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Title: jee mains solution 2013
Description: It is solution of jee main exam 2013
Description: It is solution of jee main exam 2013
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JEE (MAIN)-2013
Date : 07-04-2013
Duration : 3 Hours
Max
...
Immediately fill the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen
...
2
...
When you are directed to open the Test Booklet, take
out the Answer Sheet and fill in the particulars carefully
...
The test is of 3 hours duration
...
The Test Booklet consists of 90 questions
...
5
...
Each question is allotted 4 (four) marks for correct
response
...
Candidates will be awarded marks as stated above in Instructions No
...
¼ (one fourth) marks will be deducted for indicating incorrect response of each question
...
7
...
Filling up more than one response in any question will
be treated as wrong response and marks for wrong response will be deducted accordingly as per instructions
6 above
...
Use Blue/Black Ball Point Pen only for writing particulars/marking responses on Side-1 and Side-2 of the
Answer Sheet
...
9
...
, except the Admit Card inside the examination hall/room
...
Rough work is to be done on the space provided for this purpose in the Test Booklet only
...
11
...
However, the candidates are allowed to take away this Test Booklet with them
...
The CODE for this Booklet is P
...
In case of discrepancy, the condidate should immediately report the matter to
the Invigilator for replacement of both the Test Booklet and the Answer Sheet
...
Do not fold or make any stray marks on the Answer Sheet
...
examrace
...
A uniform cylinder of length L and mass M having cross - sectional area A is suspended, with its length
vertical, from a fixed point by a massless spring such that it is half submerged in a liquid of density at
equilibrium position
...
Mg
k
(2)
LA
Mg
1 –
M
k
(3)
LA
Mg
1 –
2M
k
(4)
Mg LA
1
k
M
kx0 + FB = mg
kx0 +
L
Ag = Mg
2
Mg
x0 =
=
LAg
2
k
Mg
LA
1
k
2M
Ans (3)
2
...
If there is a vertical magnetic field ‘B’ in the region, the e
...
f
...
e=
(x )Bdx
= B
2
=
3Bl 2
2
(3)
4Bl2
2
(4)
5Bl 2
2
[(3 )2 (2 )2 ]
2
5B 2
2
Ans
...
examrace
...
This question has statement and Statement
...
Statement - : A point particle of mass m moving with speed collides with stationary point particle of mass
1
m
2
...
If the maximum energy loss possible is given as ƒ m then ƒ =
2
Mm
Statement - : Maximum energy loss occurs when the particles get stuck together as a result of the
collision
...
(2) Statement - is true, Statment - is true, Statement - is not the correct explanation of Statement -
...
(4) Statement - is false, Statment - is true
...
Maximum energy loss =
=
P2
P2
2m 2(m M)
P2 M 1
2 M
mv
2m (m M) 2
m M
M
f
m M
Hence Statement -1 is wrong and statement 2 is correct Hence
Ans (4)
4
...
Let [ 0] denote the dimensional formula of the permittivity of vacuum
...
m
[ AT ] 2
2
=
MLT 2
...
(2)
5
...
i
j
j
i
Sol
...
(i)
y = 2t –
1
(10t2)
2
...
examrace
...
The amplitude of a damped oscillator decreases to 0
...
In another 10s it
will decrease to times its original magnitude, where equals
...
7
(2) 0
...
729
(4) 0
...
A = A0 e
bt
2m
after 5 second
0
...
(i)
After 10 more second
A = A0 e
b(15)
2m
...
729 A0
7
...
(3)
Two capacitors C1 and C2 are charged to 120 V and 200 V respectively
...
Then :
(1) 5C1 = 3C2
(2) 3C1 = 5C2
(3) 3C1 + 5C2 = 0
(4) 9C1 = 4C2
Sol
...
(2)
8
...
5 m is made of steel
...
What
is the fundamental frequency of steel if density and elasticity of steel are 7
...
2 × 1011 N/m2
respectively ?
(1) 188
...
2 Hz
(3) 200
...
f=
v
1 T
1
2 2 2
Also
Y=
T
A
= 1
...
01, d = 7
...
2 × 1011 N/m2
After solving
f=
2 10 3
Hz
7
3
f 178
...
(2)
www
...
com
9
...
A circular loop of radius 0
...
