cl
− −+ → + ....... (1)
2 2
2 2 3 4 6I S O I S O
− − −+ → + ....... (2)
meq. of 2
2 3S O
−= meq. of 2I liberated
48 0.25 1× × = m. mols of 2 2I ×
m.mols of 2I = 24 0.25×
m. mols of 2I liberated = m. moles of Bleach sol.
24 0.25= ×So, 24 0.25 25 M× = ×
.24M = 0
SECTION III : Multiple Correct Answer(s) Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE are correct.
35. The given graphs /data I, II, III and IV represent general trends observed for different physisorption andchemisorption processes under mild conditions of temperature and pressure. Which of the following choice(s)about I, II, III and IV is (are) correct ?
Am
ount o
f gas
adso
rbed
P constantI
T
Am
ount o
f gas
adso
rbed
P constantII
T
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 18]
Am
ount o
f gas
adso
rbed
III
Po
tential E
nerg
y
IV
0
1150ads
H KJmol-=
adsE
P
200k
250k
(A) I is physisorption and II is chemisorption(B) I is physisorption and III is chemisorption(C) IV is chemisorption and II is chemisorption(D)IV is chemisorption and III is chemisorption
35.(A,C)36. For the given aqueous reactions, which of the statement (s) is (are) true?
excess ( )3 6[ ]KI K Fe CN+
dilute 2 4H SO
4ZnSO
2 2 3Na S O
brownish-yellow solution
white precipitate + brownish -yellow filtrate
colourless solution
(A) The first reaction is a redox reaction
(B) white precipitate is ( )3 6 2Zn Fe CN
(C) Addition of filtrate to starch solution gives blue colour
(D) White precipitate is soluble in NaOH solution
36(A,C,D)
Excess ( )2 4
.
3 6
dil
H SOKI K Fe CN + →
( )4 36[ ]K Fe CN I
-+
+24ZnSO→
( )2 3 2 26
[ ]K Zn Fe CN I
white ppt Brownish yellow filtrate
+
−
NaOH 2 2 3Na S O
( )
( )
2
4
4
6
[ ]
[ ]
Zn OH
Fe CN
−
−
+ ( )
2
4 6I S O
Colourless
− −+
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 19]
37. With respect to graphite and diamond, which of the statement(s) given below is (are) correct?(A) Graphite is harder than diamond(B) Graphite has higher electrical conductivity than diamond(C) Graphite has higher thermal conductivity than diamond
(D) Graphite has higher C C− bond order than diamond.
37. (C,D) diamond is harder than graphitedue to delocalisation graphite showconductivity.graphite has partial double bond character so greater bond order.
38. With reference to the scheme given, which of the given statment(s) about T, U, V and W is (are) correct?
O
T3H C
4LiAlH
3 /CrO H 3 2( )CH CO O
V U W
O
+
(A) T is soluble in hot aqueous NaOH(B) U is optically active
(C) Molecular formula of W is 10 18 4C H O
(D) V gives effervescence on treatment with aqueous 3
NaHCO
38.(C,D)
O
3H C T
4LiAlH
OH
3H C U
OH
Opticallyinactive
3CH C O- -
O
3CH C-
O
3O C CH- -
3H C W
O
3O C CH- -
O
10 18 4C H O
3 /CrO H+
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 20]
3H CO
O
OO H
3H C
V
OH
O
OH
O
Give effervescences
on treatment with aq. 3NaHCO
∴ (C,D) both are correct
39. Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ?
3CH
HO
Cl
HO H
H
M
3CH
OHCl
HO H
H
N3CH
OH
Cl
HOH
H
O
3CH
ClHHO
H OH
P
3CH
ClHHO
H
Q
HO
(A) M and N are non-image stereoisomers(B) M and O are identical(C) M and P are enantiomers
(D) M and Q are identical
39.(A,B,C)
HO H
H
M
3H C
Cl
HO H
O 3CH
OHHO
H
HCl
HO H
M O=
3CH
Cl
HO H
.
