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JEE MAIN-FULL TEST 03
SOLUTIONS OF MOCK TEST
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Ques No. Question & Solution
Q No. - 1
JEE MAIN-FULL TEST 03 - PHYSICS
A swimmer can swim in still water with a speed of . While
crossing a river his average speed is
cross the river in the shortest possible time, what is the speed
of flow of water ?
(a).
(b).
(c).
(d).
CORRECT OPTION: A
SOLUTION
Avg. speed
Q No. - 2
JEE MAIN-FULL TEST 03 - PHYSICS
A car starting from rest is accelerated at constant until it
attains a constant speed v. It is then retarded at a
rate until it comes to rest. Considering that the car moves with
constant speed for half of the time of total jo
average speed of the car for the journey is
(a).
(b).
(c).
(d). Data insufficient
CORRECT OPTION: B
SOLUTION
Q No. - 3
JEE MAIN-FULL TEST 03 - PHYSICS
A smooth ring of mass can slide on a fixed horizontal rod. A
string tied to the ring passes over a fixed
carries a block of mass as shown in the figure. At an instant,
the string between the ring and the
makes an angle with the rod. The initial acceleration of the
ring is
√5m
s3m
s
2m
s
4m
s
6m
s
8m
s
3 =√(Vrt)2 + (Vmrt)2
t⇒ V 2r + 5 = 9
⇒ Vr = 2m
s
v
43v
43v
2
Vavg = =× × v + × v12
t2
12
t
3v
4
P m
Q ( )m2
60 ∘
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(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
�..(i)
�.(ii)
Solving (i) and (ii) acceleration of ring
Q No. - 4
JEE MAIN-FULL TEST 03 - PHYSICS
A block of mass is hanging over a smooth and light pulley
through a light string. The other end of the s
pulled by a constant force . If kinetic energy of the block
increases by in . Then
(a). tension in the string is .
2g
32g
62g
9g
3
g − T = am
2
m
2Tcos 60 ∘ =
ma
cos 60 ∘
=2g
g
M
F 20J 1s
Mg
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(b). tension in the string is
(c). Work done by the tension on the block is 20 J in 1 sec.
(d). Work done by the force of gravity is 20 J in 1 sec.
CORRECT OPTION: B
SOLUTION
Work done by all the forces on the block equal to change in
kinetic energy.
Q No. - 5
JEE MAIN-FULL TEST 03 - PHYSICS
A circular ring is fixed in a gravity free space and point of
the ring is earthed. Now a magnet is placed alon
the ring at a distance from its centre such that the nearer pole
is north pole as shown in figure. A sharp im
applied on the magnet so that it starts to move towards the
ring. Then,
(a). Initially magnet experiences an acceleration and then it
retards to come to an instantaneous rest.
(b). Magnet starts to oscillate about centr of the ring
(c). Magnet continous to move along the axis with constant
velocity
(d). The magnet retards and comes to rest finally.
CORRECT OPTION: D
SOLUTION
N/A
F
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Q No. - 6
JEE MAIN-FULL TEST 03 - PHYSICS
A block of mass is suspended by means of an ideal spring of
force constant from ceiling of a car whic
moving along a circular path of radius with acceleration . The
time period of oscillation of the block whe
displeced along the spring, will be
(a).
m k
r a
2π√mg + mak
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(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
No effect of \'a\' and \'g\' on time period of spring
pendulum.
Q No. - 7
JEE MAIN-FULL TEST 03 - PHYSICS
In the hydrogen atom spectrum and represent wavelengths emitted
due to transition from sec
first excited states to the ground state respectively. The value
of is
(a).
(b).
(c).
(d).
CORRECT OPTION: A
SOLUTION
√m
k√g2 + a2
2π√m
k
2π√m
√k2 + g2 + a2
λ3 − 1 λ2 − 1λ3 − 1
λ2 − 127
3232
274
99
4
= R( − ) =1λ3 − 1
1
121
328R
9
= R( − ) =1λ2 − 1
1
121
223R
4
⇒ =λ3 − 1
λ2 − 1
27
32
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Q No. - 8
JEE MAIN-FULL TEST 03 - PHYSICS
A block of mass 1 kg is pulled along the curve path ACB by a
tangential force as shown in figure. The wor
the frictional force when the block moves from A to B is
(a). 5 J
(b). 10 J
(c). 20 J
(d). none of these
CORRECT OPTION: C
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SOLUTION
Work done by friction
Q No. - 9
JEE MAIN-FULL TEST 03 - PHYSICS
In a capillary tube placed inside the liquid of density in a
container, the rise of liquid is h. When block o
is placed on the liquid as shown in figure, liquid in the tube
is h\'. If then
(a).
= ∫→F .
→ds = ∫
x
0μmg cos θ
dx
cos θ
= μmg × = 20J
(ρ)
σ < ρ
h' = h
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(b).
(c).
(d). insufficient data
CORRECT OPTION: A
SOLUTION
There will be no change.
Q No. - 10
JEE MAIN-FULL TEST 03 - PHYSICS
A light rod of length , is hanging from the vertical smooth wall
of a vehicle moving with acceleration
small mass attached at it\'s one end is free to to rotate about
an axis passing through the other end. The m
velocity given to the mass at it\'s equilibrium possition so
that it can complete vertical circular motion is
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: C
h' < h
h' > h
∴ H' = h
L √3g
√5gL
√4gL
√8gL
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SOLUTION
Conservation of energy
Q No. - 11
JEE MAIN-FULL TEST 03 - PHYSICS
Energy stored in the capacitor in it\'s steady state is
mv2 − m = m + m√3(2l )g12
g(l)
2
g(l)
2
√3
2
v = √8gl
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(a).
(b).
(c).
(d). zero
CORRECT OPTION: D
SOLUTION
Potential across capacitor is zero, hence energy stored is
zero.
Q No. - 12
JEE MAIN-FULL TEST 03 - PHYSICS
The potential difference between two points A and B which are
separated by a distance of 1m along the fie
uniform electric field of is
(a). zero
(b). volt
(c). volt
(d). volt
CORRECT OPTION: C
SOLUTION
.
CV
CV
2
QV1
2
10NC − 1
100
10
0.1
VA − VB = ∫E. dx = 10(1)
= 10vo
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Q No. - 13
JEE MAIN-FULL TEST 03 - PHYSICS
The power factor of a circuit in which a box having unknown
electrical devices connected in series with a r
resistance is . The reactance of the box is
(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
but cos
.
