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–1–
FIITJEE
SOLUTION TO AIEEE-2005
CHEMISTRY
76. Which of the following oxides is amphoteric in character?
(1) CaO (2) CO2
(3) SiO2 (4) SnO2
-1-
FIITJEE
SOLUTION TO AIEEE-2005
PHYSICS
1. A projectile can have the same range R for two angles of projection. If t1 and t2 be the
times of flights in the two cases, then the product of the two time of flights is
proportional to
(1) R2 (2) 1/R2
(3) 1/R (4) R
1. (4)
2
1 2 2
t t 2u sin2 2R
g g
θ
= =
2. An annular ring with inner and outer radii R1 and R2 is rolling without slipping with a
uniform angular speed. The ratio of the forces experienced by the two particles
situated on the inner and outer parts of the ring, F1/F2 is
(1) 2
1
R
R
(2)
2
1
2
R
R
(3) 1 (4) 1
2
R
R
2. (4)
2
1 1 1
2
2 2 2
F R R
F R R
ω
= =
ω
3. A smooth block is released at rest on a 45° incline and then slides a distance d. The
time taken to slide is n times as much to slide on rough incline than on a smooth
incline. The coefficient of friction is
(1) k 2
1 1
n
μ = − (2) k 2
1 1
n
μ = −
(3) s 2
1 1
n
μ = − (4) s 2
1 1
n
μ = −
3. (1)
2
1
d 1 g t
2 2
=
( ) 2
k 2
d 1 g 1 t
2 2
= −μ
2
2 221 k
t n 1
t 1
= =
− μ
4. The upper half of an inclined plane with inclination φ is perfectly smooth while the
lower half is rough. A body starting from rest at the top will again come to rest at the
bottom if the coefficient of friction for the lower half is given by
-2-
(1) 2sinφ (2) 2cosφ
(3) 2tanφ (4) tanφ
4. (3)
mg s sin φ = mgcos . s
2
μ φ
S/2
S/2
φ
5. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How
much further it will penetrate before coming to rest assuming that it faces constant
resistance to motion?
(1) 3.0 cm (2) 2.0 cm
(3) 1.5 cm (4) 1.0 cm
5. (4)
F.3
2
1mv2 1m v
2 2 4
= −
F(3+x) 1mv2
2
=
x = 1 cm
6. Out of the following pair, which one does NOT have identical dimensions is
(1) angular momentum and Planck’s constant
(2) impulse and momentum
(3) moment of inertia and moment of a force
(4) work and torque
6. (3)
Using dimension
7. The relation between time t and distance x is t=ax2+bx where a and b are constants.
The acceleration is
(1) −2abv2 (2) 2bv3
(3) −2av3 (4) 2av2
7. (3)
t = ax2 + bx
by differentiating acceleration = – 2av3
8. A car starting from rest accelerates at the rate f through a distance S, then continues
at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the
total distance traversed is 15 S, then
(1) S=ft (2) S = 1/6 ft2
(3) S = 1/2 ft2 (4) S = 1/4 ft2
8. (none)
2
1 S ft
2
=
0v = 2Sf
During retardation
t1 t2 t
V0
V
S2 = 2S
-3-
During constant velocity
15S -3S = 12S = vot
ft2 S
72
⇒ =
9. A particle is moving eastwards with a velocity of 5 m/s in 10 seconds the velocity
changes to 5 m/s northwards. The average acceleration in this time is
(1) 1 m/ s2
2
towards north-east (2) 1 m/ s2
2
towards north.
(3) zero (4) 1 m/ s2
2
towards north-west
9. (4)
f i a V V
t
−
=
r r
r
( ) 5ˆj 5ˆi 1 ˆ ˆ j i
10 2
−
= = −
a 1 ms 2
2
∴ = − towards north west
(S)
(W) (E)
(N) ˆj
ˆi
10. A parachutist after bailing out falls 50 m without friction. When parachute opens, it
decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height,
did he bail out?
(1) 91 m (2) 182 m
(3) 293 m (4) 111 m
10. (3)
s = 50 +
( )
( )
32 2 10 50
2 2
− × ×
−
= 293 m.
11. A block is kept on a frictionless inclined surface with angle
of inclination α. The incline is given an acceleration a to
keep the block stationary. Then a is equal to
(1) g/tanα (2) g cosecα
(3) g (4) g tanα
a
α
11. (4)
mg sinα = ma cos α
∴a = gtanα
ma
mg
α N
12. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls
down a smooth surface to the ground, then climbs up another hill of height 30 m and
finally rolls down to a horizontal base at a height of 20 m above the ground. The
velocity attained by the ball is
(1) 40 m/s (2) 20 m/s
(3) 10 m/s (4) 10 30 m/s
-4-
12. (1)
mgh = ½ mv2
v = 2gh
= 2 ×10 × 80 = 40 m/s
13. A body A of mass M while falling vertically downwards under gravity breaks into two
parts; a body B of mass 1/3 M and a body C of mass 2/3 M. The centre of mass of
bodies B and C taken together shifts compared to that of body A towards
(1) depends on height of breaking (2) does not shift
(3) body C (4) body B
13. (2)
No horizontal external force is acting
cm ∴a = 0
since cm v = 0
cm ∴Δx = 0
14. The moment of inertia of a uniform semicircular disc of mass M and radius r about a
line perpendicular to the plane of the disc through the centre is
(1) 1Mr2
4
(2) 2Mr2
5
(3) Mr2 (4) 1Mr2
2
14. (4)
R2 2I 2M
2
=
MR2 I
2
∴ =
15. A particle of mass 0.3 kg is subjected to a force F=−kx with k=15 N/m. What will be
its initial acceleration if it is released from a point 20 cm away from the origin?
(1) 3 m/s2 (2) 15 m/s2
(3) 5 m/s2 (4) 10 m/s2
15. (4)
a kx 10m/ s2
m
= =
16. The block of mass M moving on the frictionless
horizontal surface collides with a spring of spring
constant K and compresses it by length L. The
maximum momentum of the block after collision is
(1) MK L (2)
KL2
2M
(3) zero (4)
ML2
K
M
16. (1)
2
1KL2 P
2 2m
= ∴P = MK L
-5-
17. A mass ‘m’ moves with a velocity v and collides
inelastically with another identical mass. After
collision the 1st mass moves with velocity v/√3 in a
direction perpendicular to the initial direction of
motion. Find the speed of the 2nd mass after
collision
m m
v
Before
collision
After
collision
v/√3
(1) v (2) √3 v
(3) 2v/√3 (4) v/√3
17. (3)
1 mv = mv cosθ
1
0 mv mv sin
3
= − θ
1
v 2 v
3
∴ =
m
V1
θ
v
3
m
v
Before
collision
After
collision
18. A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the
entire arrangement is put in a freely falling elevator the length of water column in the
capillary tube will be
(1) 8 cm (2) 10 cm
(3) 4 cm (4) 20 cm
18. (4)
Water will rise to the full length of capillary tube
19. If S is stress and Y is Young’s modulus of material of a wire, the energy stored in the
wire per unit volume is
(1) 2S2Y (2) S2/2Y
(3) 2Y/S2 (4) S/2Y
19. (2)
U 1
2
= stress × strain
S2
2Y
=
20. Average density of the earth
(1) does not depend on g (2) is a complex function of g
(3) is directly proportional to g (4) is inversely proportional to g
20. (3)
g = av
G4 R
3
π
ρ
21. A body of mass m is accelerated uniformly from rest to a speed v in a time T. The
instantaneous power delivered to the body as a function time is given by
(1)
2
2
mv t
T
⋅ (2)
2
2
2
mv t
T
⋅
(3)
2
2
1mv t
2 T
⋅ (4)
2
2
2
1mv t
2 T
⋅
21. (1)
P= (ma).v
-6-
= m a2 t
= m
2
2
V t
T
22. Consider a car moving on a straight road with a speed of 100 m/s. The distance at
which car can be stopped is [μk = 0.5]
(1) 800 m (2) 1000 m
(3) 100 m (4) 400 m
22. (2)
2
k
mgs 1mu
2
μ =
2
k
s u 1000m
2 g
= =
μ
23. Which of the following is incorrect regarding the first law of thermodynamics?
(1) It is not applicable to any cyclic process
(2) It is a restatement of the principle of conservation of energy
(3) It introduces the concept of the internal energy
(4) It introduces the concept of the entropy
23. (none)
More than one statements are incorrect
24. A ‘T’ shaped object with dimensions shown in
the figure, is lying on a smooth floor. A force F is
applied at the point P parallel to AB, such that
the object has only the translational motion
without rotation. Find the location of P with
respect to C
P
l
2 l F
r
A B
C
(1) 2
3
l (2) 3
2
l
(3) 4
3
l (4) l
24. (3)
P will be the centre of mass of system
25. The change in the value of g at a height ‘h’ above the surface of the earth is the
same as at a depth ‘d’ below the surface of earth. When both ‘d’ and ‘h’ are much
smaller than the radius of earth, then which one of the following is correct?
(1) d h
2
= (2) d 3h
2
=
(3) d = 2h (4) d = h
25. (3)
( )2 3
GM GM
R h R
=
+
(R − d)
⇒ d = 2h
-7-
26. A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg
and radius 10 cm. Find the work to be done against the gravitational force between
them to take the particle far away from the sphere
(you may take G = 6 . 67× 10-11 Nm2 / kg2)
(1) 13.34 ×10−10 J (2) 3.33 ×10−10 J
(3) 6.67 ×10−9J (4) 6.67 ×10−10J
26. (4)
w = GMm / R = 6.67 × 10-10 J
27. A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio P
v
C
C
of the
mixture is
(1) 1.59 (2) 1.62
(3) 1.4 (4) 1.54
27. (2)
1 v1 2 v2
v
1 2
c n c n c
n n
+
=
+
= 29R
18
P
c 47R
18
= , P
v
c 1.62
c
=
28. The intensity of gamma radiation from a given source is I. On passing through 36 mm
of lead, it is reduced to I
8
. The thickness of lead which will reduce the intensity to
I
2
will be
(1) 6 mm (2) 9 mm
(3) 18 mm (4) 12 mm
28. (4)
Use I = I0 e-μx
29. The electrical conductivity of a semiconductor increases when electromagnetic
radiation of wavelength shorter than 2480 nm is incident on it. The band gap in (eV)
for the semiconductor is
(1) 1.1 eV (2) 2.5 eV
(3) 0.5 eV (4) 0.7 eV
29. (3)
g
E = hc = 0.5eV
λ
30. A photocell is illuminated by a small bright source placed 1 m away. When the same
source of light is placed 1
2
m away, the number of electrons emitted by photo
cathode would
(1) decrease by a factor of 4 (2) increase by a factor of 4
(3) decrease by a factor of 2 (4) increase by a factor of 2
-8-
30. (2)
2
I 1
r
∝
31 Starting with a sample of pure 66Cu, 7/8 of it decays into Zn in 15 minutes. The
corresponding half-life is
(1) 10 minutes (2) 15 minutes
(3) 5 minutes (4) 7 1
2
minutes
31. (3)
t1/ 2 0 t1/ 2 0 t1/ 2 0
0
N N N N
2 4 8
→ → →
1/ 2 3t = 15 1/ 2 ∴t = 5
32. If radius of 27
13 Al nucleus is estimated to be 3.6 Fermi then the radius 125
52 Te nucleus
be nearly
(1) 6 fermi (2) 8 fermi
(3) 4 fermi (4) 5 fermi
32. (1)
1
R 125 3 R 6
3.6 27
= ⇒ =
fermi
33. The temperature-entropy diagram of a reversible engine
cycle is given in the figure. Its efficiency is
(1) 1/2 (2) 1/4
(3) 1/3 (4) 2/3
2So
2To
T
To
S
S
33. (3)
BC
W
Q
Δ
η =
0 0
0 0
S T
2 1/ 3
3S T
2
= =
S
T
A C
B
S0 2S0
2T0
T0
34. The figure shows a system of two concentric spheres of radii
r1 and r2 and kept at temperatures T1 and T2 respectively. The
radial rate of flow of heat in a substance between the two
concentric sphere is proportional to
T1
T2
r1
r2
(1) 2 1
1 2
r r
r r
− (2) 2
1
r
ln
r
(3) 1 2
2 1
r r
r − r
(4) ( ) 2 1 ln r − r
-9-
34. (3)
( )( )
1 2
1 2
2 1
dQ T T 4 r r K
dt r r
π = − −
35. A system goes from A to B via two processes I and II as shown
in the figure. If ΔU1 and ΔU2 are the changes in internal energies
in the processes I and II respectively, the
(1) ΔU1 = ΔU2
(2) relation between ΔU1 and ΔU2 can not be determined
(3) ΔU2 > ΔU1
(4) ΔU2 < ΔU1
P
V
II
I
A B
35. (1)
Internal energy is state function
36. The function sin2(ωt) represents
(1) a periodic, but not simple harmonic motion with a period 2π/ω
(2) a periodic, but not simple harmonic motion with a period π/ω
(3) a simple harmonic motion with a period 2π/ω
(4) a simple harmonic motion with a period π/ω.
