IIT JEE 2011 Solved Paper by Prabhat Gaurav

Solutions of IIT-JEE-2011
CODE
PAPER 2
Time: 3 Hours
Time: 3 Hours Maximum Marks: 240
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
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13. The question paper consists of 3 parts (Chemistry, Physics and Mathematics). Each part consists of four
sections.
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bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases,
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bubble(s) corresponding to the correct answer(s) ONLY and zero marks other wise. There are no negative
marks in this section.
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bubble corresponding to the correct answer and zero marks otherwise There are no negative marks in this
section.
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you have darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks
otherwise. Thus each question in this section carries a maximum of 8 Marks. There are no negative marks
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IITJEE2011-Paper 2-CPM-2
PAPER-2 [Code – 5]
IITJEE 2011
SECTION – I (Total Marks : 24)
(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.
1. The freezing point (in o
C) of a solution containing 0.1 g of K3[Fe(CN)6 (Mol. Wt. 329) in 100 g of water
(Kf = 1.86 K kg mol−1
) is
(A) −2.3 × 10−2
(B) −5.7 × 10−2
(C) −5.7 × 10−3
(D) −1.2 × 10−2
Sol. (A)
( ) ( )
3
3 6 6
K Fe CN 3K Fe CN
−+
   → +   
i = 4
2
f f
m 1000 0.1 1000
T K i 1.86 4 2.3 10
M W 329 100
−
∆ = × × × = × × × = ×
' 2
fT 2.3 10−
= − ×
2. Amongst the compounds given, the one that would form a brilliant colored dye on treatment with NaNO2 in
dil. HCl followed by addition to an alkaline solution of β-naphthol is
(A)
N(CH3)2
(B)
NHCH3
(C)
NH2
CH3
(D)
CH2NH2
Sol. (C)
CH3 NH2
2
o o
NaNO HCl
O C 5 C
+
−
→ CH3 N N Cl
CH3 N N Cl +
H
OH
( )Neptholβ −
CH3 N N
OH
( )
OH
coupling
−
→
Brilliant Colour Dye
PART - I: CHEMISTRY

3. The major product of the following reaction is
O
( )
2RCH OH
H anhydrous⊕→
(A) a hemiacetal (B) an acetal
(C) an ether (D) an ester
Sol. (B)
O
+ H
+
O O O CH2 R
R
O H
Acetal
4. The following carbohydrate is
O
H
OH
OH
H
H
OH
OH
H
H
OH
(A) a ketohexose (B) an aldohexose
(C) an α-furanose (D) an α-pyranose
Sol. (B)
5. Oxidation states of the metal in the minerals haematite and magnetite, respectively, are
(A) II, III in haematite and III in magnetite (B) II, III in haematite and II in magnetite
(C) II in haematite and II, III in magnetite (D) III in haematite and II, III in magnetite
Sol. (D)
Haematite : Fe2O3 : 2x + 3 × (-2) = 0
x =3
Magnetite : Fe3O4 [an equimolar mixture of FeO and Fe2O3]
FeO : x – 2 = 0 ⇒ x = 2
Fe2O3 : x = 3
6. Among the following complexes (K−−−−P)
K3[Fe(CN)6] (K), [Co(NH3)6]Cl3 (L), Na3[Co(oxalate)3] (M), [Ni(H2O)6]Cl2 (N), K2[Pt(CN)4] (O) and
[Zn(H2O)6](NO3)2 (P)
(A) K, L, M, N (B) K, M, O, P
(C) L, M, O, P (D) L, M, N, O
6. (C)
Following compounds are diamagnetic.
L : [Co(NH3)6]Cl3
M : Na3[Co(Ox)3]
O : K2[Pt(CN)4]
P : [Zn(H2O)6](NO3)2

7. Passing H2S gas into a mixture of Mn2+
, Ni2+
, Cu2+
and Hg2+
ions in an acidified aqueous solution
precipitates
(A) CuS and HgS (B) MnS and CuS
(C) MnS and NiS (D) NiS and HgS
Sol. (A)
H2S in presence of aqueous acidified solution precipitates as sulphide of Cu and Hg apart from Pb+2
, Bi+3
,
Cd+2
, As+3
, Sb+3
and Sn+2
.
8. Consider the following cell reaction:
( ) ( ) ( ) ( ) ( )
2 o
2 2s g aq aq
2Fe O 4H 2Fe 2H O E 1.67 V+ +
+ + → + =ℓ
At [Fe2+
] = 10-3
M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25o
C is
(A) 1.47 V (B) 1.77 V
(C) 1.87 V (D) 1.57 V
Sol. (D)
( ) ( ) ( ) ( ) ( )2
2 22Fe s O g 4H aq 2Fe aq 2H O+ +
+ + → + ℓ
N = 4 (no. of moles of electron involved)
From Nernst’s equation,
Ecell = o
cell
0.0591
E log Q
n
−
=
( )
( )
23
43
10.0591
1.67 log
4 0.1 10
−
−
0
−
×
{ }pH
H 10+ −
  = ∵
= 1.67 – 0.106
= 1.57 V
SECTION – II (Total Marks : 16)
(Multiple Correct Answer(s) Type)
This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE OR MORE may be correct.
9. Reduction of the metal centre in aqueous permanganate ion involves
(A) 3 electrons in neutral medium (B) 5 electrons in neutral medium
(C) 3 electrons in alkaline medium (D) 5 electrons in acidic medium
Sol. (A, D)
In acidic medium
2
4 2M nO 8H 5e M n 4H O− + − +
+ + → +
In neutral medium
4 2 2MnO 2H O 3e MnO 4OH− − −
+ + → +
Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.
10. The correct functional group X and the reagent/reaction conditions Y in the following scheme are
( ) ( )i Y
2 4
X CH X condensation polymer− − →
C (CH2)4 C
O
OHOH
O
( )ii
heat
(A) X = COOCH3, Y = H2/Ni/heat (B) X = CONH2, Y = H2/Ni/heat
(C) X = CONH2, Y = Br2/NaOH (D) X = CN, Y = H2/Ni/heat