Q No. - 14
JEE MAIN-FULL TEST 03 - PHYSICS
Two points A and B are at distances of \'a\' and \'b\'
respectively from an infinite conducting plate having ch
density . The work done in moving charge from A to B is
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: A
SOLUTION
Q No. - 15
JEE MAIN-FULL TEST 03 - PHYSICS
A point charge of is placed on the circumference of a
non-conducting ring of radius 1m which is rota
constant angular acceleration of . If ring starts it\'s motion
at the magnetic field at the cent
ring at sec, is
(a).
(b).
(c).
3Ω3
55Ω
Ω5
34Ω
Ω4
3
Z = √R2 + X2 = √9 + X2
ϕ = =R
Z
3
5X = 4Ω
σ Q
(b − a)Qσ
ε0
Qσ
(b − a)Qσ
(b − a)ε0
ε =σ
ε0
VA − VB = (b) − (a)σ
ε0
σ
ε0
∴ W = Q(VA − VB) = (b − a)Qσ
ε0
0.1C
1rad
sec2t = 0
t = 10
10− 6T
10− 7T
10− 8T
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(d).
CORRECT OPTION: B
SOLUTION
Q No. - 16
JEE MAIN-FULL TEST 03 - PHYSICS
The wavelength corresponding to maximum spectral of a black body
A is . Consider anoth
body B whose surface area is twice of that of A and total
radiant energy emitted by B is 16 times that emitt
The wavelength corresponding to maximum spectrum radiancy for B
will be
(a).
(b).
(c).
(d).
CORRECT OPTION: D
SOLUTION
107T
ω = 0 + 1 × 10 = 10rad
sec2
∴ V = rω = 1 × 10 = 10m
s
→B = ⇒
∣∣∣
→B
∣∣∣
=μ0
4π
q(→v × →r )
r3μ0qv
4πr2
B = = 10− 7T10− 7 × 0.1 × 10
(1)2
λA = 5000 �
5000(8) �14
2500 �10, 000 �
�5000(8)
14
P = σAT 4
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Since = constant,
Q No. - 17
JEE MAIN-FULL TEST 03 - PHYSICS
During an adiabatic process, the density of a gas is found to be
proportional to cube of temperature. The d
freedom of gas molecule is
(a). 6
(b). 5
(c). 4
(d). 3
CORRECT OPTION: A
SOLUTION
For adiabatic process, = constant
= constant
= constant
Q No. - 18
JEE MAIN-FULL TEST 03 - PHYSICS
Two fixed charge and are located at the point of co-ordinates
and respectively
plane. Then all the points in plane where potential is zero lies
on a
(a). straight line parallel to x-axis
(b). straight line parallel to y-axis
(c). a circle of radius 4a
(d). circle of radius 2a
CORRECT OPTION: C
SOLUTION
According to equation
⇒ = ( )4
⇒ 16 = ( )4
⇒ TB = TA(8)PA
PB
AB
AA
TB
TA
2
1
TB
TA
14
λmT = = (8)λA
λB
TB
TA
14
⇒ λB = = �λA
(8)14
5000
(8)14
TV γ− 1
T( )γ− 1
m
ρT
ργ− 1
ρ ∝ T ⇒ = 3 ⇒ γ =1
γ− 11
γ − 1
4
3
f = = = 62
γ − 1
2
( − 1)43
−2Q Q ( − 3a, 0) (3a, 0)
(x − y) x − y
P(x, y), A( − 3a, 0), B = (3a, 0)
VPA = , VPB =1
4πε0
−2Q
PA
1
4πε0
Q
PB
( ) + = 014πε0
−2Q
PA
1
4πε0
Q
PB
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Q No. - 19
JEE MAIN-FULL TEST 03 - PHYSICS
In an circuit shown in the figure, . At time , charge in the
capacitor is
decre4asing at a rate of . Choose the correct statements.
(a). maximum charge in the capacitor can be 6C
(b). maximum charge in the capacitor can be 8C
(c). charge in the capacitor will be maximum after time sec
(d). None of these
CORRECT OPTION: A
SOLUTION
Q No. - 20
= ⇒ 4PB2 = PA22
PA
1
PB(x − 5a)2 + y2 = (4a)2
L − C C = 1F , L = 4H t = 0 4C
√5C
s
2 sin− 1( )23
I = √5A
= + Li2 ⇒ qmax = 6Cq2m
2C
q2
2C
1
2
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JEE MAIN-FULL TEST 03 - PHYSICS
A plank of mass is placed over smooth inclined plane and a
sphere is also placed over the plank. Fricti
sufficient between the sphere and the plank to revent slipping.
If the plank and the sphere are released fro
frictional force on the sphere is
(a). up the plane
(b). down the plane
(c). zero
(d). horizontal
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 21
M
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JEE MAIN-FULL TEST 03 - PHYSICS
A disc of mass and radius is placed over a plank of same mass .
There is sufficient friction betwee
and the plank to prevent slipping. A force is appliled at the
centre of the disc. Choose the correct statem
(a). Acceleration of the plank is
(b). Acceleration of the plank is
(c). Force of friction between disc and plank is
(d). Force of friction between disc and plank is
CORRECT OPTION: A
SOLUTION
m R m
F
F
4mF
2mF
6F
2
f = ma2
α = =τ
I
2f
mR
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Q No. - 22
JEE MAIN-FULL TEST 03 - PHYSICS
A certain amount of ideal monoatomic gas undergoes, process
given by where is the intern
of the gas. The molar specific heat of the gas for the process
will be
(a).
(b).
(c).
(d).
CORRECT OPTION: D
⇒ a2 = m, f =F
4
F
4
UV = C12 U
R
23R
5R
2
−R
2
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SOLUTION
The process is equivalent to
Compare with
Q No. - 23
JEE MAIN-FULL TEST 03 - PHYSICS
An isotropic sound is moving in a circle of radius with a small
speed . An observer is hearing this
(See figure). The intensity of the sound heard by will be
maximum when the source is at point.
(a). 1
(b). 2
(c). 6
(d). none of these
CORRECT OPTION: A
SOLUTION
Intensity will be highest at the nearest point.
