AIEEE 2005 PHYSICS PAPER

PHYSICS
1. A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time
f
t and then decelerates at the rate to come to rest. If the total distance traversed is 15 S, then
2
12 12 12
1) S= ft 2) S = ft 3) S= ft 4) S= ft
6 4 2
f fl2
Sol : t1 t 2t1
m
15 S
o
.c
12 1f 2
f1 + f1t + x 2t1 = 15S
2 22
e
S + ft1t + 2 S = 15S
⇒
g
ft1t = 12 S (1)
⇒
a
12
ft1 = S (2)
p
2
Dividing (1) by (2), we get
e
1
e
t1 =
6
ie
2
ft 2
1 t
S= f  =
⇒
.a
2 6 72
None of the given options is correct
w
2. A particle is moving eastwards with a velocity of 5 ms-1. In 10 seconds the velocity charges to 5 ms-1 northwards.
The average acceleration in this time is
w 1
1
1) ms-2 towards north 2) ms-2 towards north-east
w
2
2
1
3) ms-2 towards north-west 4) zero
2
r
Change in velocity ∆v
Sol: (2) Average acceleration = =
Time interval t v2
r
∆v = v 2 + (− v1 ) N
r r r
∆v = ( v 2 − v1 ) = v1 + v 2 + 2v1v 2 cos90
2
2
[As | v1 | = | v2 | = 5m / s]
= 52 + 52 + 0
90°
E
W
= 5 2 m/s v1
− v1
r
∆v 5 2 1
m/s 2 towards North - West
Average acceleration = = =
t 10 2
S
3. Out of the following pair, which one does NOT have identical dimensions is
1) impulse and momentus 2) angular momentum and Planck’s constant
3) work and torque 4) moment of inertia and moment of force
Sol: (4)
The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is
4.
1) 2 b v3 2) -2 a b v2 3) 2 a v2 4) - 2 a v3
Sol: (4)
t = ax2 + bx; diff. with respect to time (t)
d dx
d dx
t = a (x2 ) + b = a.2 x + b.v.
dt dt
dt dt
1 = 2axv + bv = v (2ax + b)
1
.. diff.
2ax + b =
v
m
1
o
f
2av = -
v2
.c
f = -2av3
5. A smooth block is released at rest on a 45 incline and then slides a distance ‘d’. The time taken to slide is ‘n’
0
e
times as much a slide on rough incline than on a smooth incline. The coefficient of friction is
g
1 1 1 1
μk = 1- μk = 1 - μ8 = 1 - μ8 = 1 -
1) 2) 3) 4)
2 2 2
n2
a
n n n
p
Sol: 2)
eg s in θ − μ g c o s θ
e
ie
d
.a
smooth rough
w
1
d= (g sin θ )t2
2
w
1
g sinθ . t 2
d=
2
w 2d
2d
t1 = t2 =
g sin θ g sin θ - μg cos θ
2d 2d
=
n
g sin θ g sin θ - μg cosθ
1
n= 1− μ
1
n2 =
1− μ
1
1− μ =
n2
1
μ = 1-
n2
6. 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?
a) 182 m 2) 91 m 3) 111 m 4) 293 m
Sol : (4)
50m
v
a = -2 m/s
3 m/s
v = 2gh
v = 2 × 9.8 × 50 = 14 5
m
v 2 - u 2 32 - 980
o
S= = ≈ 244 m
2× 2 4
.c
Initially he has fallen 50 m
∴ Total hight from where he bailed out = 244 + 50 = 294 m
e
7. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate
g
before coming to rest further it will penetrate before coming to rest assuming that it faces consistant resistance to
a
motion ?
p
1) 2.0 m 2) 3.0 m 3) 1.0 m 4) 1.5 m
Sol: (3)
e
e
x
3cm
u
ie
u v=0
2
.a
2
u 2
  - u = 2.a .3
2

w
3u2
= 2.a.3
w
u
u2
w
a=
8
2
u
0 -  = 2.a .x
2
u2 u2
= 2. ×x
8
4
x = 1 cm
An annular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The
8.
F1
ration of the forces experienced by the two particles situated on the inner and outer parts of the ring, F is
2
2
 R1  R2 R1
  2) 3) 4) 1
1) R  R1 R2
2 
Sol: 3)
a2
R2
V2 = ω R 2
a1 R1
V1 = ω R 1
v12 ω 2 R12
= ω 2 R1
a1 = =
R1 R1
v22
= ω 2 R2
a2 =
R2
F1 ma 1 a1 R 1
m
= = =
taking particle masses equal F2 ma 2 a 2 R 2
o
9. A projectile can have the same range ‘R’ for two angles of projection. If ‘t1’ and ‘t2’ be the times of flights in the
.c
two cases, then the product of the two time of flights is proportional to
1 1
e
2) 3) R 4)
1) R2
R2 R
g
Sol: 3)
a
2R
T1 T2 = g (It is a formula)
p
e
T, T2 ∞ R
e
10. 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
ie
given by
2 cos φ 2 sin φ tan φ 2 tan φ
1) 2) 3) 4)
.a
Sol: 4)
Acclaration of block while sliding down upper half = g sin θ ;
w
retardation of block while sliding down lower half = - ( g sin θ - μg cosθ )
w
for the block to come to rest at the bottom, acclaration in I half = retardation in II half.
g sin θ = - ( g sin θ - μg cosθ )
w
⇒ μ = 2 tan φ
11. A mass ‘m’ moves with a velocity ‘v’ and collides inelastically with another identical mass. After collision the Ist
v
mass moves with velocity in a direction perpendicular to the initial direction of motion. Find the speed of the
3
2nd mass after collision.
v
3 after
collision
m m
before
collision
v 2
1) 2) v 3) 4) v
3v 3 3
Sol: 4)
v
=v
3
u2 = 0
u1 = V
m m
In x direction : mv + O = m (O) + m(v2) x
v
In y direction : O + O = m   + m (v 2 ) y is
 
 3
V
(V2 ) y = (V2 ) x = V
3
m
2
V V2
o
4
∴V2 =   + V2 = + V2 = v
  3
3
 3
.c
2V
=
e
3
g
12. 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 ?
a
1) 15 m/s2 2) 3 m/s2 3) 10 m/s2 4) 5 m/s2
p
Sol: 3)
e
m = 0.3 ⇒ F = m.a. = -15 x
e
15 - 150
a=− x= x = - 50x
0.3 3
ie
m
a = - 50 x 0.2 = 10
.a
s2
13. The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K
and compresses it by length L. The maximum momentum of the block after collision is
w
w M
w
KL2 ML2
1) 2) 3) 4) Zero
MK L
2M K
Sol: 2)
M
K
1
1
mv 2 = kl 2 : V = .L
m
2
2
K
.L
Momentam = m x v = m x
m
= km . L
14. A block is kept on a frictionless inclined surface with angle of inclination ‘α ’. The incline is given an acceleration
‘α ’ to keep the block stationary. Then α is equal to
1) g cosec α 2) g / tan α 3) g tan α 4) g
Sol: 3)
a cos
a
m
a
g sin α
o
.c
g sin α = a cos α
a = g tan α
e
15. 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
g
the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20
a
m above the ground. The velocity attained by the ball is
p
10 30 m/s 4) 10 m/s
1) 20 m/s 2) 40 m/s 3)
e
Sol: 2)
e
ie
.a
30
100
20
w
mgh 1
mv 2 + mgh
w
2
w
12
10 x 100 = v + 10 x 20
2
12 m
v = 800, v = 1600 = 40
2 s