36. (4)
(1 cos2 t)
y
2
− ω
=
37. A Young’s double slit experiment uses a monochromatic source. The shape of the
interference fringes formed on a screen is
(1) hyperbola (2) circle
(3) straight line (4) parabola
37. (3)
Straight line
Note: If instead of young’s double slit experiment, young’s double hole experiment
was given shape would have been hyperbola.
38. Two simple harmonic motions are represented by the equation y1 = 0.1
sin 100 t
3
π π +
and y2 = 0.1 cosπt. The phase difference of the velocity of particle 1
w.r.t. the velocity of the particle 2 is
(1) −π/6 (2) π/3
(3) −π/3 (4) π/6
38. (1)
Phase difference (φ) = 99πt + π/3 −π/2
at t = 0 φ = −π/6.
39. A fish looking up through the water sees the outside world contained in a circular
horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface,
the radius of this circle in cm is
(1) 36 7 (2) 36 / 7
(3) 36 5 (4) 4 5
-10-
39. (2)
2
r h 36
1 7
= =
μ −
40. Two point white dots are 1 mm apart on a black paper. They are viewed by eye of
pupil diameter 3 mm. Approximately, what is the maximum distance at which these
dots can be resolved by the eye? [ Take wavelength of light = 500 nm ]
(1) 5 m (2) 1m
(3) 6 m (4) 3m
40. (1)
( )
1.22 (1mm) Resolution limit
3mm R
λ
= =
∴ R = 5 m
41. A thin glass (refractive index 1.5) lens has optical power of – 5D in air. Its optical
power in a liquid medium with refractive index 1.6 will be
(1) 1 D (2) -1D
(3) 25 D (4) – 25 D
41. (none)
m a
air
m
1
P
P
1
μ
− μ =
μ
− μ
l
l
Pm=5/8 D
42. The diagram shows the energy levels for an electron
in a certain atom. Which transition shown represents
the emission of a photon with the most energy ?
(1) III (2) IV
(3) I (4) II
n=4
n=3
n=2
n=1
I II III IV
42. (1)
ΔE ∝ 2 2
1 2
1 1
n n
−
43. If the kinetic energy of a free electron doubles. Its deBroglie wavelength changes by
the factor
(1) 1
2
(2) 2
(3) 1
2
(4) 2
43. (3)
h
2Km
λ =
-11-
44. In a common base amplifier, the phase difference between the input signal voltage
and output voltage is
(1)
4
π (2) π
(3) 0 (4)
2
π
44. (3)
No phase difference between input and output signal.
45. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental
frequency in the ripple would be
(1) 50 Hz (2) 25 Hz
(3) 100 Hz (4) 70.7 Hz
45. (3)
frequency = 2 (frequency of input signal).
46. A nuclear transformation is denoted by X(n, α) 7
3Li . Which of the following is the
nucleus of element X ?
(1) 12C6 (2) 10
5 B
(3) 9
5B (4) 11
4 Be
46. (2)
1 4 7
X +0 n →2 He +3 Li
47. A moving coil galvanometer has 150 equal divisions. Its current sensitivity is
10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In
order that each division reads 1 volt, the resistance in ohms needed to be connected
in series with the coil will be
(1) 103 (2) 105
(3) 99995 (4) 9995
47. (4)
Ig=15mA Vg = 75mV
g
g g
V V R
I I
= −
48. Two voltameters one of copper and another of silver, are joined in parallel. When a
total charge q flows through the voltameters, equal amount of metals are deposited.
If the electrochemical equivalents of copper and silver are z1 and z2 respectively the
charge which flows through the silver voltameter is
(1)
1
2
q
1 z
z
+
(2)
2
1
q
1 z
z
+
(3) q 1
2
z
z
(4) q 2
1
z
z
48. (2)
q1Z1=q2Z2
q=q1+q2
-12-
∴ 2
2
1
q q
1 Z
Z
=
+
49. In the circuit, the galvanometer G shows
zero deflection. If the batteries A and B
have negligible internal resistance, the
value of the resistor R will be
2V
12V
500Ω
R
A
G
B
(1) 200 Ω (2) 100 Ω
(3) 500 Ω (4) 1000 Ω
49. (2)
12R 2
500 R
=
+
50. Two sources of equal emf are connected to an external resistance R. The internal
resistance of the two sources are R1 and R2 (R2 > R1). If the potential difference
across the source having internal resistance R2 is zero, then
(1) R = R2 × (R1 + R2)/R2 – R1) (2) R = R2 – R1
(3) R = R1R2 / (R1 + R2) (4) R = R1R2 / (R2 – R1)
50. (2)
1 2
I 2E
R R R
=
+ +
2 E −R I = 0
⇒ R = R2−R1
51. A fully charged capacitor has a capacitance ‘C’ it is discharged through a small coil of
resistance wire embedded in a thermally insulated block of specific heat capacity ‘s’
and mass ‘m’. If the temperature of the block is raised by ‘ΔT’. The potential
difference V across the capacitance is
(1) 2mC T
s
Δ (2) mC T
s
Δ
(3) ms T
C
Δ (4) 2ms T
C
Δ
51. (4)
Dimensionally only 4th option is correct.
52. One conducting U tube can slide inside another as
shown in figure, maintaining electrical contacts
between the tubes. The magnetic field B is
perpendicular to the plane of the figure. if each tube
moves towards the other at a constant speed V,
then the emf induced in the circuit in terms of B, l
and V where l is the width of each tube will be
A
D x
B
x x x x
x x x x
x x x x
x x
D C
V V
(1) BlV (2) – BlV
(3) zero (4) 2 BlV
52. (4)
d 2B v
dt
φ
= l
-13-
53. A heater coil is cut into two equal parts and only one part is now used in the heater.
The heat generated will now be
(1) doubled (2) four times
(3) one fourth (4) halved
53. (1)
V2 t H
R
Δ
=
2
‘
H’ V t
R
= Δ Given R’ = R/2
54. Two thin, long parallel wires separated by a distance ‘d’ carry a current of ‘i’ A in the
same direction. They will
(1) attract each other with a force of μ0i2/(2πd)
(2) repel each other with a force of μ0i2 / (2πd)
(3) attract each other with a force of μ0i2(2πd2)
(4) repel each other with a force of μ0i2/(2πd2)
54. (1)
Using the definition of force per unit length due to two long parallel wires carrying
currents.
55. When an unpolarized light of intensity I0 is incident on a polarizing sheet, the intensity
of the light which does not get transmitted is
(1) 0
1I
2
(2) 0
1I
4
(3) zero (4) I0
55. (1)
When unpolarised light of intensity Io is incident on a polarizing sheet, only Io/2 is
transmitted.
56. A charged ball B hangs from a silk thread S which makes an
angle θ with a large charged conducting sheet P, as show in
the figure. The surface charge density σ of the sheet is
proportional to
(1) cos θ (2) cot θ
(3) sin θ (4) tan θ
++++++
+
S
B
θ
P
56. (4)
tanθ = ( ) o
q
2 mg
σ
ε
Mg
o
q
2
σ
ε
θ
57. Two point charges + 8q and – 2q are located at x = 0 and x = L respectively. The
location of a point on the x axis at which the net electric field due to these two point
charges is zero is
(1) 2L (2) L/4
-14-
(3) 8L (4) 4L
57. (1)
( )2 2
k2q k8q 0
x L x
− + =
−
⇒ x = 2L
58. Two thin wires rings each having a radius R are placed at a distance d apart with
their axes coinciding. The charges on the two rings are +q and –q. The potential
difference between the centres of the two rings is
(1) QR/4πε0d2 (2) 2 2
0
Q 1 1
2 R R d
− πε +
(3) zero (4) 2 2
0
Q 1 1
4 R R d
− πε +
58. (2)
1 2 2
v kq kq
R R d
= −
+
2 2 2
v kq kq
R R d
−
= +
+
59. A parallel plate capacitor is made by stacking n equally spaced plates connected
alternatively. If the capacitance between any two adjacent plates is C then the
resultant capacitance is
(1) (n − 1)C (2) (n + 1)C
(3) C (4) nC
59. (1)
Ceq=(n−1) C (Q all capacitors are in parallel)
60. When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per
second are heard. Now, some tape is attached on the prong of the fork 2. When the
tuning forks are sounded again, 6 beats per seconds are heard. If the frequency of
fork 1 is 200 Hz, then what was the original frequency of fork 2?
(1) 200 Hz (2) 202 Hz
(3) 196 Hz (4) 204 Hz
60. (3)
|f1−f2| =4
Since mass of second tuning fork increases so f2 decrease and beats increase so
f1>f2
⇒ f2=f1−4 = 196
61. If a simple harmonic motion is represented by
2
2
d x
dt
+ αx = 0, its time period is
(1) 2π
α
(2) 2π
α
(3) 2πα (4) 2π α
-15-
61. (2)
ω2=α
ω = √α
T 2
π
=
α
62. The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged
hole near the bottom of the oscillation bob gets suddenly unplugged. During
observation, till water is coming out, the time period of oscillation would
(1) first increase and then decrease to the original value.
(2) first decreased then increase to the original value.
(3) remain unchanged.
(4) increase towards a saturation value.
62. (1)
First CM goes down and then comes to its initial position.
63. An observer moves towards a stationary source of sound, with a velocity one fifth of
the velocity of sound. What is the percentage increase in the apparent frequency?