TV = C12
TV x− 1 = C ⇒ x =3
2
⇒ C = + = + = R − 2R = − RR
γ − 1
R
1 − X
R23
R
1 − ( )32
3
2
1
2
A R v B
B
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Q No. - 24
JEE MAIN-FULL TEST 03 - PHYSICS
A stable species can be obtained through unstable and . The half
life period for conversion
that for conversion is . Initially a sample contains unclides of
unclides of and no nu
. After how much will the nuclides of species be equal to
(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 25
JEE MAIN-FULL TEST 03 - PHYSICS
A certain thermodynamic cycle comprises of two isothermal and
adiabatic processes. The highest tempera
obtained in the entire cycle is while the lowest temperature is
. When the cycle represented o
curve the sence of the cycle is counter clockwise. The
efficiency of the cycle represented on a curv
sence of the cycle is counter clockwise. The efficiency of the
cycle is
(a).
(b).
(c).
(d). None of these
CORRECT OPTION: D
C A B A → C
B → C 2T N A, N B
C ( )N2716
2T
3T
4T
5T
600K 300K
P − V
50 %
75 %
25 %
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SOLUTION
N/A
Q No. - 26
JEE MAIN-FULL TEST 03 - PHYSICS
A man wearing spectales with diverging lenses is viewing a
distance object. The image formed on his retin
(a). real and erect.
(b). real and inverted
(c). virtual and erect
(d). virtual and inverted.
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 27
JEE MAIN-FULL TEST 03 - PHYSICS
A thin uniform hemispherical bowl of mass and radius is lying on
a smooth horizontal surface. A horiz
is now applied perpendicular to the rim of the bowl (see
figure). The instantaneous angular acceleration
will be
(a).
(b).
(c).
m R
F
20
3
F
MR10
3
F
MR40
3
F
MR
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(d).
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 28
JEE MAIN-FULL TEST 03 - PHYSICS
For a stone thrown from a tower of unkown height, the maximum
range for a projection peed of is o
a projection angle of . The correcsponding distance between the
foot of the tower and the point landin
stone is
(a).
(b).
(c).
(d).
CORRECT OPTION: D
SOLUTION
Q No. - 29
JEE MAIN-FULL TEST 03 - PHYSICS
Consider the system shown in the figure. The system is so
arranged that both and are in pure trans
a certain instant is moving down with a speed of a speed of then
speed of will be
5
3
F
MR
10m
s30 ∘
10m
20m
( )m20√3
( )m10√3
tan θ = ⇒ = = =u2
Rg
u2
g tan θ
100
100 × √3
10
√3
m M
M 1m
s1m
sm
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(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 30
JEE MAIN-FULL TEST 03 - PHYSICS
Consider the circuit in the adjacent figure. What will be
potential difference between and in the steady
(a).
(b).
(c).
(d). zero
CORRECT OPTION: D
SOLUTION
N/A
7m
s
8m
s
1m
s
15m
s
A B
εε
2ε
3
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Q No. - 31
JEE MAIN-FULL TEST 03 - CHEMISTRY
For the equilibrium, and
hence, for the equilibrium:
equilibrium constant is:
(a).
(b).
(c).
(d).
CORRECT OPTION: A
SOLUTION
N/A
Q No. - 32
JEE MAIN-FULL TEST 03 - CHEMISTRY
2-Pentyne, on reduction with lithium in liq. yields
(a). pentane
(b). cis-2-pentene
(c). trans-2-pentene
(d). 1-pentene
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 33
JEE MAIN-FULL TEST 03 - CHEMISTRY
Ag+ + 2NH3 ⇔ [Ag(NH3)2]+K1 = 1.8 × 10
7
Ag+ + CI − ⇔ AgCI, K2 = 5.6 × 109 AgCI + 2NH3 ⇔ Ag(NH3)
+2
0.32 × 10− 2
0.31 × 10− 21
1.01 × 1017
1.01 × 10− 17
NH3
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Chlordobenzene. A forms only one nitro derivative. Hence is,
(a). Benzyl chloride
(b). p-Chlorotoluene
(c). o-Chlorotoulene
(d). m-Chlorotoulene
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 34
JEE MAIN-FULL TEST 03 - CHEMISTRY
Clasien rearrangement of allyl phenyl ether gives a mixture
of
(a). -and p-allyl phenols.
(b). -and -allyl phenols.
(c). -and -allyl phenols.
(d). -and -allyl phenols.
CORRECT OPTION: D
SOLUTION
N/A
Q No. - 35
JEE MAIN-FULL TEST 03 - CHEMISTRY
If in diamond, there is a unit cell fo carbon atoms as fcc and
if carbon atom is hybridized, what fraction
ae occupied by carbon atom.
(a). tetrahedral
(b). tetrahedral
(c). octahedral
(d). octahedral
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 36
JEE MAIN-FULL TEST 03 - CHEMISTRY
Out of boiling point (I), entropy (II), (III) and standard
e.m.f. of a cell (IV), intensive properties are-
(a).
(b).
(c).
(d). all of these
CORRECT OPTION: C
A(C7H7CI)( i )KMnO4−−−−−−→( ii ) Sodalime
Δ
A
o − , m
o m
m p
o p
sp3
25 %
50 %
25 %
50 %
pH
I, II
I, II, III
I, III, IV
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SOLUTION
N/A
Q No. - 37
JEE MAIN-FULL TEST 03 - CHEMISTRY
Which metal liberates with very dilute nitric acid:
(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 38
JEE MAIN-FULL TEST 03 - CHEMISTRY
Which of the following statements are not correct?
(a). The number of and -particles emitted in neptunium series
and 7 and 4, respe
(b). The number of and -particles emitted in uranium series are
8 and 7, respective
(c). The number of and -particles emitted in actinium series are
4 and 7, respective
(d). The half-life of a radiactive decay is given as , where is
decay constant
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 39
JEE MAIN-FULL TEST 03 - CHEMISTRY
Reactivity of borazole is greater than that of benzene
because:
(a). borazole is non-polar compound
(b). borazole is polar compound
(c). borazole has electron in it
(d). of localized electrons in it
CORRECT OPTION: B
SOLUTION
N/A
H2
Zn
Cu
Mn
Hg
α β (.23793 Np →20983 Bi)
α β (.23892 U →20683 Pb)
α β (.23592 U →20783 Bi)
t =12
0.693
λλ
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Q No. - 40
JEE MAIN-FULL TEST 03 - CHEMISTRY
When acetone and chloroform are mixed, hydrogen bonding takes
place between them. Such a liquid pair
(a). No deviations from Raoult\'s law
(b). Negative deviations from Raoult\'s law
(c). Positive deviations from Raoult\'s law
(d). None
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 41
JEE MAIN-FULL TEST 03 - CHEMISTRY
On heating ammonium dichromate, the gas evolved is
(a). Oxygen
(b). Ammonia
(c). Nitrous oxide
(d). Nitrogen
CORRECT OPTION: D
SOLUTION
N/A
Q No. - 42
JEE MAIN-FULL TEST 03 - CHEMISTRY
When ammonia is added to a cupric salt solution, the deep blue
colour, so observed is due to the formatio
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(a).