1
16. A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mas
3
2
M and a body C of mass M. The centre of mass of bodies B and C taken together shifts compared to that of
3
body A towards
1) does not shift 2) depends on height of breaking
3) body B 4) body C
Sol: 1)
does not shift as no external face acts.
17. 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
2 1 1
Mr 2 Mr 2 Mr 2
1) 2) 3) 4) Mr 2
m
5 4 2
Sol: 3)
o
1
.c
Mr 2
2
18. 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
e
the surface of earth. When both ‘d’ and ‘h’ are much smaler than the radius of earth, then which one of the
g
following is correct?
a
3h h
1) d= 2) d= 3) d=h 4) d=2h
p
2 2
Sol. 4)
e
 2h   d
e
g h1 = g 1 -  ; g d1 = g 1 - 
 R  R
ie
∴ d = 2h
.a
19. 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 x 10-11 Nm2 / Kg2)
w 13.34 x 10-10 J 3) 6.67 x 10-10 J 4) 6.67 x 10-9 J
3.33 x 10-10 J
1) 2)
w
Sol. 3)
w
GM
V=−
R
GM 6.67 x 10-11 x 100
V =- =-
R 0.1
= - 6.67 x 10 -8
‘V’ at infinity (at large distance) = 0
10
x 6.67 x 10 -8
ω= m x v =
1000
= 6.67 x 10-10 J
20. 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.
l
B
A
P
2l
F
C
2 4
3
l l
l
1) 2) 3) l 4)
3 3
2
Sol. 4)
m
o
l
B
A
.c
P
2l
(0, l )
e
F
g
C
a
To have linear motion. The force F has 0.2l to be applied at centre of mass.
p
i.e. the paint ‘p’ has to centre of mass
e
4l 2 4l
2l x l + l x 2l
e
=
= Y= =
l + 2l 3
3l
ie
21. Which of the following is incorrect regarding the first law of thermodynamics?
.a
1) It is a restatement of the principle of conservation of energy
2) It is not applicable to any cyclic process
3) It introduces the concept of the entropy
w
4) It introduces the concept of the internal energy
w
Sol. 2)
w
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
[μ k = 0.5]
1) 1000 m 2) 800 m 3) 400 m 4) 100 m
Sol. 1)
v 2 − u 2 = 2as
0 − u 2 = 2 (-μg) s
1
− 100 2 = 2 x - x 10 x s
2
s = 1000 m
23. A body of mass m is accelerated uniformly from rest to a speed v is a time T. The instantaneous power delivered
to the body as a function of time is given by
mv 2 1 mv 2 2
mv 2 1 mv 2
.t 2 .t .t .t
2) 3) 4)
1)
T2 2 T2
T2 2 T2
Sol: 2)
u = 0; v = u + aT; v = a.T.
Instantaneous power = F x v = m.a.v = m.a.at = m.a2.t
v2
∴ instantaneous power = m t
T2
24. Average density of the earth
1) is a complex function of g 2) does not depend on g
3) is inversely proportional to g 4) is directly proportional to g
Sol: 4)
Gm G ρ x v
g= =
R2 R2
m
4
o
G × P × πR 3
3
.c
g=
R2
e
4
g= ρπ G.R.
g
3
a
ρ → average density
p
 3g 
ρ= 
e
 4π G R 
e
‘ ρ ’ is directly profactional to (g)
ie
25. If ‘S’ is stress and ‘Y’ is Young’s modulus of material of a wire, the enrgy stored in the wire per unit volume is
.a
S2 2Y
S
2S2Y
2) 3) 4)
1)
S2
2Y
2Y
w
Sol. 1)
Energy stored per unit volume
w
Stress 

Q Y = Strain 
w


 Stress 
Q Strain = Y 
 
1
x stress x strain
=
2
1 strain
x stress x
=
2 Y
1 S2
= .
2Y
26. 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) 10 cm 2) 8 cm 3) 20 cm 4) 4 cm
Sol. 3)
Water fills the tube entirely in gravity less condition.
∴ 20 cm.
27. The figure shows a system of two concentric spheres of radii r1 and r2 kept at temperatures T1 and T2, respectively.
The radial rate of flow of heat in a substance between the two concentric spheres is proportional is
r1
T1
T2
r2
r  r1 r2
In  r 
2
1) 2) (r2 - r1) / (r1r2) 3) (r2 - r1) 4)
 (r2 - r1 )
 1
m
Sol: 4)
o
.c
T − ∆T
dr
e
T1
g
r1
a
T2
p
r2
e
Consider a shall of thickness (dr) and of radii (r).
e
It the temperature of inner and outer surfaces of this shell be
T, (T -dT)
ie
dθ
= rate of flow of heat though it
.a
dt
kA((T − dT ) − T )
=
w
dr
−kA - dt dT
w
= - 4π kr 2
=
dr dr
w
- (dr/dT) dr
dt =
4πk r 2
integrating between the proper limits
T2 r2
 dQ  1 dr
∫ ∫r
dt = -  
 dt  4πK 2
T r1
1
dQ  1 1 

-
(T2 - T1 ) = -
dt. 4πk  r1 r2 
 
dQ 4πK r1r2 (T1 − T2 )
=
dt r2 - r1
dQ r1 r2
∴ x
dt r2 - r1
28. A system goes from A to B via two processes I and II as shown in figure. If ∆ U1 and ∆ U2 are the changes in
internal energies in the processes I and II respectively, then
p
II
B
A
I
V
1) relation between ∆ U1 and ∆ U2 can not be determined 2) ∆ U1 = ∆ U2
3) ∆ U2 < ∆ U1 3) ∆ U2 > ∆ U1
Sol: 2)
As in a cyclic process, net internal energy is zero
∆U1 = ∆U 2
m
29. The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
o
.c
T
e
< q1 =" To" toso =" To" 2 =" To" q3 =" 0" 21 =" 1−" 2 =" 1−" r2 =" x" 6 =" 6" undecayed =" N" 3=" n="" t =" half" period =" ="" e ="" 1 ="" e =" " i =" I" d =" log" 36 =" log" d =" log" 3 =" 1" d=" ="" 5 =" .a" w =" 1.63" n="4" n="3" n="2" n="1" 11 =" hc" e =" hc" h =" 4" 5 =" hc" h =" a" e =" hc" h =" L" e =" hc" h ="1→3" a =" 10" z =" 5" t=" Input" f =" 50" t1 =" ="" f1 =" 100" y1 =" 0.1" y2 =" cos" v1 =" 0.1×100π" v2 =" −0.1" t =" 0.1π" diff =" φ1" c =" ="" c =" ="" r=" cm" light =" 500" max =" 5" m =" 160" pm =" ="" f =" sin" 2 =" 2" i =" Io" y =" 0" n2 =" 6" n2 =" 6" 2 =" 6" 2 =" 1944" x =" 0" 2x =" dx" t=" ="" n ="n" 5 =";" 100 =" 20%" x =" 0" x =" L" x="0" x="L" 2q =" 2" 1 =" 2x" x =" L" x =" 0," distance =" L" l =" 2L" i =" I" light =" 2" light =" I" q =" −" v =" v1" 2 =" .+" 2q =" .−" f =" Eq" velocity =" V" v =" 2V" 2 =" m.s" v =" Sol:" p =" 0" i2 =" (attractive)" h=" (as" 2 =" (5" i=" 50" i=" ="" r =" 6" r =" 100" y =" ⇒G" 150 =" 15mA" g =" Full" current =" 10" 5 =" 9995Ω" v =" voltage" measured =" 15"> R1). If the potential difference across the source having internal resistance R2 is zero, then
R = R2 - R1 2) R = R2 x (R1 + R2) / (R2 - R1)
1)
R = R1 R2 / (R2 - R1) 4) R = R1 R2 / (R1 + R2)
3)
Sol: 1)
ε ε
R1 R2 I
2v
R
2ε
I=
R + R 1 + R2
Potential difference across L cell
= V = ε − i R2 = 0
m
2ε
o
ε− .R2 = 0
R + R1 + R2
.c
R + R1 + R2 − 2 R2 = 0
e
R + R1 − R2
g
∴ R = R2 − R1
a
64. Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows through
p
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
e
e
q q
z2 z1
z z1
ie
q q
2) 3) 4)
1) 1+ 2 1+ z1 z2
z1 z2
.a
Sol: 1)
z q
1
⇒ 1 = 2 .......(i )
m = zq ⇒ z ∝
w
q z 2 q1
w
q
q q
= 1 +1⇒ q2 =
also q = q1 + q 2 ⇒ ...(ii )
q
q 2 q2
1+ 1
q2
w q
q2 =
z
From equation (1) and (ii) 1+ 2
z1
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
( μ 0 = 4π x 10-7 Wb / A..m.)
10-5 12 x 10-5 7 x 10-5 5 x 10-5
1) 2) 3) 4)
Sol: 4)
(1)
(2)
μ 0 i1 μ 0 × 3
B1 = =
2(2π ) 4π
μ0
2 2
B2 = B1 + B2 = .5
4π
= 10 −7 × 5 × 10 2
= 5 ×10−5 ωb /n 2
m
66. In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance
o
of 2Ω , the balancing length becomes 120 cm. The internal resistance of the cell is
.c
2) 3) 4)
1) 0 .5 Ω 1Ω 2Ω 4Ω
e
Sol: 3)
g
l −l  240 − 120
r =  1 2 × R = ×2
l  120
a
2
p
= 2Ω
e
67. A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic
field B. The time taken by the particle to complete one revolution is
e 2π m
ie
2π m q 2πq B
2π q 2 B
1) 2) 3) 4)
qB
B m
m
.a
Sol: 3)
mv 2
= qvB
w
r
2π r 2π m
w
T= =
v qB
w
68. A magnetic needle is kept in a non-uniform magnetic field. It experiences
1) neither a force for a torque 2) a torque but not a force
3) a force but not a torque 4) a force and a torque
Sol: 4)
A magnetic needle kept in non-uniform magnetic field experience face and torque due to unequal forces
acting on poles
69. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100
W and 200 V lamp when not in use ?
1) 2) 3) 4)
20 Ω 40 Ω 200 Ω 400 Ω
Sol: 2)
V2
P = Vi =
R
V 2 200 × 200
R= = = 400 Ω
100
P
400
R cold = = 40Ω
10
70. 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
8μ F 4μ F 2μ F 1μ F
1) 2) 3) 4)
Sol: 4)
m
1
P=n=
o
2π 2c
.c
1
50 =
2π 50 × c
e
C = 1 uf.
g
71. An energy source will supply a constant current into the load if its internal resistance is
a
1) very large as compared to the load resistance 2) equal to the resistance of the load
p
3) non-zero but less than the resistance of the load 4) zero
e
E
e
Sol: 4) I = , Internal resistance (r) is
R+r
ie
E
zero, I = = cons tan t
R
.a
π
72. The phase difference between the alternating current and emf is . Which of the following cannot be the
2
w
constituent of the circuit?
w
1) R, L 2) C alone 3) L alone 4) L, C
Sol: 4)
w
73. A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be -
1) 0.4 2) 0.8 3) 0.125 4) 1.25
Sol:.2)
R 12 4
Power factor = cos φ = = = = 0 .8
Z 15 5
74. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an
electron is projected along the direction of the fields with a certain velocity then -
1) its velocity will increase 2) its velocity will decrease
3) it will turn towards left of direction of motion 4) it will turn towards right of direction of motion
Sol:.2)
due to electricfield, it experiences force and accelerates i.e, its velocity decreases.
75. A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage 2 V. The current reaches half
of its steady state value in
1) 0.1 s 2) 0.05 s 3) 0.3 s 4) 0.15 s
Sol: 1)
 − −
It
i = i0  1 − e L  