(1) zero. (2) 0.5%
(3) 5% (4) 20%
63. (4)
f v v / 5 f 6f
v 5
+
= =
% increase in frequency = 20%
64. If I0 is the intensity of the principal maximum in the single slit diffraction pattern, then
what will be its intensity when the slit width is doubled?
(1) 2I0 (2) 4I0
(3) I0 (4) I0/2
64. (3)
Maximum intensity is independent of slit width.
65. Two concentric coils each of radius equal to 2π cm are placed at right angles to each
other. 3 Ampere and 4 ampere are the currents flowing in each coil respectively. The
magnetic induction in Weber/m2 at the centre of the coils will be (μ0 = 4π × 10−7
Wb/A-m)
(1) 12 × 10−5 (2) 10−5
(3) 5 × 10−5 (4) 7 × 10−5
65. (3)
o 2 2
1 2 B I I
2r
μ
= +
7
2
B 4 10 5
2 2 10
−
−
π ×
= ×
× π×
B = 5×10−5
66. A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage
2V. The current reaches half of its steady state value in
(1) 0.05 s (2) 0.1 s
-16-
(3) 0.15 s (4) 0.3 s
66. (2)
Rt
L
o I I 1 e−
= −
0.693 R t
L
=
t .3 0.693 0.1sec
2
×
= =
67. The self inductance of the motor of an electric fan is 10 H. In order to impart
maximum power at 50 Hz, it should be connected to a capacitance of
(1) 4μF (2) 8μF
(3) 1μF (4) 2μF
67. (3)
f 1
2 LC
=
π
2 2
C 1
4 f 10
=
× π ×
C = 1μF
68. An energy source will supply a constant current into the load of its internal resistance
is
(1) equal to the resistance of the load.
(2) very large as compared to the load resistance.
(3) zero.
(4) non-zero but less than the resistance of the load.
68. (2)
o E E I if R r
R r r
= <<
+
69. A circuit has a resistance of 12 Ω and an impedance of 15 Ω. The power factor of the
circuit will be
(1) 0.8 (2) 0.4
(3) 1.25 (4) 0.125
69. (1)
cos R 12 4 0.8
Z 15 5
φ = = = =
70. The phase difference between the alternating current and emf is π/2. Which of the
following cannot be the constituent of the circuit?
(1) C alone (2) R.L
(3) L. C (4) L alone
70. (2)
01, +ve (2) +ve, >1, -ve
(3) -ve, <1, -ve (4) -ve, >1, -ve
82. (1)
83. Which of the following is a polyamide?
(1) Teflon (3) Nylon – 66
(3) Terylene (4) Bakelite
83 (2)
Amide→Nylon 66
N
O
H
84. Which one of the following types of drugs reduces fever?
(1) Analgesic (2) Antipyretic
(3) Antibiotic (4) Tranquiliser
84. (2)
85. Due to the presence of an unpaired electron, free radicals are:
(1) Chemically reactive (2) Chemically inactive
(3) Anions (4) Cations
85. (1)
86. Lattice energy of an ionic compounds depends upon
(1) Charge on the ion only (2) Size of the ion only
(3) Packing of ions only (4) Charge on the ion and size of the ion
86. (4)
87. The highest electrical conductivity of the following aqueous solutions is of
(1) 0.1 M acetic acid (2) 0.1 M chloroacetic acid
(3) 0.1 M fluoroacetic acid (4) 0.1 M difluoroacetic acid
87. (4)
88. Aluminium oxide may be electrolysed at 1000oC to furnish aluminium metal (Atomic
mass = 27 amu; 1 Faraday = 96,500 Coulombs). The cathode reaction is
Al3+ + 3e− →Alo
To prepare 5.12 kg of aluminium metal by this method would require
(1) 5.49 ×107 C of electricity (2) 1.83 ×107 C of electricity
(3) 5.49 ×104 C of electricity (4) 5.49 ×101 C of electricity
88. (1)
Q =
mFZ 5.12 105 96500 3
M 27
× × ×
=
= 5.49×107 C
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89. Consider an endothermic reaction, X→Y with the activation energies Eb and Ef
for the backward and forward reactions, respectively. In general
(1) Eb < Ef
(2) Eb > Ef
(3) Eb = Ef
(4) There is no definite relation between Eb and Ef
89. (1)
ΔH = Ef – Eb
For ΔH = Positive, Eb < Ef
90. Consider the reaction: 2 2 3 N + 3H →2NH carried out at constant temperature and
pressure. If ΔH and ΔU are the enthalpy and internal energy changes for the
reaction, which of the following expressions is true?
(1) ΔH = 0 (2) ΔH = ΔU
(3) ΔH < ΔU (4) ΔH > ΔU
90. (3)
ΔH = ΔU + ΔnRT
Δn = -2
ΔH = ΔU – 2RT
ΔH < ΔU
91. Which one of the following statements is NOT true about the effect of an increase in
temperature on the distribution of molecular speeds in a gas?
(1) The most probable speed increases
(2) The fraction of the molecules with the most probable speed increases
(3) The distribution becomes broader
(4) The area under the distribution curve remains the same as under the lower
temperature
91. (2)
Most probable velocity increase and fraction of molecule possessing most probable
velocity decreases.
92. The volume of a colloidal particle, VC as compared to the volume of a solute particle
in a true solution VS, could be
(1) C
S
V 1
V
(2) C 23
S
V 10
V
(3) C 3
S
V 10
V
− (4) C 3
S
V 10
V
92. (4)
93. The solubility product of a salt having general formula MX2, in water is: 4 ×10−12. The
concentration of M2+ ions in the aqueous solution of the salt is
(1) 2.0 ×10−6 M (2) 1.0 ×10−4 M
(3) 1.6 ×10−4 M (4) 4.0 ×10−10 M
93. (2)
MX2 M+2 + 2XS
2S
Ksp = 4s3 3 sp 4 K
,S 1 10
4
= = × −
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94. Benzene and toluene form nearly ideal solutions. At 20oC, the vapour pressure of
benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of
benzene at 20oC for a solution containing 78 g of benzene and 46 g of toluene in
torr is
(1) 50 (2) 25
(3) 37.5 (4) 53.5
94. (1)
PB = B
P B 75 1
1.5
° × = × = 50 torr
95. The exothermic formation of ClF3 is represented by the equation:
2(g) 2(g) 3(g) Cl + 3F 2ClF ; ΔrH = −329 kJ
Which of the following will increase the quantity of ClF3 in an equilibrium mixture of
Cl2, F2 and ClF3?
(1) Increasing the temperature (2) Removing Cl2
(3) Increasing the volume of the container (4) Adding F2
95. (4)
M3V3 = M1V2 + M2V2
M = 480(1.5) 520(1.2) 1.344M
1000
+
=
96. Two solutions of a substance (non electrolyte) are mixed in the following manner.
480 ml of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity
of the final mixture?
(1) 1.20 M (2) 1.50 M
(3) 1.344 M (4) 2.70 M
96. (3)
97. For the reaction
2(g) (g) 2(g) 2NO 2NO + O ,
6 o
c (K = 1.8 ×10− at 184 C)
(R = 0.0831 kJ / (mol.K)
When Kp and Kc are compared at 184oC, it is found that
(1) Kp is greater than Kc
(2) Kp is less than Kc
(3) Kp = Kc
(4) Whether Kp is greater than, less than or equal to Kc depends upon the total gas
pressure
97. (1)
Kp = Kc RTΔn, Δn =1
Kp > Kc
98. Hydrogen ion concentration in mol / L in a solution of pH = 5.4 will be
(1) 3.98 ×108 (2) 3.88 ×106
(3) 3.68 ×10−6 (4) 3.98 ×10−6
98. (4)
pH = – log (H+)
99. A reaction involving two different reactants can never be
(1) Unimolecular reaction (2) First order reaction
(3) second order reaction (4) Bimolecular reaction
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99. (1)
100. If we consider that 1,
6
in place of 1 ;
12
mass of carbon atom is taken to be the relative
atomic mass unit, the mass of one mole of a substance will
(1) Decrease twice
(2) Increase two fold
(3) Remain unchanged
(4) Be a function of the molecular mass of the substance
100. (3)
101. In a multi – electron atom, which of the following orbitals described by the three
quantum numbers will have the same energy in the absence of magnetic acid and
electric fields?
(a) n = 1, l = 0, m = 0
(b) n = 2, l = 0, m = 0
(c) n = 2, l = 1, m = 1
(d) n = 3, l = 2, m = 1
(e) n = 3, l = 2, m = 0
(1) (a) and (b) (2) (b) and (c)
(3) (c) and (d) (4) (d) and (e)
101. (4)
n = same
102. During the process of electrolytic refining of copper, some metals present as impurity
settle as ‘anode mud’ These are
(1) Sn and Ag (2) Pb and Zn
(3) Ag and Au (4) Fe and Ni
102. (3)
103. Electrolyte KCl KNO3 HCl NaOAc NaCl
∧∞(S cm2mol-
1)
149.9 145.0 426.2 91.0 126.5
Calculate HOAc
∧∞ Using appropriate molar conductances of the electrolytes listed
above at infinite dilution in H2O at 25°C
(1) 517.2 (2) 552.7
(3) 390.7 (4) 217.5
103. (3)
AcOH HCl AcONa NaCl
∧∞ = ∧∞ + ∧∞ − ∧∞
= 390.7
104. A schematic plot of In Keq versus inverse of temperature for a reaction is shown
below
2.0 In Keq
6.0
1.5 10 3 1 (K 1) 2.0 10 3
T
× − − × −
The reaction must be
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(1) exothermic (2) endothermic
(3) one with negligible enthalpy change (4) highly spontaneous at ordinary
temperature
104. (1)
Keq = A e H
RT
− Δ
105. The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and
negatively charged, respectively, which of the following statements is NOT correct?
(1) magnesium chloride solution coagulates, the gold sol more readily than the iron
(III) hydroxide sol.
(2) sodium sulphate solution causes coagulation in both sols
(3) mixing the sols has no effect
(4) coagulation in both sols can be brought about by electrophoresis
105. (3)
106. Based on lattice energy and other considerations which one of the following alkali
metal chlorides is expected to have the highest melting point.