3CH
ClHHO
H
P
OH
H
3CH
Cl
OHH
HO
H
3CH
Cl
H
OH
P
OH
P & M are non-super imposable
mirror image C
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 21]
40. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure.Which of the following statement(s) is (are) correct ?
isotermal
adiabatic
( )2 2 2,P V T
( )3 2 3,P V T
P
V
(A) 1 2T T= (B) 3 1T T>
(C) isotermal adiabatic
W w> (D) isotermal adiabatic
U U∆ > ∆40.(A,C,D) In expansion temperature of adiabatic will decrease but of Isothermal reamains same
So 1 2T T= for Isothermal
2 3T T>Internal energy for adiabatic system will decrease
So, iso adi
V V∆ > ∆
| | | |iso adi
w w>
PPPPARARARART III :T III :T III :T III : MA MA MA MATHEMATHEMATHEMATHEMATICSTICSTICSTICS
SECTION I : Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONLY ONE is correct.
41. Let PQR be a triangle of area ∆ with 7
2,2
a b= = and 5
2c = , where ,a b and c are the lengths of the
sides of the triangle opposite to the angles at ,P Q and R respectively. Then 2sin sin 2
2sin sin 2
P P
P P
−+
equals
(A) 3
4∆(B)
45
4∆(C)
23
4
∆
(D)
245
4
∆
41.(C)
P
Q
2a =R
7 / 2b =5 / 2c =
2sin 2sin cos
2sin 2sin cos
p p p
p p p
−
+
1 cos
1 cos
p
p
−=
+
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 22]
2 2 27 5 4cos
2.7.5p
+ −=
where cos 58 29
1 2.75 35
p= =
1 cos 35 29 3
1 cos 35 29 32
p
p
− −∴ = =
+ +
( )( )( )s s a s b s c∆ = − − −1 3
4 2 62 2
= × × × =
23 9 3
4 16 62 32
= = ∆ ×
42. If a
and b
are vectors such that 29a b+ =
and ( )ˆˆ ˆ2 3 4a i j k× + + ( )ˆˆ ˆ2 3 4i j k b= + + ×
, then a possible
value of ( ) ( )ˆˆ ˆ7 2 3a b i j k+ ⋅ − + +
is
(A) 0 (B) 3 (C) 4 (D) 8
42. (C) 0a v v b× − × =
0a v b v× + × =
( ) 0a b v+ × =
(2 3 4 )a b k i j k+ = + +
| | | | 29a b k+ =
29 | | 29 1k k= ⇒ = ± ( )( 7 2 3 )a b i j k∴ + − + +
(2 3 4 ).( 7 2 3 )i j k i j k= ± + + − + + ( 14 6 12) 4= ± − + + = ±
43. If P is a 3 3× matrix such that 2 ,TP P I= + where T
P is the transpose of P and I is the 3 3× identity
matrix, then there exists a column matrix
0
0
0
x
X y
z
= ≠
such that
(A)
0
0
0
PX
=
(B) PX X= (C) 2PX X= (D) PX X= −
43.(D) 2TP P I= +
( ) (2 )T T TP P I= +
2 TP P I= +1
2 ( )2
P I P I∴ + = −
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 23]
4 2P I P I+ = − 3 3P I⇒ = −
P I= − 0P I∴ + =
( ) 0P I X∴ + = 0PX X⇒ + =
44. Let ( )aα and ( )aβ be the roots of the equation
( ) ( ) ( )23 61 1 1 1 1 1 0a x a x a+ − + + − + + − =
where 1a > − .