 
i0
= i0 (1 − e)
2
L
× 0.693 = t
R
300 × 10 −3
= 0.693 ×
m

AIEEE 2005 MATH PAPER

Maths
1. If A2 - A+ I = 0, then the inverse of A is
1) A + I 2) A 3) A -1 4) I - A
Ans.(4) A2 - A+ I = 0
⇒ A2 = I − A
⇒ A A = I−A
∴ A -1 = I − A
2. If the cube roots of unity are I, W, W2 then the roots of the equation (x-1)3+8 = 0, are
2) -1, -1, -1
1) -1, -1+2w, -1 -2w2
m
3) -1, 1-2w, 1-2w2 4) -1, 1+2w, 1+2w2
Ans.(3) (x-1)3 + 8 = 0
o
Put x -1 = y
.c
→ y3 + 8 = 0
e
⇒ y = - 2, - 2 ω , - 2 ω 2
g
⇒ x = - 1, 1 - 2 ω , 1 - 2 ω 2
a
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}. The
relation is
p
1) reflexive and transitive is 2) reflexive only
e
3) an equivalence relation 4) reflexive and symmetric only
e
Ans.(1) R is reflexive and transitive but not symmetric because (9, 3) ∉ R
ie
x2 y2
+ = 1 is
4. Area of the greatest rectangle that can be inscribed in the ellipse 2
b2
a
.a
a
1) 2ba 2) ab 3) 4)
ab b
w
Ans.(1) Area of greatest rectangle
1
× 4ab
=
w
2
= 2ab
w ( ), where c > 0, is a parameter, is of order
5. The differential equation representing the family of curves y 2 = 2c x + c
and degree as follows :
1) order 1, degree 2 2) order 1, degree1
3) order 1, degree 3 4) order 2, degree 2
Ans.(3) y2 = 2c(x + c ) c < 0
since there is only one arbitary constant
order = 1
dy
= 2c
2y
dx
dy
⇒c= y
dx
3/ 2
 dy   dy 
⇒ y 2 = 2 y  x + 2  y 
 dx 
 dx 
Elimainating radical powers,
We get, degree = 3
1 
1 2 4 1
Lim  2 sec 2 2 + 2 sec 2 2 + ... + sec 2 1 is
6. n
n →∞  n 
n n n
1 1 1
sec 1 cosec 1 tan 1
2) 3) tan 1 4)
1)
2 2 2
1 n2 
1 2 4 n
1 Lt  2 sec 2 2 + 2 . sec 2 2 + ..... + 2 sec 2 2 
tan 1
Ans.(4) n →∞  n n
n n n n
2  
2
n
2 r 
r
1
∑
⇒ Lt   sec  
n n n
n →∞
r =1
1
∫ x sec 2 (x 2 )dn
⇒
m
0
o
1
1
∫ sec 2 t dt let x 2 = t
⇒
.c
2
0
2 x dx = dt
e
1
⇒ tan 1
g
2
a
7. ABC is a triangle. Forces P, Q, R acting along IA, IB and IC respectively are in equilibrium, where I is the incentre
p
of ∆ ABC . Then P : Q : R is
e
A B C
2) sin : sin : sin
1) sin A : sin B : sin c
2 2 2
e
ie
A B C
3) cos : cos : cos 4) cos A : cos B : cos C
2 2 2
.a
B
w
P
w
Ans.(3)
w
 B+c R
Qπ −  
2
B/2 C/2
Applying Lami’s theorem
P Q R
=
=
 A+ B 
 A+C 
 B+C 

 sin 
 sin 
sin 
2 2 2
A B C
⇒ P :Q : R = cos : cos : cos
2 2 2
8. If in a frequency 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
Ans.(4) Mode = 3 median - 2 mean
= 66 - 42
= 24
Let P be the point (1, 0) and Q a point on the locus y 2 = 8 x . The locus of mid point PQ is
9.
1) y 2 − 4 x + 2 = 0 2) y 2 + 4 x + 2 = 0 x2 + 4 y + 2 = 0 4) x 2 − 4 y + 2 = 0
3)
Ans.(1) Let the mid point be = (h, k)
Let Q = ( x1 y1 )
1 + x1 0 + y1
=h =k
2 2
⇒ x1 = 2 h − 1
y1 = 2k
x1 y1 lies on y 2 = 8x
m
⇒ 4k 2 = 8(2h − 1)
o
⇒ k 2 = 2 ( 2h − 1)
.c
∴ The eqn. of locus is
e
y 2 − 4x + 2 = 0
g
10. If C is the mid point of AB and P is any point outside AB, then
a
1) PA + PB = 2 PC 2) PA + PB = PC
p
r r
3) PA + PB + 2 PC = 0 4) PA + PB + PC = 0
e
Ans.(1) A
e
P
C B
ie
..(1)
PA = PC + CA
...(2)
.a
PA = PC + BC
Adding 2 PA = 2 PC + CA + BC
w
⇒ PA + PB = 2 PC
w
11. If the coeffieients 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
w
1) m 2 − m(4r − 1) + 4r 2 − 2 = 0 2) m 2 − m(4r + 1) + 4r 2 + 2 = 0
3) m 2 − m(4r + 1) + 4r 2 − 2 = 0 4) m 2 − m(4r − 1) + 4r 2 + 2 = 0
Ans.(3) Coefficients of r, (r+1)th or (r+2)th terms are in A.P.
⇒ 2 m C r = m C r -1 + m Cr +1
2 1 1
⇒ = +
r(m - r) (m - r + 1) (m - r) (r + 1) r
⇒ m 2 − m(4r + 1) + 4r 2 − 2 = 0
P Q
π
. If tan  2  and tan  2  are the roots of ax 2 + bx + c = 0, a ≠ 0 then
12. In a triangle PQR, ∠ R =
 