(1) LiCl (2) NaCl
(3) KCl (4) RbCl
106. (2)
Although lattice energy of LiCl higher than NaCl but LiCl is covalent in nature and
NaCl ionic there after , the melting point decreases as we move NaCl because the
lattice energy decreases as a size of alkali metal atom increases (lattice energy ∝ to
melting point of alkali metal halide)
107. Heating mixture of Cu2O and Cu2S will give
(1) Cu + SO2 (2) Cu + SO3
(3) CuO + CuS (4) Cu2SO3
107. (1)
2Cu2O + Cu2S → 6Cu + SO2
108. The molecular shapes of SF4, CF4 and XeF4 are
(1) the same with 2,0 and 1 lone pairs of electrons on the central atom, respectively
(2) the same with 1, 1 and 1 lone pair of electrons on the central atoms, respectively
(3) different with 0, 1 and 2 lone pair of electrons on the central atoms, respectively
(4) different with 1, 0 and 2 lone pairs of electron on the central atoms respectively
108 (4)
109. The number and type of bonds between two carbon atoms in calcium carbide are
(1) One sigma, one pi (2) One sigma, two pi
(3) Two sigma, one pi (4) Two sigma, two pi
109. (2)
2
CaC2 Ca+ C
One C
Two
σ
π
110. The oxidation state of chromium in the final product formed by the reaction between
KI and acidified potassium dichromate solution is
(1) +4 (2) +6
(3) +2 (4) +3
110. (4)
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2 3
Cr2O7 14H 6I 2Cr 7H2O 3I2 − + + + − → + + +
+6 +3
111. The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous
acid is
(1) zero (2) two
(3) one (4) three
111. (2)
H P
O
H
O
H
112. What is the conjugate base of OH-?
(1) O2 (2) H2O
(3) O- (4) O-2
112. (4)
OH- → O-2 + H+
113. The correct order of the thermal stability of hydrogen halides (H – X) is
(1) HI > HBr > HCl > HF (2) HF > HCl > HBr > HI
(3) HCl < HF > HBr < HI (4) HI > HCl < HF < HBr
113. (2)
114. Heating an aqueous solution of aluminium chloride to dryness will give
(1) AlCl3 (2) Al2Cl6
(3) Al2O3 (4) Al(OH)Cl2
114. (3)
Al2Cl6 6H2O → Al2O3 + + 6HCl + 3H2O↑
115. Calomel (Hg2Cl2) on reaction with ammonium hydroxide gives
(1) HgNH2Cl (2) NH2 – Hg – Hg – Cl
(3) Hg2O (4) HgO
115. (1)
2 2 4 2 4 2 Hg Cl + 2NH OH→Hg +Hg(NH )Cl +NH Cl + 2H O
116. In which of the following arrangements the order is NOT according to the property
indicated against it?
(1) Al3+ < Mg2+ < Na+ < FIncreasing
ionic size
(2) B < C < N < O
Increasing first ionization enthalpy
(3) I < Br < F < Cl
Increasing electron gain enthalpy (with negative sign)
(4) Li < Na < K < Rb
Increasing metallic radius
116. (2)
B < C < O < N
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117. In silicon dioxide
(1) Each silicon atom is surrounded by four oxygen atoms and each oxygen atom is
bonded to two silicon atoms
(2) Each silicon atom is surrounded by two oxygen atoms and each oxygen atom is
bonded to two silicon atoms
(3) Silicon atoms is bonded to two oxygen atoms
(4) there are double bonds between silicon and oxygen atoms
117. (1)
O Si
O
O
O Si
O
O
O
118. Of the following sets which one does NOT contain isoelectronic species?
(1) 3 2
PO4 ,SO4 ,ClO4 − − − (2) 2
CN ,N2,C2 − −
(3) 2 2
SO3 ,CO3 ,NO3 − − − (4) 3 2
BO3 ,CO3 ,NO3 − − −
118. (3)
119. The lanthanide contraction is responsible for the fact that
(1) Zr and Y have about the same radius (2) Zr and Nb have similar oxidation
state
(3) Zr and Hf have about the same radius (4) Zr and Zn have the same oxidation
119. (3)
Due to Lanthanide contraction.
120. The IUPAC name of the coordination compound K3[Fe(CN)6] is
(1) Potassium hexacyanoferrate (II) (2) Potassium hexacyanoferrate (III)
(3) Potassium hexacyanoiron (II) (4) tripotassium hexcyanoiron (II)
120. (2)
121. Which of the following compounds shows optical isomerism?
(1) [Cu(NH3)4]+2 (2) [ZnCl4]-2
(3) [Cr(C2O4)3]-3 (4) [Co(CN)6]-3
121. (3)
OX Cr Cr OX
OX
OX
OX
OX
−3 −3
122. Which one of the following cyano complexes would exhibit the lowest value of
paramagnetic behaviour?
(1) [Cr(CN)6]-3 (2) [Mn(CN)6]-3
(3) [Fe(CN)6]-3 (4) [Co(CN)6]-3
(At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27)
122. (4)
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123. 2 methylbutane on reacting with bromine in the presence of sunlight gives mainly
(1) 1 – bromo -2 - methylbutane (2) 2 – bromo -2 - methylbutane
(3) 2 – bromo -3 - methylbutane (4) 1 – bromo -3 – methylbutane
123. (2)
H3C CH
CH3
CH2 CH3 H3C C
CH3
Br
2 CH2 CH3 + Br →
Major
124. The photon of hard gamma radiation knocks a proton out of 24
12Mg nucleus to form
(1) the isotope of parent nucleus (2) the isobar of parent nucleus
(3) the nuclide 23
11Na (4) the isobar of 23
11Na
124. (3)
125. The best reagent to convert pent -3- en-2-ol into pent -3-en-2-one is
(1) Acidic permanganate (2) Acidic dichromate
(3) Chromic anhydride in glacial acetic acid (4) Pyridinium chloro – chromate
125. (3)
126. Tertiary alkyl halides are practically inert to substitution by SN
2 mechanism because
of
(1) insolubility (2) instability
(3) inductive effect (4) steric hindrance
126. (4)
127. In both DNA and RNA, heterocyclic base and phosphate ester linkages are at-
(1) l
5 C and l
2 C respectively of the sugar molecule
(2) l
2 C and l
5 C respectively of the sugar molecule
(3) l
1C and l
5 C respectively of the sugar molecule
(4) l
5 C and l
1C respectively of the sugar molecule
127. (3)
N
O N
H
O
NH2
HO
H
OH
H
H
CH2
O P - O
O
O-
5I
1I
128. Reaction of one molecule of HBr with one molecule of 1,3-butadiene at 400C gives
predominantly
(1) 3-bromobutene under kinetically controlled conditions
(2) 1-bromo-2-butene under thermodymically controlled conditions
(3) 3-bromobutene under thermodynamically controlled conditions
(4) 1-bromo-2-butene under kinetically controlled conditions
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128. (2)
129. Among the following acids which has the lowest pKa value?
(1) CH3COOH
(2) HCOOH
(3) (CH3)2COOH
(4) CH3CH2COOH
129. (2)
130. The decreasing order of nucleophilicity among the nucleophiles
(a) 3 ||
O
CH − C−O−
(b) CH3O-
(c) CN-
(d)
H3C S
O
O
O-
(1) (a), (b), (c), (d) (2) (d), (c), (b), (a)
(3) (b), (c), (a), (d) (4) (c), (b), (a), (d)
130. (4)
131. Which one of the following methods is neither meant for the synthesis nor for
separation of amines?
(1) Hinsberg method
(2) Hofmann method
(3) Wurtz reaction
(4) Curtius reaction
131. (3)
132. Which of the following is fully fluorinated polymer?
(1) Neoprene (2) Teflon
(3) Thiokol (4) PVC
132. (2)
133. Of the five isomeric hexanes, the isomer which can give two monochlorinated
compounds is
(1) n-hexane
(2) 2, 3-dimethylbutane
(3) 2,2-dimethylbutane
(4) 2-methylpentane
133. (2)
H3C C
H
CH3
C
H
CH3
CH3 Cl2→ H3C C
H
CH2Cl
C
H
CH3
CH3 H3C C
Cl
CH3
C
H
CH3
CH3
134. Alkyl halides react with dialkyl copper reagents to give
(1) alkenes (2) alkyl copper halides
(3) alkanes (3) alkenyl halides
134. (3)
2 R CuLi +R′X→R −R′ +R − Cu + LiX
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135. Acid catalyzed hydration of alkenes except ethene leads to the formation of
(1) primary alcohol
(2) secondary or tertiary alcohol
(3) mixture of primary and secondary alcohols
(4) mixture of secondary and tertiary alcohols
135. (4)
136. Amongst the following the most basic compound is
(1) benzylamine (2) aniline
(3) acetanilide (4) p-nitroaniline
136. (1)
-NH2 group is not linked with benzene ring.
137. Which types of isomerism is shown by 2,3-dichlorobutane?
(1) Diastereo (2) Optical
(3) Geometric (4) Structural
137. (2)
H
CH3
CH3
Cl
H Cl , H
CH3
CH3
Cl
Cl H , Cl
CH3
CH3
H
H Cl
138. The reaction
R C
X
O
Nu R C
Nu
O
X
is fastest when X is
(1) Cl (2) NH2
(3) OC2H5 (4) OCOR
138. (1)
Conjugated acid of Cl- is a stronger acid i.e. HCl.
139. Elimination of bromine from 2-bromobutane results in the formation of-
(1) equimolar mixture of 1 and 2-butene (2) predominantly 2-butene
(3) predominantly 1-butene (4) predominantly 2-butyne
139. (2)
Saytzeffs product.
140. Equimolar solutions in the same solvent have
(1) Same boiling point but different freezing point
(2) Same freezing point but different boiling point
(3) Same boiling and same freezing points
(4) Different boiling and different freezing points
140. (3)
141. Which of the following statements in relation to the hydrogen atom is correct?
(1) 3s orbital is lower in energy than 3p orbital
(2) 3p orbital is lower in energy than 3d orbital
(3) 3s and 3p orbitals are of lower energy than 3d orbital
(4) 3s, 3p and 3d orbitals all have the same energy
141. (4)
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142. The structure of diborane (B2H6) contains
(1) four 2c-2e bonds and two 3c-2e bonds
(2) two 2c-2e bonds and four 3c-2e bonds
(3) two 2c-2e bonds and two 3c-3e bonds
(4) four 2c-2e bonds and four 3c-2e bonds
142. (1)
B
H H
H H
B
H
H
143. The value of the ‘spin only’ magnetic moment for one of the following configurations
is 2.84 BM. The correct one is
(1) d4 (in strong ligand filed)
(2) d4 (in weak ligand field)
(3) d3 (in weak as well as in strong fields)
(4) d5 (in strong ligand field)
143. (1)
d4 in strong field, so unpaired electrons = 2.
144. Which of the following factors may be regarded as the main cause of lanthanide
contraction?
(1) Poor shielding of one of 4f electron by another in the subshell
(2) Effective shielding of one of 4f electrons by another in the subshell
(3) Poorer shielding of 5d electrons by 4f electrons
(4) Greater shielding of 5d electrons by 4f electrons
144. (1)
145. Reaction of cyclohexanone with dimethylamine in the presence of catalytic amount of
an acid forms a compound if water during the reaction is continuously removed. The
compound formed is generally known as
(1) a Schiff’s base (2) an enamine
(3) an imine (4) an amine
145. (2)
N
CH3
CH3
146. p-cresol reacts with chloroform in alkaline medium to give the compound A which
adds hydrogen cyanide to form, the compound B. The latter on acidic hydrolysis
gives chiral carboxylic acid. The structure of the carboxylic acid is
(1)
CH3
CH(OH)COOH
OH
(2)
CH3
OH
CH(OH)COOH
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(3)
CH3
CH2COOH
OH
(4)
CH3
OH
CH2COOH
146. (2)
CH3
OH
CHCl3
+OH− →
CH3
O
CHCl2
OH− →
CH3
O
CHO
A
HCN→
CH3
OH
CH
OH
CN
H3O+ →
CH3
OH
CH
OH
COOH
147. An organic compound having molecular mass 60 is found to contain C = 20%, H =
6.67% and N = 46.67% while rest is oxygen. On heating it gives NH3 alongwith a
solid residue. The solid residue give violet colour with alkaline copper sulphate
solution. The compound is
(1) CH3NCO (2) CH3CONH2
(3) (NH2)2CO (4) CH3CH2CONH2
147. (3)
148. If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the
ratio of 1:1:0.5 and fΔ H for the formation of XY is -200 kJ mole-1. The bond
dissociation energy of X2 will be
(1) 100 kJ mol-1 (2) 200 kJ mol-1
(3) 300 kJ mol-1 (4) 400 kJ mol-1
148. -
(None of the options is correct.)