Then ( )0
lima
aα+→ and ( )
0lim
aaβ+→
are
(A) 5
2− and 1 (B)
1
2− and 1− (C)
7
2− and 2 (D)
9
2− and 3
44.(B) 3
1 1
1 1
a
aα β
+ −+ = − + −
6 1 1
3 1 1
a
aαβ
+ −= + −
( )( )
1/ 32 / 3(1 ) 1 1
1 1
a a
aα β
+ + + + ⇒ + = − + +
( )1
6 1 1aαβ =
+ +
( )0
3lim
2xα β
+→
−+ = ( )
0
1lim
2xαβ
+→=
45. Four fair dice 1 2 3, ,D D D and
4,D each having six faces numbered 1,2,3,4,5 and 6 , are rolled
simultaneously. The probabillity that 4
D shows a number appearing on one of 1 2,D D and
3D is
(A) 91
216(B)
108
216(C)
125
216(D)
127
216
45. (A) Favrable out comes 46 6 5 5 5= − × × ×
Total outcomes 46=
Probability
4
4
6 6 5 5 5 91
2166
− × × ×= =
46. The value of the integral
/ 2
2
/ 2
ln cosx
x x dxx
π
π
ππ−
+ + − ∫
(A) 0 (B)
2
42
π− (C)
2
42
π+ (D)
2
2
π
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 24]
46.(B)
/ 22
/ 2
ln cosx
x xdxx
π
π
ππ
−
+ + − ∫
/ 22
/ 2
cosx x dx
π
π−
= ∫/ 2
2
0
2 cosx x dx
π
= ∫2
42
π= −
47. Let 1 2 3, , ,....a a a be in harmonic progression with
1 5a = and 20 25a = . The least positive integer n for
which 0na < is
(A) 22 (B) 23 (C) 24 (D) 25
47. (D)1
5a
=
125
19a d=
+
1 1 1
19 25 5d
∴ = −
1 20 4
19 25 5 19 25
− − = = × ×
1
1 4( 1)
5 19 25
na
n
=− + − ×
0na∴ <
1 4( 1) 0
5 19 25n
−⇒ + − <
×
1 4( 1)
5 19 25
n −<
×
19 251
4 5n
×⇒ + <
×
951
4n∴ > +
324
4n > 25n∴ =
48. The equation of a plane passing through the line of intersection of the planes 2 3 2x y z+ + = and
3x y z− + = and at a distance 2
3 from the point ( )3,1, 1− is
(A) 5 11 17x y z− + = (B) 2 3 2 1x y+ = − (C) 3x y z+ + = (D) 2 1 2x y− = −
48.(A)
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 25]
2 3 ( 3) 0x y z x y zλ+ + + − + − =
(1 ) (2 ) (3 ) ( 2 3 ) 0x y zλ λ λ λ+ + − + + + − − =
2 2 2
| 3(1 ) (2 ) (3 ) ( 2 3 ) | 2
3(1 ) (2 ) (3 )
λ λ λ λ
λ λ λ
+ + − − + + − −=
+ + − + +
2
2
( 2 ) 4
33 4 14
λ
λ λ
−=
+ +2 23 3 4 14λ λ λ⇒ = + +
7 / 2λ = −
2( 2 3 2) 7( 3) 0x y z x y z∴− + + − + − + − =
5 11 17 0x y z− + − =
SECTION II : Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on eachparagraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for questions 49 and 50
A tangent PT is drawn to the circle 2 2 4x y+ = at the point ( )3,1P . A straight line L , perpendicular to
PT is a tangent to the circle ( )2 23 1x y− + = .
49. A common tangent of the two circles is
(A) 4x = (B) 2y = (C) 3 4x y+ = (D) 2 2 6x y+ =
49. (D) Meeting point of direct common tangent 2 3 1 0
,02 1
P× − × = −
(0,0) (3,0)
2x =
(6,0)=
If slope m=
0 ( 6)y m x− = −
is tangent to 2 2 4x y+ =
2 29 1m m= +
2 16 2 1
2 2m m m− = ± + ⇒ = ±
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 26]
50. A possible equation of L is
(A) 3 1x y− = (B) 3 1x y+ = (C) 3 1x y− = − (D) 3 5x y+ =
50.(A) Equation of PT is
3 4 0x y+ − = ∴ Slope of 1
3L =
possible equation is
21 1
0 ( 3) 1 13 3
y x
− = − ± +
3 ( 3) 2y x= − ±
3 1x y− = or 5
Paragraph for questions 49 and 50
Let ( ) ( )2 2 21 sinf x x x x= − + for all x IR∈ , and let ( ) ( ) ( )1
2 1ln
1
x tg x t f t dt
t
− = − + ∫ for all ( )1,x ∈ ∞ .