2
1) a = b + c 2) c = a + b 3) b = c 4) b = a + c
P
Ans.(2)
π
2
Q
R
⇒ P +Q = π/2
P+Q
=π /4
⇒
2
m
P Q
⇒ tan  +  = 1
2 2
o
.c
tan p + tan Q
2 2 =1
1 − tan p tan Q
e
2 2
g
∴ tan P & tan Q
2 are the root of the eqn. ax + bx + c = 0
2
2
a
-b
a = 1 ⇒ - b = 1− c
p
⇒
a
c a
1-
a
e
⇒ -b =a -c
e
⇒ c =a+b
ie
13. The systems of equations
αx+y+z= α-1
.a
x + α y + z = α -1
x+y+ αz= α-1
w
has no solution, if α is
w
1) -2 2) either -2 or 1
3) not -2 4) 1
w
Ans.(1) ∆ =0
α11
1 α 1 =0
11α
From options, it can be easily verified
that α = -2
The value of α for which the sum of the squares of the roots of the equation x 2 − (a − 2) x − a − 1 = 0 assume the least
14.
value is
1) 1 2) 0 3) 3 4) 2
Ans.(1) Let α , β be the roots
α + β = + (a - 2) = a - 2
α β = − (a + 1)
α 2 + β2 = (α + β) 2 − 2 α β
= (a − 2) 2 + 2 (a + 1)
= a 2 − 4a + 4 + 2a + 2
= a 2 − 2a + 6
= (a − 1) 2 + 5
∴α 2 + β 2 is least value a = 1
m
If the roots of the equation x 2 − bx + c = 0 be two consecutive integers, then b2 - 4c equals
15.
o
1) -2 2) 3 3) 2 4) 1
.c
Ans.(4) Let α , α +1 be the roots
α+ α+1=b
e
⇒ 2α + 1 = b
g
α (α + 1) = c
a
b 2 − 4c
p
= (2α +1)2 − 4 (α2 + α)
e
( )
= 4α2 + 4α +1− 4α2 + 4α = 1

16. If the letters of the words SACHIN are arrange in all possible ways and these words are written out as in dictionary,
then the words are SACHIN appears at serial number.
1) 601 2) 600 3) 603 4) 602
Ans.(1) S A C H I N
A - order : - A C H I N S
Rank = (5!) + 0 (4!) + 0(3!)
+ 0 (2!) + 0(1!) + 1
= 5 × 120 + 1
= 601
6
∑ 56 - r
50
C4 + C 3 is
17. The value of
r =1
m
55 55 56 56
1) 2) 3) 4)
C4 C3 C3 C4
o
6
∑
.c
50
C4 + 56 - rc3
Ans.(4)
r =1
e
50 55 54
C 3 + 53 C 3 + 52
C 3 + 51C 3 + 50 C 3
C4 + C3 +
g
56
= C4
a
1 0 
1 0
If A = 1 1 and I = 0 1 , then which of the following holds for all n ≥ 1, by principle of mathematical induction
18.
p




e
2) A 2 = 2 n −1 A − (n − 1)I
1) A n = nA − (n − 1)I
e4) A n = 2 n −1 A + (n − 1)I
3) A n = nA + (n − 1)I
ie
1 0 
Ans.(1) A = 1 1  

.a
 
1 0
I = 0 1
w


 
w
1 0 1 0
A2 =  1 1 1 1
 
  
w
1 + 0 0 + 0 

= 

1+1 0 +1
1 0
 2 1
= 
 
 1 0 1 0   1 0 
A3 =   2 1 1 1  =  3 1 
  
   
A 2 = 2A − I
 2 0 1 0  1 0

= − = 
  
 2 2 0 1  2 1
A 3 = 3A − 2I
3 0  2 0 1 0
= 3 3 −  0 2 = 3 1
  
   
11 11
 2  1   
1
If the coefficient of x in ax +   in ax −  2 
19. equals the coefficient of x then a and b satisfy the
−7
7
 bx   bx 
 
relation
a
=1 4) ab = 1
1) a - b = 1 2) a + b = 1 3)
b
11
 1
2
Ans.(4)  x + 
bx 

2 × 11 − 7
r= =5
3
1
7 11 6
coefficient of x = C 5 a
m
b5
o
11
 1
 ax − 2 
.c
 bx 
11 + 7
e
r= =6
3
g
1
−7 11 5
Coefficeint of x = C 6 a
a
b6
p
1 1
11
C6 a 6 = 11C 6 a 5
5
b6
b
e
⇒ ab = 1
e 2x
ie
−1
Let f : (-1, 1) → B, be a function defined by f(x) = tan
20. then f is both one-one and onto when B is the interval
1− x2
.a
 π  π  π π  π π
− 2 , 2  4)  − 2 , 2 
1)  0, 2  2) 0, 2  3)
   
   
w
Ans.(4) Since f → (−1 ,1) → B &
 2x 
w
f(x) = tan −1  
1− x2 
w
if is both one - one & onto i , e ( bijective )
 π π
⇒ if ( x) inverse tangent of n whose Range is always  − 2 , 2 
 
If z1 and z 2 are two non-zero comple numbers such that z1 + z 2 = z1 + z 2 , then arg z1 - arg z 2 is equal to
21.
π −π
2) −π 3) 0 4)
1)
2 2
Ans.(3) z1 + z 2 2 = [ z1 + z 2 ]2
2 2
⇒ z1 + z 2 + 2 z1 z 2 cos (θ1 − θ 2 )
= z1 2 + z 2 2 + 2 z1 z 2
cos (θ1 − θ 2 ) = 1
⇒ θ1 − θ 2 = 0
⇒ Arg (z1 ) - Arg (z 2 ) = 0
z
w=
1
z − i and w = 1, then z lies on
22. If
3
1) an ellipse 2) a circle 3) a straight line 4) a parabola
z
w=
1
Ans.(3) z− i
3
z
w=
1
z− i
3
1
⇒ z- i = z
m
3
o
 1
⇒ z lies on the ⊥ r bisector of the line segment joining  0, 3  and (0, 0 )
.c


23. If a 2 + b 2 + c 2 = −2 and
e
1+ a2x (1 + b 2 ) x (1 + c 2 ) x
g
f ( x) = (1 + a 2 ) x 1+ b2 x (1 + c 2 ) x
a
2 2
1+ c2x
(1 + a ) x (1 + b ) x
p
then f (x) is a polynomial of degree
e
1) 1 2) 0 3) 3 4) 2
e
Ans.(4) a 2 + b 2 + c 2 = −2
ie
In f ( x ), c1 ⇒ c1 + c 2 + c 3
.a
1 + (a 2 + b 2 + c 2 )x + 2x (1 + b 2 ) x (1 + c 2 ) x
⇒ f(x) = 1 + (a 2 + b 2 + c 2 ) x + 2x 1+ b2 x (1 + c 2 ) x
1 + (a 2 + b 2 + c 2 ) x + 2x (1 + b 2 ) x 1 + c2 x
w
1 (1 + b 2 ) x (1 + c 2 ) x
w
1 + b2 x (1 + c 2 ) x
⇒1
1 (1 + b 2 ) x 1+ c2x
w
R1 → R 1 − R2
0 x -1 0
1 + b 2 x (1 + c 2 ) x
⇒1
1 (1 + b 2 ) x 1 + c 2 x
[ ]
⇒ - (x - 1) 1 + c 2 x − x − c 2 x
⇒ (1 - x) 2 ⇒ degree = 2
The normal to the curve x = α (cos θ + θ sin θ), y = a (sin θ − θ cos θ) at any point ‘ θ ’ is such that
24.
1) It passes through the origin
π
+ θ with the x - axis
2) It makes angle
2
π 
3) It passes through  a 2 ,−a 
 