Sol. (A, B, C, D)
Condensation polymers are formed by condensation of a diols or diamine with dicarboxylic acids.
Hence, X may be
||
O
C OR− − or
||
2
O
C NH− − or C N− ≡
H3CO C
O
(CH2)4 C
O
OCH3 ( )2H /Ni
2 2 24
OH H C CH CH OH→ − − −
OH C
O
(CH2)4 C
O
OH
Poly ester−
NH2 C
O
(CH2)4 C NH2
O
( )2H /Ni
2 2 2 2 24
H N CH CH CH NH→ − − −
OH C
O
(CH2)4 C
O
OH
Polyamide
2Br / OH ,−
∆
Hoffmann Bromamide Reaction
(CH2)4 NH2NH2
OH C
O
(CH2)4 C
O
OH
Polyamide
( ) ( )2H /Ni
2 2 2 2 2 24
N C CH C N H N CH CH CH NH≡ − − ≡ → − − − −
OH C
O
(CH2)4 C
O
OH
Polyamides
11. For the first order reaction
( ) ( ) ( )2 5 2 22N O g 4NO g O g→ +
(A) the concentration of the reactant decreases exponentially with time
(B) the half-life of the reaction decreases with increasing temperature
(C) the half-life of the reaction depends on the initial concentration of the reactant
(D) the reaction proceeds to 99.6% completion in eight half-life duration
Sol. (A, B, D)
For first order reaction
[A] = [A]0e–kt
Hence concentration of [NO2] decreases exponentially.
Also, t1/2 =
0.693
K
. Which is independent of concentration and t1/2 decreases with the increase of
temperature.
99.6
2.303 100
t log
K 0.4
 
=  
 
( )99.6 1/2
2.303 0.693
t 2.4 8 8 t
K K
= = × =
12. The equilibrium
I o II
2Cu Cu Cu+⇌
in aqueous medium at 25o
C shifts towards the left in the presence of
(A) 3NO−
(B) Cl−
(C) SCN−
(D) CN−
Sol. (B, C, D)
Cu2+
ions will react with CN−
and SCN−
forming [Cu(CN)4]3−
and [Cu(SCN)4]3−
leading the reaction in the
backward direction.
( )2
2
Cu 2CN Cu CN+ −
+ →
( ) ( )2 2
2Cu CN 2CuCN CN→ +

( )
3
4
CuCN 3CN Cu CN
−−
 + →  
( )
32
4
Cu 4SCN Cu SCN
−+ −
 + →  
Cu2+
also combines with CuCl2 which reacts with Cu to produce CuCl pushing the reaction in the backward
direction.
2CuCl Cu 2CuCl+ → ↓
SECTION-III (Total Marks : 24)
(Integer Answer Type)
This section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9.
The bubble corresponding to the correct answer is to be darkened in the ORS.
13. The maximum number of isomers (including stereoisomers) that are possible on monochlorination of the
following compound is
C
CH3CH2 CH2CH3
CH3
H
Sol. (8)
CH3CH2C
H
CH3
CH2CH2Cl Enantiomeric pair = 2
CH3CH2C
H
CH3
CHCH3 Two Enantiomeric pairs = 4
CH3CH2C
Cl
CH3
CH2CH3 1
CH3CH2C
H
CH2Cl
CH2CH3 1
Total = 2 + 4 + 1 + 1 = 8
14. The total number of contributing structure showing hyperconjugation (involving C−H bonds) for the
following carbocation is
CH3 CH 2CH 3