51. Consider the statements :
P : There exists some x IR∈ such that ( ) ( )22 2 1f x x x+ = +
Q : There exists some x IR∈ such that ( ) ( )2 1 2 1f x x x+ = +
Then
(A) both P and Q are true (B) P is true and Q is false
(C) P is false and Q is true (D) both P and Q are false
51. (C) 2 2 2( ) (1 ) sinf x x x x x R= − + ∀ ∈ 2 2 2 2:(1 )sin 2 2(1 )P x x x x x∴ − + + = +
2 2 2 2(sin )(1 ) 2 2 2x x x x x− = + − −
2 2 2(1 ) sin 2 2x x x x− = − +
2 2 2(1 ) sin ( 1) 1x x x− = − +
2 2(1 ) (sin 1) 1x x− − =
2
2
1cos
(1 )x
x
−=
−∴ no values of x exists so P false
Now, 2 2 22 (1 ) 1 2(1 ) sin 2x x x x x+ − = − +
2 22 1 2(1 ) sinx x x− = −
2
2
1 1sin
( 1) 2( 1)x
x x= +
− −
Now the value of R.H.S. will be from some large negative quantity to some large positive value, and value
of L.H.S. will be from 0 to 1. ∴ some value of x will exists.
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 27]
52. Which of the following is true?
(A) g is increasing on ( )1, ∞ (B) g is decreasing on ( )1, ∞
(C) g is increasing on ( )1,2 and decreasing on ( )2,∞
(D) g is decreasing on ( )1,2 and increasing on ( )2,∞
52.(B) 2 2 2( ) (1 ) sinf x x x x x R= − + ∀ ∈
1
2( 1)( ) ln ( ) (1, )
1
x tg x t f t x
t
− = − ∀ ∈ ∞ + ∫
2( 1)'( ) ln ( ) (1, )
1
xg x x f x x
x
− = − ∀ ∈ ∞ +
( )f x always (+ve)
Let 2( 1)
( ) ln1
xh x x
x
−= −
+
2
4 1'( )
( 1)h x
xx= −
+
2
2
( 1)'( ) 0 (1, )
( 1)
xh x x
x x
− −= < ∀ ∈ ∞
+
( )h x⇒ is decreasing (1, )x∈ ∞ ( ) (1) (1, )h x h x⇒ < ∀ ∈ ∞
( ) 0h x⇒ < '( ) 0 0 (1, )g x x< ∀ ∈ ∞ ( )g x⇒ is decreasing (1, )x∀ ∈ ∞
Paragraph for questions 53 and 54
Let n
a denote the number of all n -digit positive integers formed by the digits 0,1 or both such that no
consecutive digits in them are 0 . Let n
b = the number of such n -digit integers ending with digit 1 and n
c =
the number of such n - digit integers ending with digit 0 .
53. The value of 6
b is
(A) 7 (B) 8 (C) 9 (D) 11
53. (B) 46 2 (5 2 1) 8b = − + + =
54. Which of the follwing is correct?
(A) 17 16 15
a a a= + (B) 17 16 15
c c c≠ + (C) 17 16 16
b b c≠ + (D) 17 17 16
a c b= +
54. (A) 1 1a = 2 2a = 3 3a = 4 5a = 5 8a = 6 13a =
5 3 4a a a= +
6 4 5a a a= +............................................................
17 15 16a a a= +
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 28]
SECTION III : Multiple Correct Answer(s) Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) outof which ONE or MORE are correct.
55. If ( ) ( )( )2
02 3
xt
f x e t t dt= − −∫ for all ( )0,x ∈ ∞ , then
(A) f has a local maximum at 2x =
(B) f is decreasing on ( )2,3
(C) there exists some ( )0,c ∈ ∞ such that ( )" 0f c =
(D) f has a local minimum at 3x =
55.(A,B,C,D)2
'( ) ( 2)( 3) (0, )xf x e x x x− − − ∀ ∈ ∞
'( ) 0 (2,3)f x x< ∀ ∈
( )f x⇒ is increasing (2,3)x∀ ∈
sing of '( )f x
2 3
f⇒ has local maxima at 2x =
f⇒ has local minima at 2x =
'(2) '(3) 0f f= =∵
∴ from roolle’s Theorem ''( ) 0f c = for some ' 'in (0, )c ∞
56. If the straight line 1 1
2 2
x y z
k
− += = and
1 1
5 2
x y z
k
+ += = are coplanar, then the plane(s) containing these
two lines is (are)
(A) 2 1y z+ = − (B) 1y z+ = − (C) 1y z− = − (D) 2 1y z− = −
56. (B,C) lines are coplanar
1 1 1 1 0 0
2 2 0
5 2
k
k
+ − + −
=
22( 4) 0 2, 2k k− = ⇒ = −
0.( 1) 1( 1) 1( 0) 0x y z⇒ − + + − − = 1 0y z⇒ − + =
0.( 1) 1( 1) 1( 0) 0x y z− + + + − =
1 0y z+ + =
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 29]
57. Let X and Y be two events such that 1 1
( | ) , ( | )2 3
P X Y P Y X= = and 1
( )6
P X Y∩ = which of the follow-
ing is (are) correct?