4) It is at a constant distance from the origin
Ans.(2 , 4) x = a(cosθ + θsinθ)
y = a(sin θ − θ cos θ)
a (cosθ + θ sin θ − cosθ sin θ
dy
= =
m
dx a (− sin θ + θ cos θ + sin θ cosθ
o
∴ slope of the normal
.c
cos θ
dx
⇒− =−
sin θ
dy
e
π
+θ
g
⇒ Slope of the normal is
2
a
∴ Eqn. of the normal
p
cosθ
⇒ y − a sinθ + a θ cosθ = - (x − a cos θ − a θ sin θ )
sin θ
e
⇒ y sin θ − a sin 2θ + a θ cos θ sin θ = - x cos θ + a cos 2θ + a θ cos θ sin θ
e
ie
⇒ x cos θ + y sin θ = a
⇒ It is at a constant distance from the origin.
.a
25. A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs
is incorrectly matched?
w
Interval Function
(− ∞, ∞ )
1) x 3 − 3x 2 + 3 x + 3
w
[2, ∞ )
2) 2 x 3 − 3 x 2 + 12 x + 6
w
1

 − ∞, 
3) 3x3 − 2 x 2 + 1
3

(− ∞,−4]
4) x3 − 6 x 2 + 6
Ans.(3)
f (x) = 3x 3 − 2x 2 +1
f ' ( x ) = 6x − 2 ≥ 0 ⇒ x ≥ 1 / 3
Option (3) is incorrect. Checking other function similarly we find that they are correctly matched.
26. Let α and β be the distinct roots of ax 2 + bx + c = 0, then
( ) is euql to
1 - cos ax 2 + bx + c
lim
(x − a ) 2
x →α
a2
(α − β )2
1)
2
2) 0
− a2
(α − β )2
3)
2
1
(α − β )2
4)
2
m
1 − cos(ax 2 + bx + c)
o
lim
Ans.(1) We have x →α
( x − α) 2
.c
 
 (ax 2 + bx + c   a ( x − α) ( x − β) 
sin 2 
e
sin 2 
 
 
2 2
  = 2 lim  
= 2 lim
g
( x − α) 2 ( x − α) 2
x →α x →α
a
Q α, β are roots of ax 2 + bx + c = 0
p
 
2
∴ ax + bx + c = a ( x − α)( x − β) 
 
e
e
2
a ( x − α)(x − β) 

 sin  a2
2 2
= 2 lim   . ( x − β)
ie
 a ( x − α)( x − β)  4
x →a
 
 
2
.a
a2 a2
= 2(1) 2 (α − β) 2 = (α − β) 2
4 2
w 1
f (1 + h) = 5 then f(1) equals
Suppose f(x) is differentiable at x = 1 and lim
27.
w
h
h→0
1) 3 2) 4 3) 5 4) 6
w
Ans.(4) Applying L’ Hospital rule
f ' (1 + h)
=5
Lim
1
h→0
⇒ f' (1) = 5
Let f be differentiable for all x. If f(1) = -2 and f ' ( x) ≥ 2 for x ∈ [1, 6], then
28.
1) f(6) ≥ 8 2) f(6) < 8
3) f(6) < 5 4) f(6) = 5
Ans.(2) Here from LMVT
f (6) − f (1)
≥2
5
f (6) ≥ 8.
If f is a real-valued differentiable function satisfying f ( x) − f ( y ) ≤ ( x − y ) 2 , x, y ∈ and f(0) = 0, then f(1) equals
29.
1) -1 2) 0 3) 2 4) 1
f(x) - f(y)
≤ lim x - y
lim
Ans.(2) x → y x-y x→y
or f ' (x) ≤ 0
⇒ f ' (x) = 0
⇒ f (x) is constant, As ⇒ f (0) = 0
∴ f(1) = 0
3 3
 1
(1 + x) 2 − 1 + x 
m
 2
30. If x is so small that x3 and higher powers of x may be neglected, then may be approximated as
1
(1 − x )2
o
.c
3 3
2 2
1) 1 − x 2) 3x + x
8 8
e
3 x 32
2
3) − x −x
4)
g
8 28
a
3
(1 + x )3 / 2 − 1 + 1 x 
 
p
 2
Ans.(3)
(1 − x )1/ 2
e
e
31
.  3 x 3.2 x 2 
3
1 + x + 2 2 x 2 − 1 +
ie

+
2 4
 2
2 2 

=
(1 − x )1/ 2
.a
3
− x2
3
w
= − x 2 (1 − x )−1/ 2
8
=
(1 − x ) 1/ 2
8
w
3x  3
= − x 2 1 + + ...  = − x 2
82 8

w
∞ ∞ ∞
∑ ∑ ∑c
an, y = bn , z = n
If x = where a, b, c are in A.P and a < 1, b < 1, c < 1 then x, y, z are in
31.
n =0 n =0 n =0
1) GP 2) AP
3) Arithmetic - Geometric Progression 4) HP
1
1
1
Ans.(4) x = 1 − a y = 1 - b z = 1 - c
a, b, c, are in A.P
1 1 1
∴ , , are in H -P
1- a 1− b 1− c
⇒ x, y, z are in H-P
m
π
In a triangle ABC, let ∠C = . If r is the inradius and R is the circumradius of the triangle ABC, then 2(r+R) equials
32.
o
2
1) b + c 2) a + b 3) a + b + c 4) c + a
.c
A
e
g
a
c
p
b
Ans.(2)
e
e
ie
c a B
Let a = b = 1
.a
⇒c= 2
r = (s-c) tan c/2
w
 2+ 2 
= − 2  tan Π/4
w
2 
 
w
2+ 2 −2 2
⇒
2
2− 2 1
r⇒ = 1−
2 2
a b c 1× 1 × 2
R= =
4 ∆ 4 × 1 × 1× 1
2
1
=
2
 1
1
∴ 2 (r + R) = 2 1 - =2
+
 
2 2

=a+b
y
−1 -1
= α , then 4 x 2 − 4 xy cos α + y 2 is equal to
If cos x - cos
33.
2
1) 2 sin 2 α 2) 4
3) 4 sin 2 α 4) - 4 sin 2 α
y
−1 −1
=α
Ans.(3) cos x − cos
2
xy y2 
⇒ cos −1   =α
+ 1− x2 1-
2 4
 