The centre of the
small loop is on the axis of the bigger loop
...
If a current of 2
...
1 × 10–11 weber
(2) 6 × 10–11 weber
(3) 3
...
6 × 10–9 weber
0 (2)(20 10 2 )2
2 (0
...
15)2
× (0
...
216 × 10–11
9
...
Diameter of a plano - convex lens is 6 cm and thickness at the centre is 3 mm
...
n
3
2
32 + (R – 3mm)2 = R2
32 + R2 – 2R(3mm) + (3mm)2 = R2
R 15 cm
1 3
1
1
f = 30cm
f 2 15
Ans (3)
11
...
5GmM
6R
(2)
2GmM
3R
(3)
GmM
2R
Ef
GmM
3R
1
GMm 1 GM GMm GMm 1 GMm
2
m
mv 0
1
2
3R
2 3R
3R
3R 2
6R
Ei
(4)
GMm
+K
R
Ei = Ef
K
5GMm
6R
Ans (1)
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...
com
12
...
Find the maximum modulated
frequency which could be detected by it
...
62 MHz
(2) 10
...
31 MHz
(4) 5
...
C
Signal
R
= RC = 100 × 103 × 250 × 10–12 sec
= 2
...
5 × 10–5 sec
The higher frequency which can be detected with tolerable distortion is
f
1
1
Hz
2m aRC 2 0
...
5 10 5
=
100 10 4
Hz
25 1
...
2
= 10
...
Ans (2)
13
...
A beam of unpolarised light of intensity 0 is passed through a polaroid A and then through another polaroid B
which is oriented so that its principal plane makes an angle of 45º relative to that of A
...
(3)
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...
com
14
...
The resistance of the lead wires is 6
...
What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the
bulb ?
(1) zero Volt
(2) 2
...
3 Volt
(4) 10
...
120 120
= 240
60
240
120V
Req
...
073 volt
246
60
120V
V2
48
120 106
...
04 Volt
Ans (4)
120Volt
15
...
The amount of heat, extracted from the source in a single cycle is :
(1) p0v0
Sol
...
examrace
...
A hoop of radius r and mass m rotating with an angular velocity 0 is placed on a rough horizontal surface
...
What will be the velocity of the centre of the hoop when
it ceases to slip ?
(1)
r0
4
(2)
r0
3
(3)
r0
2
(4) r0
V
R
Sol
...
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M
...
When the piston is in equilibrium, the volume of
the gas is V0 and its pressure is P0
...
Assuming that the system is completely isolated from its surrounding, the piston executes a
simple harmonic motion with frequency :
(1)
Sol
...
(1)
P0 Ax0 = PA(x0– x)
let piston is displaced by x
P
(4)
P x
Mg 0 0
x x
0
P0V0 Pv'
P0 x 0
x
x0
(x 0 x)
= Frestoring
[ x0 – x x0 ]
P0 Ax
x0
1
f 2
=
MV0
AP0
A = Frestoring
1 x 0
P0 A
x 0 x
F
1
2
1
2
P0 A
x 0M
P0 A 2
MV0
Ans (3)
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...
com
18
...
According to Newtons cooling law option (3) is correct Answer
...
This questions has Statement and Statement
...
Statement - : Higher the range, greater is the resistance of ammeter
...
Sol
...
(1) Statement - is true, Statment - is true, Statement - is the correct explanation of Statement -
...
(3) Statement - is true, Statment - is false
...
Statements I is false and Statement II is true
Ans (4)
In an LCR circuit as shown below both switches are open initially
...
(q
is charge on the capacitor and = RC is Capacitive time constant)
...
examrace
...
q = CV (1– et/)
at t = 2
q = CV (1– e–2)
Ans (3)
21
...
The fringes obtained
on the screen will be :
(1) points
(2) straight lines
(3) semi-circles
(4) concentric circles
Sol
...
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT
...
| C | = 20 × 10–9 × 3 × 108
Sol
...
Ans (2)
23
...
The wavelength of the light falling on the cathode is gradually
changed
...
(2)
(4)
As is increased, there will be a value of above which photoelectrons will be cease to come out so
photocurrent will become zero
...
Ans (4)
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...
com
24
...
For same value of current higher value of voltage is required for higher frequency hance (1) should be correct
answers
...
Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains
unchanged
...
(1) L/T
Sol
...
(4)
26
...
If n>>1, the frequency of radiation emitted is proportional to :
1
n
E= h
(1)
Sol
...
(4)
www
...
com
27
...
Ans
...
Two charges, each equal to q, are kept at x = – a and x = a on the x-axis
...
If charge q0 is given a small displacement (y <
force acting on the particle is proportional to :
(1) y
(2) –y
(3)
1
y
(4)
1
y
Sol
...
y
2
y a2
Fnet
q
2kq y
2
=
2
(y a2 )3 / 2
kq2 y
a3
y
Ans
...
examrace
...
Two short bar magnets of length 1 cm each have magnetic moments 1
...
00 Am2 respectively
...
They have a common magnetic equator and are separated by a distance of 20
...
The value of the
resultand horizontal magnetic induction at the mid - point O of the line joining their centres is close to
(Horizontal component of earth’s magnetic induction is 3
...
6 × 10–5 Wb/m2
Sol
...
56 × 10–4 Wb/m2
(3) 3
...
80 × 10–4 Wb/m2
Bnet = B1 + B2 + BH
0 (M1 M2 )
BH
4
r3
Bnet =
10 7 (1
...
1)3
+ 3
...
56 × 10–4 wb/m2
Ans
...
A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure
...
2L
V=
L
kdq
x
2L
=
L
q
dx
1 L
4 0
x
q
= 4 L n(2)
0
Ans
...
examrace
...
Ans
...
32
...
Sol
...
Which of the following complex species is not expected to exhibit optical isomerism ?
(1) [Co(en)3]3+
(2) [Co(en)2 Cl2]+
(3) [Co(NH3)3 Cl3]
(4) [Co(en) (NH3)2 Cl2]+
(3)
[Co(NH3)3Cl3] show facial as well as meridional isomerism
...
So, the answer is (3)
...
A solution of ( – ) – 1 – chloro–1–phenylethane in toluene racemises slowly in the presence of a small
amount of SbCl5, due to the formation of :
(1) carbanion
(3) carbocation
Ans
...
34
...
74 V ; EMnO / Mn2 = 1
...
33 V ; E Cl / Cl = 1
...
Sol
...
35
...
Sol
...
04 mol of an ideal gas expands reversibly from 50
...
0ºC
...
The values of q and w for the process will be:
(R = 8
...
5 = 2
...
examrace
...
The molarity of a solution obtained by mixing 750 mL of 0
...
(1) 0
...
37
...
00 M
M V M2 V2
Mf 1 1
V1 V2
0
...
975 M
3
1
2
4
4 = 0
...
1
Arrange the following compounds in order of decreasing acidity :
;
Ans
...
75 M
;
;
(1) II > IV > I > III
(3) III > I > II > IV
(3)
(2) I > II > III > IV
(4) IV > III > I > II
Sol
...
38
...
225 : 1
...
128 : 1
...
128 : 1
...
225 : 1
...
(3)
Sol
...
128 : 1
...
It should be root
means square speed
So, Ans is (3)
www
...
com
39
...
Activation energy of such
a reaction will be : (R = 8
...
301)
Ans
...
6 kJ mol–1
(3) 58
...
log
(2) 48
...
5 kJ mol–1
Ea 1
K2
1
T T
K 1 2
...
303 8
...
6 J / mol = 53
...
A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass
390
...
(2)
Sol
...
Since the mass increases by (390 – 180) = 210 hence the number of –NH2 groups is 5
...
Which of the following arrangements does not represent the correct order of the property stated against it ?
(1) V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour
(2) Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size
(3) Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution
(4) Sc < Ti < Cr < Mn : number of oxidation states
Ans
...
(1)
(1)
V2+ = 3 unpaired electrons
Cr2+ = 4 unpaired electrons
Mn2+ = 5 unpaired electrons
Fe2+ = 4 unpaired electrons
Hence the order of paramagnetic behaviour should be
V2+ < Cr2+ = Fe2+ < Mn2+
(2)
ionic size decrease from left to right in same period
(3)
As per data from NCERT
...
97 ; Fe3+/Fe2+ =0
...
41
Sc3+ is highly stable (It does not show +2)
(4)
The oxidation states increases as we go from group 3 to group 7 in same period
...
examrace
...