Sol. (6)
6 × H−atoms are there
15. Among the following, the number of compounds than can react with PCl5 to give POCl3 is
O2, CO2, SO2, H2O, H2SO4, P4O10
Sol. (5)
16. The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL
of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to
Sol. (6)
Number of ionisable Cl−
in [Cr(H2O)5Cl]Cl2 is 2
∴ Millimoles of Cl−
= 30 × 0.01 × 2 = 0.6
∴ Millimoles of Ag+
required = 0.6
∴ 0.6 = 0.1 V
V = 6 ml
17. In 1 L saturated solution of AgCl [Ksp(AgCl) = 1.6 × 10−10
], 0.1 mol of CuCl [Ksp(CuCl) = 1.0 × 10−6
] is
added. The resultant concentration of Ag+
in the solution is 1.6 × 10−x
. The value of “x” is
Sol. (7)
Let the solubility of AgCl is x mollitre-1
and that of CuCl is y mollitre-1
AgCl Ag Cl
x x
+ −
+??⇀↽??
CuCl Cu Cl
y y
+ −
+??⇀↽??
∴Ksp of AgCl = [Ag+
][Cl–1
]
1.6 × 10 -10
= x(x + y) … (i)
Similarly Ksp of CuCl = [Cu+
][Cl-
]
1.6 × 10-6
= y(x + y) … (ii)
On solving (i) and (ii)
[Ag+
] = 1.6 × 10-7
∴ x = 7
18. The number of hexagonal faces that are present in a truncated octahedron is
Sol. (8)
SECTION-IV (Total Marks : 16)
(Matrix-Match Type)
This section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five
statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with ONE
or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the
statements given q and r, then for the particular question, against statement B, darken the bubbles corresponding to q
and r in the ORS.
19. Match the transformations in column I with appropriate options in column II
Column I Column II
(A) ( ) ( )2 2CO s CO g→ (p) phase transition
(B) ( ) ( ) ( )3 2CaCO s CaO s CO g→ + (q) allotropic change
(C) ( )22H H g→i (r) H∆ is positive
(D) ( ) ( )white, solid red, solid
P P→ (s) S∆ is positive
(t) S∆ is negative

Sol. (A) →→→→ (p, r, s) (B) →→→→ (r, s) (C) →→→→ (t) (D) →→→→ (p, q, t)
20. Match the reactions in column I with appropriate types of steps/reactive intermediate involved in these
reactions as given in column II
Column I Column II
(A)
CH3
O
O
aq. NaOH
→
O
(p) Nucleophilic substitution
(B)
3CH MgI
→
CH2CH2CH2Cl
O O
CH3
(q) Electrophilic substitution
(C)
2 4H SO
→
CH2CH2CH2OH
O O
18
18
(r) Dehydration
(D)
2 4H SO
→
CH2CH2CH2C(CH3)2
OH
CH3 CH3
(s) Nucleophilic addition
(t) Carbanion
Sol. (A) →→→→ (r, s, t) (B) →→→→ (p, s) (C) →→→→ (r, s) (D) →→→→ (q, r)

21. Which of the field patterns given below is valid for electric field as well as for magnetic field?
(A) (B)
(C) (D)
Sol. (C)
22. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet
of mass 0.01 kg, traveling with a velocity V m/s in a horizontal
direction, hits the centre of the ball. After the collision, the ball and
bullet travel independently. The ball hits the ground at a distance of
20 m and the bullet at a distance of 100 m from the foot of the post.
The velocity V of the bullet is 200 100
V m/s
(A) 250 m/s (B) 250 2 m/s
(C) 400 m/s (D) 500 m/s
Sol. (D)
5 = 21
(10)t
2
⇒ t = 1 sec
vball = 20 m/s
vbullet = 100 m/s
0.01 V = 0.01 × 100 + 0.2 × 20
v = 100 + 400 = 500 m/s
23. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with
a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the
main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a
relative error of 2 %, the relative percentage error in the density is
(A) 0.9 % (B) 2.4 %
(C) 3.1 % (D) 4.2 %
Sol. (C)
diameter = 2.5 +
0.5
20
50
×
PART - II:
PHYSICS

= 2.70 mm
% error =
dm dr
3 100
m r
 
+ × 
 
= 2 + 3 ×
0.01
100
2.70
×
= 3.1 %
24. A wooden block performs SHM on a frictionless surface
with frequency, ν0. The block carries a charge +Q on its
surface. If now a uniform electric field E
?
is switched-on
as shown, then the SHM of the block will be
+Q
E
(A) of the same frequency and with shifted mean position.
(B) of the same frequency and with the same mean position
(C) of changed frequency and with shifted mean position.
(D) of changed frequency and with the same mean position.
Sol. (A)
25. A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence θ. The
reflected (R) and transmitted (T) intensities, both as function of θ, are plotted. The correct sketch is
(A)
T
Intensity
R
0 90°
100%
θ
(B)
Intensity
R
0 90°
100%
θ
T
(C)
Intensity
R
0 90°
100%
θ
T
(D)
Intensity
R
0 90°
100%
θ
T
Sol. (C)
After total internal reflection, there is no refracted ray.
26. A satellite is moving with a constant speed ‘V’ in a circular orbit about the earth. An object of mass ‘m’ is
ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its
ejection, the kinetic energy of the object is
(A) 21
mV
2
(B) mV2
(C) 23
mV
2
(D) 2mV2

Sol. (B)
2
2
mV GMm
r r
= ∴ mV2
=
GMm
r
27. A long insulated copper wire is closely wound as a spiral of
‘N’ turns. The spiral has inner radius ‘a’ and outer radius ‘b’.
The spiral lies in the XY plane and a steady current ‘I’ flows
through the wire. The Z-component of the magnetic field at
the centre of the spiral is
(A) 0 NI b
ln
2(b a) a
µ  
 