(g) (g) XY→X + Y ; ΔH = +a kJ /mole ............(i)
2 X →2X; ΔH = +a kJ/mole ...............(ii)
2 Y →2Y; ΔH = +0.5a kJ/mole ............(iii)
1 (ii) 1 (iii) (i), Gives
2 2
× + × −
2 2
1 X 1 Y XY; H a 0.5 a a kJ / mole
2 2 2 2
+ → Δ = + + −
a 0.5a a 200
2 2
+ + − = −
a = 800.
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–14–
149. t1/4 can be taken as the time taken for the concentration of a reactant to drop to 3
4
of
its initial value. If the rate constant for a first order reaction is K, the t1/4 can be
written as
(1) 0.10 / K (2) 0.29 / K
(3) 0.69 / K (4) 0.75 / K
149. (2)
1/ 4
t 2.303 log 1
K 1 1
4
=
−
0.29
K
= .
150. An amount of solid NH4HS is placed in a flask already containing ammonia gas at a
certain temperature and 0.50 atm. Pressure. Ammonium hydrogen sulphide
decomposes to yield NH3 and H2S gases in the flask. When the decomposition
reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The
equilibrium constant for NH4HS decomposition at this temperature is
(1) 0.30 (2) 0.18
(3) 0.17 (4) 0.11
150. (4)
4 3(g) 2 (g) NH HS NH H S
a 0.5atm
a x 0.5 x x
+
− +
Total pressure = 0.5 + 2x = 0.84
i.e., x = 0.17
p NH3 H2S K = p . p
= (0.67). (0.17)
= 0.1139.
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FIITJEE
SOLUTION TO AIEEE-2005
MATHEMATICS
1. If A2 – A + I = 0, then the inverse of A is
(1) A + I (2) A
(3) A – I (4) I – A
1. (4)
Given A2 – A + I = 0
A–1A2 – A–1A + A–1 – I = A–1⋅0 (Multiplying A–1 on both sides)
⇒ A - I + A-1 = 0 or A–1 = I – A.
2. If the cube roots of unity are 1, ω, ω2 then the roots of the equation
(x – 1)3 + 8 = 0, are
(1) -1 , - 1 + 2ω, - 1 - 2ω2 (2) -1 , -1, - 1
(3) -1 , 1 - 2ω, 1 - 2ω2 (4) -1 , 1 + 2ω, 1 + 2ω2
2. (3)
(x – 1)3 + 8 = 0 ⇒ (x – 1) = (-2) (1)1/3
⇒ x – 1 = -2 or -2ω or -2ω2
or n = -1 or 1 – 2ω or 1 – 2ω2.
3. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on
the set A = {3, 6, 9, 12} be a relation on the set A = {3, 6, 9, 12}. The relation is
(1) reflexive and transitive only (2) reflexive only
(3) an equivalence relation (4) reflexive and symmetric only
3. (1)
Reflexive and transitive only.
e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive]
(3, 6), (6, 12), (3, 12) [Transitive].
4. Area of the greatest rectangle that can be inscribed in the ellipse
2 2
2 2
x y 1
a b
+ = is
(1) 2ab (2) ab
(3) ab (4) a
b
4. (1)
Area of rectangle ABCD = (2acosθ)
(2bsinθ) = 2absin2θ
⇒ Area of greatest rectangle is equal to
2ab
when sin2θ = 1.
(-acosθ, bsinθ)
B
(-acosθ, -bsinθ)C D(acosθ, -bsinθ)
A(acosθ, bsinθ)
X
Y
5. The differential equation representing the family of curves y2 = 2c (x + c ) , where c
> 0, is a parameter, is of order and degree as follows:
(1) order 1, degree 2 (2) order 1, degree 1
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–2–
(3) order 1, degree 3 (4) order 2, degree 2
5. (3)
y2 = 2c(x + √c) …(i)
2yy′ = 2c⋅1 or yy′ = c …(ii)
⇒ y2 = 2yy′ (x + yy′ ) [on putting value of c from (ii) in (i)]
On simplifying, we get
(y – 2xy′)2 = 4yy′3 …(iii)
Hence equation (iii) is of order 1 and degree 3.
6. 2 2 2
n 2 2 2 2 2
lim 1 sec 1 2 sec 4 …. 1 sec 1
→∞ n n n n n
+ + +
equals
(1) 1 sec1
2
(2) 1 cosec1
2
(3) tan1 (4) 1 tan1
2
6. (4)
2 2 2 2
n 2 2 2 2 2 2
lim 1 sec 1 2 sec 4 3 sec 9 …. 1 sec 1
→∞ n n n n n n n
+ + + +
is equal to
2 2
2 2
n 2 2 n 2
lim r sec r lim 1 r sec r
→∞ n n →∞ n n n
= ⋅
⇒ Given limit is equal to value of integral
1
2 2
0
∫ x sec x dx
or
1 1
2 2
0 0
1 2x sec x dx 1 sec tdt
2 2
∫ = ∫ [put x2 = t]
= ( )1
0
1 tant 1 tan1
2 2
= .
7. ABC is a triangle. Forces P, Q, R
acting along IA, IB and IC respectively are in
equilibrium, where I is the incentre of ΔABC. Then P : Q : R is
(1) sinA : sin B : sinC (2) sin A : sinB : sin C
2 2 2
(3) cos A : cosB : cos C
2 2 2
(4) cosA : cosB : cosC
7. (3)
Using Lami’s Theorem
∴P:Q:R cos A : cosB : cosC
2 2 2
= .
A
B C
I
P
Q
R
8. If in a frequently distribution, the mean and median are 21 and 22 respectively, then
its mode is approximately
(1) 22.0 (2) 20.5
(3) 25.5 (4) 24.0
8. (4)
Mode + 2Mean = 3 Median
⇒ Mode = 3 × 22 – 2 × 21= 66 – 42= 24.
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–3–
9. Let P be the point (1, 0) and Q a point on the locus y2 = 8x. The locus of mid point of
PQ is
(1) y2 – 4x + 2 = 0 (2) y2 + 4x + 2 = 0
(3) x2 + 4y + 2 = 0 (4) x2 – 4y + 2 = 0
9. (1)
P = (1, 0)
Q = (h, k) such that k2 = 8h
Let (α, β) be the midpoint of PQ
h 1
2
+
α = , k 0
2
+
β =
2α – 1 = h 2β = k.
(2β)2 = 8 (2α – 1) ⇒ β2 = 4α – 2
⇒ y2 – 4x + 2 = 0.
10. If C is the mid point of AB and P is any point outside AB, then
(1) PA + PB = 2PC
(2) PA + PB = PC
(3) PA + PB + 2PC = 0
(4) PA + PB + PC = 0
10. (1)
PA + AC + CP = 0
PB + BC + CP = 0
Adding, we get
PA + PB + AC + BC + 2CP = 0
Since AC = −BC
& CP = −PC
⇒PA + PB − 2PC = 0
.
P
A C B
11. If the coefficients of rth, (r+ 1)th and (r + 2)th terms in the binomial expansion of (1 +
y)m are in A.P., then m and r satisfy the equation
(1) m2 – m(4r – 1) + 4r2 – 2 = 0 (2) m2 – m(4r+1) + 4r2 + 2 = 0
(3) m2 – m(4r + 1) + 4r2 – 2 = 0 (4) m2 – m(4r – 1) + 4r2 + 2 = 0
11. (3)
Given m m m
r 1 r r 1 C , C, C − + are in A.P.
m m m
r r 1 r 1 2 C C C − + = +
⇒
m m
r 1 r 1
m m
r r
2 C C
C C
= − + +
= r m r
m r 1 r 1
−
+
− + +
⇒ m2 – m (4r + 1) + 4r2 – 2 = 0.
12. In a triangle PQR, ∠R =
2
π
. If tan P
2
and tan Q
2
are the roots of
ax2 + bx + c = 0, a ≠ 0 then
(1) a = b + c (2) c = a + b
(3) b = c (4) b = a + c
12. (2)
tan P , tan Q
2 2
are the roots of ax2 + bx + c = 0
tan P tan Q b
2 2 a
+ = −
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–4–
tan P tan Q c
2 2 a
=
tan P tan Q
2 2 tan P Q 1
1 tan P tan Q 2 2
2 2
+ = + = −
⇒
b
a 1
1 c
a
−
=
−
⇒ b a c
a a a
− = − ⇒ −b = a − c
c = a + b.
13. The system of equations
αx + y + z = α – 1,
x + αy + z = α – 1,
x + y + αz = α – 1
has no solution, if α is
(1) -2 (2) either – 2 or 1
(3) not -2 (4) 1
13. (1)
αx + y + z = α – 1
x + αy + z = α – 1
x + y + zα = α – 1
1 1
1 1
1 1
α
Δ = α
α
= α(α2 – 1) – 1(α – 1) + 1(1 – α)
= α (α – 1) (α + 1) – 1(α – 1) – 1(α – 1)
⇒ (α – 1)[α2 + α – 1 – 1] = 0
⇒ (α – 1)[α2 + α – 2] = 0
[α2 + 2α – α – 2] = 0
(α – 1) [α(α + 2) – 1(α + 2)] = 0
(α – 1) = 0, α + 2 = 0 ⇒ α = –2, 1; but α ≠ 1.
14. The value of α for which the sum of the squares of the roots of the equation
x2 – (a – 2)x – a – 1 = 0 assume the least value is
(1) 1 (2) 0
(3) 3 (4) 2
14. (1)
x2 – (a – 2)x – a – 1 = 0
⇒ α + β = a – 2
α β = –(a + 1)
α2 + β2 = (α + β)2 – 2αβ
= a2 – 2a + 6 = (a – 1)2 + 5
⇒ a = 1.
15. If roots of the equation x2 – bx + c = 0 be two consectutive integers, then b2 – 4c
equals
(1) – 2 (2) 3
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–5–
(3) 2 (4) 1
15. (4)
Let α, α + 1 be roots
α + α + 1 = b
α(α + 1) = c
∴ b2 – 4c = (2α + 1)2 – 4α(α + 1) = 1.
16. If the letters of word SACHIN are arranged in all possible ways and these words are
written out as in dictionary, then the word SACHIN appears at serial number
(1) 601 (2) 600
(3) 603 (4) 602
16. (1)
Alphabetical order is
A, C, H, I, N, S
No. of words starting with A – 5!
No. of words starting with C – 5!
No. of words starting with H – 5!
No. of words starting with I – 5!
No. of words starting with N – 5!
SACHIN – 1
601.