(b).
(c).
(d).
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 43
JEE MAIN-FULL TEST 03 - CHEMISTRY
Arrange the following in order of their decreasing thermal
conductivity
(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 44
JEE MAIN-FULL TEST 03 - CHEMISTRY
In a measurement of quantum efficiency of photo-synthesis in
green plants, it was found that 10 quanta of
wavelength 6850 were needed to release one molecule of . The
average energy storage in this pro
112 kcal/mol evolved. What is the energy conversion efficiency
in this experiment? Given
.
(a). 23.5
(b). 26.9
(c). 66.34
(d). 73.1
CORRECT OPTION: B
SOLUTION
Total energy of 10 quanta
Energy stored for process
efficiency
Q No. - 45
JEE MAIN-FULL TEST 03 - CHEMISTRY
The rate of effusion of two gases \'a\' and \'b\' under
identical condition of temperature and pressure are in
[Cu(OH)4]2 −
[Cu(NH3)4]2 +
[Cu(OH)2(NH3)2]
[Cu(H2O)4]2 +
AI, Ag, Cu
Cu, Ag, AI
Ag, Cu, AI
AI, Cu, Ag
� O2O2
cal = 4.18J, NA = 6 × 1023, h = 6.64 × 10− 34JS
E = = 2.9 × 10− 19Jhc
λ= 10 × 2.9 × 10− 19J
= = 7.8 × 10− 19J112 × 4.18 × 103
6 × 1023
% = × 1007.8 × 10− 19
29 × 10− 19= 27.9 %
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2: 1. What is the ratio of rms velocity of their molecules if
and are in the ratio of 2:1 ?
(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
As
Q No. - 46
JEE MAIN-FULL TEST 03 - CHEMISTRY
For first order parallel reaction and are 4 and respectively at
. If the activation energie
formation of and are respectively and respectively. The
Temperature at which
will be obtained in equimolar ratio is:
(a).
Ta Tb
2: 1
√2: 1
2√2: 1
1: √2
= √rarb
2
1
Mb
Ma
Vrms ∝ √T
MVrms = √
3RT
M
= √ = × =Vrms (a )
Vrms ( b )
Ta × MbTb × Ma
2
1
√2
1
2√2
2
k1 k2 2− 1
min 300K
B C 30, 000 33, 314 olJ
m
757.48K
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(b).
(c).
(d). None of these
CORRECT OPTION: B
SOLUTION
�(i)
�(ii)
Eqn. (ii) -Eqn.(i)
(For equimolar formation of and
.
Q No. - 47
JEE MAIN-FULL TEST 03 - CHEMISTRY
One mole of was placed in a two litre vessel at a certain
temperature. The following equilibrium was
in the vessel.
The equilibrium mixture reacted with 0.2 mole in acidc medium.
Hence is:
(a). 0.5
(b). 0.25
(c). 0.125
(d). none of these
CORRECT OPTION: C
SOLUTION
Only will oxidized.
Equivalent of Equivalent of
329.77K
600K
ln( ) = [ − ]k'1k1
E1
R
1
T1
1
T1
ln( ) = [ − ]k'2k2
E2
R
1
T1
1
T2
ln( ) = [ − ]k'2 × k1k2 × k'1
(E2 − E1)
R
1
T1
1
T2
B C, k'2 = k'1 )
ln( ) = ( )( )k1k2
8.314
8.314
T2 − 300
300 × T2
ln 2 = = ( )( )83148.314
T2 − 300
300 × T2T2 = 329.77K
SO3
2SO3(g) ⇔ 2SO2(g) + O2(g)
KMnO4 Kc
2SO3(g)g ⇔ 2SO2(g) + O2(g)
at equilibrium (1 − 2x) (2x) x
SO2
SO2 = KMnO4
2x × 2 = 0.2 × 5
2x = 0.5
Kc = = 0.125[ ]2[ ]0.52
0.252
[ ]20.52
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Q No. - 48
JEE MAIN-FULL TEST 03 - CHEMISTRY
A complex of iron and cyanide ions is ionized at 1 molal. If its
elevation in boiling point is , th
complex is: (Given
(a).
(b).
(c).
(d).
CORRECT OPTION: B
SOLUTION
It means salt on dissociation gives 4 ions. Thus the salt the
gives 4 ions is .
Q No. - 49
JEE MAIN-FULL TEST 03 - CHEMISTRY
Find the standard cell potential involving the cell
reaction:
, at
Given:
(a).
(b).
(c).
(d).
CORRECT OPTION: C
100 % 2.08∘
Kb = 0.52∘Cmol− 1kg)
Fe[Fe(CN)6]3K3[Fe(CN)6]K4[Fe(CN)6]Fe3[Fe(CN)6]2
ΔTb = i × kb × m
2.08 = 0.52 × 1 × i
i = 4
K3[Fe(CN)6]
In2 + + Cu+ 2 → In+ 3 + Cu+ 298K
E0C u+
= X1V E0
I n+= X2FV E
0
I n+= X3Vu+ 2
Cn+ 3
In+ 2
I
X1 + X3 − X2X1 + X3 − 2X2
3X1 + X3 − 2X2
X1 + X3 + 2X2
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-
SOLUTION
From Eqn. (i) + (iii) -(ii)
Q No. - 50
JEE MAIN-FULL TEST 03 - CHEMISTRY
is a covalent liquid because
(a). clouds of the ions are weakly polarized to envelop the
cation
(b). electron clouds of the ions are strongly polarized to
envelop the cation
(c). its molcules are attracted to one another by strong vander
Waals\' forces
(d). shown inter-pair effect
CORRECT OPTION: B
SOLUTION
Due to Fajan\'s rule.