−  
(B) 0 NI b a
ln
2(b a) b a
µ + 
 
− − 
(C) 0 NI b
ln
2b a
µ  
 
 
(D) 0 NI b a
ln
2b b a
µ + 
 
− 
a
b
I
Y
X
Sol. (A)
b
0
a
IN
dr
2r(b a)
µ
−∫
=
b
0 0
a
IN INdr b
ln
2(b a) r 2(b a) a
µ µ  
=  
− −  
∫
28. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1(t) = A sin ωt and
x2(t) = A sin
2
t
3
π 
ω + 
 
. Adding a third sinusoidal displacement x3(t) = B sin (ωt + φ) brings the mass to a
complete rest. The values of B and φ are
(A) 2 A,
3
4
π
(B) A,
4
3
π
(C) 3 A,
5
6
π
(D) A,
3
π
Sol. (B)
4π/3
2π/3
2π/3

29. Two solid spheres A and B of equal volumes but of different densities dA and
dB are connected by a string. They are fully immersed in a fluid of density dF.
They get arranged into an equilibrium state as shown in the figure with a
tension in the string. The arrangement is possible only if
(A) dA < dF (B) dB > dF
(C) dA > dF (D) dA + dB = 2dF
A
B
Sol. (A, B, D)
A
B
2dF Vg
(dA +dB)Vg
30. A thin ring of mass 2 kg and radius 0.5 m is rolling without on
a horizontal plane with velocity 1 m/s. A small ball of mass
0.1 kg, moving with velocity 20 m/s in the opposite direction
hits the ring at a height of 0.75 m and goes vertically up with
velocity 10 m/s. Immediately after the collision
1 m/s
10 m/s
20 m/s
0.75 m
(A) the ring has pure rotation about its stationary CM.
(B) the ring comes to a complete stop.
(C) friction between the ring and the ground is to the left.
(D) there is no friction between the ring and the ground.
Sol. (C)
During collision friction is impulsive and immediately after collision the ring will have a clockwise angular
velocity hence friction will be towards left.
31. Which of the following statement(s) is/are correct?
(A) If the electric field due to a point charge varies as r−2.5
instead of r−2
, then the Gauss law will still be
valid.
(B) The Gauss law can be used to calculate the field distribution around an electric dipole
(C) If the electric field between two point charges is zero somewhere, then the sign of the two charges is
the same.
(D) The work done by the external force in moving a unit positive charge from point A at potential VA to
point B at potential VB is (VB − VA).
Sol. (C) or (C, D*)
(D) is correct if we assume it is work done against electrostatic force
32. A series R-C circuit is connected to AC voltage source. Consider two cases; (A) when C is without a
dielectric medium and (B) when C is filled with dielectric of constant 4. The current IR through the resistor
and voltage VC across the capacitor are compared in the two cases. Which of the following is/are true?
(A) A B
R RI I> (B) A B
R RI I<
(B) A B
C CV V> (D) A B
C CV V<

Sol. (B, C)
I =
V
Z
2 2 2
R CV V V= + =
2
2 I
(IR)
C
 
+  
ω 
2
2 2 1
z R
C
 
= +  
ω 
33. A series R-C combination is connected to an AC voltage of angular frequency ω = 500 radian/s. If the
impedance of the R-C circuit is R 1.25 , the time constant (in millisecond) of the circuit is
Sol. (4)
2
2 1
R 1.25 R
C
 
= +  
ω 
RC = 4 ms
34. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-
space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the
sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the
sphere is A × 10z
(where 1 < A < 10). The value of ‘Z’ is
Sol. (7)
Stopping potential =
hc
W−
λ
= 6.2 eV − 4.7 eV
= 1.5 eV
V =
Kq
1.5
r
=
2
9 19
1.5 10
n
9 10 1.6 10
−
−
×
=
× × ×
= 1.05 × 107
Z = 7
35. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a
ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by
1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in m/s2
, is
Sol. (5)
23 1
0 10 t 10t
2 2
= − …(i)
t = 3 sec
⇒ 1.15 =
3
5 3 a
2
− …(ii)
⇒ a ≈ 5 m/s2

36. A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. The
coefficient of friction between the block and the floor is 0.1. Initially the
block is at rest and the spring is un-stretched. An impulse is given to the
block as shown in the figure. The block slides a distance of 0.06 m and
comes to rest for the first time. The initial velocity of the block in m/s is
V = N/10. Then N is
Sol. (4)
Applying work energy theorem
2 21 1
kx mgx mV
2 2
− − µ = −
⇒
4
V
10
=
N = 4
37. Two batteries of different emfs and different internal
resistances are connected as shown. The voltage across AB in
volts is
A B
6V
3V 2ΩΩΩΩ
1ΩΩΩΩ
Sol. (5)
3
i
3
= = 1 ampere
A BV 6 1 V 0− + − =
A BV V 5− =
38. Water (with refractive index =
4
3
) in a tank is 18 cm deep. Oil of
refractive index
7
4
lies on water making a convex surface of radius of
curvature ‘R = 6 cm’ as shown. Consider oil to act as a thin lens. An
object ‘S’ is placed 24 cm above water surface. The location of its image
is at ‘x’ cm above the bottom of the tank. Then ‘x’ is
S
µµµµ = 1.0
µµµµ = 7/4
µµµµ = 4/3
R = 6 cm
Sol. (2)
1
7
1
7 1 4
4V 24 6
−
− =
−
⇒ V1 = 21 cm
2
4 / 3 7 / 4
0
V 21
− =
V2 = 16 cm
x = 18 − 16 = 2 cm