17. The value of 50C4 +
6
56 r
3
r 1
− C
= Σ
is
(1) 55C4 (2) 55C3
(3) 56C3 (4) 56C4
17. (4)
50C4 +
6
56 r
3
r 1
− C
= Σ
50 55 54 53 52 51 50
4 3 3 3 3 3 3 ⇒ C + C + C + C + C + C + C
(50 50 ) 51 52 53 54 55
4 3 3 3 3 3 3 = C + C + C + C + C + C + C
⇒ (51 51 ) 52 53 54 55
4 3 3 3 3 3 C + C + C + C + C + C
⇒ 55C4 + 55C3 = 56C4.
18. If A =
1 0
1 1
and I =
1 0
0 1
, then which one of the following holds for all n ≥ 1, by
the principle of mathematical indunction
(1) An = nA – (n – 1)I (2) An = 2n-1A – (n – 1)I
(3) An = nA + (n – 1)I (4) An = 2n-1A + (n – 1)I
18. (1)
By the principle of mathematical induction (1) is true.
19. If the coefficient of x7 in
11
ax2 1
bx
+
equals the coefficient of x-7 in
11
ax2 1
bx
−
,
then a and b satisfy the relation
(1) a – b = 1 (2) a + b = 1
(3) a
b
= 1 (4) ab = 1
19. (4)
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–6–
Tr + 1 in the expansion ( ) 11 r
2 11 2 11 r
r
ax 1 C ax 1
bx bx
+ = −
= 11Cr (a)11 – r (b)-r (x)22 – 2r – r
⇒ 22 – 3r = 7 ⇒ r = 5
∴ coefficient of x7 = 11C5(a)6 (b)-5 ……(1)
Again Tr + 1 in the expansion ( )
11 r
11 11 r
2 r 2
ax 1 C ax 1
bx bx
− = − −
= 11Cr a11 – r (-1)r × (b)-r (x)-2r (x)11 – r
Now 11 – 3r = -7 ⇒ 3r = 18 ⇒ r = 6
∴ coefficient of x-7 = 11C6 a5 × 1 × (b)-6
⇒ 11 ( )6 ( ) 5 11 5 ( ) 6
5 6 C a b Ca b − − = ×
⇒ ab = 1.
20. Let f : (-1, 1) → B, be a function defined by f(x) = 1
2
tan 2x
1 x
−
−
, then f is both one-one
and onto when B is the interval
(1) 0,
2
π
(2)0,
2
π
(3) ,
2 2
π π −
(4) ,
2 2
π π −
20. (4)
Given f(x) = 1
2
tan 2x
1 x
−
−
for x∈(-1, 1)
clearly range of f(x) = ,
2 2
π π −
∴ co-domain of function = B = ,
2 2
π π −
.
21. If z1 and z2 are two non-zero complex numbers such that |z1 + z2| = |z1| + |z2| then
argz1 – argz2 is equal to
(1)
2
π
(2) – π
(3) 0 (4) –
2
π
21. (3)
|z1 + z2| = |z1| + |z2| ⇒ z1 and z2 are collinear and are to the same side of origin;
hence arg z1 – arg z2 = 0.
22. If ω = z
z 1 i
3
−
and |ω| = 1, then z lies on
(1) an ellipse (2) a circle
(3) a straight line (4) a parabola.
22. (3)
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–7–
As given w z | w | | z |
z 1 i | z 1 i |
3 3
= ⇒ =
− −
= 1 ⇒ distance of z from origin and point
0, 1
3
is same hence z lies on bisector of the line joining points (0, 0) and (0, 1/3).
Hence z lies on a straight line.
23. If a2 + b2 + c2 = -2 and f(x) =
( ) ( )
( ) ( )
( ) ( )
2 2 2
2 2 2
2 2 2
1 a x 1 b x 1 c x
1 a x 1 b x 1 c x
1 a x 1 b x 1 c x
+ + +
+ + +
+ + +
then f(x) is a
polynomial of degree
(1) 1 (2) 0
(3) 3 (4) 2
23. (4)
( )
( ) ( ) ( )
( ) ( )
( ) ( )
2 2 2 2 2
2 2 2 2 2
2 2 2 2 2
1 a b c 2 x 1 b x 1 c x
f x 1 a b c 2 x 1 b x 1 c x
1 a b c 2 x 1 b x 1 c x
+ + + + + +
= + + + + + +
+ + + + + +
, Applying C1 → C1 + C2 + C3
=
( ) ( )
( )
( )
2 2
2 2
2 2
1 1 b x 1 c x
1 1 b x 1 c x
1 1 b x 1 c x
+ +
+ +
+ +
∵ a2 + b2 + c2 + 2 = 0
f(x) =
( 2 ) 2
0 x 1 0
0 1 x x 1
1 1 b x 1 c x
−
− −
+ +
; Applying R1 → R1 – R2 , R2 → R2 – R3
f(x) = (x – 1)2
Hence degree = 2.
24. The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ – θ cosθ) at any point ‘θ’ is
such that
(1) it passes through the origin
(2) it makes angle
2
π
+ θ with the x-axis
(3) it passes through a , a
2
π −
(4) it is at a constant distance from the origin
24. (4)
Clearly dy
dx
= tan θ ⇒ slope of normal = – cot θ
Equation of normal at ‘θ’ is
y – a(sin θ – θ cos θ) = – cot θ(x – a(cos θ + θ sin θ)
⇒ y sin θ – a sin2 θ + a θ cos θ sin θ = -x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
Clearly this is an equation of straight line which is at a constant distance ‘a’ from
origin.
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–8–
25. A function is matched below against an interval where it is supposed to be
increasing. Which of the following pairs is incorrectly matched?
Interval Function
(1) (-∞, ∞) x3 – 3×2 + 3x + 3
(2) [2, ∞) 2×3 – 3×2 – 12x + 6
(3) , 1
3
−∞
3×2 – 2x + 1
(4) (- ∞, -4] x3 + 6×2 + 6
25. (3)
Clearly function f(x) = 3×2 – 2x + 1 is increasing when
f′(x) = 6x – 2 ≥ 0 ⇒ x∈[1/3, ∞)
Hence (3) is incorrect.
26. Let α and β be the distinct roots of ax2 + bx + c = 0, then
( )
( )
2
x 2
1 cos ax bx c
lim
→α x
− + +
− α
is
equal to
(1) ( )
2
a 2
2
α − β (2) 0
(3) ( )
2
a 2
2
− α − β (4) 1 ( )2
2
α − β
26. (1)
Given limit =
( )( )
( )
( )( )
( )
2
x 2 x 2
x x
2sin a
1 cosa x x 2
lim lim
→α x →α x
− α − β
− − α − β =
− α − α
( )
( )( )
( ) ( )
( ) ( )
2
2 2 2
x 2 2 2 2
x x
sin a
2 2 a x x lim
x a x x 4
4
→α
− α − β
− α − β = × ×
− α − α − β
=
a2 ( )2
2
α − β
.
27. Suppose f(x) is differentiable x = 1 and ( )
h 0
lim 1 f 1 h 5
→ h
+ = , then f′(1) equals
(1) 3 (2) 4
(3) 5 (4) 6
27. (3)
( ) ( ) ( )
h 0
f 1 h f 1
f 1 lim
→ h
+ −
′ = ; As function is differentiable so it is continuous as it is given
that
( )
h 0
f 1 h
lim 5
→ h
+
= and hence f(1) = 0
Hence f′(1)
( )
h 0
f 1 h
lim 5
→ h
+
= =
Hence (3) is the correct answer.
28. Let f be differentiable for all x. If f(1) = – 2 and f′(x) ≥ 2 for x ∈ [1, 6] , then
(1) f(6) ≥ 8 (2) f(6) < 8
(3) f(6) < 5 (4) f(6) = 5
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–9–
28. (1)
As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6]
Applying Lagrange’s mean value theorem
f (6) f (1) ( )
f c 2
5
−
= ′ ≥
⇒ f(6) ≥ 10 + f(1)
⇒ f(6) ≥ 10 – 2
⇒ f(6) ≥ 8.
29. If f is a real-valued differentiable function satisfying |f(x) – f(y)| ≤ (x – y)2, x, y ∈ R and
f(0) = 0, then f(1) equals
(1) -1 (2) 0
(3) 2 (4) 1
29. (2)
f′(x) =
( ) ( )
h 0
f x h f x
lim
→ h
+ −
( ) ( ) ( ) ( )2
h 0 h 0
f x h f x h
| f x | lim lim
→ h → h
+ −
′ = ≤
⇒ |f′(x)| ≤ 0 ⇒ f′(x) = 0 ⇒ f(x) = constant
As f(0) = 0 ⇒ f(1) = 0.
30. If x is so small that x3 and higher powers of x may be neglected, then
( )
( )
3
3/2
1/ 2
1 x 1 1 x
2
1 x
+ − +
−
may be approximated as
(1)1 3 x2
8
− (2) 3x 3 x2
8
+
(3) 3 x2
8
− (4) x 3 x2
2 8
−
30. (3)
(1 – x)1/2 ( )
2
1 3 x 3 3 1 x2 1 3 1 x 3 2 1 x
2 2 2 2 2
+ + − − − −
= (1 – x)1/2 3 x2
8
−
= - 3 x2
8
.
31. If x = n n n
n 0 n 0 n 0
a , y b , z c
∞ ∞ ∞
= = =
Σ =Σ =Σ where a, b, c are in A.P. and |a| < 1, |b|<1, |c|< 1,
then x, y, z are in
(1) G.P. (2) A.P.
(3) Arithmetic − Geometric Progression (4) H.P.
31. (4)
x = n
n 0
a 1
1 a
∞
=
=
− Σ a = 1 1
x
−
y = n
n 0
b 1
1 b
∞
=
=
− Σ b = 1 1
y
−
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–10–
z = n
n 0
c 1
1 c
∞
=
=
− Σ c = 1 1
z
−
a, b, c are in A.P.
2b = a + c
2 1 1 1 1 1 1
y x y
− = − + −
2 1 1
y x z
= +
⇒ x, y, z are in H.P.
32. In a triangle ABC, let ∠C =
2
π
. If r is the inradius and R is the circumradius of the the
triangle ABC, then 2 (r + R) equals
(1) b + c (2) a + b
(3) a + b + c (4) c + a
32. (2)
2r + 2R = c + ( )
( ) ( )
( )
2 2ab a b c a b a b
a b c a b c
+ + +
= = +
+ + + +
( since c2 = a2 + b2).
33. If cos−1 x − cos−1 y
2
= α, then 4x2 − 4xy cos α + y2 is equal to
(1) 2 sin 2α (2) 4
(3) 4 sin2 α (4) − 4 sin2 α
33. (3)
cos-1x – cos-1
y
2 = α
( ) 2
cos 1 xy 1 x2 1 y
2 4
−
+ − − = α
2 2 22
1 xy 4 y 4x x y
cos
2
−
+ − − +
= α
⇒ 4 – y2 – 4x2 + x2y2 = 4 cos2α + x2y2 – 4xy cosα
⇒ 4x2 + y2 – 4xy cosα = 4 sin2α.
34. If in a triangle ABC, the altitudes from the vertices A, B, C on opposite sides are in
H.P., then sin A, sin B, sin C are in
(1) G.P. (2) A.P.
(3) Arithmetic − Geometric Progression (4) H.P.
34. (2)
Δ = 1 2 3
1p a 1p b 1p b
2 2 2
= =
p1, p2, p3 are in H.P.
⇒ 2 , 2 , 2
a b c
Δ Δ Δ
are in H.P.