Q No. - 51
JEE MAIN-FULL TEST 03 - CHEMISTRY
Boron nitride obtained by heating borazole is:
(a). white solid with a diamoned like structure
(b). slippery white solid with layered structure similar to that
of graphite
(c). covalent liquid and is structurally similar to carbon
monoxide
(d). low melting solid with rock-salt like structure
molecules
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 52
JEE MAIN-FULL TEST 03 - CHEMISTRY
If a metal has low oxygen affinity then purification of metal
may be carried out by
(a). Liquation
(b). Distillation
(c). Zone refining
(d). Cupellation
CORRECT OPTION: B
(Cu+ 2 + e− → Cu+ , , E ∘ = X1V , ΔG ∘1 = − FX1, ...(i)),
(In+ 3 + 2e− → In+ 1, E ∘ = X2V , ΔG ∘2 = − 2FX2, . . (ii)),
(In+ 2 + e− → In+ , E ∘ = X2V , ΔG ∘3 = − FX3, ...(iii))
In+ 2 + Cu2 + → In+ 3 _ Cu+ ΔG ∘ = − FE ∘
ΔG ∘ = − F(X1 + X3 − 2X2) = − FE∘
E ∘ = (X1 + X3 − 2X2)
SnCI4
e− CI −
CI −
Sn
-
SOLUTION
N/A
Q No. - 53
JEE MAIN-FULL TEST 03 - CHEMISTRY
An original salt solution in acidic medium did not give any
precipitate on passing gas. Such a solution
obiled, reboiled after dilution 3 times. To such a solutions two
drops of conc. were added, then hea
water was added. To this resulting solution, was first added
followed by excess of . fina
ppt. was obtained. Hence the cation may be.
(a).
(b).
(c).
(d).
CORRECT OPTION: D
SOLUTION
Green ppt. is
Q No. - 54
JEE MAIN-FULL TEST 03 - CHEMISTRY
Which of the following statement is wrong?
(a). orbital taking part in is
(b). orbital taking part in is
(c). orbitals taking part in are and
(d). orbitals taking part in are and
CORRECT OPTION: B
SOLUTION
N/A
Q No. - 55
JEE MAIN-FULL TEST 03 - CHEMISTRY
Here is:
(a). Glycollic acid
(b). -hydroxypropionic acid
(c). succinic acid
(d). malonic acid
CORRECT OPTION: D
H2S
HNO3
NH4CI NH4OH
AI + 3
Fe+ 2
Fe+ 3
Cr+ 3
−Cr(OH)3
d dsp2 dx2 − y2
d sp3d dyz
d sp3d2 dx2 − y2 dz2
d sp3d3 dxy, dz2 dz2 − y2
H3CCOOHB
−−→ (Y )( i )KCN
−−−−−→( ii )H3O+
(X)
r2P
(X)
α
-
SOLUTION
Q No. - 56
JEE MAIN-FULL TEST 03 - CHEMISTRY
In order of distinguish between and , which of the following
reagents is useful
(a). Heinsberg reagent
(b). p-napthol
(c). Benzene diazonium chloride
(d). None of these
CORRECT OPTION: B
SOLUTION
aromatic amine on diazotisation followed by cuplting with
-napthol gives azo dye test.
Q No. - 57
JEE MAIN-FULL TEST 03 - CHEMISTRY
The weakest interparticle forces are present in
(a). thermosetting polymers
(b). thermoplastic polymers
(c). fibres
(d). elastomers
CORRECT OPTION: D
SOLUTION
N/A
H3C −
O
∣ ∣
C − OHHVZ−−→B
BrCH2 −
O
∣ ∣
C − OHKCN−−−→H3O+
(Y )
OH −
O
∣ ∣
C − CH2 − COOH(X ) Malonic acid
r2P
C2H5NH2 C6H5NH2
1∘ β
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Q No. - 58
JEE MAIN-FULL TEST 03 - CHEMISTRY
Which of the following statements is false?
(a).
(b).
(c).
(d). in exists in the cage form due to intermolecular hydrogen
bonding.
CORRECT OPTION: D
SOLUTION
(A) Viscosity intermolcular attraction
intermolecular bonding tendency
(B) No intramoecular -bonding in the cis-isomer due to loss in
planarity as a result of steric and electron
repulsion.
stability symmetry.
(C ) Even though -nitrophenol shows intramolecular -bonding but
has greater and effect than
(D) .
No cage lattice structure of .
Q No. - 59
JEE MAIN-FULL TEST 03 - CHEMISTRY
Which among the following compounds will not give effervescence
with sodium bicarbonate?
(a).
(b).
(c).
(d). picric acid
CORRECT OPTION: C
SOLUTION
Because phenol is less acidic than .
Q No. - 60
JEE MAIN-FULL TEST 03 - CHEMISTRY
Which of the following is most reactive for reaction?
H3BO3 C2H5OH
α
α H −
H
∴ α
o H −R −I
H3BO3 + 3C2H5OH → B(OC2H5)3 + 3H2O
∴ H3BO3
C6H5COOH
C6H5SO3H
C6H5OH
H2CO3
SN2
-
(a).
(b).
(c).
-
(d).
CORRECT OPTION: D
SOLUTION
is less electronegative and therefore more reactive towards
.
Q No. - 61
JEE MAIN-FULL TEST 03 - MATHS
Sum of the roots of the equation is
(a). 0
(b). 2
(c). 1
(d). 5
CORRECT OPTION: B
SOLUTION
Given equatio can be written as
or
or
Q No. - 62
JEE MAIN-FULL TEST 03 - MATHS
if are points on the circle such that and the value of the
ex
is least then,
(a). and
(b).
(c). must be real
(d).
CORRECT OPTION: A
S SN2
x2 − 2x + |x − 1| − 5 = 0
( ∣ x − 1)2 + |x − 1| − 6 = 0
t2 + t − 6 = 0
⇒ t = 2 −3 ⇒ |x − 1| = 2
⇒ x − 1 = ± 2 x = 3, − 1
z1, z2, z3, z4 |z| = 1 z1 + z2 + z3 + z4 = 0
|z1 − z2|2 + |z2 − z3|
2 + |z3 − z4|2 + |z4 + z1|
2
z1 + z3 = 0 z2 + z4 = 0
z1 + z2 = z3 + z4
z1, z2, z3, z4
=z1
z3
z2
z4
-
SOLUTION
Q No. - 63
JEE MAIN-FULL TEST 03 - MATHS
if and , then the value of is equal to
(a). 13
(b).
(c).
(d).