1- x 2 4 - y2
xy
= cos α
⇒ +
2 2
m
( )( )
⇒ 1 - x 2 4 - y 2 = (2 cos α − xy )2
o
.c
4 − 4 x 2 − y 2 + x 2 y 2 = 4 cos 2 α + x 2 y 2 − 4 xy cos α
⇒ 4x 2 − 4 xy cosα + y 2 = 4 - 4 cos 2 α = 4sin 2 α
e
34. If in a ∆ 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
g
1) G.P. 2) A.P.
a
3) Arithmetic - Geometric - Progression 4) H . P.
p
Ans.(2) In a ∆ ABC ,
e
Let the altitudes be p1 , p 2 , p 3 from the vertices A, B, C respectively then
e
2∆ 2∆ 2∆
P1 = , P2 = , P3 =
ie
a b c
2∆ 2∆ 2∆
⇒ , ,
.a
are in HP
a bc
111
⇒ , , are in HP
w
abc
⇒ a, b, c are in AP
w
⇒ 2 R sin A, 2 R sin B, 2 R sin C are in AP
⇒ sin A, Sin B, sin C in AP.
w
1 1
∫ ∫
2 3
2 x dx, I 2 = 2 x dx,
If I1 =
35.
0 0
2 2
∫ ∫
2 3
2 x dx and I 4 = 2 x dx, then
I3 =
1 1
1) I 2 > I1 2) I1 > I 2 I3 = I4 4) I 3 > I 4
3)
1 1
∫ ∫
2 3
2 x dx, 2 x dx
Ans.(2) I1 = I2 =
0 0
1 2
∫ ∫
2 3
2 x dx, 2 x dx
I3 = I4 =
0 0
For x ∈ (0, 1)
x2 > x3
2 3
⇒ 2 x > 2x
1 1
∫ ∫
2 3
2 2 dx > 2 x dx
⇒
0 0
= I1 > I 2
For x ∈ (1, 2)
x3 > x 2
3 2
2x > 2x
2 2
∫ ∫
3 2
2 x dx > 2 x dx
1 1
I 4 > I3
36. The area enclosed between the curve y = loge (x+e) and the coordinate axes is
m
1) 1 2) 2 3) 3 4) 4
o
Ans.(1) Required area
.c
0
∫
= log( x + e) dx
e
1− e
Put x + e = t
g
a
e
∫
= log t dt
p
1
e
e
= [ x log x - x]1 = 1 sq. units
e
The parabolas y 2 = 4 x and x 2 = 4 y divide the square region bounded by the lines x = 4, y = 4 and the coordinate
37.
ie
axes. If S1 , S2 , S3 are respectively the area of these parts numbered from top to bottom; then S1 : S2 : S3 is
1) 1 : 2 : 1 2) 1 : 2 : 1 3) 2: 1 : 2 4) 1: 1 : 1
.a
Ans.(4)
w
y=4 (4 , 4)
w
S1
S2
x=4
w
S3
(0, 0)
 2
 2 x − x  dx
4
∫
S2 =  4
0
 
= 16 / 3
∴ S 2 + S 2 + S 3 = 4 × 4 = 16
16 32
∴ S 2 + S 3 = 16 − =
3 3
⇒ S1 = 16 , = S3 16
3
3
∴ S1 : S 2 : S3 = 1 :1 :1
dy
= y (log y - log x + 1), then the solution of the equation is
If x
38.
dx
x y
1) y log  y  = cx 2) x log  x  = cy



x
y
log   = cy
log   = cx
3) 4) y
x 
dy
= y (log y - log x + 1)
Ans.(3) If x
dx
dy y  y
⇒ =  log + 1
dx x  x
m
Let y = vx
o
dv
⇒ v+x = V (log v + 1)
dx
.c
dv
⇒ v+x = V (log v + v)
dx
e
dv dx
∫ ∫
g
⇒ =
v log v x
a
⇒ log (log v) = log x + log c
p
log (log v) = log (cx)
e
⇒ log y =cx
x
e
39. The line parallel to the x-axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx - 2ay - 3a=
ie
0 where (a, b) ≠ (0, 0) is
3
from it
1) below the x-axis at a distance of
.a
2
2
2) below the x-axis at a distance of from it
3
w 3
3) above the x-axis at a distance of from it
2
w 2
4) above the x-axis at a distance of form it
w
3
Ans.(1) ax + 2by = -3 b
bx - 2ay = 3a
3
Solving these two equations we get y = −
2
40. A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that 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 1 5
cm/min. 2) cm/min. 3) cm/min. 4) cm/min.
1)
36π 18π 54π 6π
4
π (x + 10) 3 Where x is thickness of ice.
Ans.(2) V =
3
dV dx
= 4π (10 + x) 2
∴
dt dt
 dx  1
∴  at x = 5 ; = cm/min.
 dt  18 π
2
 (log x − 1)  

∫  dx is equal to

41.
1 + (log x) 2  

log x x
1) (log x) 2 + 1 + C +C
2) 2
x +1
x
xe x +C
+C
3) 4) (log x) 2 + 1
2
1+ x
2
 log x − 1  

∫
Ans.(4) Let f ( x)   dx
1 + (log x) 2  

Differentiating from the optins Q(x)
m
 
x
d + c

o
dx  (log x) + 1  2
 
.c
[(log x) + 1]− x [(2 log x) 1 x ]
2
[(log x) + 1]
⇒
e
2
g
(log x - 1)2
⇒
[(log x) + 1]
a
2
2
p
f(x)
4t 3
1
∫
e
Let f : R → R be a differentiable function having f(2) = 6, f ' (2) =  48  . Then xLt2 dt equals
42. x−2
 →
6
e
1) 24 2) 36 3) 12 4) 18
ie
f ( x)
∫ 4t 3 dt
.a
Ans.(4) 6
lim (0/0 form)
x-2
x→ 2
w
4 (f (x) ) 3 × f ' ( x)
= 4 ( f (2)) 3 × f ' (2) = 18
= lim
1
x →2
w
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
w
π 
 
π
π π
and x = β > is  β sin β + 4 cos β + 2 β  . Then f  2  is
ordinates x =
  
4 4
π π
 
1)  4 + 2 − 1 2)  4 − 2 + 1
 
 
π π
 
 
3) 1 − 4 − 2  4) 1 − 4 + 2 
 
 
β
π
∫ f ( x)dx = β sin β + cos β + 2 β
Ans.(4) 4
π/4
π
∴ f ' (β ) = sin β + β cos β - cos β + 2
4
π π
f   = 1− + 2
2 4
The locus of a point P (α , β ) moving under the condition that the line y = αx + β is a tangent to the hyperbola
44.
x2 y2
− = 1 is
a2 b2
1) an ellipse 2) a circle 3) a parabola 4) a hyperbola
Ans.(4) ⇒ β 2 = a 2 α 2 − b 2
∴ the locus is a 2 x2 − y2 = b2
x2 y2
= b2
⇒ −
1 1
a2
∴ the locus is a- hyperbola
m
x +1 y −1 z − 2 1
and the plane 2x − y + λ z + 4 = 0 is such that sin θ = the value
= =
If the angle θ between the line
45.
o
3
1 2 2
of λ is
.c
5 -3
1) 2)
e
3 5
g
-4
3
3) 4)
3
4
a
x + 1 y −1 z −1
p
= = (a1 , b1 , c1 ) = (1, 2, 2)
Ans.(1)
1 2 2
e
2x − y + λ z + 4 = 0
e
(a 2 , b2 , c 2 ) = (2,−1, λ )
ie
2-2+ 2 λ
sin θ =
.a
3 5+λ
2λ
1
w
⇒ =
3 3 5+λ
w
⇒ 5 + λ = 4λ
⇒ λ = 5/3
w

46. The angle between the lines 2x = 3y = -z and 6x = - y = - 4z is
1) 00 2) 90 0 3) 4) 30 0
45 0
Ans.(2) 2 x = 3 y = − z,
x y z
⇒ = =
1 \ 2 1/ 3 −1
1 1 
(a1 , b1 , c1 ) =  , ,−1
2 3 
6x = -y = -4z
x y z
⇒ = =
1 −1 −1/ 4
6
m
1 1
o
(a 2 b 2 c 2 ) =  ,−1,− 
6 4