⇒ 1, 1, 1
a b c
are in H.P
⇒ a, b, c are in A.P.
⇒ sinA, sinB, sinC are in A.P.
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35. If I1 = 2 3
1 1
x x
2
0 0
∫ 2 dx, I = ∫ 2 dx , I3 = 2 3
2 2
x x
4
1 1
∫ 2 dx and I = ∫ 2 dx then
(1) I2 > I1 (2) I1 > I2
(3) I3 = I4 (4) I3 > I4
35. (2)
I1 = 2
1
x
0
∫ 2 dx , I2 = 3
1
x
0
∫ 2 dx , I3 = 2
1
x
0
∫ 2 dx , I4 = 3
1
x
0
∫ 2 dx
∀ 0 < x < 1, x2 > x3
⇒ 2
1
x
0
∫ 2 dx > 3
1
x
0
∫ 2 dx
⇒ I1 > I2.
36. The area enclosed between the curve y = loge (x + e) and the coordinate axes is
(1) 1 (2) 2
(3) 3 (4) 4
36. (1)
Required area (OAB) = ( )
0
1 e
ln x e dx
−
∫ +
= ( )
1
0
xln x e 1 x dx
x e
+ − + ∫ = 1.
37. The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x =
4, y = 4 and the coordinate axes. If S1, S2, S3 are respectively the areas of these
parts numbered from top to bottom; then S1 : S2 : S3 is
(1) 1 : 2 : 1 (2) 1 : 2 : 3
(3) 2 : 1 : 2 (4) 1 : 1 : 1
37. (4)
y2 = 4x and x2
= 4y are symmetric about line y = x
⇒ area bounded between y2 = 4x and y = x is ( ) 4
0
2 x x dx 8
3
∫ − =
⇒ s2 A = 16
3
and s1 s3 A = A = 16
3
⇒ s1 A : s2 A : s3 A :: 1 : 1 : 1.
38. If x dy
dx
= y (log y − log x + 1), then the solution of the equation is
(1) y log x cx
y
=
(2) x log y cy
x
=
(3) log y cx
x
=
(4) log x cy
y
=
38. (3)
xdy y
dx
= (log y – log x + 1)
dy y log y 1
dx x x
= +
Put y = vx
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dy v x dv
dx dx
= +
v xdv v (log v 1)
dx
⇒ + = +
xdv v logv
dx
=
dv dx
v logv x
⇒ =
put log v = z
1 dv dz
v
=
dz dx
z x
⇒ =
ln z = ln x + ln c
z = cx
log v = cx
log y cx
x
=
.
39. The line parallel to the x−axis and passing through the intersection of the lines ax +
2by + 3b = 0 and bx − 2ay − 3a = 0, where (a, b) ≠ (0, 0) is
(1) below the x−axis at a distance of 3
2
from it
(2) below the x−axis at a distance of 2
3
from it
(3) above the x−axis at a distance of 3
2
from it
(4) above the x−axis at a distance of 2
3
from it
39. (1)
ax + 2by + 3b + λ(bx – 2ay – 3a) = 0
⇒ (a + bλ)x + (2b – 2aλ)y + 3b – 3λa = 0
a + bλ = 0 ⇒ λ = -a/b
⇒ ax + 2by + 3b – a
b
(bx – 2ay – 3a) = 0
⇒ ax + 2by + 3b – ax +
2a2 3a2 y 0
b b
+ =
2a2 3a2 y 2b 3b 0
b b
+ + + =
2b2 2a2 3b2 3a2 y
b b
+ +
= −
( )
( )
2 2
2 2
3 a b 3 y
2 b a 2
− + −
= =
+
y 3
2
= − so it is 3/2 units below x-axis.
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40. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness
than melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at
which the thickness of ice decreases, is
(1) 1
36π
cm/min (2) 1
18π
cm/min
(3) 1
54π
cm/min (4) 5
6π
cm/min
40. (2)
dv 50
dt
=
4πr2 dr 50
dt
=
⇒
( )2
dr 50
dt 4 15
=
π
where r = 15
= 1
16π
.
41.
2
2
(logx 1) dx
(1 (log x)
−
+ ∫ is equal to
(1) 2
log x C
(logx) 1
+
+
(2) 2
x C
x 1
+
+
(3)
x
2
xe C
1 x
+
+
(4) 2
x C
(logx) 1
+
+
41. (4)
( )
( ( ) )
2
2 2
logx 1
dx
1 logx
−
+
∫
=
( ( ) ) ( ( ) )2 2 2
1 2logx dx
1 logx 1 logx
−
+ +
∫
= ( )
t t
2 2 2
e 2te dt
1 t 1 t
−
+ +
∫ put logx = t ⇒ dx = et dt
( )
t
2 2 2
e 1 2t dt
1 t 1 t
−
+ +
∫
=
t
2
e c
1 t
+
+
=
( )2
x c
1 logx
+
+
42. Let f : R → R be a differentiable function having f (2) = 6, f′ (2) = 1
48
. Then
f(x) 3
x 2
6
lim 4t dt
→ x − 2 ∫ equals
(1) 24 (2) 36
(3) 12 (4) 18
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42. (4)
f (x) 3
x 2
0
lim 4t dt
→ x − 2 ∫
Applying L Hospital rule
( )2 ( )
x 2
lim 4f x f x
→
′
= 4f(2)3 f′(2)
= 4 × 63 × 1
48
= 18.
43. Let f (x) be a non−negative continuous function such that the area bounded by the
curve y = f (x), x−axis and the ordinates x =
4
π
and x = β >
4
π
is sin cos 2
4
π β β + β + β
. Then f
2
π
is
(1) 2 1
4
π + −
(2) 2 1
4
π − +
(3)1 2
4
π − −
(4) 1 2
4
π − +
43. (4)
Given that ( )
/ 4
f x dx sin cos 2
4
β
π
π
∫ = β β + β + β
Differentiating w. r. t β
f(β) = β cosβ + sinβ –
4
π
sinβ + 2
f 1 sin 2 1 2
2 4 2 4
π π π π = − + = − +
.
44. The locus of a point P (α, β) moving under the condition that the line y = αx + β is a
tangent to the hyperbola
2 2
2 2
x y 1
a b
− = is
(1) an ellipse (2) a circle
(3) a parabola (4) a hyperbola
44. (4)
Tangent to the hyperbola
2 2
2 2
x y 1
a b
− = is
y = mx ± a2m2 − b2
Given that y = αx + β is the tangent of hyperbola
⇒ m = α and a2m2 – b2 = β2
∴ a2α2 – b2 = β2
Locus is a2x2 – y2= b2 which is hyperbola.
45. If the angle θ between the line x 1 y 1 z 2
1 2 2
+ − −
= = and the plane 2x − y + λ z + 4 =
0 is such that sin θ = 1
3
the value of λ is
(1)5
3
(2) 3
5
−
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(3)3
4
(4) 4
3
−
45. (1)
Angle between line and normal to plane is
cos 2 2 2
2 3 5
π − + λ − θ = × + λ
where θ is angle between line & plane
⇒ sinθ = 2 1
3 5 3
λ
=
+ λ
⇒ λ = 5
3
.
46. The angle between the lines 2x = 3y = − z and 6x = − y = − 4z is
(1) 00 (2) 900
(3) 450 (4) 300
46. (2)
Angle between the lines 2x = 3y = – z & 6x = -y = -4z is 90°
Since a1a2 + b1b2 + c1c2 = 0.
47. If the plane 2ax − 3ay + 4az + 6 = 0 passes through the midpoint of the line joining
the centres of the spheres
x2 + y2 + z2 + 6x − 8y − 2z = 13 and
x2 + y2 + z2 − 10x + 4y − 2z = 8, then a equals
(1) − 1 (2) 1
(3) − 2 (4) 2
47. (3)
Plane
2ax – 3ay + 4az + 6 = 0 passes through the mid point of the centre of spheres
x2 + y2 + z2 + 6x – 8y – 2z = 13 and x2 + y2 + z2 – 10x + 4y – 2z = 8 respectively
centre of spheres are (-3, 4, 1) & (5, – 2, 1)
Mid point of centre is (1, 1, 1)
Satisfying this in the equation of plane, we get
2a – 3a + 4a + 6 = 0 ⇒ a = -2.
48. The distance between the line r = 2ˆi − 2ˆj + 3kˆ + λ(ˆi − ˆj + 4kˆ)
and the plane
r ⋅ (ˆi + 5ˆj + kˆ) = 5
is
(1)10
9
(2) 10
3 3
(3) 3
10
(4) 10
3
48. (2)
Distance between the line
r = 2ˆi − 2ˆj + 3kˆ + λ(ˆi − ˆj + 4kˆ )
and the plane r ⋅ (ˆi + 5ˆj + kˆ )
= 5 is
equation of plane is x + 5y + z = 5
∴ Distance of line from this plane
= perpendicular distance of point (2, -2, 3) from the plane
i.e.
2
2 10 3 5 10
1 5 1 3 3
− + −
=
+ +
.
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49. For any vectora
, the value of (a × ˆi)2 + (a × ˆj)2 + (a × kˆ)2
is equal to
(1) 2 3a (2) 2 a
(3) 2a2 (4) 4a2
49. (3)
Let a = xˆi + yˆj + zkˆ
a × ˆi = zˆj − ykˆ
⇒ ( )ˆ 2 2 2 a × i = y + z
similarly ( )ˆ 2 2 2 a × j = x + z
and ( )ˆ 2 2 2 a × k = x + y
⇒ ( )ˆ 2 2 2 a × i = y + z
similarly ( )ˆ 2 2 2 a × j = x + z
and ( )ˆ 2 2 2 a × k = x + y
⇒ ( ) ( ) ( ) ( ) ˆ 2 ˆ 2 ˆ 2 2 2 2 a × i + a × j + a × k = 2 x + y + z
= 2 2 a
.
50. If non-zero numbers a, b, c are in H.P., then the straight line x y 1 0
a b c
+ + = always
passes through a fixed point. That point is
(1) (-1, 2) (2) (-1, -2)
(3) (1, -2) (4) 1, 1
2
−
50. (3)
a, b, c are in H.P.
⇒ 2 1 1 0
b a c
− − =
x y 1 0
a b c
+ + =
x y 1
1 2 1
⇒ = =
− −
∴ x = 1, y = -2
51. If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex
are (-1, 2) and (3, 2), then the centroid of the triangle is
(1) 1, 7
3
−
(2) 1, 7
3 3
−
(3) 1, 7
3
(4) 1, 7
3 3
51. (3)
Vertex of triangle is (1, 1) and midpoint of sides
through this vertex is (-1, 2) and (3, 2)
⇒ vertex B and C come out to be
(-3, 3) and (5, 3)
∴ centroid is 1 3 5 , 1 3 3
3 3
− + + +
⇒ (1, 7/3)
A(1, 1)
(-1, 2)
(3, 2)
B C
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52. If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect in two
distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for
(1) exactly one value of a (2) no value of a
(3) infinitely many values of a (4) exactly two values of a
52. (2)
S1 = x2 + y2 + 2ax + cy + a = 0
S2 = x2 + y2 – 3ax + dy – 1 = 0
Equation of radical axis of S1 and S2
S1 – S2 = 0
⇒ 5ax + (c – d)y + a + 1 = 0
Given that 5x + by – a = 0 passes through P and Q
a c d a 1
1 b a
− +
⇒ = =−
⇒ a + 1 = -a2
a2 + a + 1 = 0
No real value of a.