(A) 2
( )3
P X Y∪ = (B) X and Y are independent
(C) X and Y are not independent (D) 1
( )3
cP X Y∩ =
57. (A,B)
( ) 1 1( )
( ) 2 3
X P X YP P Y
Y P Y
∩ = = ⇒ =
( ) 1 1' ( )
( ) 3 2
Y P Y XP P X
X P X
∩ = = ⇒ =
( ) ( ) ( )P X Y P X P Y∩ =∵ &X Y∴ are independent
( ) ( ) ( ) ( )P X Y P X P Y P X Y∪ = + − ∩1 1 1 2
2 3 6 3= + − =
( ) ( ) ( )cP X Y P Y P X Y∩ = − ∩1 1 1
3 6 6= − =
58. If the adjoint of a 3 3× matrix P is
1 4 4
2 1 7
1 1 3
, then the possible value(s) of the determinant of P is (are)
(A) 2− (B) 1− (C) 1 (D) 2
58. (A,D)
1 4 4
2 1 7
1 1 3
adj P
=
2
1 4 4
| | 2 1 7 4 4 4 4
1 1 3
P = = − + + =
| | 2, 2P = −
59. Let : ( 1,1)f IR− → be such that 2
2(cos 4 )
2 secf θ
θ=
− for 0, ,
4 4 2
π π πθ ∈ ∪
Then the value(s) of
1
3f
is (are)
(A) 3
12
− (B) 3
12
+ (C) 2
13
− (D) 2
13
+
JABALPUR - (0761) 2400022/8/9, NAGPUR - (0712) 3221105, GWALIOR - (0751) 3205976, AMRAVATI - (0721) 2572602,
HISAR - (01662) 296006, REWA - (07662) 421444, BALAGHAT - 324766
www.momentumacademy.com [ 30]
59. (A,B) 2
2(cos 4 )
2 secf θ
θ=
−
2
2
2cos(cos 4 )
2cos 1f
θθ
θ=
−
1 cos2(cos 4 )
cos2f
θθ
θ+
=
2 1(2cos 2 1) 1
cos2f θ
θ⇒ − = + 2 1
(2 1) 1f xx
⇒ − = +
2
3x = ±
1 11
3
2 / 3
f
⇒ = + ±
31
2= ±
60. For every integer n , let na and nb be real numbers. Let function :R IR IR→ be given by
sin , for [2 ,2 1]( ) ,
cos , for (2 1,2 )
n
n
a x x n nf x
b x x n n
π
π
+ ∈ +=
+ ∈ − for all integers n . If f is continuous, then which of the following
hold(s) for all n
(A) 1 1 0n na b− −− = (B) 1n na b− = (C) 1 1n na b +− = (D) 1 1n na b− − = −
60. (B,D) sin [2 ,2 1]
( )cos (2 1,2 )
n
n
a x x n nf x
b x x n n
π
π
+ ∀ ∈ +
+ ∀ ∈ −
Let 'f in continuous at 2x n= ⇒ L.H.L. = R.H.L (2 )f n=
2 2lim (2 ) lim (2 ) (2 )
h n h nf n h f n h f n
+ +→ →⇒ − = + = cos 2 sin 2 sin 2n n nb n a n a nπ π π⇒ + = + = +
1n nb a⇒ + = 1n na b⇒ − =
R.H.L of (2 1) cos (2 1)nn b nπ− = + − 1nb= −
In given function replasing ‘n’ by 1n −
[ ]1 sin 2 2,2 1na x x n nπ− + ∀ ∈ − −
[ ]1 cos 2 3,2 2nb x x n nπ− + ∀ ∈ − −
Now, L.H.L at 1(2 1) sin (2 1)nn a nπ−− = + − 1 0na −= +
1(2 1) nf n a −− =
11n nb a −− =