CORRECT OPTION: A
SOLUTION
Q No. - 64
(|z1 − z2|)2 + (|z2 − z3|)
2 + (|z3 − z4|)2 + (|z4 − z1|)
2
= 2(|z1|2 + |z2|2 + |z3|3 + |z4|2) + |z1 + z3|2 + |z2 + z4|2
> 8 + |z1 + z3|2 + |z2 + z4|2
Tr =r
r4 + 4Sn =
n
∑r= 1
tr 37S5 −7
26
25
261
27
26
Tr = = =r
r4 + 4
r
(r2 + 2)2 − 4r2r
(r2 + 2r + 2)(r2 _ 2r + 2)
= [ − ] = [ − ]14
1
r2 − 2r + 2
1
(r2 + 2r + 2)
1
4
1
(r − 1)2 + 1
1
(r + 1)2 + 1
⇒ Sn = [1 + − − ]1
4
1
2
1
n2 − 1
1
(n + 1)2 + 1
S5 = [1 + − − ] = [ ] = × =1
4
1
2
1
26
1
37
1
4
38 × 37 − 26 × 1
26 × 37
1
4
1380
26 × 37
345
26 × 37
⇒ 37S5 − = = 137
26
338
26
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JEE MAIN-FULL TEST 03 - MATHS
Number of distinct terms in the expansion of is equal to
(a).
(b).
(c).
(d).
CORRECT OPTION: A
SOLUTION
Number of terms number of non-negative integral solution
Q No. - 65
JEE MAIN-FULL TEST 03 - MATHS
if is symmetric about the lines and then it must be symm
(a). (1,1)
(b).
(c).
(d).
CORRECT OPTION: B
SOLUTION
if a function is symmetric about two mutually perpendicular
lines, it must be symmetric about their point of
intersection.
Q No. - 66
JEE MAIN-FULL TEST 03 - MATHS
if and then,
(a).
(b).
(c).
(d).
CORRECT OPTION: B
SOLUTION
(x + y + z + w)50
.53 C3
51
512
.51 C3
=
y = f(x) 3x + 4y + 1 = 0 4x − 3y − 7 = 0
(1, − 1)
(0, 0)
(1, 0)
√1 − x6 + √1 − y6 = a(x3 − y6) = f(x, y)√dydx
1 − y6
1 − x6
f(xky) =y
x
f(x, y) =x2
y2
f(x, y) =2y2
x2
f(x, y) =y2
x2
− − = a(x2 − y2 )x5
√1 − x6y5
√1 − y6dy
dx
dy
dx
⇒ (ay2 − ) = ax2 +y5
√1 − y6dy
dx
x5
√1 − x6
-
...(1)
Also, or
Hence, from equation (1),
Q No. - 67
JEE MAIN-FULL TEST 03 - MATHS
If with standard notations are four co-normal points of the
hyperbola then the orthoc
the triangle formed by joining the points is given by
(a).
(b).
(c).
(d).
CORRECT OPTION: B
SOLUTION
For a triangle formed by joining any three co-normal points,
orthocentre consides with the fourth point.
Q No. - 68
JEE MAIN-FULL TEST 03 - MATHS
If be the focus of a parabla and be the focal chord, such that
and , then the lengt
rectum of the parabola is
(a). 4
(b). 2
(c). 8
(d). 16
CORRECT OPTION: C
SOLUTION
of and semi latus rectum.
Q No. - 69
JEE MAIN-FULL TEST 03 - MATHS
if a circle having centre at cut the circles and
orthogonaly, then is equal to
(a). 1
⇒ = ×dy
dx
√1 − y6
√1 − x6
x2(a√1 − x6 + x3)
y2(a√1 − y6 − y3)
= a(x3 − y3)(1 − x6) − (1 − y6)
√1 − x6 − √1 − y6x3 + y3 = a(√1 − y6 − √1 − x6)
⇒ a√1 − y6 − y3 = a√1 − x6 + x3
= ×dy
dx
√1 − y6
√1 − x6x2
y2
t1, t2, t3, t4 xy = x2
t1, t2, t3
(0, 0)
(ct4, )c
t4
(ct1 + t2 + t3 , )c
t1 + t2 + t3
( , ct4)c
t4
S PQ SP = 3 SQ = 6
HM SP SQ = = 4 =2(3)(6)
3 + 6
(α, β) x2 + y2 − 2x − 2y − 7 = 0 x2 + y2 + 4x − 6y∣∣∣
α −∣∣∣
3
4
β
2
-
(b).
(c).
(d).
CORRECT OPTION: B
SOLUTION
If a circle cuts two circles orthogonally, radical axis of the
two sicles pass through the centre of the first circ
Here radical axis of given circles is or
Hence
Q No. - 70
JEE MAIN-FULL TEST 03 - MATHS
Two circles with radii and . , touch each other externally if be
the angle between the d
common tangents, then
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: B
SOLUTION
1
21
40
6x − 4y + 40 3x − 2y + 2 = 0
3α − 2β + 2 = 0 ⇒ α − = −3
4
β
2
1
2
r1 r2 r1 > r2 ≥ 2 θ
θ = sin− 1( )r1 + r2r1 − r2
θ = 2 sin− 1( )r1 − r2r1 + r2
θ = sin− 1( )r1 − r2r1 + r2
-
Q No. - 71
JEE MAIN-FULL TEST 03 - MATHS
A unit vector is plane which makes an angle of with the vector
and an angle of w
vector is
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: D
SOLUTION
Let �(1)
Also,
�(2)
.(3)
There exists no real values of a and b satisfying (1), (2) and
(3)
Hence no such unit vector exists.
sinα =r1 − r2r1 + r2
⇒ θ = 2 sin− 1( )r1 − r2r1 + r2
xy − 45 ∘→i +
→j 60 ∘
→3i −
→4j
→i→i +
→j
√2→i −
→j
√2
→r = a
→i + b
→j ⇒ a2 + b2 = 1
cos 45 ∘ = ⇒ a + b = 1
→r . (
→i +
→j )
∣∣→r ∣∣
∣∣∣
→i +
→j
∣∣∣
cos 60 ∘ = ⇒ 3a − 4b =
→r . (3
→i − 4
→j )
∣∣→r ∣∣
∣∣∣3
→i − 4
→j
∣∣∣
5
2
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Q No. - 72
JEE MAIN-FULL TEST 03 - MATHS
Let �.. . Then the sum of the series
�. Is
(a).
(b).
(c).
(d).
CORRECT OPTION: B
SOLUTION
.... �
On comparison and
Q No. - 73
JEE MAIN-FULL TEST 03 - MATHS
There are three points (a,x),(b,y) and (x,z) such that the
straight lines joining any two of them are not equa
to the corrdinate axes where a,b,c,x,y,z .
if and , then are in.
(a). A.P.
(b). G.P.
(c). H.P.