53. A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius
2. The locus of the centre of the circle is
(1) an ellipse (2) a circle
(3) a hyperbola (4) a parabola
53. (4)
Equation of circle with centre (0, 3) and radius 2 is
x2 + (y – 3)2 = 4.
Let locus of the variable circle is (α, β)
∵It touches x-axis.
∴ It equation (x – α)2 + (y – β)2 = β2
Circles touch externally
∴ α2 + (β − 3)2 = 2 + β
α2 + (β – 3)2 = β2 + 4 + 4β
α2 = 10(β – 1/2)
∴ Locus is x2 = 10(y – 1/2) which is parabola.
(α, β)
54. If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally,
then the equation of the locus of its centre is
(1) x2 + y2 – 3ax – 4by + (a2 + b2 – p2) = 0 (2) 2ax + 2by – (a2 – b2 + p2) = 0
(3) x2 + y2 – 2ax – 3by + (a2 – b2 – p2) = 0 (4) 2ax + 2by – (a2 + b2 + p2) = 0
54. (4)
Let the centre be (α, β)
∵It cut the circle x2 + y2 = p2 orthogonally
2(-α) × 0 + 2(-β) × 0 = c1 – p2
c1 = p2
Let equation of circle is x2 + y2 – 2αx – 2βy + p2 = 0
It pass through (a, b) ⇒ a2 + b2 – 2αa – 2βb + p2 = 0
Locus ∴ 2ax + 2by – (a2 + b2 + p2) = 0.
55. An ellipse has OB as semi minor axis, F and F′ its focii and the angle FBF′ is a right
angle. Then the eccentricity of the ellipse is
(1) 1
2
(2) 1
2
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(3)1
4
(4) 1
3
55. (1)
∵∠FBF′ = 90o
( ) ( ) 2 2
∴ a2e2 + b2 + a2e2 + b2 = (2ae)2
⇒ 2(a2 e2 + b2) = 4a2e2
⇒ e2 = b2/a2
Also e2 = 1- b2/a2 = 1 – e2
B(0, b)
F′(-ae, 0) O F(ae, 0)
2e2 1, e 1
2
⇒ = = .
56. Let a, b and c be distinct non-negative numbers. If the vectors aˆi + aˆj + ckˆ, ˆi + kˆ and
cˆi + cˆj + bkˆ lie in a plane, then c is
(1) the Geometric Mean of a and b (2) the Arithmetic Mean of a and b
(3) equal to zero (4) the Harmonic Mean of a and b
56. (1)
Vector aˆi + aˆj + ckˆ , ˆi + kˆ and cˆi + cˆj + bkˆ are coplanar
a a c
1 0 1 0
c c b
= ⇒ c2 = ab
∴ a, b, c are in G.P.
57. If a, b, c
are non-coplanar vectors and λ is a real number then
λ (a + b)λ2b λc = a b + c b
for
(1) exactly one value of λ (2) no value of λ
(3) exactly three values of λ (4) exactly two values of λ
57. (2)
λ (a + b) λ2b λc = a b + c b
2
0 1 0 0
0 0 0 1 1
0 0 0 1 0
λ λ
λ =
λ
⇒ λ4 = -1
Hence no real value of λ.
58. Let a = ˆi − kˆ, b = xˆi + ˆj + (1− x)kˆ
andc = yˆi + xˆj + (1+ x − y)kˆ . Then a, b, c
depends on
(1) only y (2) only x
(3) both x and y (4) neither x nor y
58. (4)
a = ˆi − kˆ , b = xˆi + ˆj + (1− x)kˆ
and c = yˆi + xˆj + (1+ x − y)kˆ
a b c = a⋅(b × c)
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ˆi ˆj kˆ
b c x 1 1 x
y x 1 x y
× = −
+ −
= ˆi (1 + x – x –x2) – ˆj (x + x2- xy – y + xy) + ˆk (x2 – y)
a.(b × c)
= 1
which does not depend on x and y.
59. Three houses are available in a locality. Three persons apply for the houses. Each
applies for one house without consulting others. The probability that all the three
apply for the same house is
(1)2
9
(2) 1
9
(3)8
9
(4) 7
9
59. (2)
For a particular house being selected
Probability = 1
3
Prob(all the persons apply for the same house) = 1 1 1 3
3 3 3
× ×
= 1
9
.
60. A random variable X has Poisson distribution with mean 2. Then P(X > 1.5) equals
(1) 2
2
e
(2) 0
(3) 2
1 3
e
− (4) 2
3
e
60. (3)
P(x = k) =
k
e
k!
−λ λ
P(x ≥ 2) = 1 – P(x = 0) – P(x = 1)
= 1 – e-λ – e-λ
1!
λ
= 1 – 2
3
e
.
61. Let A and B be two events such thatP(A B) 1
6
∪ = , P(A B) 1
4
∩ = andP(A) 1
4
= ,
where A stands for complement of event A. Then events A and B are
(1) equally likely and mutually exclusive
(2) equally likely but not independent
(3) independent but not equally likely
(4) mutually exclusive and independent
61. (3)
P(A B) 1
6
∪ = , P(A ∩ B) = 1
4
and P(A) 1
4
=
⇒ P(A ∪ B) = 5/6 P(A) = 3/4
Also P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(B) = 5/6 – 3/4 + 1/4 = 1/3
P(A) P(B) = 3/4 – 1/3 = 1/4 = P(A ∩ B)
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Hence A and B are independent but not equally likely.
62. A lizard, at an initial distance of 21 cm behind an insect, moves from rest with an
acceleration of 2 cm/s2 and pursues the insect which is crawling uniformly along a
straight line at a speed of 20 cm/s. Then the lizard will catch the insect after
(1) 20 s (2) 1 s
(3) 21 s (4) 24 s
62. (3)
1 2t2
2
= 21 + 20t
⇒ t = 21.
63. Two points A and B move from rest along a straight line with constant acceleration f
and f′ respectively. If A takes m sec. more than B and describes ‘n’ units more than B
in acquiring the same speed then
(1) (f – f′)m2 = ff′n (2) (f + f′)m2 = ff′n
(3) 1 (f f )m ff n2
2
+ ′ = ′ (4) (f f )n 1 ff m2
2
′ − = ′
63. (4)
v2 = 2f(d + n) = 2f′d
v = f′(t) = (m + t)f
eliminate d and m we get
(f′ – f)n = 1 ff m2
2
′ .
64. A and B are two like parallel forces. A couple of moment H lies in the plane of A and
B and is contained with them. The resultant of A and B after combining is displaced
through a distance
(1) 2H
A −B
(2) H
A + B
(3) ( )
H
2 A + B
(4) H
A − B
64. (2)
(A + B) = d = H
d = H
A B
+
.
65. The resultant R of two forces acting on a particle is at right angles to one of them and
its magnitude is one third of the other force. The ratio of larger force to smaller one is
(1) 2 : 1 (2) 3 : 2
(3) 3 : 2 (4) 3 : 2 2
65. (4)
F′ = 3F cos θ
F = 3F sin θ
⇒ F′ = 2 2 F
F : F′ : : 3 : 2 2.
3F
F
F′
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66. The sum of the series 1 1 1 1 ………
4.2! 16.4! 64.6!
+ + + + ad inf. is
(1) e 1
e
−
(2) e 1
e
+
(3) e 1
2 e
−
(4) e 1
2 e
+
66. (4)
ex e x x2 x4 x6 1 …….
2 2! 4! 6!
+ −
= + + + +
putting x = 1/2 we get
e 1
2 e
+ .
67. The value of
2
x
cos x
1 a
π
−π + ∫ dx, a > 0, is
(1) a π (2)
2
π
(3)
a
π
(4) 2 π
67. (2)
2
2
x
0
cos x dx cos x dx
1 a 2
π π
−π
π
= =
+ ∫ ∫ .
68. The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of
radius
(1) 3 (2) 1
(3) 2 (4) 2
68. (2)
Perpendicular distance of centre 1, 0, 1
2 2
−
from x + 2y – 2 = 4
1 1 4
2 2 3
6 2
+ −
=
radius = 5 3 1
2 2
− = .
69. If the pair of lines ax2 + 2(a + b)xy + by2 = 0 lie along diameters of a circle and divide
the circle into four sectors such that the area of one of the sectors is thrice the area
of another sector then
(1) 3a2 – 10ab + 3b2 = 0 (2) 3a2 – 2ab + 3b2 = 0
(3) 3a2 + 10ab + 3b2 = 0 (4) 3a2 + 2ab + 3b2 = 0
69. (4)
( )2 2 a b ab
1
a b
+ −
=
+
⇒ (a + b)2 = 4(a2 + b2 + ab)
⇒ 3a2 + 3b2 + 2ab = 0.
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70. Let x1, x2, …,xn be n observations such that 2
Σxi = 400 and i Σx = 80 . Then a
possible value of n among the following is
(1) 15 (2) 18
(3) 9 (4) 12
70. (2)
2 2
i i x x
n n
≥
Σ Σ
⇒ n ≥ 16.
71. A particle is projected from a point O with velocity u at an angle of 60o with the
horizontal. When it is moving in a direction at right angles to its direction at O, its
velocity then is given by
(1) u
3
(2) u
2
(3) 2u
3
(4) u
3
71. (4)
u cos 60o = v cos 30o
v = 4
3
.
60o
30o
30o
72. If both the roots of the quadratic equation x2 – 2kx + k2 + k – 5 = 0 are less than 5,
then k lies in the interval
(1) (5, 6] (2) (6, ∞)
(3) (-∞, 4) (4) [4, 5]
72. (3)
b 5
2a
− <
f(5) > 0
⇒ k∈(-∞, 4).
73. If a1, a2, a3,…, an,… are in G.P., then the determinant
n n1 n 2
n 3 n 4 n 5
n 6 n 7 n 8
loga loga loga
loga loga loga
loga loga loga
+ +
+ + +
+ + +
Δ = is equal to
(1) 1 (2) 0
(3) 4 (4) 2
73. (2)
C1 – C2, C2 – C3
two rows becomes identical
Answer: 0.
74. A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x)
f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to
(1) –f(x) (2) f(x)
(3) f(a) + f(a – x) (4) f(-x)
FIITJEE Ltd. ICES House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi – 110016, Ph : 26515949, 26569493, Fax : 26513942
–23–
74. (1)
f(a – (x – a)) = f(a) f(x – a) – f(0) f(x)
= -f(x) ∵ x = 0, y = 0, f (0) = f2 (0) − f2 (a) ⇒ f2 (a) = 0 ⇒ f (a) = 0 .
75. If the equation
n n1
n n1 1 a x a x − …… a x 0
− + + + = , a1 ≠ 0, n ≥ 2, has a positive root x = α, then the
equation n 1 ( ) n 2
n n1 1 na x − n 1 a x − ….. a 0
− + − + + = has a positive root, which is
(1) greater than α (2) smaller than α
(3) greater than or equal to α (4) equal to α
75. (2)
f(0) = 0, f(α) = 0
⇒ f′(k) = 0 for some k∈(0, α).