The order of stability of the following carbocations :
;
;
is :
(1) III > II > I
(2) II > III > I
(3) I > II > III
(4) III > I > II
Ans
...
(4)
The order of stability of carbocation will be
43
...
(3)
Sol
...
44
...
(2) O3 molecule is bent
(3) Ozone is violet-black in solid state
(4) Ozone is diamagnetic gas
...
(All statement are correct there is no answer)
...
(1) ONCl = 8 + 7 + 17 = 32e–
ONO– = 8 + 7 + 8 + 1 = 24e– (correct)
...
(Ref
...
examrace
...
A gaseous hydrocarbon gives upon combustion 0
...
08 g
...
The empirical formula of
the hydrocarbon is :
(1) C2H4
(3) C6H5
Ans
...
(2) C3H4
(4) C7H8
(4)
18g H2O contains 2g H
0
...
08 gH
...
08 g CO2 contains 0
...
84 0
...
07 : 0
...
In which of the following pairs of molecules/ions, both the species are not likely to exist ?
2(2) H 2 , He
2
+
2(1) H , He
2
2
(3) H
2+
(4) H 2 , He
2
2+ He
,
2
2
Ans
...
H2 : Bond order = 0
2
He2 : Bond order =
22
0
2
So, both H2 & He2 do not exist
...
Which of the following exists as covalent crystals in the solid state ?
(1) Iodine
(2) Silicon
(3) Sulphur
(4) Phosphorus
Ans
...
Silicon exists as covalent crystal in solid state
...
48
...
Synthesis of each molecule of glucose in photosynthesis involves :
(1) 18 molecules of ATP
(2) 10 molecules of ATP
(3) 8 molecules of ATP
(4) 6 molecules of ATP
(1) Fact
Sol
...
The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in the
order :
(1) Al3+ < Ba2+ < Na+
(2) Na+ < Ba2+ < Al3+
(3) Ba2+ < Na+ < Al3+
(4) Al3+ < Na+ < Ba2+
(2)
According to Hardy Schulze rule, greater the charge on cation, greater is its coagulating power for negatively
charged solution
...
Ans
...
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...
com
50
...
Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se
and Ar ?
(1) Ca < S < Ba < Se < Ar
(2) S < Se < Ca < Ba < Ar
(3) Ba < Ca < Se < S < Ar
(4) Ca < Ba < S < Se < Ar
(3)
Sol
...
So ionisation energy decreases
...
51
...
178 10–18J 2
...
62 10–34 Js and c = 3
...
214 10–7 m
(2) 2
...
500 10–7 m
(4) 8
...
(1)
Sol
...
178 × 10–18 2 2
2
1
1 6
...
0 108
1
2
...
214 × 10–7m
52
...
Oxidation of (A) gives
an acid (B), C8H6O4
...
Identify the compound (A)
...
CH3
CH3
(4)
Sol
...
examrace
...
Four successive members of the first row transition elements are listed below with atomic numbers
...
(4)
Sol
...
41 V ; Eº 3+/Mn2+ = + 1
...
77 V ; Eº 3+/Co2+ = + 1
...
Some SRP value are
exceptionally higher due to stability of product ion
...
g
...
57 V ; Eº 3+/Co2+ = + 1
...
Mn
Co
54
...
1 L
(3) 2
...
9 L
(4) 9
...
Sol
...
1 M
pH = 2 [H+] = 10–2 = 0
...
1 × 1 = 0
...
55
...
1 eV
...
(1) –2
...
2 eV
(2)
Sol
...
E
...
1 eV
(4) +2
...
Heg = – 5
...
56
...
On heating B gives C
...
A is :
(1) CH3COOH
(2) CH3CH2CH2COOH
(3) CH3
CH
COOH
(4) CH3CH2COOH
CH3
Ans
...
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...
com
57
...
(2)
Sol
...
An unknown alochol is treated with the “Lucas reagent” to determine whether the alcohol is primary, secondary
or tertiary
...
5
Bond order = 0
...
Sol
...
59
...
Sol
...
Experimentally it was found that a metal oxide has formula M0 98O
...
Fraction of the metal which exists as M3+ would be :
(1) 7
...
08%
(3) 6
...
08%
Ans
...
M0
...