(1 + x)n = 1 + nx + +n(n − 1)x2
2x, n ∈ R
3 + + + +8
3
80
33240
34
9
27
12
101
(1 + x)n = 3 + + + +8
3
80
33240
34= 1 + nx + +
n(n − 1)x2
2
n = − 3 x = −2
3
∈ R∣∣ ∣ ∣∣
x + a y + b z + c
y + b z + c x + a
z + c x + a y + b
∣∣ ∣ ∣∣
= 0 a + c = − b x, − , zy
2
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-
(d). none of these
CORRECT OPTION: A
SOLUTION
From the given conditions
determinant is a symmetric one. The determinant will be equal to
zero is
but (given)
are in A.P.
Q No. - 74
JEE MAIN-FULL TEST 03 - MATHS
if at is tangent to the parabola , then respective values of
a,b,c are
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: A
SOLUTION
For
Tangent at is
Comparing with
Q No. - 75
JEE MAIN-FULL TEST 03 - MATHS
if : then is increasing on
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: B
SOLUTION
so is increasing in
is increasing on
Q No. - 76
JEE MAIN-FULL TEST 03 - MATHS
Coordinates of the point on the straight line , which is nearest
to the parabola
≠ ± 1, ≠ ± 1, ≠ ± 1 ⇒ x + a ≠ y + b ≠ z +y − x
b − a
z − y
c − b
z − x
c − ax + a + y + b + z + c = 0
a + b + c = 0
⇒ x + y + z = 0
⇒ x + z = 2( − ) ⇒ x, − , zy2
y
2
x = 1, y = 2x y = ax2 + bx + c
, 1,1
2
1
2
1, ,1
2
1
2
, , 11
2
1
2
x = 1, y = a + b + c
(1, a + b + c) (y + a + b + c) = ax + (x + 1) + c1
2
b
2y = 2x, c = a, b = 2(1 − a)
f' (x2 − 4x + 3) > 0, ∀x ∈ (2, 3) f(sinx)
⋃n∈ l
(2nπ, (4n + 1) )π2
⋃n∈ l
((4n − 1) , 2nπ)π2
R
x ∈ (2, 3) ⇒ − 1 < x24x + 3 < 0 f(x) ( − 1, 0)
⇒ f(sinx) ⋃n∈ l
((4n − 1) , 2nπ)π2
x + y = 4 y2 = 4(x − 10
-
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: A
SOLUTION
Let point on the straight line be , this will be nearest to the
parabola if at th
the straight line becomes to the parabola.
let it is normal at
Perpendicular to at is ..(1)
Normal at parabola at is ....(2)
(1) and (2) are same so required point is
Q No. - 77
JEE MAIN-FULL TEST 03 - MATHS
if be three vectos of magnitude sch that if is the angle
between
then is equal to
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: A
SOLUTION
Q No. - 78
JEE MAIN-FULL TEST 03 - MATHS
if �..then B is equal to
(a). 0
(b). 1
(c). 2
(d). none of these
CORRECT OPTION: A
( , − )172
9
2
(2, 2)
( , )32
5
2
P x + y = 4 (m, 4 − m) ⊥
x − 10 = t2, y = 2t
x + y = 4 (m, 4 − m) y − (4 − m) = (x − m)
(t2 + 10, 2t) y + t(x − 10) = 12t + t3
⇒ t = − 1, m =17
2( , − )17
2
9
2
ā, b̄, c̄ √3, 1, 2 ā × (ā × c̄) + 3b̄ = 0 θ
cos2 θ3
41
21
4
|ā × (ā × c̄)| = ∣∣3b̄∣∣ = 3∣∣b̄∣∣
|ā|. |(ā × c̄)|sin( ) = 3.1 ⇒ 3 = |ā|. (||ā||c̄ ∣ ∣ sin
θ)π2
⇒ 3 = 3.2 sin θ ⇒ sin θ = ⇒ cos2 θ =1
2
3
4
△ (x) =
∣∣ ∣ ∣∣
ex sin 2x tanx2
ln(1 + x) cosx sinx
cosx2, ex − 1 sinx2
∣∣ ∣ ∣∣
= A + bx + Cx2 +
-
SOLUTION
...
put
Q No. - 79
JEE MAIN-FULL TEST 03 - MATHS
if the function is defined by , then is
(a).
(b).
(c).
(d). does not exist
CORRECT OPTION: A
SOLUTION
Let be the inverse of then
since
so
△ ' (x) =
∣∣ ∣ ∣∣
ex 2 cos 2x 2x sec2 x2
ln(1 + x) cosx sinx
cosx2, ex − 1 sinx2
∣∣ ∣ ∣∣
+
∣∣ ∣ ∣∣
ex sin 2x tanx2
−sinx cosx
cosx2, ex − 1 sinx2
∣∣ ∣ ∣∣
+
∣∣ ∣ ∣∣
ex sin 2x tanx2
ln(1 + x) cosx sinx
−2xsinx2, ex 2xcosx2
∣∣ ∣ ∣∣
1( 1 +x )
= B + 2Cx +
x = 0, B =
∣∣ ∣∣
1 2 0
0 1 0
1 0 0
∣∣ ∣∣
+
∣∣ ∣∣
1 0 0
1 0 −1
1 0 0
∣∣ ∣∣
+
∣∣ ∣∣
1 0 0
0 1 0
0 1 0
∣∣ ∣∣
= 0
f : [2, ∞) → [1, ∞) f(x) = 3x (x− 2 ) f − 1(x)
1 + √1 + log3 x1 − √1 + log3 x1 + √1 − log3 x
g(x) f f(g(x)) = x
⇒ 3g (x ) ( g (x ) − 2 ) = x
⇒ (g(x))2 − 2g(x) − log3 x = 0
⇒ g(x) = = 1 ± √1 + log3 x2 ± √4 + 4 log3 x
2g : [1, ∞] → [2, ∞]
g(x) = 1 ± √1 + log3 x
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Q No. - 80
JEE MAIN-FULL TEST 03 - MATHS
The number of real solutions of the equations
(a). one
(b). two
(c). zero
(d). infinite
CORRECT OPTION: C
SOLUTION
Since
Since
The given equation has no solution.
Q No. - 81
JEE MAIN-FULL TEST 03 - MATHS
let then denotes the fractional part of x) is equal to
(a).
(b).
(c).
(d).