(Matrix-Match Type)
and r in the ORS.
39. One mole of a monatomic gas is taken through a
cycle ABCDA as shown in the P-V diagram.
Column II give the characteristics involved in the
cycle. Match them with each of the processes
given in Column I.
0 1V 3V 9V V
P
1P
3P
AB
C D
Column I Column II
(A) Process A → B (p) Internal energy decreases
(B) Process B → C (q) Internal energy increases.
(C) Process C → D (r) Heat is lost
(D) Process D → A (s) Heat is gained
(t) Work is done on the gas
Sol. (A) →→→→ (p, r, t) (B) →→→→ (p, r) (C) →→→→ (q, s) (D) (r, t)
Process A → B → Isobaric compression
Process B → C → Isochoric process
Process C → D → Isobaric expansion
Process D → A → Polytropic with TA = TD
40. Column I shows four systems, each of the same length L, for producing standing waves. The lowest
possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as
λf. Match each system with statements given in Column II describing the nature and wavelength of the
standing waves.
Column I Column II
(A) Pipe closed at one end (p) Longitudinal waves
O L
(B) Pipe open at both ends (q) Transverse waves
O L
(C) Stretched wire clamped at both ends (r) λf = L
O L
(D) Stretched wire clamped at both ends
and at mid-point
(s) λf = 2L
O LL/2
(t) λf = 4L
Sol. (A) →→→→ (p, t) (B) →→→→ (p, s) (C) →→→→ (q, s) (D) (q, r)

41. If ( )
1/
2
0
lim 1 ln 1
x
x
x b
→
 + +  = 2bsin2
θ, b > 0 and θ ∈ (−π, π], then the value of θ is
(A)
4
π
± (B)
3
π
±
(C)
6
π
± (D)
2
π
±
Sol. (D)
( )2
ln 1 b 2
e 2bsin
+
= θ
⇒ sin2
θ =
2
1 b
2b
+
⇒ sin2
θ = 1 as
2
1 b
1
2b
+
≥
θ = ±π/2.
42. Let f : [−1, 2] → [0, ∞) be a continuous function such that f(x) = f(1 − x) for all x ∈[−1, 2]. Let
2
1
1
( )R xf x dx
−
= ∫ , and R2 be the area of the region bounded by y = f(x), x = −1, x = 2, and the x-axis. Then
(A) R1 = 2R2 (B) R1 = 3R2
(C) 2R1 = R2 (D) 3R1 = R2
Sol. (C)
( ) ( ) ( )
2 2
1
1 1
R xf x dx 2 1 x f 2 1 x dx
− −
= = − − − −∫ ∫
= ( ) ( ) ( ) ( )
2 2
1 1
1 x f 1 x dx 1 x f x dx
− −
− − = −∫ ∫
Hence 2R1 = ( )
2
2
1
f x dx R
−
=∫ .
43. Let f(x) = x2
and g(x) = sinx for all x ∈ R. Then the set of all x satisfying (f o g o g o f) (x) = (g o g o f) (x),
where (f o g) (x) = f(g(x)), is
(A) { }, 0,1, 2, ....n n± π ∈ (B) { }, 1, 2, ....n n± π ∈
(C) { }2 , ..., 2, 1,0,1, 2, ....
2
n n
π
+ π ∈ − − (D) { }2 , ..., 2, 1,0,1, 2, ....n nπ ∈ − −
Sol. (A)
(fogogof) (x) = sin2
(sin x2
)
(gogof) (x) = sin (sin x2
)
∴ sin2
(sin x2
) = sin (sin x2
)
⇒ sin (sin x2
) [sin (sin x2
) − 1] = 0
PART - III:
MATHEMATICS