Moles of M = 0
...
98 – x
Doing charge balance
(0
...
96 – 2x + 3x – 2 = 0
x = 0
...
04
100 = 4
...
98
www
...
com
PART C – MATHEMATICS
61
...
3
2
(2)
5
2
(3)
2x + y + 2z – 8 = 0
2x + y + 2z +
(3)
5
=0
2
9
2
...
It is estimated that the rate of change of production P w
...
t
...
(4)
...
7
2
dP
= 100 – 12 x
...
Sol
...
Let A and B two sets containing 2 elements and 4 elements respectively
...
+ 8C8 = 28 – 8C0 – 8C1 – 8C2
= 256 – 1 – 8 – 28
= 219
If the lines
x–2 y–3 z–4
x –1 y – 4 z – 5
=
=
and
=
=
are coplanar, then k can have
1
1
–k
k
2
1
(1) any value
(2) exactly one value
(3) exactly two values
(4) exactly three values
www
...
com
Sol
...
ˆ
ˆ
If the vectors AB 3ˆ 4k and AC 5ˆ – 2ˆ 4k are the sides of a triangle ABC, then the length of the
i
i
j
median through A is
(1)
Sol
...
The real number k for which the equation, 2x3 + 3x + k = 0 has two distinct real roots in [0, 1]
(1) lies between 1 and 2
(2) lies between 2 and 3
(3) lies between –1 and 0
(4) does not exist
...
The sum of first 20 terms of the sequence 0
...
77, 0
...
, is
66
...
examrace
...
(3)
7
77 777
+
...
up to 20 terms
= 7
10 100 10 3
=
79
99
999
10 100 1000
...
up to 20 terms
10 10
20
1 1
1 –
10 10
20
7
1
7
1
20 –
20 – 1 –
1
= 9
= 9
1–
9 10
10
7 179 1 1
= 9 9 9 10
68
...
A ray of light along x +
(2)
3
3yx– 3
(3) y =
3x– 3
(4)
3 y x –1
(2)
Take any point B(0, 1) on given line
Equation of AB
y–0=
–1 – 0
0– 3
x – 3
– 3y – x 3
x – 3y 3
3y x – 3
69
...
The number of values of k, for which the system of equations :
(k + 1)x + 8y = 4k
kx + (k + 3)y = 3k – 1
has no solution, is
(1) infinite
(2) 1
(3) 2
(2)
(4) 3
k 1
8
4k
k
k 3 3k – 1
k2 + 4k + 3 = 8k
k2 – 4k + 3 = 0
k = 1, 3
If k = 1
8
4
...
3
True
6 9 –1
therefore k = 3
Hence only one value of k
...
examrace
...
Sol
...
(i)
ax2 + bx + c = 0
...
Thus
a b c
1 2 3
Hence 1 : 2 : 3
71
...
72
...
a = , b = 2, c = 3
The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the point
(1) (–5, 2)
(2) (2, – 5)
(3) (5, – 2)
(4) (–2, 5)
(3)
Let the equation of circle be
(x – 3)2 + (y – 0)2 + y = 0
As it passes through (1, – 2)
A(3, 0)
(1 – 3)2 + (– 2)2 + (–2) = 0
=4
equation of circle is
A
(x – 3)2 + y2 – 8 = 0
(1, –2)
so (5, – 2) satisfies equation of circle
If x, y, z are in A
...
and tan–1x, tan–1y and tan–1z are also in A
...
, then
(1) x = y = z
(2) 2x = 3y = 6z
(3) 6x = 3y = 2z
(1)
2y = x + z
2 tan–1 y = tan–1 x + tan–1 (z)
(4) 6x = 4y = 3z
2y
xz
–1
tan–1
1 – y 2 = tan 1 – xz
xz
1– y2
=
xz
1 – xz
y2 = xz or
x+z=0
x=y=z
73
...
...
...
(1) Statement-I is true; Statement-II is true; Statement-II is a correct explanation for Statement-I
...
(3) Statement-I is true; Statement-II is false
...