.c
a1 a 2 + b1 b2 + c1 c 2
e
111
= −+
12 3 4
g
=0
a
47. If the plane 2ax - 3ay +4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres
p
x 2 + y 2 + z 2 + 6x − 8x − 2z = 13 and x 2 + y 2 + z 2 − 10x + 4 y − 2z = 8 then α equals
e
1) -1 2) 1 3) -2 4) 2
e
Ans.(3) Centre of S1 = (−3, 4, 1)
ie
Centre S 2 = (5, − 2, 1)
Mid pt. of S1S 2 (1, 1, 1)
.a
2ax - 3ay + 4az + 6 = 0
passes through (1, 1, 1)
w
2a- 3a + 4a = -6
⇒ 3a = - 6
w
⇒a =-2
w
r r
The distance between the line r = 2ˆ - 2ˆ + 3k + λ (ˆ - ˆ + 4k) and the plane r . (ˆ + 5ˆ + k) = 5 is
ijˆ ij ˆ i jˆ
48.
10
10 3 10
1) 2) 3) 4)
9 10 3
33
r
Ans.(2) r = 2 i - 2j + 3k + λ (i - j +4k)
r
r . ( i + 5j + k) = 5
the line passes through (2, -2, 3)
2 − 10 + 3 − 5
∴ the reqd distance =
1 + 25 + 1
10
=
27
10
=
33
r r rˆ
r
For any vector a , the value (a × ˆ) 2 + (a × ˆ) 2 + (a × k ) 2 is equal to
49. i j
r r
2) a 2
1) 3a 2
r r
3) 2a 2 4) 4a 2
r
a = xi + yj + zk
Ans.(3) Let
rˆ
a × i = ( xi + yj + zk ) × i
= −y k + z j
(a × iˆ)
r 2
= y2 + z2
(r ) 2
m
ˆ = z 2 + x2
|| ly a × i
o
r
(a × k )2 = x 2 + y 2
.c
∴ The required sum
e
= 2( x 2 + y 2 + z 2 )
g
r
= 2a 2
a
xy1
+ + = 0 always passes through a fixed point. That
50. If non-zero numbers a, b, c are in H.P., then the straight line
abc
p
point is
e
1) (-1, 2) 2) (-1, -2)
e
 1
4) 1,− 2 
3) (1, -2)
ie
 
.a
11
,
Ans.(3) Let a, b and c be 1,
23
xy1
w
+ + =0
abc
w
⇒ x + 2y + 3 = 0
(1, -2) satisfies the equation
w
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 traingle is
7

1)  − 1, 3 


 −1 7 
2)  3 , 3 
 
 7
3) 1, 3 
 
1 7
4)  3 , 3 
 
(1, 1)
Ans.(3) A
(-1, 2) (3, 2)
F
D
B E C
(1, 3)
E = -1 + 3 -1, 2+ 2 -1
= (1, 3)
Centried of ∆ ABC = centroid of ∆ DEF
 -1 + 1 + 3 2 + 2 + 3 
= 
,
m
3 3
 
o
 7
= 1, 
.c
 3
e
If the circles x 2 + y 2 + 2ax + cy + a = 0 and x 2 + y 2 − 3ax + dy − 1 = 0 intersect in two distinct points P and Q then the
52.
g
line 5x + by - a = 0 passes through P and Q for
1) exactly one value of α
a
2) no value of α
p
3) infinitely many values of α
e
4) exactly two values of α
e
Ans.(2) Radical axis is
ie
5a x + (c-d) y + (a+1) = 0 ....(1)
5 x + by - a = 0 .......(2)
.a
a +1
5a
=
5 -a
w
⇒ - a2 = a +1
⇒ a 2 + a +1 = 0
w
No real value of a satisfies
w
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
(0,3)
2
(h, k)
Ans.(4) k
( x − h) 2 + ( y − k ) 2 = k 2
( x − 0) 2 + ( y − 3) 2 = 4
d [(0,3), (h, k )] = 2 + k
⇒ h 2 + ( k − 3) 2 = (k + 2)
⇒ h 2 + (k − 3)2 = (k + 2)2
⇒ h 2 + k 2 + 9 − 6k = k 2 + 4k + 4
⇒ h 2 − 10k + 5 = 0
∴ locus of its centre is
h 2 + 10k + 5 = 0
i.e., x 2 + 10 y + 5 = 0
x 2 = −10 y − 5
x2 = −5 (2y +1)
m
∴ The locus is a parabola
o
If a circle passes through the point (a, b) and cuts the circle x 2 + y 2 = p 2 orthogonally, then the equation of the locus
54.
.c
of its centre is
( )
1) x 2 + y 2 − 3ax − 4by + a 2 + b 2 − p 2 = 0
e
( )
g
2) 2ax + 2by - a 2 − b 2 + p 2 = 0
( )
a
3) x 2 + y 2 - 2ax - 3by a 2 − b 2 − p 2 = 0
p
( )
4) 2ax + 2by − a 2 + b 2 + p 2 = 0
e
Ans.(4) It the equation be
e
x 2 + y 2 + 2 gx + 2 fy + c = 0
ie
passes through (a, b)
a 2 + b 2 + 2ag + 2bf + c = 0 ....(1)
.a
x 2 + y 2 = p 2 orthogonally
It cuts
2(g.0 + f.0) = c-p2
w
⇒ c − p2 = 0
w
c = p2
Substituting the value of c in 1
w
we get
a 2 + b 2 + 2ag + 2bf + p 2 = 0
⇒ 2ag + 2bf + a 2 + b 2 + p 2 = 0
locus of its centre (-g, -f) is
− 2ag − 2bf + a 2 + b 2 + p 2 = 0
⇒ 2ag + 2bf - (a 2 + b 2 + p 2 ) = 0
2ax + 2by − (a 2 + b 2 + P 2 ) = 0
55. An ellipse has OB as semi minor axis, F and F' its focil and the angle FBF' is a right angle. Then the eccentricity of
the ellipse is
1 1
2)
1)
2 2
1
1
3) 4)
3
4
Ans.(1) (0,1) B
I
(-ae, o) (a e,o)
1
F
F
2ae
b
Slope of F B =
ae
b
1
Slope of F B =
− ae
b
b
× - = −1
ae
ae
m
⇒ b 2 = a 2e2
o
b 2 = a 2 − a 2e2
.c
2a 2 e 2 = a 2
e
1
⇒ e2 =
2
g
1
=e=
a
2
p
Let a, b, and c be distinct non-negative numbers. If the vectors aˆ + aˆ + ck, ˆ + k and cˆ + cˆ + bk lie in a plane, then c is
i j ˆi ˆ ijˆ
56.
e
1) the Geometric Mean of a and b
2) the Arithmetic Mean of a and b
e
3) equal to zero
ie
4) the Harmonic Mean of a and b
.a
aac
Ans.(1) 1 0 1 = 0
ccb
w
⇒ c 2 = ab
w
⇒ c is the G.M. of a & b
[ ][ ]
rrr
57. If a , b, c are non-coplanar vectors and λ is a real number then λ (a + b) λ2 b λ c = a b + c b for
w
1) exactly one value of λ
2) no value of λ
3) exactly three values of λ
4) exactly two values of λ
Ans.(2)
rr r r rrrr
λ(a + b). [λ2 b × λc] = a. [(b + c) × b]
rrr rrr rrrrr
⇒ λ 4 [a. (b × c) + b. (b × c) = a. [b × b + c × b]
rrr rrr
⇒ λ 4 [a. (b × c)] = a. (c × b)
⇒ λ 4 = −1
no real value of λ is possible
58. Let a = i − k , b = x i + j + (1 = x)k and c = y i + x j + (1 + x − y )k . then [a, b, c] depends on
1) only y 2) only x
3) both x and y 4) neither x nor y
rrr
Ans.4) [a b c]
−1
10
=x 1 1− x
y x 1+ x − y
= [(1 + x − y) − x(1 − x)]- (x 2 − y)
= 1+ x − y − x + x2 − x2 + y
m
=1
59. Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without
o
consulting others. The probability that all the three apply for the same house is
.c
2 1 8 7
1) 2) 3) 4)
9 9 9 9
e
Ans.(2) Three person A, B, C
g
& 3 houses a, b, c
a
3 1 1 1
⇒  × × =
p
3 3 3 9
e
60. A random variable X has Poisson distribution with mean 2. Then P(x > 1.5) equals
e
2
2) 0
1)
e2
ie
3 3
3) 1 − 4)
.a
2
e2
e
e −λ λ r
Ans.(3) P( x = r ) =
w
r!
λ=2
w
P( x > 1.5) = 1 - [ p(x = 0) + P(x = 1)]
w
p(x= 0) +p (x=1)
 e −2 2 0 e −2 21 
= + 
 0! 1! 
 