CORRECT OPTION: D
SOLUTION
Let
and integer an integer
tan− 1 √x2 − 3x + 2 + cos− 1 √4x − x2 − 3 = π
√x2 − 3x + 2 ≥ ⇒ 0 tan− 1 √x2 − 3x + 2 ≥π
2√4x − x2 − 3 ≥ 0 ⇒ 0 ≤ cos− 1 √4x − x2 − 3 ≤
π
2⇒ 0 ≤ L. H. S. ≤ π ⇒
x = (5√2 + 7)19
x{x}({. }
219
319
0
1
f = (5√2 − 7)19
x − f = ⇒ [x] + {x} − f =
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and integer but
So,
Q No. - 82
JEE MAIN-FULL TEST 03 - MATHS
A flagstaff stands vertically on a pillar, the height of the
flagstaff being double the height of the pillar.A man
ground at a distance finds that both the pillar and te flagstaff
subtend equal angles at his eyes.the ratio of
of the pillar ad the distance of the man from the pillar, is
(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
N/A
Q No. - 83
JEE MAIN-FULL TEST 03 - MATHS
� then find
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: D
SOLUTION
...
�
so,
Q No. - 84
JEE MAIN-FULL TEST 03 - MATHS
if lines and third line passing through form triangle of area
units t
⇒ {x} − f = −1 < {x} − f < 1 ⇒ {x} = f
x{x} = x. f = 119 = 1
√3: 1
1: 3
1: √3
√3: 2
= + + +1
x
2e
3!
4e
5!
6e
7!∞ ∫
x
0f(y)logy xdy, y > 1
[f(e)]2
2[f( )]
21e
2[f(e2)]92
2
= 2e[ + + +1x
1
3!
2
5!
3
7!∞] = 2eS
S =∞
∑r= 1
=∞
∑r= 1
=∞
∑r= 1
[ − ]r(2r + 1)!
1
2
(2r − 1) − 1
(2r + 1)!
1
2
1
(2r) !
1
(2r + 1)1
= [ − + − +12
1
2!
1
3!
1
4!
1
5!∞]
= e− 1 =1
2
1
2e
⇒ = 2eS = = 11
x
2e
2e
∫1
0f(y)logy xdy = 0
x = y = z, x = =y
2
z
3(1, 1, 1) √6
-
intersection of third line with second line will be
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: B
SOLUTION
let any point on second line be
So, B is
Q No. - 85
JEE MAIN-FULL TEST 03 - MATHS
if and be vertices of a triangle whose circumcentre is te
origin, then orthocentre is given
(1, 2, 3)
(2, 4, 6)
( , , )43
8
3
12
3
(λ, 2λ, 3λ)
cos θ = sin θ =6
√42
6
√42
△OAB = (OA). Ob sin θ = √3. λ√14 × = √61
2
1
2
6
√42⇒ λ = 2
(2, 4, 6)
A(ā), B(b̄) C(c̄)
-
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: C
SOLUTION
Centroid triangle will be
Now line joining the orthocentre and the circumcentre is divided
by centroid in ratio internally, so ortho
be
Q No. - 86
JEE MAIN-FULL TEST 03 - MATHS
Let . Then number of roots of the equation in is
(a). 2
(b). 4
(c). 0
(d). infinite
CORRECT OPTION: C
SOLUTION
If we draw the graph of tan x ad cot x, we observes that range
of is
So does not have any root.
Q No. - 87
JEE MAIN-FULL TEST 03 - MATHS
ā + b̄ + c̄
3ā + b̄ + c̄
2ā + b̄ + c̄
ā + b̄ + c̄
32: 1
ā + b̄ + c̄
f(x) = max {tanx, cotx} f(x) =1
√3(0, 2π)
f(x) [ − 1, 0) ∪ [1, ∞)
f(x) =1
√3
-
if denote the sum of infinity and the sum of terms of the series
� such tha
, then the least value of is
(a).
(b).
(c).
(d).
CORRECT OPTION: C
SOLUTION
least value of
Q No. - 88
JEE MAIN-FULL TEST 03 - MATHS
if , the maximum value of f(x) is
(a).
(b).
(c).
(d). none of these
CORRECT OPTION: A
S Sn n 1 + + + +1
3
1
9
1
27
S − Sn <1
300n
4
5
6
7
S − Sn <1
300
⇒ − < ⇒ [1 − 1 + ] 4501
3n1
300
2
3
1
3n1
450∵ n = 6
f(x) = ∫4
0e | t−x | dt(0 < x < 4)
e4 − 1
2(e2 − 1)e2 − 1
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-
SOLUTION
, so maximum value of is .
Q No. - 89
JEE MAIN-FULL TEST 03 - MATHS
Let then number of solutions of the inequation is
(a). 4
(b). 2
(c). 1
(d). infinite
CORRECT OPTION: A
SOLUTION
Given inequation can be rewritten as
now and
So, we should have
Q No. - 90
JEE MAIN-FULL TEST 03 - MATHS
A triangle is inscribed in a circle. The vertices of the
triangle divide the circle in to three arcs of length 3,4 a
then area of the triangle is equal to
(a).
(b).
(c).
(d).
CORRECT OPTION: A
SOLUTION
let radius of the circle be
f(x) = ∫x
0e | t−x | dt + ∫
4
x
e | t−x | dt = ∫x
0ex− tdt + ∫
4
x
et−xdt = − ex− t∣∣.x0 + et−x∣∣
x
4= ex +
f(x) = ex − e4 −x = 0 ⇒ x = 4 − ξmpliex = 2
f(0) = f(4) = e4 − 1, f(2) = 2(e2 − 1) f(x) e4 − 1
x, y ∈ [0, 10] (x, y)∣∣∣3
sec2 x− 1√9y2 − 6y + 2 < 1
3sec2 x√y2 + < 1−2y
3
2
9
⇒ 3sec2 x√(y − )
2
+ < 11
3
1
9
3sec2 x > 3 √(y − )
2
+ >1
3
1
9
1
3
sec2 x = 1, y =1
3⇒ x = 0, π, 2π, 3π
9√3(1 + √3)
π2
9√3(√3 − 1)
π2
9√3(1 + √3)
2π2
9√3(√3 − 1)
2π2
arc(AC) = 3, arc(AB) = 4, arc(BC) = 5
r
-
Now,
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⇒ rα = 3 ⇒ α = , β = , γ =3
r
4
r
5
r
3 + 4 + 5 = 2πr ⇒ r = ⇒ =6
π
1
r
π
6△ABC = △OAC + △OAB + △OBC
= r2(sin( ) + sin( ) + sin( ))12
3
r
4
r
5
r
= . (sin( ) + sin( ) + sin( ))12
36
π2π
2
2π
3
5π
6
= (1 + + ) =18π2
√3
2
1
2
9√3(1 + √3)
π2
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