⇒ sin (sin x2
) = 0 or 1
⇒ sin x2
= nπ or 2mπ + π/2, where m, n ∈ I
⇒ sin x2
= 0
⇒ x2
= nπ ⇒ x = ± nπ , n ∈ {0, 1, 2, …}.
44. Let (x, y) be any point on the parabola y2
= 4x. Let P be the point that divides the line segment from (0, 0)
to (x, y) in the ratio 1 : 3. Then the locus of P is
(A) x2
= y (B) y2
= 2x
(C) y2
= x (D) x2
= 2y
Sol. (C)
y2
= 4x
and Q will lie on it
⇒ (4k)2
= 4 × 4h
⇒ k2
= h
⇒ y2
= x (replacing h by x and k by y)
O P Q
(h, k) (4h, 4k)
(0, 0)
45. Let P(6, 3) be a point on the hyperbola
2 2
2 2
1
x y
a b
− = . If the normal at the point P intersects the x-axis at (9,
0), then the eccentricity of the hyperbola is
(A)
5
2
(B)
3
2
(C) 2 (D) 3
Sol. (B)
Equation of normal is (y − 3) = ( )
2
2
a
x 6
2b
−
− ⇒
2
2
a
1
2b
= ⇒
3
e
2
= .
46. A value of b for which the equations
x2
+ bx − 1 = 0
x2
+ x + b = 0,
have one root in common is
(A) 2− (B) 3i−
(C) 5i (D) 2
Sol. (B)
x2
+ bx − 1 = 0
x2
+ x + b = 0 … (1)
Common root is
(b − 1) x − 1 − b = 0
⇒ x =
b 1
b 1
+
−
This value of x satisfies equation (1)
⇒
( )
( )
2
2
b 1 b 1
b 0
b 1b 1
+ +
+ + =
−−
⇒ b = 3i, 3i, 0− .

47. Let ω ≠ 1 be a cube root of unity and S be the set of all non-singular matrices of the form
2
1
1
1
a b
c
 
 ω 
 ω ω 
,
where each of a, b, and c is either ω or ω2
. Then the number of distinct matrices in the set S is
(A) 2 (B) 6
(C) 4 (D) 8
Sol. (A)
For being non-singular
2
1 a b
1 c
1
ω
ω ω
≠ 0
⇒ acω2
− (a + c)ω + 1 ≠ 0
Hence number of possible triplets of (a, b, c) is 2.
i.e. (ω, ω2
, ω) and (ω, ω, ω).
48. The circle passing through the point (−1, 0) and touching the y-axis at (0, 2) also passes through the point
(A)
3
, 0
2
 
− 
 
(B)
5
, 2
2
 
− 
 
(C)
3 5
,
2 2
 
− 
 
(D) (−4, 0)
Sol. (D)
Circle touching y-axis at (0, 2) is (x − 0)2
+ (y − 2)2
+ λx = 0
passes through (− 1, 0)
∴ 1 + 4 − λ = 0 ⇒ λ = 5
∴ x2
+ y2
+ 5x − 4y + 4 = 0
Put y = 0 ⇒ x = − 1, − 4
∴ Circle passes through (− 4, 0)
49. If ( )
,
2 2
cos , 0
2
1, 0 1
ln , 1
x x
x xf x
x x
x x
π π
− − ≤ −

π− − < ≤= 

− < ≤
 >
, then
(A) f(x) is continuous at x = − π/2 (B) f(x) is not differentiable at x = 0
(C) f(x) is differentiable at x = 1 (D) f(x) is differentiable at x = −3/2
Sol. (A, B, C, D)
( ) ( )
x
2
lim f x 0 f / 2−
π
→−
= = −π
( )
x
2
lim f x cos 0
2+
π
→−
π 
= − = 
 

( )
1, x / 2
sin x, / 2 x 0
f x
1, 0 x 1
1/ x, x 1
− ≤ −π
 −π < ≤
′ = 
< ≤
 >
Clearly, f (x) is not differentiable at x = 0 as f´(0−
) = 0 and f´(0+
) = 1.
f (x) is differentiable at x = 1 as f′ (1–
) = f′ (1+
) = 1.
50. Let f : (0, 1) → R be defined by ( )
1
b x
f x
bx
−
=
−
, where be is a constant such that 0 < b < 1. Then
(A) f is not invertible on (0, 1) (B) f ≠ f−1
on (0, 1) and ( )
( )
1
0
f b
f
′ =
′
(C) f = f−1
on (0, 1) and ( )
( )
1
0
f b
f
′ =
′
(D) f−1
is differentiable on (0, 1)
Sol. (A)
f(x) =
b x
1 bx
−
−
Let y =
b x
1 bx
−
−
⇒ x =
b y
1 by
−
−
0 < x < 1 ⇒ 0 <
b y
1 by
−
−
< 1
b y
1 by
−
−
> 0 ⇒ y < b or y >
1
b
b y
1 by
−
−
– 1 < 0 ⇒ – 1 < y <
1
b
∴ – 1 < y < b.
51. Let L be a normal to the parabola y2
= 4x. If L passes through the point (9, 6), then L is given by
(A) y − x + 3 = 0 (B) y + 3x − 33 = 0
(C) y + x − 15 = 0 (D) y − 2x + 12 = 0
Sol. (A, B, D)
y2
= 4x
Equation of normal is y = mx – 2m – m3
.
It passes through (9, 6)
⇒ m3
– 7m + 6 = 0
⇒ m = 1, 2, – 3
⇒ y – x + 3 = 0, y + 3x – 33 = 0, y – 2x + 12 = 0.
52. Let E and F be two independent events. The probability that exactly one of them occurs is
11
25
and the
probability of none of them occurring is
2
25
. If P(T) denotes the probability of occurrence of the event T,
then
(A)
4 3
( ) , ( )
5 5
P E P F= = (B)
1 2
( ) , ( )
5 5
P E P F= =
(C)
2 1
( ) , ( )
5 5
P E P F= = (D)
3 4
( ) , ( )
5 5
P E P F= =