3
= e − 2 (1 + 2) =
e2
3
P ( x > 1 .5 ) = 1 -
e2

( ) ()
1 1 1
, P(A ∩ B ) = and P A = , Where A stands for complement
Let A and B be two events such that P A ∪ B =
61.
6 4 4
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
Ans.(3) P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
53 1
⇒ = + P( B) −
64 4
m
1
P ( B) =
3
o
.c
1
P ( A ∩ B) = P ( A ) × P ( B) =
4
e
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
g
the insect after
a
1) 20 s 2) 1 s 3) 21 s 4) 24 s
p
1 2
Ans.(3) 0.t + .(2).t = 20 × t + 21
2
e
⇒ t 2 − 20t − 21 = 0
e
⇒ t + 21 s.
ie
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
.a
1) ( f − f ' )m 2 = ff ' n 2) ( f + f ' )m 2 = ff ' n
w
1 1
( f + f ' )m = ff ' n 2 ff ' m 2
4) ( f − f ' ) n =
3)
2 2
w
Ans.(4) Here f ' t = f (t + m)
w
mf
⇒t = .....(1)
f '− f
1 1
f ' t 2 + n = f (t + m) 2 ..(2)
Also,
2 2
using (1) and (2)
 mf 
n1
'
 f '-f 
=f 
m2  
12
∴( f ' - f ) n = m ff '
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
H
2H H H
a) b) 3) 4)
2( A + B)
A− B A+ B A− B
R
Ans.(2)
l2
l1
R
A
A+B B
Al1 = Bl 2
H = (l1 + l 2 ) R
Let shift distance = x
( A + R )(l1 − x ) = ( B − R ) (l 2 + x)
m
H
o
⇒ x=
A+B
.c
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 :
e
1) 2 :1 2) 3 : 2 3) 3 : 2 4) 3 : 2 2
g
R
Ans. (4) F2
a
p
e
e
F1
ie
R 2 + F12 = F2 2
.a
F2 2
+ F12 = F2 2
⇒
9
82
w
⇒ F12 = F2
9
w
F1 2 2
⇒ =
3
F2
w
⇒ F2 : F1 = 3 : 2 2
1 1 1
The sum of the series 1 + + + + .... ad inf. is
66.
4 : 2 ! 16.4 ! 64.6 !
e −1 e +1 e −1 e +1
1) 2) 3) 4)
e e 2e 2e
2 4
1 1 1 1 1 1
Ans.(4) 1 +   +   +   + 64....
2! 2  4!  2  6! 2 
1
e 2 + e −1/ 2
=
2
1
e+
e
=
2
e +1
=
2e
π
cos 2 x
∫ 1+ a dx, a > 0, is
67. the value of x
−π
π
2)
1) απ
2
π
3) 4) 2 π
α
cos 2 x
π
∫ dx
Ans.(2)
1+ a x
−π
( )
 cos 2 x cos 2 x 
π
∫
⇒ +a x  dx
 x
 1+ a a +1 
m
0
 
o
(1 + a x ) cos 2 x
π
∫
⇒ dx
.c
(1 + a x )
0
π/2
∫
e
cos 2 x dx
⇒2
0
g
1 π π
⇒ 2 ×  =
a
2 2 2
p
The plane x + 2y - z = 4 cuts the sphere x 2 + y 2 + z 2 − x + z − 2 = 0 in a circle of radius
68.
e
1) 3 2) 1
e
3) 2 4) 2
ie
1 1
Ans.(2) Centre of the sphere =  2 , 0, - 2 
 
.a
1
1
+0+ +2
radius =
w
4
4
10
w
=
2
w
⊥ r distance from centre to the plane
1 1
+0+ −4
=2 2
6
3
=
6
10 9
−
radius of the reqd. circle =
46
30 − 18
=
12
12
= =1
12
If the pair of lines ax 2 + 2(a + b) xy + by 2 = 0 lie along diameters of a circle and divide the circle into four sectors such
69.
that the area of one of the sectors is thrice the area of another sector then
1) 3a 2 − 10ab + 3b 2 = 0 2) 3a 2 − 2ab + 3b 2 = 0
3) 3a 2 + 10ab + 3b 2 = 0 4) 3a 2 + 2ab + 3b 2 = 0
π
Ans.(4) Here angle between the given lines is
4
2 (a + b) 2 − ab
⇒1=
a+b
⇒ 3a 2 + 2ab + 3b 2 = 0
∑x ∑x
m
2
= 400 and = 80 . Then a possible value of n among the
70. Let x1 , x2 ,..., xn be n observations such that i i
following is
o
1) 15 2) 18 3) 9 4) 12
.c
∑ xi2 = 400
Ans.(2)
e
∑x = 80
i
g
σ2 ≥0
a
∑x ∑x 
2
 
2
p
−
i i
≥0
⇒


n n
e
 
e
400 6400
⇒ − 2 ≥0
n n
ie
⇒ n ≥ 16
∴ (2)
.a
A particles is projected from a 0 with velocity u at an angle of 60 0 with the horizontal. When it is moving in a
71.
direction at right angles to its direction at 0, its velocity then is given by
w
u
u u
u
2
2) 3) 4)
1)
3 3 3
2
w
Ans.(4) Horizontal component of velocity remains same at any point
w
∴ u x = vx
u cos 60 0 = v cos 30 0
u
⇒v=
3
72. If both the roots of the quadratic equation x 2 − 2kx + k 2 + k = 5 = 0 are less than 5, then k lies in the interval
2) (6, ∞) ( − ∞ , 4)
3) 4) [4, 5]
1) (5, 6]
Ans.(3) x 2 − 2kx + k 2 + k − 5 = 0
let f(x) = x 2 − 2kx + k 2 + k − 5
since f(x) has real roots less than 5, then
D > or f(5) > 0
⇒ k 2 − (k 2 + k − 5) > 0 or k 2 − 9k + 20 > 0
⇒ k<5 k < 4 or k > 5
or
⇒ k<4
73. If a1 , a2 , a3 ,..., an ,... are in G.P., then the determinant
loga n loga n +1 loga n + 2
∆ = loga n + 3 loga n + 4 loga n + 5
is equal to
loga n + 6 loga n + 7 loga n +8
1) 1 2) 0 3) 4 4) 2
Ans.(2) Since a n , a n +1 , a n + 2 are in G.P,
a 2 +1 = a n , a n + 2
n
2 log a n +1 − log a n − log a n + 2 = 0
Similarly 2 log a n + 4 − log a n + 3 − log a n +5 = 0
m
2 log a n + 7 − log a n + 6 − log a n +8 = 0
o
Applying c1 + c 3 − 2c 2 ,
.c
We have ∆ = 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 α is a given
e
constant and f(0) = 1, f(2 α -x) is equal to
1) -f(x) 2) f(x)
g
3) f( α ) + f( α -x) 4) f(x)
a
Ans.(1) f (a − ( x − a)) = f (a) f ( x − a) − f (0) f ( x)
p
Put x = 0, y = 0
e
f (0) = f (0)) 2 − [ f (a)]2
e
⇒ f(a) = 0 [Q f(0) = 1]
ie
From (1)
.a
⇒ f(2a - x) = f(x)
If the equation a n x n + an −1 x n −1 + .... + a1 x = 0 ,
75.
w
a1 ≠ 0, n ≥ 2, ahs a positive root x = α , then the equation nan x n −1 + (n − 1)a n −1 x n −1 + ... + a1 = 0 has a positive root,
w
which is
1) greater than α 2) smaller than α
w
3) greater than or equal to α 4) equal to α
Ans.(2) Let f(x) = a n x n + a n −1x n −1 + ... + a1x
f(0) = 0
f (α ) = 0
⇒ f ' (x) = 0 has at least one root between (0, α )
⇒ f ' ( x ) = 0 has a positive root less than α

w