(2)
Statement-II : (p q) (~ q ~p)
(p q) (p q)
which is always true
so statement -II is true
Statement-I :
(p ~q) (~ p q)
=p ~q ~ p q
=p ~p ~ q q
=f f
=f
so statement -I is true
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...
com
Alternate
Statement-II : (p q) (~ q ~p)
~ q ~ p is contrapositive
of p q hence (p q) (p q)
will be a tautology
statement -II
(p ~q) (~ p q)
p
T
T
F
F
q
T
F
T
F
P^~q
F
T
F
F
~p ^ q
F
F
T
F
(p ^ ~ q) ^ (~ p ^ q)
F
F
F
F
It is a fallacy
74
...
(3)
f (x )dx = (x)
I=
5
x f (x
3
)dx
put x3 = t
x2dx =
=
dt
3
1
tf (t)dt
3
1
t( t) – ( t)dt
3
=
=
1 3
x ( x 3 ) – 3 x 2(x 3 )dx c
3
=
1 3
x (x 3 ) – x 2(x 3 )dx c
3
75
...
1
4
(2)
1
2
(3) 1
(4) 2
(4)
im
I = 0
x
(1 – cos 2x ) (3 cos x )
x
...
3 cos x
...
4
...
examrace
...
Statement-I : The value of the integral
dx
1
/6
b
b
a
a
Sol
...
is true; Statement-II is true; Statement-II is not a correct explanation for Statement-I
...
is false; Statement-II is true
...
f (x )dx f (a b x ) dx
...
The equation of the circle passing through the foci of the ellipse
Sol
...
examrace
...
A multiple choice examination has 5 questions
...
The probability that a student will get 4 or more correct answers just by guessing is :
(1)
Sol
...
+ 5 C5
3 3
3
= 5
...
3
(3)
5
(3)
p=
79
...
0 2
...
Sol
...
examrace
...
The area (in square units) bounded by the curves y =
x , 2y – x + 3 = 0, x-axis, and lying in the first
quadrant is :
(1) 9
Sol
...
(1)
x
and
2y – x + 3 = 0
On solving both y = – 1, 3
...
82
...
83
...
If
Tn+1 – Tn = 10, then the value of n is :
(1) 7
(2) 5
(3) 10
(4) 8
(2)
Tn = nC3
Tn + 1 = n + 1C3
Tn + 1 – Tn = n + 1C3 – nC3
n
C2 = 10
n = 5
...
(2)
(3)
| z | = 1, arg z =
z = ei
z
2
(3)
(4) –
1
z
1 z
= arg (z) =
...
ABCD is a trapezium such that AB and CD are parallel and BC CD
...
examrace
...
(1)
Let
AB = x
p
p
tan ( + ) =
xq
q x
q – x = p cot ( + )
x = q – p cot ( + )
tan ( – – ) =
cot cot 1
=q–p
cot cot
q
cot 1
q cot p
q cos p sin
p
=q–p
=q–p q
=q–p
q p cot
q sin p cos
cot
p
x=
(p 2 q2 ) sin
q2 sin pq cos pq cos p 2 sin
AB =
...
p cos q sin
Sol
...
86
...
x
Sol
...
www
...
com
87
...
Statement-I : An equation of a common tangent to these curves is y = x +
Statement-II : If the line, y = mx +
Sol
...
5
(m 0) is their common tangent, then m satisfies m4 – 3m2 + 2=0
...
(2) Statement-I is true; Statement-II is true; Statement-II is not a correct explanation for Statement-I
...
(4) Statement-I is false; Statement-II is true
...
5 ), both statements are correct as m = ± 1 satisfies the given equation of statement-2
...
(2)
2
1
2
(3) 1
(4)
2
(1)
y = sec (tan–1 x)
Let
tan–1 x =
x = tan
y = sec
y=
1 x 2
1
dy
· 2x
=
dx
2 1 x2
at
x=1
1
dy
=
...
The expression
tan A
cot A
can be written as :
1 cot A 1 tan A
(1) sinA cosA + 1
(3) tanA + cotA
(2) secA cosecA + 1
(4) secA + cosecA
www
...
com
Sol
...
sin 2 A sin A cos A cos 2 A
= 1 + sec A cosec A
sin A cos A
All the students of a class performed poorly in Mathematics
...
Which of the following statistical measures will not change even after the grace
marks were given ?
(1) mean
Sol
...
22
=
www
...
com
Title: jee mains solution 2013
Description: It is solution of jee main exam 2013
Description: It is solution of jee main exam 2013