Sol. (A, D)
Let P (E) = e and P (F) = f
P (E ∪ F) − P (E ∩ F) =
11
25
⇒ e + f − 2ef =
11
25
… (1)
( )P E F∩ =
2
25
⇒ (1 − e) (1 − f) =
2
25
⇒ 1 − e − f + ef =
2
25
… (2)
From (1) and (2)
ef =
12
25
and e + f =
7
5
Solving, we get
e =
4 3
, f
5 5
= or
3 4
e , f
5 5
= = .
53. Let ˆ ˆˆ,a i k b i j= − − = − +
??
and ˆ ˆ ˆ2 3c i j k= + +
?
be three given vectors. If r
?
is a vector such that r b c b× = ×
? ?? ?
and 0r a⋅ =
? ?
, then the value of r b⋅
??
is
Sol. (9)
r b c b× = ×
? ?? ?
taking cross with a
( ) ( )a r b a c b× × = × ×
? ?? ? ? ?
( ) ( ) ( )a b r a r b a c b⋅ − ⋅ = × ×
? ? ?? ? ? ? ? ?
⇒ ˆ ˆ ˆr 3i 6j 3k= − + +
?
r b 3 6 9⋅ = + =
??
.
54. The straight line 2x − 3y = 1 divides the circular region x2
+ y2
≤ 6 into two parts. If
3 5 3 1 1 1 1
2, , , , , , ,
4 2 4 4 4 8 4
S
        
= −        
        
,
then the number of point(s) in S lying inside the smaller part is
Sol. (2)
L: 2x – 3y – 1
S: x2
+ y2
– 6
If L1 > 0 and S1 < 0
Then point lies in the smaller part.
∴
3 1 1
2, and ,
4 4 4
   
−   
   
lie inside.
L: 2x –3y –1
S: x2
+ y2
– 6

55. Let ω = eiπ/3
, and a, b, c, x, y, z be non-zero complex numbers such that
a + b + c = x
a + bω + cω2
= y
a + bω2
+ cω = z.
Then the value of
2 2 2
2 2 2
x y z
a b c
+ +
+ +
is
Sol. (3)
The expression may not attain integral value for all a, b, c
If we consider a = b = c, then
x = 3a
y = a (1 + ω + ω2
) = a ( )1 i 3+
z = a (1 + ω2
+ ω) = a ( )1 i 3+
∴ |x|2
+ |y|2
+ |z|2
= 9 |a|2
+ 4 |a|2
+ 4 |a|2
= 17 |a|2
∴
2 2 2
2 2 2
| x | | y | | z | 17
3| a | | b | | c |
+ +
=
+ +
Note: However if ω =
( )i 2 /3
e π
, then the value of the expression = 3.
56. The number of distinct real roots of x4
− 4x3
+ 12x2
+ x − 1 = 0 is
Sol. (2)
Let f (x) = x4
− 4x3
+ 12x2
+ x − 1 = 0
f′ (x) = 4x3
− 12x2
+ 24x + 1= 4 (x3
− 3x2
+ 6x) + 1
f″ (x) = 12x2
− 24x + 24 = 12 (x2
− 2x + 2)
f″(x) has 0 real roots
f (x) has maximum 2 distinct real roots as f (0) = –1.
57. Let y′(x) + y(x)g′(x) = g(x)g′(x), y(0) = 0, x ∈ R, where f′(x) denotes
( )df x
dx
and g(x) is a given non-
constant differentiable function on R with g(0) = g(2) = 0. Then the value of y(2) is
Sol. (0)
y′(x) + y(x) g′(x) = g(x) g′(x)
⇒ eg(x)
y′(x) + eg(x)
g′(x) y(x) = eg(x)
g(x) g′(x)
⇒ ( ) ( )
( )g xd
y x e
dx
= eg(x)
g(x) g′(x)
∴ y(x) = eg(x)
=
( )
( ) ( )g x
e g x g x dx′
∫
= t
e tdt∫ , where g(x) = t
= (t – 1) et
+ c
∴ y(x) eg(x)
= (g(x) – 1) eg(x)
+ c
Put x = 0 ⇒ 0 = (0 – 1) . 1 + c ⇒ c = 1
Put x = 2 ⇒ y(2) . 1 = (0 – 1) . (1) + 1
y(2) = 0.
58. Let M be a 3 × 3 matrix satisfying
0 1 1 1 1 0
1 2 , 1 1 , 1 0
0 3 0 1 1 12
M M and M
−           
           = − = =           
           −           
.
Then the sum of the diagonal entries of M is

Sol. (9)
Let M =
a b c
d e f
g h i
 
 
 
  
0 1
M 1 2
0 3
−   
   =   
      
⇒ b = − 1, e = 2, h = 3
1 1
M 1 1
0 1
   
   − =   
   −   
⇒ a = 0, d = 3, g = 2
1 0
M 1 0
1 12
   
   =   
      
⇒ g + h + i = 12 ⇒ i = 7
∴ Sum of diagonal elements = 9.
(Matrix-Match Type)
and r in the ORS.
59. Match the statements given in Column I with the values given in Column II
Column −−−− I Column −−−− II
(A) If ˆ ˆˆ ˆ3 , 3a j k b j k= + = − +
??
and ˆ2 3c k=
?
form a triangle, then
the internal angle of the triangle between a
?
and b
?
is
(p)
6
π
(B)
If ( )( ) 2 2
3
b
a
f x x dx a b− = −∫ , then the value of
6
f
π 
 
 
is (q)
2
3
π
(C)
The value of ( )
5/62
7/6
sec
ln3
x dx
π
π∫ is (r)
3
π
(D) The maximum value of
1
1
Arg
z
 
 
 − 
for |z| = 1, z ≠ 1 is given
by
(s) π
(t)
2
π
Sol. (A) →→→→ (q) (B) →→→→ (p) (C) →→→→ (s) (D) →→→→ (t)
(A). a b 1 3 2− = − + =
??
a 2, b 2= =
??
cosθ =
2 1
2 2 2
=
×

θ =
2
,
3 3
π π
but its
2
3
π
as its opposite to side of maximum length.
(B). ( )( )
b
2 2
a
f x 3x dx a b− = −∫
( ) ( )
b 2 2
2 2 2 2
a
3 a b
f x dx b a a b
2 2
− +
= − + − =∫
⇒ f(x) = x.
(C).
( )
5/6
2
7/6ln sec x tan x
ln3
 π + ππ   
π 
=
5 5 7 7
ln sec tan ln sec tan
ln3 6 6 6 6
π  π π π π 
+ − + 
 
= π.
(D). Let u =
1
1 z−
⇒ z = 1 –
1
u
|z| = 1 ⇒
1
1 1
u
− =
⇒ |u – 1| = |u|
∴ locus of u is perpendicular bisector of line segment joining 0 and 1
⇒ maximum arg u approaches
2
π
but will not attain.
60. Match the statements given in Column I with the intervals/union of intervals given in Column II
Column −−−− I Column −−−− II
(A)
The set 2
2
Re : , 1, 1
1
iz
z is a complex number z z
z
  
= ≠ ±  
 −  
is (p) (−∞, −1) ∪ (1, ∞)
(B)
The domain of the function
2
1
2( 1)
8(3)
( ) sin
1 3
x
x
f x
−
−
−
 
=  
− 
is (q) (−∞, 0) ∪ (0, ∞)
(C)
If
1 tan 1
( ) tan 1 tan
1 tan 1
f
θ
θ = − θ θ
− − θ
, then the set
( ){ }: 0
2
f
π
θ ≤ θ < is
(r) [2, ∞)
(D)
If f(x) = x3/2
(3x − 10), x ≥ 0, then f(x) is increasing in (s) (−∞, −1] ∪ [1, ∞)
(t) (−∞, 0] ∪ [2, ∞)
Sol. (A) →→→→ (s) (B) →→→→ (t) (C) →→→→ (r) (D) →→→→ (r)
(A). z = 2 2 2
2i(x iy) 2i(x iy)
1 (x iy) 1 (x y 2ixy)
+ +
=
− + − − +
Using 1 − x2
= y2
Z = 2
2ix 2y 1
y2y 2ixy
−
= −
−
.

∵ –1 ≤ y ≤ 1 ⇒ –
1 1
1 or 1
y y
≤ − − ≥ .
(B). For domain
− 1 ≤
x 2
2(x 1)
8 3
1
1 3
−
−
⋅
≤
−
⇒ − 1≤
x x 2
2x 2
3 3
1
1 3
−
−
−
≤
−
.
Case −I:
x x 2
2x 2
3 3
1 0
1 3
−
−
−
− ≤
−
⇒
x x 2
2x 2
(3 1)(3 1)
0
(3 1)
−
−
− −
≥
−
⇒ x ∈ (− ∞, 0] ∪ (1, ∞).
Case −II:
x x 2
2x
3 3
1 0
1 3 2
−
−
+ ≥
− −
⇒
x 2 x
x x 2
(3 1)(3 1)
0
(3 3 1)
−
−
− +
≥
⋅ −
⇒ x ∈ (− ∞, 1) ∪ [2, ∞)
So, x ∈ (− ∞, 0] ∪ [2, ∞).
(C). R1 → R1 + R3
f (θ) =
0 0 2
tan 1 tan
1 tan 1
− θ θ
− − θ
= 2 (tan2
θ + 1) = 2 sec2
θ.
(D). f´(x) = ( ) ( ) ( )
1/2 3/23
x 3x 10 x 3
2
− + × = ( ) ( )
1/215
x x 2
2
−
Increasing, when x ≥ 2.

IIT JEE 2011 Solved Paper by Prabhat Gaurav

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IIT JEE 2011 Solved Paper by Prabhat Gaurav